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File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 INITIAL BENCHMARKING OF CLAUSE 12 (DRAFT G) Report No: L22909/NDC/YH 22.01.2008 1 Final Issue for Client Review - minor corrections and added Appendix C YH RWPS 11.01.2008 O Initial Issue for Client Review YH RWPS 29 11.2007 Draft 1 Draft For Internal Review YH RWPS Date Rev. Description Prepared by Authorised by Distribution Shell UK Exploration and Production Attn: Mr Rupert Hunt WG7 P10 Convenor, c/o ABS Consulting Mr John Stiff Internal (A, E, CA, W/S No. 05/04/2803)

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Page 1: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

ISO TC 67 / SC 7 / WG 7 / Panel 10

ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12 (DRAFT G) Report No: L22909/NDC/YH

22.01.2008 1 Final Issue for Client Review - minor corrections and added Appendix C

YH RWPS

11.01.2008 O Initial Issue for Client Review YH RWPS

29 11.2007 Draft 1 Draft For Internal Review YH RWPS

Date Rev. Description Prepared by Authorised by

Distribution Shell UK Exploration and Production Attn: Mr Rupert Hunt WG7 P10 Convenor, c/o ABS Consulting Mr John Stiff Internal (A, E, CA, W/S No. 05/04/2803)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

CONTENTS SECTION PAGE NO.

1 EXECUTIVE SUMMARY 5 2 INSTRUCTIONS AND SCOPE 8

2.1 BACKGROUND AND OBJECTIVES: 8 2.2 SCOPE OF WORK: 8 2.3 TUBULARS: 9 2.4 NON-TUBULARS: 10 2.5 REPORT 10

3 SELECTED JACK-UP DESIGNS 12 3.1 INTRODUCTION 12 3.2 DETAILS OF UNITS 12 3.3 SELECTED LOAD CASES 13 3.4 MEMBER LOADS CHOSEN FOR INVESTIGATION 14

4 TUBULAR MEMBERS 15 4.1 INTRODUCTION 15 4.2 CASE STUDY RESULTS AND COMPARISONS 16 4.3 GENERAL FORM ASSUMPTIONS AND COMPARISONS 19 4.4 COMMENTS ON EQUATIONS AND LIMITS 29

5 PRISMATIC (CHORD) MEMBERS 32 5.1 INTRODUCTION 32 5.2 CASE STUDY RESULTS AND COMPARISONS 32 5.3 GENERAL FORM ASSUMPTIONS AND COMPARISONS 38 5.4 STRENGTH UTILISATION COMPARISONS 42 5.5 COMMENTS ON EQUATIONS AND LIMITS 48

6 SUMMARY OF ISO CLAUSE 12 AND COMMENTS RAISED 51 6.1 SUMMARY OF ISO CLAUSE 12 51 6.2 COMMENTS ON TUBULAR STRENGTH CHECK OF CLAUSE 12 51 6.3 COMMENTS ON PRISMATIC STRENGTH CHECK OF CLAUSE 12 52

REFERENCES 54 APPENDICES 55

A : NOTES AND COMMENTS ON ISO-19905 CLAUSE 12 SECTIONS 55 B : SPREADSHEETS FOR THE STUDIED CASES 62 C : ISO 19905-1 CLAUSES 12-13-12A-&-13A ONLY - WORKING

COPY @ CITY U.DOC 163

FIGURES

Figure 3-1 Triangular chord of Case 1 rig design 12 Figure 3-2 Compact tubular chord of Case 2 rig design 13 Figure 3-3 Moderate tubular chord of Case 3 rig design 13 Figure 4-1 Comparison of the axial compression strength checks of the ISO-

Clause 12 and the SNAME 5-5A when (Fy D) / (2 Cx E t) ≤ 0.170 (tubular) 22

Figure 4-2 Comparison of the axial compression strength checks of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (2 Cx E t) ≤ 0.170 (tubular, close form) 22

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Figure 4-3 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (E t) ≤ 0.0517 (tubular) 25

Figure 4-4 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (E t) = 0.076 (tubulars, Case 1) 25

Figure 4-5 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (E t) = 0.059 (horizontal tubular, Case 3) 26

Figure 5-1 The LTB moment options against different Lb for the case 2 doubly symmetry chord section 40

Figure 5-2 The LTB moment options against different Lb for the case 1 singly symmetry chord section 41

Figure 5-3 Comparison of the axial compression strength checks of the ISO-Clause 12 and the SNAME 5-5A when Fymin = Fy = Fyeff (prismatic) 43

Figure 5-4 Comparison of the axial compression strength checks of the ISO-Clause 12 and the SNAME 5-5A when Fymin = Fy = Fyeff (prismatic, close form) 43

Figure 5-5 The (ISOc - SNAMEc) / SNAMEc curves deduced from the ISO-Clause 12 and the SNAME 5-5A compression strength comparison when Fymin = Fy = Fyeff (prismatic) 45

Figure 5-6 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when Fymin = Fy = Fyeff (prismatic) 46

Figure 5-7 The axial compression and bending strength checks of the ISO-Clause 12 for Case 1 triangular and Case 2 compact tubular chords 48

TABLES Table 4-1 Effective length factors used 16 Table 4-2 Tubular member properties used for utilisation check in SNAME and

ISO 16 Table 4-3 Tubular brace factored loading cases (resulting in SNAME utilisation

values of unity and subsequently used for the ISO checks) 17 Table 4-4 The ISO Clause 12 utilisation checks for the SNAME checked unity

loading cases of the selected jack-up braces 18 Table 4-5 Comparison of the axial compression strength check of the SNAME

and the ISO methods regarding to L/r (tubular) 23 Table 4-6 Usages of the factored combined loads from the SNAME unity

utilisation check and the resulted ISO utilisations (tubular) 28 Table 5-1 Chord properties used for utilisation check in SNAME and ISO Clause

12 33 Table 5-2 Loading cases used for comparing the ISO chord utilisation check with

SNAME (the loads calculated or factored to bring the SNAME check to unity) 34

Table 5-3 The ISO Clause 12 utilisation checks for the SNAME checked unity loading cases of the selected jack-up chords 34

Table 5-4 The ISO classification for lateral torsional buckling check of the selected chords 36

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Table 5-5 The lateral torsional buckling strength and the plastic buckling strength of the chord case 2 (calculated by the ISO) 37

Table 5-6 Comparison of the separated stiffener / plate classification and the reinforced plate classification from the ISO 37

Table 5-7 Length/thickness ratios of the ISO classification for a fully compressed prismatic member section (ref to Appendix C) 38

Table 5-8 Comparison of the axial compression strength check of the SNAME and the ISO methods regarding to KL/r (prismatic) 44

Table 5-9 Usages of the factored combined loads from the SNAME unity utilisation check and the resulted ISO utilisations (prismatic) 48

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1 EXECUTIVE SUMMARY 1.1 The purpose of this work is to provide an initial “High Level” screening

benchmarking of Clause 12 of the proposed ISO 19905-1 document for Site Specific Assessment of Mobile Offshore Units. Clause 12 of that document deals with “Structural Strength”. The screening study presented in the present report was to ensure that there are no significant discrepancies within the requirements of the Clause, and to gauge the influence the Clause will have on the approvability of existing jack up designs when compared to SNAME 5-5A.

1.2 To this end, two methodologies are adopted: a case study of selected design types and general comments on the ability of the Clause to be used by a structural engineer.

1.3 For the case study, three typical jack-up designs were selected based on three generic chord types and realistic loading cases were estimated from selected realistic environmental conditions. The utilisation checks were undertaken using the two methods for each chord and brace section.

1.4 For the general investigation, the equations used for the ISO check and the SNAME check are compared directly over ranges of key parameters such as column and beam slenderness. These results were cross-checked with the case study comparisons where appropriate.

1.5 Both the case studies and the general investigation showed that, for the limiting load cases investigated, the SNAME method is more conservative than the proposed ISO Clause 12 except for some specific combinations of axial and bending load. The beam column strength utilisation checks calculated by the ISO equations were as low as 67% of the SNAME utilisation results in some cases.

1.6 The ISO lateral torsional buckling check, quoted as an adaptation from TABLE A-F1.1 by AISC (Ref [3]), is inconsistent with its origin. The limit set for the check is much more conservative than the detailed checks that are applied next. This LTB check may be redundant for the cases studied. For instance, the beam lengths at which the lateral torsional buckling bending moment reached the plastic section capacity is between 67.86m and 679.3m for the cases investigated. These appear to indicate that a lateral torsional buckling check is not required for typical chord sections and a more realistic screening check should be employed.

1.7 The ISO classification combines parallel plates in a section into a single plate with an equivalent thickness. Using this feature all the sections checked in this study were classed as “plastic”. However if the individual plates had been considered some elements of the sections investigated would have been classed as “slender”. The actual requirements in this regard should be made clearer.

1.8 As no sections were classified as “slender” during this study the complications of computing effective section properties based on the requirements of Tables A.12.2-1 through 3 and Tables A.12.2-2(a) through (c) were avoided. However these sections are likely to prove difficult to interpret and use consistently and clarification or simplification is recommended.

1.9 The ISO checking methodology introduces a requirement to check beam-columns (such as chords and braces) for direct and torsional shear strength. For the typical members checked (all chosen because they had the largest axial and bending loads) the shear stress utilisations were all quite small for both the tubular and prismatic

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members studied. For the checks undertaken in this study these shear and torsional checks do not result in a limiting condition. It is possible that if members had been selected for the worst shear loads then these might have shown larger shear utilisation values.

1.10 Some inconsistencies were found in the strength equations which resulted in step changes in strength with variation in element slenderness. These inconsistencies should be addressed.

1.11 Some work is required to improve the consistency of the equation numbering and definition of variables in Clause 12.

1.12 The hydrostatic load check introduced into ISO Clause 12 did not require detailed checking for the typical braces examined as the largest D/t ratio was 24.9 and for a water depth of 100m the limiting D/t ratio is 45.2.

1.13 Where possible, given the restricted nature of the present study, the ISO Clause 12 and SNAME results have been compared over ranges of beam-column slenderness in order to illustrate the differences that result from the different methods of assessment. The following general conclusions have been drawn:

• The utilisation check for the maximum axial tension (no bending) calculated by the ISO is always 94.5% of the SNAME result for any tubular or prismatic section due to the different partial resistance factor used.

• When Fymin = Fyeff ≤ 450MPa, the utilisation check for the maximum axial compression (no bending) by the ISO is constantly 93.5% of the SNAME for prismatic members due to the different partial resistance factor used. When Fymin > 450MPa, this percentage is even lower because of the high strength steel behaviour considered in the ISO method.

• For tubular members, the utilisation check for the maximum axial compression by the ISO is always lower (< 93.5%) than the SNAME when (Fy D) / (2 Cx E t) ≤ 0.170. This is not only due to the different partial resistance factors used in the two methods but also, and more importantly, to the lower effective length factor K for standard bracing configurations given by the ISO (0.7~0.8) as opposed to those specified by SNAME (0.8~0.9).

• The utilisation checks for the maximum bending (no axial load) calculated by the ISO are also lower than those for SNAME for prismatic members under the conditions considered. For tubular members, however, the maximum bending utilisation check by the ISO method tends to be slightly larger than the SNAME for values of (Fy D)/(E t) which are larger than say 0.076, (i.e. D/t >22 for a yield stress of 690 MPa). For tubes with a smaller D/t ratio, the ISO maximum bending utilisation is lower than the SNAME.

• The ISO axial compression strength check formulae for tubular members depend on the value of D/t. The braces studied have all been within the first slenderness condition: (Fy D) / (2 Cx E t) ≤ 0.170 which deals with relatively thick walled tubulars. Assuming Cx = 0.3, E = 200GPa and Fy = 600MPa, a tubular size ratio D/t > 34 is required for the second slenderness condition and D/t > 382.2 for the third one. Such large D/t ratios are unlikely to be found in typical braces of trussed legs. Further research has found that in ISO 19902, from which these formulae were adapted, the third condition of D/t > 382.2 (or (Fy D) / (2 Cx E t) > 1.911) of the axial compression strength

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check is not presented. So this limit must have come from elsewhere, or from an earlier draft of ISO 19902. Although these checks might apply to tubular leg jack-ups, they have not been investigated in the present study.

• Similarly for the ISO bending strength check formulae for tubular member, the bending strength curve is divided into four sections depending on shell slenderness. For the case when E = 200GPa and Fy = 600MPa, the D/t values at which the equations change are 17.2, 34.5 and 120. No expression is provided for the bending strength of tubulars above the D/t=120 limit. As compressive strength can be defined for tubulars with D/t>382.2, it appears inconsistent to omit an expression for bending strength of such tubulars. It is surprising that the limit for bending is more severe that the limit for axial compression. That the D/t limit for bending is 120 implies that the tubulars with D/t > 120 are not permitted.

1.14 The referred version of the ISO 19905-1 - Clause 12 has been attached to the report as Appendix C.

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2 INSTRUCTIONS AND SCOPE

2.1 BACKGROUND AND OBJECTIVES: 2.1.1 Clause 12 (“Structural Integrity”, as presently titled) of the proposed ISO 19905-1

(Appendix C) provides strength checks for structural elements in such items as the braced legs of Jack-up units.

2.1.2 The clause incorporates significant changes from the document on which the ISO was to be based: SNAME T&R Bulletin 5-5A. Some of these changes have been almost mandated (the new structural assessment of tubulars) and others have been included for the purposes of completeness (additional applicability checks on non-tubular members). In addition, the member classifications have been changed, and are in a state of flux since there was no interpolation between class 2 (compact) and class 3 (semi-compact). This interpolation is now being incorporated.

2.1.3 The majority of the checks within Clause 12 were written by PAFA, due in large part to the largesse of UK HSE. Undertaking a review of these modifications is not an insignificant task, but certainly needs to be accomplished before accepting them as appropriate for retention in ISO 19905-1.

2.2 SCOPE OF WORK: The scope given in our proposal was as follows:

2.2.1 The intent of this work is to undertake an initial “High Level” screening benchmarking, to ensure that there are no significant discrepancies in Clause 12, and to ascertain the effects its incorporation will have on the approvability of jack-ups as compared to SNAME 5-5A. This is not intended to be a full benchmarking, but a relatively low level of effort to ensure that there are no obvious inconsistencies.

2.2.2 To this end, typical chord and brace geometries will be kept in mind during the review and calculations - in particular the three generic chord types that make up so many modern jack-up braced-legs designs will be considered:

• Tear-drop (triangular) chords

• Chords using compact diameter split tubulars with racks

• Chords using moderate diameter split tubulars with racks

2.2.3 The requirement is to check the limiting utilisation for each type based on the existing SNAME 5-5A formulations and to compare these results against those predicted using the revised formations.

2.2.4 The RFQ calls for typical loads to be taken for two classes of jack-up, one with and one without fixation system to ascertain how the checks change when the new tubular formulation is applied to braces at the ends of the KL/r spectrum. The loads from selected cases from file will be scaled to bring the utilization to SNAME 5-5A Rev 2 to unity, to ensure that the results are not skewed by the non-linearities inherent in the formulations. The checks will be repeated using the Clause 12 formulations and the results compared.

2.2.5 We note that it is anticipated that more detailed benchmarking, or calibration, will be undertaken at a later date to ensure that the complete document is consistent, one part of which will be the Clause 12 formulations, however we doubt that will be the most

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appropriate time to exercise the formulations over a range of geometries and load levels/combinations. Therefore, in order to enhance the value of the benchmarking it is also proposed that the general form of the utilisation results will be investigated as a function of the section geometry (KL/r and local buckling) and axial-to-moment ratios for both compression and tension, in order to obtain a more comprehensive perspective on the performance of the proposed ISO formations.

2.2.6 The scope is divided into two parts and a report:

• Checks on Tubulars

• Checks on Non-Tubulars

• Report of Findings

2.3 TUBULARS: 2.3.1 The tubular check formulations are completely new, and while based on those

contained within the fixed structure ISO 19902, they have been reformulated to be based on limiting loads not stresses. In addition, the nomenclature has significantly changed from both SNAME and 19902. This changing nomenclature needs to be carefully followed to ensure no errors.

2.3.2 We will take the bracing loads from two classes of jack-up, one jack-supported, and one with a fixation system, and ascertain if applying the new checks alters the acceptability of the braces. The reason for using the two jack-up types would be to assess tubulars at the ends of the realistic KL/r spectrum. We will also run other cases to check on KL/r sensitivity. In order to achieve this, it will be necessary to:

• Determine a realistic set of bracing loads that are close to their acceptability limits. For a jack-supported unit this should not be difficult, as long as the correct site specific location is used. For a chock supported unit, where the operational bracing stresses tend to be low, it may be necessary to find an unusual set of loads e.g. from and accidental case, or artificially increase the operational loads to those close to the allowable limits. In any event, the loads will be adjusted so that they are close to the presently acceptable limits under 5-5A and are, where possible, realistic for the survival condition.

• Factor the loads such that the unity checks come out to be 1.0 under the existing SNAME 5-5A formulations.

• Determine the bracing unity checks under the factored loads using the draft Clause 12.

• Investigate the general form of the utilisation results as a function of the section geometry (KL/r and D/t) and axial-to-moment ratios, for both compression and tension, in order to obtain a more comprehensive perspective on the performance of the proposed ISO formations

• Compare and comment on any differences, to ensure that the major causes of differences are identified, particularly if there are competing factors (one lowering unity checks, one increasing them).

• Comment on any anomalies or inconsistencies in the Clause 12 methods, particularly if there are factors or classifications, e.g. D/t ratios, that seems out of place or unusual, and identify areas where changes in wording may be of value.

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• Note any nomenclature or other discrepancies so they can be corrected

2.4 NON-TUBULARS: 2.4.1 The non-tubular checks within 19905-1 should be as originally contained within

SNAME 5-5A, but again, there have been some changes in member classification and nomenclature. The objective is to check that the changes do not change the results, and to ensure that the additional compatibility/ classification checks included within Clause 12 do not impose unrealistic limits on the non-tubular members used in typical jack-ups. In areas where the non-tubular checks are as used in SNAME 5-5A, no detailed strength checks will be needed, however, it is important to ensure that the new member classifications do not have an unexpected adverse effect and to check the formulations. In order to achieve this, we will:

• Choose three chord types that cover the range of independent leg jack-ups (a “tear-drop” using multiple materials and yield strengths, a moderate diameter split tube and rack and also a compact split tube and rack).

• Decide on suitable member lengths, orientations, and end constraint conditions from real units using the chosen chord types.

• Determine if the chosen chords fail, or come close to failing, any of the new checks contained within Clause 12 member classifications, and that the classifications are as expected, referring to what has been supplied for inclusion within the Informative to clause 12, as appropriate.

• Check that the strength checks within clause 12 are as contained within SNAME 5-5A, and that there have been no transposing problems, or omissions.

• Undertake some limited sensitivity checks to determine at what point members change their classification (e.g. by changing side plate thickness on a 116 chord and checking the new classification).

• Check that the new class 2 to class 3 interpolation system works as expected (probably again by changing side plate thickness)

• Investigate the general form of the utilisation results as a function of the section geometry (KL/r and B/t) and axial-to-moment ratios, for both compression and tension, in order to obtain a more comprehensive perspective on the performance of the proposed ISO formations

• Note any nomenclature or other discrepancies so they can be corrected, particularly if there are inconsistencies between the tubular and prismatic nomenclature

2.5 REPORT 2.5.1 The deliverables will include a report that details the findings of the study. The

report will be directly usable in improving the ISO document. It will clearly set out where problems within the ISO document exist and, if possible, how they can be corrected. It will not always be possible to present a simple solution to the problems, but wherever possible, one will be proposed.

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2.5.2 It is proposed that the majority of the report will be in tabular form detailing sub-clause number, problem and proposed correction. However, it will still be necessary to present the numerical results of the analyses, and explanations for discrepancies.

2.5.3 For better coverage, we propose a brief review of the range of jack-up scantlings to ensure that the kL/r range is properly covered. We also propose to consider sufficient load cases such that the following are covered:

2.5.4 Axial tension and compression.

2.5.5 High, medium and low Axial : Moment ratios (say Axial part of the SNAME UC in the region of 0.1, 0.3, 0.5, 0.7 and 0.9).

2.5.6 Note: At least one of the above 5 cases to be scaled from a real case. The other cases will most likely be artificial.

2.5.7 The effect of hydrostatic pressure in the Clause 12 checks due to, say, a submergence of 100m (assuming the pressure term is still to be found in clause 12). This would be a simple repeat of the Clause 12 in-air checks already carried out.

2.5.8 The results will be compared graphically to facilitate the readers understanding.

2.5.9 Prepare report section comparing the results and comment on any differences. This will not be a treatise, but sufficient to ensure that the major causes of differences are identified, particularly if there are competing factors (one lowering unity checks, one increasing them).

2.5.10 Prepare report section commenting on any anomalies or inconsistencies in the Clause 12 methods, particularly if there are factors or classifications, e.g. D/t ratios, that seems out of place or unusual, and identify areas where changes in wording may be of value.

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3 SELECTED JACK-UP DESIGNS

3.1 INTRODUCTION 3.1.1 In order to undertake the high level assessment of the ISO 19905-1 Clause 12

requirements for strength three different jack-up leg designs with their general chord types and braces, have been considered in this benchmarking study:

• Case 1 - triangular tear-drop chords, (leg with no rack chocks)

• Case 2 - compact diameter split tubular chords (leg with rack chocks)

• Case 3 - moderate diameter split tubular chords (leg with no rack chocks).

3.1.2 Representative environmental loads were estimated for each rig using the Noble Denton in-house software JUSTAS and the finite element (FE) detailed model tool PAFEC to calculate the reaction forces in the rig components under the selected environmental loading. The resulted loading combinations in the leg members from the selected cases were then scaled to bring the utilization check of SNAME 5-5A Rev 2 to unity, to ensure that the results were not skewed by the non-linearities inherent in the formulations. The checks were then repeated using the Clause 12 formulations and the results compared.

3.2 DETAILS OF UNITS 3.2.1 The chords used in Case 1 jack-up design are typical triangular tear-drop type as

shown in Figure 3-1. For this design the hull is supported on pinions when elevated and as a result a significant part of the leg bending load is carried by shear in the leg between the upper and lower guides. As a result the braces on legs of this type are relatively stocky.

Figure 3-1 Triangular chord of Case 1 rig design

3.2.2 The chords used in Case 2 jack-up design are typical split tubular type of compact diameter, as shown in Figure 3-2. This design has a rack-chock fixation system and when this system is active during the elevated operations the leg bending loads will be mainly be carried through the chocks. As a result the braces for legs of this type are relatively slender.

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Figure 3-2 Compact tubular chord of Case 2 rig design

3.2.3 The chords used in Case 3 are slightly modified split tubular type of moderate diameter, as shown in Figure 3-3. In the variant considered the two opposing rack bars are connected by thinner plates as shown.

Figure 3-3 Moderate tubular chord of Case 3 rig design

3.3 SELECTED LOAD CASES 3.3.1 This is not a site assessment for the rig operation so there is no necessity to use site-

specific metocean data. However an attempt was made to use realistic environmental conditions for the leg length and leg capacities under study. Thus, the loading cases used for each rig in the study were all carefully selected from the Noble Denton database.

3.3.2 An airgap of 25 meters was assumed for all the three rigs as a realistic value. A spudcan penetration of 2 meters with 1 meter effective penetration was assumed for all the cases. The water depths for each rig were calculated from the assumed airgap and penetration data and the leg bay heights in order to locate the hull towards the top of the leg and thus in (typically) the strongest section of the leg at which the chords and bracings would be expected to be operating close to their design limits.

3.3.3 For the selected Case 1 design, which might typically operate in the Persian Gulf, a set of metocean data generated for that region was selected A water depth of 77 meters was assumed to position the hull near the top of the leg.

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3.3.4 For the Case 2 design a water depth of 100 meters was assumed and the severity of the Metocean data - in particular wave height - was increased until the limiting leg strength capacity was reached.

3.3.5 For the Case 3 design case, a set of Metocean data appropriate for the west coast of India was selected. A water depth of 94 meters was assumed as the hull was jacked to its maximum height.

3.4 MEMBER LOADS CHOSEN FOR INVESTIGATION 3.4.1 The JUSTAS program was used to produce detailed PAFEC FE leg models and the

associated environmental and boundary loadings. These were used to calculate the reaction loads on the leg chords and braces. The combined loads in those leg members were then checked. Considering the members with the maximum utilisation checks (SNAME 5-5A Rev 2) for chord, horizontal brace and diagonal brace, respectively, the loads were scaled to bring the utilization to unity. These scaled load combinations were then used as ultimate loading cases for comparing the SNAME code check results and Clause 12 formulation results.

3.4.2 For leg member checks, the SNAME code considers only axial loading and bending about principle axes. Shear and torsion and hydrostatic pressure are not considered in the SNAME checks. The ISO code does account for these additional load components. Thus these items were only checked by ISO Clause 12 using the shear load and torsional moment calculated from PAFEC for each selected leg member. In the cases considered the highest member stresses occurred in the leg close to the hull guides and thus hydrostatic pressure was not present. A general assessment for these shear loading formulations can be undertaken by simply checking if they were significant in comparison to the other loading utilisations.

3.4.3 The maximum loading components were also checked and compared using the two codes. This was achieved by adjusting each loading component (and making the other two components zero) for the overall utilisation check to reach unity in the SNAME code and then using these values in the ISO check. The comparison of these extreme cases highlights the difference of the utilisation range of the two codes. These are explored further in the general graphs drawn from the two formulations.

3.4.4 Two groups of comprehensive spreadsheets have been developed to undertake the calculations coded in ISO Clause 12. The calculations using these spreadsheets for all the cases have been attached to this report as appendixes. In order for the calculating process to be followed in detail, all the phased results and formulations were shown explicitly in the sheets although it was known that only one set of data was useful for each circumstance.

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4 TUBULAR MEMBERS

4.1 INTRODUCTION 4.1.1 As mentioned before, the tubular check formulations in Clause 12 are quite different

from those used in SNAME (Ref [1]). While based on those contained within the fixed steel structures ISO 19902, they have been reformulated to be based on limiting loads, not stresses. In addition, the nomenclature has significantly changed from both SNAME and ISO 19902. Care has been taken to follow this changing nomenclature through the report.

4.1.2 Three jack-up designs have been selected and each has two different tubular members (horizontal and diagonal braces) to be checked and compared using the SNAME and ISO codes. Load cases studied for each tubular member were selected using the method described in Section 3.4 and summarised below:

• Maximum axial tension, the pure axial tensile force which brings the SNAME utilisation check to unity for the given tubular section properties;

• Maximum axial compression, the pure axial compressive force which brings the SNAME utilisation (with buckling) check to unity for the given tubular section and length;

• Maximum section bending, the pure section bending moment which brings the SNAME utilisation check to unity for the given tubular section properties. Only one directional case was checked due to the symmetry of the tubular section.

• Factored axial load and bending combination, the reaction forces resulting from a structural analysis of the leg under realistic environmental loads and scaled to bring the SNAME utilisation (overall) check to unity;

• Factored shear force and torsional moment, the other two reaction force components resulting from the structural analysis models and taken from the same model element used for the axial load and bending combination case with the same scaling factor. These were used to check the beam shear capacity and torsional shear capacity, respectively, using ISO only as the SNAME check does not consider shear and torsion.

4.1.3 The three selected jack-up designs fell into two classes: for case 1 & case 3 the hull is supported by pinions only while in case 2 a rack-chock fixation system is provided. For the two pinion-supported units, the realistic set of bracing loads that are close to their acceptability limits have been obtained by using realistic site specific environmental data. For the chock supported unit, the bracing stresses under realistic storm loading are far below their capacity. For this kind of unit the leg braces are typically designed to be capable of supporting loads that may occur when the unit is being placed on location. The brace wall thicknesses used for the Case 2 study were based on an early understanding of that leg design. In order to adopt a sensible load set however it was decided to factor up the axial and bending loads found from the site specific storm case until they reached the acceptable limit under SNAME 5-5A.

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4.1.4 The loading cases from the SNAME unity check were then used to determine the bracing utilisation under the draft Clause 12. Any differences between the two code check results were identified for discussion.

4.1.5 In order to obtain a more comprehensive perspective on the performance of the proposed ISO formations, the general forms of the utilisation results have been investigated for both SNAME and Clause 12, as a function of the section geometry (KL/r and D/t) and axial-to-moment ratios, etc. Anomalies or inconsistencies in the Clause 12 methods were reported and a comparison of the graphical results for the two coding systems made.

4.1.6 Notes on the nomenclature, wordings or other discrepancies regarding the tubular member check in the draft Clause 12 were summarised and are presented in Appendix A.

4.2 CASE STUDY RESULTS AND COMPARISONS 4.2.1 The brace details for the three cases studied are listed in Table 4-1 and Table 4-2.

Table 4-1 Effective length factors used

Case 1 Case 2 Case 3

Bracing Type K-brace X-brace X-brace

SNAME ISO SNAME ISO SNAME ISO

Effect. length factor K[1] 0.8 0.7 0.9 0.8 0.9 0.8 [1] Effective length factors are same for both horizontal and diagonal braces in one leg. For SNAME, K refers to Table C-C2.1 in [1]; for ISO 19905 - Clause 12, K refers to Table A.12.4-1 in Appendix C.

Table 4-2 Tubular member properties used for utilisation check in SNAME and ISO

Case 1 Case 2 Case 3

Horizontal Diagonal Horizontal Diagonal Horizontal Diagonal

Unbraced length L[1], m 4.379 5.550 5.684 6.657 7.725 6.625

Outer Diameter D[1], m 0.324 0.324 0.220 0.220 0.324 0.324

Thickness t[1], m 0.013 0.013 0.032 0.032 0.021 0.016

D/t Ratio 24.92 24.92 6.88 6.88 15.43 20.25

Yield stress Fy[1], MPa 607 607 448 448 586 586

[1] Section properties of each tubular member are the same whether being used in either SNAME or ISO 19905 - Clause 12.

4.2.2 The effective length factors are specified based on whether the bracing type is “K-brace” or “X-brace” and are given in Table 4-1. Note that the effective length factors prescribed by ISO are always lower than those prescribed in SNAME for the same

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class of bracing.

4.2.3 Section properties for the brace are listed in Table 4-2. For the detailed section properties of each brace, such as section area and radius of gyration, please refer to the ISO code check spreadsheets for each case provided in Appendix B and the corresponding SNAME code check spreadsheets in Appendix B.3. It is worth noting that the shear area used for the ISO code check has been simplified and taken as half of the section area for both tubular and prismatic members in all the selected cases. This assumption is conservative for tubulars and will also typically be conservative for stocky chord sections.

4.2.4 The loading cases used for comparing the two checking methods are listed in Table 4-3. The SNAME code check spreadsheets, being used to generate the maximum loading components for each brace, and to provide the realistic loading combinations from the PAFEC models, are attached as Appendix B.3 to this report. It can be seen that the factored combined loads resulting for the six braces all have high axial compressive load, taking 77.7% ~ 94.6% of the corresponding maximum axial compression. This indicates the importance of ensuring that the ISO formulations are checked particularly for load combinations in which axial compression dominates.

Table 4-3 Tubular brace factored loading cases (resulting in SNAME utilisation values of unity and subsequently used for the ISO checks)

Combined loads Unit: kN, kN·m

Maximum axial

tension

Maximum axial

compression

Maximum section bending

Shear force

Section torque

Axial Bdg y Bdg z

Horizontal 6941[1] -5750 687 31 6.5 -4465 -167 0.6 Case 1

Diagonal 6941 -5315 687 7.8 2.4 -5026 33 5.1

Horizontal 7570 -4135 453 20 14 -3640 31 -1.1 Case 2

Diagonal 7570 -3373 453 5.8 3.7 -3116 -11 -10

Horizontal 8165 -5320 800 6.2 5.2 -4978 7.0 -34 Case 3

Diagonal 10540 -5915 1017 8.5 2.7 -5376 -52 5.8 [1] Tensile load is positive

4.2.5 It has been noted that the SNAME formulations have no code to check shear capacity. Thus, the shear force and torsional moment listed in Table 4-3 were taken directly from the PAFEC model results for the same elements, scaled to the same level of the combined load components and used only for the ISO code check in this report. It can be seen from Table 4-3 that the shear forces and section torques found in bracing members were small compared to the other load components and especially the axial loads.

4.2.6 The ISO Clause 12 formulation utilisation check results using the SNAME “calibrated” loading cases (which resulted in a SNAME code check of unity) are listed in Table 4-4.

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4.2.7 Comparing the utilisation checks of the two systems, it can be seen that the proposed ISO-Clause 12 utilisations are smaller than the SNAME values for almost all the cases; the exception is the maximum section bending checks for the brace tubulars in Case 1. In other words, the SNAME checks were more conservative in almost all cases with the SNAME 1.0 utilisation reducing to less 1.0 for the ISO check. For the two exceptional cases, the maximum bending check in the ISO check was 3.3% higher than the SNAME check, which is considered small since these are already extreme bending cases assumed specially for the comparison and not representative of the typical loading that braces would be designed for.

4.2.8 The lowest utilisation checks resulted from the ISO method were 0.760 and 0.761 for the combined loading case and the maximum axial compression strength of the diagonal brace of Case 2, respectively.

4.2.9 The maximum axial tension checks for all the six braces are the same ratio of 0.945 when comparing the ISO to the SNAME. This was due to the same formulae but different resistance factors (fixed) used in these two code systems for the axial tension checks of tubular members. This trend will be further confirmed by the general form comparison presented in the next section.

Table 4-4 The ISO Clause 12 utilisation checks for the SNAME checked unity loading cases of the selected jack-up braces

Maximum

axial tension

Maximum axial

compression

Maximum section bending

Shear force

Section torque

Combined loads

Horizontal 0.945 0.939 1.033 0.015 0.010 0.960 Case 1

Diagonal 0.945 0.887 1.033 0.004 0.004 0.896

Horizontal 0.945 0.794 0.945 0.009 0.037 0.798 Case 2

Diagonal 0.945 0.761 0.945 0.003 0.010 0.760

Horizontal 0.945 0.848 0.982 0.002 0.006 0.843 Case 3

Diagonal 0.945 0.799 0.945 0.003 0.004 0.802

4.2.10 The combined loads check utilisation for each brace has resulted in similar values as found for the corresponding maximum axial compression check. This is consistent with the observation from the selected loading cases listed in Table 4-3, in which a dominatingly high axial compressive load (77.7% ~ 94.6% of the corresponding maximum axial compression) was found in the factored combined load combinations. It is worth noting that the moment application parameter Cm used in both methods has been defined in exactly the same way and thus has been taken as the default value of 0.85 in both code checks.

4.2.11 The largest ISO code utilisation found for the factored shear force was 0.015 and for the factored torsion was 0.037. SNAME utilisations are not required for these load components but these results indicate that in typical bracing members they are not significant.

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4.2.12 Clause 12 specifies three checks which depend on the D/t ratio of tubular members. For all tubulars which meet the requirements of Equation A12.5-1 the strength checks included in the Clause are applicable. All the tubulars checked were found to meet this requirement. Note that if tubulars do not meet this requirement the Clause requires that they should be assessed using alternative methods that result in similar levels of reliability; however no specific formulations are provided. For tubulars which meet the more stringent check of Equation A.12.2-1 the member is deemed to comply with Class 1 classification but this is only relevant when undertaking earthquake, accidental or alternative strength analyses. Finally the D/t ratio of the tubular determines whether hydrostatic loading needs to be assessed. For tubulars that meet the requirements of Equation A.12.5-2 hydrostatic loading may be ignored. If hydrostatic loading may not be ignored then reference is made to ISO 19902. As the maximum brace loads were all found at the hull level there were no cases for which hydrostatic pressure had to be considered. The largest D/t ratio for bracing members investigated was 24.92 (see Table 4-2). For the nominal 100m water depth given in the scope of work (see Section 2.5.7 of this report) the maximum allowable tubular D/t for a 100m pressure head (taken from ISO Clause 12 - Table A.12.5-1) is 45.2 hence all the tubulars in the three cases can be considered for this water depth without special calculation. Note however that Equation 12.5-2 given in ISO Clause 12 has a quadratic form and for water depths greater than about 170m the maximum tubular D/t starts to increase which is not rational. Either the equation needs to be modified or some limits need to be applied.

