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STAAD.Pro

V8i (SELECTseries 4)

Release STAAD.Pro 20.07.09.21

Document Revised on 25 January 2013

Verification ManualDAA038960-1/0003

STAAD Verification ProblemsRelease STAAD.Pro 20.07.09.21

Date last updated: 25 January 2013

Quality Assurance ProgramSTAAD.Pro V8i (SELECTseries 4) is a suite of proprietary computer programs of ResearchEngineers, a Bentley Solutions Center. Although every effort has been made to ensure thecorrectness of these programs, neither REI nor Bentley Systems will accept responsibility for anymistake, error or misrepresentation in or as a result of the usage of these programs.

Verification Manual — iii

Table of Contents

STAAD Verification Problems

1 Plane Truss: Support Reactions Due to a Joint Load 1

2 Natural Frequency Calculation 9

3 Deflection and Moments for a Plate-Bending Finite Element 17

4 Support Reactions for a Simple Frame 27

5 Deflection and stress for a simple beam 35

6 Plane Frame with no Sidesway 43

7 Plane Truss Joint Deflection 51

8 Unequal Plane frame with Sidesway 59

9 Multiple level plane frame with horizontal loads 67

10 Forces in a Space Frame 75

11 Temperature load in different materials 85

12 Deflection and stress of a truss system 93

13 Steel Design per 1989 AISC Code 103

14 Concrete Design per ACI 318 Code 119

15 Natural Modes of a Simple Beam 129

16 Natural Frequencies of a Circular Plate 139

17 Natural Frequencies of a Rectangular Plate 151

18 Natural Modes of a Framework 163

19 Response of a Simply Supported Beam to a Shock Spectrum 177

20 Thermal Loading on a Simply Supported Rectangular Plate 189

21 Deflection Calculation Through Pushover Analysis 207

22 Fixed – Fixed Beam Excited by Harmonic Distributed Load 219

iv — STAAD.Pro

Copyright Information

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Verification Manual — v

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1 Plane Truss: Support Reactions Due to a Joint Load

ObjectiveTo find the support reactions due to a joint load in a plane truss.

ReferenceTimoshenko, S., Strength of Materials, Part 1, D. Van Nostrand Co., Inc., 3rd edition, 1956, p.346,problem 3.

ProblemDetermine the horizontal reaction at support 4 of the system.

Figure 1-1: Plane truss

Comparison

Result Type TheorySTAAD

Basic SolverSTAAD

Advanced Solver

R4 8.77 8.77 8.77

Table 1-1: Comparison of Support Reaction, in kips, for verification problem no.1

STAAD InputSTAAD TRUSS VERIFICATION PROBLEM NO. 1** REFERENCE `STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO* PAGE 346 PROBLEM NO. 3. THE ANSWER IS REACTION = 0.877P.

Verification Manual — 1

* THEREFORE IF P=10, REACTION = 8.77*UNITS INCH KIPJOINT COORD1 0. 0. ; 2 150. 100. ; 3 150. 50. ; 4 300. 0.MEMBER INCI1 1 2 ; 2 1 3 ; 3 2 3 ; 4 2 4 ; 5 3 4MEMB PROP1 4 PRIS AX 5.0 ; 2 5 PRIS AX 3.0 ; 3 PRIS AX 2CONSTANTE 30000. ALLPOISSON STEEL ALLSUPPORT ; 1 4 PINNEDLOADING 1JOINT LOAD ; 2 FY -10.PERFORM ANALYSISPRINT REACTIONFINISH

STAAD OutputPAGE NO. 1

***************************************************** ** STAAD.Pro V8i SELECTseries4 ** Version 20.07.09.21 ** Proprietary Program of ** Bentley Systems, Inc. ** Date= JAN 25, 2013 ** Time= 16:58:18 ** ** USER ID: Bentley Systems, Inc. *****************************************************

1. STAAD TRUSS VERIFICATION PROBLEM NO. 1INPUT FILE: VER01.STD

2. *3. * REFERENCE `STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO4. * PAGE 346 PROBLEM NO. 3. THE ANSWER IS REACTION = 0.877P.5. * THEREFORE IF P=10, REACTION = 8.776. *7. UNITS INCH KIP8. JOINT COORD9. 1 0. 0. ; 2 150. 100. ; 3 150. 50. ; 4 300. 0.10. MEMBER INCI11. 1 1 2 ; 2 1 3 ; 3 2 3 ; 4 2 4 ; 5 3 412. MEMB PROP13. 1 4 PRIS AX 5.0 ; 2 5 PRIS AX 3.0 ; 3 PRIS AX 214. CONSTANT15. E 30000. ALL16. POISSON STEEL ALL17. SUPPORT ; 1 4 PINNED18. LOADING 119. JOINT LOAD ; 2 FY -10.20. PERFORM ANALYSIS

P R O B L E M S T A T I S T I C S-----------------------------------

NUMBER OF JOINTS 4 NUMBER OF MEMBERS 5NUMBER OF PLATES 0 NUMBER OF SOLIDS 0NUMBER OF SURFACES 0 NUMBER OF SUPPORTS 2

SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 4

2— STAAD.Pro


VERIFICATION PROBLEM NO. 1 -- PAGE NO. 2*21. PRINT REACTION

REACTION



VERIFICATION PROBLEM NO. 1 -- PAGE NO. 3*SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = TRUSS-----------------

JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z1 1 8.77 5.00 0.00 0.00 0.00 0.004 1 -8.77 5.00 0.00 0.00 0.00 0.00

************** END OF LATEST ANALYSIS RESULT **************22. FINISH

*********** END OF THE STAAD.Pro RUN *************** DATE= JAN 25,2013 TIME= 16:58:31 ****

4 — STAAD.Pro


VERIFICATION PROBLEM NO. 1 -- PAGE NO. 4*

************************************************************* For questions on STAAD.Pro, please contact ** Bentley Systems or Partner offices ** ** Telephone Web / Email ** USA +1 (714) 974-2500 ** UK +44 (0) 808 101 9246 ** SINGAPORE +65 6225-6158 ** FRANCE +33 (0) 1 55238400 ** GERMANY +49 0931 40468 ** INDIA +91 (033) 4006-2021 ** JAPAN +81 (03)5952-6500 http://www.ctc-g.co.jp ** CHINA +86 21 6288 4040 ** THAILAND +66 (0)2645-1018/19 [email protected]** ** Worldwide http://selectservices.bentley.com/en-US/ ** *************************************************************




Information about the key files in the current distributionModification Date CRC Size (Bytes) File Name-----------------------------------------------------------------------------------------12/21/2012 0x880 19054592 SProStaad.exe12/21/2012 0x6c00 08679424 SProStaadStl.exe10/18/2012 0x3940 00373248 AISCSteelDesignNew.dll09/19/2003 0x2fc0 00081970 CMesh.dll11/29/2012 0x15c1 04476998 dbSectionInterface.dll10/18/2012 0x5680 00315392 EC3Annex.dll10/18/2012 0x91c1 00421888 EuroCode3.dll10/18/2012 0xf981 00102400 IS800SteelDesign.dll01/24/2001 0x9b40 00073728 LoadGen.dll06/21/2012 0xcb81 00040960 MathExtensions.dll09/25/2003 0x6340 00704512 MeshEngine.dll11/27/2012 0x9a40 00059392 NORSOKSteelDesign.dll09/22/2003 0xce00 00069632 QuadPlateEngine.dll10/18/2012 0x6381 00079872 SteelCollection.dll10/18/2012 0xf240 00030720 SteelDesign.dll10/18/2012 0xac1 00061440 SteelSections.dll05/28/2008 0x77c1 00106572 SurfMeshEngine.dll10/18/2012 0x1dc0 00017408 UnitConversions.dll10/19/2012 0x280 00197120 WrapperAISCSteelDesignNew.dll10/19/2012 0x57c0 00019968 WrapperSteelDesign.dll10/19/2012 0x2e80 00108032 WrapperSteelSections.dll10/19/2012 0x6480 00008704 WrapperSystem.dll10/19/2012 0xe141 00016384 WrapperUnitConversions.dll08/22/2011 0x9741 00974848 AISCSections.mdb06/13/2006 0xd4c1 01204224 AISCSectionsRCeco.mdb06/05/2007 0xbc00 00331776 aiscsections_all_editions.mdb01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb01/05/2005 0xcac1 03651584 aitctimbersections.mdb05/01/2008 0xb001 00552960 aluminumsections.mdb01/13/2010 0x40 01019904 AustralianColdFormedSections.mdb10/02/2008 0x40 00520192 AustralianSections.mdb10/18/2012 0x4d41 00765952 BrazilSections.mdb05/19/2011 0x1641 00643074 BritishSections.mdb09/12/2006 0xf4c0 00466944 bscoldformedsections.mdb06/28/2005 0x8201 00327680 butlercoldformedsections.mdb02/25/2008 0x2e01 00329728 CanadianSections.mdb09/26/2006 0x5ec1 00280576 CanadianSections_old.mdb05/31/2005 0x9e81 00450560 canadiantimbersections.mdb02/10/2010 0x94c1 00483328 ChineseSections.mdb01/05/2005 0xd6c0 00600064 dutchsections.mdb08/08/2011 0xc880 02699264 EuropeanColdFormedSections.mdb10/07/2010 0xd2c0 03596290 EuropeanSections.mdb01/25/2011 0xc901 00692224 EuropeanSectionsLegacy.mdb01/05/2005 0xd301 00202752 frenchsections.mdb01/05/2005 0x11c1 00233472 germansections.mdb10/06/2010 0x1bc1 00393216 GOSTColdFormedSections.mdb12/08/2009 0x6ac0 00569344 IndianLegacySections.mdb08/13/2012 0x6281 00942082 IndianSections.mdb01/05/2005 0xd540 00180224 iscoldformedsections.mdb02/21/2008 0xc380 00327680 JapaneseColdFormedSections.mdb08/02/2011 0x2b80 00593920 JapaneseSections.mdb07/15/2011 0xce40 00532480 JindalSections.mdb11/09/2005 0x9081 00376832 Kingspancoldformedsections.mdb08/24/2007 0x3ec0 00454656 Koreansections.mdb04/24/2007 0x99c0 00114688 lysaghtcoldformedsections.mdb02/07/2005 0x9a00 00243712 mexicansteeltables.mdb04/19/2007 0x9ec1 00421888 RCecoColdFormedSections.mdb11/20/2009 0xf700 00428034 RussianSections.mdb05/12/2009 0xe001 00237570 SouthAfricanSections.mdb01/06/2005 0x9341 00194560 spanishsections.mdb11/20/2009 0x5100 00368640 STO_ASChMSections.mdb

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12/20/2011 0x2b41 00409600 TATAStructuraSections.mdb04/19/2007 0x7480 00223232 uscoldformedsections.mdb10/31/2007 0x3700 00133120 VenezuelanSections.mdb-----------------------------------------------------------------------------------------



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Notes

2 Natural Frequency Calculation

ObjectiveTo find the period of free vibration for a beam supported on two springs with a point mass.

ReferenceTimoshenko, S., Young, D., and Weaver, W., Vibration Problems in Engineering, John Wiley &Sons, 4th edition, 1974. page 11, problem 1.1-3.

ProblemA simple beam is supported by two spring as shown in the figure. Neglecting the distributed massof the beam, calculate the period of free vibration of the beam subjected to a load of W.

EI = 30,000.0 ksi

A = 7.0 ft

B = 3.0 ft.

W = 1,000 lbfK = 300.0 lb/in.

Figure 2-1: Beam supported on springs

Comparison


Basic SolverSTAAD

Advanced Solver

Period, sec 0.533 0.533 0.533

Table 2-1: Comparison of period, in sec., for verification problem no. 2

STAAD InputSTAAD PLANE VERIFICATION PROBLEM NO 2** REFERENCE 'VIBRATION PROBLEMS IN ENGINEERING' BY* TIMOSHENKO,YOUNG,WEAVER. (4TH EDITION, PAGE 11, PROB 1.1-3)* THE ANSWER IN THE BOOK IS T = 0.533 sec., viz., F = 1.876 CPS*UNIT POUND FEETJOINT COORD ; 1 0. 0. ; 2 7. 0. ; 3 10. 0.MEMB INCI ; 1 1 2 2


UNIT INCHSUPPORT1 3 FIXED BUT MZ KFY 300.MEMB PROP ; 1 2 PRIS AX 1. IZ 1.CONSTANTE 30E6 ALLPOISSON STEEL ALLCUT OFF MODE SHAPE 1LOADING 1 1000 LB LOAD AT JOINT 2JOINT LOAD ; 2 FY -1000.MODAL CALCULATIONPERFORM ANALYSFINISH



1. STAAD PLANE VERIFICATION PROBLEM NO 2INPUT FILE: VER02.STD

2. *3. * REFERENCE 'VIBRATION PROBLEMS IN ENGINEERING' BY4. * TIMOSHENKO,YOUNG,WEAVER. (4TH EDITION, PAGE 11, PROB 1.1-3)5. * THE ANSWER IN THE BOOK IS T = 0.533 SEC., VIZ., F = 1.876 CPS6. *7. UNIT POUND FEET8. JOINT COORD ; 1 0. 0. ; 2 7. 0. ; 3 10. 0.9. MEMB INCI ; 1 1 2 210. UNIT INCH11. SUPPORT12. 1 3 FIXED BUT MZ KFY 300.13. MEMB PROP ; 1 2 PRIS AX 1. IZ 1.14. CONSTANT15. E 30E6 ALL16. POISSON STEEL ALL17. CUT OFF MODE SHAPE 118. LOADING 1 1000 LB LOAD AT JOINT 219. JOINT LOAD ; 2 FY -1000.20. MODAL CALCULATION21. PERFORM ANALYS




10 — STAAD.Pro


VERIFICATION PROBLEM NO 2 -- PAGE NO. 2*

NUMBER OF MODES REQUESTED = 1NUMBER OF EXISTING MASSES IN THE MODEL = 1NUMBER OF MODES THAT WILL BE USED = 1*** EIGENSOLUTION : ADVANCED METHOD ***




CALCULATED FREQUENCIES FOR LOAD CASE 1MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 1.876 0.53317

MODAL WEIGHT (MODAL MASS TIMES g) IN POUN GENERALIZEDMODE X Y Z WEIGHT

1 0.000000E+00 9.999999E+02 0.000000E+00 9.999999E+02PARTICIPATION FACTORS

MASS PARTICIPATION FACTORS IN PERCENT--------------------------------------

MODE X Y Z SUMM-X SUMM-Y SUMM-Z1 0.00100.00 0.00 0.000 100.000 0.000

22. FINISH*********** END OF THE STAAD.Pro RUN *************** DATE= JAN 25,2013 TIME= 16:58:49 ****

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14 — STAAD.Pro


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Notes

3 Deflection and Moments for a Plate-Bending Finite Element

ObjectiveTo find the deflection and moments for plate-bending finite element due to a pressure load.

ReferenceSimple hand calculation by considering the entire structure as a cantilever beam.

ProblemA simple cantilever plate is divided into 12 4-noded finite elements. A uniform pressure load isapplied and the maximum deflection at the tip of the cantilever and the maximum bending at thesupport are calculated.

Plate thickness = 25 mm

Uniform pressure= 5 N/mm2

Plate length = 6 spaces at 50 mm = 300 mm

Plate width = 2 space at 50 mm = 100 mm

Figure 3-1: Finite element mesh of cantilevered plate.

Theoretical SolutionMaximum deflection is equal to WL3/8EI, where:

= = =

⋅

⋅∆ 18.51 mm

( )( )max

5 300 100 (300)

8 210 10

4050(10)

218.75(10)

3

3 100 253

12

9

9

Maximum Moment:


= = = ⋅M 22.5(10) NmmWL

max 2

5(300)(100)(300)

2

6

Comparison

Result TypeHand

Calculation

STAADBasicSolver

STAADAdvancedSolver

δmax (mm) 18.51 18.20 18.20

Mmax (KN·m) 22.50 22.50 22.50

Table 3-1: Comparison of max. defl. and max. moment for verification problemno. 3



1. STAAD SPACE FINITE ELEMENT VERIFICATIONINPUT FILE: ver03.STD

2. *3. * FILE : VER03.STD4. *5. * DEFLECTION OF A CANTILEVER PLATE UNDER UNIFORM PRESSURE.6. * COMPARISON WITH ESTABLISHED FORMULA (WL^3/8EI)7. *8. UNIT KNS MMS9. JOINT COORDINATES10. 1 0 0 0 7 300 0 011. REPEAT 2 0 50 012. *13. ELEMENT INCIDENCE14. 1 1 2 9 8 TO 615. REPEAT 1 6 716. *17. ELEMENT PROP18. 1 TO 12 THICK 25.019. *20. CONSTANT21. E 210.0 ALL22. POISSON STEEL ALL23. *24. SUPPORT25. 1 8 15 FIXED26. *27. UNIT NEWTON28. LOAD 1 5N/SQ.MM. UNIFORM LOAD29. ELEMENT LOAD30. 1 TO 12 PRESSURE 5.031. *

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32. PERFORM ANALYSIS



FINITE ELEMENT VERIFICATION -- PAGE NO. 2*



SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 10833. *34. PRINT DISPLACEMENT LIST 14

DISPLACE LIST 14

20 — STAAD.Pro


FINITE ELEMENT VERIFICATION -- PAGE NO. 3*JOINT DISPLACEMENT (CM RADIANS) STRUCTURE TYPE = SPACE------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN14 1 0.0000 0.0000 1.8159 0.0000 -0.0813 0.0000

************** END OF LATEST ANALYSIS RESULT **************35. *36. UNIT KN METER37. PRINT REACTION

REACTION



FINITE ELEMENT VERIFICATION -- PAGE NO. 4*SUPPORT REACTIONS -UNIT KN METE STRUCTURE TYPE = SPACE-----------------

JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z1 1 0.00 0.00 -18.91 -1.54 5.47 0.008 1 0.00 0.00 -112.19 0.00 11.56 0.0015 1 0.00 0.00 -18.91 1.54 5.47 0.00


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*********** END OF THE STAAD.Pro RUN *************** DATE= JAN 25,2013 TIME= 16:59: 5 ****






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Notes

4 Support Reactions for a Simple Frame

ObjectiveTo find support reactions due to a load at the free end of a cantilever bent plate with anintermediate support.

ReferenceTimoshenko, S., Strength of Materials, Part 1, D. Van Nostrand Co., Inc., 3rd edition, 1956, page 346,problem 2.

ProblemDetermine the reaction of the system as shown in the figure.

P = 1 kip

L = 10 in

Figure 4-1: Cantilever model

Comparison


Basic SolverSTAAD

Advanced Solver

Rx 1.5 1.5 1.5

Table 4-1: Comparison of reaction, in kips, for verification problem no. 4


STAAD InputSTAAD PLANE VERIFICATION PROBLEM NO. 4** REFERENCE 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO* PAGE 346 PROBLEM NO. 2. THE ANSWER IN THE BOOK AFTER* RECALCULATION = 1.5*UNIT INCH KIPJOINT COORD1 0. 0. ; 2 0. 10. ; 3 0. 20. ; 4 10. 20.MEMB INCI1 1 2 3MEMB PROP ; 1 2 3 PRIS AX 10. IZ 100.CONSTANTE 3000. ALLPOISSON CONCRETE ALLSUPPORT1 FIXED ; 2 FIXED BUT FY MZLOADING 1JOINT LOAD ; 4 FY -1.PERFORM ANALYSPRINT REACTIONFINI


***************************************************** ** STAAD.Pro V8i SELECTseries4 ** Version 20.07.09.21 ** Proprietary Program of ** Bentley Systems, Inc. ** Date= JAN 25, 2013 ** Time= 16:59: 9 ** ** USER ID: Bentley Systems, Inc. *****************************************************

1. STAAD PLANE VERIFICATION PROBLEM NO. 4INPUT FILE: VER04.STD

2. *3. * REFERENCE 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO4. * PAGE 346 PROBLEM NO. 2. THE ANSWER IN THE BOOK AFTER5. * RECALCULATION = 1.56. *7. UNIT INCH KIP8. JOINT COORD9. 1 0. 0. ; 2 0. 10. ; 3 0. 20. ; 4 10. 20.10. MEMB INCI11. 1 1 2 312. MEMB PROP ; 1 2 3 PRIS AX 10. IZ 100.13. CONSTANT14. E 3000. ALL15. POISSON CONCRETE ALL16. SUPPORT17. 1 FIXED ; 2 FIXED BUT FY MZ18. LOADING 119. JOINT LOAD ; 4 FY -1.20. PERFORM ANALYS



28— STAAD.Pro


VERIFICATION PROBLEM NO. 4 -- PAGE NO. 2*21. PRINT REACTION

REACTION

30 — STAAD.Pro


VERIFICATION PROBLEM NO. 4 -- PAGE NO. 3*SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = PLANE-----------------

JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z1 1 1.50 1.00 0.00 0.00 0.00 -5.002 1 -1.50 0.00 0.00 0.00 0.00 0.00

************** END OF LATEST ANALYSIS RESULT **************22. FINI






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34 — STAAD.Pro


5 Deflection and stress for a simple beam

ObjectiveTo find deflections and stress at the center of a locomotive axle.

ReferenceTimoshenko, S., Strength of Materials, Part- 1, D. Van Nostrand Co., 3rd edition, 1956. page 97,problems 1, 2.

ProblemDetermine the maximum stress in a locomotive axle (as shown in the figure) as well as thedeflection at the middle of the axle.

Diameter = 10 in.,

P = 26000 lbf

E = 30E6 psi

L1 = 13.5 in., L2 = 59 in.

Figure 5-1: Locomotive axle model

Comparison


Basic SolverSTAAD

Advanced Solver

σ 3575.* 3575. 3575.

δ 0.01040 0.01037 0.01037

Table 5-1: Comparison of stress (σ), psi, and Deflection (δ), in. for verificationmodel no. 5

* The value is recalculated.


