ocument . r str-calc-548 0 · project amen date sample project in the middle east title revision...

LETTENWEG 118, 4123 ALLSCHWIL, SWITZERLAND TEL.: +41 (0)61 501 8210 / +41 (0)79 508 1651 REGISTERED IN SWITZERLAND │VAT REGISTRATION CHE-437.605.665

PROJECT ENGINEER

SAMPLE PROJECT IN THE MIDDLE EAST DOCUMENT NO. REVISION

STR-CALC-548 0 TITLE Pages

UNITISED CURTAIN WALL 117

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SAMPLE PROJECT IN THE MIDDLE EAST TITLE REVISION PAGES

UNITISED CURTAIN WALL 2 of 117


Table of Contents

1 Summary 3

2 Basic Data 4 2.1 Standards and References 4

2.2 Materials 5

2.3 Performance Criteria 5

2.4 Programs used for the structural analysis 5

3 Layout and Dimensions 6 3.1 Key Location: Wall Type 3 6

3.2 Façade Elevation: Wall Type 3 6

4 Design Narrative 7 4.1 Load Path 7

4.2 Coupled Mullion 8

4.3 Mullion Structural Systems 10

4.4 Critical Panel Evaluation 13

4.5 CW In-plane Performance 14

5 Loading 15 5.1 Dead Load, QD 15

5.2 Wind Load, QW 16

6 Section Properties – Structural System 1 17 6.1 Mullion Profiles 17

6.2 Stack Joint Transom Profiles 26

6.3 Transom – 1 Profile 34


6.5 Sword Profile 42

7 Section Properties – Structural System 2 43 7.1 Mullion Profiles 43

7.2 Stack Joint Transom Profiles 44



7.5 Sword Profile 56

8 Analysis & Code Check – Structural System 1 57 8.1 Mullion Check 57

8.2 Check Stack Joint – Header and Sill Transom 67

8.3 Transom – 1 (Type 1) 70


8.5 Check Stainless Steel Spandrel Panel 74

9 Analysis & Code Check – Structural System 2 77 9.1 Mullion Check 77

9.2 Check Stack Joint – Header and Sill Transom 87



9.5 Check Stainless Steel Spandrel Panel 94

10 Joints & Connections 97 10.1 Mullion Shear Connectors 97

10.2 Header Transom Connection Check 103

10.3 Sill Transom Connection Check 105

10.4 Transom - 1 Connection Check 107

10.5 Transom - 2 (System – 1) Connection Check 109

10.6 Transom - 2 (System – 2) Connection Check 111

10.7 Sword Connection Check 113

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10.8 Back Pan Check 114

1 SUMMARY

This set of structural calculation covers only Wall Type 3 curtain wall system. This includes the typical units except

those located at the corners in which the structural calculations are submitted under a separate cover.

The results of the system analyses and all associated structural calculations in the succeeding sections of this report

are summarized as follows:

Table 1 Results Summary

Element Material Stress

Utilization Deflection Utilization

Critical Section

Referenc

e

Male Mullion 6063-T6 98 % 98 % Biaxial bending §8.1

Female Mullion 6063-T6 43 % 62 % Bending §8.1

Stack Joint - Header 6063-T6 65 % 38 % Biaxial bending §9.2.2

Stack Joint - Sill 6063-T6 35 % 38 % Bending §9.2.1

Transom 1 (Type1) 6063-T6 43 % 45 % Biaxial bending §8.3.1




Sword (Type 1) 6063-T6 8 % - No bending (only

shear)

§8.1.5

Sword (Type 2) 7022-T651 51 % - Biaxial bending §9.1

Thermal Break Polyamide 95 % - Shear §9.3.2

Connections - Screw ST4.8 A4-80 87 % - Shear §10.7.3

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2 BASIC DATA

2.1 Standards and References

2.1.1 Codes and Standards

Table 2.1-1 Codes and Standards

Nr. Code/Standard Title

1 ASTM F593 Standard Specification for Stainless Steel Bolts, Hex Cap Screws, and Studs

2 ASTM 1300 Standard Practice for Determining Load Resistance of Glass in Buildings

3 ADM 2005 The Aluminum Association Inc. "Aluminum Design Manual",

4 AAMA TIR-A8-04 Structural Performance of Composite Thermal Barrier Framing Systems

5 AAMA TIR-A9-1991 Metal Curtain Wall Fasteners

6 EN ISO 1478:1999 Tapping screw thread

7 ISO 3506-1:1997 Mechanical properties of corrosion-resistant stainless-steel fasteners -Part 1: Bolts,

screws and studs

8 ISO 7049: 1983 Cross recessed pan head tapping screws

2.1.2 Document reference

Table 2.1-2 Project Specification by Skidmore, Owings & Merrill LLP (SOM)