4.2.13 The ISO spreadsheets in Appendix B.1 have incorporated these classifications for all the tubulars investigated and found they were all within the Clause 12 applicable range.

4.2.14 The ISO classification results and the other intermediate results including the sectional formulation results, which are not shown in the body of this report, have been detailed in Appendix B.1.

4.3 GENERAL FORM ASSUMPTIONS AND COMPARISONS 4.3.1 Some general forms of the utilisation results for both SNAME and Clause 12 have

been deduced and compared in this section, as a function of the section geometry, using ratios such as KL/r and D/t and axial-to-moment ratios. For cases where the sectional formulations are too complicated to be clearly presented, some “conditioned” general forms of these utilisation formulations were deduced and compared instead. The general form results and conclusions were also compared with the previous case study results to ensure there was no obvious error in either forms.

4.3.2 Axial Tension Strength Check 4.3.2.1 The axial tension strength check formulated by the ISO - Clause 12 is shown as

Equation A.12.5-3 (equation numbers as in Appendix C):

Put ≤ A Fy / γt (A.12.5-3)

Where A is the section area, Fy is the material yield stress and γt is a partial resistance factor.

This expression is in the same fashion of the SNAME formula except that in the SNAME formula, the partial resistance factor φt is a multiply factor and taking the

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value of 0.9 while in this ISO equation, the partial resistance factor γt is a divider with the default value of 1.05. Thus, for any pure axial tension strength check, the SNAME utilisation result upon the ISO utilisation result would be

0.9/(1/1.05) = 0.945

This number is the ratio of ISO to SNAME strength for members in simple tension as listed in Table 4-4.

4.3.3 Axial Compression Strength Check 4.3.3.1 The axial compression strength check formulae used by ISO - Clause 12 is defined

(Appendix C) as follows:

Puc ≤ Pa / γc (A.12.5-4)

where

Pa = [1.0 - 0.278λ2] Pyc for λ ≤ 1.34 (A.12.5-5a)

= 0.9 Pyc/λ2 for λ > 1.34 (A.12.5-5b)

λ = [Pyc/PE]0.5

PE = π2 E I/ (KL)2

Pyc = A Fy for A Fy / Pxe ≤ 0.170 (A.12.5-7a)

= [1.047 – 0.274 A Fy/Pxe] A Fy

for 0.170 < A Fy / Pxe ≤ 1.911 (A.12.5-7b)

= Pxe for 1.911 < A Fy / Pxe (A.12.5-7c)

Pxe = 2 Cx E A (t / D) (A.12.5-8)

The definitions of the variables in the formulae above can be found in the proposed Clause 12.

4.3.3.2 The axial compression strength check formulated in SNAME 5-5A is as followed (considering column buckling, equation numbers as in Ref [1]):

Fcr = (0.658λc2) Fy for λc ≤ 1.5 [Eq. E2.2]

Fcr = 0 877

2

.λc

⎧⎨⎩

⎫⎬⎭

Fy for λc > 1.5 [Eq. E2.3]

Where

λc = 2

1

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

Again, the full explanations of the variables can be found in Ref [1].

4.3.3.3 When comparing SNAME with ISO this was done using a non-dimensional form, Pa / A Fy and Pn/ A Fy = Fcr / Fy to represent the general forms of the two methods.

4.3.3.4 For the general form functions of the two methods to be compared, the more complicated ISO codes were examined first. To reduce the variables involved for a simpler general form expression, the following transformation between the ISO formulae was adopted:

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• By modifying the equation Pxe = 2 Cx E A (t / D) using the section condition A Fy / Pxe the later can be rewritten as (Fy D) / (2 Cx E t). Assuming Cx = 0.3, as recommended by the Clause (Appendix C)[1], and E = 200 GPa (as used in all our selected cases), the unknown variables in the formulae are thus reduced to three: Fy, D and t, or a further reduction as Fy and (D/t).

• Using the sectional equations of Pyc and the expression for PE in the equations of λ and Pa, it was found that for all values of λ and Pyc, a combined variable (KL/r) can be extracted from the final expression of Pa. Thus, this (KL/r) was selected as the main independent variable of the general form functions in this case. A range of values of Fy and (D/t) were considered.

• Further investigation has indicated that when A Fy / Pxe ≤ 0.170, i.e., (Fy D) / (2 Cx E t) ≤ 0.170, the final expression of Pa will be independent of (D/t) while Fy would always be a variable in the calculation of Pa (i.e. Fy can not removed as a divider of Pa).

• It was found that for the representative jack-up cases selected in this report all the tubular braces fell in the range of A Fy / Pxe ≤ 0.170. This indicates that these bracing members are not sensitive to local compression buckling instability and for general robustness this is a desirable feature for jack-up leg bracing members.

4.3.3.5 The SNAME formulae can also use the (KL/r) as the main independent variable of its general form function. In fact, (KL/r), and Fy, are the only variables determining Fcr / Fy for any tubular section.

4.3.3.6 Thus, the general forms for the axial compression strength check of the two methods were compared by plotting (Pa/AFy), or (Fcr/Fy), against (KL/r) for certain specific yield stress values. This comparison assumed that local buckling instability was not significant, by keeping A Fy / Pxe ≤ 0.170. The comparison between ISO and SNAME codes, is shown in Figure 4-1. Figure 4-2 shows the same plots as in Figure 4-1 but with the (KL/r) in a smaller range of 0~100, which gives a more detailed view of the part of the curve which is representative for typical tubular bracing members.

4.3.3.7 The two yield stresses, 450 MPa and 600 MPa, were selected because they represented the realistic jack-up cases studied in this report. Note that for checking tubular members only one formulation is provided and this covers all values of yield stress. With the known values of Cx and E, the (Fy D) / (2 Cx E t) ≤ 0.170 condition categorised the (D/t) range as D/t ≤ 45.3 when Fy = 450 MPa and D/t ≤ 34 when Fy = 600 MPa. The maximum (D/t) in our selected cases is 24.9 for the tubulars in Case 1, where the steel yield stress is 607MPa as shown in Table 4-2.

4.3.3.8 The partial resistance factor γc was also considered in this comparison. For the axial compression strength check, the ISO code has a partial resistance factor (dividing) of γt = 1.15 and SNAME has a multiplying factor of φt = 0.85. In this case, they are not very different (1/0.85 = 1.18). Simple comparison of resistance factors shows that the utilisation value for the ISO check will be smaller than that using a SNAME check (i.e. 1.15 x 0.85 = 0.98).

4.3.3.9 Comparing the two ISO curves with different yield stresses in Figure 4-1, it can be seen that their curvatures are very similar with increasing KL/r and the higher relative capacity Pa/AFy was obtained from the lower yield stress (450MPa) case.

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This illustrates the fact that enhancing the strength of the material increases the tendency of the column to buckle and this in turn reduces the comparative strength advantage provided by the higher yield material. The two curves of the SNAME check results show the same trends.

Fy = 600MPaFy = 450MPa

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200

KL/r

Colu

mn

buck

ling

stre

ngth

, Pa/

AFy

or F

cr/F

y

SNAME, Fy=600MPa

ISO, Fy=600MPa, D/t<=34

SNAME, Fy=450MPa

ISO, Fy=450MPa, D/t<=45.3

Figure 4-1 Comparison of the axial compression strength checks of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (2 Cx E t) ≤ 0.170 (tubular)

Fy = 600MPa

Fy = 450MPa

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 10 20 30 40 50 60 70 80 90 100

KL/r

Col

umn

buck

ling

stre

ngth

, Pa/A

F y o

r F c

r/Fy

SNAME, Fy=600MPa

ISO, Fy=600MPa, D/t<=34

SNAME, Fy=450MPa

ISO, Fy=450MPa, D/t<=45.3

Figure 4-2 Comparison of the axial compression strength checks of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (2 Cx E t) ≤ 0.170 (tubular, close form)

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4.3.3.10 Comparing the ISO curves and the SNAME curves with the same yield stress, it was found that the ISO strengths are always higher than the SNAME strengths for a given KL/r. Especially in the range of KL/r = 30~100, as illustrated in Figure 4-2, the difference can be up to 13% for the ISO result over the SNAME data (when KL/r = 60). This range of member slenderness is representative of the braces examined in this study.

4.3.3.11 To illustrate the results from the general forms, the ISO and SNAME utilisation checks of the maximum axial compression for each brace, which have been listed in Table 4-4, are also listed in Table 4-5, together with the KL/r and the yield stress data for each brace.

4.3.3.12 Difficulty has been encountered in presenting the results of Table 4-1 and Figure 4-2 within one plot because the ISO and SNAME codes assign different values of effective length factor K to braces. As listed in Table 4-1, the ISO K is always one level smaller than the SNAME K. For instance the effective length factor value is set at 0.7 in ISO for a K-braced member while the same member is defined as 0.8 in SNAME, an increase of 14%. Since the L/r is the same in both calculations, the effective slenderness values will differ by the value of effective length assigned. A smaller effective length results in a reduced value of kL/r and hence a higher value of column buckling strength. Bearing this in mind, the rather large percentage differences shown in Table 4-5, were not unexpected high but rather consistent with the general form of comparison. The differences between the ISO and SNAME strength results demonstrated here are up to 31.4% in some cases.

Table 4-5 Comparison of the axial compression strength check of the SNAME and the ISO methods regarding to L/r (tubular)

Axial Compression Strength Check Fy (MPa) L/r ISOu / SNAME u

[1] ISOc/SNAME c[2] %[3]

Horizontal 607 39.79 0.939 1.065 6.5 Case 1

Diagonal 607 50.43 0.887 1.127 12.7

Horizontal 448 84.33 0.794 1.259 25.9 Case 2

Diagonal 448 98.75 0.761 1.314 31.4

Horizontal 586 60.77 0.848 1.179 17.9 Case 3

Diagonal 586 71.96 0.799 1.252 25.2

[1] Relative utilisation of the ISO check to the SNAME check. [2] Relative capacity of the ISO check to SNAME check, equal to 1/relative utilisation. [3] Percentage change of the ISO strength capacity over the SNAME strength capacity.

4.3.4 Axial Tension and Bending Strength Check 4.3.4.1 The axial tension strength check for the two codes has been compared in Section

4.3.2 and the difference from the two methods has been found to be a fixed factor (=0.945). In this section, bending capacity envelopes with different axial tension strength usages (selected typical Put/AFy values) are checked and compared.

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4.3.4.2 Again, the axial tension and bending strength check formulae used by the ISO - Clause 12 is defined as follows. The bending capacity expressions and related , equation numbers are taken from the Clause as shown in Appendix C):

γt Put / Pp + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

where

Mb = Mp for (Fy D)/(E t) ≤ 0.0517 (A.12.5-9a)

= [1.13 – 2.85 (Fy D)/(E t)] Mp

for 0.0517 < (Fy D)/(E t) ≤ 0.1034 (A.12.5-9b)

= [0.94 – 0.76 (Fy D)/(E t)] Mp

for 0.1034 < (Fy D)/(E t) ≤ 120 (Fy / E) (A.12.5-9c)

Mp = Fy (1/6) [D3 – (D - 2t)3]

Pp = A Fy

4.3.4.3 The axial tension and bending strength check formulated in SNAME 5-5A is simpler and is quoted below. The expressions and equation numbers have been taken from Ref [1]):

If Put/φtAFy > 0.2

1.0M

MM

M98

PP

1

bz

uz

by

uy

p

ut ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

φφφ bbt

[Eq. H1-1a]

else

1.0M

MM

MP2

P1

bz

uz

by

uy

p

ut ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

φφφ bbt

[Eq. H1-1b]

where

Mb = Mp = Fy (1/6) [D3 – (D - 2t)3]

η =2 for tubular members.

4.3.4.4 It was found that when (Fy D)/(E t) ≤ 0.0517, the formula of the bending capacity Mb is the same in both the ISO calculation and the SNAME prescription. For this case, the general form of the two methods for the bending capacity can be compared independent of either tubular geometry or material properties, as shown in Figure 4-3. However, when (Fy D)/(E t) > 0.0517, a general form function for the bending capacity can not be deduced without knowing the full information of the tubular section properties.

4.3.4.5 In our selected jack-up cases, there are three tubular braces having their (Fy D)/(E t) slightly more than 0.0517 and the maximum value is 0.076 (Case 1), yet lower than 0.1034, the second sectional threshold. Therefore, the general form function for the 0.0517 < (Fy D)/(E t) ≤ 0.1034 condition will be developed based on the actual braces, i.e., both horizontal and diagonal braces of Case 1 with (Fy D) / (E t) = 0.076 and the horizontal brace of Case 3 with (Fy D) / (E t) = 0.059, for both the ISO and

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the SNAME methods. These conditioned general form comparisons, as shown in Figure 4-4 and Figure 4-5, help to compare the results from the two codes.

Put/Pp=0.8Put/Pp=0.6

Put/Pp=0.4

Put/Pp=0.2

Put/Pp=0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Muey/Mby

Mue

z/Mbz

ISO, 80%Axial Tension

SNAME, 80%Axial Tension

ISO, 60%Axial Tension

SNAME, 60%Axial Tension

ISO, 40%Axial Tension

SNAME, 40%Axial Tension

ISO, 20%Axial Tension

SNAME, 20%Axial Tension

ISO, 0%Axial Tension

SNAME, 0%Axial Tension

Figure 4-3 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (E t) ≤ 0.0517 (tubular)

Put/Pp=0.8Put/Pp=0.6

Put/Pp=0.4

Put/Pp=0.2

Put/Pp=0

(FyD)/(Et)=0.076

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Muey/Mby

Mue

z/Mbz

ISO, 80%Axial Tension

SNAME, 80%Axial Tension

ISO, 60%Axial Tension

SNAME, 60%Axial Tension

ISO, 40%Axial Tension

SNAME, 40%Axial Tension

ISO, 20%Axial Tension

SNAME, 20%Axial Tension

ISO, 0%Axial Tension

SNAME, 0%Axial Tension

Figure 4-4 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (E t) = 0.076 (tubulars, Case 1)

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4.3.4.6 The different partial resistance factors, 1.05 for both tension and bending in the ISO (γt and γb) and 0.9 for both tension and bending in the SNAME, were considered in all the plots.

4.3.4.7 Figure 4-3 shows the general form comparison of the ISO and the SNAME's bending capacities for the (Fy D) / (E t) ≤ 0.0517 tubular with varying axial tension level. It is obvious that the bending capacity envelops are circles due to the symmetry of tubular section and their radiuses increase with the decrease of the axial tension usage. It is also observed that when the axial tension levels are high (e.g. Put / Pp = 0.8, 0.6), the ISO bending envelopes are greater than the SNAME ones and when the axial tension levels are low (Put / Pp = 0.4, 0.2), the ISO bending envelops are smaller than the SNAME ones.

Put/Pp=0.8

Put/Pp=0.6

Put/Pp=0.4

Put/Pp=0.2

Put/Pp=0

(FyD)/(Et)=0.059

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Muey/Mby

Mue

z/M

bz

ISO, 80%Axial Tension

SNAME, 80%Axial Tension

ISO, 60%Axial Tension

SNAME, 60%Axial Tension

ISO, 40%Axial Tension

SNAME, 40%Axial Tension

ISO, 20%Axial Tension

SNAME, 20%Axial Tension

ISO, 0%Axial Tension

SNAME, 0%Axial Tension

Figure 4-5 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when (Fy D) / (E t) = 0.059 (horizontal tubular,

Case 3)

4.3.4.8 According to these four pairs of the plots, it would be easy to conclude that the biased differences tend to be bigger when the axial tension level gets closer to either ends. This was confirmed by one extreme deduction that when the Put / Pp reaches its maximum in the SNAME (1.0 after multiplying the factor φt = 0.9), the bending capacity in the SNAME would shrink into the centre, i.e., reduce to zero, but the bending capacity in the ISO would still be a small circle with the factored Put / γt Pp = 0.945, as concluded in Section 4.3.2. The radius of this small circle can be easily computed from the formulae:

Muy(z) / Mby(z) = (1-Put / γt Pp) γb = (1-0.945) x 0.9 = 0.0495.

It is very small though and thus not shown in Figure 4-3.

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4.3.4.9 However, this conclusion would fail to satisfy the other end where the Put / Pp = 0, as shown in Figure 4-3. It is because this extreme case eliminates the axial tension resistance factor γt which has played an important role in the other four cases together with the bending resistance factor γb. The maximum bending capacity Muy(z) / Mby(z) is now governed by the resistance factor used for each method, which is 0.9 for the SNAME code and 1/1.05=0.952 for the ISO code, as the radiuses of the two biggest circles shown in Figure 4-3. Furthermore, similar to the axial tension strength check, the ratio of the maximum bending capacity for ISO to SNAME would be 0.952/0.9 = 0.945 for all the (Fy D) / (E t) ≤ 0.0517 tubular braces. This is the very number obtained for the tubulars of Case 2 and Case 3, as shown in both Table 4-4 and Appendix B.1.

4.3.4.10 Nevertheless, the comparison of the axial tension and bending strength check of the two methods for the (Fy D) / (E t) ≤ 0.0517 tubulars has shown a relative small difference, noting the sudden change of the trend between the two bending capacity checks at Put / Pp = 0.

4.3.4.11 Horizontal tubulars from Case 1 and Case 3 are represented by Figure 4-4 and Figure 4-5 and are generally similar to Figure 4-3 with the same trend of reduction in moment capacity with increasing axial tension. Comparing Figure 4-4 and Figure 4-5 the difference between the ISO and the SNAME seems enlarged with the increase of the (Fy D) / (E t). Take the Put / Pp = 0.2 pair as example. For the (Fy D) / (E t) = 0.059 case in Figure 4-5, the SNAME bending capacity is about 8% greater than the ISO bending capacity. For the (Fy D) / (E t) = 0.076 case in Figure 4-4, the SNAME bending capacity is up to 15% greater than that for ISO. Following this trend, the threshold of a axial tension usage, which would make the SNAME bending radius equals to the ISO bending radius, is expected to increase with the increase of (Fy D) / (E t).

4.3.4.12 Next we compare the two bending capacities with the Put / Pp = 0 for these two cases. It was found that the maximum ISO bending capacity is 3.3% more than the maximum SNAME bending capacity for the Case 3 horizontal brace in Figure 4-5 while 1.8% less for Case 1 in Figure 4-4. Again, the maximum bending capacity relationship between the two methods has fallen out of the general (though not simple) trend when Put / Pp > 0. In these cases, the general form function will depend on not only the resistance factor but the specific section properties. However, comparing these two conditioned general forms with their maximum bending utilisation checks listed in Table 4-4, perfect matches were once more demonstrated.

4.3.5 Axial Compression and Bending Strength Check 4.3.5.1 The axial compression and bending strength check formulae used by the ISO -

Clause 12 and the SNAME 5-5A are quite similar to the axial tension and bending strength check formulae quoted in Section 4.3.4. However, besides the different resistance factor used for the axial compression component in each method, the relationship between the two general forms for the axial compression strength check is also complicated and could not be simplified in one or two graphs. This has been discussed in Section 4.3.3. On the other side, when (Fy D) / (2 Cx E t) ≤ 0.170, the difference of the axial compression capacity plotted in Figure 4-1 is not very significant. Thus, the similar trends, observed from the three general formed comparisons made for the axial tension and bending strength check in the previous section, are expected to be followed approximately in the compression and bending

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case and could then be taken as the reference for the combined loading case results from the case study since they were all bending mixed with axial compression.

4.3.5.2 The usages of the factored combined loading components to their corresponding maximum capacities for each brace with the SNAME utilisation check =1 have been calculated based on Table 4-3 and are listed in Table 4-6. The ISO utilisation checks resulted from the same applied loading and the brace (Fy D) / (E t) values are also listed in Table 4-6.

Table 4-6 Usages of the factored combined loads from the SNAME unity utilisation check and the resulted ISO utilisations (tubular)

(Fy D) / (E t)

Puc[1]

/ Pp Muy / Mby Muz / Mbz ISO

Utilisation

Horizontal 77.7% 24.3% 0.1% 0.960 Case 1

Diagonal 0.076

94.6% 4.8% 0.7% 0.896

Horizontal 88.0% 6.8% 0.2% 0.798 Case 2

Diagonal 0.016

92.4% 2.4% 2.2% 0.760

Horizontal 93.6% 0.9% 4.3% 0.843 Case 3

Diagonal 0.059

90.9% 5.1% 0.6% 0.802 [1] Puc

is axial compression.

4.3.8.1 Because a tubular brace in compression is subject to column and local tubular buckling it is not possible to develop general expressions and plots for the interaction of compression and moment capacity in the way that was done for bending and tension in Section 4.3.4. However based on the preceding discussion, the following approximate comparisons are provided:

• Comparing Case 1 data in Table 4-6 with Figure 4-4 in Section 4.3.4, the case with the axial capacity usage of 77.7% for the horizontal brace fell into the field between the 60% and 80% moment capacities while the axial usage of 94.6% for the diagonal brace fell into the field for the moment capacity in excess of 80%. These are the zone where the ISO utilisation would be smaller (i.e. the capacity is bigger) than SNAME and the difference would be bigger when the axial percentage is bigger. It can be seen that the ISO utilisation results of 0.960 and 0.896 for each case respectively fit this conclusion very well. The same (matched) conclusions have been drawn when comparing Case 2 data in Table 4-6 with Figure 4-5 and Case 3 data in Table 4-6 with Figure 4-3.

• Another trend being observed from the axial tension and bending general forms in the previous section is that the ISO strength capacity will be smaller, (i.e., its utilisation check will be bigger) with the same axial tension usage when the (Fy D) / (E t) is bigger. With the (Fy D) / (E t) value in the three cases being Case 1 > Case 3 > Case 2, we have the three average utilisations of the ISO to the SNAME as Case 1 > Case 3 > Case 2.

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4.3.8.2 The above comparisons using the case study results indicated that the difference between the ISO and the SNAME code checks for the axial compression and bending were quite similar to that shown in the axial tension and bending strength check general forms.

4.4 COMMENTS ON EQUATIONS AND LIMITS 4.4.1 Based on the case study of the three carefully selected jack-ups and the general form

investigation for tubular members the following comments have been prepared:

• In nearly all the cases studied, the ISO calculations for tubular brace members gave lower utilisation checks than the SNAME method for identical loadings.

• Typical jack-up leg bracing members are designed to carry mainly axial load and this was reflected in the three load cases considered. In all of them the axial load was dominant as can be seen by the results listed in Table 4-3. Consequently the resulting ISO Clause 12 unity check is heavily dependent on the formulation of the axial compression check.

• The axial compression strength check used by ISO is consistently smaller than that produced by the SNAME compression check because ISO recommends the use of a smaller value of effective length factor and in addition the column buckling curve used by ISO Clause 12 is less conservative than that used by SNAME. A further contribution to this trend is the fact that the partial resistance factor used by ISO Clause 12 is smaller (1.15) than that used by SNAME (1/0.85 = 1.176).

• The largest difference found for the cases checked was for the diagonal brace in Case 2. For this comparison the ISO utilisation check was 76% that of the SNAME check for the same member and loads.

• The axial tension strength checks for tubulars using ISO Clause 12 were also found to be constantly smaller than those using SNAME. The main difference was again due to the different values of partial resistance factor for tension used by ISO-Clause 12 (1.05) and SNAME (1/0.9 = 1.111). The ratio between the ISO and SNAME utilisation is therefore 0.945 which explains the tension values reported in Table 4-4.

• The axial compression strength check (including column buckling) formula in the ISO is a complex expression which depends on column buckling slenderness and local buckling sensitivity. The actual jack-up braces checked all fell within the local buckling range of (Fy D) / (2 Cx E t) ≤ 0.170. Given Cx= 0.3, E = 200GPa and Fy = 600MPa, which are typical for a high strength tubular member used in common jack-ups, a tubular size ratio D/t > 34 is required to move the section into the next local buckling slenderness range. To move the tubular into the next part of the curve the limit is (Fy D) / (2 Cx E t) > 1.911, which increases D/t to more than 382.2. For tubulars of this local slenderness the axial capacity is entirely dependent on elastic shell buckling and as these geometries are not realistic for typical jack-up brace members the exact form of the equation is academic. Such large D/t values might occur for tubular legs but as these would have either penetrating holes (for elevating pins) or welded rack bars they would be considered as a “prismatic member” and therefore not be covered by the tubular strength formulations. A further research has found that in the ISO 19902, where these formulae are originally

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adapted from, the third condition of D/t > 382.2 (or (Fy D) / (2 Cx E t) > 1.911) of the axial compression strength check does not exist.

• The bending strength check formulae in the ISO - Clause 12 also depends on the local buckling slenderness and is divided into four ranges depending on whether (Fy D)/(E t) is greater or smaller than 0.0517, 0.1034 and 0.36. For E = 200GPa and Fy = 600MPa these limits become (D/t) greater or smaller than 17.2, 34.5 and 120. Equations are provided for the bending buckling capacities for all values of (D/t) up to 120 but the bending strength for more slender shells is not defined. A general form can be plotted independent of the tubular geometry and material property when (Fy D)/(E t) ≤ 0.0517 for both methods and under this condition. For bending only loading (i.e. when Put/Pp=0) for tubes with this range of slenderness the ratio between the ISO and SNAME utilisation is 0.945. This is a result of the different partial resistance factors used by ISO-Clause 12 and SNAME. The braces in Case 2 are classed in this tube slenderness range and thus the results for bending of Case 2 only follow these results.

• For tubular slenderness that exceeds the above limits the bending capacity is adjusted accordingly by the ISO formulations. The braces in Case 1 with (Fy D) / (E t) = 0.076 and the brace in Case 3 with (Fy D) / (E t) = 0.059 both fall into the second condition of the ISO bending strength function where 0.0517 < (Fy D)/(E t) ≤ 0.1034. Two specific plots were prepared for the Case 1 and Case 3 tube slenderness (see Figure 4-4 and Figure 4-5) Under bending only loading the ratio of ISO to SNAME utilisation for the stockier of the two tubes ((Fy D) / (E t) = 0.059) was 1.033 while the ratio for the more slender tube ((Fy D) / (E t) = 0.076) was 0.982. These small differences highlight the different ways in which local tube slenderness is accounted for by ISO-Clause 12 and SNAME.

• The third section of the bending strength function in the ISO checks covers tube slenderness in the zone 0.1034 < (Fy D)/(E t) ≤ 120 (Fy / E). Again, assuming typically E = 200GPa and Fy = 600MPa for the jack-ups, a tubular size ratio 34.5 < D/t ≤ 120 is required. Notice the maximum threshold 120 for D/t in this case and thus no bending strength can be assigned for tubulars which exceed this limit.

• The axial tension combined bending strength check formulae has been investigated by plotting the bending capacity envelops at varied axial tension usage levels (Put / Pp = 0.2, 0.4, 0.6, 0.8) for the (Fy D)/(E t) ≤ 0.0517 and the other two specific (Fy D)/(E t) cases. It was found that there would be one Put/Pp level for each plot where the ISO and the SNAME bending capacity magnitudes are equivalent. Identifying this load as the “balance level” (or BL) when Put / Pp > BL, the ISO utilisation is smaller (i.e. the capacity is bigger) than the SNAME and the difference ratio would be bigger when Put / Pp is higher. Similar when Put / Pp < BL, the ISO utilisation is bigger (i.e. the capacity is smaller) than the SNAME and the difference ratio is bigger when Put / Pp is lower. It was also found that the ISO axial tension and bending strength check tend to be more conservative - the BL envelop shrinks - with an increased (Fy D) / (E t) value. This could be a problem for a tubular carrying large bending loads where the axial loading is rather low as can be seen in the graphs.

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• The axial compression and bending strength check formulae of the ISO are more complicated than the axial tension and bending formulae but their general form plots are similar. The case study results using combined compression and bending loads determined from the realistic environmental loading have shown the same trends as concluded from the axial tension and bending general form investigations.

• The beam shear strength check and the torsional shear strength check were only undertaken for the ISO since they were not formulated in the SNAME. The resulting utilisations were all below 1.5%. This could be a justification for such checks being ignored in the SNAME code.

• The ISO - Clause 12 has also appended three geometry classification checks for tubular member which have been verified in the detailed spreadsheets attached as Appendix B. These classifications are considered to have made the code more comprehensive.

• The expression for limiting D/t ratio with respect to water depth needs clarification.

4.4.2 Notes on the nomenclature, wording and other discrepancies regarding this section are summarised in Appendix A.

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5 PRISMATIC (CHORD) MEMBERS

5.1 INTRODUCTION 5.1.1 The prismatic members prescribed in the proposed ISO - Clause 12 include brace and

chord members which do not meet the “tubulars” categorisation. For a typical jack up the chords will fall into the “prismatic” category.

5.1.2 The same loading cases and the same processes, used to compare SNAME and ISO-Clause 12 results for tubular member in the previous section, have been used to check the prismatic members, i.e. chords, used for the selected jack-ups in this section. The member classification and the lateral torsional buckling check, prescribed by the ISO only, were also applied to the selected cases when appropriate.

5.1.3 The prismatic checks do not assign a limiting pressure head below which hydrostatic loading can be ignored thus by implication all prismatic members should account for hydrostatic loading. The ISO-Clause 12 code gives no guidance on how this should be done. For the present jack-up design Cases checked in this section it has been assumed that there is no hydrostatic loading as the most highly stressed elements were above water level.

5.1.4 The selected chords, which are representative of three of the main types of chord used in jack-up designs, have all been classified as Class 1 - the “plastic” section - by the strictest slenderness classification condition of the ISO. However, in the ISO calculation for the Case 1 chord (see Figure 3-1), the concept of the reinforced plate is used. This combines the side plate with the side-plate reinforcement as shown in the ISO formula A.12.2-2 and the ISO figure A.12.2-1(Appendix C) and does not require local bucking checks for separated components. An extra classification check on each component of the selected chords has thus been undertaken although for the utilisation calculation the Class 1 section resulting from the proposed ISO has been used i.e. considering the combined section only.

5.1.5 A general review of the ISO prismatic section classification was also undertaken but within a limited range: that of the fully compressed section condition. The general form comparison of the ISO and the SNAME prismatic member utilisation checks has focused on only the Class 1 section as all the selected chords were in this class. We note however that due to the difficulty in the calculation of the slender section properties (Table A.12.3-1 in Appendix C) checks made on more slender sections would require considerable effort. It appears that some of these calculations will be iterative in nature. A general check for the lateral torsional buckling has been made at the critical length/thickness ratio and compared with the Class 1 “plastic" section results.

5.1.6 Comments on nomenclature, wordings and other discrepancies noted in the prismatic member checks found in the draft version of Clause 12 are summarised in Appendix A.

5.2 CASE STUDY RESULTS AND COMPARISONS

5.2.1 The utilisation check and comparison 5.2.1.1 The selected chords representing the three different types are shown in Figure 3-1,

Figure 3-2 and Figure 3-3. The section properties for these sections were calculated from their geometries either by hand or using the in-house software PLASTIC.

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Results are listed in Table 5-1. The detailed description and size of each chord component can be found in the relevant ISO chord spreadsheets in Appendix B.2. The orientation of the y and z axis definitions for these single symmetrical chords is also illustrated in the spreadsheets.

Table 5-1 Chord properties used for utilisation check in SNAME and ISO Clause 12

Case 1 Case 2 Case 3

Chord Type Triangle Compact tubular Moderate tubular

Unbraced Chord Span (m) 3.41 4.26 3.97

Effect. length factor K 1.0 1.0 1.0

Check exponent η 1.8 1.8 1.8

Chord, Fy1 483 Yield stress (MPa) Rack, Fy2 587

690 677

Plastic, Ap 0.107 Section area[1] (m2)

Effect., Af 0.102 0.123 0.159

Effective shear area (m2) 0.051 0.062 0.080

Plastic, Zpy 0.0133 0.0126 0.0371 Y section modulus[1]

(m3) Effect., Sfy 0.0154 0.0083 0.0263

Plastic, Zpz 0.0232 0.0131 0.0265 Z section modulus[1]

(m3) Effect., Sfz 0.0067 0.0081 0.0176

KL/ry 14.696 36.698 15.513 KL/r

KL/rz 21.173 31.805 19.380

[1] Plastic area and modulus are calculated by considering different yield stress of the chord material and the rack material.

5.2.1.2 The effective length factor K was taken as 1 for all the chords for both the SNAME and ISO-clause 12 methods. The modifying formulae exponent η has been found to have little effect on the comparison of the utilisation checks between the two methods so a general value of 1.8 was assumed for all checks using both methods in this review.

5.2.1.3 The shear area used for the ISO code check has been simplified as half of the section area for both tubular and prismatic members in all the selected cases. This assumption is conservative for tubulars and is also conservative for most typical chord sections. The chord sections have all been found to meet the conditions for Class 1 - plastic. Only the compact tubular chord of case 2 required a lateral torsional buckling check by not meeting the requirements of Section A.12.2.3.2. However the detailed lateral torsional buckling check in Section A.12.6.2.6 indicated that this

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member was not sensitive to lateral torsional buckling. The details for these cases can be found in Appendix B.2 and the results will be discussed later in this section.

5.2.1.4 The loading cases for the chords were obtained from the leg PAFEC FE analyses described in Section 3.3. The highest member loads were selected from the realistically estimated environmental conditions that were considered and then factored up until the SNAME utilisation checks reached unity. The results are listed in Table 5-2. The SNAME code check spreadsheets for both tubular and prismatic members are attached as Appendix B.3. It can be seen that the components of factored combined loads for the three chords when compared to their corresponding maximum component capacities have very different proportions: Case 1 was 47% maximum compression plus 59% maximum y bending, Case 2 was 90% maximum compression dominating and Case 3 was 98% maximum y bending dominating. These were quite different from the loading cases obtained for the tubular brace members where all the combined loading was compression dominated.

5.2.1.5 Again, it is worth noting that the shear force and torsional moment listed in Table 5-2 were taken directly from the PAFEC model results for the same chord elements, scaled to the same level of the combined load components. These were only used for the ISO code check in this report since the SNAME formulations take no account of shear capacity.

Table 5-2 Loading cases used for comparing the ISO chord utilisation check with SNAME (the loads calculated or factored to bring the SNAME check to unity)

Combined loads Unit: kN, kN·m

Max axial tension

Max axial compression

Max y bending

Max z bending

Shear force Torque

Axial Bdg y Bdg z

Case 1 4.58x104[1] -4.12 x104 5780 9830 26 1.9 -1.95x104 -3433 112

Case 2 7.63x104 -5.92x104 7790 8100 219 27 -5.33x104 -729 -0.7

Case 3 9.70x104 -8.68x104 2.26x104 1.61x104 295 0.4 1850 -2.21x104 -2030 [1] Tensile load is positive

5.2.1.6 The ISO utilisation check results using the SNAME factored loading cases from Table 5-2 and the proposed Clause 12 prismatic formulations are listed in Table 5-3. As noted above all the chords studied were classified as “plastic” sections and passed the lateral torsional bucking strength check, so that Table 5-3 shows only the utilisation checks for a plastic section chord.

Table 5-3 The ISO Clause 12 utilisation checks for the SNAME checked unity loading cases of the selected jack-up chords

Max axial tension

Max axial compression

Max y bending

Max z bending

Shear force

Section torque

Combined loads

Case 1 0.927 0.922 0.944 0.923 0.019 0.004 0.992

Case 2 0.945 0.870 0.945 0.945 0.013 0.001 0.877

Case 3 0.945 0.935 0.945 0.945 0.074 0.017 0.957

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5.2.1.7 It was found that the SNAME code check was more conservative than the proposed ISO code in all the combined and component loading cases in Table 5-3 for the selected jack-up chords. Compared to the SNAME checks having a utilisation of unity the lowest utilisation checks from the ISO method were 0.870 for the maximum axial compression strength check of the compact tubular chord in Case 2.