STAAD InputSTAAD PLANE VERIFICATION PROBLEM NO. 5** REFERENCE 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO* PAGE 97 PROBLEM NO. 1 AND 2. ANSWERS ARE 3580 FOR MAX. STRESS* AND 0.104 INCH FOR MAX. DEFLECTION.*UNIT INCH POUNDJOINT COORD1 0. 0. ; 2 13.5 0. ; 3 43. 0. ; 4 72.5 0. ; 5 86. 0.MEMB INCI ; 1 1 2 4MEMB PROP ; 1 TO 4 TABLE ST PIPE OD 10. ID 0.CONSTANTE 30E6 ALLPOISSON STEEL ALLSUPPORT ; 2 4 PINNEDLOADING 1JOINT LOAD ; 1 5 FY -26000.PERFORM ANALYSISPRINT MEMBER STRESSESPRINT DISPLACEMENTSFINISH




2. *3. * REFERENCE 'STRENGTH OF MATERIALS' PART-1 BY S. TIMOSHENKO4. * PAGE 97 PROBLEM NO. 1 AND 2. ANSWERS ARE 3580 FOR MAX. STRESS5. * AND 0.104 INCH FOR MAX. DEFLECTION.6. *7. UNIT INCH POUND8. JOINT COORD9. 1 0. 0. ; 2 13.5 0. ; 3 43. 0. ; 4 72.5 0. ; 5 86. 0.10. MEMB INCI ; 1 1 2 411. MEMB PROP ; 1 TO 4 TABLE ST PIPE OD 10. ID 0.12. CONSTANT13. E 30E6 ALL14. POISSON STEEL ALL15. SUPPORT ; 2 4 PINNED16. LOADING 117. JOINT LOAD ; 1 5 FY -26000.18. PERFORM ANALYSIS




36— STAAD.Pro


VERIFICATION PROBLEM NO. 5 -- PAGE NO. 2*19. PRINT MEMBER STRESSES

MEMBER STRESSES




MEMBER STRESSES---------------ALL UNITS ARE POUN/SQ INCHMEMB LD SECT AXIAL BEND-Y BEND-Z COMBINED SHEAR-Y SHEAR-Z

1 1 .0 0.0 0.0 0.0 0.0 441.4 0.01.00 0.0 0.0 3575.3 3575.3 441.4 0.0

2 1 .0 0.0 0.0 3575.3 3575.3 0.0 0.01.00 0.0 0.0 3575.3 3575.3 0.0 0.0

3 1 .0 0.0 0.0 3575.3 3575.3 0.0 0.01.00 0.0 0.0 3575.3 3575.3 0.0 0.0

4 1 .0 0.0 0.0 3575.3 3575.3 441.4 0.01.00 0.0 0.0 0.0 0.0 441.4 0.0

************** END OF LATEST ANALYSIS RESULT **************20. PRINT DISPLACEMENTS

DISPLACE

38— STAAD.Pro


VERIFICATION PROBLEM NO. 5 -- PAGE NO. 4*JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = PLANE------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN1 1 0.00000 -0.01138 0.00000 0.00000 0.00000 0.000862 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000703 1 0.00000 0.01037 0.00000 0.00000 0.00000 0.000004 1 0.00000 0.00000 0.00000 0.00000 0.00000 -0.000705 1 0.00000 -0.01138 0.00000 0.00000 0.00000 -0.00086







40 — STAAD.Pro



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6 Plane Frame with no Sidesway

ObjectiveTo find the maximum moment due to a uniform load on the horizontal member in a 1x1 bay planeframe.

ReferenceMcCormack, J. C., Structural Analysis, Intext Educational Publishers, 3rd edition, 1975, page 383,example 22 - 5.

ProblemDetermine the maximum moment in the frame.

E and I same for all members.

Figure 6-1: 1x1 bay plane frame


Comparison


Basic SolverSTAAD

Advanced Solver

MMax 44.40 44.44 44.44

Table 6-1: Comparison of moment, in kip-ft, for verification problem no. 6

STAAD InputSTAAD PLANE VERIFICATION PROBLEM NO. 6** REFERENCE 'STRUCTURAL ANALYSIS' BY JACK C. McCORMACK,* PAGE 383 EXAMPLE 22-5, PLANE FRAME WITH NO SIDESWAY* ANSWER - MAX BENDING = 44.4 FT-KIP*UNIT FT KIPJOINT COORD1 0. 0. ; 2 0. 20. ; 3 20. 20. ; 4 20. 0.MEMB INCI ; 1 1 2 3MEMB PROP ; 1 2 3 PRIS AX 1. IZ 0.05CONSTANTE 4132E3 ALLPOISSON STEEL ALLSUPPORT ; 1 4 FIXEDLOADING 1 ; MEMB LOAD ; 2 UNI Y -2.0PERFORM ANALPRINT FORCESFINISH




2. *3. * REFERENCE 'STRUCTURAL ANALYSIS' BY JACK C. MCCORMACK,4. * PAGE 383 EXAMPLE 22-5, PLANE FRAME WITH NO SIDESWAY5. * ANSWER - MAX BENDING = 44.4 FT-KIP6. *7. UNIT FT KIP8. JOINT COORD9. 1 0. 0. ; 2 0. 20. ; 3 20. 20. ; 4 20. 0.10. MEMB INCI ; 1 1 2 311. MEMB PROP ; 1 2 3 PRIS AX 1. IZ 0.0512. CONSTANT13. E 4132E3 ALL14. POISSON STEEL ALL15. SUPPORT ; 1 4 FIXED16. LOADING 1 ; MEMB LOAD ; 2 UNI Y -2.017. PERFORM ANAL

44 — STAAD.Pro




SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 618. PRINT FORCES




FORCES

46— STAAD.Pro


VERIFICATION PROBLEM NO. 6 -- PAGE NO. 3*MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KIP FEET (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

1 1 1 20.00 -3.33 0.00 0.00 0.00 -22.212 -20.00 3.33 0.00 0.00 0.00 -44.44

2 1 2 3.33 20.00 0.00 0.00 0.00 44.443 -3.33 20.00 0.00 0.00 0.00 -44.44

3 1 3 20.00 3.33 0.00 0.00 0.00 44.444 -20.00 -3.33 0.00 0.00 0.00 22.21







48— STAAD.Pro



50 — STAAD.Pro


7 Plane Truss Joint Deflection

ObjectiveTo find the joint deflection due to joint loads in a plane truss.

ReferenceMcCormack, J. C., “Structural Analysis,” Intext Educational Publishers, 3rd edition, 1975, page 271,example 18 - 2.

ProblemDetermine the vertical deflection at point 5 of plane truss structure shown in the figure.

Figure 7-1: Plane truss model

Given

P = 20 kip

Truss width = 4 spaces at 15 ft = 60 ft

Truss height = 15 ft

AX 1-4 = 1 in2, AX 5-6 = 2 in

2, AX 7-8 =1.5 in2,

AX 9-11B = 3 in2P, AX 12-13 = 4 in

2

E = 30E3 ksi


Comparison


Basic SolverSTAAD

Advanced Solver

δ5 (in.) 2.63 2.63 2.63

Table 7-1: Comparison of deflection (δ) for verification problem no. 7

STAAD InputSTAAD TRUSS VERIFICATION PROBLEM NO. 7** REFERENCE 'STRUCTURAL ANALYSIS' BY JACK McCORMACK, PAGE* 271 EXAMPLE 18-2. ANSWER - Y-DISP AT JOINT 5 = 2.63 INCH*UNIT FT KIPJOINT COORD1 0 0 0 5 60 0 06 15. 7.5 ; 7 30. 15. ; 8 45. 7.5MEMB INCI1 2 6 ; 2 3 4 ; 3 4 8 ; 4 4 5 ; 5 1 26 2 3 ; 7 3 6 ; 8 3 8 ; 9 3 710 1 6 ; 11 5 8 ; 12 6 7 13UNIT INCHMEMB PROP1 TO 4 PRI AX 1.05 6 PRIS AX 2.7 8 PRI AX 1.59 10 11 PRI AX 3.12 13 PRI AX 4.CONSTANTE 30E3 ALLPOISSON STEEL ALLSUPPORT1 PINNED ; 3 FIXED BUT FX MZLOAD 1 VERTICAL LOADJOINT LOAD2 4 5 FY -20.0PERFORM ANALYSISPRINT DISPLACEMENTSFINISH



1. STAAD TRUSS VERIFICATION PROBLEM NO. 7INPUT FILE: VER07.STD

2. *3. * REFERENCE 'STRUCTURAL ANALYSIS' BY JACK MCCORMACK, PAGE4. * 271 EXAMPLE 18-2. ANSWER - Y-DISP AT JOINT 5 = 2.63 INCH5. *

52— STAAD.Pro


6. UNIT FT KIP7. JOINT COORD8. 1 0 0 0 5 60 0 09. 6 15. 7.5 ; 7 30. 15. ; 8 45. 7.510. MEMB INCI11. 1 2 6 ; 2 3 4 ; 3 4 8 ; 4 4 5 ; 5 1 212. 6 2 3 ; 7 3 6 ; 8 3 8 ; 9 3 713. 10 1 6 ; 11 5 8 ; 12 6 7 1314. UNIT INCH15. MEMB PROP16. 1 TO 4 PRI AX 1.017. 5 6 PRIS AX 2.18. 7 8 PRI AX 1.519. 9 10 11 PRI AX 3.20. 12 13 PRI AX 4.21. CONSTANT22. E 30E3 ALL23. POISSON STEEL ALL24. SUPPORT25. 1 PINNED ; 3 FIXED BUT FX MZ26. LOAD 1 VERTICAL LOAD27. JOINT LOAD28. 2 4 5 FY -20.029. PERFORM ANALYSIS






SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 1330. PRINT DISPLACEMENTS

DISPLACE

54 — STAAD.Pro


VERIFICATION PROBLEM NO. 7 -- PAGE NO. 3*JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = TRUSS------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN1 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000002 1 -0.12000 0.18000 0.00000 0.00000 0.00000 0.000003 1 -0.24000 0.00000 0.00000 0.00000 0.00000 0.000004 1 -0.48000 -0.89516 0.00000 0.00000 0.00000 0.000005 1 -0.72000 -2.63033 0.00000 0.00000 0.00000 0.000006 1 -0.00820 0.24000 0.00000 0.00000 0.00000 0.000007 1 0.29758 -0.12000 0.00000 0.00000 0.00000 0.000008 1 0.06578 -0.83516 0.00000 0.00000 0.00000 0.00000


*********** END OF THE STAAD.Pro RUN *************** DATE= JAN 25,2013 TIME= 17: 0: 8 ****





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58— STAAD.Pro


8 Unequal Plane frame with Sidesway

ObjectiveTo find the maximum moment due to a concentrated load on the horizontal member in a 1x1 bayplane frame.

ReferenceMcCormack, J. C., Structural Analysis, Intext Educational Publishers, 3rd edition, 1975, page 385,problem 22 - 6.

ProblemDetermine the maximum moment in the structure.

P = 30 kip

H1 = 20 ft, H3 = 30 ft

L = 30 ft, with P occurring at 10 ft from Leg #1

E and I same for all members

Figure 8-1: Unequal leg bay model

Comparison


Basic SolverSTAAD

Advanced Solver

MMax 69.40 69.44 69.44

Table 8-1: Comparison of max. moment, in kip-ft, for verification model no. 8


STAAD InputSTAAD PLANE VERIFICATION PROBLEM NO. 8** PLANE FRAME WITH SIDESWAY. REFERENCE 'STRUCTURAL ANALYSIS'* BY JACK McCORMACK. PAGE 385 PROB 22-6.* ANSWER - MAX BENDING IN MEMB 1 = 69.4 KIP-FT*UNIT FT KIPJOINT COORD1 0. 10. ; 2 0 30 ; 3 30 30 ; 4 30 0MEMB INCI1 1 2 3MEMB PROP AMERICAN1 2 3 TAB ST W12X26CONSTANTE 4176E3POISSON STEEL ALLSUPPORT ; 1 4 FIXEDLOAD 1 VERTICAL LOADMEMBER LOAD2 CON Y -30. 10.PERFORM ANALYSISPRINT FORCESFINISH


***************************************************** ** STAAD.Pro V8i SELECTseries4 ** Version 20.07.09.21 ** Proprietary Program of ** Bentley Systems, Inc. ** Date= JAN 25, 2013 ** Time= 17: 0:13 ** ** USER ID: Bentley Systems, Inc. *****************************************************


2. *3. * PLANE FRAME WITH SIDESWAY. REFERENCE 'STRUCTURAL ANALYSIS'4. * BY JACK MCCORMACK. PAGE 385 PROB 22-6.5. * ANSWER - MAX BENDING IN MEMB 1 = 69.4 KIP-FT6. *7. UNIT FT KIP8. JOINT COORD9. 1 0. 10. ; 2 0 30 ; 3 30 30 ; 4 30 010. MEMB INCI11. 1 1 2 312. MEMB PROP AMERICAN13. 1 2 3 TAB ST W12X2614. CONSTANT15. E 4176E316. POISSON STEEL ALL17. SUPPORT ; 1 4 FIXED18. LOAD 1 VERTICAL LOAD19. MEMBER LOAD20. 2 CON Y -30. 10.21. PERFORM ANALYSIS


NUMBER OF JOINTS 4 NUMBER OF MEMBERS 3

60 — STAAD.Pro


NUMBER OF PLATES 0 NUMBER OF SOLIDS 0NUMBER OF SURFACES 0 NUMBER OF SUPPORTS 2




VERIFICATION PROBLEM NO. 8 -- PAGE NO. 2*22. PRINT FORCES

FORCES

62— STAAD.Pro


VERIFICATION PROBLEM NO. 8 -- PAGE NO. 3*MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KIP FEET (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

1 1 1 20.09 -3.74 0.00 0.00 0.00 -5.342 -20.09 3.74 0.00 0.00 0.00 -69.44

2 1 2 3.74 20.09 0.00 0.00 0.00 69.443 -3.74 9.91 0.00 0.00 0.00 -66.66

3 1 3 9.91 3.74 0.00 0.00 0.00 66.664 -9.91 -3.74 0.00 0.00 0.00 45.51


*********** END OF THE STAAD.Pro RUN *************** DATE= JAN 25,2013 TIME= 17: 0:26 ****





64 — STAAD.Pro



66— STAAD.Pro


9 Multiple level plane frame with horizontal loads

ObjectiveTo find the maximum moment due to lateral joint loads in a 1x2 bay plane frame.

ReferenceMcCormack, J. C., Structural Analysis, Intext Educational Publishers, 3rd edition, 1975, page 388,example 22 - 7.

ProblemDetermine the maximum moment in the frame.

L = 20 ft

H3 = 20 kip, H5 = 10 kip

E and I same for all members.

Figure 9-1: Two story frame model


Comparison


Basic SolverSTAAD

Advanced Solver

MMax 176.40 178.01 178.01

Table 9-1: Comparison of max. moment, in kip-ft, for verification problem no.9

STAAD InputSTAAD PLANE VERIFICATION PROB NO. 9** MULTIPLE LEVEL PLANE FRAME WITH HORIZONTAL LOAD.* REFERENCE 'STRUCTURAL ANALYSIS' BY JACK McCORMACK,* PAGE 388, PROB 22-7. ANSWER - MAX MOM IN MEMB 1 = 176.4 K-F*UNIT FT KIPJOINT COORD1 0 0 0 5 0 40 0 2 ; 2 20 0 0 6 20 40 0 2MEMB INCI1 1 3 2 ; 3 3 5 4 ; 5 3 4 ; 6 5 6MEMB PROP1 TO 6 PRI AX .2 IZ .1CONSTANTE 4176E3POISSON STEEL ALLSUPPORT ; 1 2 FIXEDLOAD 1 HORIZONTAL LOADJOINT LOAD3 FX 20 ; 5 FX 10PERFORM ANALYSPRINT FORCESFINISH



1. STAAD PLANE VERIFICATION PROB NO. 9INPUT FILE: VER09.STD

2. *3. * MULTIPLE LEVEL PLANE FRAME WITH HORIZONTAL LOAD.4. * REFERENCE 'STRUCTURAL ANALYSIS' BY JACK MCCORMACK,5. * PAGE 388, PROB 22-7. ANSWER - MAX MOM IN MEMB 1 = 176.4 K-F6. *7. UNIT FT KIP8. JOINT COORD9. 1 0 0 0 5 0 40 0 2 ; 2 20 0 0 6 20 40 0 210. MEMB INCI11. 1 1 3 2 ; 3 3 5 4 ; 5 3 4 ; 6 5 6

68— STAAD.Pro


12. MEMB PROP13. 1 TO 6 PRI AX .2 IZ .114. CONSTANT15. E 4176E316. POISSON STEEL ALL17. SUPPORT ; 1 2 FIXED18. LOAD 1 HORIZONTAL LOAD19. JOINT LOAD20. 3 FX 20 ; 5 FX 1021. PERFORM ANALYS






VERIFICATION PROB NO. 9 -- PAGE NO. 2*22. PRINT FORCES

FORCES

70 — STAAD.Pro


VERIFICATION PROB NO. 9 -- PAGE NO. 3*MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KIP FEET (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

1 1 1 -22.26 15.06 0.00 0.00 0.00 178.013 22.26 -15.06 0.00 0.00 0.00 123.16

2 1 2 22.26 14.94 0.00 0.00 0.00 176.734 -22.26 -14.94 0.00 0.00 0.00 122.10

3 1 3 -6.51 4.97 0.00 0.00 0.00 34.495 6.51 -4.97 0.00 0.00 0.00 64.93

4 1 4 6.51 5.03 0.00 0.00 0.00 35.346 -6.51 -5.03 0.00 0.00 0.00 65.24

5 1 3 9.91 -15.75 0.00 0.00 0.00 -157.654 -9.91 15.75 0.00 0.00 0.00 -157.44

6 1 5 5.03 -6.51 0.00 0.00 0.00 -64.936 -5.03 6.51 0.00 0.00 0.00 -65.24





VERIFICATION PROB NO. 9 -- PAGE NO. 4*


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74 — STAAD.Pro


10 Forces in a Space Frame

ObjectiveTo find the maximum axial force and moment due to load and moment applied at a joint in aSpace frame.

ReferenceWeaver Jr., W., Computer Programs for Structural Analysis, page 146, problem 8.

ProblemDetermine the maximum axial force and moment in the space structure.

F = 2 kip, P = 1 kip, M = 120 in·kip

L = 120 in.

E = 30E3 ksi,

AX = 11 in2

IX = 83 in4

IY = 56 in4

IZ = 56 in4

Figure 10-1: Space frame model

Comparison


Basic SolverSTAAD

Advanced Solver

FMax (kips) 1.47 1.47 1.47

Table 10-1: Comparison of max. Force, F, and max. Moments,M, for verificationproblem no. 10



Basic SolverSTAAD

Advanced Solver

MY,Max (in·kip) 84.04 84.04 84.04

MZ,Max (in·kip) 95.319 96.120 96.120

STAAD InputSTAAD SPACE VERIFICATION PROB NO. 10** REFERENCE 'COMPUTER PROGRAMS FOR STRUCTURAL ANALYSIS'* BY WILLIAM WEAVER JR. PAGE 146 STRUCTURE NO. 8.* ANSWER - MAX AXIAL FORCE= 1.47 (MEMB 3)* MAX BEND-Y= 84.04, MAX BEND-Z= 95.319 (BOTH MEMB 3)*UNIT INCH KIPJOINT COORD1 0 120 0 ; 2 240 120 03 0 0 0 ; 4 360 0 120MEMB INCI1 1 2 ; 2 3 1 ; 3 2 4MEMB PROP1 2 3 PRIS AX 11. IX 83. IY 56. IZ 56CONSTANT ; E 30000. ALLPOISS .25 ALLSUPPORT3 4 FIXEDLOAD 1 JOINT LOADJOINT LOAD1 FX 2. ; 2 FY -1. ; 2 MZ -120.PERFORM ANALPRINT ANALYSIS RESULTFINISH



1. STAAD SPACE VERIFICATION PROB NO. 10INPUT FILE: VER10.STD

2. *3. * REFERENCE 'COMPUTER PROGRAMS FOR STRUCTURAL ANALYSIS'4. * BY WILLIAM WEAVER JR. PAGE 146 STRUCTURE NO. 8.5. * ANSWER - MAX AXIAL FORCE= 1.47 (MEMB 3)6. * MAX BEND-Y= 84.04, MAX BEND-Z= 95.319 (BOTH MEMB 3)7. *8. UNIT INCH KIP9. JOINT COORD10. 1 0 120 0 ; 2 240 120 011. 3 0 0 0 ; 4 360 0 12012. MEMB INCI13. 1 1 2 ; 2 3 1 ; 3 2 4

76— STAAD.Pro


14. MEMB PROP15. 1 2 3 PRIS AX 11. IX 83. IY 56. IZ 5616. CONSTANT ; E 30000. ALL17. POISS .25 ALL18. SUPPORT19. 3 4 FIXED20. LOAD 1 JOINT LOAD21. JOINT LOAD22. 1 FX 2. ; 2 FY -1. ; 2 MZ -120.23. PERFORM ANAL






SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 1224. PRINT ANALYSIS RESULT

ANALYSIS RESULT

78— STAAD.Pro


VERIFICATION PROB NO. 10 -- PAGE NO. 3*JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = SPACE------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN1 1 0.22267 0.00016 -0.17182 -0.00255 0.00217 -0.002132 1 0.22202 -0.48119 -0.70161 -0.00802 0.00101 -0.004353 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000004 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000



VERIFICATION PROB NO. 10 -- PAGE NO. 4*SUPPORT REACTIONS -UNIT KIP INCH STRUCTURE TYPE = SPACE-----------------

JOINT LOAD FORCE-X FORCE-Y FORCE-Z MOM-X MOM-Y MOM Z3 1 -1.10 -0.43 0.22 48.78 -17.97 96.124 1 -0.90 1.43 -0.22 123.08 47.25 -11.72

80 — STAAD.Pro


VERIFICATION PROB NO. 10 -- PAGE NO. 5*MEMBER END FORCES STRUCTURE TYPE = SPACE-----------------ALL UNITS ARE -- KIP INCH (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

1 1 1 0.90 -0.43 0.22 22.71 -17.97 -36.372 -0.90 0.43 -0.22 -22.71 -34.18 -67.36

2 1 3 -0.43 1.10 0.22 -17.97 -48.78 96.121 0.43 -1.10 -0.22 17.97 22.71 36.37

3 1 2 1.47 -0.71 -0.48 -37.02 15.69 -53.284 -1.47 0.71 0.48 37.02 84.04 -95.32







82— STAAD.Pro



84 — STAAD.Pro


11 Temperature load in different materials

ObjectiveA rigid bar is suspended by two copper wires and one steel wire. Find the stresses in the wires dueto a rise in temperature.

ReferenceTimoshenko, S., Strength of Materials, Part 1, D. Van Nostrand Co., 3rd edition, 1956, page 30,problem 9.

ProblemAssuming the horizontal member to be very rigid, determine the stresses in the copper and steelwires if the temperature rise is 10º F.

Esteel = 30E6 psi, Ecopper = 16E6 psi

αsteel = 70E-7 in/in/°F, αcopper = 92E-7 in/in/°F

AX = 0.1 in2

w = 400 lbf/in.

L = 20 in.

d= 5 in.

Hint:When modeling, assume a large moment of inertia for the horizontal rigid member anddistribute of the concentrated load as uniform.

Figure 11-1: Model of a rigid wire suspended by wires


Comparison


Basic SolverSTAAD

Advanced Solver

σSteel (psi) 19,695 19,698 19,698

σCopper (psi) 10,152 10,151 10,151

Table 11-1: Comparison of stress (σ) for verification model #11

STAAD InputSTAAD PLANE VERIFICATION PROB NO 11** THIS EXAMPLE IS TAKEN FROM 'STRENGTH OF MATERIALS' BY* TIMOSHENKO (PART 1), PAGE 30, PROB 9.* THE ANSWERS ARE 19700 PSI AND 10200 PSI.*UNIT INCH POUNDJOINT COORD1 0. 20. ; 2 5. 20. ; 3 10. 20.4 0. 0. ; 5 5. 0. ; 6 10. 0.MEMB INCI1 1 4 3 ; 4 4 5 5MEMB PROP1 2 3 PRI AX 0.1 ; 4 5 PRI AX 1. IZ 100.CONSTANT ; E 30E6 MEMB 2 4 5E 16E6 MEMB 1 3POISSON 0.15 ALLALPHA 92E-7 MEMB 1 3 ; ALPHA 70E-7 MEMB 2MEMB TRUSS ; 1 2 3SUPPORT ; 1 2 3 PINNEDLOADING 1 VERT LOAD + TEMP LOADMEMB LOAD ;4 5 UNI Y -400.TEMP LOAD ; 1 2 3 TEMP 10.PERFORM ANALYSISPRINT STRESSESFINISH


***************************************************** ** STAAD.Pro V8i SELECTseries4 ** Version 20.07.09.21 ** Proprietary Program of ** Bentley Systems, Inc. ** Date= JAN 25, 2013 ** Time= 17: 1: 8 ** ** USER ID: Bentley Systems, Inc. *****************************************************

1. STAAD PLANE VERIFICATION PROB NO 11INPUT FILE: VER11.STD

2. *3. * THIS EXAMPLE IS TAKEN FROM 'STRENGTH OF MATERIALS' BY4. * TIMOSHENKO (PART 1), PAGE 30, PROB 9.5. * THE ANSWERS ARE 19700 PSI AND 10200 PSI.6. *7. UNIT INCH POUND

86— STAAD.Pro


8. JOINT COORD9. 1 0. 20. ; 2 5. 20. ; 3 10. 20.10. 4 0. 0. ; 5 5. 0. ; 6 10. 0.11. MEMB INCI12. 1 1 4 3 ; 4 4 5 513. MEMB PROP14. 1 2 3 PRI AX 0.1 ; 4 5 PRI AX 1. IZ 100.15. CONSTANT ; E 30E6 MEMB 2 4 516. E 16E6 MEMB 1 317. POISSON 0.15 ALL18. ALPHA 92E-7 MEMB 1 3 ; ALPHA 70E-7 MEMB 219. MEMB TRUSS ; 1 2 320. SUPPORT ; 1 2 3 PINNED21. LOADING 1 VERT LOAD + TEMP LOAD22. MEMB LOAD ;4 5 UNI Y -400.23. TEMP LOAD ; 1 2 3 TEMP 10.24. PERFORM ANALYSIS





VERIFICATION PROB NO 11 -- PAGE NO. 2*

SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 12ZERO STIFFNESS IN DIRECTION 6 AT JOINT 1 EQN.NO. 1

LOADS APPLIED OR DISTRIBUTED HERE FROM ELEMENTS WILL BE IGNORED.THIS MAY BE DUE TO ALL MEMBERS AT THIS JOINT BEING RELEASED OREFFECTIVELY RELEASED IN THIS DIRECTION.

ZERO STIFFNESS IN DIRECTION 6 AT JOINT 2 EQN.NO. 2ZERO STIFFNESS IN DIRECTION 6 AT JOINT 3 EQN.NO. 3

***WARNING - INSTABILITY AT JOINT 5 DIRECTION = FXPROBABLE CAUSE SINGULAR-ADDING WEAK SPRINGK-MATRIX DIAG= 1.2000001E+04 L-MATRIX DIAG= 0.0000000E+00 EQN NO 7***NOTE - VERY WEAK SPRING ADDED FOR STABILITY25. PRINT STRESSES

STRESSES

88— STAAD.Pro




1 1 .0 10150.8 T 0.0 0.0 10150.8 0.0 0.01.00 10150.8 T 0.0 0.0 10150.8 0.0 0.0

2 1 .0 19698.3 T 0.0 0.0 19698.3 0.0 0.01.00 19698.3 T 0.0 0.0 19698.3 0.0 0.0

3 1 .0 10150.8 T 0.0 0.0 10150.8 0.0 0.01.00 10150.8 T 0.0 0.0 10150.8 0.0 0.0

4 1 .0 0.0 0.0 0.0 0.0 1522.6 0.01.00 0.0 0.0 3.8 3.8 1477.4 0.0

5 1 .0 0.0 0.0 3.8 3.8 1477.4 0.01.00 0.0 0.0 0.0 0.0 1522.6 0.0







90 — STAAD.Pro



92— STAAD.Pro


12 Deflection and stress of a truss system

ObjectiveTo find the joint deflection and member stress due to a joint load in a plane truss.

ReferenceTimoshenko, S., Strength of Materials, Part 1, D. Van Nostrand Co., Inc., 3rd edition, 1956, page 10,problem 2.

ProblemDetermine the vertical deflection at point A and the member stresses

AX = 0.5 in2

E = 30E6 psi

P = 5000 lbf

L = 180 in.

angle = 30°

Figure 12-1: Model of two member truss

Comparison


Basic SolverSTAAD

Advanced Solver

σA (psi) 10,000. 10,000. 10,000.

δA (in) 0.12 0.12 0.12

Table 12-1: Comparison of stress (σ) and Deflection (δ) for verification problem12


STAAD InputSTAAD TRUSS VERIFICATION PROBLEM NO 12** THIS EXAMPLE IS TAKEN FROM 'STRENGTH OF MATERIALS'* (PART 1) BY TIMOSHENKO, PAGE 10 PROB 2.* THE ANSWER IN THE BOOK , DEFLECTION = 0.12 INCH* AND STRESS =10000 PSI*UNIT INCH POUNDJOINT COORD1 0. 0. ; 2 155.88457 -90. ; 3 311.76914 0.MEMB INCI ; 1 1 2 2MEMB PROP1 2 PRIS AX 0.5CONSTANTE 30E6POISSON 0.15 ALLSUPPORT ; 1 3 PINNEDLOAD 1 VERT LOADJOINT LOAD ; 2 FY -5000.PERFORM ANALYSISPRINT DISPLACEMENTSPRINT STRESSESFINISH



1. STAAD TRUSS VERIFICATION PROBLEM NO 12INPUT FILE: VER12.STD

2. *3. * THIS EXAMPLE IS TAKEN FROM 'STRENGTH OF MATERIALS'4. * (PART 1) BY TIMOSHENKO, PAGE 10 PROB 2.5. * THE ANSWER IN THE BOOK , DEFLECTION = 0.12 INCH6. * AND STRESS =10000 PSI7. *8. UNIT INCH POUND9. JOINT COORD10. 1 0. 0. ; 2 155.88457 -90. ; 3 311.76914 0.11. MEMB INCI ; 1 1 2 212. MEMB PROP13. 1 2 PRIS AX 0.514. CONSTANT15. E 30E616. POISSON 0.15 ALL17. SUPPORT ; 1 3 PINNED18. LOAD 1 VERT LOAD19. JOINT LOAD ; 2 FY -5000.20. PERFORM ANALYSIS


NUMBER OF JOINTS 3 NUMBER OF MEMBERS 2NUMBER OF PLATES 0 NUMBER OF SOLIDS 0

94 — STAAD.Pro


NUMBER OF SURFACES 0 NUMBER OF SUPPORTS 2SOLVER USED IS THE IN-CORE ADVANCED SOLVER

TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 2



VERIFICATION PROBLEM NO 12 -- PAGE NO. 2*21. PRINT DISPLACEMENTS

DISPLACE

96— STAAD.Pro


VERIFICATION PROBLEM NO 12 -- PAGE NO. 3*JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = TRUSS------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN1 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.000002 1 0.00000 -0.12000 0.00000 0.00000 0.00000 0.000003 1 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000

************** END OF LATEST ANALYSIS RESULT **************22. PRINT STRESSES

STRESSES





1 1 .0 10000.0 T 0.0 0.0 10000.0 0.0 0.01.00 10000.0 T 0.0 0.0 10000.0 0.0 0.0

2 1 .0 10000.0 T 0.0 0.0 10000.0 0.0 0.01.00 10000.0 T 0.0 0.0 10000.0 0.0 0.0



98— STAAD.Pro




100 — STAAD.Pro


102— STAAD.Pro


Notes

13 Steel Design per 1989 AISC Code

TypeSteel Design.

ReferenceAttached step by step hand calculation as per 1989 AISC ASD code. Ninth Edition.

ProblemDetermine the allowable stresses (per 1989 AISC code) for the members of the structure as shownin figure. Also, perform a code check for these members based on the results of the analysis.

Figure 13-1: Steel braced frame frame

Members 1, 2 = W12X26, Members 3, 4 = W14X43

Members 5, 6, 7 = W16X36, Memb 8= L40404, Memb 9 = L50506

F = 15 kip (wind load)

w = 2 kip/ft (dead load + liveload)

Comparison

Member No. 1 2 3 4 5 6 7 8 9

Hand Calculation 1.1576 .914 1.117 0.936 .579 1.119 .668 1.024 .823

Table 13-1: Comparison of governing ratios for the member checks inverification problem no. 13


Member No. 1 2 3 4 5 6 7 8 9

STAAD BasicSolver

1.157 .916 1.117 0.936 .582 1.120 .668 1.024 .823

STAAD AdvancedSolver

1.157 .916 1.117 0.936 .582 1.120 .668 1.024 .823

Hand CalculationManual / Code refers to AISC Manual of Steel Construction, Allowable Stress Design, ninthedition.

Member 1

Size W12x26, L = 10 ft., a = 7.65 in2, Sz = 33.39 in3

From clause F1-2, page 5-45 of Manual, Lc = 6.85 ft.

From observation Load case 1 will govern,

Fx = 25.0 kip (compression), Mz = 56.5 ft·kip

Area of compressive flange = 6.49*0.38 = 2.466 in2

Allowable bending stress, per Clause F1-8 (page 5-47 of Manual)

fb = 12. · 1000 · 1.0/[10 · 12 · (12.22/2.466)] = 20.1817 ksi

(kl/r)y = 120/1.5038 = 79.8, so fa = 15.38 ksi

Stresses per Table C-36 (page 3-16 of manual)

fa = 25./7.65 = 3.268

fb = 56.5 · 12/33.39 = 20.31 ksi

(kl/r)z = 120/5.1639 = 23.238, so F'ez = = 276.54 ksi

Equation H1-1, page 5-54 of Manual

3.268/15.38 + 0.85 · 20.31 / [(1-3.268/276.54) · 20.1817] = 1.078

Equation H1-2, page 5-54 of Manual.

3.268/(0.6 · 36) +20.31/20.1817 = 1.1576

Therefore, equation H1-2 governs and ratio = 1.1576

Member 2

Size W12x26, L = 5 ft., a = 7.65 in2, Sz = 33.39 in3

From observation load case 1 will govern,


Since L is less than Lc = 6.85ft, per Clauses F1-1 & F1-2 (page 5-45 of Manual)

fb = 0.66 · 36 = 23.76 ksi

(kl/r)y = 60/1.5038 = 39.90, so fa = 19.19 ksi

Per Table C-36 (page 3-16 of Manual)

104 — STAAD.Pro


fa = 8.71/7.65 = 1.1385, fb = 56.5 · 12/33.39 = 20.31 ksi

Since fa/Fa less than 0.15, apply equation H1-3 (page 5-54 of Manual)

1.1385/19.19 + 20.31/23.76 = .0593 + .8548 = 0.9141

Member 3

Size W14X43, L = 11ft., a = 12.6 in2, Sz = 62.7 in3

From observation load case 3 will govern,


Referring to clause F1-2, page 5-45 of Manual.

Lc = 8.4 ft. Therefore

fb = 0.6 · 36 = 21.6 ksi

(kl/r)y = 132/1.8941 = 69.69,

so fa = 16.46 ksi per Table C-36 (page 3-16 of Manual)

fa = 25.5/12.6 = 2.024

fb = 112.173 · 12/62.66 = 21.48 ksi

since fa/Fa less than 0.15, use equation H1-3, page 5-54 of Manual

2.024/16.46 + 21.48/21.6 = 0.123 + 0.994 = 1.117

Member 4

Size W14x43, L = 4ft, a = 12.6 in2, Sz = 62.7 in3

From observation, load case 3 will govern,

Fx = 8.75 kip (tension), Mz = 112.173 ft·kip

Since L is less than Lc = 8.4 ft

fb = 0.66 · 36 = 23.76 ksi

Clause F1-1 (page 5-45 of Manual)

fa = 8.75/12.6 = 0.694

fb = 112.73x12/62.66 = 21.48 ksi

Combined tension and bending, use equation H2-1, page 5-55 of Manual.

0.694/(0.6 · 36) + 21.48/23.76 = 0.032 + 0.904 = 0.936

Member 5


From observation, load case 3 will govern.

Fx = 14.02 kip (compression), Mz =57.04 ft·kip

Since L is less than Lc = 7.37 (per Clause F1-2, page 5-45 of Manual)

fb = 0.66 · 36 = 23.76 ksi

(kl/r)y = 60./1.52 = 39.47

so fa = 19.23 ksi per Table C-36 (page 3-16 of Manual)



fa = 14.02/10.6 = 1.32

fb = 57.04 · 12/56.5 = 12.12 ksi

Since fa/Fa less than 0.15, use equation H1-3, page 5-54 of Manual

1.32/19.23 + 12.12/23.76 = 0.069 + 0.510 = 0.579

Member 6


From observation, load case 1 will govern. Forces at midspan are


From Chapter F of the AISC ASD 9PthP ed. specs., with Cb = 1.0,

(102,000Cb/Fy)1/2 =53.229

(510,000Cb/Fy)1/2 =119.02

L/rT = 192/1.79 = 107.26

53.229 < 107.26 < 119.02

Therefore Fb (as per F1-6, page 5-47 of Manual)

[(2/3) – 36 · 107.26 · 107.26/(1,530,000)] · 36 = 14.25 ksi

(Kl/r)y = 192/1.5203 = 126.29

so fa = 9.36 per Table C-36 (page 3-16 of Manual)

fa = 5.65/10.6 = 0.533

fb = 71.25x12/56.49 = 15.14 ksi

Since fa/Fa less than 0.15 use formula [H1-3, page 5-54 of Manual]

0.533/9.36 + 15.14/14.25 = 0.057 + 1.062 = 1.119

Member 7

Size W16x36, L =4ft, a = 10.6in2, Sz =56.49in3

Lc = 7.37ft (Clause F1-2 page 5-45 of Manual)

From observation load case 3 will govern, Fx = 24.06 kip (tension), Mz = 62.96 ft·kip

From Clause F1-1, the allowable compressive stress is

fb = 0.66 Fy = 23.76 ksi

Since section is in tension, per Clause F1-5 (page5-45 of Manual)

fb = 0.60 · 36 = 21.60 Ksi

Choosing the larger of above 2 values, fb = 23.76 Ksi

fa = 24.06/10.6 = 2.2698

fb = 62.96 · 12/56.49 = 13.37

Since combined tension and bending, use equation H 2-1, page 5-55 of the AISC ASD 9PthP ed.specs.

2.2698/(0.6 · 36) + 13.37(23.76) = 0.105 + 0.5627 = 0.6677

106— STAAD.Pro


Member 8

Size L4x4x1/4, L = 7.071 ft, a = 1.94 in2

From observation load case 1 will govern, Fx = 23.04 kip (Comp.)

Fa is computed as per page 5-310 of the AISC ASD 9PthP ed.specs.

Qs = 1.34 – 0.00447 · (4/0.25)·(36)1/2 = 0.9108

Qa = 1.0

Q = Qs · Qa = 0.9108

Cc= 92.0·π2·E/(Q·Fy)]1/2 = [2.0·π2 · 29000/(0.9108 · 36)]1/2 = 132.1241

Kl/r = 7.071 · 12/0.795 = 106.73 < Cc.

Hence, fa = 11.6027 ksi (computed per equation 4-1)

Actual compressive stress

fa = 23.04/1.94 = 11.876 ksi

Therefore, Ratio =

fa/fa = 11.876/11.602 = 1.024

Member 9

Size L5x5x3/8, L = 5.657 ft, a = 3.61 in2

From observation, load case 1 governs, Fx = 48.44 kip (Comp.)

Fa is computed as per page 5-310 of the AISC ASD 9PthP ed.specs.

Qs = 1.34 – 0.00447 · (5/0.375)·(36)1/2 = 0.9824

Qa = 1.0

Q = Qs · Qa = 0.9824

Cc= 92.0·π2·E/(Q·Fy)]1/2 = [2.0·π2 · 29000/(0.9824 · 36)]1/2 = 127.2238

(Kl/r)min = 5.657 · 12/0.99 = 68.57 < Cc.

Hence, fa = 16.301 ksi (computed per equation 4-1)

Actual compressive stress

fa = 48.44/3.61 = 13.418 ksi

Therefore Ratio =

fa/fa = 13.418/16.301 = 0.823


***************************************************** ** STAAD.Pro V8i SELECTseries4 ** Version 20.07.09.21 ** Proprietary Program of ** Bentley Systems, Inc. ** Date= JAN 25, 2013 ** Time= 17: 3:10 *



* ** USER ID: Bentley Systems, Inc. *****************************************************

1. STAAD PLANE VERIFICATION PROBLEM NO 13INPUT FILE: VER13.STD

2. *3. * THIS DESIGN EXAMPLE IS VERIFIED BY HAND CALCULATION4. * FOLLOWING AISC-89 CODE.5. *6. UNIT FEET KIP7. JOINT COORD8. 1 0 0 ; 2 25 0 ; 3 0 10 ; 4 25 119. 5 0 15 ; 6 25 15 ; 7 5 15 ; 8 21 1510. MEMB INCI11. 1 1 3 ; 2 3 5 ; 3 2 4 ; 4 4 612. 5 5 7 ; 6 7 8 ; 7 8 6 ; 8 3 7 ; 9 4 813. MEMB PROP AMERICAN14. 1 2 TA ST W12X26 ; 3 4 TA ST W14X4315. 5 6 7 TA ST W16X36 ; 8 TA ST L40404 ; 9 TA ST L5050616. MEMB TRUSS ; 8 917. CONSTANT18. E 4176E3 ALL19. POISSON STEEL ALL20. SUPPORT ; 1 2 PINNED21. LOADING 1 DL + LL22. MEMB LOAD ; 5 6 7 UNI Y -2.023. LOADING 2 WIND FROM LEFT24. JOINT LOAD ; 5 FX 15.25. LOAD COMB 3 ; 1 0.75 2 0.7526. PERFORM ANALYSIS

108— STAAD.Pro





SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 2, TOTAL DEGREES OF FREEDOM = 2027. LOAD LIST 1 328. PRINT FORCES

FORCES



VERIFICATION PROBLEM NO 13 -- PAGE NO. 3*MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KIP FEET (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

1 1 1 25.00 -5.65 0.00 0.00 0.00 0.003 -25.00 5.65 0.00 0.00 0.00 -56.50

3 1 12.00 1.05 0.00 0.00 0.00 0.003 -12.00 -1.05 0.00 0.00 0.00 10.52

2 1 3 8.71 10.64 0.00 0.00 0.00 56.505 -8.71 -10.64 0.00 0.00 0.00 -3.29

3 3 15.83 -2.77 0.00 0.00 0.00 -10.525 -15.83 2.77 0.00 0.00 0.00 -3.34

3 1 2 25.00 5.65 0.00 0.00 0.00 0.004 -25.00 -5.65 0.00 0.00 0.00 62.15

3 2 25.50 10.20 0.00 0.00 0.00 0.004 -25.50 -10.20 0.00 0.00 0.00 112.17

4 1 4 6.50 -12.85 0.00 0.00 0.00 -62.156 -6.50 12.85 0.00 0.00 0.00 10.76

3 4 -8.75 -24.06 0.00 0.00 0.00 -112.176 8.75 24.06 0.00 0.00 0.00 15.95

5 1 5 -10.64 8.71 0.00 0.00 0.00 3.297 10.64 1.29 0.00 0.00 0.00 15.25

3 5 14.02 15.83 0.00 0.00 0.00 3.347 -14.02 -8.33 0.00 0.00 0.00 57.04

6 1 7 5.65 15.00 0.00 0.00 0.00 -15.258 -5.65 17.00 0.00 0.00 0.00 -0.75

3 7 10.20 4.50 0.00 0.00 0.00 -57.048 -10.20 19.50 0.00 0.00 0.00 -62.96

7 1 8 -12.85 1.50 0.00 0.00 0.00 0.756 12.85 6.50 0.00 0.00 0.00 -10.76

3 8 -24.06 14.75 0.00 0.00 0.00 62.966 24.06 -8.75 0.00 0.00 0.00 -15.95

8 1 3 23.04 0.00 0.00 0.00 0.00 0.007 -23.04 0.00 0.00 0.00 0.00 0.00

3 3 -5.41 0.00 0.00 0.00 0.00 0.007 5.41 0.00 0.00 0.00 0.00 0.00

9 1 4 26.16 0.00 0.00 0.00 0.00 0.008 -26.16 0.00 0.00 0.00 0.00 0.00

110 — STAAD.Pro


VERIFICATION PROBLEM NO 13 -- PAGE NO. 4*MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KIP FEET (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