5.2.1.8 A value of 0.945 for the same comparison was been found in the maximum axial tension and bending strength checks, respectively, for the compact tubular chord in Case 2 and the moderate diameter tubular chord in Case 3. The reason is the same as for tubular cases, i.e., the somewhat similar formulae but different resistance factors (fixed) used in the two code-checking formulations for the two kinds of utilisation checks. The exception for the triangular chord in Case 1 was due to the requirements for a chord section with different yield stresses in rack and main chord in the ISO formulae: Fymin instead of Fy and plastic section area and modulus, Ap and Zpy(z), instead of geometrical (effective) area and modulus Af and Sfy(z), as shown in Table 5-1 and Appendix C. In the SNAME check, an effective yield stress Fyeff = (Fy,rack x Arack + Fy,chord x Achord)/(Arack + Achord) rather than Fymin and the geometrical Af and Sfy(z) rather than Ap and Zpy(z) are used (ref [1]). Using Fymin is more conservative than using Fyeff while the use of Af and Sfy(z) is more conservative than the use of Ap and Zpy(z).

5.2.1.9 It is worth noting that the moment application parameter Cm used in both methods has been defined as the same for both codes. For simplification of comparisons a value of 0.85 was taken for both codes.

5.2.1.10 For the Case 1 and Case 2 chords the largest factored shear/torsional utilisation found in those members examined was no more than 1.7% using the ISO code. The result for the Case 3 chord gave a somewhat higher value of 7.4%.

5.2.2 The lateral torsional buckling check for Case 2 5.2.2.1 As prescribed in the ISO (section A. 12.2.3.2 in Appendix C), prismatic members

which does not satisfy the following criterion need to be checked for lateral torsional buckling:

b y yL / r 1.51 E / F≤ min

A reference “TABLE A-F1.1(c)” is quoted for this check in the ISO draft. If this refers to “TABLE A-F1.1(cont’d) - Nominal Strength Parameters” in the “ Load & Resistance Factor Design, 1st Ed., 1986” [3], the ISO formulations for the slenderness Lb/ry is taken from elsewhere.

5.2.2.2 The LTB formulation used in SNAME is also adapted from “TABLE A-F1.1” in Ref [3], which is:

pyb MJArL /)(25860/ ≤

where J is the torsional constant for the section and Mp is the section plastic moment. The units used here is m and MPa. It is worth noting that the above formula has been incorrectly printed in the SNAME 5-5A Rev-2 document (Ref [1]) as pyb MJArL /)(25860/ ≤ .

The LTB check results calculated from the three selected chords using the Clause and the SNAME formulations, respectively, are list in Table 5-4.

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Table 5-4 The ISO classification for lateral torsional buckling check of the selected chords

Code LTB Check Equation Case 1 (triangular)

Case 2 (compact tubular)

Case 3 (moderate tubular)

ISO (Lb/ry)/(1.51√(E/Fymin)) <1.00

0.69 1.43 (>1, fail) 0.75

SNAME (after

correction) (Lb/ry)/(25860√(JA)/Mp)

< 1.00 0.20 0.73 0.31

The uniformed check results of the ISO are all larger than the corresponding SNAME ones. Based on the SNAME calculations, all the three chords were exempted from the LTB check. Based on the ISO, Case 2 has to take the LTB strength check followed in the draft.

5.2.2.3 For the compact tubular chord in Case 2 it was found necessary to consider the lateral torsional buckling strength for the doubly symmetric member in the following way:

When Lb ≤ Lr: Mlt= Sf Fcr

Fcr = 6.895

21

22

211

21

2

⎭⎬⎫

⎩⎨⎧

+λλXXXCb

where

Cb = 1.75 + 1.05(M1/M2) + 0.3(M1/M2)2 ≤ 2.3

X1 = (π/Sf) )/EGJA( 2

X2 = (4Cw/Iz)(Sy/GJ)2

and Cw is warping constant.

Lr = see AISC (1999 version) section F2.2c 9 (ref [2])

When Lb > Lr:

Mlt= Sf Fcr

where

Fcr = the lowest value of the critical buckling stress from (where appropriate)

The full definitions for other variables are in Appendix C.

5.2.2.4 Some checks were completed using the lateral torsional buckling equations provided in section A.12.6.2.6 of the Clause 12 document. These checks considered the chords as a closed section, for which the warping displacement can be ignored, it follows that Cw = 0, X2 = 0. As a conservative simplification, M1/M2 can be taken as -1, thus Cb = 1. The lateral torsional buckling strengths calculated for the Case 2 chord section are listed in Table 5-5. The axial plastic column buckling strengths for the same section are also listed in Table 5-5 for comparison. These results confirm that

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for the given chord section lateral torsional buckling will only occur at very large unbraced spans and that for realistic spans the plastic section bending capacity will govern.

Table 5-5 The lateral torsional buckling strength and the plastic buckling strength of the chord case 2 (calculated by the ISO)

Lateral torsional buckling Plastic column buckling

Fcr, MPa Mby, MN·m Mbz, MN·m Fcr, MPa Mby, MN·m Mbz, MN·m

153643.8 1271.99 1241.91 636.2 8.66 9.0

5.2.2.5 It can be seen from Table 5-5 that the lateral torsional buckling stresses and moments of the studied section were very considerably higher than the plastic buckling stress and moment values. Therefore the normal plastic strength stresses have been used for the utilisation check (as discussed in the previous sub-section 5.2.1).

5.2.2.6 A general form investigation for this LTB - lateral torsional buckling check based on the selected chord sections has been undertaken and is discussed later in Section 5.3.

5.2.3 Classification for the separated stiffener and plate components 5.2.3.1 It has already been noted that using the ISO classification method the stiffener

components in the triangular chord of case 1 (see Figure 3-1) and the moderate diameter tubular chord of case 3 (see Figure 3-3) are combined with the base plates to produce an effective reinforced plates . The ISO check does not require a buckling check of the individual plate components. An extra check on each plate component in the two chords has been undertaken and the results are compared with the previous ISO calculations in Table 5-6. In this check the sections were assumed to be uniformly compressed in order to avoid the complicated formulations for such sections subjected to non-uniform compression. The chord component geometries, the ISO classification results and the other intermediate results including the sectional formulation results for prismatic member, which are not shown in the main report, are detailed in Appendix B.2

Table 5-6 Comparison of the separated stiffener / plate classification and the reinforced plate classification from the ISO

Chord Component √Fy(d/tf) Class Note (Appendix C)

Wall plate 537.61 3

Wall stiffener 263.73 1 Case 1 triangular chord

Reinforced wall plate 301.13 1

Rack plate 113.21 1

Rack stiffener 732.50 4 Case 3

moderate diameter tubular chord

Reinforced rack plate 366.25 1

√Fy(d/tf) ≤ 465, Class 1 - plastic;

465<√Fy(d/tf) ≤ 530, Class 2- compact;

530<√Fy(d/tf) ≤ 650, Class 3 - semi-compact;

Class 4 - slender

5.2.3.2 It can be seen from Table 5-6 that if the component plates on the chord sections are calculated separately, the triangular chord of case 1 would be classified into Class 3 -

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“semi-compact”, and the moderate diameter tubular chord of case 3 would be classified as a Class 4 - “slender” section. The utilisation checks using Class 3 and Class 4 formulations for these classified components have not been undertaken in this study because the present version of Clause 12 requires that only the combined effective plate be checked. As this study has shown that the components of typical sections are likely to be classed as “semi-compact” and “slender” there will be a real benefit in simplifying the formulations that are used for such elements.

5.3 GENERAL FORM ASSUMPTIONS AND COMPARISONS

5.3.1 General geometry classification for fully compressed chord section 5.3.1.1 A general investigation of the ISO classification has been undertaken for the

uniformly compressed section condition. This avoided the iterative << see previous comment >>calculations required by the ISO Clause 12 methodology for sections where the stress varies across the section as a result of bending. Under this assumption, the critical length/thickness ratios for the classification with the three typical yield stresses are listed in Table 5-7. The definitions of flange and web can be seen in Appendix C.

Table 5-7 Length/thickness ratios of the ISO classification for a fully compressed prismatic member section (ref to Appendix C)

Flange outstand component Flange & web internal component Yield stress b/t √Fy(b/tf)-welded d/t √Fy(d/tf)

Class

≤ 21.92 √Fy(d/tf) ≤ 465 1 “Plastic” ≤ 6.36 √Fy(b/tf) ≤ 135

≤ 24.98 465<√Fy(d/tf) ≤ 530 2 “Compact”

≤ 10.61 135<√Fy(b/tf) ≤ 225 ≤ 30.64 530<√Fy(d/tf) ≤ 650 3 “Non” or

“Semi-Compact”

450 MPa

> 10.61 225<√Fy(b/tf) > 30.64 650<√Fy(d/tf) 4 “Slender”

≤ 18.98 √Fy(d/tf) ≤ 465 1 “Plastic” ≤ 5.51 √Fy(b/tf) ≤ 135

≤ 21.64 465<√Fy(d/tf) ≤ 530 2 “Compact”

≤ 9.19 135<√Fy(b/tf) ≤ 225 ≤ 26.54 530<√Fy(d/tf) ≤ 650 3 “Semi-Compact”

600 MPa

> 9.19 225<√Fy(b/tf) > 26.54 650<√Fy(d/tf) 4 “Slender”

≤ 15.95 √Fy(d/tf) ≤ 465 1 “Plastic” ≤ 4.63 √Fy(b/tf) ≤ 135

≤ 18.18 465<√Fy(d/tf) ≤ 530 2 “Compact”

≤ 7.72 135<√Fy(b/tf) ≤ 225 ≤ 22.29 530<√Fy(d/tf) ≤ 650 3 “Semi-Compact”

850 MPa

> 7.72 225<√Fy(b/tf) > 22.29 650<√Fy(d/tf) 4 “Slender”

5.3.1.2 It is noted that these definitions are not as satisfactory for the chord sections used in jack-ups as they would be for beams used in framed building structures where gravity loading dominates and the direction of loading does not vary substantially during the life of the structure. For a jack-up the loading direction in a chord will

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vary considerably depending on the direction of the storm and the position of the leg within the guides.

5.3.2 General LTB (lateral torsional buckling) curves for the selected chord sections 5.3.2.1 A lateral torsional buckling (LTB) strength check is included in ISO Clause 12 as

item A.12.6.2.6. A general investigation is undertaken in this section.

5.3.2.2 The requirements of item A.12.6.2.6 (no equation numbers given) were found to be confusing to interpret. For beam spans below a limiting length Lr , the beam strength is given as Mlt = Sf Fcr but this should presumably be either Sf.Fy or Zp Fy.

5.3.2.3 The reference for Lr is given as “AISC Section F2.2c” and this was interpreted to mean a reference to the nominal strength parameter tables given in Ref [2].

5.3.2.4 It was found in the case study that the ISO Clause 12 lateral torsional buckling (LTB) strength checks calculated for the case 2 chord section members produced section bending moments far in excess of the elastic or plastic bending capacity of the section. This was as expected as lateral torsional buckling is not a failure mode to be expected for closed sections having a rather small span and with rather similar bending capacity about the two principal axes.

5.3.2.5 The LTB bending moment for the case 2 chord is plotted against span dimension in Figure 5-1. The LTB moment curve, Mlb = SfFcr.LTB, in Figure 5-1 was calculated by the doubly symmetric formulation described in Section 5.2.2 (and Appendix C) with the same assumptions of Cw =0 and Cb = 1 and for the minor axis of the case 2 chord section. The horizontal pink line (Mp) is the full section plastic moment calculated as ZpFymin in Appendix C. This intersects the buckling curve at the unbraced span dimension Lp. The horizontal green line is set to represent the moment at which the maximum bending stress in the section reaches the critical axial buckling stress Fcr,axial . This intersects the buckling curve at an unbraced length of La.

5.3.2.6 It was not clear how to calculate the Lr for a non-I shaped member section from the “AISC Section F2.2c” as in Ref [2].

5.3.2.7 According to the ISO classification formulation given in Section A.12.2.3.2 (see also Section 5.2.2 of this report) the length given by b yL / r 1.51 E / F≤ y min can be used to identify members for which lateral torsional buckling checks are required. For the chord in case 2 this length is given by L1.51 = 2.99m. As the actual span of the chord in this case is about 4.3m (see Table 5-1) according to ISO Clause 12 this chord must be checked for LTB.

5.3.2.8 However, based on Figure 5-1, the resulting first limit Lp = 679.3m, represents the member length at which the LTB moment Mlb equals the plastic moment for the section. Clearly this particular result indicates that for this doubly symmetric chord with a span of about 4.3m (see Table 5-1) lateral torsional buckling is not a worthwhile consideration.

5.3.2.9 In Figure 5-1 the intersection of the LTB curve and the SfFcr.axial line indicates a critical length La = 1189m at which LTB will occur when the maximum bending stress in the section reaches the critical buckling stress. Interpreting the meaning of this value is difficult. In the ISO referred AISC (1999 version) table A-F1.1 [2], only the I-shape beam formulation for the second limit Lr is given. This is not particularly suitable for a jack-up chord section because a jack-up chord section usually has similar bending strength for the two axis directions. Nevertheless, in the ISO LTB

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check definition, when Lr > La, the blue curve (Mlb = SfFcr.LTB) will continue after the Lp and when Lr ≤ La, the green line of SfFcr.axial will connect the blue curve (Mlb = SfFcr.LTB) between the Lp and the Lr.

Case 2 compact tubular chord section (double symmetry)

Mlb = SfFcr.LTB

Mp

SfFcr,axial

Lp LaL1.510

5

10

15

20

0 1000 2000 3000 4000 5000

Unbraced member length Lb, m

Buc

klin

g m

omen

t M, M

N•m

Figure 5-1 The LTB moment options against different Lb for the case 2 doubly symmetry chord section

5.3.2.10 The moderate diameter tubular chord of case 3 has also a doubly symmetrical section and it gave a very similar LTB curve as shown in Figure 5-1. The curve and discussion has thus been omitted in the report. The triangular chord section of case 1 is singly symmetry and used the following formulations for the lateral torsional buckling check:

Fcr = bf

b

LSC000,393 {B1+ )BB( 2

121 ++ } )( JI z

where

B1 = 2.25 2

1

12⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛JI

Lh

II z

bz

c

B2 = 25 ⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−

JI

Lh

II c

bz

c

2

1

h = web depth.

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Ic = second moment of area of compression flange about the section minor axis.

5.3.2.11 Again, the web depth and compressed flange indicated in the above formulations are not quite agreeable with the triangular chord section. To apply these equations to the selected triangular chord, conservatively, Ic was assumed to equal to Iz and h was taken as the base plate thickness t1, as detailed in Appendix B.2. The calculated LTB moment curve, Mlb = SfFcr.LTB, with the Mp and the SfFcr.axial lines for the studied section are shown in Figure 5-2.

5.3.2.12 Based on Figure 5-2, the first limit Lp = 67.86m, which was much lower than the result obtained for the case 2 section but still too large for a jack-up chord member. This indicated that such a LTB check is probably redundant for closed section jack-up chords which are not well represented by checks that are more applicable for I-shaped sections. The ISO classification limit L1.51 = 4.81m which, again, was within the plastic moment Mp dominating region. The cross point of the LTB curve and the SfFcr.axial line gave La = 240.7m, which was far beyond the possible range of an unbraced jack-up chord length.

5.3.2.13 Based on the two limited general investigations above, the lateral torsional buckling check prescribed in the ISO was considered redundant for the commonly used closed section shaped chords typically used for jack-ups.

Case 1 triangular chord section (single symmetry)

Mlb = SfFcr.LTB

SfFcr,axial

L1.51

Mp

Lp La0

5

10

15

20

0 500 1000 1500 2000

Unbraced member length Lb, m

Buc

klin

g m

omen

t M, M

N•m

Figure 5-2 The LTB moment options against different Lb for the case 1 singly symmetry chord section

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5.4 STRENGTH UTILISATION COMPARISONS 5.4.1 The following general form investigation for the utilisation check of the ISO and the

comparison with the SNAME has been focused on Class 1 (Class 2 utilisation check is found same to Class 1) prismatic member section.

5.4.2 Axial Tension Strength Check 5.4.2.1 The chord axial tension strength check formulated in the ISO - Clause 12 is:

Put ≤ A Fymin / γt The only difference from the tubular member formula (A.12.5-3) is Fy replaced by Fymin in this case. In the SNAME, Fyeff takes place of Fy. It has been discussed and demonstrated in the chord case study, the relationship for the axial tension strength checks of the ISO prismatic code and the SNAME chord code is the same as for their tubular member checks when all the components have same yield stress.

5.4.3 Axial Compression Strength Check 5.4.3.1 The axial compression strength check formulae used by the ISO - Clause 12 is

expressed by different equations depending on the value of yield stress and column slenderness (see equation numbers as in Appendix C[1]):

For low strength steel (Fy ≤ 450 MPa):

Fcr = (0.658λc2) Fymin for λc ≤ 1.5 [Eq. E2.2]

Fcr = ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

2

877.0

cλ Fymin for λc > 1.5 [Eq. E2.3]

For high strength steel (Fy > 450MPa):

Fcr = (0.7625λc3.22) Fymin for λc ≤ 1.2 (A.12.7-1a)

= (0.8608/λc1.854) Fymin for λc > 1.2 (A.12.7-1b)

where

λc = 2

1

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π (for max KL/r from all directions)

Note that when Fy (or Fymin) ≤ 450 MPa, the ISO formulations are very similar to the SNAME formulations [1] (they are exactly same when Fymin = Fy = Fyeff). Their equation numbering is actually the same as that in SNAME. However, their partial resistance factor γc would be different when used for utilisation: a divisor of φt = 0.85 is used by SNAME and a multiplier of γt =1.1 (for Fy ≤ 450 MPa) and 1.15 (for Fy > 450 MPa) in the case of the ISO Clause 12 assessment. Thus, when Fy ≤ 450 MPa and Fymin = Fy = Fyeff, the difference between the two utilisation checks for the axial compression strength of the chords is a constant factor of Uc(ISO) = 0.85/(1/1.1) Uc(SNAME) = 0.935 Uc(SNAME).

5.4.3.2 With the assumption of Fymin = Fy = Fyeff, the general form for the axial compression strength check of chords using the two methods are compared in Figure 5-3 and Figure 5-4. Figure 5-4 shows the same plots as in Figure 5-3 but with the (KL/r) in a smaller range of 0~80.

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Fy = 450MPa

Fy = 850MPa

0

0.2

0.4

0.6

0.8

1

0 50 100 150 200

KL/r

Colu

mn

buck

ling

stre

ngth

, Fcr

/Fy

ISO Eq. E2.2-E2.3 (whenFy<=450MPa), Fy=450MPa

ISO A.12.7(whenFy>450MPa), Fy=450MPa

SNAME, Fy=450MPa

ISO A.12.7(Fy>450MPa),Fy=850MPa

SNAME, Fy=850MPa

Figure 5-3 Comparison of the axial compression strength checks of the ISO-Clause 12 and the SNAME 5-5A when Fymin = Fy = Fyeff (prismatic)

Fy = 450MPa

Fy = 850MPa

0.4

0.5

0.6

0.7

0.8

0.9

1

0 10 20 30 40 50 60 70 8

KL/r

Col

umn

buck

ling

stre

ngth

, Fcr

/Fy

0

ISO Eq. E2.2-E2.3 (whenFy<=450MPa), Fy=450MPaISO A.12.7(whenFy>450MPa), Fy=450MPaSNAME, Fy=450MPa

ISO A.12.7(Fy>450MPa),Fy=850MPaSNAME, Fy=850MPa

Figure 5-4 Comparison of the axial compression strength checks of the ISO-Clause 12 and the SNAME 5-5A when Fymin = Fy = Fyeff (prismatic, close form)

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5.4.3.3 In the comparison, the two specific yield stress values used for the tubulars, 450 MPa and 850 MPa, were also selected to give a range of assessment. In addition the yield stress value of 450 MPa represents one of the expression limits used by the ISO formulae.

5.4.3.4 It can be seen in Figure 5-3 that the ISO Fy ≤ 450 MPa formulation (blue) curve lies above the SNAME strength check (green) curve for Fy = 450 MPa case, as expected, but the ISO functions have given quite different results (blue and pink) when the material yield stress changes from just below to just above 450 MPa. The step difference which occurs in column strength at the threshold in the yield stress is clearly undesirable and the committee may wish to consider modifying these formulations to remove the inconsistency.

5.4.3.5 It can be seen in both Figure 5-3 and Figure 5-4, the ISO strength curves are always above the SNAME ones. Especially for KL/r = 10 ~ 70, the difference between the higher ISO strength checks and those of SNAME is quite significant and the maximum percentage is about 18.4% when Fy = 450 MPa for KL/r = 60 and 18.6% when Fy = 850 MPa for KL/r = 45. This might be acceptable due to the fact that the SNAME code has been quite conservative by using the low strength steel formula for the high strength steel sections. The difference curves, (ISOc - SNAMEc) / SNAMEc, are also plotted against column slenderness at three yield stress levels in Figure 5-5 for further comparison.

5.4.3.6 Comparing the two high strength ISO curves with different yield stresses in Figure 5-3, it can be seen that their curvatures are very similar along the KL/r and the higher relative capacity Fcr/Fy was obtained from the lower yield stress (450MPa) case. The two curves of the SNAME check results show the same trends.

5.4.3.7 To demonstrate the case study results from the general forms, the ISO and SNAME utilisation checks of the maximum axial compression for each chord, which have been listed in Table 5-3, are re-listed in Table 5-8, together with the KL/r and the yield stress data used for each chord. The resulting difference in capacity (%) for the three cases is plotted in Figure 5-5 together with the general forms at the same yield stress level. It can be seen that the case study results agree quite well with the general form plots (noting the slight difference of the yield stress in Case 1 & Case 3 and also the fact in Case 1 that Fymin ≠ Fy ≠ Fyeff).

Table 5-8 Comparison of the axial compression strength check of the SNAME and the ISO methods regarding to KL/r (prismatic)

Axial Compression Strength Check

Fymin (MPa) (KL/r)max ISOu / SNAME u ISOc/SNAME c %

Case 1 483 21.173 0.922 1.085 8.5

Case 2 690 36.698 0.870 1.149 14.9

Case 3 677 19.380 0.935 1.070 7.0

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0

0.05

0.1

0.15

0.2

0 10 20 30 40 50 60 70 8

KL/r

Col

umn

buck

ling

stre

ngth

com

paris

on,

(ISO

c - S

NAM

E c) /

SN

AME c

0

Fy=450MPaFy=690MPaFy=850MPaCase 1, Fy=483MPaCase 2, Fy=690MPaCase 3, Fy=677MPa

Figure 5-5 The (ISOc - SNAMEc) / SNAMEc curves deduced from the ISO-Clause 12 and the SNAME 5-5A compression strength comparison when Fymin = Fy = Fyeff

(prismatic)

5.4.3.8 Note that all the curves shown in Figure 5-5 have been deduced from the ISO high

strength steel formulae (A.12.7 in Appendix C) and the corresponding equations used in SNAME since all the selected chords were made of steel with Fy > 450MPa as shown in Table 5-8.

5.4.4 Axial Tension and Bending Strength Check 5.4.4.1 The axial tension strength check for the two codes has been compared in Section

5.4.2 and the difference between the two methods has been found as a fixed factor (=0.945) when Fymin = Fy = Fyeff. In this section, bending capacity envelopes with different axial tension strength usages (selected typical Put/AFy values) are checked and compared.

5.4.4.2 The axial tension and bending strength check formulae used by the ISO - Clause 12 are similar to the SNAME formulations for prismatic member case except for the different partial resistance factors:

If γtPut/AFy > 0.2

1.0MM

MM

98

PP

1

bz

uz

by

uy

p

ut ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bbt [Eq. H1-1a]

else

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1.0MM

MM

2PP

1

bz

uz

by

uy

p

ut ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bbt [Eq. H1-1b]

where

Mb = Zp Fymin

Zp = (Σ Fyi di Ai)/ Fymin

Zp is fully plastic section modulus as being listed in Table 5-1. For the studied cases 2 & 3, Zp = Sp as Fymin = Fy = Fyeff but for case 1, Zp ≠ Sp as Fymin ≠ Fy ≠ Fyeff,.

5.4.4.3 The general form comparison of the ISO and the SNAME formulations for the axial tension and bending strength check at varying axial tension levels is shown in Figure 5-6, with the same assumption of Fymin = Fy = Fyeff. The different partial resistance factors, 1.05 for both tension and bending in the ISO (γt and γb) and 0.9 for both tension and bending in the SNAME, were considered in all the plots. The modifying formulae exponent η has been found to have little effect on the comparison of the utilisation checks from the two methods so that a general number of 1.8 was assumed for both methods in this review.

Put/Pp=0.8

Put/Pp=0.5

Put/Pp=0.3Put/Pp=0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Muey/Mby

Mue

z/M

bz

ISO, 80%AxialCapacity

SNAME, 80%AxialCapacity

ISO, 50%AxialCapacity

SNAME, 50%AxialCapacity

ISO, 30%AxialCapacity

SNAME, 30%AxialCapacity

ISO, 0%AxialCapacity

SNAME, 0%AxialCapacity

Figure 5-6 Comparison of the axial tension and bending strength check of the ISO-Clause 12 and the SNAME 5-5A when Fymin = Fy = Fyeff (prismatic)

5.4.4.4 In Figure 5-6 it is seen that the ISO relative bending capacity Μuy(z)/Mby(z) are always larger than the SNAME envelops and the difference of the radius is a constant value of about 0.059 for the cases of Put/Pp > 0. A further examination has found this

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constant difference of 0.059 is universal for any γtPut/AFy > 0.2. This can be demonstrated by the following equation transformation:

When Μuy/Mby = 0, γt = γb = 1.05 for ISO, φt = φb = 0.9 for SNAME

(Μuy - Mby)iso = 98 (1- γt Put/Pp)/γb =

98 (1/1.05- Put/Pp)

while

(Μuy - Mby)sname = 98 (1- Put/φt Pp) φb =

98 (0.9- Put/Pp)

thus

(Μuy - Mby)iso - (Μuy - Mby)sname = 98 (1/γb - φb) = 0.05893

Similar for γtPut/AFy ≤ 0.2, When Μuy/Mby = 0,

(Μuy - Mby)iso - (Μuy - Mby)sname = (1/γb - φb) = 0.05238

This is the same value obtained from the two radiuses of the Put/Pp = 0 curves shown in Figure 5-6.

5.4.4.5 However, if the two general forms are compared at, for instance, the Put/Pp = 0.8 level, the capacity envelope of the ISO curve permits a 50% bigger moment to be supported than the SNAME one. This large difference may suggest the need for a more detailed examination in order to consider whether the new ISO proposed results are sufficiently safe. The ISO utilisation proportional to the SNAME utilisation at Put/Pp = 0 is a constant ratio of 0.945, which has been demonstrated during the case study discussion and it is also the minimum ratio for Μuy(z)/Mby(z), according to the general forms shown in Figure 5-6

5.4.4.6 The realistic loading case for the moderate diameter tubular chord in Case 3 has been an axial tension and bending combined case with Put/Pp = 0.02 and Μuy/Mby = 0.98. The resulted relative utilisation of 0.957 for the ISO to the SNAME seems reasonable based on the above investigation.

5.4.5 Axial Compression and Bending Strength Check 5.4.5.1 The axial compression and bending strength check formulae used by the ISO Clause

12 and the SNAME 5-5A for prismatic member would be quite similar to the axial tension and bending strength check formulae quoted previously when Fy ≤ 450 MPa, with a consideration of the different resistance factor used for the axial compression component in each method. However, the distances between the bending envelopes of the ISO and the SNAME at different axial usage levels of γtPut/AFy > 0.2 would not be same because the partial resistance factors for axial compression and bending are different. A detailed general form investigation on this Fy ≤ 450 MPa section will not be further discussed in this report because the representative cases selected all had Fy > 450 MPa.

5.4.5.2 The axial compression strength formulae differ between ISO and SNAME when Fy > 450MPa and involve both geometry and material information. Thus, following the tubular cases, the general form plots for this formula section have been created under the Case 1 and Case 2 conditions, which have axial compression and bending combined loading cases as listed in Table 5-2. Based on Table 5-2 and Table 5-3,

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the usages of the factored combined loading components to their corresponding maximum capacities for the two chords using the methods were calculated and are listed in Table 5-9. The conditioned general forms of the ISO bending envelop with the specific axial compression usages and the resulted bending utilisations are cross-checked in Figure 5-7. It should be noted that the load component usages are different in the two methods due to the different strength check formulae and parameters.

Table 5-9 Usages of the factored combined loads from the SNAME unity utilisation check and the resulted ISO utilisations (prismatic)

SNAME load component usage ISO load component usage

Puc/ Pp Muy / Mby Muz / Mbz Puc/ Pp Muy / Mby Muz / Mbz

ISO Utilisation

Case 1 47.3% 59.4% 1.1% 47.3% 59.4% 1.1% 0.992

Case 2 90.0% 9.4% 0.0% 78.3% 9.4% 0.0% 0.877

rcPuc/Pp=0.436

rcPuc/Pp=0.783

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

Muey/Mby

Mue

z/Mbz

Case 1, triangular chord

Case 2, compact tubular chord

Figure 5-7 The axial compression and bending strength checks of the ISO-Clause 12 for Case 1 triangular and Case 2 compact tubular chords

5.5 COMMENTS ON EQUATIONS AND LIMITS 5.5.1.1 As a result of the case studies of typical jack-up chords described above we have the

following comments on the proposed ISO Clause 12 prismatic member checks:

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• The ISO chord classification combines parallel welded plates into one effective plate thickness but does not require checking of the component plate elements. For the cases checked it appears there should also be a requirement to check each component. Consideration may be required as to whether the resulting re-classification of a section from “plastic” to “compact” or “non-compact” and the implications this has on the capacity of these sections is too conservative.

• A fully compressed section has been assumed in the general investigation of the chord classification. This avoided the iterative calculation of the complicated formulations involving bending, which demands not only section property information but also a knowledge of the axial and bending force components. In view of the variations in axial and bending moments that a chord section is subjected to the present formulations are extremely complex and will be difficult to implement.

• A general lateral torsional buckling check has been investigated for both doubly symmetric and singly symmetric cases based on the selected chord sections. The first limit length Lp of 679.3m and 67.86m, resulted from the doubly (compact tubular) and singly (triangular) symmetric cases respectively. These indicate that typical chords will not suffer LTB restrictions. However the presently proposed test for whether an LTB check is required is set such that some typical chords will need to be checked in this way. A further study shows that the LTB requirement check formulation in the ISO is not consistent with the formulation in AISC (1st version, 1986, Ref [3]), which the ISO claimed to be referred to and which does be used by the SNAME.

• The axial tension strength check formulae for prismatic member in the ISO - Clause 12 is found very similar to the formulae used for tubular member but with a Fymin used instead of Fy. When Fymin = Fy, the axial tension strength check by the ISO for the same chord is constantly 94.5% of the utilisation resulted from the SNAME.

• The axial compression strength check (including column buckling) formulae in the ISO is a 2-section conditioned function with an enhanced formula for the high strength steel when Fy > 450MPa. It is found that at the Fy = 450MPa threshold, the two functions give different capacity curves when plotted against KL/r. It is also found that when Fy ≤ 450MPa, the ISO axial compression strength formulae are similar to the SNAME's except for the different partial resistance factors used and the different effective yield stress (Fymin and Fyeff) adopted for non-uniformed chord materials. In cases where Fymin = Fyeff ≤ 450MPa, the axial compression strength check by ISO Clause 12 for the same chord is 93.5% of the utilisation from SNAME.

• For Fymin = Fyeff = Fy = 850MPa > 450MPa, a 18.6% higher compression capacity was found along the Puc/Pp against KL/r curve using the ISO Clause 12 formulae when compared with those of SNAME. Note however that the SNAME code includes a conservatism by using the low strength steel formula for the high strength materials. The materials used in the selected chords in the report are Fy = 483 / 587 MPa (Case 1), 690 MPa (Case 2), and 677 MPa (Case 3), respectively.

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• The bending strength check formulae in the ISO Clause 12 is similar to that in SNAME except for the different partial resistance factors used and the different effective yield stress (Fymin and Fyeff) adopted for the section modulus calculation. When Fymin = Fyeff = Fy, the pure bending strength utilisation checks (i.e. when axial loading = 0) from the ISO is 5.5% lower than the utilisation results from the SNAME.

• For the axial tension and bending combined cases the relative bending capacity envelopes Muy(z)/Mby(z) from ISO Clause 12 are 0.059 larger than the SNAME bending envelopes when γtPut/AFy > 0.2 and 0.052 larger when γtPut/AFy ≤ 0.2. When comparing the relative radiuses of the bending capacity by the proportion, the capacity envelope of the ISO is more than 50% bigger than the SNAME envelope at the Put/Pp = 0.8 level.

• For the combined axial compression and bending cases the strength check formulae used for prismatic members by the ISO - Clause 12 and SNAME 5-5A are similar to the axial tension and bending strength checks when Fy ≤ 450 MPa. The relative bending capacity envelopes Muy(z)/Mby(z) of the ISO are always larger than the SNAME bending ones but the bending magnitudes at different axial usage levels are no longer constant due to the difference of the partial resisstance factors for axial compression and bending.

• The selected chords in this report all had Fy > 450 MPa and the combined loading case results from the realistic environment estimates are of two kinds: axial tension and bending for Case 3 and axial compression and bending for Case 1 and Case 2. The utilisation comparison from the case study has been cross-checked using the limited general form plots and the ISO method gave the lower utilisation checks for all the studied loading cases in the chords.

• The beam shear strength check and the torsional shear strength check were only undertaken for the ISO since they were not formulated in SNAME. The resulted utilisations were small.

• The selected jack-up cases, which are considered representative, have all had chord sections that could be classed as Class 1 - “plastic” sections. As a result the strength check formulae for the other classes, especially Class 4 - slender section - have not been reviewed in this report.

5.5.2 Notes on the nomenclature, wordings or other discrepancies regarding this section are summarised in Appendix A.

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6 SUMMARY OF ISO CLAUSE 12 AND COMMENTS RAISED

6.1 SUMMARY OF ISO CLAUSE 12 6.1.1 Based on the three representative chord types, three typical jack-up designs have

been selected for comparing the checks on the structural integrity of leg components from ISO Clause 12 against those of SNAME 5-5A. The ISO Clause 12 has a series of formulations to cover all the chord types and tubular sections used in the selected jack-ups. Extreme and realistic loading cases were applied to each chord or tubular member and the utilisation results given by the ISO were nearly always lower than the checks from the SNAME. The lowest was for the diagonal brace in Case 2 where the ISO utilisation check was 76% of the SNAME utilisation usage.

6.1.2 ISO Clause 12 includes a large number of different formulations to cover different loading and leg designs. However, since the SNAME code and all or most of our carefully selected general cases have fallen into only one of the classifications, for both tubular and prismatic member checks, additional checking may be needed to justify all the formulations presented in the Clause 12 document. Further comments on this topic can be found in the notes of Appendix A.

6.1.3 Under a few special conditions, some general form investigation has been undertaken of the ISO Clause 12 formulae against those provided by SNAME. Both the maximum axial tension and compression capacity calculated using the ISO have been found to be larger than those from SNAME results for both tubular and prismatic members under these conditions. For prismatic member, the maximum bending capacity calculated using the ISO is also larger than the SNAME results under considered conditions. For tubular member, the maximum bending capacity calculated using the ISO tends to be slightly smaller than the SNAME results when (Fy D)/(E t) becomes large (say, >0.076 or D/t >22 for Fy = 690 MPa) but otherwise, the ISO maximum bending strength (axial load = 0) is higher than the SNAME check for the same brace. Five out of the six braces studied in this report have a higher ISO maximum bending strength than the value from SNAME.

6.1.4 The hydrostatic pressure check, the beam shear strength check and the torsional shear strength check were only required for ISO. The resulted utilisations were all quite small for both the tubular and prismatic members studied. This could be for the reason why such checks were ignored in the SNAME code.

6.2 COMMENTS ON TUBULAR STRENGTH CHECK OF CLAUSE 12 6.2.1 The axial compression strength check formulae in the ISO is a 3-section conditioned

function with the condition formula expressed as C1: (Fy D) / (2 Cx E t) ≤ 0.170

C2: 0.170 < (Fy D) / (2 Cx E t) ≤ 1.911

C3: (Fy D) / (2 Cx E t) > 1.911

It has been found that when typical values are assumed for Cx = 0.3, E = 200GPa and Fy = 600MPa, a quite large tubular ratio D/t > 34 would be required for the C2 condition and an extreme high number of 382.2 for the C3. Thus most tubular braces will fall within the range of C1 with some cases only occurring in the range C2. It is unlikely that tubular braces falling in the range C3 will be required. Tubular legs

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having these dimensions will presumably be classed as “prismatic” on account of features such as pin-holes and welded rack bars.

6.2.2 For the maximum axial compression strength check of tubular member, the general form investigation with the (Fy D) / (2 Cx E t) ≤ 0.170 has given lower utilisation against the same varied KL/r when using the ISO as against SNAME. Moreover, the actual effective length factor K suggested by ISO for certain brace configurations are lower than those proposed by SNAME. This combined effect increases the difference in brace strength between the ISO and the SNAME code checks.

6.2.3 The bending strength check in the ISO - Clause 12 for tubular member is also a 3-section conditioned function. Half of the braces studied have been categorised into the first condition of (Fy D)/(E t) ≤ 0.0517 and half of them satisfied the second condition of 0.0517 < (Fy D)/(E t) ≤ 0.1034. To make the third section of 0.1034 < (Fy D)/(E t) ≤ 120 (Fy / E), again, assuming typically E = 200GPa and Fy = 600MPa, a tubular size ratio 34.5 < D/t ≤ 120 is required. Notice that the maximum threshold 120 for D/t of tubulars is not consistent with the maximum D/t (= 382.2) ratio allowed for the axial compression strength check discussed previously. It is also found that this D/t > 382.32 (or (Fy D)/(E t) > 1.911) limit is not coming from the ISO 19902 which is the main reference of the tubular formulations of the ISO 19905.

6.3 COMMENTS ON PRISMATIC STRENGTH CHECK OF CLAUSE 12 6.3.1 The ISO classification combines the chord or rack plate with any stiffeners into one

reinforced plate, as shown in the ISO formula A.12.2-2 and the ISO figure A.12.2-1 (Appendix C). This excludes a local bucking check for individual components. Using the effective thickness for these reinforced plates, all the chords studied were classified into Class 1 - “plastic” sections and this classification was used for all the utilisation check calculation. An extra check on each component of the reinforced plates was also undertaken which indicated that some individual components would be classified as Class 3 and Class 4 section. This implies that the reinforced plate method used in the ISO classification may not be adequate as it allows the thick chord plates and two thin internal webs of the Case 3 chord (see Figure 3-3) to be combined to give a stocky equivalent web plate. As a result the relatively slender web plates of this design are not checked as a slender section.

6.3.2 A general lateral torsional buckling check has been investigated for both doubly symmetry and singly symmetry cases based on the selected chord sections. The first limit length Lp of 679.3m and 67.86m, resulted from the doubly (compact tubular) and singly (triangular) symmetry cases respectively, indicate that lateral torsional buckling is not a consideration for these sections. The ISO LTB screening classification assigned rather small limits, 2.99m and 4.81m for the two cases respectively, which was considered inconsistent with the general LTB curves and the corresponding plastic bending moment results. A more realistic filter is required to avoid wasting effort in undertaking LTB capacity checks on typical chord members that are not realistically limited by this effect.

6.3.3 The axial compression strength check in the ISO is a 2-section conditioned function with an enhanced formula for the high strength steel when Fy > 450MPa. This enhancement is considered appropriate since chords are normally made of high strength steels. However, the general form which results from the two functions at the Fy = 450MPa threshold do not match and it is suggested that the formulae should be adjusted to remove this inconsistency.

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REFERENCES

[1] Recommended Practice for Site Specific Assessment of Mobile Jack-Up Units, 1st Ed., Rev 2., SNAME Technical and Research Bulletin 5-5A, 2002.

[2] APPENDIX F, LRFD, Manual of Steel Construction, 2rd Ed., AISC, 1994.

[3] APPENDIX F, LRFD, Manuel of Steel Construction, 1st Ed., AISC, 1986.

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

APPENDICES

APPENDIX A

A : NOTES AND COMMENTS ON ISO-19905 CLAUSE 12 SECTIONS

Section Notes and comments

A.12 Structural Strength

A.12.1 Applicability

A.12.1.1 General

Line 5, grammar, “are include” should be “are including” or “include”

In 3rd para reference is made to the components combined in a typical section. It refers to Table A.12.2-1 as showing examples. The examples shown are typical of chord sections but the diagrams of plastic and elastic stress distribution are clearly more appropriate to building sections where bending is predominantly in one direction. We foresee difficulty in analysing these sections when the neutral axis is likely to be different for each load case considered, - particularly for those elements which are considered as slender.

A.12.1.2 Truss type legs

A.12.1.3 Other leg types

A.12.1.4 Fixation system and/or elevating system

A.12.1.5 Spudcan strength including connection to the leg

A.12.1.6 Overview of the assessment procedure

A.12.2 Classification of member cross-sections

A.12.2.1 Member type

A.12.2.2 Material yield strength

Fyeff defined here is not used anywhere else in Clause 12

A.12.2.3 Classification definitions

A.12.2.3.1 Tubular member classification

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A.12.2.3.2 Prismatic member classification

Equation numbers are missing.

When defining “Internal Components” a distinction is drawn between “flange internal” and “web internal” components. For the purposes of a jack-up chord this is a confusing distinction since bending may occur about any axis through the section.

It is not clear what procedure should be adopted if not all the elements in a section are of the same slenderness class. (but see also A.12.3.1)

The equation supplied to check for need to consider torsional buckling is missing an equation reference number:

b y yL / r 1.51 E / F≤ min

Several typical chords do not pass this test and therefore require checking for Lateral Torsional Buckling (LTB) - see notes for Section A.12.6.2.6.

Considering the usual types of chords, as shown in Table A.12.2-1~3, concepts of base plate, side plate, rack plate, split tubular components and stiffeners, etc., may be better than web and flange when used for classification.

In Table A.12.2-1~3, Fy has not been non-dimensionalised and has no units defined (MPa was assumed in the review though)

In Table A.12.2-1~3, stress distribution pattern drawn as I -shaped beam which is difficult to associate with real chords. Moreover, the true stress reacted across the chord section may not be about the major and minor bending axes assumed in Tables and thus very difficult to be used for the classification.

Classification for class 4 - slender section needs to be counted in this section.

A.12.2.3.3 Reinforced components

The illustration under this section (Figure A.12.2-1: “Definitions for reinforced plate”) shows a reinforcement typical for LeTourneau style chords. This is similar to the example of case 1 and allowed the reinforced plate to be classified as an element which placed the whole section into the “Plastic - Class 1” category.

The present requirement is for one check using an effective plate thickness. Checks of typical chords have shown that quite different results occur if the individual elements are considered. Base plate and reinforcing plate may be susceptible to local plate buckling when separated as discussed in the report.

The engineer given this task used the same procedure to take the two chord plates and two web plates found in the case 3 chord (see Figure 3-3) an classified the resulting reinforced plate as “Plastic - Class 1” even though consideration of the individual web plates would have classified the section as “Non-compact - Class 3”. As this may not be the intention of the Clause 12 authors some additional clarification may be required in this part of the document.

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A.12.3 Section properties

A.12.3.1 General

This clause stipulates “the properties appropriate for the stiffness assessment of prismatic members shall be based on elastic considerations” by which it is understood that the full elastic section properties (without any allowance for local buckling) should be used for all prismatic sections in order to undertake an elastic analysis of the overall structure (leg) to determine the distribution of forces and moments through the structure (see also para 2 of A.12.3.4.1). Is this interpretation correct?

It has been assumed that if any one element in a prismatic section is classified as “non-compact” or “slender” then the whole section must be assigned accordingly under the most slender assessment. Is this correct?

A.12.3.2 Plastic and compact sections

A.12.3.2.1 Axial properties

A.12.3.2.2 Flexural properties

A.12.3.3 Semi-compact sections

Some clarification of If and Sf would be helpful. Does the term “fully effective section” mean that all the section can be taken to be fully effective or does it mean that only those effective parts of the section which are fully effective should be included? If the later then how are the effective parts to be determined?

The terms “gross section properties” , “gross web area” etc are used in various places (A.12.3.4.1, A.12.6.2.4 etc) - do these have the same meaning as “fully effective section” for the Class 1, 2 and 3 sections - i.e. all except slender?

A.12.3.4 Slender sections

As noted in A.12.2.3.3 all sections checked were shown to have avoided “slender” categorisation by use of the specified method for dealing with reinforced plates. However if the various plates were considered by themselves then one section (case 3 - see Figure 3-3) which has internal plates which would be classified as “slender” would require the section to be assessed under these provisions.

A.12.3.4.1 General

In Table A.12.3-1, Fy has not been non-dimensionalised and the true stress reacted across the chord section, required for the calculation, may not be about the major and minor bending axes assumed. The iterative calculation and the multi-conditioned formulations will increase the difficulty when these recommendations are used in practice.

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INITIAL BENCHMARKING OF CLAUSE 12

A.12.3.4.2 Effective areas for compressive loading conditions

See also comments on Table A.12.3-1 given in paragraph A.12.3.4.3 below.

A.12.3.4.3 Effective moduli for flexural loading conditions

This section refers to Table A.12.2.3-1(a), (b) and (c). Assessment of these section will require a knowledge of the loading condition (axial and both axes of bending) before the effective widths can be assigned.

A.12.3.5 Cross section properties for assessment

A.12.3.5.1 Tension

A.12.3.5.2 Compression

A.12.3.5.3 Flexure

A.12.4 Member moment amplification and effective lengths

A.12.5 Strength of Tubular Members

A.12.5.1 Applicability

Equation numbering is not consistent (e.g. one equation is prefixed “A” and the other is not).

For tubulars that do not meet the requirements of equation (A.12.5-1) no advice is offered on where to obtain formulations for strength which result in comparable “levels of reliability”.

Formula A.12.5-2 has not been non-dimensionalised.

Equation (12.5-2) [sic] should presumably be (A.12.5-2)

For tubular members that require checking for hydrostatic loading because they do not meet the requirements of equation (A.12.5-2) must be checked using ISO 19902. Note however that in ISO 19902 the utilisation are computed on the basis of stresses rather than member strengths.

For prismatic members there is no limiting hydrostatic head for which hydrostatic loading can be ignored. Thus by implication all prismatic members will require a check for hydrostatic pressure.

Note that due to the quadratic form of this equation when values of d exceed about 170 the limiting D/t value, for which hydrostatic loading may be ignored, starts to increase.

A.12.5.2 Tension, compression and bending strength of tubular members

Some formula numbering in this section is missing and re-numbering is needed.

A.12.5.2.1 Axial tensile strength check

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INITIAL BENCHMARKING OF CLAUSE 12

A.12.5.2.2 Axial compressive strength check

In ISO 19902, from which these formulae were adapted, the third condition of D/t > 382.2 (or (Fy D) / (2 Cx E t) > 1.911) of the axial compression strength check is not presented.

The limits allow more slender D/t ratios than are permitted by A.12.5.2.5

A.12.5.2.3 Column buckling strength

A.12.5.2.4 Local buckling strength

A.12.5.2.5 Bending strength check

The bending strength is given as a function of (FyD/Et) over three ranges. There a step changes at the range limits, i.e. at (FyD/Et) = 0.0517 the two values returned by the two expressions are 1 and 0.9827 (2% difference) and at (FyD/Et) = 0.1034 the two values are 0.8353 and 0.8614 (3% difference). These step changes are relatively small but presumably could be reduced.

The bending strength is not defined for (D/t) > 120.

A.12.5.3 Tubular members combined strength checks

A.12.5.3.1 Axial tension and bending strength check

A.12.5.3.2 Axial compression and bending strength check

A.12.5.3.3 Beam shear strength check

A.12.5.3.4 Torsional shear strength check

A.12.6 Strength of prismatic members

Formula numbering in this section either missing or not consistent with that used previously - this needs to be re-written.

A.12.6.1 General

A.12.6.2 Prismatic members subject to tension, compression, bending or shear

A.12.6.2.1 General

A.12.6.2.2 Axial tensile strength check

Equation numbers missing.

A.12.6.2.3 Axial compressive strength check

Equation numbers missing.

The actual yield stress threshold needs to be specified for γc (450MPa as being used for this review as suggested in comments in Section A.12.6.2.4).

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A.12.6.2.4 Column buckling strength

Equation numbers are prefixed by “E” for no clear reason.

A precise definition is needed for the “gross area of section”.

A.12.6.2.5 Bending strength

A.12.6.2.5.1 General

A.12.6.2.5.2 Class 1 plastic and class 2 compact sections

Equation numbers are missing.

A.12.6.2.5.3 Class 3 semi-compact sections

The first equation number is prefixed by “A-F” for no clear reason but thereafter equation numbers are missing.

Mp has not been defined anywhere (assumed = Mb in Section A.12.6.2.5.2)

A.12.6.2.5.4 Class 4 slender sections

Equation number missing.

A.12.6.2.6 Lateral torsional buckling strength check

Equation numbers missing.

The reference supplied for Lr is AISC Section F2.2c.

The current edition of “Steel Construction Manual - AISC- 13th Edition” (see Ref [3] contains a section F2.2 in the Specifications and this has an equation (eqn F2-6) which specifies Lr and which is presumably the relevant expression.

Note that the equation provided for Fcr of lateral torsional buckling of doubly symmetric members in sub-paragraph (i) is not referenced by an equation number. It does however include a reference to “[Table A-F1.1(b)]” which is ambiguous as the table supplied in Chapter F of Ref [3] is designated as “TABLE user Note F1.1 - Selection Table for the Application of Chapter F Sections”. We note that there is a Table A-F1.1 in “Appendix F - Beams and other Flexural Members” of Ref [2] but which we understand is now superseded.

The expression for Lr for I-shaped beam section in Ref [2] was considered not suitable for chords. The general investigation based on the two chords shows that this LTB check was confusing to use and probably redundant for typical chord types.

A.12.6.2.7 Bending strength check

Equation numbers are missing.

A.12.6.3 Prismatic members combined strength checks

Formula numbering in this section needs to be re- written.

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A.12.6.3.1 General

A.12.6.3.2 Interaction equation approach

Equation numbering is missing in some cases and there are some equation numbers prefixed by “H” for no clear reason.

Pu/γaPp > 0.2 is considered to be γaPu/Pp compared to SNAME 5-5A.

Mry, Muay, and Muey are considered wrong positioned.

γc = partial resistance factor …is considered to be γa = γc partial resistance factor …

A.12.6.3.3 The interaction surface approach

Equation numbers are missing.

Not reviewed in this report

A.12.6.3.4 Beam shear

Equation numbers are incomplete and the prefix “A” is not understood.

Fy, or Fymin ?

A.12.6.3.5 Torsional shear

Fy, or Fymin ?

A.12.7 Assessment of member joints

(no detail provided and no comments offered)

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INITIAL BENCHMARKING OF CLAUSE 12

APPENDIX B

B : SPREADSHEETS FOR THE STUDIED CASES

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1 THE ISO CLAUSE 12 SPREADSHEETS FOR TUBULAR MEMBERS

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B.1.1 CASE 1, HORIZONTAL BRACE B.1.1.1 Combined loads:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 4.379 Measured between centrelines

Effective length factor K = 0.7 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (combined) Notes

Axial force Pu = -4.46E+06Bending moment in y Muy = -1.67E+05Bending moment in z Muz = 6.19E+02

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.26E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.859 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.960 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.916 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 3.23E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.488 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.20E+06 When λ <= 1.34, Eqn A.12.5-5a

2.91E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 0.986 B =

Amplified moment in y Muay = -1.64E+05Amplified moment in z Muaz = 6.10E+02 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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INITIAL BENCHMARKING OF CLAUSE 12

B.1.1.2 Maximum axial compression:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 4.379 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial compression) Notes

Axial force Pu = -5.75E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.12E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.783 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.939 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.858 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.48E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.558 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.04E+06 When λ <= 1.34, Eqn A.12.5-5a

2.23E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.107 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.1.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 4.379 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial tension) Notes

Axial force Pu = 6.94E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.12E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.134 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.035 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.48E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.558 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.04E+06 When λ <= 1.34, Eqn A.12.5-5a

2.23E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.1.4 Maximum bending:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 4.379 Measured between centrelines

Effective length factor K = 0.7 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (bending) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 6.87E+05

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.26E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 1.033 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.033 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.033 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 3.23E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.488 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.20E+06 When λ <= 1.34, Eqn A.12.5-5a

2.91E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 6.87E+05 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.1.5 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 4.379 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (shear) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 3.10E+04Torsional moment T = 6.47E+03

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.12E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.000 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.000 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.000 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.015 Eqn A.12.5-13

Torsional Shear Strength Check: 0.010 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.48E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.558 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.04E+06 When λ <= 1.34, Eqn A.12.5-5a

2.23E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.2 CASE 1, DIAGONAL BRACE B.1.2.1 Combined loads:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 5.55 Measured between centrelines

Effective length factor K = 0.7 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (combined) Notes

Axial force Pu = -5.03E+06Bending moment in y Muy = 3.28E+04Bending moment in z Muz = 5.11E+03

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.99E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.734 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.896 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.800 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.01E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.619 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 6.89E+06 When λ <= 1.34, Eqn A.12.5-5a

1.81E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.133 B =

Amplified moment in y Muay = 3.71E+04Amplified moment in z Muaz = 5.79E+03 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.2.2 Maximum axial compression:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 5.55 Measured between centrelines

Effective length factor K = 0.7 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial compression) Notes

Axial force Pu = -5.32E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.99E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.724 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.887 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.793 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.01E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.619 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 6.89E+06 When λ <= 1.34, Eqn A.12.5-5a

1.81E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.155 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.2.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 5.55 Measured between centrelines

Effective length factor K = 0.7 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial compression) Notes

Axial force Pu = 6.94E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.99E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.159 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.035 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.01E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.619 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 6.89E+06 When λ <= 1.34, Eqn A.12.5-5a

1.81E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.2.4 Maximum bending:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 5.55 Measured between centrelines

Effective length factor K = 0.7 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (bending) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 6.87E+05

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.99E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 1.033 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.033 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.033 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.01E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.619 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 6.89E+06 When λ <= 1.34, Eqn A.12.5-5a

1.81E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 6.87E+05 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.2.5 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.324Wall thickness t = 0.013

Unbraced length L = 5.55 Measured between centrelines

Effective length factor K = 0.7 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 6.07E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.076 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 24.923 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (shear) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 7.75E+03Torsional moment T = 2.40E+03

Axial tensile strength AFy/γt = 7.34E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.99E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 6.65E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.12E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 6.34E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.000 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.000 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.000 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.004 Eqn A.12.5-13

Torsional Shear Strength Check: 0.004 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.27E-02Second moment of area Iyy = Izz = 1.54E-04

Polar moment of inertia Ip = 3.08E-04

Euler buckling strength PE = 2.01E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 7.64E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.23E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 6.66E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 6.12E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.126Local buckling strength Pyc = 7.71E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

7.81E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b6.12E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.619 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 6.89E+06 When λ <= 1.34, Eqn A.12.5-5a

1.81E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.076 2nd bending parameter 120Fy/E = 0.364

Bending moment strength Mb = 7.64E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

6.98E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b6.74E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.3 CASE 2, HORIZONTAL BRACE B.1.3.1 Combined loads:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 5.684 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (combined) Notes

Axial force Pu = -3.64E+06Bending moment in y Muy = 3.10E+04Bending moment in z Muz = -1.10E+03

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.21E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.519 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.798 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.563 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 8.14E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.016 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.99E+06 When λ <= 1.34, Eqn A.12.5-5a

7.32E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.538 B =

Amplified moment in y Muay = 4.77E+04Amplified moment in z Muaz = -1.70E+03 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.3.2 Maximum axial compression:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 5.684 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial compression) Notes

Axial force Pu = -4.14E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.21E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.516 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.794 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.566 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 8.14E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.016 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.99E+06 When λ <= 1.34, Eqn A.12.5-5a

7.32E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.728 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.3.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 5.684 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial tension) Notes

Axial force Pu = 7.57E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.21E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.453 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.036 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 8.14E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.016 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.99E+06 When λ <= 1.34, Eqn A.12.5-5a

7.32E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.3.4 Maximum bending:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 5.684 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (bending) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 4.53E+05

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.21E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.945 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.945 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 8.14E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.016 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.99E+06 When λ <= 1.34, Eqn A.12.5-5a

7.32E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 4.53E+05 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.3.5 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 5.684 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (shear) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 2.01E+04Torsional moment T = 1.42E+04

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 5.21E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.000 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.000 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.000 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.009 Eqn A.12.5-13

Torsional Shear Strength Check: 0.037 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 8.14E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.016 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.99E+06 When λ <= 1.34, Eqn A.12.5-5a

7.32E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.4 CASE 2, DIAGONAL BRACE B.1.4.1 Combined loads:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 6.657 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (combined) Notes

Axial force Pu = -3.12E+06Bending moment in y Muy = -1.12E+04Bending moment in z Muz = 1.04E+04

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 4.43E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.421 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.760 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.458 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 5.93E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.190 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.10E+06 When λ <= 1.34, Eqn A.12.5-5a

5.34E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.790 B =

Amplified moment in y Muay = -2.00E+04Amplified moment in z Muaz = 1.87E+04 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.4.2 Maximum axial compression:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 6.657 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial compression) Notes

Axial force Pu = -3.37E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 4.43E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.421 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.761 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.461 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 5.93E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.190 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.10E+06 When λ <= 1.34, Eqn A.12.5-5a

5.34E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.970 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

− )/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.4.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 6.657 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial tension) Notes

Axial force Pu = 7.57E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 4.43E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.708 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.036 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 5.93E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.190 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.10E+06 When λ <= 1.34, Eqn A.12.5-5a

5.34E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.08E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

− )/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.4.4 Maximum bending:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 6.657 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (bending) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 4.53E+05Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 4.43E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.945 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.945 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 5.93E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.190 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.10E+06 When λ <= 1.34, Eqn A.12.5-5a

5.34E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 4.53E+05Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

− )/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.4.5 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.21975Wall thickness t = 0.03175

Unbraced length L = 6.657 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 4.48E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.016 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 6.921 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (shear) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 5.82E+03Torsional moment T = 3.74E+03

Axial tensile strength AFy/γt = 8.01E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 4.43E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 5.25E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.31E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 3.82E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.000 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.000 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.000 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.003 Eqn A.12.5-13

Torsional Shear Strength Check: 0.010 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.88E-02Second moment of area Iyy = Izz = 8.52E-05

Polar moment of inertia Ip = 1.70E-04

Euler buckling strength PE = 5.93E+06 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 5.08E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.43E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 4.01E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 3.25E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.026Local buckling strength Pyc = 8.41E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

8.74E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b3.25E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 1.190 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 5.10E+06 When λ <= 1.34, Eqn A.12.5-5a

5.34E+06 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.016 2nd bending parameter 120Fy/E = 0.269

Bending moment strength Mb = 5.04E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

5.51E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b4.71E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.5 CASE 3, HORIZONTAL BRACE B.1.5.1 Combined loads:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.016

Unbraced length L = 6.625 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 20.244 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (combined) Notes

Axial force Pu = -4.98E+06Bending moment in y Muy = 6.97E+03Bending moment in z Muz = -3.39E+04

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.64E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.35E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 8.15E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.49E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 7.32E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.619 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.843 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.674 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.55E-02Second moment of area Iyy = Izz = 1.84E-04

Polar moment of inertia Ip = 3.68E-04

Euler buckling strength PE = 1.29E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 8.90E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.62E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 7.69E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 9.18E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.099Local buckling strength Pyc = 9.07E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

9.25E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b9.18E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.838 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.30E+06 When λ <= 1.34, Eqn A.12.5-5a

1.16E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.059 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 8.90E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

8.55E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b7.97E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.382 B =

Amplified moment in y Muay = 9.64E+03Amplified moment in z Muaz = -4.69E+04 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.5.2 Maximum axial compression:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.016

Unbraced length L = 6.625 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 20.244 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial compression) Notes

Axial force Pu = -5.32E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.64E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.35E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 8.15E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.49E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 7.32E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.616 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.838 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.674 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.55E-02Second moment of area Iyy = Izz = 1.84E-04

Polar moment of inertia Ip = 3.68E-04

Euler buckling strength PE = 1.29E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 8.90E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.62E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 7.69E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 9.18E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.099Local buckling strength Pyc = 9.07E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

9.25E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b9.18E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.838 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.30E+06 When λ <= 1.34, Eqn A.12.5-5a

1.16E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.059 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 8.90E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

8.55E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b7.97E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.444 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.5.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.016

Unbraced length L = 6.625 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 20.244 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial tension) Notes

Axial force Pu = 8.17E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.64E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.35E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 8.15E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.49E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 7.32E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.286 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.035 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.55E-02Second moment of area Iyy = Izz = 1.84E-04

Polar moment of inertia Ip = 3.68E-04

Euler buckling strength PE = 1.29E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 8.90E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.62E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 7.69E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 9.18E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.099Local buckling strength Pyc = 9.07E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

9.25E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b9.18E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.838 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.30E+06 When λ <= 1.34, Eqn A.12.5-5a

1.16E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.059 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 8.90E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

8.55E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b7.97E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.5.4 Maximum bending:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.016

Unbraced length L = 6.625 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 20.244 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (bending) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 8.00E+05

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 8.64E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.35E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 8.15E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.49E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 7.32E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.982 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.982 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.982 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.55E-02Second moment of area Iyy = Izz = 1.84E-04

Polar moment of inertia Ip = 3.68E-04

Euler buckling strength PE = 1.29E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 8.90E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.62E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 7.69E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 9.18E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.099Local buckling strength Pyc = 9.07E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

9.25E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b9.18E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.838 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.30E+06 When λ <= 1.34, Eqn A.12.5-5a

1.16E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.059 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 8.90E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

8.55E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b7.97E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 8.00E+05 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.5.5 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.016

Unbraced length L = 6.625 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Out of Class 1 range (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.059 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 20.244 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (shear) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 8.47E+03Torsional moment T = 2.65E+03

Axial tensile strength AFy/γt = 8.64E+06 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 6.35E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 8.15E+05 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 2.49E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 7.32E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.000 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.000 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.000 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.003 Eqn A.12.5-13

Torsional Shear Strength Check: 0.004 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 1.55E-02Second moment of area Iyy = Izz = 1.84E-04

Polar moment of inertia Ip = 3.68E-04

Euler buckling strength PE = 1.29E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 8.90E+05 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 2.62E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 7.69E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 9.18E+07 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.099Local buckling strength Pyc = 9.07E+06 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

9.25E+06 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b9.18E+07 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.838 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 7.30E+06 When λ <= 1.34, Eqn A.12.5-5a

1.16E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.059 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 8.90E+05 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

8.55E+05 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b7.97E+05 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.6 CASE 3, DIAGONAL BRACE B.1.6.1 Combined loads:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.021

Unbraced length L = 7.725 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 15.424 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (combined) Notes

Axial force Pu = -5.38E+06Bending moment in y Muy = -5.22E+04Bending moment in z Muz = 5.81E+03

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 1.12E+07 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 7.40E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 1.08E+06 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 3.22E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 9.17E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.531 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.802 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.577 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 2.00E-02Second moment of area Iyy = Izz = 2.30E-04

Polar moment of inertia Ip = 4.61E-04

Euler buckling strength PE = 1.19E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 1.13E+06 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 3.38E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 9.63E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 1.56E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.075Local buckling strength Pyc = 1.17E+07 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

1.20E+07 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b1.56E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.992 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 8.51E+06 When λ <= 1.34, Eqn A.12.5-5a

1.07E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.045 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 1.13E+06 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

1.13E+06 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b1.02E+06 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.550 B =

Amplified moment in y Muay = -8.10E+04Amplified moment in z Muaz = 9.00E+03 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.6.2 Maximum axial compression:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.021

Unbraced length L = 7.725 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 15.424 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial compression) Notes

Axial force Pu = -5.92E+06Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 1.12E+07 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 7.40E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 1.08E+06 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 3.22E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 9.17E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.530 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.799 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.581 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 2.00E-02Second moment of area Iyy = Izz = 2.30E-04

Polar moment of inertia Ip = 4.61E-04

Euler buckling strength PE = 1.19E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 1.13E+06 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 3.38E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 9.63E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 1.56E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.075Local buckling strength Pyc = 1.17E+07 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

1.20E+07 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b1.56E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.992 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 8.51E+06 When λ <= 1.34, Eqn A.12.5-5a

1.07E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.045 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 1.13E+06 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

1.13E+06 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b1.02E+06 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.689 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

− )/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.6.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.021

Unbraced length L = 7.725 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 15.424 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (axial tension) Notes

Axial force Pu = 1.05E+07Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 1.12E+07 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 7.40E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 1.08E+06 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 3.22E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 9.17E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 1.424 When P < 0, Eqn A.12.5-11

Local Strength Check: 1.035 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 2.00E-02Second moment of area Iyy = Izz = 2.30E-04

Polar moment of inertia Ip = 4.61E-04

Euler buckling strength PE = 1.19E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 1.13E+06 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 3.38E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 9.63E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 1.56E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.075Local buckling strength Pyc = 1.17E+07 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

1.20E+07 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b1.56E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.992 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 8.51E+06 When λ <= 1.34, Eqn A.12.5-5a

1.07E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.045 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 1.13E+06 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

1.13E+06 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b1.02E+06 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

− )/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.6.4 Maximum bending:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.021

Unbraced length L = 7.725 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 15.424 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (bending) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 1.02E+06Bending moment in z Muz = 0.00E+00

Shear force V = 0.00E+00Torsional moment T = 0.00E+00

Axial tensile strength AFy/γt = 1.12E+07 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 7.40E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 1.08E+06 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 3.22E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 9.17E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.945 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.945 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.945 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.000 Eqn A.12.5-13

Torsional Shear Strength Check: 0.000 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 2.00E-02Second moment of area Iyy = Izz = 2.30E-04

Polar moment of inertia Ip = 4.61E-04

Euler buckling strength PE = 1.19E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 1.13E+06 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 3.38E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 9.63E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 1.56E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.075Local buckling strength Pyc = 1.17E+07 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

1.20E+07 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b1.56E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.992 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 8.51E+06 When λ <= 1.34, Eqn A.12.5-5a

1.07E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.045 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 1.13E+06 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

1.13E+06 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b1.02E+06 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 1.02E+06Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

− )/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.1.6.5 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Tubular Section Properties Notes

Outer diameter D = 0.3239Wall thickness t = 0.021

Unbraced length L = 7.725 Measured between centrelines

Effective length factor K = 0.8 Table A.12.4-1Moment amplification parameter Cm = 0.85 Table A.12.4-1

Young's modulus E = 2.00E+11Yield stress Fy = 5.86E+08

Water depth d = 0 Depth below the water surface

Section Property Check Notes

Class 1 classification Check: Class 1 (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.0537, Class 1, Eqn A.12.2-1

Only relevant when analyse earthquake etc.

General applicabiity Check: Clause 12 applicable (D/t)/(E/Fy) = 0.045 When (D/t)/(E/Fy) <= 0.244, applicable, Eqn A.12.5-1

`Hydrostatic pressure (HP) Check: HP can be ignored, clause 12 applicableMaximum D/t ratio for d (D/t)m = 74.500 When D/t <= (D/t)m, HP can be ignored

D/t = 15.424 Thus clause 12 applicable, Eqn A.12.5-2

Member Strength Check (shear) Notes

Axial force Pu = 0.00E+00Bending moment in y Muy = 0.00E+00Bending moment in z Muz = 0.00E+00

Shear force V = 6.21E+03Torsional moment T = 5.24E+03

Axial tensile strength AFy/γt = 1.12E+07 Resistance factor γt = 1.05, Eqn A.12.5-3

Axial compressive strength Pa/γc = 7.40E+06 Resistance factor γc = 1.15, Eqn A.12.5-4

Purely bending strength Mb/γb = 1.08E+06 Resistance factor γb = 1.05, Eqn A.12.5-9 ?