3 4 48.44 0.00 0.00 0.00 0.00 0.008 -48.44 0.00 0.00 0.00 0.00 0.00

************** END OF LATEST ANALYSIS RESULT **************29. PARAMETER30. CODE AISC31. TRACK 1.0 ALL32. CHECK CODE ALL

STEEL DESIGN




STAAD.Pro CODE CHECKING - (AISC 9TH EDITION) v1.0***********************

ALL UNITS ARE - KIP FEET (UNLESS OTHERWISE NOTED)MEMBER TABLE RESULT/ CRITICAL COND/ RATIO/ LOADING/

FX MY MZ LOCATION=======================================================================* 1 ST W12X26 (AISC SECTIONS)

FAIL AISC- H1-2 1.157 125.00 C 0.00 56.50 10.00

-----------------------------------------------------------------------| MEM= 1, UNIT KIP-INCH, L= 120.0 AX= 7.65 SZ= 33.4 SY= 5.3|| KL/R-Y= 79.8 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 20.18 || FTZ= 21.60 FCY= 27.00 FTY= 27.00 FC= 15.26 FT= 21.60 FV= 14.40 |-----------------------------------------------------------------------

2 ST W12X26 (AISC SECTIONS)PASS AISC- H1-3 0.916 1

8.71 C 0.00 56.50 0.00-----------------------------------------------------------------------| MEM= 2, UNIT KIP-INCH, L= 60.0 AX= 7.65 SZ= 33.4 SY= 5.3|| KL/R-Y= 39.9 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 || FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 18.61 FT= 21.60 FV= 14.40 |-----------------------------------------------------------------------* 3 ST W14X43 (AISC SECTIONS)

FAIL AISC- H1-3 1.117 325.50 C 0.00 -112.17 11.00



8.75 T 0.00 -112.17 0.00-----------------------------------------------------------------------| MEM= 4, UNIT KIP-INCH, L= 48.0 AX= 12.60 SZ= 62.7 SY= 11.3|| KL/R-Y= 25.3 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 || FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 20.26 FT= 21.60 FV= 14.40 |-----------------------------------------------------------------------


14.02 C 0.00 -57.04 5.00

112— STAAD.Pro




FX MY MZ LOCATION=======================================================================-----------------------------------------------------------------------| MEM= 5, UNIT KIP-INCH, L= 60.0 AX= 10.60 SZ= 56.5 SY= 7.0|| KL/R-Y= 39.5 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 || FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 18.39 FT= 21.60 FV= 14.40 |-----------------------------------------------------------------------* 6 ST W16X36 (AISC SECTIONS)

FAIL AISC- H1-3 1.120 15.65 C 0.00 -71.25 8.00



24.06 T 0.00 62.96 0.00-----------------------------------------------------------------------| MEM= 7, UNIT KIP-INCH, L= 48.0 AX= 10.60 SZ= 56.5 SY= 7.0|| KL/R-Y= 31.6 CB= 1.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 23.76 || FTZ= 23.76 FCY= 27.00 FTY= 27.00 FC= 18.87 FT= 21.60 FV= 14.40 |-----------------------------------------------------------------------* 8 ST L40404 (AISC SECTIONS)

FAIL AISC- H1-1 1.024 123.04 C 0.00 0.00 0.00

-----------------------------------------------------------------------| MEM= 8, UNIT KIP-INCH, L= 84.9 AX= 1.94 SZ= 0.8 SY= 1.7|| KL/R- = 106.7 CB= 0.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 0.00 || FTZ= 0.00 FCY= 0.00 FTY= 0.00 FC= 11.60 FT= 21.60 FV= 0.00 |-----------------------------------------------------------------------

9 ST L50506 (AISC SECTIONS)PASS AISC- H1-1 0.823 3

48.44 C 0.00 0.00 0.00





FX MY MZ LOCATION=======================================================================-----------------------------------------------------------------------| MEM= 9, UNIT KIP-INCH, L= 67.9 AX= 3.61 SZ= 1.8 SY= 3.9|| KL/R- = 69.0 CB= 0.00 YLD= 36.00 ALLOWABLE STRESSES: FCZ= 0.00 || FTZ= 0.00 FCY= 0.00 FTY= 0.00 FC= 16.30 FT= 21.60 FV= 0.00 |-----------------------------------------------------------------------33. FINISH


114 — STAAD.Pro




116— STAAD.Pro


118— STAAD.Pro


Notes

14 Concrete Design per ACI 318 Code

TypeConcrete design as per ACI code.

ReferenceCRSI Handbook and Notes on ACI-318-99.

ProblemA plane frame is created with such loading as to create 138 Kip-Ft moment on beam and 574 Kip ofaxial load coupled with above moment on column.

Figure 14-1: Concrete moment frame

GivenSize of the beam is 10" x 16"and size of the column 14" x 16".

P = 521.32 kip

w = 5.268 kip/ft

L = 20 ft, H = 15 ft


Comparison

Result TypeTheoryACINotes

TheoryCRSI

Handbook

STAADBasicSolver

STAADAdvancedSolver

Area of Steel inbeam

2.78 in2 X 2.792 in2 2.792 in2

Area of Steel incolumn

X 4.01% 4.09% required

4.23% provided

4.09% required

4.23% provided

Table 14-1: Comparison of concrete designs for verifcation problem no. 14



1. STAAD PLANE VERIFICATION FOR CONCRETE DESIGNINPUT FILE: VER14.STD

2. UNIT KIP FEET3. JOINT COORDINATES4. 1 0. 0. ; 2 0. 15. ; 3 20. 15. ; 4 20. 0.5. MEMBER INCIDENCE6. 1 1 2 ; 2 2 3 ; 3 3 47. UNIT INCH8. MEMBER PROPERTY9. 1 3 PRISMATIC YD 16. ZD 14.10. 2 PRISM YD 16. ZD 10.11. CONSTANTS12. E CONCRETE ALL13. POISSON CONCRETE ALL14. SUPPORT15. 1 4 FIXED16. UNIT FT17. LOADING 1 DEAD + LIVE18. JOINT LOAD19. 2 3 FY -521.3220. MEMBER LOAD21. 2 UNI GY -5.26822. PERFORM ANALYSIS



SOLVER USED IS THE IN-CORE ADVANCED SOLVER

120 — STAAD.Pro


VERIFICATION FOR CONCRETE DESIGN -- PAGE NO. 2TOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 623. PRINT MEMBER FORCES

MEMBER FORCES



VERIFICATION FOR CONCRETE DESIGN -- PAGE NO. 3MEMBER END FORCES STRUCTURE TYPE = PLANE-----------------ALL UNITS ARE -- KIP FEET (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

1 1 1 574.00 -13.69 0.00 0.00 0.00 -67.442 -574.00 13.69 0.00 0.00 0.00 -137.87

2 1 2 13.69 52.68 0.00 0.00 0.00 137.873 -13.69 52.68 0.00 0.00 0.00 -137.87

3 1 3 574.00 13.69 0.00 0.00 0.00 137.874 -574.00 -13.69 0.00 0.00 0.00 67.44

************** END OF LATEST ANALYSIS RESULT **************24. UNIT INCH25. START CONC DESIGN

CONCRETE DESIGN26. CODE ACI 199927. TRACK 1.0 MEMB 228. FYMAIN 60.0 ALL29. FC 4.0 ALL30. CLB 1.4375 ALL31. DESIGN BEAM 2

122— STAAD.Pro


VERIFICATION FOR CONCRETE DESIGN -- PAGE NO. 4=====================================================================

BEAM NO. 2 DESIGN RESULTS - FLEXURE PER CODE ACI 318-99LEN - 20.00FT. FY - 60000. FC - 4000. SIZE - 10.00 X 16.00 INCHESLEVEL HEIGHT BAR INFO FROM TO ANCHOR

FT. IN. FT. IN. FT. IN. STA END_____________________________________________________________________1 0 + 2-5/8 2-NUM.10 0 + 0-0/0 20 + 0-0/0 YES YES

|----------------------------------------------------------------|| CRITICAL POS MOMENT= 125.53 KIP-FT AT 10.00 FT, LOAD 1|| REQD STEEL= 2.48 IN2, RHO=0.0185, RHOMX=0.0214 RHOMN=0.0033 || MAX/MIN/ACTUAL BAR SPACING= 10.16/ 2.54/ 4.73 INCH || REQD. DEVELOPMENT LENGTH = 48.52 INCH ||----------------------------------------------------------------|Cracked Moment of Inertia Iz at above location = 1837.17 inch^4

*** A SUITABLE BAR ARRANGEMENT COULD NOT BE DETERMINED.REQD. STEEL = 2.792 IN2, MAX. STEEL PERMISSIBLE = 2.873 IN2MAX NEG MOMENT = 137.87 KIP-FT, LOADING 1*** A SUITABLE BAR ARRANGEMENT COULD NOT BE DETERMINED.REQD. STEEL = 2.792 IN2, MAX. STEEL PERMISSIBLE = 2.873 IN2MAX NEG MOMENT = 137.87 KIP-FT, LOADING 1___ 2J____________________ 240.X 10.X 16_____________________ 3J____| || || || || 2#10H 3. 0.TO 240. |||=========================================================================||| ||___________________________________________________________________________|_________ _________ _________ _________ _________ _________| | | | | | | | | | | || | | | | | | | | | | || | | | | | | | | | | || | | | | | | | | | | || 2#10 | | 2#10 | | 2#10 | | 2#10 | | 2#10 | | 2#10 || OO | | OO | | OO | | OO | | OO | | OO || | | | | | | | | | | ||_________| |_________| |_________| |_________| |_________| |_________|



VERIFICATION FOR CONCRETE DESIGN -- PAGE NO. 5********************END OF BEAM DESIGN**************************32. DESIGN COLUMN 1

124 — STAAD.Pro


VERIFICATION FOR CONCRETE DESIGN -- PAGE NO. 6====================================================================

COLUMN NO. 1 DESIGN PER ACI 318-99 - AXIAL + BENDINGFY - 60000 FC - 4000 PSI, RECT SIZE - 14.00 X 16.00 INCHES, TIED

AREA OF STEEL REQUIRED = 9.164 SQ. IN.BAR CONFIGURATION REINF PCT. LOAD LOCATION PHI----------------------------------------------------------12 - NUMBER 8 4.232 1 END 0.700(PROVIDE EQUAL NUMBER OF BARS ON EACH FACE)TIE BAR NUMBER 4 SPACING 14.00 IN********************END OF COLUMN DESIGN RESULTS********************33. END CONC DESIGN34. FINISH




VERIFICATION FOR CONCRETE DESIGN -- PAGE NO. 7************************************************************* For questions on STAAD.Pro, please contact ** Bentley Systems or Partner offices ** ** Telephone Web / Email ** USA +1 (714) 974-2500 ** UK +44 (0) 808 101 9246 ** SINGAPORE +65 6225-6158 ** FRANCE +33 (0) 1 55238400 ** GERMANY +49 0931 40468 ** INDIA +91 (033) 4006-2021 ** JAPAN +81 (03)5952-6500 http://www.ctc-g.co.jp ** CHINA +86 21 6288 4040 ** THAILAND +66 (0)2645-1018/19 [email protected]** ** Worldwide http://selectservices.bentley.com/en-US/ ** *************************************************************

126— STAAD.Pro


VERIFICATION FOR CONCRETE DESIGN -- PAGE NO. 8Information about the key files in the current distributionModification Date CRC Size (Bytes) File Name-----------------------------------------------------------------------------------------12/21/2012 0x880 19054592 SProStaad.exe12/21/2012 0x6c00 08679424 SProStaadStl.exe10/18/2012 0x3940 00373248 AISCSteelDesignNew.dll09/19/2003 0x2fc0 00081970 CMesh.dll11/29/2012 0x15c1 04476998 dbSectionInterface.dll10/18/2012 0x5680 00315392 EC3Annex.dll10/18/2012 0x91c1 00421888 EuroCode3.dll10/18/2012 0xf981 00102400 IS800SteelDesign.dll01/24/2001 0x9b40 00073728 LoadGen.dll06/21/2012 0xcb81 00040960 MathExtensions.dll09/25/2003 0x6340 00704512 MeshEngine.dll11/27/2012 0x9a40 00059392 NORSOKSteelDesign.dll09/22/2003 0xce00 00069632 QuadPlateEngine.dll10/18/2012 0x6381 00079872 SteelCollection.dll10/18/2012 0xf240 00030720 SteelDesign.dll10/18/2012 0xac1 00061440 SteelSections.dll05/28/2008 0x77c1 00106572 SurfMeshEngine.dll10/18/2012 0x1dc0 00017408 UnitConversions.dll10/19/2012 0x280 00197120 WrapperAISCSteelDesignNew.dll10/19/2012 0x57c0 00019968 WrapperSteelDesign.dll10/19/2012 0x2e80 00108032 WrapperSteelSections.dll10/19/2012 0x6480 00008704 WrapperSystem.dll10/19/2012 0xe141 00016384 WrapperUnitConversions.dll08/22/2011 0x9741 00974848 AISCSections.mdb06/13/2006 0xd4c1 01204224 AISCSectionsRCeco.mdb06/05/2007 0xbc00 00331776 aiscsections_all_editions.mdb01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb01/05/2005 0xcac1 03651584 aitctimbersections.mdb05/01/2008 0xb001 00552960 aluminumsections.mdb01/13/2010 0x40 01019904 AustralianColdFormedSections.mdb10/02/2008 0x40 00520192 AustralianSections.mdb10/18/2012 0x4d41 00765952 BrazilSections.mdb05/19/2011 0x1641 00643074 BritishSections.mdb09/12/2006 0xf4c0 00466944 bscoldformedsections.mdb06/28/2005 0x8201 00327680 butlercoldformedsections.mdb02/25/2008 0x2e01 00329728 CanadianSections.mdb09/26/2006 0x5ec1 00280576 CanadianSections_old.mdb05/31/2005 0x9e81 00450560 canadiantimbersections.mdb02/10/2010 0x94c1 00483328 ChineseSections.mdb01/05/2005 0xd6c0 00600064 dutchsections.mdb08/08/2011 0xc880 02699264 EuropeanColdFormedSections.mdb10/07/2010 0xd2c0 03596290 EuropeanSections.mdb01/25/2011 0xc901 00692224 EuropeanSectionsLegacy.mdb01/05/2005 0xd301 00202752 frenchsections.mdb01/05/2005 0x11c1 00233472 germansections.mdb10/06/2010 0x1bc1 00393216 GOSTColdFormedSections.mdb12/08/2009 0x6ac0 00569344 IndianLegacySections.mdb08/13/2012 0x6281 00942082 IndianSections.mdb01/05/2005 0xd540 00180224 iscoldformedsections.mdb02/21/2008 0xc380 00327680 JapaneseColdFormedSections.mdb08/02/2011 0x2b80 00593920 JapaneseSections.mdb07/15/2011 0xce40 00532480 JindalSections.mdb11/09/2005 0x9081 00376832 Kingspancoldformedsections.mdb08/24/2007 0x3ec0 00454656 Koreansections.mdb04/24/2007 0x99c0 00114688 lysaghtcoldformedsections.mdb02/07/2005 0x9a00 00243712 mexicansteeltables.mdb04/19/2007 0x9ec1 00421888 RCecoColdFormedSections.mdb11/20/2009 0xf700 00428034 RussianSections.mdb05/12/2009 0xe001 00237570 SouthAfricanSections.mdb01/06/2005 0x9341 00194560 spanishsections.mdb11/20/2009 0x5100 00368640 STO_ASChMSections.mdb12/20/2011 0x2b41 00409600 TATAStructuraSections.mdb



04/19/2007 0x7480 00223232 uscoldformedsections.mdb10/31/2007 0x3700 00133120 VenezuelanSections.mdb-----------------------------------------------------------------------------------------

128— STAAD.Pro


15 Natural Modes of a Simple Beam

PurposeCompare theoretical answers to the STAAD solution.

ReferenceTimoshenko, S., “Vibration Problems in Engineering”, Third Edition, D. Van Nostrand Company,Inc., 1955, page 322

ProblemThe first five natural frequencies and the associated mode shapes are computed for the flexuralmotion of a simply supported beam.

Figure 15-1: Simple beam diagram

ModelFigure 15-2: Finite element model

The simply supported beam is divided into twenty spanwise beam elements. At nodes 1 and 21, alldegrees of freedom except the rotation about the Z axis are restrained. For the remaining nodes,only the translation along Y and the rotation about Z are permitted. Both shear deformation androtary inertia have been excluded from the model. The mass matrix is a diagonal matrix.

Cross-section Properties

Rectangular Section: 1 inch Width x 2 inch Depth


Area = 2 in2

J = b3a/16 {16/3 - 3.36(b/a)[1 - 1/12(b/a)4]}

where

a = 2

b = 1

J = 0.4578 inch4

I2 = 1x23/12 inch4

I3 = 2x13/12 inch4

Theoretical Results

The natural bending frequencies, for a uniform beam with hinged ends, are given by:

fn = πn2/212(EIg/Aγ)1/2

where

fn = natural frequency for mode n, in cycles per second

l = span of the beam

E = elastic modulus

I = cross-section moment of inertia

g = gravitational constant

A = cross-section area

γ = weight density

The parameters used in the frequency equation are:

l = 20 inches

E = 10x106 psi

I = 0.6667 inches4

g = 386.4 inches per second2

A = 2.0 sq. inches

γ = 0.1 pounds per cubic inch

from which:

fn = n2x 445.685674

The table below shows the natural frequencies computed from the theoretical equation and thesubspace iteration method available within STAAD. Frequencies are in cycles per second.

130 — STAAD.Pro


Comparison

Mode Number TheoreticalSTAAD

Basic SolverSTAAD

Advanced Solver

1 445.686 445.506 445.506

2 1,782.74 1,782.01 1,782.01

3 4,011.17 4,009.41 4,009.41

4 7,130.97 7,127.25 7,127.25

5 11,142.1 11,134.25 11,134.25

Table 15-1: Comparison of frequency values for verification problem no. 15

Input filestaad plane STAAD Ver. PROB. 15

UNIT POUND INCH

JOINT COORD

1 0 0 0 21 20 0 0

MEMB INCI

1 1 2 20

MEMB PROP

1 TO 20 PRIS AX 2 IZ 0.6667

CONST

E 10E6 ALL

POISSON 0.3 ALL

DENS 0.1 ALL

CUT OFF MODE SHAPE 5

SUPP

1 21 FIXED BUT MZ

2 TO 20 FIXED BUT FY MZ

LOAD 1

SELF X 1.0

SELF Y 1.0

MODAL CALC REQ

PERF ANALY

finish





1. STAAD PLANE STAAD VER. PROB. 15INPUT FILE: ver15.STD

2. * NATURAL MODES OF A SIMPLE BEAM4. UNIT POUND INCH5. JOINT COORD6. 1 0 0 0 21 20 0 07. MEMB INCI8. 1 1 2 209. MEMB PROP10. 1 TO 20 PRIS AX 2 IZ 0.666711. CONST12. E 10E6 ALL13. POISSON 0.3 ALL14. DENS 0.1 ALL15. CUT OFF MODE SHAPE 516. SUPP17. 1 21 FIXED BUT MZ18. 2 TO 20 FIXED BUT FY MZ19. LOAD 120. SELF X 1.021. SELF Y 1.022. MODAL CALC REQ23. PERF ANALY



SOLVER USED IS THE IN-CORE ADVANCED SOLVER

132— STAAD.Pro


STAAD VER. PROB. 15 -- PAGE NO. 2* NATURAL MODES OF A SIMPLE BEAMTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 40




STAAD VER. PROB. 15 -- PAGE NO. 3* NATURAL MODES OF A SIMPLE BEAM

CALCULATED FREQUENCIES FOR LOAD CASE 1MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 445.506 0.002242 1782.012 0.000563 4009.409 0.000254 7127.250 0.000145 11134.253 0.00009


1 0.000000E+00 3.228953E+00 0.000000E+00 2.000000E+002 0.000000E+00 2.763825E-27 0.000000E+00 2.000000E+003 0.000000E+00 3.469944E-01 0.000000E+00 2.000000E+004 0.000000E+00 9.994771E-32 0.000000E+00 2.211146E+005 0.000000E+00 1.165685E-01 0.000000E+00 2.000000E+00

PARTICIPATION FACTORSMASS PARTICIPATION FACTORS IN PERCENT--------------------------------------

MODE X Y Z SUMM-X SUMM-Y SUMM-Z1 0.00 84.97 0.00 0.000 84.972 0.0002 0.00 0.00 0.00 0.000 84.972 0.0003 0.00 9.13 0.00 0.000 94.104 0.0004 0.00 0.00 0.00 0.000 94.104 0.0005 0.00 3.07 0.00 0.000 97.171 0.000

24. FINISH

134 — STAAD.Pro




136— STAAD.Pro


138— STAAD.Pro


Notes

16 Natural Frequencies of a Circular Plate


ReferenceBlevins, Robert D., Formulas for Natural Frequency and Mode Shape, Van Nostrand ReinholdCompany, 1979, Page 240.

ProblemA flat circular plate is simply supported around the entire perimeter. The first six modes andtheir associated natural frequencies are to be computed using the subspace iteration methodoffered by STAAD. This problem demonstrates that the natural frequencies of an axi-symmetricstructure can be accurately computed utilizing a 180 degree model with the appropriate boundaryconditions.

Figure 16-1: Diagram of circular plate

ModelThe 180 degree sector was modeled using radial lines at intervals of 15 degrees. Tangential lineswere then located utilizing a relationship such that the aspect ratio of the quad-plate elementswas approximately 1.0. All rotations normal to the plane of the plate were restrained. In-planetranslations for all nodes were restrained because the theoretical solution does not consider in-plane effects. Rotations about the global Y axis for the nodes at X=0.0 were restrained because thisis a symmetry boundary. Z translation of all nodes on the outside radius were restrained toprovide for the simply-supported condition.