Factored shear strength Pv/γv = 3.22E+06 Resistance factor γv = 1.05, Eqn A.12.5-13

Factored torsional strength Tv/γv = 9.17E+05 Resistance factor γv = 1.05, Eqn A.12.5-14

Tension&Bending Strength Check: 0.000 When P > 0, Eqn A.12.5-10Compression&Bending Strength Check: 0.000 When P < 0, Eqn A.12.5-11

Local Strength Check: 0.000 Only when P < 0. Eqn A.12.5-12Beam Shear Strength Check: 0.002 Eqn A.12.5-13

Torsional Shear Strength Check: 0.006 Eqn A.12.5-14

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Cross-section area A = 2.00E-02Second moment of area Iyy = Izz = 2.30E-04

Polar moment of inertia Ip = 4.61E-04

Euler buckling strength PE = 1.19E+07 PE = π2 E I/ (KL)2

Plastic moment strength Mp = 1.13E+06 Mp = Fy (1/6) [D3 – (D - 2t)3]

Shear strength Pv = 3.38E+06 Pv = A Fy/(2√3)

Torsional strength Tv = 9.63E+05 Tv = 2 Ip Fy/(D √3)

Elastic buckling strength Pxθ = 1.56E+08 Pxe = 0.6EA(t/D), Eqn A.12.5-6

Local buckling parameter AFy/Pxθ = 0.075Local buckling strength Pyc = 1.17E+07 When AFy/Pxθ <= 0.170, Eqn A.12.5-7a

1.20E+07 When 0.170 < AFy/Pxθ <= 1.911, Eqn A.12.5-7b1.56E+08 When AFy/Pxθ >1.911, Eqn A.12.5-7c

Column slenderness λ = 0.992 λ = [Pyc/Pxθ]0.5, Eqn A.12.5-6

Column buckling strength Pa = 8.51E+06 When λ <= 1.34, Eqn A.12.5-5a

1.07E+07 When λ > 1.34, Eqn A.12.5-5b

1st bending parameter FyD/(Et) = 0.045 2nd bending parameter 120Fy/E = 0.352

Bending moment strength Mb = 1.13E+06 When FyD/(Et) <= 0.0517, Eqn A.12.5-9a

1.13E+06 When 0.0517 < FyD/(Et) <= 0.1034, Eqn A.12.5-9b1.02E+06 When 0.1034 < FyD/(Et) <= 120 (Fy/E), Eqn A.12.5-9c

Moment amplification factor B = 1.000 B =

Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Some ISO Equations

)/1( Eu

m

PPC

− )/1( Eu

m

PPC

Axial tension and bending strength check

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

Axial compression and bending strength check

Beam-column check:

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

Local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2 THE ISO CLAUSE 12 SPREADSHEETS FOR PRISMATIC MEMBERS

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.1 CASE 1, TRIANGULAR CHORD B.2.1.1 Combined loads: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Base plate length L1 = 0.7112 Cross-section area A = 0.10154Base plate thickness t1 = 0.0508 Centroid to y axis Yc = 0.276

Wall plate length L2 = 0.466 Centroid to z axis Zc = 0Wall plate thickness t2 = 0.01905 2nd moment of area Iyy = 5.46E-03

Rack plate length L3 = 0.2127 2nd moment of area Izz = 2.63E-03Rack plate thickness t3 = 0.127 Fully plastic moment Mpy = 1.09E+07

Fully plastic moment Mpz = 6.42E+06Rack stiffener length L4 = 0.1016 Plastic neutral axis Yp = 0.277

Rack stiffener thickness t4 = 0.0254 Plastic neutral axis Zp = 0Wall stiffener length L5 = 0.3048

Wall stiffener thickness t5 = 0.0254

Base plate internal offset X1 = 0.236Rack internal offset Y1 = 0.457

Rack stiffener internal offset Y2 = 0.524Wall stiffener internal offset Y3 = 0.118

Bay height H = 3.408

Young's modulus E = 2.00E+11

Base plate yield stress Fy1 = 4.83E+08Wall plate yield stress Fy2 = 4.83E+08

Rack plate yield stress Fy3 = 5.87E+08 Chord Section ForcesRack stiffener yield stress Fy4 = 4.83E+08Wall stiffener yield stress Fy5 = 4.83E+08 Axial force Pu = -1.94E+07

Bending moment in y Muy = -3.43E+06Effective length factor K = 1.00 (Table A.12.4 Bending moment in z Muz = 1.13E+05

Moment amplification Cm = 0.85 (Table A.12.4 Shear force V = 0.00E+00Strength check exponent ή = 1.80 (A.12.6.3.2) Torsional moment T = 0.00E+00

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jac

Flange internal components: Class 1 Table A.12.2-1Base plate: Class 1 (Fy)

0.5(b/t) <= 465, Class 1-plastic (Fy)

0.5b/t = 204.198 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Wall plate: Class 1 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2Base plate: Class 1 (Fy)

0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic (Fy)

0.5b/t = 43.500 Weld 135(w)/150(r) < (Fy)0.5(b/t) <= 150(w)/166(r), Class 2-compact

Rack plate: Class 1 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact

(Fy)0.5b/t = 30.563 Weld (Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3Base plate: Class 1 α is calculated by elastic assumption

Distance of bending axis to tip α = 17.976 Considering only α = 1.0 (fully compression) (Fy)

0.5b/t = 204.198 (Fy)0.5(b/t) <= 465, Class 1-plastic

Wall plate: Class 1 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Distance of bending axis to tip α = 0.291 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 1.040 Check needed only when (Lb/rmin)(Fymin/E)0.5 <=1.51, A.12.2.3.2

Member Strength Check (combined) Notes

Tensile strength AtFymin/γt = 4.94E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 4.51E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 5.15E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 6.12E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.06E+07Factored shear strength Pv/γv = 1.35E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 4.78E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.899 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.992 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.867 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

θ

X

YPLASTIC/JUSTAS

PAFEC/CODE-86Z

Y

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 126 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Wall inclination θ = 1.192 radiansStiffened wall check thickness tw = 0.034 tcheck=tw,Eqn A.12.2-2

Stiffened rack check thickness tr = 0.147 tcheck=tr,Eqn A.12.2-2

Min component yeild stress Fymin = 4.83E+08

Section bending strength Mby = 6.43E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.12E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.07E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.33E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.32E-02Fully effective cross-seciton Af = 1.02E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 1.54E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 6.68E-03Fully effective cross-seciton Ify = 5.46E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.63E-03Radius of gyration ry = 2.32E-01 For gross area ? In A.12.6.2.4

Radius of gyration rz = 1.61E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.331 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.230 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.331Critical buckling force Fcr = 4.61E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

3.86E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

4.79E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.23E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 5.45E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.01E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 6.01E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

6.81E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 7.42E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

9.01E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 3.23E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

3.92E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 5.08E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 1.42E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 8.09E-03Torsional radius r = 4.50E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 5.01E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 9.28E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 4.47E+08Moment amplification factor By = 0.868 B = , A.12.4

Moment amplification factor Bz = 0.889Amplified moment in y Muay = -2.98E+06Amplified moment in z Muaz = 1.00E+05 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.1.2 Maximum axial compression: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Base plate length L1 = 0.7112 Cross-section area A = 0.10154Base plate thickness t1 = 0.0508 Centroid to y axis Yc = 0.276

Wall plate length L2 = 0.466 Centroid to z axis Zc = 0Wall plate thickness t2 = 0.01905 2nd moment of area Iyy = 5.46E-03

Rack plate length L3 = 0.2127 2nd moment of area Izz = 2.63E-03Rack plate thickness t3 = 0.127 Fully plastic moment Mpy = 1.09E+07

Fully plastic moment Mpz = 6.42E+06Rack stiffener length L4 = 0.1016 Plastic neutral axis Yp = 0.277

Rack stiffener thickness t4 = 0.0254 Plastic neutral axis Zp = 0Wall stiffener length L5 = 0.3048

Wall stiffener thickness t5 = 0.0254

Base plate internal offset X1 = 0.236Rack internal offset Y1 = 0.457

Rack stiffener internal offset Y2 = 0.524Wall stiffener internal offset Y3 = 0.118

Bay height H = 3.408

Young's modulus E = 2.00E+11

Base plate yield stress Fy1 = 4.83E+08Wall plate yield stress Fy2 = 4.83E+08

Rack plate yield stress Fy3 = 5.87E+08 Chord Section ForcesRack stiffener yield stress Fy4 = 4.83E+08Wall stiffener yield stress Fy5 = 4.83E+08 Axial force Pu = -4.12E+07

Bending moment in y Muy = 0.00E+00Effective length factor K = 1.00 (Table A.12.4 Bending moment in z Muz = 0.00E+00

Moment amplification Cm = 0.85 (Table A.12.4 Shear force V = 0.00E+00Strength check exponent ή = 1.80 (A.12.6.3.2) Torsional moment T = 0.00E+00

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jac

Flange internal components: Class 1 Table A.12.2-1Base plate: Class 1 (Fy)

0.5(b/t) <= 465, Class 1-plastic (Fy)

0.5b/t = 204.198 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Wall plate: Class 1 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2Base plate: Class 1 (Fy)

0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic (Fy)

0.5b/t = 43.500 Weld 135(w)/150(r) < (Fy)0.5(b/t) <= 150(w)/166(r), Class 2-compact

Rack plate: Class 1 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact

(Fy)0.5b/t = 30.563 Weld (Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3Base plate: Class 1 α is calculated by elastic assumption

Distance of bending axis to tip α = 1.000 Considering only α = 1.0 (fully compression) (Fy)

0.5b/t = 204.198 (Fy)0.5(b/t) <= 465, Class 1-plastic

Wall plate: Class 1 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Distance of bending axis to tip α = 1.000 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 1.040 Check needed only when (Lb/rmin)(Fymin/E)0.5 <=1.51, A.12.2.3.2

Member Strength Check (compression) Notes

Tensile strength AtFymin/γt = 4.94E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 4.51E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 5.15E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 6.12E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.06E+07Factored shear strength Pv/γv = 1.35E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 4.78E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.875 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.915 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.922 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

θ

X

YPLASTIC/JUSTAS

PAFEC/CODE-86Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 127 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 128 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Wall inclination θ = 1.192 radiansStiffened wall check thickness tw = 0.034 tcheck=tw,Eqn A.12.2-2

Stiffened rack check thickness tr = 0.147 tcheck=tr,Eqn A.12.2-2

Min component yeild stress Fymin = 4.83E+08

Section bending strength Mby = 6.43E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.12E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.07E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.33E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.32E-02Fully effective cross-seciton Af = 1.02E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 1.54E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 6.68E-03Fully effective cross-seciton Ify = 5.46E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.63E-03Radius of gyration ry = 2.32E-01 For gross area ? In A.12.6.2.4

Radius of gyration rz = 1.61E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.331 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.230 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.331Critical buckling force Fcr = 4.61E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

3.86E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

4.79E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.23E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 5.45E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.01E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 6.01E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

6.81E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 7.42E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

9.01E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 3.23E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

3.92E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 5.08E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 1.42E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 8.09E-03Torsional radius r = 4.50E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 5.01E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 9.28E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 4.47E+08Moment amplification factor By = 0.890 B = , A.12.4

Moment amplification factor Bz = 0.936Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

Page 129: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.1.3 Maximum axial tension: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Base plate length L1 = 0.7112 Cross-section area A = 0.10154Base plate thickness t1 = 0.0508 Centroid to y axis Yc = 0.276

Wall plate length L2 = 0.466 Centroid to z axis Zc = 0Wall plate thickness t2 = 0.01905 2nd moment of area Iyy = 5.46E-03

Rack plate length L3 = 0.2127 2nd moment of area Izz = 2.63E-03Rack plate thickness t3 = 0.127 Fully plastic moment Mpy = 1.09E+07

Fully plastic moment Mpz = 6.42E+06Rack stiffener length L4 = 0.1016 Plastic neutral axis Yp = 0.277

Rack stiffener thickness t4 = 0.0254 Plastic neutral axis Zp = 0Wall stiffener length L5 = 0.3048

Wall stiffener thickness t5 = 0.0254

Base plate internal offset X1 = 0.236Rack internal offset Y1 = 0.457

Rack stiffener internal offset Y2 = 0.524Wall stiffener internal offset Y3 = 0.118

Bay height H = 3.408

Young's modulus E = 2.00E+11

Base plate yield stress Fy1 = 4.83E+08Wall plate yield stress Fy2 = 4.83E+08

Rack plate yield stress Fy3 = 5.87E+08 Chord Section ForcesRack stiffener yield stress Fy4 = 4.83E+08Wall stiffener yield stress Fy5 = 4.83E+08 Axial force Pu = 4.58E+07

Bending moment in y Muy = 0.00E+00Effective length factor K = 1.00 (Table A.12.4 Bending moment in z Muz = 0.00E+00

Moment amplification Cm = 0.85 (Table A.12.4 Shear force V = 0.00E+00Strength check exponent ή = 1.80 (A.12.6.3.2) Torsional moment T = 0.00E+00

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jac

Flange internal components: Class 1 Table A.12.2-1Base plate: Class 1 (Fy)

0.5(b/t) <= 465, Class 1-plastic (Fy)

0.5b/t = 204.198 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Wall plate: Class 1 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2Base plate: Class 1 (Fy)

0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic (Fy)

0.5b/t = 43.500 Weld 135(w)/150(r) < (Fy)0.5(b/t) <= 150(w)/166(r), Class 2-compact

Rack plate: Class 1 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact

(Fy)0.5b/t = 30.563 Weld (Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3Base plate: Class 1 α is calculated by elastic assumption

Distance of bending axis to tip α = 0.500 Considering only α = 1.0 (fully compression) (Fy)

0.5b/t = 204.198 (Fy)0.5(b/t) <= 465, Class 1-plastic

Wall plate: Class 1 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Distance of bending axis to tip α = 0.500 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 1.040 Check needed only when (Lb/rmin)(Fymin/E)0.5 <=1.51, A.12.2.3.2

Member Strength Check (tension) Notes

Tensile strength AtFymin/γt = 4.94E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 4.51E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 5.15E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 6.12E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.06E+07Factored shear strength Pv/γv = 1.35E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 4.78E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.927 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 1.016 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 1.024 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

θ

X

YPLASTIC/JUSTAS

PAFEC/CODE-86Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 129 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 130: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 130 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Wall inclination θ = 1.192 radiansStiffened wall check thickness tw = 0.034 tcheck=tw,Eqn A.12.2-2

Stiffened rack check thickness tr = 0.147 tcheck=tr,Eqn A.12.2-2

Min component yeild stress Fymin = 4.83E+08

Section bending strength Mby = 6.43E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.12E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.07E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.33E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.32E-02Fully effective cross-seciton Af = 1.02E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 1.54E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 6.68E-03Fully effective cross-seciton Ify = 5.46E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.63E-03Radius of gyration ry = 2.32E-01 For gross area ? In A.12.6.2.4

Radius of gyration rz = 1.61E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.331 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.230 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.331Critical buckling force Fcr = 4.61E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

3.86E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

4.79E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.23E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 5.45E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.01E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 6.01E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

6.81E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 7.42E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

9.01E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 3.23E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

3.92E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 5.08E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 1.42E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 8.09E-03 Ip = Iyy+Izz

Torsional radius r = 4.50E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 5.01E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 9.28E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 4.47E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

Page 131: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.1.4 Maximum bending y: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Base plate length L1 = 0.7112 Cross-section area A = 0.10154Base plate thickness t1 = 0.0508 Centroid to y axis Yc = 0.276

Wall plate length L2 = 0.466 Centroid to z axis Zc = 0Wall plate thickness t2 = 0.01905 2nd moment of area Iyy = 5.46E-03

Rack plate length L3 = 0.2127 2nd moment of area Izz = 2.63E-03Rack plate thickness t3 = 0.127 Fully plastic moment Mpy = 1.09E+07

Fully plastic moment Mpz = 6.42E+06Rack stiffener length L4 = 0.1016 Plastic neutral axis Yp = 0.277

Rack stiffener thickness t4 = 0.0254 Plastic neutral axis Zp = 0Wall stiffener length L5 = 0.3048

Wall stiffener thickness t5 = 0.0254

Base plate internal offset X1 = 0.236Rack internal offset Y1 = 0.457

Rack stiffener internal offset Y2 = 0.524Wall stiffener internal offset Y3 = 0.118

Bay height H = 3.408

Young's modulus E = 2.00E+11

Base plate yield stress Fy1 = 4.83E+08Wall plate yield stress Fy2 = 4.83E+08

Rack plate yield stress Fy3 = 5.87E+08 Chord Section ForcesRack stiffener yield stress Fy4 = 4.83E+08Wall stiffener yield stress Fy5 = 4.83E+08 Axial force Pu = 0.00E+00

Bending moment in y Muy = 5.78E+06Effective length factor K = 1.00 (Table A.12.4 Bending moment in z Muz = 0.00E+00

Moment amplification Cm = 0.85 (Table A.12.4 Shear force V = 0.00E+00Strength check exponent ή = 1.80 (A.12.6.3.2) Torsional moment T = 0.00E+00

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jac

Flange internal components: Class 1 Table A.12.2-1Base plate: Class 1 (Fy)

0.5(b/t) <= 465, Class 1-plastic (Fy)

0.5b/t = 204.198 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Wall plate: Class 1 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2Base plate: Class 1 (Fy)

0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic (Fy)

0.5b/t = 43.500 Weld 135(w)/150(r) < (Fy)0.5(b/t) <= 150(w)/166(r), Class 2-compact

Rack plate: Class 1 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact

(Fy)0.5b/t = 30.563 Weld (Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3Base plate: Class 1 α is calculated by elastic assumption

Distance of bending axis to tip α = 0.500 Considering only α = 1.0 (fully compression) (Fy)

0.5b/t = 204.198 (Fy)0.5(b/t) <= 465, Class 1-plastic

Wall plate: Class 1 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Distance of bending axis to tip α = 0.500 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 1.040 Check needed only when (Lb/rmin)(Fymin/E)0.5 <=1.51, A.12.2.3.2

Member Strength Check (bending y) Notes

Tensile strength AtFymin/γt = 4.94E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 4.51E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 5.15E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 6.12E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.06E+07Factored shear strength Pv/γv = 1.35E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 4.78E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.944 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.944 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.944 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

θ

X

YPLASTIC/JUSTAS

PAFEC/CODE-86Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 131 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 132 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Wall inclination θ = 1.192 radiansStiffened wall check thickness tw = 0.034 tcheck=tw,Eqn A.12.2-2

Stiffened rack check thickness tr = 0.147 tcheck=tr,Eqn A.12.2-2

Min component yeild stress Fymin = 4.83E+08

Section bending strength Mby = 6.43E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.12E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.07E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.33E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.32E-02Fully effective cross-seciton Af = 1.02E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 1.54E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 6.68E-03Fully effective cross-seciton Ify = 5.46E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.63E-03Radius of gyration ry = 2.32E-01 For gross area ? In A.12.6.2.4

Radius of gyration rz = 1.61E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.331 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.230 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.331Critical buckling force Fcr = 4.61E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

3.86E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

4.79E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.23E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 5.45E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.01E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 6.01E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

6.81E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 7.42E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

9.01E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 3.23E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

3.92E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 5.08E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 1.42E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 8.09E-03Torsional radius r = 4.50E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 5.01E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 9.28E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 4.47E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 5.78E+06Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

Page 133: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.1.5 Maximum bending z: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Base plate length L1 = 0.7112 Cross-section area A = 0.10154Base plate thickness t1 = 0.0508 Centroid to y axis Yc = 0.276

Wall plate length L2 = 0.466 Centroid to z axis Zc = 0Wall plate thickness t2 = 0.01905 2nd moment of area Iyy = 5.46E-03

Rack plate length L3 = 0.2127 2nd moment of area Izz = 2.63E-03Rack plate thickness t3 = 0.127 Fully plastic moment Mpy = 1.09E+07

Fully plastic moment Mpz = 6.42E+06Rack stiffener length L4 = 0.1016 Plastic neutral axis Yp = 0.277

Rack stiffener thickness t4 = 0.0254 Plastic neutral axis Zp = 0Wall stiffener length L5 = 0.3048

Wall stiffener thickness t5 = 0.0254

Base plate internal offset X1 = 0.236Rack internal offset Y1 = 0.457

Rack stiffener internal offset Y2 = 0.524Wall stiffener internal offset Y3 = 0.118

Bay height H = 3.408

Young's modulus E = 2.00E+11

Base plate yield stress Fy1 = 4.83E+08Wall plate yield stress Fy2 = 4.83E+08

Rack plate yield stress Fy3 = 5.87E+08 Chord Section ForcesRack stiffener yield stress Fy4 = 4.83E+08Wall stiffener yield stress Fy5 = 4.83E+08 Axial force Pu = 0.00E+00

Bending moment in y Muy = 0.00E+00Effective length factor K = 1.00 (Table A.12.4 Bending moment in z Muz = 9.83E+06

Moment amplification Cm = 0.85 (Table A.12.4 Shear force V = 0.00E+00Strength check exponent ή = 1.80 (A.12.6.3.2) Torsional moment T = 0.00E+00

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jac

Flange internal components: Class 1 Table A.12.2-1Base plate: Class 1 (Fy)

0.5(b/t) <= 465, Class 1-plastic (Fy)

0.5b/t = 204.198 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Wall plate: Class 1 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2Base plate: Class 1 (Fy)

0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic (Fy)

0.5b/t = 43.500 Weld 135(w)/150(r) < (Fy)0.5(b/t) <= 150(w)/166(r), Class 2-compact

Rack plate: Class 1 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact

(Fy)0.5b/t = 30.563 Weld (Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3Base plate: Class 1 α is calculated by elastic assumption

Distance of bending axis to tip α = 0.500 Considering only α = 1.0 (fully compression) (Fy)

0.5b/t = 204.198 (Fy)0.5(b/t) <= 465, Class 1-plastic

Wall plate: Class 1 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Distance of bending axis to tip α = 0.500 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 1.040 Check needed only when (Lb/rmin)(Fymin/E)0.5 <=1.51, A.12.2.3.2

Member Strength Check (bending z) Notes

Tensile strength AtFymin/γt = 4.94E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 4.51E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 5.15E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 6.12E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.06E+07Factored shear strength Pv/γv = 1.35E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 4.78E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.923 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.923 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.923 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

θ

X

YPLASTIC/JUSTAS

PAFEC/CODE-86Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 133 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 134: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 134 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Wall inclination θ = 1.192 radiansStiffened wall check thickness tw = 0.034 tcheck=tw,Eqn A.12.2-2

Stiffened rack check thickness tr = 0.147 tcheck=tr,Eqn A.12.2-2

Min component yeild stress Fymin = 4.83E+08

Section bending strength Mby = 6.43E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.12E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.07E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.33E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.32E-02Fully effective cross-seciton Af = 1.02E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 1.54E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 6.68E-03Fully effective cross-seciton Ify = 5.46E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.63E-03Radius of gyration ry = 2.32E-01 For gross area ? In A.12.6.2.4

Radius of gyration rz = 1.61E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.331 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.230 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.331Critical buckling force Fcr = 4.61E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

3.86E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

4.79E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.23E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 5.45E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.01E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 6.01E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

6.81E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 7.42E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

9.01E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 3.23E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

3.92E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 5.08E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 1.42E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 8.09E-03Torsional radius r = 4.50E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 5.01E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 9.28E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 4.47E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 9.83E+06 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

Page 135: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.1.6 Beam shear and torsional shear: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Base plate length L1 = 0.7112 Cross-section area A = 0.10154Base plate thickness t1 = 0.0508 Centroid to y axis Yc = 0.276

Wall plate length L2 = 0.466 Centroid to z axis Zc = 0Wall plate thickness t2 = 0.01905 2nd moment of area Iyy = 5.46E-03

Rack plate length L3 = 0.2127 2nd moment of area Izz = 2.63E-03Rack plate thickness t3 = 0.127 Fully plastic moment Mpy = 1.09E+07

Fully plastic moment Mpz = 6.42E+06Rack stiffener length L4 = 0.1016 Plastic neutral axis Yp = 0.277

Rack stiffener thickness t4 = 0.0254 Plastic neutral axis Zp = 0Wall stiffener length L5 = 0.3048

Wall stiffener thickness t5 = 0.0254

Base plate internal offset X1 = 0.236Rack internal offset Y1 = 0.457

Rack stiffener internal offset Y2 = 0.524Wall stiffener internal offset Y3 = 0.118

Bay height H = 3.408

Young's modulus E = 2.00E+11

Base plate yield stress Fy1 = 4.83E+08Wall plate yield stress Fy2 = 4.83E+08

Rack plate yield stress Fy3 = 5.87E+08 Chord Section ForcesRack stiffener yield stress Fy4 = 4.83E+08Wall stiffener yield stress Fy5 = 4.83E+08 Axial force Pu = 0.00E+00

Bending moment in y Muy = 0.00E+00Effective length factor K = 1.00 (Table A.12.4 Bending moment in z Muz = 0.00E+00

Moment amplification Cm = 0.85 (Table A.12.4 Shear force V = 2.59E+05Strength check exponent ή = 1.80 (A.12.6.3.2) Torsional moment T = 1.90E+04

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jac

Flange internal components: Class 1 Table A.12.2-1Base plate: Class 1 (Fy)

0.5(b/t) <= 465, Class 1-plastic (Fy)

0.5b/t = 204.198 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Wall plate: Class 1 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2Base plate: Class 1 (Fy)

0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic (Fy)

0.5b/t = 43.500 Weld 135(w)/150(r) < (Fy)0.5(b/t) <= 150(w)/166(r), Class 2-compact

Rack plate: Class 1 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact

(Fy)0.5b/t = 30.563 Weld (Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3Base plate: Class 1 α is calculated by elastic assumption

Distance of bending axis to tip α = 0.500 Considering only α = 1.0 (fully compression) (Fy)

0.5b/t = 204.198 (Fy)0.5(b/t) <= 465, Class 1-plastic

Wall plate: Class 1 465 < (Fy)0.5(b/t) <= 530, Class 2-compact

Distance of bending axis to tip α = 0.500 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact

(Fy)0.5b/t = 301.127 (Fy)

0.5(b/t) > 650, Class 4-slender section

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 1.040 Check needed only when (Lb/rmin)(Fymin/E)0.5 <=1.51, A.12.2.3.2

Member Strength Check (shear) Notes

Tensile strength AtFymin/γt = 4.94E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 4.51E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 5.15E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 6.12E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.06E+07Factored shear strength Pv/γv = 1.35E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 4.78E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.000 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.000 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.000 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.019 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.004 Eqn A.12.6-x Numbering?

θ

X

YPLASTIC/JUSTAS

PAFEC/CODE-86Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 135 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 136 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Wall inclination θ = 1.192 radiansStiffened wall check thickness tw = 0.034 tcheck=tw,Eqn A.12.2-2

Stiffened rack check thickness tr = 0.147 tcheck=tr,Eqn A.12.2-2

Min component yeild stress Fymin = 4.83E+08

Section bending strength Mby = 6.43E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.12E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.07E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.33E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.32E-02Fully effective cross-seciton Af = 1.02E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 1.54E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 6.68E-03Fully effective cross-seciton Ify = 5.46E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.63E-03Radius of gyration ry = 2.32E-01 For gross area ? In A.12.6.2.4

Radius of gyration rz = 1.61E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.331 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.230 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.331Critical buckling force Fcr = 4.61E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

3.86E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

4.79E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.23E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 5.45E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.01E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 6.01E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

6.81E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 7.42E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

9.01E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 3.23E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

3.92E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 5.08E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 1.42E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 8.09E-03Torsional radius r = 4.50E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 5.01E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 9.28E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 4.47E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

Page 137: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.2 CASE 2, COMPACT TUBULAR CHORD B.2.2.1 Combined loads:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Radius r = 0.1905 Cross-section area A = 0.12303Wall plate thickness t = 0.0508 Centroid to y axis Yc = 0

Rack plate length Lr = 0.401 Centroid to z axis Zc = 0Rack plate thickness tr = 0.1778 2nd moment of area Iyy = 1.66E-03

Inset d = 0.006 2nd moment of area Izz = 2.21E-03Fully plastic moment Mpy = 8.66E+06

Rack stiffener length Ls = 0 Fully plastic moment Mpz = 9.00E+06Rack stiffener thickness ts = 0 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 4.267

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.90E+08 Axial force Pu = -5.33E+07Rack plate yield stress Fy2 = 6.90E+08 Bending moment in y Muy = -7.28E+05

Rack stiffener yield stress Fy3 = 6.90E+08 Bending moment in z Muz = -7.04E+02Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 41.269077 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 15.546 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 41.269077 α is calculated by elastic assumption

Lateral torsional buckling check: Needed(Lb/rmin)(Fymin/E)0.5 = 2.157 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (combined) Notes

Tensile strength AtFymin/γt = 8.08E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 7.38E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 7.83E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 8.25E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 8.57E+06Factored shear strength Pv/γv = 2.33E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 7.32E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.691 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.810 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.877 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 137 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 138: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 138 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Stiffened wall check thickness tw = 0.178 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.90E+08

Section bending strength Mby = 8.66E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 9.00E+06 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.23E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.26E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 1.31E-02Fully effective cross-seciton Af = 1.23E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 8.28E-03 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 8.08E-03Fully effective cross-seciton Ify = 1.66E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.21E-03Radius of gyration ry = 1.16E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 1.34E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.687 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.687 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.595Critical buckling force Fcr = 5.66E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

1.28E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.36E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 1.19E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.52E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.52E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.51E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.51E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 5.71E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.71E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 5.57E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.57E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 6.15E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 2.45E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 3.87E-03Torsional radius r = 2.01E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 7.69E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.80E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 2.40E+08Moment amplification factor By = 1.207 B = , A.12.4

Moment amplification factor Bz = 1.093Amplified moment in y Muay = -8.79E+05Amplified moment in z Muaz = -7.69E+02 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling Mlty = 1.27E+09 Mit = Sf FcF

Mltz = 1.24E+09 Mlt = Mlt ,min for L > = Lr - how to calculate Lr for non I-shape chord?

Fcr = 1.54E+11 Fcr = 6.895 , Cb =1(conservative assumption), Incomplete section A.12.6.2.6

A = 1.81E-01 , A = total closed areaJ = 2.33E-03λ = 36.736 λ = Lb/rmin

X1 = 5.79E+11 X1 = (π/Sf) , J = Ip, G = 77200MPa

X2 = 0 X2 = (4Cw/Iz)(Sy/GJ)2, Cw = 0, ignore warping for close section

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

)/EGJA( 2

21

22

211

21

2

⎭⎬⎫

⎩⎨⎧

+λλXXXCb

∫=

tdsAJ

24

Page 139: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.2.2 Maximum axial compression: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Radius r = 0.1905 Cross-section area A = 0.12303Wall plate thickness t = 0.0508 Centroid to y axis Yc = 0

Rack plate length Lr = 0.401 Centroid to z axis Zc = 0Rack plate thickness tr = 0.1778 2nd moment of area Iyy = 1.66E-03

Inset d = 0.006 2nd moment of area Izz = 2.21E-03Fully plastic moment Mpy = 8.66E+06

Rack stiffener length Ls = 0 Fully plastic moment Mpz = 9.00E+06Rack stiffener thickness ts = 0 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 4.267

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.90E+08 Axial force Pu = -5.92E+07Rack plate yield stress Fy2 = 6.90E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.90E+08 Bending moment in z Muz = 0.00E+00Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 41.269077 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 15.546 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 41.269077 α is calculated by elastic assumption

Lateral torsional buckling check: Needed(Lb/rmin)(Fymin/E)0.5 = 2.157 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (compression) Notes

Tensile strength AtFymin/γt = 8.08E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 7.38E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 7.83E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 8.25E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 8.57E+06Factored shear strength Pv/γv = 2.33E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 7.32E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.767 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.802 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.870 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 139 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 140: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 140 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Stiffened wall check thickness tw = 0.178 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.90E+08

Section bending strength Mby = 8.66E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 9.00E+06 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.23E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.26E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 1.31E-02Fully effective cross-seciton Af = 1.23E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 8.28E-03 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 8.08E-03Fully effective cross-seciton Ify = 1.66E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.21E-03Radius of gyration ry = 1.16E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 1.34E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.687 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.687 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.595Critical buckling force Fcr = 5.66E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

1.28E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.36E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 1.19E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.52E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.52E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.51E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.51E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 5.71E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.71E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 5.57E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.57E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 6.15E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 2.45E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 3.87E-03Torsional radius r = 2.01E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 7.69E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.80E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 2.40E+08Moment amplification factor By = 1.266 B = , A.12.4

Moment amplification factor Bz = 1.129Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling Mlty = 1.27E+09 Mit = Sf FcF

Mltz = 1.24E+09 Mlt = Mlt ,min for L > = Lr - how to calculate Lr for non I-shape chord?

Fcr = 1.54E+11 Fcr = 6.895 , Cb =1(conservative assumption), Incomplete section A.12.6.2.6

A = 1.81E-01 , A = total closed areaJ = 2.33E-03λ = 36.736 λ = Lb/rmin

X1 = 5.79E+11 X1 = (π/Sf) , J = Ip, G = 77200MPa

X2 = 0 X2 = (4Cw/Iz)(Sy/GJ)2, Cw = 0, ignore warping for close section

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.2.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Radius r = 0.1905 Cross-section area A = 0.12303Wall plate thickness t = 0.0508 Centroid to y axis Yc = 0

Rack plate length Lr = 0.401 Centroid to z axis Zc = 0Rack plate thickness tr = 0.1778 2nd moment of area Iyy = 1.66E-03

Inset d = 0.006 2nd moment of area Izz = 2.21E-03Fully plastic moment Mpy = 8.66E+06

Rack stiffener length Ls = 0 Fully plastic moment Mpz = 9.00E+06Rack stiffener thickness ts = 0 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 4.267

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.90E+08 Axial force Pu = 7.64E+07Rack plate yield stress Fy2 = 6.90E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.90E+08 Bending moment in z Muz = 0.00E+00Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 41.269077 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 15.546 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 41.269077 α is calculated by elastic assumption

Lateral torsional buckling check: Needed(Lb/rmin)(Fymin/E)0.5 = 2.157 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (tension) Notes

Tensile strength AtFymin/γt = 8.08E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 7.38E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 7.83E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 8.25E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 8.57E+06Factored shear strength Pv/γv = 2.33E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 7.32E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.945 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 1.035 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 1.122 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 141 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 142: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 142 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Stiffened wall check thickness tw = 0.178 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.90E+08

Section bending strength Mby = 8.66E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 9.00E+06 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.23E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.26E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 1.31E-02Fully effective cross-seciton Af = 1.23E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 8.28E-03 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 8.08E-03Fully effective cross-seciton Ify = 1.66E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.21E-03Radius of gyration ry = 1.16E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 1.34E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.687 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.687 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.595Critical buckling force Fcr = 5.66E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

1.28E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.36E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 1.19E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.52E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.52E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.51E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.51E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 5.71E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.71E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 5.57E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.57E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 6.15E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 2.45E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 3.87E-03Torsional radius r = 2.01E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 7.69E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.80E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 2.40E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling Mlty = 1.27E+09 Mit = Sf FcF

Mltz = 1.24E+09 Mlt = Mlt ,min for L > = Lr - how to calculate Lr for non I-shape chord?

Fcr = 1.54E+11 Fcr = 6.895 , Cb =1(conservative assumption), Incomplete section A.12.6.2.6

A = 1.81E-01 , A = total closed areaJ = 2.33E-03λ = 36.736 λ = Lb/rmin

X1 = 5.79E+11 X1 = (π/Sf) , J = Ip, G = 77200MPa

X2 = 0 X2 = (4Cw/Iz)(Sy/GJ)2, Cw = 0, ignore warping for close section

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Page 143: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.2.4 Maximum bending y:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Radius r = 0.1905 Cross-section area A = 0.12303Wall plate thickness t = 0.0508 Centroid to y axis Yc = 0

Rack plate length Lr = 0.401 Centroid to z axis Zc = 0Rack plate thickness tr = 0.1778 2nd moment of area Iyy = 1.66E-03

Inset d = 0.006 2nd moment of area Izz = 2.21E-03Fully plastic moment Mpy = 8.66E+06

Rack stiffener length Ls = 0 Fully plastic moment Mpz = 9.00E+06Rack stiffener thickness ts = 0 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 4.267

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.90E+08 Axial force Pu = 0.00E+00Rack plate yield stress Fy2 = 6.90E+08 Bending moment in y Muy = 7.79E+06

Rack stiffener yield stress Fy3 = 6.90E+08 Bending moment in z Muz = 0.00E+00Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 41.269077 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 15.546 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 41.269077 α is calculated by elastic assumption

Lateral torsional buckling check: Needed(Lb/rmin)(Fymin/E)0.5 = 2.157 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (bending y) Notes

Tensile strength AtFymin/γt = 8.08E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 7.38E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 7.83E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 8.25E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 8.57E+06Factored shear strength Pv/γv = 2.33E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 7.32E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.945 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.945 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.945 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 143 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 144: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 144 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Stiffened wall check thickness tw = 0.178 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.90E+08

Section bending strength Mby = 8.66E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 9.00E+06 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.23E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.26E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 1.31E-02Fully effective cross-seciton Af = 1.23E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 8.28E-03 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 8.08E-03Fully effective cross-seciton Ify = 1.66E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.21E-03Radius of gyration ry = 1.16E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 1.34E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.687 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.687 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.595Critical buckling force Fcr = 5.66E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

1.28E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.36E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 1.19E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.52E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.52E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.51E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.51E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 5.71E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.71E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 5.57E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.57E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 6.15E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 2.45E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 3.87E-03Torsional radius r = 2.01E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 7.69E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.80E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 2.40E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 7.79E+06Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling Mlty = 1.27E+09 Mit = Sf FcF

Mltz = 1.24E+09 Mlt = Mlt ,min for L > = Lr - how to calculate Lr for non I-shape chord?

Fcr = 1.54E+11 Fcr = 6.895 , Cb =1(conservative assumption), Incomplete section A.12.6.2.6

A = 1.81E-01 , A = total closed areaJ = 2.33E-03λ = 36.736 λ = Lb/rmin

X1 = 5.79E+11 X1 = (π/Sf) , J = Ip, G = 77200MPa

X2 = 0 X2 = (4Cw/Iz)(Sy/GJ)2, Cw = 0, ignore warping for close section

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.2.5 Maximum bending z:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Radius r = 0.1905 Cross-section area A = 0.12303Wall plate thickness t = 0.0508 Centroid to y axis Yc = 0

Rack plate length Lr = 0.401 Centroid to z axis Zc = 0Rack plate thickness tr = 0.1778 2nd moment of area Iyy = 1.66E-03

Inset d = 0.006 2nd moment of area Izz = 2.21E-03Fully plastic moment Mpy = 8.66E+06

Rack stiffener length Ls = 0 Fully plastic moment Mpz = 9.00E+06Rack stiffener thickness ts = 0 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 4.267

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.90E+08 Axial force Pu = 0.00E+00Rack plate yield stress Fy2 = 6.90E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.90E+08 Bending moment in z Muz = 8.10E+06Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 41.269077 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 15.546 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 41.269077 α is calculated by elastic assumption

Lateral torsional buckling check: Needed(Lb/rmin)(Fymin/E)0.5 = 2.157 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (bending z) Notes

Tensile strength AtFymin/γt = 8.08E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 7.38E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 7.83E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 8.25E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 8.57E+06Factored shear strength Pv/γv = 2.33E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 7.32E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.945 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.945 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.945 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 145 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 146: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 146 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Stiffened wall check thickness tw = 0.178 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.90E+08

Section bending strength Mby = 8.66E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 9.00E+06 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.23E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.26E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 1.31E-02Fully effective cross-seciton Af = 1.23E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 8.28E-03 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 8.08E-03Fully effective cross-seciton Ify = 1.66E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.21E-03Radius of gyration ry = 1.16E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 1.34E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.687 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.687 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.595Critical buckling force Fcr = 5.66E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

1.28E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.36E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 1.19E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.52E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.52E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.51E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.51E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 5.71E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.71E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 5.57E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.57E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 6.15E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 2.45E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 3.87E-03Torsional radius r = 2.01E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 7.69E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.80E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 2.40E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 8.10E+06 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling Mlty = 1.27E+09 Mit = Sf FcF

Mltz = 1.24E+09 Mlt = Mlt ,min for L > = Lr - how to calculate Lr for non I-shape chord?