139— STAAD.Pro


Figure 16-2: Finite model for half circular plate

It should be noted that the outside edge of the plate is a series of secant lines instead of a truearc. This will result in a loss of about 1% of the plate’s true mass or about .5% of the mass that iseffective for this problem.

Therefore, it is not unlikely that a few natural frequencies will be lower than the theoretical valuesinstead of higher which is typical for a finite element analysis using plate elements. In addition, atrue simple-support condition for this problem would require restraining the component ofrotation that is radial to the outside edge.

Theoretical Answers

From the reference case 2 in Table 11-1, the first six natural frequencies of the plate are described bythe following equation:

=−

fij

λ

πa

Eh

γ ν( )2 12 1

ij2

2

3

2

= dimensionless parameter associated with the mode indices i,j

i = number of nodal diameters in this mode shape

j = number of nodal circles in this mode shape not counting the boundary

ν = Poisson’s ratio

E = Elastic Modulus

h = plate thickness

γ = mass of plate per unit area

a = radius of plate

The numerical values used for this example are:



ν = 0.30

E = 10.0x10P6P psi

h = 0.10 inches

γ = (0.10 lbm/in3)(0.10 in)/(386.4 lbm-in/lbf-s

2) = 2.588(10)-5 lbf-s2/in3

a = 10.0 inches

with the numerical values used above

= =−

⋅

⋅ −

− 9.467

πa

Eh

γ ν π( )( )

( )1

2 12 1

1

2 (10.0)

10.0 10 (0.10)

12 2.588 10 1 (0.3)2

3

2 2

6 3

5 2

cycles/sec.

λ2ij is tabulated from the reference as follows:

ModeNumber

λ2ij

Number of NodalDiameters (i)

Number of NodalCircles (j)

1 4.977 0 0

2 13.94 1 0

3 25.65 2 0

4 29.76 0 1

5(l) 3 0

6 48.51 1 1

Table 16-1: Values of λ2ij

(l) not tabulated in the reference

Comparison

ModeNumbe-

rTheoretical

STAAD Basi-c Solver

STAAD Advance-d Solver

1 47.12 46.21 46.21

2 132.0 130.54 130.54

3 242.8 240.15 240.15

4 281.7 283.03 283.03

5 * 373.2 373.2

6 459.3 466.58 466.58

Table 16-2: Comparison of frequency values, Hz, for verification problem no. 16

141 — STAAD.Pro


*The reference did not tabulate a value of for the fifth mode of the structure, hence a comparisonwith the theoretical value of this mode cannot be made.

DiscussionAll anti-symmetric mode shapes for the 360 degree circular plate were captured by the 180 degreemodel with a phase angle included in the calculation. Some of the difference between thetheoretical and STAAD frequencies is attributed to the loss of mass due to the piecewise secantrepresentation of the outer radius and, since this mass is about 1 percent lower than for a truecircular plate, it is not surprising that the first few modes are lower than the theoretical solution.

Input filestaad space STAAD Ver. ProB. 16

* NATURAL FREQUENCIES OF A CIRCULAR PLATE

UNIT POUND INCH

JOINT COORD CYLINDRICAL

1 1 -90 0 13 1 90 0

REP 2 1.244 0 0

REP 1 1.051 0 0

REP 1 1.367 0 0

REP 1 1.779 0 0

REP 1 2.315 0 0

JOINT COORD

1000 0 0 0

ELEM INCI

1 1000 1 2 ; 2 1000 2 3 ; 3 1000 3 4 ; 4 1000 4 5 ; 5 1000 5 6

6 1000 6 7 ; 7 1000 7 8 ; 8 1000 8 9 ; 9 1000 9 10 ; 10 1000 10 11

11 1000 11 12 ; 12 1000 12 13

13 1 14 15 2 TO 24

REP 5 12 13

ELEM PROP

1 TO 84 TH 0.1

CONST

E 10.0E6 ALL

POISS 0.3 ALL

DENS 0.1 ALL

SUPP

* CENTRE OF CIRCLE



1000 FIXED BUT FZ MX

* INTERIOR NODES

2 TO 12 15 TO 25 28 TO 38 41 TO 51 54 TO 64 -

67 TO 77 FIXED BUT FZ MX MY

* NODES ALONG CIRCUMFERENCE

79 TO 91 FIXED BUT MX MY

* NODES ALONG DIAMETER EXCEPT THE TWO AT THE ENDS OF THE DIAMETER.

1 TO 66 BY 13 13 TO 78 BY 13 FIXED BUT FZ MX

LOAD 1

SELF X 1.0

SELF Y 1.0

SELF Z 1.0

MODAL CALC REQ

PERF ANALY

FINI



1. STAAD SPACE STAAD VER. PROB. 16INPUT FILE: ver16.STD

2. * NATURAL FREQUENCIES OF A CIRCULAR PLATE4. UNIT POUND INCH5. JOINT COORD CYLINDRICAL6. 1 1 -90 0 13 1 90 07. REP 2 1.244 0 08. REP 1 1.051 0 09. REP 1 1.367 0 010. REP 1 1.779 0 011. REP 1 2.315 0 013. JOINT COORD14. 1000 0 0 016. ELEM INCI17. 1 1000 1 2 ; 2 1000 2 3 ; 3 1000 3 4 ; 4 1000 4 5 ; 5 1000 5 618. 6 1000 6 7 ; 7 1000 7 8 ; 8 1000 8 9 ; 9 1000 9 10 ; 10 1000 10 1119. 11 1000 11 12 ; 12 1000 12 1321. 13 1 14 15 2 TO 2422. REP 5 12 1324. ELEM PROP25. 1 TO 84 TH 0.127. CONST28. E 10.0E6 ALL

143— STAAD.Pro


29. POISS 0.3 ALL30. DENS 0.1 ALL32. SUPP34. * CENTRE OF CIRCLE35. 1000 FIXED BUT FZ MX37. * INTERIOR NODES38. 2 TO 12 15 TO 25 28 TO 38 41 TO 51 54 TO 64 -39. 67 TO 77 FIXED BUT FZ MX MY41. * NODES ALONG CIRCUMFERENCE42. 79 TO 91 FIXED BUT MX MY44. * NODES ALONG DIAMETER EXCEPT THE TWO AT THE ENDS OF THE DIAMETER.45. 1 TO 66 BY 13 13 TO 78 BY 13 FIXED BUT FZ MX47. LOAD 148. SELF X 1.049. SELF Y 1.050. SELF Z 1.0



STAAD VER. PROB. 16 -- PAGE NO. 2* NATURAL FREQUENCIES OF A CIRCULAR PLATE51. MODAL CALC REQ52. PERF ANALY





145— STAAD.Pro


STAAD VER. PROB. 16 -- PAGE NO. 3* NATURAL FREQUENCIES OF A CIRCULAR PLATE

CALCULATED FREQUENCIES FOR LOAD CASE 1MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 46.211 0.021642 130.541 0.007663 240.146 0.004164 283.028 0.003535 373.198 0.002686 466.579 0.00214


1 0.000000E+00 0.000000E+00 1.053697E+00 4.537235E-012 0.000000E+00 0.000000E+00 2.213994E-10 3.944625E-013 0.000000E+00 0.000000E+00 1.886318E-07 3.935445E-014 0.000000E+00 0.000000E+00 1.275807E-01 1.752589E-015 0.000000E+00 0.000000E+00 1.060158E-08 4.222041E-016 0.000000E+00 0.000000E+00 7.773679E-10 2.165169E-01


MODE X Y Z SUMM-X SUMM-Y SUMM-Z1 0.00 0.00 86.78 0.000 0.000 86.7802 0.00 0.00 0.00 0.000 0.000 86.7803 0.00 0.00 0.00 0.000 0.000 86.7804 0.00 0.00 10.51 0.000 0.000 97.2875 0.00 0.00 0.00 0.000 0.000 97.2876 0.00 0.00 0.00 0.000 0.000 97.287

53. FINI




147— STAAD.Pro




149— STAAD.Pro


17 Natural Frequencies of a Rectangular Plate


ReferenceBlevins, Robert D., Formulas for Natural Frequency and Mode Shape, Van Nostrand ReinholdCompany, 1979, page 258.

ProblemA flat rectangular plate is simply supported on all four sides. The first six modes and theirassociated natural frequencies are to be computed for this structure using the subspace iterationmethod offered by STAAD. This problem also demonstrates that the mesh refinement can bechosen to accurately calculate modes of interest based on the expected mode shapes.

Figure 17-1: Simply supported, rectangular plate

L = 45 in., W = 30 in., t = 0.2 in.

ModelA plate with an aspect ratio of 1.5 was used so that comparison could be made with theoreticalresults tabulated for plates in the reference. An equally spaced mesh was utilized in both the x andthe y dimensions of the plate. The number of elements in each dimension was determined on thebasis of the highest mode of interest. Since the number of half-waves in the sixth mode is 3 inthe length dimension and 2 in the width dimension, a node spacing of 3.75 inches results in eachhalf- wave being represented by four elements which means that no element will be expected todeform in double curvature. The simply supported edge condition requires that translation normal


to the plane of the plate be restrained for these edge nodes. Rotations normal to the plate wererestrained for all nodes.

Figure 17-2: Model: finite element meshed

Theoretical Answers

From the reference case 16 in Table 11-4, the first six natural frequencies of the plate are describedby the following equations:

=−

fij

λ

πa

Eh

γ ν( )2 12 1

ij2

2

3

2

= dimensionless parameter associated with the mode indices i, j

i = number of half-waves in this mode shape along the horizontal axis

j = number of half-waves in this mode shape along the vertical axis


E = elastic modulus

h = plate thickness

γ = mass of material per unit area

a = length of plate

b = width of plate


152— STAAD.Pro


ν = 0.30

E =30.0x106 psi

h=0.2 inches

γ = (0.282 lbm/in3)(0.20 in)/(386.4 lbm-in/lbf-s

2) = .1460(10)-4 lbf-s2/in3

a = 45.0 inches

b = 30.0 inches

with the numerical values used above

= =−

⋅

⋅ −

− 0.9644

πa

Eh

γ ν π( )( )

( )1

2 12 1

1

2 (10.0)

30.0 10 (0.20)

12 1.460 10 1 (0.3)2

3

2 2

6 3

4 2

cycles/sec.

λ2ij is tabulated from the reference as follows:

ModeNumber

λ2ij

Number of Half-Wavesin Length (i)

Number of Half-Wavesin Width (j)

1 32.08 1 1

2 61.69 2 1

3 98.70 1 2

4 111.0 3 1

5 128.3 2 2

6 177.7 3 2

Table 17-1: Values of λ2ij

Comparison

ModeNumber

TheoreticalSTAAD Basic

SolverSTAAD Advanced

Solver

1 30.94 30.6 30.6

2 59.49 58.72 58.72

3 95.18 95.06 95.06

4 107.1 106.28 106.28

5 123.7 122.09 122.09

6 171.3 168.0 168.0

Table 17-2: Comparison of frequency values, Hz, for verificatoin problem no. 17



DiscussionAs was noted earlier, the node spacing was based on the highest mode of interest. It follows thatthe difference between the theoretical and STAAD frequencies generally increases with increasingmode sequence.

Input filestaad space STAAD Ver. PROB. 17.

* NATURAL FREQUENCIES OF A RECTANGULAR PLATE

UNIT POUND INCH

JOINT COORD

1 0 0 0 13 45 0 0

REP ALL 8 0 3.75 0

ELEM INCI

1 1 2 15 14 TO 12

REP 7 12 13

ELEM PROP

1 TO 96 TH 0.2

CONST

E 30.0E6 ALL

DENS 0.282 ALL

POISSON 0.3 ALL


cut off freq 1000

SUPPORT

* CORNER NODES

1 13 105 117 FIXED BUT MX MY

* NODES ALONG Y=0 AND Y=30

2 TO 12 106 TO 116 FIXED BUT MX MY

* NODES ALONG X=0

14 TO 92 BY 13 FIXED BUT MX MY

* NODES ALONG X=45

26 TO 104 BY 13 FIXED BUT MX MY

* INTERIOR NODES

15 TO 25 28 TO 38 41 TO 51 54 TO 64 67 TO 77 80 TO 90 -

93 TO 103 FIXED BUT FZ MX MY

154 — STAAD.Pro


LOAD 1

SELF X 1

SELF Y 1

SELF Z 1

MODAL CALC REQ

PERF ANALY

fini




2. * NATURAL FREQUENCIES OF A RECTANGULAR PLATE4. UNIT POUND INCH5. JOINT COORD6. 1 0 0 0 13 45 0 07. REP ALL 8 0 3.75 08. ELEM INCI9. 1 1 2 15 14 TO 1210. REP 7 12 1311. ELEM PROP12. 1 TO 96 TH 0.213. CONST14. E 30.0E6 ALL15. DENS 0.282 ALL16. POISSON 0.3 ALL17. CUT OFF MODE SHAPE 618. CUT OFF FREQ 100019. SUPP21. * CORNER NODES22. 1 13 105 117 FIXED BUT MX MY24. * NODES ALONG Y=0 AND Y=3025. 2 TO 12 106 TO 116 FIXED BUT MX MY27. * NODES ALONG X=028. 14 TO 92 BY 13 FIXED BUT MX MY30. * NODES ALONG X=4531. 26 TO 104 BY 13 FIXED BUT MX MY33. * INTERIOR NODES34. 15 TO 25 28 TO 38 41 TO 51 54 TO 64 67 TO 77 80 TO 90 -35. 93 TO 103 FIXED BUT FZ MX MY37. LOAD 138. SELF X 139. SELF Y 140. SELF Z 141. MODAL CALC REQ42. PERF ANALY



STAAD VER. PROB. 17 -- PAGE NO. 2* NATURAL FREQUENCIES OF A RECTANGULAR PLATE





156— STAAD.Pro



CALCULATED FREQUENCIES FOR LOAD CASE 1MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 30.599 0.032682 58.724 0.017033 95.063 0.010524 106.277 0.009415 122.092 0.008196 168.009 0.00595


1 0.000000E+00 0.000000E+00 4.833474E+01 1.915442E+012 0.000000E+00 0.000000E+00 1.438414E-16 1.920213E+013 0.000000E+00 0.000000E+00 3.063663E-15 1.925533E+014 0.000000E+00 0.000000E+00 4.837136E+00 1.927465E+015 0.000000E+00 0.000000E+00 7.328017E-14 1.927600E+016 0.000000E+00 0.000000E+00 1.883367E-11 1.936298E+01


MODE X Y Z SUMM-X SUMM-Y SUMM-Z1 0.00 0.00 79.15 0.000 0.000 79.1462 0.00 0.00 0.00 0.000 0.000 79.1463 0.00 0.00 0.00 0.000 0.000 79.1464 0.00 0.00 7.92 0.000 0.000 87.0665 0.00 0.00 0.00 0.000 0.000 87.0666 0.00 0.00 0.00 0.000 0.000 87.066

43. FINI




158— STAAD.Pro




160 — STAAD.Pro


162— STAAD.Pro


Notes

18 Natural Modes of a Framework

PurposeCompare the results from STAAD's subspace iteration method with the results from EASE2® andANSYS®.

Reference

1. Problem 1, from the ASME 1972 Program Verification and Qualification Library

2. DeSalvo, G.J., and Swanson, J.A., ANSYS Engineering Analysis System Examples Manual,Swanson Analysis Systems, Inc., 1979, Example Problem No. 2.

3. Peterson, F.E., EASE2 Elastic Analysis for Structural Engineering Example Problem Manual,Engineering Analysis Corporation, 1981, Example 2.03.

ProblemA three dimensional frame is analyzed for its natural frequencies and the associated mode shapesusing the subspace iteration method offered by STAAD.

Figure 18-1: Frame model

All dimensions are in inches.


Model

The only entity present in the model is the general purpose three dimensional beam.

Cross-section Properties

All the beam elements have the same cross section:

Outside radius, Ro = 1.1875 in.

Inside radius, Ri = 1.0335 in.

Area = π(Ro2 - Ri2) = 1.074532 in2

Moment of Inertia, I = π/4(Ro4 - Ri4)

I = 0.665747 in4

Torsion constant = twice the inertia, for closed circular sections = 1.331494 in4

The expression for the shear flexibility factor is derived from the ratio of maximum shear stress tothe average shear stress:

α = AQ/Ib

Where the four items – A, Q, I, and b are properties of the half cross-section about the centerline.For the circular section they are:

A = π/2(Ro2 - Ri2)

Q = 2/3(Ro3 - Ri3)

I = π/8(Ro4 - Ri4)

I = 2(Ro - Ri)

The final expression for the shear flexibility factor, for a circular tube section is:

= = −

− −α

AQ

Ib

R R

R R R R( )( )4

3

o i

o i o i

3 3

2 2

α = 1.993620

Hence the shear area is entered as

AY = Cross Section Area / α = 1.074532 / 1.99362 = 0.538985 in2

ComparisonThe following table compares Twenty-Four Natural Frequencies

ModeNumber

DeSalvo andSwanson2

Peterson3

STAAD BasicSolver


1 111.52 111.53 111.23 111.23

2 115.95 115.96 115.80 115.80

Table 18-1: Comparison of natural frequency values (cycles per second) forverification problem no. 18

164 — STAAD.Pro


ModeNumber

DeSalvo andSwanson2

Peterson3

STAAD BasicSolver


3 137.60 137.61 137.17 137.17

4 218.02 218.03 215.80 215.80

5 404.23 404.23 404.27 404.27

6 422.70 422.72 422.64 422.64

7 451.72 451.75 451.60 451.60

8 553.99 554.07 549.01 549.01

9 735.70 735.81 733.59 733.59

10 762.32 762.44 758.56 758.56

11 852.57 852.72 851.36 851.36

12 894.08 894.26 892.28 892.28

13 910.21 910.40 893.06 893.06

14 916.98 917.18 910.83 910.83

15 940.02 940.02 932.29 932.29

16 959.98 960.27 956.39 956.39

17 971.15 971.44 964.01 964.01

18 976.92 977.22 967.27 967.27

19 1012.2 1012.5 981.54 981.54

20 1028.4 1028.8 1009.26 1009.26

21 1123.6 1123.9 1070.56 1070.56

22 1134.5 1134.9 1123.16 1123.16

23 1164.1 1164.4 1149.37 1149.37

24 1216.7 1217.2 1199.88 1199.88

In both references by DeSalvo and Swanson2 and Peterson3, the number of dynamic degrees offreedom has been reduced from 42 to 24, by means of the Guyan method. No such reduction isperformed in the STAAD model.

Input Filestaad space STAAD Ver. PROB. 18

* NATURAL FREQUENCIES OF A SPACE FRAME



UNIT POUND INCH

JOINT COORD

1 0 0 0 ; 2 27.25 0 0

3 0 10 0 ; 4 27.25 10 0

5 0 18.625 0 ; 6 8.625 18.625 0 ; 7 18.625 18.625 0 ; 8 27.25 18.625 0

9 0 18.625 8.625 ; 10 27.25 18.625 8.625

11 0 0 17.25 ; 12 27.25 0 17.25

13 0 10 17.25 ; 14 27.25 10 17.25

15 0 18.625 17.25 ; 16 8.625 18.625 17.25

17 18.625 18.625 17.25 ; 18 27.25 18.625 17.25

MEMB INCI

1 1 3 ; 2 2 4 ; 3 3 5 ; 4 4 8

5 5 6 7

8 5 9 ; 9 8 10 ; 10 9 15 ; 11 10 18

12 11 13 ; 13 12 14 ; 14 13 15 ; 15 14 18

16 15 16 ; 17 16 17 ; 18 17 18

MEMB PROP

1 TO 18 PRIS AX 1.074532 IZ 0.665747 IY 0.665747 IX 1.331494 -

AY 0.538985 AZ 0.538985

* AY 1.88 AZ 1.88

CONST

E 27.9E6 ALL

POISS 0.3 ALL


1 2 11 12 FIXED

LOAD 1

JOINT LOAD

166— STAAD.Pro


3 4 13 14 FX 3.4517 FY 3.4517 FZ 3.4517

6 7 16 17 FX 3.4517 FY 3.4517 FZ 3.4517

9 10 FX 3.4517 FY 3.4517 FZ 3.4517

5 8 15 18 FX 9.7973 FY 9.7973 FZ 9.7973

MODAL CALC REQ

PERF ANALY

fini




2. * NATURAL MODES OF A SPACE FRAME4. UNIT POUND INCH5. JOINT COORD6. 1 0 0 0 ; 2 27.25 0 07. 3 0 10 0 ; 4 27.25 10 08. 5 0 18.625 0 ; 6 8.625 18.625 0 ; 7 18.625 18.625 0 ; 8 27.25 18.625 010. 9 0 18.625 8.625 ; 10 27.25 18.625 8.62512. 11 0 0 17.25 ; 12 27.25 0 17.2513. 13 0 10 17.25 ; 14 27.25 10 17.2514. 15 0 18.625 17.25 ; 16 8.625 18.625 17.2515. 17 18.625 18.625 17.25 ; 18 27.25 18.625 17.2517. MEMB INCI18. 1 1 3 ; 2 2 4 ; 3 3 5 ; 4 4 819. 5 5 6 721. 8 5 9 ; 9 8 10 ; 10 9 15 ; 11 10 1823. 12 11 13 ; 13 12 14 ; 14 13 15 ; 15 14 1824. 16 15 16 ; 17 16 17 ; 18 17 1826. MEMB PROP27. 1 TO 18 PRIS AX 1.074532 IZ 0.665747 IY 0.665747 IX 1.331494 -28. AY 0.538985 AZ 0.53898529. * AY 1.88 AZ 1.8831. CONST32. E 27.9E6 ALL33. POISS 0.3 ALL34. CUT OFF MODE SHAPE 2436. SUPP37. 1 2 11 12 FIXED39. LOAD 140. JOINT LOAD41. 3 4 13 14 FX 3.4517 FY 3.4517 FZ 3.451742. 6 7 16 17 FX 3.4517 FY 3.4517 FZ 3.451743. 9 10 FX 3.4517 FY 3.4517 FZ 3.451744. 5 8 15 18 FX 9.7973 FY 9.7973 FZ 9.797346. MODAL CALC REQ



47. PERF ANALY

168— STAAD.Pro


STAAD VER. PROB. 18 -- PAGE NO. 2* NATURAL MODES OF A SPACE FRAME








CALCULATED FREQUENCIES FOR LOAD CASE 1MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 111.236 0.008992 115.801 0.008643 137.169 0.007294 215.799 0.004635 404.271 0.002476 422.642 0.002377 451.600 0.002218 549.012 0.001829 733.589 0.0013610 758.563 0.0013211 851.354 0.0011712 892.281 0.0011213 893.060 0.0011214 910.829 0.0011015 932.286 0.0010716 956.391 0.0010517 964.012 0.0010418 967.267 0.0010319 981.536 0.0010220 1009.256 0.0009921 1070.562 0.0009322 1123.161 0.0008923 1149.368 0.0008724 1199.882 0.00083


1 6.984127E+01 1.585650E-31 1.600468E-29 6.127270E+012 1.586676E-29 3.061270E-32 7.005318E+01 5.765801E+013 1.530424E-27 9.787485E-34 1.243985E-27 7.213568E+014 1.120662E-28 6.042509E-32 2.161771E-32 7.915617E+015 1.703629E-34 4.244586E-03 6.051562E-31 1.532638E+016 5.415551E-31 1.825469E+01 9.149730E-33 1.469096E+017 1.534827E-31 3.211749E-29 5.112395E-03 1.450510E+018 1.793772E-32 1.069950E-32 2.966451E-01 1.825927E+019 2.958819E-32 9.521517E+00 5.195713E-31 1.177227E+0110 9.808873E-04 2.321914E-29 1.152307E-32 1.174657E+0111 1.321193E+00 6.262300E-31 7.417272E-32 1.582130E+0112 1.954924E-27 1.911258E+00 8.329941E-33 1.624941E+0113 1.722238E+00 2.372506E-27 4.394091E-33 3.802992E+0114 2.577481E-30 6.851764E-30 4.001621E-31 2.195000E+0115 5.909354E-32 3.217315E-31 2.363041E-01 1.628835E+0116 4.416769E-31 4.122944E+00 2.251072E-30 1.648078E+0117 3.524252E-32 4.118921E-32 1.054592E-28 1.622822E+0118 2.754368E-31 1.629803E-31 3.083967E+00 1.640511E+01

170 — STAAD.Pro



19 3.491379E-01 3.862104E-30 4.169724E-30 2.835636E+0120 9.398409E-33 1.122183E+01 1.044268E-33 2.560370E+0121 0.000000E+00 1.344792E-29 3.735534E-31 3.464233E+0122 3.635978E-01 2.304610E-30 3.409187E-33 3.137066E+0123 4.264358E-31 3.878643E-32 1.719066E-31 1.555328E+0124 6.381997E-02 2.651481E-34 4.242369E-33 2.520960E+01





MODE X Y Z SUMM-X SUMM-Y SUMM-Z1 94.76 0.00 0.00 94.756 0.000 0.0002 0.00 0.00 95.04 94.756 0.000 95.0443 0.00 0.00 0.00 94.756 0.000 95.0444 0.00 0.00 0.00 94.756 0.000 95.0445 0.00 0.01 0.00 94.756 0.006 95.0446 0.00 24.77 0.00 94.756 24.773 95.0447 0.00 0.00 0.01 94.756 24.773 95.0518 0.00 0.00 0.40 94.756 24.773 95.4539 0.00 12.92 0.00 94.756 37.691 95.45310 0.00 0.00 0.00 94.758 37.691 95.45311 1.79 0.00 0.00 96.550 37.691 95.45312 0.00 2.59 0.00 96.550 40.284 95.45313 2.34 0.00 0.00 98.887 40.284 95.45314 0.00 0.00 0.00 98.887 40.284 95.45315 0.00 0.00 0.32 98.887 40.284 95.77416 0.00 5.59 0.00 98.887 45.878 95.77417 0.00 0.00 0.00 98.887 45.878 95.77418 0.00 0.00 4.18 98.887 45.878 99.95819 0.47 0.00 0.00 99.360 45.878 99.95820 0.00 15.23 0.00 99.360 61.103 99.95821 0.00 0.00 0.00 99.360 61.103 99.95822 0.49 0.00 0.00 99.854 61.103 99.95823 0.00 0.00 0.00 99.854 61.103 99.95824 0.09 0.00 0.00 99.940 61.103 99.958

48. FINI*********** END OF THE STAAD.Pro RUN *************** DATE= JAN 25,2013 TIME= 17: 5:17 ****

172— STAAD.Pro




174 — STAAD.Pro


176— STAAD.Pro


Notes

19 Response of a Simply Supported Beam to a Shock Spectrum


Reference

1. Biggs, John M., Introduction to Structural Dynamics, McGraw Hill, 1964, pp. 256-263

2. Blevins, Robert D., Formulas for Natural Frequency and Mode Shape, Van Nostrand-Reinhold, 1979.

ProblemThe supports of a simply supported beam are subjected to an acceleration time history. Themaximum bending moment in the beam is computed for the first mode of the structure. Thisproblem demonstrates the capabilities of STAAD to calculate the correct modal response of astructure utilizing response spectrum data.

Figure 19-1: Simple span beam

ModelThe STAAD model consists of 11 nodes and 10 elastic beam elements. Node 1 is completelyrestrained with the exception of having rotational freedom in the Z direction, the remaining nodesare restrained except for X and Y displacements and Z rotations. Node 11 is additionally restrainedagainst displacements in the Y direction to provide for the simple support condition . Only thecontribution of the first mode of the structure is considered.


Figure 19-2: Finite element model

Theoretical SolutionMaterial Properties

L = 240.0 inches

E = 30 x 106 lb/in2

EI = 1.0 x 1010 lb-in2

m = 0.2 lb-sec2 / in2

h = 14.0 in.

From Reference 2, Table 8-1, page 108, the fundamental frequency of the beam is:

= = =f hz6.098i

λ

π

EI

m π2

9.869

2 (240)

1.0(10)

0.2

i2

2 2

10

The modal participation factor for the fundamental mode is:

= ∫∫

Γmϕ x d x

mϕ x d x( )( )0

1

0

1 2

Where the first mode shape, φ(x) = sin(πx/1)

= =∫

∫πΓ 4 /

m d x

m d x

sin

sin

πx

πx

0

1

1

0

12

1

The maximum relative modal displacement is given by:

Amax = Γu0maxWhere:

u0max = y''so/ω2(DLF)max

ω - 2π*6.098Hz

y''so = 1.0g

(DLF)max = 1.648 at 6.098 Hz

178— STAAD.Pro


therefore:

Amax = 4(1.648)(386.4)/[π(2π)2(6.098)2] = 0.5523 in.

The bending moment

M = -EIδ2u/(δx2)

Where u for the first mode = A sin(πx/l)

δ2u/(δx2) = -π2/l2A·sin(πx/l)

M = AEI·(π2/l2)·sin(πx/l)

Mmax = AmaxEI·(π2/l2)

at x = l/2

Mmax = 1(10)10·(0.5523)·π2/(2402) = 946.351(10)3 lb·in

at x=l/2

Comparison

Solution TheorySTAAD

Basic SolverSTAAD

Advanced Solver

Bending Moment (kip-inch) 946.351 947.09 947.09

Table 19-1: Comparison of max. moment in beam 5 for verification problem no.19

Input FileSTAAD SPACE STAAD VER. PROB. 19

* RESPONSE OF A SIMPLY SUPPORTED BEAM TO A SHOCK SPECTRUM

UNIT POUND INCH

JOINT COORD

1 0 0 0 11 240 0 0

MEMB INCI

1 1 2 10

MEMB PROP

1 TO 10 PRIS YD 14.0 ZD 1.45777 AX 20.4082 IX 40.0 IY 3.6139 IZ 333.333

CONST

E 30.0E6 ALL

POISS 0.3 ALL

DENS 3.78672 ALL


SUPPORT

1 FIXED BUT MZ



2 TO 10 FIXED BUT FX FY MZ

11 FIXED BUT FX MZ

LOAD 1

SELF X 1.0

SELF Y 1.0

SPECTRUM SRSS Y 1.0 ACC DAMP 0.001 SCALE 386.4

0.15 1.648 ; 0.17 1.648

PERF ANALY

PRINT MEMB FORCE LIST 5

fini



1. STAAD SPACE STAAD VER. PROB. 19INPUT FILE: Ver19.STD

2. * RESPONSE OF A SIMPLY SUPPORTED BEAM TO A SHOCK SPECTRUM3. UNIT POUND INCH4. JOINT COORD5. 1 0 0 0 11 240 0 06. MEMB INCI7. 1 1 2 108. MEMB PROP9. 1 TO 10 PRIS YD 14.0 ZD 1.45777 AX 20.4082 IX 40.0 IY 3.6139 IZ 333.33310. CONST11. E 30.0E6 ALL12. POISS 0.3 ALL13. DENS 3.78672 ALL14. CUT OFF MODE SHAPE 115. SUPPORT16. 1 FIXED BUT MZ17. 2 TO 10 FIXED BUT FX FY MZ18. 11 FIXED BUT FX MZ19. LOAD 120. SELF X 1.021. SELF Y 1.022. SPECTRUM SRSS Y 1.0 ACC DAMP 0.001 SCALE 386.423. 0.15 1.648 ; 0.17 1.64824. PERF ANALY



180 — STAAD.Pro


STAAD VER. PROB. 19 -- PAGE NO. 2* RESPONSE OF A SIMPLY SUPPORTED BEAM TO A S






CALCULATED FREQUENCIES FOR LOAD CASE 1MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 6.069 0.16476

RESPONSE SPECTRUM LOAD 1RESPONSE LOAD CASE 1


1 1.709965E-23 1.478714E+04 0.000000E+00 9.273617E+03SRSS MODAL COMBINATION METHOD USED.DYNAMIC WEIGHT X Y Z 1.761987E+04 1.669251E+04 0.000000E+00 POUNMISSING WEIGHT X Y Z -1.761987E+04 -1.905373E+03 0.000000E+00 POUNMODAL WEIGHT X Y Z 1.709965E-23 1.478714E+04 0.000000E+00 POUN

MODE ACCELERATION-G DAMPING---- -------------- -------

1 1.64933 0.00100MODAL BASE ACTIONSMODAL BASE ACTIONS FORCES IN POUN LENGTH IN INCH-----------------------------------------------------------

MOMENTS ARE ABOUT THE ORIGINMODE PERIOD FX FY FZ MX MY MZ

1 0.165 0.00 24388.86 0.00 0.00 0.00 2926662.97

182— STAAD.Pro



PARTICIPATION FACTORSMASS PARTICIPATION FACTORS IN PERCENT BASE SHEAR IN POUN-------------------------------------- ------------------

MODE X Y Z SUMM-X SUMM-Y SUMM-Z X Y Z1 0.00 88.59 0.00 0.000 88.585 0.000 0.00 24388.86 0.00

---------------------------TOTAL SRSS SHEAR 0.00 24388.86 0.00TOTAL 10PCT SHEAR 0.00 24388.86 0.00TOTAL ABS SHEAR 0.00 24388.86 0.00

25. PRINT MEMB FORCE LIST 5MEMB FORCE LIST 5



STAAD VER. PROB. 19 -- PAGE NO. 5* RESPONSE OF A SIMPLY SUPPORTED BEAM TO A SMEMBER END FORCES STRUCTURE TYPE = SPACE-----------------ALL UNITS ARE -- POUN INCH (LOCAL )MEMBER LOAD JT AXIAL SHEAR-Y SHEAR-Z TORSION MOM-Y MOM-Z

5 1 5 0.00 1931.41 0.00 0.00 0.00 900734.256 0.00 1931.41 0.00 0.00 0.00 947088.00

************** END OF LATEST ANALYSIS RESULT **************26. FINI


184 — STAAD.Pro




186— STAAD.Pro


188— STAAD.Pro


Notes

20 Thermal Loading on a Simply Supported Rectangular Plate


ReferenceTimoshenko, S., and Woinowsky-Krieger, S., Theory of Plates and Shells, Second Edition, McGraw-Hill, 1959, pages 162 - 165.

ProblemA rectangular plate is simply supported on all four sides. The transverse and longitudinal bendingmoments as well as the deflections at several points on the plate are computed.

Figure 20-1: Rectangular plate model

L = 12 in., W = 16 in.

ModelThe plate is modeled using 1 in. X 1 in. size elements. At the corner nodes, all the degrees offreedom are considered restrained. For the nodes along the four edges, rotation is permitted aboutthat edge.

Theoretical SolutionFrom the Reference, equation (j), the expression for deflection normal to the plate surface is:

= − ∑

−

+= …

∞w 1αt ν

π hm

m α

( )1 4

1, 3, 5

sin cosh

cosh

mπx mπy

m

2

3 3

where


=αmmπb

2

From the Reference, equation (k), the expressions for bending moment per unit width are

= ∑−

= …∞

⋅Mx

Dαt ν

πh m m α

( )4 1

1, 3, 5

sin cosh

cosh

mπx mπy

m

2

= − ∑− −

= …∞

⋅My

αt ν D

h

Dαt ν

πh m m α

( ) ( )1 4 1

1, 3, 5

sin cosh

cosh

mπx mπy

m

2 2

Where:

α = Coefficient of Thermal Expansion

t = Difference between the temperatures of the upper and lower surfaces of the plate


h = Plate thickness

a = Dimension of the plate along the x1 axis

b = Dimension of the plate along the x2 axis

E = Elastic Modulus

D = Eh3/[12(1 - ν2)


α = 12.0E-06 / °F

t = 450 °F

ν = 0.3

h = 0.3 in.

a = 12 in.

b = 16 in.

E= 10.0E6 psi

Comparison

Deflections along X = 6

NodeNumber

X YTheoreticalDeflection

STAAD BasicSolver

Deflection

STAADAdvancedSolver

Deflection

7 6 -8 0.00 0.00 0.00

20 6 -7 0.0897 0.0897 0.0897

Table 20-1: Comparison of deflections along X = 6, in inches, for verificationproblem no. 20

190 — STAAD.Pro


NodeNumber


STAAD BasicSolver

Deflection

STAADAdvancedSolver

Deflection

33 6 -6 0.1597 0.1597 0.1597

46 6 -5 0.2132 0.2132 0.2132

59 6 -4 0.2531 0.2531 0.2531

72 6 -3 0.2818 0.2818 0.2818

85 6 -2 0.3011 0.3011 0.3011

98 6 -1 0.3122 0.3122 0.3122

111 6 0 0.3158 0.3158 0.3158

124 6 1 0.3122 0.3122 0.3122

Deflections along Y = 1

NodeNumber


STAAD BasicSolver

Deflection

STAADAdvancedSolver

Deflection

118 0 1 0 0 0

119 1 1 0.1004 0.1004 0.1004

120 2 1 0.1794 0.1794 0.1794

121 3 1 0.2387 0.2387 0.2387

122 4 1 0.2799 0.2799 0.2799

123 5 1 0.3042 0.3042 0.3042

124 6 1 0.3122 0.3122 0.3122

125 7 1 0.3042 0.3042 0.3042

126 8 1 0.2799 0.2799 0.2799

127 9 1 0.2387 0.2387 0.2387

128 10 1 0.1794 0.1794 0.1794

Table 20-2: Comparison of deflections along Y = 1, in inches, for verificationproblem no. 20



NodeNumber


STAAD BasicSolver

Deflection

STAADAdvancedSolver

Deflection

129 11 1 0.1004 0.1004 0.1004

130 12 1 0 0 0

Bending Moments along Y = 0.5

ElementNumber

X Y

TheoreticalMoment

(Pound-in/in)

STAAD BASICSOLVERMoment

(Pound-in/in)

STAADADVANCEDSOLVER

Moment (Pound-in/in)

Mx My Mx My Mx My

97 0.5 0.5 16.74 388.26 16.85 388.37 16.85 388.37

98 1.5 0.5 48.93 356.07 49.25 356.37 49.25 356.37

99 2.5 0.5 77.45 327.55 77.92 328.01 77.92 328.01

100 3.5 0.5 100.36 304.64 100.93 305.20 100.93 305.20

101 4.5 0.5 116.31 288.69 116.93 289.32 116.93 289.32

102 5.5 0.5 124.47 280.54 125.11 281.19 125.11 281.19

103 6.5 0.5 124.47 280.54 125.11 281.19 125.11 281.19

104 7.5 0.5 116.31 288.69 116.93 289.32 116.93 289.32

105 8.5 0.5 100.36 304.64 100.93 305.20 100.93 305.20

106 9.5 0.5 77.45 327.55 77.92 328.01 77.92 328.01

107 10.5 0.5 48.93 356.07 49.25 356.37 49.25 356.37

108 11.5 0.5 16.74 388.26 16.85 388.37 16.85 388.37

Table 20-3: Comparison of bending moments along Y = 0.5, in in-lbs/in, forverification problem no. 20

192— STAAD.Pro


Bending Moments along X = 5.5

ElementNumber

X Y

TheoreticalMoment(Pound-in/in)

STAAD BASICSOLVERMoment

(Pound-in/in)

STAADADVANCEDSOLVER

Moment (Pound-in/in)

Mx My Mx My Mx My

6 5.5 -7.5

373.88 31.12 373.27 32.17 373.27 32.17

18 5.5 -6.5

311.64 93.36 312.19 93.95 312.19 93.95

30 5.5 -5.5

256.83 148.17 257.59 148.94 257.59 148.94

42 5.5 -4.5

211.14 193.87 211.95 194.66 211.95 194.66

54 5.5 -3.5

175.41 229.60 176.20 230.35 176.20 230.35

66 5.5 -2.5

149.47 255.53 150.19 256.25 150.19 256.25

78 5.5 -1.5

132.68 272.32 133.36 273.00 133.36 273.00

90 5.5 -0.5

124.47 280.54 125.11 281.19 125.11 281.19

102 5.5 0.5 124.47 280.54 125.11 281.19 125.11 281.19

114 5.5 1.5 132.68 272.32 133.36 273.00 133.36 273.00

126 5.5 2.5 149.47 255.53 150.19 256.25 150.19 256.25

138 5.5 3.5 175.41 229.59 176.20 230.35 176.20 230.35

150 5.5 4.5 211.14 193.87 211.95 194.66 211.95 194.66

162 5.5 5.5 256.83 148.17 257.59 148.94 257.59 148.94

Table 20-4: Comparison of bending moments along X = 5.5, in in-lbs/in, forverification problem no. 20

Input FileSTAAD SPACE STAAD VER. PROB. 20

* THERMAL LOADING ON A RECTANGULAR PLATE.



UNIT POUND INCH

JOINT COORD

1 0 -8 0 13 12 -8 0

REP 16 0 1 0

ELEM INCI

1 1 2 15 14 TO 12

REP 15 12 13

ELEM PROP

1 TO 192 TH 0.3

CONST

E 10.0E6 ALL

POISS 0.3 ALL

ALPHA 12.0E-6 ALL

SUPP

1 13 209 221 FIXED

2 TO 12 210 TO 220 FIXED BUT MX

14 TO 196 BY 13 26 TO 208 BY 13 FIXED BUT MY

LOAD 1

TEMP LOAD

1 TO 192 TEMP 0 450

PERF ANALY

PRINT JOINT DISP LIST 119 TO 129 7 TO 215 BY 13

PRINT ELEM FORCE LIST 97 TO 108 6 TO 186 BY 12

fini




2. * THERMAL LOADING ON A SIMPLY SUPPORTED RECTANGULAR PLATE.4. UNIT POUND INCH5. JOINT COORD

194 — STAAD.Pro


6. 1 0 -8 0 13 12 -8 07. REP 16 0 1 08. ELEM INCI9. 1 1 2 15 14 TO 1210. REP 15 12 1311. ELEM PROP12. 1 TO 192 TH 0.313. CONST14. E 10.0E6 ALL15. POISS 0.3 ALL16. ALPHA 12.0E-6 ALL17. SUPP18. 1 13 209 221 FIXED19. 2 TO 12 210 TO 220 FIXED BUT MX20. 14 TO 196 BY 13 26 TO 208 BY 13 FIXED BUT MY21. LOAD 122. TEMP LOAD23. 1 TO 192 TEMP 0 45024. PERF ANALY





STAAD VER. PROB. 20 -- PAGE NO. 2* THERMAL LOADING ON A SIMPLY SUPPORTED RECT

SOLVER USED IS THE IN-CORE ADVANCED SOLVERTOTAL PRIMARY LOAD CASES = 1, TOTAL DEGREES OF FREEDOM = 104225. PRINT JOINT DISP LIST 119 TO 129 7 TO 215 BY 13

JOINT DISP LIST 119

196— STAAD.Pro


STAAD VER. PROB. 20 -- PAGE NO. 3* THERMAL LOADING ON A SIMPLY SUPPORTED RECTJOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = SPACE------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN119 1 0.00000 0.00000 0.10040 -0.00193 -0.08944 0.00000120 1 0.00000 0.00000 0.17937 -0.00371 -0.06893 0.00000121 1 0.00000 0.00000 0.23869 -0.00521 -0.05010 0.00000122 1 0.00000 0.00000 0.27991 -0.00635 -0.03262 0.00000123 1 0.00000 0.00000 0.30417 -0.00705 -0.01607 0.00000124 1 0.00000 0.00000 0.31218 -0.00729 0.00000 0.00000125 1 0.00000 0.00000 0.30417 -0.00705 0.01607 0.00000126 1 0.00000 0.00000 0.27991 -0.00635 0.03262 0.00000127 1 0.00000 0.00000 0.23869 -0.00521 0.05010 0.00000128 1 0.00000 0.00000 0.17937 -0.00371 0.06893 0.00000129 1 0.00000 0.00000 0.10040 -0.00193 0.08944 0.000007 1 0.00000 0.00000 0.00000 0.10098 0.00000 0.0000020 1 0.00000 0.00000 0.08974 0.07940 0.00000 0.0000033 1 0.00000 0.00000 0.15969 0.06132 0.00000 0.0000046 1 0.00000 0.00000 0.21318 0.04637 0.00000 0.0000059 1 0.00000 0.00000 0.25310 0.03406 0.00000 0.0000072 1 0.00000 0.00000 0.28181 0.02380 0.00000 0.0000085 1 0.00000 0.00000 0.30109 0.01506 0.00000 0.0000098 1 0.00000 0.00000 0.31218 0.00729 0.00000 0.00000111 1 0.00000 0.00000 0.31579 0.00000 0.00000 0.00000124 1 0.00000 0.00000 0.31218 -0.00729 0.00000 0.00000137 1 0.00000 0.00000 0.30109 -0.01506 0.00000 0.00000150 1 0.00000 0.00000 0.28181 -0.02380 0.00000 0.00000163 1 0.00000 0.00000 0.25310 -0.03406 0.00000 0.00000176 1 0.00000 0.00000 0.21318 -0.04637 0.00000 0.00000189 1 0.00000 0.00000 0.15969 -0.06132 0.00000 0.00000202 1 0.00000 0.00000 0.08974 -0.07940 0.00000 0.00000215 1 0.00000 0.00000 0.00000 -0.10098 0.00000 0.00000************** END OF LATEST ANALYSIS RESULT **************26. PRINT ELEM STRESS LIST 97 TO 108 6 TO 186 BY 12