Fcr = 1.54E+11 Fcr = 6.895 , Cb =1(conservative assumption), Incomplete section A.12.6.2.6

A = 1.81E-01 , A = total closed areaJ = 2.33E-03λ = 36.736 λ = Lb/rmin

X1 = 5.79E+11 X1 = (π/Sf) , J = Ip, G = 77200MPa

X2 = 0 X2 = (4Cw/Iz)(Sy/GJ)2, Cw = 0, ignore warping for close section

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Page 147: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.2.6 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Radius r = 0.1905 Cross-section area A = 0.12303Wall plate thickness t = 0.0508 Centroid to y axis Yc = 0

Rack plate length Lr = 0.401 Centroid to z axis Zc = 0Rack plate thickness tr = 0.1778 2nd moment of area Iyy = 1.66E-03

Inset d = 0.006 2nd moment of area Izz = 2.21E-03Fully plastic moment Mpy = 8.66E+06

Rack stiffener length Ls = 0 Fully plastic moment Mpz = 9.00E+06Rack stiffener thickness ts = 0 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 4.267

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.90E+08 Axial force Pu = 0.00E+00Rack plate yield stress Fy2 = 6.90E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.90E+08 Bending moment in z Muz = 0.00E+00Shear force V = 2.95E+05

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 3.97E+03Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 41.269077 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 15.546 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 41.269077 α is calculated by elastic assumption

Lateral torsional buckling check: Needed(Lb/rmin)(Fymin/E)0.5 = 2.157 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (shear) Notes

Tensile strength AtFymin/γt = 8.08E+07 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 7.38E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 7.83E+07 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 8.25E+06 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 8.57E+06Factored shear strength Pv/γv = 2.33E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 7.32E+06 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.000 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.000 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.000 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.013 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.001 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 147 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 148: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

L22909/NDC/YH Rev: 1 22/01/2008 Page 148 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Calculated Data Notes

Stiffened wall check thickness tw = 0.178 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.90E+08

Section bending strength Mby = 8.66E+06 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 9.00E+06 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.23E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 1.26E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 1.31E-02Fully effective cross-seciton Af = 1.23E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 8.28E-03 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 8.08E-03Fully effective cross-seciton Ify = 1.66E-03 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 2.21E-03Radius of gyration ry = 1.16E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 1.34E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.687 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.687 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.595Critical buckling force Fcr = 5.66E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

1.28E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.36E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 1.19E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.52E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.52E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.51E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.51E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 5.71E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.71E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 5.57E+06 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

5.57E+06 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 6.15E-02 Half of strength area for both directions, according to Justus

Shear strength Pv = 2.45E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 3.87E-03Torsional radius r = 2.01E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 7.69E+06 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.80E+08 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 2.40E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling Mlty = 1.27E+09 Mit = Sf FcF

Mltz = 1.24E+09 Mlt = Mlt ,min for L > = Lr - how to calculate Lr for non I-shape chord?

Fcr = 1.54E+11 Fcr = 6.895 , Cb =1(conservative assumption), Incomplete section A.12.6.2.6

A = 1.81E-01 , A = total closed areaJ = 2.33E-03λ = 36.736 λ = Lb/rmin

X1 = 5.79E+11 X1 = (π/Sf) , J = Ip, G = 77200MPa

X2 = 0 X2 = (4Cw/Iz)(Sy/GJ)2, Cw = 0, ignore warping for close section

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Page 149: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.3 CASE 3, MODERATE DIAMETER TUBULAR CHORD B.2.3.1 Combined loads:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Wall outside radius r = 0.381 Cross-section area A = 0.15921Wall plate thickness t = 0.044 Centroid to y axis Yc = 0

Inset d = 0.0826 Centroid to z axis Zc = 0Rack plate length Lr = 0.1524 2nd moment of area Iyy = 1.05E-02

Rack plate thickness tr = 0.1651 2nd moment of area Izz = 6.72E-03Rack plate offset dr = 0.2456 Fully plastic moment Mpy = 2.51E+07

Rack stiffener length Ls = 0.5912 Fully plastic moment Mpz = 1.79E+07Rack stiffener thickness ts = 0.021 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 3.972

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.77E+08 Axial force Pu = 1.85E+06Rack plate yield stress Fy2 = 6.77E+08 Bending moment in y Muy = -2.21E+07

Rack stiffener yield stress Fy3 = 6.77E+08 Bending moment in z Muz = -2.03E+06Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 366.250 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 2.683 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 366.250 α is calculated by elastic assumption

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 0.703 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (combined) Notes

Tensile strength AtFymin/γt = 1.03E+08 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 9.37E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 1.07E+08 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 2.39E+07 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.71E+07Factored shear strength Pv/γv = 2.96E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 1.61E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.019 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.956 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.946 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Stiffened rack check thickness tw = 0.203 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.77E+08

Section bending strength Mby = 2.51E+07 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.79E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.59E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 3.71E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.65E-02Fully effective cross-seciton Af = 1.59E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 2.63E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 1.76E-02Fully effective cross-seciton Ify = 1.05E-02 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 6.72E-03Radius of gyration ry = 2.57E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 2.05E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.358 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.287 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.358Critical buckling force Fcr = 6.42E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

4.63E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.70E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.91E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.46E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.46E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.43E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.43E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 1.78E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.78E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 1.19E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.19E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 7.96E-02 Half of strength area for both directions, according to Justas

Shear strength Pv = 3.11E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 1.72E-02Torsional radius r = 3.98E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 1.69E+07 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.31E+09 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 8.41E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = -2.21E+07Amplified moment in z Muaz = -2.03E+06 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.3.2 Maximum axial compression: Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Wall outside radius r = 0.381 Cross-section area A = 0.15921Wall plate thickness t = 0.044 Centroid to y axis Yc = 0

Inset d = 0.0826 Centroid to z axis Zc = 0Rack plate length Lr = 0.1524 2nd moment of area Iyy = 1.05E-02

Rack plate thickness tr = 0.1651 2nd moment of area Izz = 6.72E-03Rack plate offset dr = 0.2456 Fully plastic moment Mpy = 2.51E+07

Rack stiffener length Ls = 0.5912 Fully plastic moment Mpz = 1.79E+07Rack stiffener thickness ts = 0.021 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 3.972

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.77E+08 Axial force Pu = -8.68E+07Rack plate yield stress Fy2 = 6.77E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.77E+08 Bending moment in z Muz = 0.00E+00Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 366.250 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 2.683 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 366.250 α is calculated by elastic assumption

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 0.703 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (compression) Notes

Tensile strength AtFymin/γt = 1.03E+08 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 9.37E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 1.07E+08 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 2.39E+07 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.71E+07Factored shear strength Pv/γv = 2.96E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 1.61E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.886 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.926 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.935 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

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L22909/NDC/YH Rev: 1 22/01/2008 Page 151 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Stiffened rack check thickness tw = 0.203 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.77E+08

Section bending strength Mby = 2.51E+07 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.79E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.59E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 3.71E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.65E-02Fully effective cross-seciton Af = 1.59E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 2.63E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 1.76E-02Fully effective cross-seciton Ify = 1.05E-02 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 6.72E-03Radius of gyration ry = 2.57E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 2.05E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.358 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.287 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.358Critical buckling force Fcr = 6.42E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

4.63E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.70E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.91E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.46E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.46E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.43E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.43E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 1.78E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.78E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 1.19E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.19E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 7.96E-02 Half of strength area for both directions, according to Justas

Shear strength Pv = 3.11E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 1.72E-02Torsional radius r = 3.98E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 1.69E+07 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.31E+09 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 8.41E+08Moment amplification factor By = 0.910 B = , A.12.4

Moment amplification factor Bz = 0.948Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

L22909/NDC/YH Rev: 1 22/01/2008 Page 152 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.3.3 Maximum axial tension:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Wall outside radius r = 0.381 Cross-section area A = 0.15921Wall plate thickness t = 0.044 Centroid to y axis Yc = 0

Inset d = 0.0826 Centroid to z axis Zc = 0Rack plate length Lr = 0.1524 2nd moment of area Iyy = 1.05E-02

Rack plate thickness tr = 0.1651 2nd moment of area Izz = 6.72E-03Rack plate offset dr = 0.2456 Fully plastic moment Mpy = 2.51E+07

Rack stiffener length Ls = 0.5912 Fully plastic moment Mpz = 1.79E+07Rack stiffener thickness ts = 0.021 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 3.972

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.77E+08 Axial force Pu = 9.70E+07Rack plate yield stress Fy2 = 6.77E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.77E+08 Bending moment in z Muz = 0.00E+00Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 366.250 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 2.683 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 366.250 α is calculated by elastic assumption

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 0.703 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (tension) Notes

Tensile strength AtFymin/γt = 1.03E+08 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 9.37E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 1.07E+08 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 2.39E+07 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.71E+07Factored shear strength Pv/γv = 2.96E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 1.61E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.945 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 1.035 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 1.045 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

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L22909/NDC/YH Rev: 1 22/01/2008 Page 153 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 154: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Stiffened rack check thickness tw = 0.203 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.77E+08

Section bending strength Mby = 2.51E+07 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.79E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.59E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 3.71E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.65E-02Fully effective cross-seciton Af = 1.59E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 2.63E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 1.76E-02Fully effective cross-seciton Ify = 1.05E-02 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 6.72E-03Radius of gyration ry = 2.57E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 2.05E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.358 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.287 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.358Critical buckling force Fcr = 6.42E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

4.63E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.70E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.91E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.46E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.46E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.43E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.43E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 1.78E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.78E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 1.19E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.19E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 7.96E-02 Half of strength area for both directions, according to Justas

Shear strength Pv = 3.11E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 1.72E-02Torsional radius r = 3.98E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 1.69E+07 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.31E+09 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 8.41E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

L22909/NDC/YH Rev: 1 22/01/2008 Page 154 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.3.4 Maximum bending y:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Wall outside radius r = 0.381 Cross-section area A = 0.15921Wall plate thickness t = 0.044 Centroid to y axis Yc = 0

Inset d = 0.0826 Centroid to z axis Zc = 0Rack plate length Lr = 0.1524 2nd moment of area Iyy = 1.05E-02

Rack plate thickness tr = 0.1651 2nd moment of area Izz = 6.72E-03Rack plate offset dr = 0.2456 Fully plastic moment Mpy = 2.51E+07

Rack stiffener length Ls = 0.5912 Fully plastic moment Mpz = 1.79E+07Rack stiffener thickness ts = 0.021 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 3.972

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.77E+08 Axial force Pu = 0.00E+00Rack plate yield stress Fy2 = 6.77E+08 Bending moment in y Muy = 2.26E+07

Rack stiffener yield stress Fy3 = 6.77E+08 Bending moment in z Muz = 0.00E+00Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 366.250 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 2.683 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 366.250 α is calculated by elastic assumption

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 0.703 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (bending y) Notes

Tensile strength AtFymin/γt = 1.03E+08 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 9.37E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 1.07E+08 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 2.39E+07 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.71E+07Factored shear strength Pv/γv = 2.96E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 1.61E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.945 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.945 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.945 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 155 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 156: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Stiffened rack check thickness tw = 0.203 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.77E+08

Section bending strength Mby = 2.51E+07 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.79E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.59E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 3.71E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.65E-02Fully effective cross-seciton Af = 1.59E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 2.63E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 1.76E-02Fully effective cross-seciton Ify = 1.05E-02 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 6.72E-03Radius of gyration ry = 2.57E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 2.05E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.358 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.287 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.358Critical buckling force Fcr = 6.42E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

4.63E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.70E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.91E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.46E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.46E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.43E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.43E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 1.78E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.78E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 1.19E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.19E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 7.96E-02 Half of strength area for both directions, according to Justas

Shear strength Pv = 3.11E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 1.72E-02Torsional radius r = 3.98E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 1.69E+07 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.31E+09 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 8.41E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 2.26E+07Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

L22909/NDC/YH Rev: 1 22/01/2008 Page 156 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 157: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.3.5 Maximum bending z:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Wall outside radius r = 0.381 Cross-section area A = 0.15921Wall plate thickness t = 0.044 Centroid to y axis Yc = 0

Inset d = 0.0826 Centroid to z axis Zc = 0Rack plate length Lr = 0.1524 2nd moment of area Iyy = 1.05E-02

Rack plate thickness tr = 0.1651 2nd moment of area Izz = 6.72E-03Rack plate offset dr = 0.2456 Fully plastic moment Mpy = 2.51E+07

Rack stiffener length Ls = 0.5912 Fully plastic moment Mpz = 1.79E+07Rack stiffener thickness ts = 0.021 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 3.972

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.77E+08 Axial force Pu = 0.00E+00Rack plate yield stress Fy2 = 6.77E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.77E+08 Bending moment in z Muz = 1.61E+07Shear force V = 0.00E+00

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 0.00E+00Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 366.250 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 2.683 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 366.250 α is calculated by elastic assumption

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 0.703 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (bending y) Notes

Tensile strength AtFymin/γt = 1.03E+08 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 9.37E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 1.07E+08 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 2.39E+07 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.71E+07Factored shear strength Pv/γv = 2.96E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 1.61E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.945 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.945 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.945 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.000 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

L22909/NDC/YH Rev: 1 22/01/2008 Page 157 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

Page 158: ISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 - DNV GL · PDF fileISO TC 67 / SC 7 / WG 7 / Panel 10 ISO 19905-1 ... \projects\05-04-2803 ... the shear stress utilisations were all

DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Stiffened rack check thickness tw = 0.203 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.77E+08

Section bending strength Mby = 2.51E+07 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.79E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.59E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 3.71E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.65E-02Fully effective cross-seciton Af = 1.59E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 2.63E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 1.76E-02Fully effective cross-seciton Ify = 1.05E-02 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 6.72E-03Radius of gyration ry = 2.57E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 2.05E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.358 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.287 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.358Critical buckling force Fcr = 6.42E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

4.63E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.70E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.91E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.46E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.46E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.43E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.43E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 1.78E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.78E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 1.19E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.19E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 7.96E-02 Half of strength area for both directions, according to Justas

Shear strength Pv = 3.11E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 1.72E-02Torsional radius r = 3.98E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 1.69E+07 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.31E+09 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 8.41E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 1.61E+07 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

L22909/NDC/YH Rev: 1 22/01/2008 Page 158 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.2.3.6 Beam shear and torsional shear:

Yellow cells for user input Unit: N, m

Chord Section Properties PLASTIC Output

Wall outside radius r = 0.381 Cross-section area A = 0.15921Wall plate thickness t = 0.044 Centroid to y axis Yc = 0

Inset d = 0.0826 Centroid to z axis Zc = 0Rack plate length Lr = 0.1524 2nd moment of area Iyy = 1.05E-02

Rack plate thickness tr = 0.1651 2nd moment of area Izz = 6.72E-03Rack plate offset dr = 0.2456 Fully plastic moment Mpy = 2.51E+07

Rack stiffener length Ls = 0.5912 Fully plastic moment Mpz = 1.79E+07Rack stiffener thickness ts = 0.021 Plastic neutral axis Yp = 0

Plastic neutral axis Zp = 0Bay height H = 3.972

Chord Section ForcesYoung's modulus E = 2.00E+11

Wall plate yield stress Fy1 = 6.77E+08 Axial force Pu = 0.00E+00Rack plate yield stress Fy2 = 6.77E+08 Bending moment in y Muy = 0.00E+00

Rack stiffener yield stress Fy3 = 6.77E+08 Bending moment in z Muz = 0.00E+00Shear force V = 2.19E+06

Effective length factor K = 1.00 (Table A.12.4 Torsional moment T = 2.73E+05Moment amplification Cm = 0.85 (Table A.12.4-1)

Strength check exponent ή = 1.80 (A.12.6.3.2)

Section Classification Notes

Selected Chord classification: Class 1 Depends on check position and load condition which is not quite suitable for jack-ups

Flange internal components: Class 1 Table A.12.2-1, (Fy)0.5(b/t) <= 465, Class 1-plastic; 465 < (Fy)

0.5(b/t) <= 530, Class 2-compact; (Fy)

0.5b/t = 366.250 530 < (Fy)0.5(b/t) <= 650, Class 3-semi compact ; (Fy)

0.5(b/t) > 650, Class 4-slender section

Flange outstand components: Class 1 Table A.12.2-2, (Fy)0.5(b/t) <= 135(weld)/150(rolled), Class 1-plastic; 135(w)/150(r) < (Fy)

0.5(b/t) <= 150(w)/166(r), Class 2-compa (Fy)

0.5b/t = 2.683 Weld 150(w)/166(r) < (Fy)0.5(b/t) <= 225(w)/250(r), Class 3-semi compact;(Fy)

0.5(b/t) > 225(w)/250(r), Class 4-slender section

Web internal components: Class 1 Table A.12.2-3, considering only α = 1.0 (fully compression) and then same to Table A.12.2.-3 (Fy)

0.5b/t = 366.250 α is calculated by elastic assumption

Lateral torsional buckling check: No Need(Lb/rmin)(Fymin/E)0.5 = 0.703 Check needed only when (Lb/rmin)(Fymin/E)0.5 >1.51, A.12.2.3.2

Member Strength Check (shear) Notes

Tensile strength AtFymin/γt = 1.03E+08 Resistance factor γt = 1.05, A.12.6.2.2

Compression strength AcFymin/γc = 9.37E+07 Resistance factor γc = 1.15 (Fy>450MPa)/1.1(Fy<=450MPa), A.12.6.2.3

Column buckling strength Pn = 1.07E+08 Eq. E2.1 ?-Equation No

Purely bending strength Mby/γb = 2.39E+07 Resistance factor γb = 1.05, A.12.6.2.7

Purely bending strength Mbz/γb = 1.71E+07Factored shear strength Pv/γv = 2.96E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Factored torsional strength Tv/γv = 1.61E+07 Resistance factor γv = 1.05, Eqn A.12.6-x ?

Tension&Bending Check: 0.000 When P > 0, γt = 1.1, A.12.6.3.2, Eq No ?Compression&Bending Check: 0.000 When P < 0, A.12.6.3.2, Eq No ?Beam-Column Strength Check: 0.000 Eq.H1-1 Numbering?

Beam Shear Strength Check: 0.074 Eqn A.12.6-x Numbering?Torsional Shear Strength Check: 0.017 Eqn A.12.6-x Numbering?

X

YPLASTIC/JUSTAS

PAFEC/CODE-86

Z

Y

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

Calculated Data Notes

Stiffened rack check thickness tw = 0.203 tcheck=tw,Eqn A.12.2-2

Min component yeild stress Fymin = 6.77E+08

Section bending strength Mby = 2.51E+07 Mb = ZpFymin for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Section bending strength Mbz = 1.79E+07 Mb = ZpFymin, for class 1 or 2, Eqn No ?

? Mb = Mp-(Mp-MR) for class 3, Eq. A-F1.3 ?, depends on check position? Mb = SeFy for class 4, A.12.6.5.4, Eqn No ?

Plastic cross-section area Ap = 1.59E-01 For class 1 or 2, Eqn A.12.3-1

Plastic section modulus Zpy = 3.71E-02 For class 1 or 2, Eqn A.12.3-2

Plastic section modulus Zpz = 2.65E-02Fully effective cross-seciton Af = 1.59E-01 For class 3 or 4 in tension, Eqn A.12.3-3

Fully elastic seciton modulus Sfy = 2.63E-02 For class 3 or 4 in tension

Fully elastic seciton modulus Sfz = 1.76E-02Fully effective cross-seciton Ify = 1.05E-02 For class 1, 2 or 3, in A.12.3.5.3

Fully effective cross-seciton Ifz = 6.72E-03Radius of gyration ry = 2.57E-01 For class 1or 2, in A.12.6.2.4

Radius of gyration rz = 2.05E-01

Slender effective area Ae = ? For class 4 in compression, Eqn A.12.3-4

Slender effective moment Iey = ?Slender effective moment Iez = ?

Slender effective modulus Sey = ?Slender effective modulus Sey = ?

Column slenderness λc = 0.358 λc = max(λy,λz), for class 1 or 2

y slenderness λy = 0.287 λ = , for class 1 or 2, Eq. E2.4 ?-Equation No

z slenderness λz = 0.358Critical buckling force Fcr = 6.42E+08 Fcr = (0.658λc2) Fymin? for Fymin <= 450?Mpa, λ <= 1.5, for class 1 or 2, Eq. E2.2?

4.63E+09 Fcr = (0.877/λc2) Fymin? for Fymin <= 450?Mpa, λ > 1.5, for class 1 or 2, Eq. E2.3?

6.70E+08 Fcr = (0.7625λc3.22) Fymin? for Fymin > 450?Mpa, λ <= 1.2, for class 1 or 2, Eqn A.12.7-1a 3.91E+09 Fcr = (0.8608/λc

1.854) Fymin? for Fymin > 450?Mpa, λ > 1.2, for class 1 or 2, Eqn A.12.7-1b

Class 3 bending parameter λ = ? λ = b/t depends on check position, A.12.6.2.5.3

Class 3 bending parameter λp = 6.46E-02 λp = 1.11/ , chord, for class 3, A.12.6.2.5.3

6.46E-02 λp = 1.11/ , rack, for class 3, A.12.6.2.5.3Class 3 bending parameter λr = 7.43E-02 λr = 1.4/ , Fr = 114 MPa, chord, A.12.6.2.5.3

7.43E-02 λr = 1.4/ , Fr = 114 MPa, rack, A.12.6.2.5.3Class 3 buckling moment MRy = 1.78E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.78E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3Class 3 buckling moment MRz = 1.19E+07 MR = Sf,Fy, chord, for class 3, A.12.6.2.5.3

1.19E+07 MR = Sf,Fy, rack, for class 3, A.12.6.2.5.3

Shear area As = 7.96E-02 Half of strength area for both directions, according to Justas

Shear strength Pv = 3.11E+07 Pv = AsFy/(√3)

Polar moment of inertia Ip = 1.72E-02Torsional radius r = 3.98E-01 Maximum distance from centroid? to an extreme fibre, A.12.6.3.5

Torsional strength Tv = 1.69E+07 Tv = Ip Fy/(r√3), A.12.6.3.5

Euler buckling strength PEy = 1.31E+09 PE = (π2AcE)/(KL/r)2, A.12.4

Euler buckling strength PEz = 8.41E+08Moment amplification factor By = 1.000 B = , A.12.4

Moment amplification factor Bz = 1.000Amplified moment in y Muay = 0.00E+00Amplified moment in z Muaz = 0.00E+00 Mua = B Mu , Eqn A.12.4-1

Lateral torsional buckling: ? Incomplete, A.12.6.2.6 & A.12.2.3.2

Some ISO Equations

Ref: A.12.6.3.2

)/1( Eu

m

PPC

21

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

)/( yiFE

)/( yiFE

)/( rywj FFE −

)/( rywj FFE −

Local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

Beam-column check: If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

L22909/NDC/YH Rev: 1 22/01/2008 Page 160 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.3 THE SNAME SPREADSHEETS FOR FACTORING LOADS

B.3.1 CASE 1 B.3.1.1 Horizontal brace

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util1132 1 -5.806E+06 -2.166E+05 8.047E+02 2.0 0.013 1.54E-04 1.54E-04 4.379 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.321

2 -5.806E+06 7.791E+04 3.945E+03 2.0 0.013 1.54E-04 1.54E-04 4.379 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.121UC Comp -5.750E+06 0.000E+00 0.000E+00 2.0 0.013 1.54E-04 1.54E-04 4.379 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000

Tension 6.941E+06 0.000E+00 0.000E+00 2.0 0.013 1.54E-04 1.54E-04 4.379 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000Bending 0.000E+00 6.870E+05 0.000E+00 2.0 0.013 1.54E-04 1.54E-04 4.379 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000

77% Comb -4.465E+06 -1.666E+05 6.188E+02 2.0 0.013 1.54E-04 1.54E-04 4.379 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000

PAFEC Results

B.3.1.2 Diagonal brace

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util2431 1 -6.129E+06 3.997E+04 6.228E+03 2.0 0.013 1.54E-04 1.54E-04 5.550 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.227

2 -6.129E+06 6.268E+04 -5.103E+03 2.0 0.013 1.54E-04 1.54E-04 5.550 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.268UC Comp -5.315E+06 0.000E+00 0.000E+00 2.0 0.013 1.54E-04 1.54E-04 5.550 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000

Tension 6.941E+06 0.000E+00 0.000E+00 2.0 0.013 1.54E-04 1.54E-04 5.550 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000Bending 0.000E+00 6.870E+05 0.000E+00 2.0 0.013 1.54E-04 1.54E-04 5.550 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000

82% Comb -5.026E+06 3.278E+04 5.107E+03 2.0 0.013 1.54E-04 1.54E-04 5.550 0.80 7.709E+06 7.632E+05 7.632E+05 6.070E+08 1.000

PAFEC Results

B.3.1.3 Triangular chord

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util531 1 -2.942E+07 -5.194E+06 1.702E+05 1.8 0.102 5.46E-03 2.63E-03 3.408 1.00 5.087E+07 6.419E+06 1.093E+07 5.010E+08 1.513

2 -2.793E+07 -4.914E+06 7.463E+04 1.8 0.102 5.46E-03 2.63E-03 3.408 1.00 5.087E+07 6.419E+06 1.093E+07 5.010E+08 1.434UC Comp -4.124E+07 0.000E+00 0.000E+00 1.8 0.102 5.46E-03 2.63E-03 3.408 1.00 5.087E+07 6.419E+06 1.093E+07 5.010E+08 1.000

Tension 4.580E+07 0.000E+00 0.000E+00 1.8 0.102 5.46E-03 2.63E-03 3.408 1.00 5.087E+07 6.419E+06 1.093E+07 5.010E+08 1.000Y bending 0.000E+00 5.780E+06 0.000E+00 1.8 0.102 5.46E-03 2.63E-03 3.408 1.00 5.087E+07 6.419E+06 1.093E+07 5.010E+08 1.000Z bending 0.000E+00 0.000E+00 9.830E+06 1.8 0.102 5.46E-03 2.63E-03 3.408 1.00 5.087E+07 6.419E+06 1.093E+07 5.010E+08 1.000

66% Comb -1.945E+07 -3.433E+06 1.125E+05 1.8 0.102 5.46E-03 2.63E-03 3.408 1.00 5.087E+07 6.419E+06 1.093E+07 5.010E+08 1.000

PAFEC Results

B.3.2 CASE 2 B.3.2.1 Horizontal brace

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util1501 1 -2.189E+06 1.865E+04 -6.643E+02 2.0 0.019 8.52E-05 8.52E-05 5.684 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 0.577

2 -2.189E+06 -2.171E+03 -4.635E+01 2.0 0.019 8.52E-05 8.52E-05 5.684 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 0.535UC Comp -4.135E+06 0.000E+00 0.000E+00 2.0 0.019 8.52E-05 8.52E-05 5.684 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000

Tension 7.570E+06 0.000E+00 0.000E+00 2.0 0.019 8.52E-05 8.52E-05 5.684 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000Bending 0.000E+00 4.530E+05 0.000E+00 2.0 0.019 8.52E-05 8.52E-05 5.684 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000

166% Comb -3.640E+06 3.101E+04 -1.105E+03 2.0 0.019 8.52E-05 8.52E-05 5.684 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000

PAFEC Results

B.3.2.2 Diagonal brace

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util3214 1 -1.308E+06 -4.687E+03 4.385E+03 2.0 0.019 8.52E-05 8.52E-05 6.657 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 0.403

2 -1.308E+06 1.552E+04 -8.608E+03 2.0 0.019 8.52E-05 8.52E-05 6.657 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 0.429UC Comp -3.373E+06 0.000E+00 0.000E+00 2.0 0.019 8.52E-05 8.52E-05 6.657 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000

Tension 7.570E+06 0.000E+00 0.000E+00 2.0 0.019 8.52E-05 8.52E-05 6.657 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000Bending 0.000E+00 4.530E+05 0.000E+00 2.0 0.019 8.52E-05 8.52E-05 6.657 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000

238% Comb -3.116E+06 -1.116E+04 1.045E+04 2.0 0.019 8.52E-05 8.52E-05 6.657 0.90 8.407E+06 5.031E+05 5.031E+05 4.483E+08 1.000

PAFEC Results

B.3.2.3 Compact tubular chord

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util523 1 -8.336E+07 -1.140E+06 -1.101E+03 1.8 0.123 1.66E-03 2.21E-03 4.267 1.00 8.486E+07 8.658E+06 9.001E+06 6.897E+08 1.614

2 -8.336E+07 8.489E+05 3.851E+03 1.8 0.123 1.66E-03 2.21E-03 4.267 1.00 8.486E+07 8.658E+06 9.001E+06 6.897E+08 1.561UC Comp -5.920E+07 0.000E+00 0.000E+00 1.8 0.123 1.66E-03 2.21E-03 4.267 1.00 8.486E+07 8.658E+06 9.001E+06 6.897E+08 1.000

Tension 7.635E+07 0.000E+00 0.000E+00 1.8 0.123 1.66E-03 2.21E-03 4.267 1.00 8.486E+07 8.658E+06 9.001E+06 6.897E+08 1.000Y bending 0.000E+00 7.790E+06 0.000E+00 1.8 0.123 1.66E-03 2.21E-03 4.267 1.00 8.486E+07 8.658E+06 9.001E+06 6.897E+08 1.000Z bending 0.000E+00 0.000E+00 8.100E+06 1.8 0.123 1.66E-03 2.21E-03 4.267 1.00 8.486E+07 8.658E+06 9.001E+06 6.897E+08 1.000

63.9% Comb -5.327E+07 -7.285E+05 -7.035E+02 1.8 0.123 1.66E-03 2.21E-03 4.267 1.00 8.486E+07 8.658E+06 9.001E+06 6.897E+08 1.000

PAFEC Results

L22909/NDC/YH Rev: 1 22/01/2008 Page 161 File: g:\projects\05-04-2803 - iso 19905-1 clause 12 high level benchmarking\report\l22909-rev 1.doc

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

B.3.3 CASE 3 B.3.3.1 Horizontal brace

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util1116 1 -6.127E+06 8.580E+03 -4.178E+04 2.0 0.015 1.84E-04 1.84E-04 6.625 0.90 9.073E+06 8.892E+05 8.892E+05 5.862E+08 1.253

2 -6.127E+06 1.014E+04 1.432E+04 2.0 0.015 1.84E-04 1.84E-04 6.625 0.90 9.073E+06 8.892E+05 8.892E+05 5.862E+08 1.194UC Comp -5.320E+06 0.000E+00 0.000E+00 2.0 0.015 1.84E-04 1.84E-04 6.625 0.90 9.073E+06 8.892E+05 8.892E+05 5.862E+08 1.000

Tension 8.165E+06 0.000E+00 0.000E+00 2.0 0.015 1.84E-04 1.84E-04 6.625 0.90 9.073E+06 8.892E+05 8.892E+05 5.862E+08 1.000Bending 0.000E+00 8.000E+05 0.000E+00 2.0 0.015 1.84E-04 1.84E-04 6.625 0.90 9.073E+06 8.892E+05 8.892E+05 5.862E+08 1.000

81% Comb -4.978E+06 6.971E+03 -3.395E+04 2.0 0.015 1.84E-04 1.84E-04 6.625 0.90 9.073E+06 8.892E+05 8.892E+05 5.862E+08 1.000

PAFEC Results

B.3.3.2 Diagonal brace

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util2917 1 -5.271E+06 -5.122E+04 5.692E+03 2.0 0.020 2.30E-04 2.30E-04 7.725 0.90 1.172E+07 1.130E+06 1.130E+06 5.862E+08 0.978

2 -5.271E+06 -3.230E+03 -1.670E+04 2.0 0.020 2.30E-04 2.30E-04 7.725 0.90 1.172E+07 1.130E+06 1.130E+06 5.862E+08 0.920UC Comp -5.915E+06 0.000E+00 0.000E+00 2.0 0.020 2.30E-04 2.30E-04 7.725 0.90 1.172E+07 1.130E+06 1.130E+06 5.862E+08 1.000

Tension 1.054E+07 0.000E+00 0.000E+00 2.0 0.020 2.30E-04 2.30E-04 7.725 0.90 1.172E+07 1.130E+06 1.130E+06 5.862E+08 1.000Bending 0.000E+00 1.017E+06 0.000E+00 2.0 0.020 2.30E-04 2.30E-04 7.725 0.90 1.172E+07 1.130E+06 1.130E+06 5.862E+08 1.000

102% Comb -5.376E+06 -5.224E+04 5.806E+03 2.0 0.020 2.30E-04 2.30E-04 7.725 0.90 1.172E+07 1.130E+06 1.130E+06 5.862E+08 1.000

PAFEC Results

B.3.3.3 Moderate diameter tubular chord

Element Node Pa MYY MZZ η X-Area Iyy Izz Length K Squash Load MPy MPz Yeff Max-Util339 1 8.344E+05 -9.964E+06 -9.155E+05 1.8 0.159 1.05E-02 6.72E-03 3.972 1.00 1.078E+08 2.512E+07 1.792E+07 6.769E+08 0.451

2 8.344E+05 -2.279E+05 -4.302E+05 1.8 0.159 1.05E-02 6.72E-03 3.972 1.00 1.078E+08 2.512E+07 1.792E+07 6.769E+08 0.033UC Comp -8.680E+07 0.000E+00 0.000E+00 1.8 0.159 1.05E-02 6.72E-03 3.972 1.00 1.078E+08 2.512E+07 1.792E+07 6.769E+08 1.000

Tension 9.700E+07 0.000E+00 0.000E+00 1.8 0.159 1.05E-02 6.72E-03 3.972 1.00 1.078E+08 2.512E+07 1.792E+07 6.769E+08 1.000Y bending 0.000E+00 2.260E+07 0.000E+00 1.8 0.159 1.05E-02 6.72E-03 3.972 1.00 1.078E+08 2.512E+07 1.792E+07 6.769E+08 1.000Z bending 0.000E+00 0.000E+00 1.613E+07 1.8 0.159 1.05E-02 6.72E-03 3.972 1.00 1.078E+08 2.512E+07 1.792E+07 6.769E+08 1.000

222% Comb 1.850E+06 -2.209E+07 -2.030E+06 1.8 0.159 1.05E-02 6.72E-03 3.972 1.00 1.078E+08 2.512E+07 1.792E+07 6.769E+08 1.000

PAFEC Results

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DRAFT ISO 19905-1

INITIAL BENCHMARKING OF CLAUSE 12

APPENDIX C

C : ISO 19905-1 CLAUSES 12-13-12A-&-13A ONLY - WORKING COPY @ CITY U.DOC

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12 Structural strength

12.1 Applicability

12.1.1 General

This clause provides the basis for the determination of the structural strength of truss type legs. Limited guidance is given for other leg types. The strength of the fixation system and/or the elevating system and the strength of the spudcan are normally provided by the manufacturer. The utilization checks shall be carried out according to the equations in Clause 13.

A suitable method of carrying out the specific calculations required by this clause can be found in A.12. The resistance factors given in Annex B are specifically tied to the calculation methods presented in Annex A and should be re-calibrated if other methods are used.