ELEM STRESS LIST 97



STAAD VER. PROB. 20 -- PAGE NO. 4* THERMAL LOADING ON A SIMPLY SUPPORTED RECTELEMENT STRESSES FORCE,LENGTH UNITS= POUN INCH----------------

STRESS = FORCE/UNIT WIDTH/THICK, MOMENT = FORCE-LENGTH/UNIT WIDTHELEMENT LOAD SQX SQY MX MY MXY

VONT VONB SX SY SXYTRESCAT TRESCAB

97 1 -2.00 -0.08 -16.85 -388.37 16.5325419.74 25419.74 0.00 0.00 0.0025939.96 25939.96

TOP : SMAX= -1074.52 SMIN= -25939.96 TMAX= 12432.72 ANGLE= 2.5BOTT: SMAX= 25939.96 SMIN= 1074.52 TMAX= 12432.72 ANGLE=-87.5

98 1 -1.69 -0.20 -49.25 -356.37 15.2722367.84 22367.84 0.00 0.00 0.0023808.31 23808.31


99 1 -1.23 -0.23 -77.92 -328.01 12.9219844.14 19844.14 0.00 0.00 0.0021911.49 21911.49


100 1 -0.78 -0.21 -100.93 -305.20 9.7617989.87 17989.87 0.00 0.00 0.0020377.92 20377.92


101 1 -0.42 -0.16 -116.93 -289.32 6.0516820.57 16820.57 0.00 0.00 0.0019302.28 19302.28


102 1 -0.13 -0.13 -125.11 -281.19 2.0516269.07 16269.07 0.00 0.00 0.0018747.84 18747.84


103 1 0.13 -0.13 -125.11 -281.19 -2.0516269.07 16269.07 0.00 0.00 0.0018747.84 18747.84

TOP : SMAX= -8338.78 SMIN= -18747.84 TMAX= 5204.53 ANGLE= -0.8BOTT: SMAX= 18747.84 SMIN= 8338.78 TMAX= 5204.53 ANGLE= 89.2

104 1 0.42 -0.16 -116.93 -289.32 -6.0516820.57 16820.57 0.00 0.00 0.0019302.28 19302.28


198— STAAD.Pro





105 1 0.78 -0.21 -100.93 -305.20 -9.7617989.87 17989.87 0.00 0.00 0.0020377.92 20377.92


106 1 1.23 -0.23 -77.92 -328.01 -12.9219844.14 19844.14 0.00 0.00 0.0021911.49 21911.49


107 1 1.69 -0.20 -49.25 -356.37 -15.2722367.84 22367.84 0.00 0.00 0.0023808.31 23808.31


108 1 2.00 -0.08 -16.85 -388.37 -16.5325419.74 25419.74 0.00 0.00 0.0025939.96 25939.96

TOP : SMAX= -1074.52 SMIN= -25939.96 TMAX= 12432.72 ANGLE= -2.5BOTT: SMAX= 25939.96 SMIN= 1074.52 TMAX= 12432.72 ANGLE= 87.56 1 0.34 -4.36 -373.27 -32.17 -31.51

24160.11 24160.11 0.00 0.00 0.0025076.87 25076.87

TOP : SMAX= -1951.93 SMIN= -25076.87 TMAX= 11562.47 ANGLE=-84.8BOTT: SMAX= 25076.87 SMIN= 1951.93 TMAX= 11562.47 ANGLE= 5.2

18 1 0.74 -2.75 -312.19 -93.95 -29.1918798.88 18798.88 0.00 0.00 0.0021068.45 21068.45


30 1 0.67 -1.03 -257.59 -148.94 -25.1915213.03 15213.03 0.00 0.00 0.0017542.96 17542.96


42 1 0.38 -0.06 -211.95 -194.66 -20.3613792.44 13792.44 0.00 0.00 0.0015028.62 15028.62







54 1 0.12 0.39 -176.20 -230.35 -15.4014020.85 14020.85 0.00 0.00 0.0015628.28 15628.28


66 1 -0.02 0.49 -150.19 -256.25 -10.6614918.57 14918.57 0.00 0.00 0.0017154.23 17154.23


78 1 -0.09 0.36 -133.36 -273.00 -6.2315779.24 15779.24 0.00 0.00 0.0018218.24 18218.24


90 1 -0.13 0.13 -125.11 -281.19 -2.0516269.07 16269.07 0.00 0.00 0.0018747.84 18747.84


102 1 -0.13 -0.13 -125.11 -281.19 2.0516269.07 16269.07 0.00 0.00 0.0018747.84 18747.84


114 1 -0.09 -0.36 -133.36 -273.00 6.2315779.24 15779.24 0.00 0.00 0.0018218.24 18218.24


126 1 -0.02 -0.49 -150.19 -256.25 10.6614918.57 14918.57 0.00 0.00 0.0017154.23 17154.23


138 1 0.12 -0.39 -176.20 -230.35 15.4014020.85 14020.85 0.00 0.00 0.0015628.28 15628.28


200 — STAAD.Pro





150 1 0.38 0.06 -211.95 -194.66 20.3613792.44 13792.44 0.00 0.00 0.0015028.62 15028.62


162 1 0.67 1.03 -257.59 -148.94 25.1915213.03 15213.03 0.00 0.00 0.0017542.96 17542.96


174 1 0.74 2.75 -312.19 -93.95 29.1918798.88 18798.88 0.00 0.00 0.0021068.45 21068.45

TOP : SMAX= -6007.83 SMIN= -21068.45 TMAX= 7530.31 ANGLE= 82.5BOTT: SMAX= 21068.45 SMIN= 6007.83 TMAX= 7530.31 ANGLE= -7.5

186 1 0.34 4.36 -373.27 -32.17 31.5124160.11 24160.11 0.00 0.00 0.0025076.87 25076.87

TOP : SMAX= -1951.93 SMIN= -25076.87 TMAX= 11562.47 ANGLE= 84.8BOTT: SMAX= 25076.87 SMIN= 1951.93 TMAX= 11562.47 ANGLE= -5.2

**** MAXIMUM STRESSES AMONG SELECTED PLATES AND CASES ****MAXIMUM MINIMUM MAXIMUM MAXIMUM MAXIMUMPRINCIPAL PRINCIPAL SHEAR VONMISES TRESCASTRESS STRESS STRESS STRESS STRESS

2.593996E+04 -2.593996E+04 1.243272E+04 2.541974E+04 2.593996E+04PLATE NO. 97 97 97 97 97CASE NO. 1 1 1 1 1********************END OF ELEMENT FORCES********************27. FINI




202— STAAD.Pro




204 — STAAD.Pro


206— STAAD.Pro


Notes

21 Deflection Calculation Through Pushover Analysis


ReferenceHand calculation.

ProblemFigure 21-1: Cantilever member model

A transverse load P is applied to a cantilever member and increased until the member fails.

Geometry & Loading

Member length = 24 inch

Member Properties

Section = Wide flange W16X77

Material = Steel

Expected yield strength = 36 ksi


Figure 21-2: Plastic Rotation (in radians) vs. Moment (in in-kips) plot for moment hinge

Comparison

Results from STAAD.Pro

The analysis results are saved at an interval of 0.1 inch deflection of the cantilever tip.

Figure 21-3: Capacity curve (Displacement at Control Joint, inches vs. Base Shear, kips) as calculated bySTAAD.Pro

208— STAAD.Pro


Load Step Displacement in Base Shear kip

1 0 0.153

2 0.001 1.153

3 0.078 57.837

4 0.136 61.953

5 0.237 63.859

6 0.338 65.765

7 0.439 67.671

8 0.54 69.578

9 0.641 71.484

10 0.742 73.39

11 0.843 75.296

12 0.945 77.202

13 1.045 79.096

14 1.057 79.32

15 1.057 21.462

Table 21-1: Tabular form of the different points in Capacitycurve as reported by STAAD.Pro

Results from hand calculation

At Load Step 3

Equation of elastic deflection at cantilever tip

δz = PL3/3EI +PL/GAv

P = 57.837 kip, l = 24 in, E = 29000 ksi, I = 138.0 in4, G = 11154 ksi, Av = 10.4323 in2

Thus elastic deformation

δz elastic = [57.837 x 243 / (3 x 29000 x 138)] + [57.837 x 24 / (11154 x 10.4323)] = 0.066 +

0.012 = 0.078 inch

At Load Step 14

Equation of elastic deflection at cantilever tip

δz = PL3/3EI +PL/GAv

P = 79.32 kip, l = 24 inch, E = 29000 ksi, I = 138.0 inP4, G = 11154 ksi, Av = 10.4323 in2



Thus elastic deformation at cantilever tip

δz elastic= [79.32 x 243 / (3 x 29000 x 138)] + [79.32 x 24 / (11154 x 10.4323)] = 0.091 +

0.016 = 0.107 inch

Plastic rotation = 0.04 radian.

Since STAAD.Pro assumes small displacements (sinθ = θ), plastic deformation at cantilever tip

δz plastic = l x θ = 24 x 0.04 = 0.96 inch

Total deflection = δz elastic + δz plastic = 0.107 + 0.96 = 1.067 inch

Comparison

LoadStep

Force P (freeend) kips

Deflection (free end) inPercentDifferenceSTAAD.Pro

Advanced SolverHand

Calculation

3 57.837 0.078 0.078 0%

14 79.32 1.057 1.067 0.94%

Table 21-2: Comparison of results for verification problem no. 21

Note: The deflection results from STAAD shown in the above table were obtained in the post-processing mode – Pushover-Node Results page.

Pushover analysis can be done through the Advanced Analysis Engine only.

Input FileSTAAD SPACE

START JOB INFORMATION

ENGINEER DATE 20-Jul-07

END JOB INFORMATION

INPUT WIDTH 79

UNIT INCHES KIP

JOINT COORDINATES

1 0 0 0; 2 0 24 0;

MEMBER INCIDENCES

1 1 2;

DEFINE MATERIAL START

ISOTROPIC STEEL

E 29000

POISSON 0.3

210 — STAAD.Pro


DENSITY 0.000283

ALPHA 6.5e-006

DAMP 0.03

END DEFINE MATERIAL

MEMBER PROPERTY AMERICAN

1 TABLE ST W16X77

CONSTANTS

MATERIAL STEEL ALL

SUPPORTS

1 FIXED

DEFINE PUSHOVER DATA

FRAME 2

FYE 36.000000 ALL

HINGE PROPERTY MOMENT

TYPE 1

A 0 0 B 1 1 C 11.0 1.30 D 11.0 0.32 E 15.0 0.32 YM 1480 YR 0.004

IO 2.0 LS 7.0 CP 10.5

HINGE TYPE 1 MEMBER 1

GNONL 1

DISP Z 2.0 JOINT 2

LOADING PATTERN 1

LDSTEP 1

SPECTRUM PARAMETERS

DAMPING 2.0000

SC 4

SS 1.1

S1 1.6

SAVE LOADSTEP RESULT DISP 0.1

END PUSHOVER DATA

LOAD 1 LOADTYPE Gravity

SELFWEIGHT Z 1

LOAD 2 LOADTYPE Push

JOINT LOAD

2 FZ 1

PERFORM PUSHOVER ANALYSIS



FINISH



1. STAAD SPACEINPUT FILE: Ver21.STD

2. START JOB INFORMATION3. ENGINEER DATE 20-JUL-074. END JOB INFORMATION5. INPUT WIDTH 796. UNIT INCHES KIP7. JOINT COORDINATES8. 1 0 0 0; 2 0 24 09. MEMBER INCIDENCES10. 1 1 211. DEFINE MATERIAL START12. ISOTROPIC STEEL13. E 2900014. POISSON 0.315. DENSITY 0.00028316. ALPHA 6.5E-00617. DAMP 0.0318. END DEFINE MATERIAL19. MEMBER PROPERTY AMERICAN20. 1 TABLE ST W16X7721. CONSTANTS22. MATERIAL STEEL ALL23. SUPPORTS24. 1 FIXED25. DEFINE PUSHOVER DATA26. FRAME 227. FYE 36.000000 ALL28. HINGE PROPERTY MOMENT29. TYPE 130. A 0 0 B 1 1 C 11.0 1.30 D 11.0 0.32 E 15.0 0.32 YM 1480 YR 0.00431. IO 2.0 LS 7.0 CP 10.532. HINGE TYPE 1 MEMBER 133. GNONL 134. DISP Z 2.0 JOINT 235. LOADING PATTERN 136. LDSTEP 137. SPECTRUM PARAMETERS38. DAMPING 2.0000

212— STAAD.Pro


STAAD SPACE -- PAGE NO. 239. SC 440. SS 1.141. S1 1.642. SAVE LOADSTEP RESULT DISP 0.143. END PUSHOVER DATA44. LOAD 1 LOADTYPE GRAVITY45. SELFWEIGHT Z 146. LOAD 2 LOADTYPE PUSH47. JOINT LOAD48. 2 FZ 149. PERFORM PUSHOVER ANALYSIS







STAAD SPACE -- PAGE NO. 3CALCULATED FREQUENCIES FOR LOAD CASE 3

MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 306.356 0.00326

MODAL WEIGHT (MODAL MASS TIMES g) IN KIP GENERALIZEDMODE X Y Z WEIGHT

1 5.821397E-63 2.398385E-61 7.674960E-02 7.674960E-02PARTICIPATION FACTORS

MASS PARTICIPATION FACTORS IN PERCENT--------------------------------------

MODE X Y Z SUMM-X SUMM-Y SUMM-Z1 0.00 0.00100.00 0.000 0.000 100.000

***WARNING : COLUMN NO. 1 HAS FAILED IN DEFORMATION-CONTROLLED ACTION.*** WARNING : STRUCTURE HAS REACHED AN UNSTABLE STATE WHERE

A VERY SMALL INCREASE IN PUSH LOAD IS CAUSING LARGE DISPLACEMENT.50. FINISH

214 — STAAD.Pro


STAAD SPACE -- PAGE NO. 4*********** END OF THE STAAD.Pro RUN *************** DATE= JAN 25,2013 TIME= 17:17:17 ****




STAAD SPACE -- PAGE NO. 5Information about the key files in the current distributionModification Date CRC Size (Bytes) File Name-----------------------------------------------------------------------------------------12/21/2012 0x880 19054592 SProStaad.exe12/21/2012 0x6c00 08679424 SProStaadStl.exe10/18/2012 0x3940 00373248 AISCSteelDesignNew.dll09/19/2003 0x2fc0 00081970 CMesh.dll11/29/2012 0x15c1 04476998 dbSectionInterface.dll10/18/2012 0x5680 00315392 EC3Annex.dll10/18/2012 0x91c1 00421888 EuroCode3.dll10/18/2012 0xf981 00102400 IS800SteelDesign.dll01/24/2001 0x9b40 00073728 LoadGen.dll06/21/2012 0xcb81 00040960 MathExtensions.dll09/25/2003 0x6340 00704512 MeshEngine.dll11/27/2012 0x9a40 00059392 NORSOKSteelDesign.dll09/22/2003 0xce00 00069632 QuadPlateEngine.dll10/18/2012 0x6381 00079872 SteelCollection.dll10/18/2012 0xf240 00030720 SteelDesign.dll10/18/2012 0xac1 00061440 SteelSections.dll05/28/2008 0x77c1 00106572 SurfMeshEngine.dll10/18/2012 0x1dc0 00017408 UnitConversions.dll10/19/2012 0x280 00197120 WrapperAISCSteelDesignNew.dll10/19/2012 0x57c0 00019968 WrapperSteelDesign.dll10/19/2012 0x2e80 00108032 WrapperSteelSections.dll10/19/2012 0x6480 00008704 WrapperSystem.dll10/19/2012 0xe141 00016384 WrapperUnitConversions.dll08/22/2011 0x9741 00974848 AISCSections.mdb06/13/2006 0xd4c1 01204224 AISCSectionsRCeco.mdb06/05/2007 0xbc00 00331776 aiscsections_all_editions.mdb01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb01/05/2005 0xcac1 03651584 aitctimbersections.mdb05/01/2008 0xb001 00552960 aluminumsections.mdb01/13/2010 0x40 01019904 AustralianColdFormedSections.mdb10/02/2008 0x40 00520192 AustralianSections.mdb10/18/2012 0x4d41 00765952 BrazilSections.mdb05/19/2011 0x1641 00643074 BritishSections.mdb09/12/2006 0xf4c0 00466944 bscoldformedsections.mdb06/28/2005 0x8201 00327680 butlercoldformedsections.mdb02/25/2008 0x2e01 00329728 CanadianSections.mdb09/26/2006 0x5ec1 00280576 CanadianSections_old.mdb05/31/2005 0x9e81 00450560 canadiantimbersections.mdb02/10/2010 0x94c1 00483328 ChineseSections.mdb01/05/2005 0xd6c0 00600064 dutchsections.mdb08/08/2011 0xc880 02699264 EuropeanColdFormedSections.mdb10/07/2010 0xd2c0 03596290 EuropeanSections.mdb01/25/2011 0xc901 00692224 EuropeanSectionsLegacy.mdb01/05/2005 0xd301 00202752 frenchsections.mdb01/05/2005 0x11c1 00233472 germansections.mdb10/06/2010 0x1bc1 00393216 GOSTColdFormedSections.mdb12/08/2009 0x6ac0 00569344 IndianLegacySections.mdb08/13/2012 0x6281 00942082 IndianSections.mdb01/05/2005 0xd540 00180224 iscoldformedsections.mdb02/21/2008 0xc380 00327680 JapaneseColdFormedSections.mdb08/02/2011 0x2b80 00593920 JapaneseSections.mdb07/15/2011 0xce40 00532480 JindalSections.mdb11/09/2005 0x9081 00376832 Kingspancoldformedsections.mdb08/24/2007 0x3ec0 00454656 Koreansections.mdb04/24/2007 0x99c0 00114688 lysaghtcoldformedsections.mdb02/07/2005 0x9a00 00243712 mexicansteeltables.mdb04/19/2007 0x9ec1 00421888 RCecoColdFormedSections.mdb11/20/2009 0xf700 00428034 RussianSections.mdb05/12/2009 0xe001 00237570 SouthAfricanSections.mdb01/06/2005 0x9341 00194560 spanishsections.mdb11/20/2009 0x5100 00368640 STO_ASChMSections.mdb12/20/2011 0x2b41 00409600 TATAStructuraSections.mdb

216— STAAD.Pro


218— STAAD.Pro


Notes

22 Fixed – Fixed Beam Excited by Harmonic Distributed Load

ObjectiveTo calculate the deflections at two points along the beam at steady-state condition.

Reference

1. Blevins, R. D., Formulas for natural Frequency and Mode Shape, Van Nostrand Reinhold,1979, pp. 108, 455-486.

2. Warburton, G. B., The Dynamical Behavior of Structures, Pergamon Press, 1964, pp. 10-15, 85,86.

ProblemDetermine the steady-state displacements of the quarter and mid-span points of a fixed-fixed beamsubjected to a parabolically varying distributed load operating at a 7.5 Hz frequency.

Figure 22-1: Beam with harmonic distributed load

Figure 22-2: Model: divide span [email protected]”


GivenE = 10.0x106 psi

L = 200 inches

I = 2/3 in4

A = 2 in2

Po = 0.1 lbf/in3

g = 386.4 in/sec2

The DYNRE2 run utilizes all 19 modes calculated by STAAD (Steady State analysis) for which avalue of 1.0(10)-10 times critical damping is assigned. A single forcing frequency equal to 7.5 Hz isspecified for the distributed load. This load is distributed to the nodes by calculating the totalintegrated load for each beam and lumping one-half of this force to the respective i and j nodes.

∫ ∫= = = −F F d x xl x d x( )i j x

x P x

x

x

l

( )

2

1

2

4 2

i

j

i

j2

= = − = −− −

F Fi jx

l

x

lx

xx x x x2

3 200 60, 000

i

jj i j i

2 3

2

2 2 3 3

Theoretical SolutionThe theoretical solution for this example is taken from:

1. Blevins, R. D., “Formulas for natural Frequency and Mode Shape,” Van Nostrand Reinhold,1979, pp 108, 455-486.

2. Warburton, G. B., “The Dynamical Behavior of Structures,” Pergamon Press, 1964, pp. 10-15,85, 86.

The natural frequencies of the system are calculated using the equations from reference 1 page 108and reference 2 page 85.

=fi

λ

π

EI

m2

1

2

2

Where:

m = ρA/g

=fi

λ

π

( )2 (200)

1.0(10) 2 / 3

0.10(2.0 / 386.4)

i2

2

6

fi = 4.5156271*10-1λi2

λisatisfies the characteristic equation:

cosλ coshλ - 1 = 0

i λi

ωi

fi

1 4.730041 63.47865 10.10294

Table 22-1: Calculated natural frequency for each mode

220 — STAAD.Pro


i λi

ωi

fi

2 7.853205 174.9814 27.84915

3 10.99561 343.0334 54.59546

4 14.13717 567.0517 90.24907

5 17.27876 847.0773 134.8165

6 20.42035 1183.108 188.2975

7 23.56194 1575.144 250.6919

8 26.70354 2023.185 321.9998

The mode shapes are:

= − − −ϕ σ ( )cosh cos sinh sini

λ x λ xi

λ x λ xi i i i

Where:

= −−

σiλ λ

λ λ

cosh cos

sinh sin

i i

i i

The response of mode i to a harmonic force:

= −∫

∫

−

−

+

ηi

ϕ x P x d x

ω ϕ x m d x

ωt ψ

( )

( ) ( )

{ ( )}

( )sin

1

l i

i i

i

ω

ω i

cω

k

2 22

2

22

Where ψi is the response phase lag relative to the applied force and c is the damping.

Since c = 0.0, ψi = 0.0

Upon substitution and rearranging terms:

=∫

∫

−

−

η ωtsini

xl x ϕ x d x

ω ω ϕ x d x{ ( )}

lPo

li

ρA

g i l i

4

2

2

2 2 2

From reference 1, page 466, case c and page 467, case 29:

∫ ∫− =

+ − − −

−

−

−xϕ x d x x ϕ x d x i i σ β l σ β l( )( ) 1 ( ) ( ) 2

P

l l i

P

l l i

P

β l

i ii i

l P

β li i

4 4 2 8 8( )o o

i

i

i2 2

02

02

Since the load is symmetric, this expression is zero for I = 2, 4, 6 …;

Therefore, i = 1,3,5……, and: (-1)i = -1

And from the reference, βi = λi/l

So:

∫ =ϕ x P x d x( ) ( )l i

P l

λ

16o

i

2

From Reference 1, page 457 case 5:

∫ =ϕ x d x{ ( )} 1l i

2



Therefore:

ω = 7.5(2π) = 47.1239 radians/sec

= = =−

−

−

⋅

−η ωtsini

P

mλ ω ω

ωt

λ ω

ωt

λ ω( )( ) ( )16 16( 1.0)sin

2, 220.661

30, 912 sin

2, 220.661

o

i i i i i i2 2 2 0.10(2)

386.4

2 2 2 2

i λi

ωi

ηi(t) σ

iφ(1/4) φ(1/2)

1 4.73004-1

63.4786-5

− −ωt7.63815(10) sin

1 0.9825022 0.8631319 1.5881463

3 10.99561 343.033-4

− −ωt2.21457(10) sin

3 0.999966-4

1.3708047 -1.405998-

4

5 17.27876 847.077-3

− −ωt1.44744(10) sin

4 0.999999-9

-0.527889-

7

1.4145675

7 23.5619-4

1575.144 − −ωt2.24623(10) sin

5 1.0000000 -1.3037973

-1.4141982

Table 22-2: Mode shapes

= ∑ =y x t η t ϕ x( ) ( ) ( ), i i i1, 3, 5, 7

i φ(¼)ηi(t) φ(1/2)η

i(t)

1 -0.6592727 -1.2130493

3 -0.0030357 0.0031137

5 0.0000764 -0.0002048

7 0.0000293 -0.0000318

Summation -0.6622028 -1.2101086

Table 22-3: Steady state displacements for 1/4 and 1/2 points (nodes 6 and 11,respectively)

Comparison

Location TheorySTAAD

Advanced Solver

Node 6 (X = 50 inches) 0.66220 0.65964

Node 11 (X = 100 inches) 1.21011 1.20545

Table 22-4: Comparison of steady-state displacements of the quarter (Node 6)and mid-span points (Node 11) for verification problem no. 22

222— STAAD.Pro


Input FileSTAAD SPACE

START JOB INFORMATION

ENGINEER DATE 29-Mar-06

END JOB INFORMATION

* FIXED BEAM SUBJECTED TO A HARMONIC LOAD WITH A PARABOLIC DISTRIBUTION X

* NUMBER OF NODES 21 X

* HIGH NODE NUMBER 21 X

* NODES FULLY RESTRAINED 2 X

* NUMBER OF BEAM ELEMENTS 20 X

* NUMBER OF EIGENVECTORS 19

SET SHEAR

UNIT INCHES POUND

JOINT COORDINATES

1 0 0 0; 2 10 0 0; 3 20 0 0; 4 30 0 0; 5 40 0 0; 6 50 0 0; 7 60 0 0;

8 70 0 0; 9 80 0 0; 10 90 0 0; 11 100 0 0; 12 110 0 0; 13 120 0 0;

14 130 0 0; 15 140 0 0; 16 150 0 0; 17 160 0 0; 18 170 0 0; 19 180 0 0;

20 190 0 0; 21 200 0 0;

MEMBER INCIDENCES

1 1 2; 2 2 3; 3 3 4; 4 4 5; 5 5 6; 6 6 7; 7 7 8; 8 8 9; 9 9 10;

10 10 11; 11 11 12; 12 12 13; 13 13 14; 14 14 15; 15 15 16; 16 16 17;

17 17 18; 18 18 19; 19 19 20; 20 20 21;

MEMBER PROPERTY AMERICAN

1 TO 20 PRIS AX 2 AY 0 AZ 0 IX 0.001 IY 0.666667 IZ 0.166667

SUPPORTS

2 TO 20 FIXED BUT FY MZ

1 21 FIXED

DEFINE MATERIAL START

ISOTROPIC MATERIAL1

E 1e+007

POISSON 0.3

DENSITY 0.0999194

END DEFINE MATERIAL

CONSTANTS



BETA 90 ALL

MATERIAL MATERIAL1 MEMB 1 TO 20


CUT OFF FREQUENCY 500

LOAD 1

SELFWEIGHT X 1

SELFWEIGHT Y 1

SELFWEIGHT Z 1

MODAL CALCULATION REQUESTED

PERFORM STEADY STATE ANALYSIS

BEGIN STEADY FORCE

STEADY FORCE FREQ 7.5 DAMP 1e-010

JOINT LOAD

2 FY 1.8666

3 FY 3.5666

4 FY 5.0666

5 FY 6.3666

6 FY 7.4666

7 FY 8.3666

8 FY 9.0666

9 FY 9.5666

10 FY 9.8666

11 FY 9.9666

12 FY 9.8666

13 FY 9.5666

14 FY 9.0666

15 FY 8.3666

16 FY 7.4666

17 FY 6.3666

18 FY 5.0666

19 FY 3.5666

20 FY 1.8666

END

PRINT JOINT DISPLACEMENTS LIST 6 11

FINISH

224 — STAAD.Pro




1. STAAD SPACEINPUT FILE: Ver22.STD

2. START JOB INFORMATION3. ENGINEER DATE 29-MAR-064. END JOB INFORMATION5. * FIXED BEAM SUBJECTED TO A HARMONIC LOAD WITH A PARABOLIC DISTRIBUTION X6. * NUMBER OF NODES 21 X7. * HIGH NODE NUMBER 21 X8. * NODES FULLY RESTRAINED 2 X9. * NUMBER OF BEAM ELEMENTS 20 X10. * NUMBER OF EIGENVECTORS 1911. SET SHEAR12. UNIT INCHES POUND13. JOINT COORDINATES14. 1 0 0 0; 2 10 0 0; 3 20 0 0; 4 30 0 0; 5 40 0 0; 6 50 0 0; 7 60 0 015. 8 70 0 0; 9 80 0 0; 10 90 0 0; 11 100 0 0; 12 110 0 0; 13 120 0 016. 14 130 0 0; 15 140 0 0; 16 150 0 0; 17 160 0 0; 18 170 0 0; 19 180 0 017. 20 190 0 0; 21 200 0 018. MEMBER INCIDENCES19. 1 1 2; 2 2 3; 3 3 4; 4 4 5; 5 5 6; 6 6 7; 7 7 8; 8 8 9; 9 9 1020. 10 10 11; 11 11 12; 12 12 13; 13 13 14; 14 14 15; 15 15 16; 16 16 1721. 17 17 18; 18 18 19; 19 19 20; 20 20 2122. MEMBER PROPERTY AMERICAN23. 1 TO 20 PRIS AX 2 AY 0 AZ 0 IX 0.001 IY 0.666667 IZ 0.16666724. SUPPORTS25. 2 TO 20 FIXED BUT FY MZ26. 1 21 FIXED27. DEFINE MATERIAL START28. ISOTROPIC MATERIAL129. E 1E+00730. POISSON 0.331. DENSITY 0.099919432. END DEFINE MATERIAL33. CONSTANTS34. BETA 90 ALL35. MATERIAL MATERIAL1 MEMB 1 TO 2036. CUT OFF MODE SHAPE 737. CUT OFF FREQUENCY 50038. LOAD 1



STAAD SPACE -- PAGE NO. 239. SELFWEIGHT X 140. SELFWEIGHT Y 141. SELFWEIGHT Z 142. MODAL CALCULATION REQUESTED43. PERFORM STEADY STATE ANALYSIS





226— STAAD.Pro


STAAD SPACE -- PAGE NO. 3CALCULATED FREQUENCIES FOR LOAD CASE 1

MODE FREQUENCY(CYCLES/SEC) PERIOD(SEC)1 10.103 0.098982 27.849 0.035913 54.591 0.018324 90.230 0.011085 134.747 0.007426 188.093 0.005327 250.166 0.00400


1 0.000000E+00 2.759074E+01 0.000000E+00 1.584634E+012 0.000000E+00 5.211133E-28 0.000000E+00 1.763389E+013 0.000000E+00 5.287477E+00 0.000000E+00 1.757978E+014 0.000000E+00 8.411011E-30 0.000000E+00 1.788088E+015 0.000000E+00 2.138447E+00 0.000000E+00 1.901368E+016 0.000000E+00 1.991622E-30 0.000000E+00 1.837854E+017 0.000000E+00 1.145139E+00 0.000000E+00 1.756698E+01


MODE X Y Z SUMM-X SUMM-Y SUMM-Z1 0.00 72.67 0.00 0.000 72.666 0.0002 0.00 0.00 0.00 0.000 72.666 0.0003 0.00 13.93 0.00 0.000 86.591 0.0004 0.00 0.00 0.00 0.000 86.591 0.0005 0.00 5.63 0.00 0.000 92.223 0.0006 0.00 0.00 0.00 0.000 92.223 0.0007 0.00 3.02 0.00 0.000 95.239 0.000

44. BEGIN STEADY FORCE45. STEADY FORCE FREQ 7.5 DAMP 1E-01046. JOINT LOAD



STAAD SPACE -- PAGE NO. 447. 2 FY 1.866648. 3 FY 3.566649. 4 FY 5.066650. 5 FY 6.366651. 6 FY 7.466652. 7 FY 8.366653. 8 FY 9.066654. 9 FY 9.566655. 10 FY 9.866656. 11 FY 9.966657. 12 FY 9.866658. 13 FY 9.566659. 14 FY 9.066660. 15 FY 8.366661. 16 FY 7.466662. 17 FY 6.366663. 18 FY 5.066664. 19 FY 3.566665. 20 FY 1.866666. END

228— STAAD.Pro


STAAD SPACE -- PAGE NO. 57 MODES (EIGENVECTORS) HAVE BEEN SELECTED.

MODE NATURAL FREQUENCY GENERALIZED WEIGHT DAMPING DAMPED FREQUENCYNO. (HZ) (RAD/SEC) (WEIGHT) (MASS) COEFFICIENT (HZ)

1 1.010292E+01 6.347852E+01 1.584634E+01 4.104327E-02 1.000000E-10 1.010292E+012 2.784865E+01 1.749782E+02 1.763389E+01 4.567317E-02 1.000000E-10 2.784865E+013 5.459144E+01 3.430081E+02 1.757978E+01 4.553302E-02 1.000000E-10 5.459144E+014 9.022966E+01 5.669297E+02 1.788088E+01 4.631289E-02 1.000000E-10 9.022966E+015 1.347470E+02 8.466406E+02 1.901368E+01 4.924693E-02 1.000000E-10 1.347470E+026 1.880927E+02 1.181821E+03 1.837854E+01 4.760188E-02 1.000000E-10 1.880927E+027 2.501660E+02 1.571840E+03 1.756698E+01 4.549988E-02 1.000000E-10 2.501660E+02

PARTICIPATION FACTORS FOR EACH MODEMODE NO. X Y Z MX MY MZ

1 0.000000E+00 0.218577E+04 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+002 0.000000E+00 0.781939E-11 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+003 0.000000E+00 0.382113E+03 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+004 0.000000E+00 -0.153455E-11 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+005 0.000000E+00 0.148269E+03 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+006 0.000000E+00 -0.357590E-12 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+007 0.000000E+00 0.829074E+02 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00

67. PRINT JOINT DISPLACEMENTS LIST 6 11JOINT DISPLACE LIST 6



STAAD SPACE -- PAGE NO. 6JOINT DISPLACEMENT (INCH RADIANS) STRUCTURE TYPE = SPACE------------------

JOINT LOAD X-TRANS Y-TRANS Z-TRANS X-ROTAN Y-ROTAN Z-ROTAN6 1 0.00000 0.65963 0.00000 0.00000 0.00000 0.0183111 1 0.00000 1.20545 0.00000 0.00000 0.00000 0.00000



230 — STAAD.Pro


STAAD SPACE -- PAGE NO. 7************************************************************* For questions on STAAD.Pro, please contact ** Bentley Systems or Partner offices ** ** Telephone Web / Email ** USA +1 (714) 974-2500 ** UK +44 (0) 808 101 9246 ** SINGAPORE +65 6225-6158 ** FRANCE +33 (0) 1 55238400 ** GERMANY +49 0931 40468 ** INDIA +91 (033) 4006-2021 ** JAPAN +81 (03)5952-6500 http://www.ctc-g.co.jp ** CHINA +86 21 6288 4040 ** THAILAND +66 (0)2645-1018/19 [email protected]** ** Worldwide http://selectservices.bentley.com/en-US/ ** *************************************************************



STAAD SPACE -- PAGE NO. 8Information about the key files in the current distributionModification Date CRC Size (Bytes) File Name-----------------------------------------------------------------------------------------12/21/2012 0x880 19054592 SProStaad.exe12/21/2012 0x6c00 08679424 SProStaadStl.exe10/18/2012 0x3940 00373248 AISCSteelDesignNew.dll09/19/2003 0x2fc0 00081970 CMesh.dll11/29/2012 0x15c1 04476998 dbSectionInterface.dll10/18/2012 0x5680 00315392 EC3Annex.dll10/18/2012 0x91c1 00421888 EuroCode3.dll10/18/2012 0xf981 00102400 IS800SteelDesign.dll01/24/2001 0x9b40 00073728 LoadGen.dll06/21/2012 0xcb81 00040960 MathExtensions.dll09/25/2003 0x6340 00704512 MeshEngine.dll11/27/2012 0x9a40 00059392 NORSOKSteelDesign.dll09/22/2003 0xce00 00069632 QuadPlateEngine.dll10/18/2012 0x6381 00079872 SteelCollection.dll10/18/2012 0xf240 00030720 SteelDesign.dll10/18/2012 0xac1 00061440 SteelSections.dll05/28/2008 0x77c1 00106572 SurfMeshEngine.dll10/18/2012 0x1dc0 00017408 UnitConversions.dll10/19/2012 0x280 00197120 WrapperAISCSteelDesignNew.dll10/19/2012 0x57c0 00019968 WrapperSteelDesign.dll10/19/2012 0x2e80 00108032 WrapperSteelSections.dll10/19/2012 0x6480 00008704 WrapperSystem.dll10/19/2012 0xe141 00016384 WrapperUnitConversions.dll08/22/2011 0x9741 00974848 AISCSections.mdb06/13/2006 0xd4c1 01204224 AISCSectionsRCeco.mdb06/05/2007 0xbc00 00331776 aiscsections_all_editions.mdb01/05/2005 0x4b81 01810432 aiscsteeljoists.mdb01/05/2005 0xcac1 03651584 aitctimbersections.mdb05/01/2008 0xb001 00552960 aluminumsections.mdb01/13/2010 0x40 01019904 AustralianColdFormedSections.mdb10/02/2008 0x40 00520192 AustralianSections.mdb10/18/2012 0x4d41 00765952 BrazilSections.mdb05/19/2011 0x1641 00643074 BritishSections.mdb09/12/2006 0xf4c0 00466944 bscoldformedsections.mdb06/28/2005 0x8201 00327680 butlercoldformedsections.mdb02/25/2008 0x2e01 00329728 CanadianSections.mdb09/26/2006 0x5ec1 00280576 CanadianSections_old.mdb05/31/2005 0x9e81 00450560 canadiantimbersections.mdb02/10/2010 0x94c1 00483328 ChineseSections.mdb01/05/2005 0xd6c0 00600064 dutchsections.mdb08/08/2011 0xc880 02699264 EuropeanColdFormedSections.mdb10/07/2010 0xd2c0 03596290 EuropeanSections.mdb01/25/2011 0xc901 00692224 EuropeanSectionsLegacy.mdb01/05/2005 0xd301 00202752 frenchsections.mdb01/05/2005 0x11c1 00233472 germansections.mdb10/06/2010 0x1bc1 00393216 GOSTColdFormedSections.mdb12/08/2009 0x6ac0 00569344 IndianLegacySections.mdb08/13/2012 0x6281 00942082 IndianSections.mdb01/05/2005 0xd540 00180224 iscoldformedsections.mdb02/21/2008 0xc380 00327680 JapaneseColdFormedSections.mdb08/02/2011 0x2b80 00593920 JapaneseSections.mdb07/15/2011 0xce40 00532480 JindalSections.mdb11/09/2005 0x9081 00376832 Kingspancoldformedsections.mdb08/24/2007 0x3ec0 00454656 Koreansections.mdb04/24/2007 0x99c0 00114688 lysaghtcoldformedsections.mdb02/07/2005 0x9a00 00243712 mexicansteeltables.mdb04/19/2007 0x9ec1 00421888 RCecoColdFormedSections.mdb11/20/2009 0xf700 00428034 RussianSections.mdb05/12/2009 0xe001 00237570 SouthAfricanSections.mdb01/06/2005 0x9341 00194560 spanishsections.mdb11/20/2009 0x5100 00368640 STO_ASChMSections.mdb12/20/2011 0x2b41 00409600 TATAStructuraSections.mdb

232— STAAD.Pro


234 — STAAD.Pro


Notes

Lists of Tables and Figures

Tables

Table 1-1: Comparison of Support Reaction, in kips, for verification problem no. 1 1

Table 2-1: Comparison of period, in sec., for verification problem no. 2 9

Table 3-1: Comparison of max. defl. and max. moment for verification problem no. 3 18

Table 4-1: Comparison of reaction, in kips, for verification problem no. 4 27

Table 5-1: Comparison of stress (σ), psi, and Deflection (δ), in. for verification model no. 5 35

Table 6-1: Comparison of moment, in kip-ft, for verification problem no. 6 44

Table 7-1: Comparison of deflection (δ) for verification problem no. 7 52

Table 8-1: Comparison of max. moment, in kip-ft, for verification model no. 8 59

Table 9-1: Comparison of max. moment, in kip-ft, for verification problem no. 9 68

Table 10-1: Comparison of max. Force, F, and max. Moments,M, for verification problemno. 10 75

Table 11-1: Comparison of stress (σ) for verification model #11 86

Table 12-1: Comparison of stress (σ) and Deflection (δ) for verification problem 12 93

Table 13-1: Comparison of governing ratios for the member checks in verification problemno. 13 103

Table 14-1: Comparison of concrete designs for verifcation problem no. 14 120

Table 15-1: Comparison of frequency values for verification problem no. 15 131

Table 16-1: Values of λ2ij 141

Table 16-2: Comparison of frequency values, Hz, for verification problem no. 16 141

Table 17-1: Values of λ2ij 153

Table 17-2: Comparison of frequency values, Hz, for verificatoin problem no. 17 153

Table 18-1: Comparison of natural frequency values (cycles per second) for verificationproblem no. 18 164

Table 19-1: Comparison of max. moment in beam 5 for verification problem no. 19 179

Table 20-1: Comparison of deflections along X = 6, in inches, for verification problem no.20 190

Table 20-2: Comparison of deflections along Y = 1, in inches, for verification problem no.20 191

Table 20-3: Comparison of bending moments along Y = 0.5, in in-lbs/in, for verificationproblem no. 20 192

Table 20-4: Comparison of bending moments along X = 5.5, in in-lbs/in, for verificationproblem no. 20 193


Table 21-1: Tabular form of the different points in Capacity curve as reported bySTAAD.Pro 209

Table 21-2: Comparison of results for verification problem no. 21 210

Table 22-1: Calculated natural frequency for each mode 220

Table 22-2: Mode shapes 222

Table 22-3: Steady state displacements for 1/4 and 1/2 points (nodes 6 and 11,respectively) 222

Table 22-4: Comparison of steady-state displacements of the quarter (Node 6) and mid-span points (Node 11) for verification problem no. 22 222

Figures

Figure 1-1: Plane truss 1

Figure 2-1: Beam supported on springs 9

Figure 3-1: Finite element mesh of cantilevered plate. 17

Figure 4-1: Cantilever model 27

Figure 5-1: Locomotive axle model 35

Figure 6-1: 1x1 bay plane frame 43

Figure 7-1: Plane truss model 51

Figure 8-1: Unequal leg bay model 59

Figure 9-1: Two story frame model 67

Figure 10-1: Space frame model 75

Figure 11-1: Model of a rigid wire suspended by wires 85

Figure 12-1: Model of two member truss 93

Figure 13-1: Steel braced frame frame 103

Figure 14-1: Concrete moment frame 119

Figure 15-1: Simple beam diagram 129

Figure 15-2: Finite element model 129

Figure 16-1: Diagram of circular plate 139

Figure 16-2: Finite model for half circular plate 140

Figure 17-1: Simply supported, rectangular plate 151

Figure 17-2: Model: finite element meshed 152

Figure 18-1: Frame model 163

Figure 19-1: Simple span beam 177

Figure 19-2: Finite element model 178

Figure 20-1: Rectangular plate model 189

236— STAAD.Pro

Figure 21-1: Cantilever member model 207

Figure 21-2: Plastic Rotation (in radians) vs. Moment (in in-kips) plot for moment hinge 208

Figure 21-3: Capacity curve (Displacement at Control Joint, inches vs. Base Shear, kips) ascalculated by STAAD.Pro 208

Figure 22-1: Beam with harmonic distributed load 219

Figure 22-2: Model: divide span [email protected]” 219


238— STAAD.Pro

Notes

Bentley Systems, Incorporated

685 Stockton Drive, Exton, PA 19341 USA

+1 (800) 236-8539

www.bentley.com

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