NOTE Classification society requirements cover the design, construction, and periodic survey of the jack-up (see Clause 1). Issues, such as material properties, fabrication tolerances, welds, and construction details, are addressed by the classification society. Similarly, other parts of the jack-up (e.g. jackhouse and hull structure) are also not addressed here.

EXAMPLE If the reactions within the fixation system are within the limits set by the manufacturer, then no additional assessment is required of the hull and jackhouse. In addition, if the spudcan vertical and rotational reactions are within the limits set by the manufacturer, it is not necessary to check the strength of the leg to spudcan connection.

12.1.2 Truss type legs

The requirements set out in this clause relate to chords and braces. Weld sizes, gusset plates, the strength of joints, etc. are covered by classification society requirements (see Clause 1), and should not control the overall structural integrity. Chords and braces are covered in 12.2 to 12.6.

12.1.3 Other leg types

Some of the checks included in this clause are applicable to either tubular or box-type legs but this clause needs to be supplemented with other documents to address stiffened sections. (e.g. API Bull 2U "Stability Design of Cylindrical Shells", API Bull 2V "Design of Flat Plate Structures", DNV-RP-C201 "Buckling Strength of Plated Structures" and DNV-RP-C202 "Buckling Strength of Shells")

12.1.4 Fixation system and/or elevating system

Strength of the fixation and/or the elevating system is normally supplied by the manufacturer. The data should represent the unfactored ultimate strength of the system, normally separately given in the vertical and horizontal directions.

12.1.5 Spudcan strength including connection to the leg

The strength of the spudcan is normally supplied by the manufacturer. The data is expected to represent the unfactored ultimate strength of the system, normally given for all applicable vertical and horizontal forces, and for moments about the horizontal axes.

12.1.6 Overview of the assessment procedure

The basic approach consists of the determination of:

⎯ classification of cross sections (12.2);

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⎯ section properties (12.3);

⎯ node-to-node Euler amplification of member forces (if not included within the structural analysis) (12.4);

⎯ strength of lattice leg members (tubular members (12.5), and non-tubular or prismatic members (12.6)); and

⎯ strength of joints (12.7).

12.2 Classification of member cross-sections

12.2.1 Member types

The methodology used to classify member cross-sections is different for circular tubulars and all other cross sections which are called prismatics. Longitudinally reinforced tubulars and tubulars with pin-holes, cut-outs, etc. shall be considered to be prismatic.

12.2.2 Material yield strength

The material yield strength used in the member classification and the calculation of the capacities shall correspond to the value at 0.2% strain offset from the initial linear stress-strain behaviour. A lesser value shall be used when the material does not exhibit sufficient work-hardening.

12.2.3 Classification definitions

The strength of a steel cross-section is affected by its potential to suffer local buckling when subjected to compression due to either a bending moment or an axial force. By classifying cross-sections, the need to explicitly calculate local buckling stresses is avoided. Components and cross-sections are classified as plastic, compact, non-compact (or semi-compact) and slender. When a cross-section is composed of components of different classes, it shall be classified according to the highest (least favourable) class of its compression components.

However, in 12.5, strength of tubular members, the presented equations account for local buckling, whether plastic or elastic. Consequently, there is no need for tubular sections to be classified to the same extent as prismatic sections other than to identify those tubulars for which plastic hinge rotation capability, i.e., class 1, is possible.

The following classification shall be applied:

Class 1 Plastic: Cross-sections with plastic hinge rotation capability. Compliance with this classification enables a plastic hinge to develop with sufficient rotation capability to allow redistribution of moments to occur within the structure. All plastic sections are inherently compact.

NOTE Compliance with this classification is only relevant when undertaking earthquake, accidental or alternative strength analyses (see 10.7, 10.8 and 10.9). In all other cases the distinction between plastic and compact is irrelevant to the assessment.

Class 2 Compact: Cross-sections with plastic moment capability. Compliance with this classification enables the full plastic moment capability of a cross-section to be developed but local buckling prevents the development of a plastic hinge with sufficient rotation capability to permit plastic assessment.

Class 3 Non-compact (or semi-compact): Cross-sections with yield moment capability. Compliance with this classification enables the yield stress to be realized at the extreme compression fibre but local buckling prevents full << EPR June 07: to discuss with PAFA >> development of the plastic moment capability.

Class 4 Slender: Cross-sections that buckle locally before yield stress can be achieved. A cross-section is classified as slender if any of the compression components of the cross-section do not comply with the limits for non-compact components.

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12.3 Section properties

12.3.1 General

The requirements in this subclause apply to rolled and welded prismatic members which may be made up of one or more components such as can be found in a chord section of a jack-up leg.

Cross-sectional properties for prismatic members shall be determined as described in this subclause.

Cross-sectional properties of tubular members are included within the determination of their strength and addressed in 12.5.

12.3.2 Plastic and compact sections

For class 1 plastic and class 2 compact sections, section properties can be determined assuming fully plastic properties.

Where elastic section properties are determined for class 1 and 2 sections in place of plastic section properties, these can be based on a fully effective cross-section and shall then be treated as for class 3 sections.

12.3.3 Semi-compact sections

Section properties for class 3 semi-compact sections shall be based on elastic properties assuming fully effective cross-sections. When considering a cross-section comprised of components having different yield strengths, the critical stress locations shall be evaluated as these may not coincide with the minimum section modulus or the principal axes.

NOTE The critical stress locations are typically at the edges of the components and are a function of the member forces, the yield strength of the component and its position within the cross-section of the member.

12.3.4 Slender sections

Cross-section properties for class 4 slender sections shall be determined using elastic principles. When the stress across the entire section is tensile, the full section may be used. If any part of the section is in compression, the sectional properties shall be reduced as required based on effective sections (see A.12.xx).

12.3.5 Cross-section properties for assessment

The nomenclature and equations required by this subclause for use in the assessment of members are summarized in A.12.3.5.

12.4 Member moment amplification and effective lengths

Member moments shall include the effects of member deformation amplified by axial compression (Euler moment amplification - p-δ).

NOTE Traditionally, the effects of Euler amplification are included in the strength checks. However, some analyses implicitly include the effects of Euler amplification while in others it is explicitly added. The assessment needs to include the effects of both the global P-Δ and the local member moment amplification (p-δ). Large displacement effects (P-Δ) are addressed in Clause 8.

Moment amplification shall be included in the strength check when it is not included in the structural analysis. The effective length factors and moment reduction factors (Cm) for use in strength checks are listed in Table A.12.4-1. Alternatively, they may be determined using a rational analysis that includes joint flexibility and side-sway.

Moment amplification associated with the eccentricity between the elastic and plastic centroids shall also be included; this can occur in cross-sections which include materials of differing yield strengths.

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12.5 Strength of tubular members

The strength of tubular members subjected to combined axial forces and bending shall be checked. The strength of members subjected to shear and torsional shear shall also be checked.

The requirements given in this subclause ignore the effects of hydrostatic pressure. The validity of this assumption shall be checked for all sealed tubular sections.

12.6 Strength of prismatic members

The strength of prismatic members subjected to combined axial forces and bending shall be checked. The strength of prismatic members subjected to beam shear and torsional shear shall also be checked.

12.7 Assessment of member joints

Tubular joints shall be assessed in accordance with the requirements of ISO 19902 subclause 24.9.2.2.2 (Connections). Rational analysis shall be used for non-tubular joints. The value of the internal forces (action effects) due to factored actions shall be determined in accordance with ISO 19905-1 subclause 8.8, rather than using the factors in ISO 19902 and 19901-3.

NOTE The intent of the joint check is to ensure that the joint is strong enough to resist the internal forces due to applied actions. There is no requirement for the joint to be sufficiently strong to withstand the full member strength. Guidance on non-tubular joint strength can be found in other clauses of ISO 19902 and 19901-3.

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13 Acceptance criteria

13.1 Applicability

13.1.1 General

This clause defines the criteria for checking the acceptability of a jack-up for operation at a specific site.

The partial action and resistance factors set out in these acceptance criteria have been developed in conjunction with the analysis methodology set out in the rest of this document and are valid only if used with this methodology. The factors do not necessarily provide adequate reliability if used with other methodologies.

NOTE This clause presently addresses the manned non-evacuated condition (L1) only and is currently based on the 50 year independent extremes as given in 6.4. Other levels will be included when they are developed << i.e GoMex L2 stuff >>.

The criteria for checking the acceptability of a jack-up are discussed in this clause, and include consideration of the following issues:

⎯ structural strength of legs, spudcan, and holding system (13.3, 13.4, and 13.5 respectively);

⎯ hull elevation (13.6);

⎯ leg length reserve (13.7);

⎯ overturning stability (13.8);

⎯ foundation integrity including preload, bearing capacity, sliding displacement, settlement resulting from exceedance of the capacity envelope (13.9); and

⎯ temperature (13.10).

<<< With ref to “Consequence Based in 1st para: jjs: present categorisation of L1 to include any other than manned evacuated needs to be re-considered. mjrh: Other return periods can be used based on 6.4. Would like it to be “consequence based exposure level criteria (life-safety PAFA).>>>

13.1.2 Ultimate limit states (ULS)

The assessment shall ensure that the acceptance criteria are not exceeded in any of the applicable assessment situations; see 5.1, 5.3 and 5.4.

Areas that are often critical on jack-up rigs are the legs at the lower guides, the legs between guides, the pinions and/or rack teeth, the chocks and/or chock supports (if chocks are fitted) and the leg to spudcan connection.

Where foundation fixity exists, the lower parts of the leg shall be checked assuming an upper bound fixity value. Foundation fixity shall only be included in the evaluation of the upper leg when an applicable and detailed foundation study has been made.

Compliance may be demonstrated through comparison with prior assessments according to the provisions of this document.

13.1.3 Serviceability and accidental limit states

Serviceability limit states and accidental limit states are discussed in 5.3.

13.1.4 Fatigue limit states

For jack-up operations with a duration less than the classification society special survey period a fatigue analysis is not required, provided that structural integrity is maintained through an appropriate programme of inspection. For long-term applications, fatigue shall be considered in accordance with Clause 11.

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NOTE The special survey period is normally 5 years.

13.2 General formulation

The assessment shall generally follow a partial safety factor format. The partial action factors shall be applied to actions as defined in other clauses and not the action effects.

NOTE Normally both partial action and partial resistance factors are greater than unity: actions are multiplied by partial action factors and resistances are divided by partial resistance factors.

The utilization of each limit state and assessment situation check shall satisfy the following general form:

0.1)()(n Utilisatio ≤= ifU α

where:

U is the utilization to one significant decimal place (see also definition of utilization 3.107)

i

dEi R

FA⎟⎟⎠

⎞⎜⎜⎝

⎛=

)( Resistance Factored)(Action Factored todue )(Effect Action α

This is a generalized form of the equation, and the form appropriate for each application can differ slightly. In the linear case the function simplifies to a summation. See specific clauses for the particular form of the equation to be used.

The factored action can be generalised as

fd AF γ=

where;

A = Assessment load

γf = The specific load factor for each type of load

Factored actions shall be determined in accordance with 8.8.

Action effects are determined by following the calculation methodology set out in 10 of this standard. The Factored Resistance can be defined as:

R

RγC

=

where;

C = representative capacity <<<Characteristic needs to be changed to Representative throughout and all lower case >>>

γR = the resistance factor for each representative capacity

The structural and foundation resistance factors are given in normative Annex B.

13.3 Leg strength

13.3.1 Partial factor approach

The methodology for undertaking checks on the strength of members is described in Clause 12. In general the formulation given in 13.2 shall be used to assess the utilization of the leg structure.

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Resistance factors for assessing the strength of members are given in normative Annex B.

NOTE Reference can be made to the applicable clause in the informative where the relevant factors are given along with the calculation methodology. However, to ensure that the most up to date values are used, reference should always be made to he Table in Normative Annex B which contains all the factors to be used in a site-specific analysis,

13.3.2 Working stress approach

<< Section to be included when completed in SNAME>>

13.4 Spudcan capacity

The forces on the top and bottom of the spudcan due to factored actions for any of the applicable assessment situations shall be checked against the factored ultimate capacity derived from the manufacturer's specification. Where limited information is available a rational approach shall be used.

<< ERP July 07: we need to devise a way of checking leg-can connection when the rest of the analysis is pinned ? use Vmax and Mlo as a start point >>

NOTE This check addresses issues such as: spudcan overburden (at maximum penetration); spudcan capacity (over the range of predicted penetration); and eccentric spudcan loading (e.g. due to sloping seabed or existing spudcan footprints).

13.5 Holding system strength

The forces applied to the holding system due to factored actions shall be checked against the factored ultimate capacity derived from the manufacturer's specification. Where limited information is available a rational approach shall be used.

13.6 Hull elevation

A hull elevation resulting in at least 1.5 m clearance between the extreme wave crest and the underside of the hull shall be provided. The extreme wave crest elevation is normally determined from the extreme water level in A.6.4.4 and the wave crest in A.6.4.2.3.

In some areas of the world an abnormal wave crest height (see 19901-1), can be greater than the extreme crest height plus 1.5 m. If the abnormal wave is likely to have a significant effect on the global response, the hull elevation shall be increased sufficiently to mitigate the effect. <<< (Maybe next text should be in definition of Abnormal Crest height). Per the NL comment, we need to clarify the text, but we do not want to change the Abnormal to Extreme: we want to keep Extreme to be consistent with the 50 year independent values (or 100 yr JP). There is input from Panel ½ and Rupert that may be of value.>>> Where appropriate metocean databases and reliability models exist, account may be taken of the joint probability of tide, surge and crest height in determining abnormal crest elevation.

The hull elevation is also intended to account for any settlement due to the extreme or abnormal storm event.

<<<The above definition and wording was agreed to by WG 7 in Oslo November 2006 - and modified by ERP July 2007 to account for NL comments. Three optional wordings were considered, but the simple requirement to exceed the “abnormal” crest height was chosen.>>>

NOTE The air gap is defined in ISO 19900 as the clearance between the highest water surface that occurs during the extreme environmental conditions and the lowest exposed part not designed to withstand wave impingement. This differs from the definition historically used by the jack-up industry.

13.7 Leg length reserve

The leg length reserve above the upper guides should reflect the uncertainty in the prediction of leg penetration and account for any settlement. The leg length reserve shall be at least 1.5 m. At locations where there is uncertainty a larger reserve should be available. A larger reserve can also be required due to the strength limitations of the top bay or leg/hull interface considerations << what are the leg/hull intgerface considerations - spell out ! >>.

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13.8 Overturning stability

The formulation given in 13.2 shall be used to assess safety against overturning of the jack-up. The utilization shall be calculated based on:

MOTM = overturning moment due to factored actions Fd

Rd OTM = the factored stabilizing moment based on the representative stabilizing moment Rr,OTM

The overturning moment shall be calculated from factored actions about the overturning axis in the most critical assessment situation. For independent leg jack-ups the overturning axes shall pass through any two or more spudcan reaction points. The reaction points are given in 8.6.2 and further described in A.8.6.2.

The representative stabilizing moment Rr,OTM shall be calculated about the same axis for the same assessment situation as used to calculate the overturning moment and shall account for the following parameters:

⎯ The stabilizing moments due to the fixed action (at the displaced position resulting from the factored actions (e.g. environmental actions)).

⎯ The stabilizing moment due to the most onerous combination of minimum action due to variable load and centre of gravity as specified in 5.4.4.

⎯ The stabilizing moments due to seabed foundation fixity. Any stabilizing moments due to fixity shall be calculated in accordance with Clause 9, taking account of reduction of the moment fixity to comply with the yield surface of the foundation.

Large deflection effects shall be included when computing the overturning utilization. When the overturning moment is calculated from the foundation reactions obtained from a large deflection analysis, the reduction in stabilizing moment due to large deflection effects is implicitly included within the overturning moment. Otherwise the reduction in stabilizing moment from fixed actions and variable load caused by the displacement resulting from the factored actions shall be explicitly considered either as an increase in the overturning moment or as a reduction in the stabilizing moment.

The resistance factor on the representative stabilizing moment shall be taken as:

γR,OTM = the resistance factor on representative stabilizing moment = 1.05 << per NL >>

Note: The overturning check is a traditional benchmark. It serves no other purpose as the foundation checks govern.

13.9 Foundation Integrity

13.9.1 Capacity check

The formulation given in 13.2 shall be used to assess the foundation. The following utilizations shall be checked:

US,vhm = the foundation (vertical) bearing capacity utilization <<<Is this correct since the bearing check is a bearing with all the other stuff in it, as per US comment that combined bearing and sliding be checked.>>>

US,hvm = the foundation sliding resistance utilization

The spudcan reactions due to factored actions shall be checked against the factored vertical capacity and the factored horizontal resistance. The spudcan reactions and foundation capacity / resistance shall be determined in accordance with Clause 9.

NOTE The foundation checks described above encompass the traditional sliding and preload checks. <<<Jan V thinks that “traditional sliding....” is unclear.>>>

13.9.2 Displacement check

If the reactions on any spudcan due to factored actions exceed the factored bearing capacity and/or the factored sliding resistance discussed in 13.9.1 a further assessment may be performed in order to show that any additional

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settlements and/or the associated additional structural action effects are within acceptable limits and the hull can be jacked. (see A.9.3.2.2).

NOTE An allowable settlement can be estimated from the hull inclination limit, if this is specified in the operations manual.

13.10 Temperatures

The 50 year lowest mean daily average air and water temperatures shall be in compliance with the limits given in the operating manual.

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A.12 Structural strength

<<<Note to the CD reader/reviewer: Clause A 12 is very much a work-in-progress and suggested for review at a high level only. The Tubular checks are based on 19902, but converted to a load basis. The prismatic checks come out of SNAME T&R 5-5A. There has had to be some combining of the two in certain places which is not complete. In addition, there are some terminology/symbol changes that Panel 10 knows will have to change. The Clause has not yet been run through a proper “laugh” test to ensure that it is complete and workable. Up to Clause A12.6 the overall checking procedure is in order and complete, although still needing work. A12.6 is in an early stage of development based on the Rose Court Scroll without any significant editorial input. References such as <<<from 8.1.4.5>>> point to the relevantly number Section in SNAME. These have been maintained temporarily to ensure traceability. >>>

Clauses that critically require work in order to undertake any reasonable level of checking. There may be critical formulae missing that are NOT labeled as green, but there are known sources for these formulae/information.

Most of the comments added at City U in September 2007. Not normally critical, but must be resolved in order to have a document that will produce the right answer.

Old and outstnding issues needing resolution. Not normally critical, but must be resolved in order to have a document that will produce the right answer.

<< General issues from PAFA May 2007 input:

May v/s can - appears Paul and Andrea have different views of the "permissive may"

Paul has changed many should's to shall's even though were are in an Informative Annex >>

A.12.1 Applicability

A.12.1.1 General

This clause applies to steel structures only. Where necessary, the equations included in this clause have been non-dimensionalized using Young's Modulus, E, of 205000 N/mm2 (or 29,070 ksi).

For the purposes of strength assessment, it is necessary to consider the structure as comprised of structural members. Typically each structural member can be represented by a single beam-column (more general) element in an appropriate analytical model of the structure. Examples of members are include braces and chords in truss type legs and box or tubular legs,all of which form a piece of structure for which the properties can readily be calculated.

The cross-section of a prismatic structural member is usually comprised of several structural components. Table A.12.2-1 shows example components for typical jack-up chords comprising split-tubulars, rack plates, side plates and back plates. A component is comprised of only one material. Therefore, where a plate component is reinforced by another piece of plating of a different yield strength (see Figure A.12.2-1) the reinforcing plate should be treated as a separate component.

Tubulars shall be assessed as members.

In this clause, subscripts y and z are used to define the two axes of bending of tubular and prismatic members, however Fy is used to define yield strength in stress units.

A.12.1.2 Truss type legs

No Guidance Offered

A.12.1.3 Other leg types

No Guidance Offered

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A.12.1.4 Fixation system and/or elevating system

No Guidance Offered

A.12.1.5 Spudcan strength including connection to the leg

No Guidance Offered

A.12.1.6 Overview of the assessment procedure

No Guidance Offered

A.12.2 Classification of member cross-sections

<< in all the following we should re-write in terms of E/Fy or sqrt (E/Fy) - using E of 205000 N/mm2 >>

A.12.2.1 Member type

No guidance offered.

A.12.2.2 Material yield strength

The value of yield strength taken from a tensile test should correspond to the 0.2% offset value. Where this is greater than 90% of the ultimate tensile strength (UTS), the yield strength used in this clause should be taken as 90% of UTS. The following variables are used in this clause:

Fy = yield strength in stress units

= minimum of the yield strength and 90% of the UTS

Fyi = yield strength of the ith component of a prismatic member cross-section, in stress units

= minimum of the yield strength and 90% of the UTS of the ith component of the cross-section

Fymin = minimum yield strength of all components in a prismatic member cross-section, in stress units

Fyeff = effective yield strength of a prismatic member cross-section, in stress units, determined from the plastic tensile axial capacity divided by the minimum cross-sectional area.

A.12.2.3 Classification definitions

A.12.2.3.1 Tubular member classification

The requirement for a tubular section for classification as a class 1 section is

D/t ≤ 0.0537 E / Fy (A.12.2-1)

where

D = outside diameter

t = wall thickness

Fy = yield strength in stress units

E = elastic modulus

<<Need to include a check for split tubes>>

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NOTE Compliance with class 1 classification is only relevant when undertaking earthquake, accidental or alternative strength analyses (see 10.7, 10.8 and 10.9). In all other cases the distinction between plastic and compact is irrelevant to the assessment.

A.12.2.3.2 Prismatic member classification

When classifying prismatic components in accordance with Table A.12.2-1 to 3, a distinction is made between internal components and outstand components as follows:

⎯ Internal components: components that are supported by other components along both longitudinal edges, i.e., the edges parallel to the direction of compression stress, and include:

⎯ Flange internal components: internal components parallel to the axis of bending

⎯ Web internal components: internal components perpendicular to the axis of bending

⎯ Outstand Components: These components are supported by other components along one longitudinal edge and at both ends of the member under consideration, with the other longitudinal edge free.

In calculating the ratios given in Table A.12.2-1 to 3, the dimensions to be used are those given in the table. The components are generally of constant thickness; for components that taper in thickness, the average thickness over the width of the component should be adopted.

Members which satisfy the following criterion need not be checked for lateral torsional buckling. Members that do not meet the criterion need to be assessed for lateral torsional buckling. Additional guidance on LTB can be found in AISC 13th Edition and BS 5400 part 3.<<<NEED TO CHECK THAT THIS WORKS FOR MOST REASONABLE SECTIONS/LENGTHS. Clause A.12.6.2.6 Lateral torsional buckling strength check will need to be deleted if this simple check works. Paul believes need additional check for non-slender members which can be copy/pasted out of original work. This was not working very well, hence try the simple limiting approach given here.>>>

[based on Singly symmetric members (lateral torsional buckling) [Table A-F1.1(c)]

b y y minL / r 1.51 E / F≤ << ERP June 07: we think this should use Fyeff or even Fymax >>

<<There is agreement that use of Fymin could lead to unconservative ratio. Paul says OK as long as consistent use of Fy. ERP to sort out the details. PAFA asks if should located under flexural checks>>

where;

Lb = Laterally unbraced length; length between points which are either braced against lateral displacement of the compression flange or braced against twist of the cross section.

ry = Radius of gyration about the minor axis.

Fymin = Minimum yield strength in the section as defined in A.12.2.2.

E = as defined in A.12.2.3.1.

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Axis of

A.12.2.3.3 Reinforced components

Reinforcement of member cross-sections is often of the form shown in Figure A.12.2-1.

Figure A.12.2-1 Definitions for reinforced plate

The reinforcing plate shall be classified as a compression flange internal component or web internal component in accordance with Table A.12.2-1 and 3 depending on the type of in-plane loading. The value of yield stress to be used is that of the reinforcing plate. <<City U 2007: Need to check both between the welds and the combination overall.>>

The composite section shall be classified as a compression flange internal component, a web internal component or a compression flange outstand component in accordance with Tables A.12.2-1 to 3 depending on the type of in-plane loading and support conditions. The value of thickness to be used with width b1 in the formulae in Table A.12.2-1 and A.12.2-3 shall be

t = tcheck (A.12.2-2)

where

tcheck = (t3eff t1)0.25 << Is this robust for any combination of thickness ?? Why tcheck not teff ?Paul to check, City U 2007>>

teff = (12 I / b1)1/3

I = b1(t1 + t2)3/3 - (b1 - b2)t23/3 - A(t1 + t2 - y1)2

y1 = [b1t12 + b2t2(2t1 + t2)]/2A

A = b1t1 + b2t2

The value of yield stress to be used in Table A.12.2-1 to 3 is the larger of the yield stress values for the reinforcing plate or the base plate.

Table A.12.2-1 - Cross-section classification - flange internal limiting components

Limiting width-to-thickness ratios for compressed internal components

Bending Class Type Section in bending Section in compression

Base plate Reinforcing plate

b1

t1

b2 t2

tf b

b

tf

b

tf

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Plastic stress distribution in component and across section (compression positive)

Plastic - Class 1

Rolled or Welded b/tf ≤ 465/√Fy b/tf ≤ 650/√Fy

Compact - Class 2

Rolled or Welded b/tf ≤ 530/√Fy b/tf ≤ 650/√Fy

Elastic stress distribution in component and across section (compression positive)

Semi-Compact - Class 3

Rolled or Welded b/tf ≤ 650/√Fy b/tf ≤ 650/√Fy

Table A.12.2-2 - Cross-section classification - flange outstand components

Limiting width-to-thickness ratios for compression flange outstand components

Flange subject to compression and bending Class Type Flange subject to compression Tip in compression Tip in tension

Plastic stress distribution in component (compression positive)

Plastic - Class 1

Rolled Welded

b/tf ≤ 150/√Fy

b/tf ≤ 135/√Fy

b/tf ≤ 150/(α√Fy)

b/tf ≤ 135/(α√Fy)

b/tf ≤ 150/(α√[αFy])

b/tf ≤ 135/(α√[αFy])

Fy Fy Fy

Fy Fy

tf

b tf

b

b

tf

Axis of bending

b b

b

b

αb

b

αb

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Fy

Fy

Fy

Fy

Fy

Fy

Compact - Class 2

Rolled Welded

b/tf ≤ 166/√Fy

b/tf ≤ 150/√Fy

b/tf ≤ 166/(α√Fy)

b/tf ≤ 150/(α√Fy)

b/tf ≤ 166/(α√[αFy])

b/tf ≤ 150/(α√[αFy]) Maximum

compression at tip Maximum

compression at connected edge

Elastic stress distribution in component (compression positive)

Semi-Compact - Class 3

Rolled Welded

b/tf ≤ 250/√Fy

b/tf ≤ 225/√Fy

b/tf ≤ 380√(kσ/Fy)

b/tf ≤ 340√(kσ/Fy)

ψ = σ2/σ1

kσ=0.57–

0.21ψ+0.07 ψ2 for 1 ≥ ψ ≥ -1

b/tf ≤ 380√(kσ/Fy)

b/tf ≤ 340√(kσ/Fy)

ψ = σ2/σ1

kσ = 0.578/(ψ + 0.34)

for 1 ≥ ψ ≥ 0 kσ =1.7–5ψ + 17.1 ψ2

for 0 > ψ ≥ -1 NOTE In the figures relating to stress distributions, the dimension b is illustrated only in the case of rolled sections. For welded sections, b should be assigned as shown in the diagrams at the top of the table..

Table A.12.2-3 - Cross-section classification - web internal components

Limiting width-to-thickness ratios for web internal components when subject to compression and/or bending

Class Web subject to Bending

Web subject to compression

Web subject to bending and compression

Plastic stress distribution in component (compression positive)

Plastic - Class 1

α = 0.5

d/tw ≤ 1160/√Fy

α = 1.0

d/tw ≤ 465/√Fy

when α > 0.5 d/tw ≤ 2345/[(6.043α-1)√Fy]

when α ≤ 0.5

d/tw ≤ 580/(α√Fy)

tw

Axis of bending

d d

tw

d

tw

b

b

σ1

σ2

b

σ1

σ2

αd d d d

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fb

fb

fc fb

ψfb

Compact - Class 2

d/tw ≤ 1400/√Fy

d/tw ≤ 530/√Fy

when α > 0.5 d/tw ≤ 2184/[(5.12α-1)√Fy] when α ≤ 0.5 d/tw ≤ 700/(α√Fy)

Elastic stress distribution in component (compression positive)

Semi-Compact - Class 3

d/tw ≤ 1875/√Fy

d/tw ≤ 650/√Fy

when ψ > -1.0 d/tw ≤

650/[(0.674+0.327ψ)√Fy] when ψ ≤ -1.0 d/tw ≤ 937.5(1-ψ)√(-ψ)/√Fy

A.12.3 Section properties

A.12.3.1 General

Cross-sectional properties appropriate for the strength assessment of prismatic members shall be determined as described in A.12.3.2 to A.12.3.4, and as summarized in A.12.3.5. The properties appropriate for the stiffness assessment of prismatic members shall be based on elastic considerations.

<<Need to add a bit stating that the plastic section properties can be calculated properly, but there are some simplifications given below that may be found appropriate.PAFA suggestion City U>>

Where elastic section properties are determined for class 1 and 2 sections in place of plastic section properties, these should be treated as for class 3 sections. << THIS MAY NEED CHANGING when we fix-up Class 3 to be capable of more than outer fibre yield >>

Cross-sectional properties are normally required in respect of both major and minor axes of a prismatic section.

Cross-sectional properties for tubular members are specified in A.12.5.

NOTE In certain situations, the stiffness properties used in determining structural deflections and natural periods can differ from those used to assess member strength. For example, leg chord properties may include approximately 10% of the rack area when determining the leg stiffness for natural period calculation, however, this additional material should not be included when calculating the section properties for strength assessment.

A.12.3.2 Plastic and compact sections

A.12.3.2.1 Axial properties

For class 1 plastic and class 2 compact sections, section properties may be determined assuming fully plastic behaviour can occur. The properties required for a strength assessment should be determined taking into account the physical distribution of components comprising the cross-section and their yield strengths. For simplicity, the following approximations may be used to determine the relevant properties.

For axial tension and compression, the fully plastic cross-sectional area for use in a strength assessment, Ap may be found as:

<< Define Ap Fyeff here >>

d d d

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Ap = (Σ Fyi Ai)/ Fymin << It would be better to use the true area and fudged Fy PAFA: – not for consistency with requirements for other sections Classes ERP June 07: Due to the fact that we already have weight, buoyancy and stiffness areas, we don't want any more - to be reverted to adjusted Fy >> (A.12.3-1)

where

Fyi = yield strength of the ith component comprising the structural member, as defined in A.12.2.2.

Ai = cross-sectional area of the ith component comprising the structural member.

Fymin = yield strength to be used in the calculation, as defined in A.12.2.2

NOTE The centroid of the plastic section (or squash centre) of a member comprising components of differing yield strength can be offset from the centroid of the elastic section.

NOTE Ap can be larger than the physical cross section of the member. << delete if we use Fyeff or go direct to Mp etc>>

A.12.3.2.2 Flexural properties

The second moment of area If should be determined using the fully effective cross-section.

The fully plastic section modulus Zp may be found as:

Zp = (Σ Fyi di Ai)/ Fymin<< It would be better to use the true area and fudged Fy OR BETTER STILL, go directly to Mp. PAFA: This is Mp (A.12.3-2)

where

di = distance between the centroid of the component under consideration and plastic neutral axis.

Note The plastic neutral axis will not necessarily coincide with the equal area axis for cross-sections composed of different yield strengths

When using this definition of Zp, the value of yield stress to be used in the calculation of plastic moment strengths shall be Fymin as defined in A.12.2.2. << to be deleted if we go direct to Mp >>

A.12.3.3 Semi-compact sections

For class 3 semi-compact sections the section properties, Af, the cross-sectional area, If, the second moment of area and Sf, the elastic section modulus should be based on elastic properties assuming a fully effective cross-section.

Af = Σ Ai (A.12.3-3)

The properties If and Sf, shall be determined assuming a fully effective cross-section for bending about both major and minor axes. When considering a cross-section comprised of components having different yield strengths, the section moduli used in the calculations should encompass all critical points on the cross-section.. << what are the full definitions for these variables >>

<<Need to get the knocked down plastic moment capacity in here (the stuff out of SNAME and reviewed by Paul prior to City U 2007). But still need Af (from above) as it is used in other places of the Clause. Also need a knockdown factor on the axial capacity. Paul suggestion would be to calculate an effective Fy at the Class 3/Class 4 transition that is backed out from the section modulus that results in the first compressive component yield due to bending of the section about the resultant bending axis. This is then used for the class 3/class 4 transition Fy for axial. There is then a linear interpolation back up to the full plastic at the class 2/class 3 transition. Could this be unconservative for a section that has high axial and very low bending in the opposite direction? Is Fymin the way to go? Also, most J-U section should be at the upper end of Class 3, so what is used at the bottom end may not be that relevant. jjs “I think we might need a diagram!!” >>

NOTE Critical stress locations are typically those at the edges of components and are a function of the member forces, the yield strength of the component and its position within the cross-section of the member.

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A.12.3.4 Slender sections

A.12.3.4.1 General

Cross-section properties for class 4 slender sections shall be determined using elastic principles. In tension, fully effective sections shall be assumed, i.e., Af and Sf. In compression, the sectional properties shall be based on effective sections as described in this clause. The bases for the determination of effective cross-sections are the limiting effective width to thickness ratios specified for semi-compact sections in Table A.12.2-1 to 3.

When analysing structures that contain class 4 sections care needs to be taken in determining the load distributions. Generally it is recommended that the structural analysis is performed using full elastic section properties and the reduced section properties only used for the member capacity checks. However, care needs to be taken when use of the reduced sections would result in significantly different load distributions, in which case iterative analyses may be required.

Effective sections shall be based on actual plating thicknesses combined with plating effective widths. The effective widths of compression flange internal or outstand components shall be determined in accordance with the formulae presented in Table A.12.3-1 (a) or (b) respectively. The effective widths of web internal components subject to compression and/or bending shall be determined as shown in Table A.12.3-1 (c) for which the following definitions apply:

ψ = compression to bending stress ratio

σ1 = compressive stress if σ2 tensile or larger compressive stress if σ2 compressive, compression positive

σ2 = tensile stress if σ2 tensile or smaller compressive stress if σ2 compressive, tension negative

k = buckling coefficient

ρ = reduction coefficient

λp = plate slenderness parameter

λplim = limiting plate slenderness ratio

λpo = plate slenderness ratio coefficient

When determining effective widths for web internal components, the stress ratio ψ used in Table A.12.3-1 may be based on compression flange internal and outstand component effective widths and web gross section properties.

The use of plating effective widths will generally lead to a shift in the neutral axis compared with that found using gross sectional properties. This shift should be taken into account when determining effective widths. The structural analysis should be performed using effective section properties. If the analysis is performed using gross section properties, the additional moment caused by the shift in the neutral axis should be found as the product of the axial force acting on the member and the shift in the neutral axis. This moment shall be treated as additional to other moments acting on the effective section unless more onerous conditions arise if it is omitted.

A.12.3.4.2 Effective areas for compressive loading conditions

The effective area Aeff of a compressed component shall be found as the product of its thickness and its effective width. The total effective area:

Ae = Σ Aeff (A.12.3-4)

A.12.3.4.3 Effective moduli for flexural loading conditions

For web or flange internal components subject to combinations of flexural and compression loading, effective widths shall be determined from Table A.12.3-1 (c). For web or flange outstand components subject to combinations of flexural and compression loading, effective widths shall be determined from Table A.12.3-1 (b).

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The effective second moment of area Ie shall be found by calculating the properties of the section based on fully effective areas for components subject to tension, effective areas as defined in A.12.3.4.2 for components subject to compression, and effective areas as defined immediately above for components subject to combinations of compression and flexure.

Application of this procedure to determine effective second moments of area when applied to cross-sections with slender components, especially when the section is not symmetric with respect to a particular axis, will lead to two values of Ie about such an axis, depending upon the sign of the bending moment. The smaller value of Ie shall be used throughout the analysis.

When considering a cross-section comprised of components having different yield strengths, the section moduli Se used in the calculations should encompass all critical points on the cross-section.

NOTE Critical stress locations are typically those at the edges of components and are a function of the member forces, the yield strength of the component and its position within the cross-section of the member.

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Table A.12.3-1 Section properties - Effective widths for Slender Sections

(a) Compression flange internal components

tf

325 tf /√Fy 325 tf /√Fy

tf

325 tf /√Fy

tf

tf

tf

tf 225/√Fy

tf 225/√Fy

tf Axis of bending

tf 340√(kσ/Fy) tf 340√(kσ/Fy)

tf 225/√Fy

(b) Outstand components under compression and/or bending

See Table A.12.2-2 for definition of kσ

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A.12.3.5 Cross section properties for assessment

A.12.3.5.1 Tension

In tension, the cross-sectional area to be used in assessment shall be At where

At = Ap for class 1 plastic or class 2 compact sections (see equation A.12.3-1)

= Af for class 3 semi-compact as defined in equation A.12.3-3.

= Af as defined in equation A.12.3-3 for class 4 slender sections in tension across the whole of the cross-section (including bending); otherwise use Ae for class 4 sections as defined by equation A.12.3-4.

Where the cross-section contains cut-outs, pin-holes, etc., At should be determined at the location of the minimum cross-section.

A.12.3.5.2 Compression

In compression, the cross-sectional area to be used in assessment shall be Ac where

(c) Web internal components under bending and/or compression

Ineffective area - to be ignored when calculating effective section properties

σ1

σ2

d

de1

de2

1 ≥ ψ ≥ 0

deff = ρ d

de1 = 2deff/(5 - ψ)

de2 = deff - de1

σ1

σ2

d

de1

de2

ψ ≤ 0

deff = ρ dc = ρ d/(1 - ψ)

de1 = 0.4 deff

de2 = 0.6 deff

dc

dt

tw

Axis of bending

d d

tw

d

tw

ψ = σ2/σ1

ρ = 1 if λp ≤ λplim

= (λp - λpo)/ λp2 if λp > λplim

λp = (d/tw) √(Fy/k)/435.4

λplim = 0.747 (1.09 - 0.09 ψ)

λpo = λplim - λplim2

k = 8.2/(1.05 + ψ) if 1 ≥ ψ > 0

= 7.81 - 6.29 ψ + 9.78 ψ2 if 0 ≥ ψ > -1

= 5.98 (1 - ψ)2 if -1 ≥ ψ > -2

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Ac = Ap for class 1 plastic or class 2 compact sections (see equation A.12.3-1)

= Af for class 3 semi-compact sections as defined in equation A.12.3-3

= Ae for class 4 slender sections as defined in equation A.12.3-4..

A.12.3.5.3 Flexure

In flexure, the second moment of area with respect to major (y) and minor (z) axis bending to be used in assessment should be determined from:

Iy, Iz = If for class 1 plastic and class 2 compact sections as defined in A.12.3.2.2

= If for class 3 semi-compact sections as defined in A.12.3.3

= Ie for class 4 slender sections as described in A.12.3.4.3 accounting for both the chosen axis and the direction of bending

<< WE DON'T USE THESE ????

The section moduli for the two bending axes should be determined from:

Sy, Sz = Zp for Class 1 Plastic or Class 2 Compact sections (see equation A.12.3-2)

= Sf for Class 3 Semi-compact sections as defined in A.12.3.3 for each critical stress location

= Se for Class 4 Slender sections as defined in A.12.3.4.3 for each critical stress location, accounting for both the chosen axis and the direction of bending.

>>>

The radius of gyration to be used for lateral-torsional buckling considerations rz should be determined from

rz = (Iz/Ac)0.5 for sections in classes 1 to 3 (A.12.3-5)

= (Ie/Ae)0.5 for sections in class 4

A.12.4 Member moment amplification and effective lengths

There are two contributors to moment amplification, member Euler amplification (p-δ) and the eccentricity between the elastic and plastic centroids. The plastic centroid or 'centre of squash' is defined as the location at which the axial load produces no moment on the fully plastic section. For chords without material or geometric symmetry (e.g. triangular or tubular with offset rack) the centre of squash is offset from the elastic centroid when the section includes materials of differing yield strengths. Before a section is checked it is necessary to correct the moments by the axial load times the offset distance between the elastic centroid (used in the structural analysis) and the 'centre of squash' as given in Eq. A.12.4-1.a.

The non-amplified and amplified moments, Mue and Mua respectively should be calculated for each axis of bending from:

Mue = Mu + ePu (A.12.4-1.a)

Mua = B Mue (A.12.4-1.b)

where;

Mu = the moment in a member determined in an analysis which includes global P-Δ/hull-sway effects.

Pu = the axial force, Puc or Put as applicable, in the member determined in an analysis which includes global P-Δ/hull-sway effects.

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e = the eccentricity between the elastic and plastic neutral axes taken orthogonal to the axis of bending under consideration, taking due account of sign. The term Pue may be omitted from equation A.12.4-1.a when the eccentricity is included in the member modelling.

Note: e = 0 for tubular members, other cross-sections with geometric and material symmetry or when an elastic strength check is used for the assessment of a member.

B = member moment amplification factor

B = 1.0 (i) for members in tension or

(ii) for members in compression where the individual member forces are determined from a second order analysis i.e. the equilibrium conditions are formulated on the elastically deformed structure so that local p-δ effects are already included in Mu.

B = )/1( Eu

m

PPC

− for members in compression where the local member forces are determined

from a first-order linear elastic analysis i.e. the equilibrium conditions are formulated on the undeformed structure and therefore Mu does not include the local member p-δ effects

where:

PE = (π2AcE)/(KL/r)2 and is to be calculated for the plane of bending. Ac is defined in A.12.3.5.2 and r is the radius of gyration for the plane of flexural buckling.

<<Define L in both y and z>>

K and Cm are given in Table A.12.4-1

NOTE For truss legs, global P-Δ effects need to be included in the determination of the member forces.

When the analysis of a jack-up with single-column tubular or box section legs has been undertaken accounting for the member moment amplification effects of global P-Δ/hull-sway, B may be taken as 1.0 as local p-δ and global P-Δ are the same. For these jack-ups, local strength due to guide reactions should be assessed in conjunction with the member forces.

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24 Error! Reference source not found.

Table A.12.4-1 - Effective length and moment reduction factors

Structural element K Cm(1)

Tubular or Box complete legs

Chords with lateral loading

Chords without lateral loading

Braces

Primary diagonals and horizontals

K-braces(3)

X-brace(3)

Longer segment length

Full length(4)

Secondary horizontals

2(2)

1.0

1.0

0.7

0.7

0.8

0.7

0.7

(a)

(c)

(b)

(b) or (c)

(c)

(c)

(c)

(c)

1. The value of Cm can be determined from rational analysis. In lieu of such analysis, the following values may be used:

(a) For members whose ends are restrained against sidesway Cm = 0.85

For members whose ends are unrestrained against sidesway Cm = 1.0

(b) for members with no transverse loading, ignoring self-weight:

Cm = 0.6 - 0.4 M1 / M2

where M1/M2 is the ratio of the smaller to the larger non-amplified end moments of the segment of the member in the plane of bending under consideration. M1/M2 is positive for the segment subject to reverse curvature and negative when subject to single curvature.

M1 = Mu + ePuc at end 1; similarly for M2

(c) for members with transverse loading, ignoring self-weight:

Cm = 1.0 - 0.4 Pu / PE << PAFA: [latest AISC recommendation] >>

PE = PEy or PEz as appropriate for the axis of bending under consideration.

2. Alternatively use effective length alignment chart <<to be taken from 19902 A.13.5>>.

3. For either in-plane or out-of-plane effective lengths, at least one pair of members framing into a K- or X-joint must be in tension if the joint is not braced out-of-plane.

4. For X-braces, when all members are in compression and the joint is not braced out-of-plane, use charts <<to be taken from 19902 A.13.5>>

<< We need at least the sym X-brace figure and the alignment chart from 19902 >>

A.12.5 Strength of tubular members

A.12.5.1 Applicability

The strength of unstiffened tubular members that satisfy the following condition should be assessed in accordance with this clause.

Any tubular with D/t < 0.244 E /Fy (A.12.5-1)

Tubulars that do not satisfy these conditions should be assessed using alternative methods that result in levels of reliability comparable to those implicit in this standard.

Where a tubular includes cross-sections with cut-outs, pin-holes, etc., it shall be treated as a prismatic member unless equipped with doubler plates that at least replace all the lost area all around the hole in which case tubular formulations may be used. The major concern is that the D/t ratios are unconservative for tubulars with reductions

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Error! Reference source not found. 25

in their cross-section. << ERP June 07: to be further considered as this could be too onerous if the missing material is replaced with stiffening. May need to do literature search on tubes with holes to find effect on allowable d/t ratios.>>

The requirements are considered applicable for steels with yield strength of up to 700 N/mm2. The yield strength used in this clause should be as specified in A.12.2.2.

NOTE The requirements for tubular members are based on ISO 19902 Clause 13. However, for use in this document, the 19902 formulations have been converted to a force base rather than a stress base.

The provisions given in this clause ignore the effect of hydrostatic pressure. The condition under which hydrostatic pressure may be ignored for a specific member is given by:

(D/t)m = 0.00123 d2 - 0.417 d + 74.5 (12.5-2)

where

d = depth below the water surface, accounting for any leg penetration, applicable to the tubular in question

(D/t)m = maximum D/t ratio possible given d.

For convenience, some typical (D/t)m values are listed in Table A.12.5-1.

Table A.12.5-1 Maximum D/t ratios for given depth

Water depth d

(m)

Maximum tubular

D/t

39.3 60.0

50 56.9

75 49.7

100 45.2

125 41.9

150 39.5

If (D/t) exceeds the limiting value (D/t)m for the depth of the tubular, refer to ISO 19902.

A.12.5.2 Tension, compression and bending strength of tubular members

A.12.5.2.1 Axial tensile strength check

Tubular members subjected to axial tensile forces, Put, shall be assessed to satisfy:

Put ≤ A Fy/γt (A.12.5-3)

where

Fy = specified minimum yield stress

A = total cross-sectional area

γt = partial resistance factor for axial tension, 1.05

A.12.5.2.2 Axial compressive strength check

Tubular members subjected to axial compressive forces, Puc, shall be assessed to satisfy:

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26 Error! Reference source not found.

Puc ≤ Pa/γc (12.5-4)

where

Pa = representative compressive strength determined in A.12.5.2.3

γc = partial resistance factor for axial compressive strength, 1.15

A.12.5.2.3 Column buckling strength

The representative axial compressive strength of tubular members, Pa, should be determined from:

Pa = [1.0 - 0.278λ2] Pyc for λ ≤ 1.34 (A.12.5-5a)

= 0.9 Pyc/λ2 for λ > 1.34 (A.12.5-5b)

λ = [Pyc/PE]0.5 (A.12.5-6)

where

Pyc = representative local buckling strength (see A.12.5.2.4)

λ = column slenderness parameter

PE = smaller of the Euler buckling strengths about the y or z direction

= π2 E I/ (KL)2

E = as defined in A.12.1.1

K = effective length factor in y or z direction, see A.12.4

L = unbraced length in y or z direction measured between centrelines

I = Second moment of area of the tubular.

A.12.5.2.4 Local buckling strength << consider moving above column buckling so Pyc is already defined >>

The representative local buckling strength, Pyc, should be determined from:

Pyc = A Fy for A Fy / Pxe ≤ 0.170 (A.12.5-7a)

= [1.047 – 0.274 A Fy/Pxe] A Fy for 0.170 < A Fy / Pxe ≤ 1.911 (A.12.5-7b)

= Pxe for 1.911 < A Fy / Pxe (A.12.5-7c)

where

Fy = specified minimum yield stress

A = total cross-sectional area as defined in A.12.5.2

Pxe = representative elastic local buckling strength

= 2 Cx E A (t / D) (A.12.5-8)

Cx = critical elastic buckling coefficient

The theoretical value of Cx for an ideal tubular is 0.6. However, a reduced value of Cx = 0.3 is recommended for use in Equation 12.5-8 to account for the effect of initial geometric

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Error! Reference source not found. 27

imperfections. A reduced value of Cx = 0.3 is also implicit in the limits for Pc/Pxe given in Equations 12.5-7. << ERP June 07: Is 0.3 too harsh for jack-up tubular dimensions?? Paul says “no” @ City U 2007>>

A.12.5.2.5 Bending strength check

Tubular members subjected to bending moments, Mb, shall be assessed to satisfy:

Mu ≤ Mb/γb

Where

Mu = Muy or Muz the bending moment about member y- and z-axes respectively due to factored actions

Mb = representative bending moment strength, determined from:

Mb = Mp for (Fy D)/(E t) ≤ 0.0517 (A.12.5-9a)

= [1.13 – 2.85 (Fy D)/(E t)] Mp for 0.0517 < (Fy D)/(E t) ≤ 0.1034 (A.12.5-9b)

= [0.94 – 0.76 (Fy D)/(E t)] Mp for 0.1034 < (Fy D)/(E t) ≤ 120 (Fy / E) (A.12.5-9c)

Mp = plastic moment strength

= Fy (1/6) [D3 – (D - 2t)3]

γb = partial resistance factor for bending, 1.05

A.12.5.3 Tubular member combined strength checks

A.12.5.3.1 Axial tension and bending strength check

Tubular members subjected to combined axial tension and bending forces should satisfy the following condition at all cross sections along their length:

γt Put / (A Fy) + γb (Muy2 + Muz

2)0.5 / Mb ≤ 1.0 (A.12.5-10)

where

Put = axial tensile force due to factored actions

A = total cross-sectional area

Fy = specified minimum yield stress

Muy, Muz= bending moments about member y- and z-axes respectively due to factored actions

Mb = representative moment strength, as defined in equations A.12.5-9

γt = partial resistance factor for axial tension, 1.05

γb = partial resistance factor for bending, 1.05

A.12.5.3.2 Axial compression and bending strength check

Tubular members subjected to combined axial compression and bending forces should satisfy the following conditions at all cross sections along their length:

beam-column check:

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28 Error! Reference source not found.

(γcPuc/Pa) + (γb/Mb) (Muay2 +Muaz)

2)0.5 ≤ 1.0 (A.12.5-11)

and local strength check:

(γc Puc/Pyc) + (γb/Mb) (Muy2 + Muz

2)0.5 ≤ 1.0 (A.12.5-12)

where

Puc = axial compressive force due to factored actions

Pyc = the representative local buckling strength in A.12.5.2.4,

Pa = as defined in A.12.5.2.3

Muy = non-amplified bending moment about member y-axis due to factored actions from A.12.4

Muz = non-amplified bending moment about member z-axis due to factored actions from A.12.4

Muay = amplified bending moment about member y-axis due to factored actions from A.12.4

Muaz = amplified bending moment about member z-axis due to factored actions from A.12.4

Mb = representative bending strength, as defined in equations A.12.5-9

γb = partial resistance factor for bending, 1.05

γc = partial resistance factor for axial compressive strength, 1.15

A.12.5.3.3 Beam shear strength check

Tubular members subjected to beam shear forces should satisfy the following:

V ≤ Pv/γv (A.12.5-13)

where

V = beam shear due to factored actions

Pv = representative shear strength

= A Fy/(2√3)

A = total cross-sectional area

γv = partial resistance factor for beam shear strength, 1.05

A.12.5.3.4 Torsional shear strength check

Tubular members subjected to torsional shear forces should satisfy the following:

T ≤ Tv/γv (A.12.5-14)

where

T = torsional moment due to factored actions

Tv = representative torsional strength

= 2 Ip Fy/(D √3)

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Error! Reference source not found. 29

Ip = polar moment of inertia

= (π / 32) [D4 – (D - 2t)4]

A.12.6 Strength of prismatic members

A.12.6.1 General

The structural strength provisions for rolled and welded prismatic members were developed from the AISC 'Load and Resistance Factor Design (LRFD) Specification for Structural Steel Buildings' [14]. The AISC LRFD specification was interpreted and, in some cases, modified for use in the assessment of mobile jack-up structures. Interpretation of the specification was necessary to enable a straight-forward method to be presented for the assessment of beam-columns with components of varying yield strength and/or with cross sections having only a single axis of symmetry. Development of the specification was necessary to provide:

a) A method to deal with member cross-sections comprising components constructed of steels with different yield strengths.

b) A method for the assessment of beam-columns under biaxial bending to overcome a conservatism which has been identified in the standard AISC LRFD interaction equations.

The yield strength used in this subclause should be as specified in A.12.2.2.

The resistance factors used in the 13th Editions of the AISC LRFD specification have been adopted. << add comments about LRFD and 13th being the same (we hope) >>

<<< Maybe insert version of Figure 8.1 here???>>>

In the following sub-sections, y and z are used to define the major and minor axes of a prismatic member.

A.12.6.2 Prismatic members subjected to tension, compression, bending or shear

A.12.6.2.1 General

Prismatic members subjected to axial tension, axial compression, bending or shear should satisfy the applicable strength and stability requirements specified in A.12.6.

A.12.6.2.2 Axial tensile strength check

Prismatic members subjected to axial tensile forces, Put, shall be assessed to satisfy:

Put ≤ FyminAt/γt

where;

At = area of section from A.12.3.5.1

Fymin = minimum yield stress of the cross section as defined in A.12.2.2.

γt = partial resistance factor for axial tension, 1.05

A.12.6.2.3 Axial compressive strength check

Prismatic members subjected to axial compressive forces, Puc, not subject to flexural buckling should be assessed to satisfy: <<JJS May 2007: Either need an L/r limit for this part, or get a smooth transition between 12.6.2.3 and 12.6.2.4. To be resolved by ERP later. See also discussion in A.12.3.3>>:

Puc ≤ FyminAc/γc

where;

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30 Error! Reference source not found.

Ac = area of section from A.12.3.5.2

Fymin = minimum yield stress of the cross section as defined in A.12.2.2.

γc = partial resistance factor for axial compressive strength, 1.15 with the 1P curve/high strength steels and 1.10 with the 2P curve/lower strength steels <<Need to insert the actual yield values>>

A.12.6.2.4 Column buckling strength

The representative compression strength of all member classifications subject to flexural buckling should be determined from the following equations:

<< ? use Ac from 12.3.5.1earlier >><< 13th edition presents ?same eqns in revised format >>

Pn = AcFcr [Eq. E2.1]

The following may be used for all grades of steel <<DRL will check! Also consider updating to 13th Ed the formulae below>>

Fcr = (0.658λc2) Fymin For λc ≤ 1.5 [Eq. E2.2]

Fcr = 0877

2

.λc

⎧⎨⎩

⎫⎬⎭

Fymin For λc > 1.5 [Eq. E2.3]

The following may only be used for high strength steels (Fy>450 ??):

Fcr = (0.7625λc3.22

) Fymin for λc ≤ 1.2 (A.12.7-1a)

= (0.8608/λc1.854) Fymin for λc > 1.2 (A.12.7-1b)

<< Maybe capture some of the following text as an explanation to start this sub-clause:The requirements in Ref. A.5 are supposedly based on an approximation to the SSRC Curve 2P. In fact, the Ref. A.5 column curve, although a reasonable approximation to Curve 2P for λ < 1.5 and λ > 1.8, actually lies closer to SSRC Curve 1P for λ = 1.3 than it does to Curve 2P. Notwithstanding, Curve 2P and thus the Ref. A.5 curve relate primarily to traditional building construction steels typically with yield stresses up to 345 N/mm² together with their corresponding column tolerances of length/1000. However, because of the high strength steel used in the construction of chords in particular, up to 700 N/mm², and possibly tighter column tolerances at around length/1500, it is likely that a higher SSRC Column Curve is appropriate for the assessment of chord members. The relevance of yield stress is that the effects of welding and rolling residual stresses reduce as the yield stress of the material increases.

SSRC Column Curve 1P was investigated as a possible candidate curve. A good approximation to this, within 0.8% of the SSRC expression from λ =0.0 to 2.0, is as shown above.

Preliminary attempts have been made to see if it is possible to relatively simply allocate typical chord sections to this 1P curve or to the Ref. A.5 curve. However, this process has not been successful primarily because there are no data (test or numerical) for chord or other high strength steel sections on which to base such allocation. Nevertheless, should a jack-up leg fabricator be able to demonstrate that a chord section consistently achieves the strengths consistent with those of SSRC Curve 1P, then (A.12.7-1) may be used in the assessment of such chords. >>

where;

Ac = area of section (excluding rack teeth of chords) from A.12.3.5.2

λc = 2

1

min

⎭⎬⎫

⎩⎨⎧

EF

rKL y

π for max. KL/r from all directions [Eq. E2.4]

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Error! Reference source not found. 31

K = effective length factor in y or z direction, see A.12.4

L = unbraced length of member in y or z direction <<This should be in 12.4 ? Also need to ensure that L does not have different meanings in different places (may be different for tubes (centre to centre, not face to face))>>:

- face to face for braces

- braced point to braced point for chords

- longer segment length of X-braces (one pair must be in tension, if not braced out of plane) See also 19902 13.5

r = radius of gyration in y or z direction, based on gross area of section.

E = Young's modulus.

Fymin = minimum yield strength of the cross section as defined in A.12.2.2, unless rational analysis shows that a higher value may be used.

A.12.6.2.5 Bending strength

A.12.6.2.5.1 General

The classification of member cross sections in A.12.2 is used to identify the potential for local buckling. The slender section properties determined in A.12.3.4 account for the local buckling of class 4 cross sections.

The bending strength of typical closed section jack-up chord members used in truss legs will not normally be limited by lateral torsional buckling. However this can be checked using the approach presented in A.12.6.2.5.4.

A.12.6.2.5.2 Class 1 plastic and class 2 compact sections

The representative bending strength, Mb, is given by the plastic bending moment of the whole section:

Mb = ZpFymin

where

Zp = fully plastic section modulus determined from equation A.12.3-2

Fymin = minimum yield strength of the cross section as defined in A.12.2.2

NOTE Hybrid sections built up from components of different yield strengths are addressed by the methodology described in A.12.3.2.

A.12.6.2.5.3 Class 3 semi-compact sections << ERP JUNE 07 : PAFA needs to check this still works Done for bending!!!!>>

The representative bending strength, Mb, is given by:

an interpolation between the plastic bending moment and the limiting buckling moment:

Mb = Mp - (Mp- MR)hpr

p

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

λλλλ

[Eq. A-F1.3]

where;

Mp = section plastic moment as calculated for a class 1 or 2 cross section.

h = subscript referring to the component which produces the smallest value of Mn.

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32 Error! Reference source not found.

λ = b/t or 2R/t as applicable for component h.

λp is determined for component h from: <<λp and λr need to be inserted from the stuff Paul supplied August 2007. The constants are the same as in Table 12.2, but all need to be non-dimentionalized. Note that the Fr is dropped in all cases.>>

i) For rectangular components stiffened along both edges λp = 1.11 / )/( yiFE << 13th uses 1.12 >>

ii) For rectangular components stiffened along one edge λp = 0.38 / )/( yiFE << 13th uses 0.38 >>

iii) For components derived from circular tubes <<Paul to check this>> λp = 0.07 E /Fyi

λr is determined for component h from: << AISC 13th may have dropped Fr >>

i) For rectangular components stiffened along both edges λr = 1.40 / )/( ryi FFE − << 13th uses 1.40 & no Fr >>

Fr = 114 MN/m2 {16.5 ksi} residual stress ii) For rectangular components stiffened along one edge

λr = 0.62 / )/( rywj FFE − << 13th uses ??1.0?? & no Fr - check how AISC LRFD 1st

Table B5.1 edition maps into 13th (F-5 Table B4.1)- in imperial from 1st the constant was 238 >> Fywj = web component yield stress. Fr = 114 MN/m2 {16.5 ksi} residual stress.

iii) For components derived from circular tubes <<Paul to check this>> λr = 0.31 E /Fyi

MR = Sf Fy < Mp

Sf = full elastic section modulus for the plane of bending under consideration, defined in A.12.3.3

<< ERP June 07: OR SHOULD THE ABOVE USE Se ?? Note to PAFA: go see: www.cnx.org/content/m10756/latest/ >>

A.12.6.2.5.4 Class 4 slender sections

The representative bending strength, Mb, of members including components which do not satisfy the above criteria for compact and noncompact sections or for lateral torsional buckling should be determined in accordance this section.

The representative bending strength of a member is given by the limiting flexural bending moment:

Mb = Se Fy

where;

Se = reduced elastic section modulus for the plane of bending under consideration, defined in A.12.3.4.3

A.12.6.2.6 Lateral torsional buckling strength check <<This clause is incomplete, but hopefully the A12.2.3.2 checks will be compromised, so we will not need to complete>>

<<Before delting any of this, check it is not needed for slender members>> Lateral torsional buckling should be checked for all members, regardless of the classes of section, that do not meet the length requirements of 12.2.3.2 Lb/ry. <<Also need to check the slenderness that allows use of up to yield section modulus. Paul to supply>>

The representative lateral torsional buckling strength for members (Mlb) with length less than Lr, defined below, is given by:

Mlt= Sf Fcr <<This is not correct format>>

where

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Error! Reference source not found. 33

Lr = <<See AISC Section F2.2c to get Lr but that is for certain sections only>>

The representative lateral torsional buckling strength for members (Mlb) with length greater than Lr is given by:

Mlt= Sf Fcr

where

Sf = full elastic section modulus for the plane of bending under consideration, defined in A.12.3.3

Fcr = the lowest value of the critical buckling stress from (where appropriate):

i) Doubly symmetric members (lateral torsional buckling) << Need to ensure that this knock down presented below is correct for classes other than just class 4. Paul suspects it is, but may need care in applying to Class 1.>>

Fcr = 6.895

21

22

211

21

2

⎭⎬⎫

⎩⎨⎧

+λλXXXCb << this metric version needs to be non-dimensionalised

>> [Table A-F1.1(b)]

where << all updated for axis change x=major to y=major >>;

Cb = 1.75 + 1.05(M1/M2) + 0.3(M1/M2)2 ≤ 2.3 where M1 is the smaller and M2 the larger end moment in the unbraced member; M1/M2 is positive when the moments cause reverse curvature. ( 1/MLT in PAFA)

X1 = (π/Sf) )/EGJA( 2 <<? Subscript for S - Check AISC &/or SNAME Rev 1 >>

X2 = (4Cw/Iz)(Sy/GJ)2

E = Modulus of elasticity (200,000 MN/m2 {29,000 ksi}). << PAFA USES (205,000 N/mm2 will be 29,070 ksi >>)

G = Shear modulus of elasticity (77,200 MN/m2 {11,200 ksi}).

J = Torsion constant for section.

A = Cross-sectional area (excluding rack teeth).

Iz = Second moment of area of section about minor axis.

Sy = Elastic section modulus for major axis bending. Cw = Warping constant.

λ = Lb/rz

rz = Radius of gyration about the minor axis

ii) Singly symmetric members (lateral torsional buckling) [Table A-F1.1(c)]

Fcr = bf

b

LSC000,393

{B1 + )BB( 2121 ++ } )( JI z ≤ Fy << this metric version needs to be

non-dimensionalised - >>

where;

B1 = 2.25 2

1

12⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

−⎟⎟⎠

⎞⎜⎜⎝

⎛JI

Lh

II z

bz

c

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34 Error! Reference source not found.

B2 = 25 ⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−

JI

Lh

II c

bz

c

2

1

h = web depth.

Ic = second moment of area of compression flange about the section minor axis

Cb = as for doubly symmetric sections.

A.12.6.2.7 Bending strength check

Tubular members subjected to bending moments, Mb, shall be assessed to satisfy:

Mu ≤ Mb/γb

where

Mu = Muy or Muz the bending moment about member y- and z-axes respectively due to factored actions

Mb = representative bending moment strength, determined from 12.6.2.5

γb = partial resistance factor for bending, 1.05

A.12.6.3 Prismatic member combined strength checks

A.12.6.3.1 General

There are two different assessment approaches to determine the utilizations of structural members:

⎯ Interaction equation approach, which is applicable to all member classifications.

⎯ The plastic interaction surface approach which is applicable to members in class 1 and 2.

Either approach can be used for members in class 1 or 2. The approaches are presented in the following clauses.

A.12.6.3.2 Interaction equation approach

Each structural member within the scope of Section 8.1.2 should satisfy the following conditions when the shear load is less than 60 percent of the shear capacity (Pvy, Pvz in A.12.6.3.4):

local strength check:

1.0MM

MM

PP

1

bz

uez

by

uey

pl

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba

and beam-column check:

If Pu/γaPp > 0.2

1.0MM

MM

98

PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1a]

else

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Error! Reference source not found. 35

1.0MM

MM

2PP

1

bz

uaz

by

uay

p

u ≤⎥⎥

⎢⎢

⎭⎬⎫

⎩⎨⎧

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+ηηη

γγγ bba [Eq. H1-1b]

where;

Pu = applied axial load

Ppl = representative axial strength for the local strength check

= At Fymin for members in tension, as defined A.12.6.2.2

= Ac Fymin for members in compression, as defined A.12.6.2.3

Pp = representative axial strength for the beam column check

= At Fymin for members in tension, as defined A.12.6.2.2

= Pn for members in compression, as defined A.12.6.2.4

Mry = unamplified bending moment about member y-axis due to factored actions from A.12.4

Mrz = unamplified bending moment about member z-axis due to factored actions from A.12.4

Muey = amplified bending moment about member y-axis due to factored actions from A.12.4

Muez = amplified bending moment about member z-axis due to factored actions from A.12.4

Mbx = representative moment strength, as defined in A.12.6.2.5

Mby = representative moment strength, as defined in A.12.6.2.5

γb = partial resistance factor for bending, 1.05

γa = partial resistance factor for axial loading

γc = partial resistance factor for axial compressive strength, 1.15 with the 1P curve/high strength steels and 1.10 with the 2P curve/lower strength steels <<Need to insert the actual yield values>>

= γt for axial tension = 1.1

η = Exponent for biaxial bending, a constant dependent on the member cross section geometry, determined as follows:

i) For solid or hollow rectangular sections η = 5/3

ii) For doubly symmetric open section members, η = 1.0

iii) For all geometries, a conservative value of η = 1.0 may be used.

NOTE 1 For purely circular tubular members, η = 2.0

NOTE 2 The TR presents alternative and conservative interaction equations and curves for generic families of chord cross sections << i.e. THE DYER EQUATIONS + CHORD SECTION DATA >>. The data also includes the offset distance between the elastic centroid (used in the structural analysis) and the 'center of squash', together with other geometric data for the members of each family of chord.

NOTE 3 The TR << 8.1.4.7 & C.8.1.4.7 from SNAME >> also presents an approach to determining the value of η by very tedious manual calculation.

When the shear is greater than 60 percent of the shear capacity the moment capacity should be reduced parabolically to zero when the shear equals the shear capacity.

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36 Error! Reference source not found.

A.12.6.3.3 The interaction surface approach

The interaction surface approach is based on Dyer [19] and was developed from the interaction approach proposed by e.g. Duan & Chen [20]. The approach is based on axial load applied at the 'centre of squash' which is defined as the location at which the axial load produces no moment on the fully plastic section.

The interaction equations are based on ultimate strength. It is therefore necessary to introduce the required resistance factors. This is achieved by defining:

Py = F1Pn / γa

Mpx = F2Mbx / γb

Mpy = F2Mby / γb

where; F1 = 1.0, unless alternative values are justified by analysis. and when the shear load is less than 60 percent of the shear capacity (Pvy, Pvz in A.12.6.3.4)

F2 = 1.0, unless alternative values are justified by analysis. and when the shear load is less than 60 percent of the shear capacity (Pvy, Pvz in A.12.6.3.4)

The user should develop the plastic interaction surface using Py Mpx Mpy and determine the point on the surface nearest to the member forces. A measure of the interaction ratio can then be obtained as the ratio between the vector lengths from the origin at the dead load position to the member forces and from the origin at the dead load position to the nearest point on the surface.

The user should note that the sign of the moment is crucially important for sections without material or geometric symmetry. The sign convention should therefore be observed with care.

Need some figures:

- Member cross-section + bulls-eye plot

- UC determination

When the shear is greater than 60 percent of the shear capacity the moment capacity should be reduced parabolically to zero when the shear equals the shear capacity.

A.12.6.3.4 Beam shear

Prismatic members subjected to beam shear forces should satisfy the following:

Vy ≤ Pvy/γv (A.12.6-x)

Vz ≤ Pvz/γv (A.12.6-x)

where

Vy, Vz = beam shear due to factored actions in the local y and z directions

Pvy, Pvz = representative shear strength in the local y and z directions

= A Fy/√3

A = effective shear area in the direction being considered (for guidance on calculating effective shear area see BS 5950-1, or equivalent <<Paul to provide the relevant text>>)

γv = partial resistance factor for beam shear strength, 1.05

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Error! Reference source not found. 37

A.12.6.3.5 Torsional shear

Closed section prismatic members subjected to torsional shear forces should satisfy the following:

T ≤ Tv/γv (A.12.6-x)

where

T = torsional moment due to factored actions

Tv = representative torsional strength

= Ip Fy/(r √3)

Ip = polar moment of inertia

r = maximum distance from centroid to an extreme fibre

Open section prismatic members subjected to torsional shear forces should be checked as appropriate.

A.12.7 Assessment of member joints

No guidance offered.

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38 Error! Reference source not found.

A.13 Acceptance checks

No guidance offered.

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Error! Reference source not found. 39

Annex B (normative)

Summary of Load and Resistance Factors

Symbol Description Factor Ref. Clause

γf,G Action factor to be applied to the Fixed Action G 1.0 8.8.1

γf,Q Action factor to be applied to the actions due to variable load U

1.0 8.8.1

γf,E Action factor to be applied to the Extreme environmental action Ee (used with 50 year independent extreme values)

1.15 8.8.1

γf,E Action factor to be applied to the Extreme Environmental Action Ee (used with 100 year joint probability values)

1.25 8.8.1

γf,D Action factor on the Dynamic Action De in combination with γf,E 1.0 8.8.1

??? Inertial action induced by the ALE ground motions 0.9 A.10.7

γR,Nt Resistance Factor for prismatic axial tensile loads << Load -> strength /strength ? >> 1.1 A.12.6

γR,Nc Resistance Factor for prismatic axial compressive loads 1.1 A.12.6

γR,Nb Resistance Factor for prismatic bending loads 1.1 A.12.6

γR,Tt Resistance Factor for tubular axial tensile loads 1.1 A.12.5

γR,Tc Resistance Factor for tubular axial compressive loads

1.15 << Per PAFA

May 07 update to A.12.5.3.2

>>

A.12.5

γR,Tb Resistance Factor for tubular bending loads << moments ? >> 1.1 A.12.5

γR,H Resistance Factor for vertical holding system between hull and leg 1.1 13.5

γR,OTN Resistance Factor for overturning 1.05 13.8

γR,P Resistance Factor for preload 1.1 A.9.3.6.1

γR,Hfc Partial resistance factor for horizontal foundation capacity.

= 1.25 (effective stress - sand/drained).

= 1.56 (total stress -

clay/undrained).

A.9.3.6.2

γR,VH.

Partial resistance factor for foundation capacity (see Commentary).

= 1.1 - Maximum bearing area not mobilized.

= 1.15 - Penetration sufficient to mobilize maximum bearing area.

A.9.3.6.3

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40 Error! Reference source not found.

NOTE Values given in this table are for normative, however reference should always be made to the specific clause for method of application. The factors are specifically tied to the calculation methodologies given in each reference clause

<<<Value for the compression resistance factor for prismatic members has been changed from 1.18 to 1.1 to be consistent with AISC 13th Edition (DRL to confirm) However, maybe the tubulars should remain at 1.18 depending on 19902, PAFA, API, Peter Marshall, and others.>>>