Nr. Document Title

1 SOM 05120 Section 05120 - Structural Steel - Part 1 - General

2 SOM 05310 Section 05310 - Steel deck - Part 1 - General

3 SOM 05460 Section 05460 - Strongback metal framing - Part 1 - General

4 SOM 08050 Section 08050 - Exterior wall - Part 1 - General

5 SOM 08800 Section 08800 - Glazing - Part 1 - General

6 SOM 08900 Section 08900 - Window wall - Part 1 - General

7 SOM 10210 Section 10210 - Wall louvers - Part 1 - General

2.1.3 Drawing reference

Table 2.1-4 Drawings by Skidmore, Owings & Merrill LLP (SOM)

Nr. Drawing Nr. Title

1 AL-00-AR-502 Rev 009 Building enclosure wind pressure diagrams (addendum 051)

Table 2.1-5 Schmidlin Drawings

Nr. Drawing Nr. Title

1 SD-001 Wall type 1: SD-001 Rev 00

2 SD-102 Wall type 1-5: SD-102 Rev 00






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2.2 Materials

2.2.1 Material Properties

Table 2.2-1 Properties of Aluminum

Material Modulus

of Elasticity,

E (N/mm2)

Modulus

of Elasticity for

deflection,

E (N/mm2)

Min. Tensile Yield

Strength,

Fty (N/mm2)

Min. Tensile

Ultimate

Strength,

Ftu (N/mm2)

Reference

Extrusion

6063-T6

69600 68900 170 205 ADM 2005 Table 3.3-

1M

Sheets

5005-H14

69600 68900 115 145 ADM 2005 Table 3.3-

1M

Sword - 1

6063-T6

69600 68900 170 205 ADM 2005 Table 3.3-

1M

Sword - 2

7022-T651

72000 72000 490 520 EN - AW7022

Table 2.2-2 Properties of Stainless Steel

Material Modulus

of Elasticity,

E (N/mm2)

Modulus

of Elasticity for

deflection,

E (N/mm2)

Min. Tensile Yield

Strength,

Fty (N/mm2)

Min. Tensile

Ultimate

Strength,

Ftu (N/mm2)

Reference

Gr. 1.4404 193 000 193 000 170 485 ASTM A240

Table 2.2-3 Properties of Screws

Material Modulus

of Elasticity,

E (N/mm2)

Modulus

of Elasticity for

deflection,

E (N/mm2)

Min. Tensile Yield

Strength,

Fty (N/mm2)

Min. Tensile

Ultimate

Strength,

Ftu (N/mm2)

Reference

A2/A4-701

(S31603)

- - 450 700 EN ISO 3506-1

(ASTM A240)

Note: 1A4 austenitic steel is equivalent to S31603 (low carbon-316L) based on similar chemical properties in ASTM A240

and EN ISO 3506-1,

2.3 Performance Criteria

The deflection limits are based on the performance requirements stated in SOM 08900 Project Specification on

Window Wall.

Table 2.3-1 Deflection Limits [§1.3.B. Deflections, SOM 08900]

Criteria Member Glass Entire Assembly

Span ≤ 4800mm 1/175 of the clear span, or 19mm 25mm 38mm

Span > 4800mm 1/250 of the clear span, or 38mm 25mm 63mm

Parallel to wall plane 75% of the clearance - -

Table 2.3-2 Deflection Limits [Others]

Criteria Member

In-plane deflection 75% clearance

2.4 Programs used for the structural analysis

Table 2.4-1 Computer Software

Criteria Usage Description

MS Office Excel 2003 Checking profile Spreadsheet

AutoCAD 2004 Obtaining section properties Drafting

ANSYS 2004 Finite Element Analysis ANSYS (R) Release 9.OA1 UP20050128, ANSYS (R)

Mechanical Toolbar Release 9.OA1 UP20050128,

Copyright 2004 SAS IP, Inc.

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3 LAYOUT AND DIMENSIONS

The typical units of Wall Type 3, has a general height of 3.70m. For the width of the units, refer to Table 4.4-2 on

Critical Panel Evaluation.

3.1 Key Location: Wall Type 3

Wall Type 3 is located at levels 7, 8, 25, 33, 37, 52 & 63 indicated by ( ) mark.

Fig. 3.1-1 Wall Type 3 Key Location

3.2 Façade Elevation: Wall Type 3

(a) Structural System - 1 (b) Structural System - 2

Fig. 3.2-2 Wall Type 3 Layout

Stack Jojnt

Header Transom

Transom – 1

(Type 1)

Transom – 2

(Type 1)

Stack Joint

Sill Transom

Mullion

Male & Female

Stack Jojnt

Header Transom

Transom – 1

(Type 2)

Transom – 2

(Type 2)

Stack Joint

Sill Transom

Mullion

Male & Female

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4 DESIGN NARRATIVE

4.1 Load Path

4.1.1 Vertical Loads

Vertical loads, due to the weight of the curtain wall unit including all attachments, are transferred equally by the

vertical mullions on both sides of the CW unit. Each mullion has a bracket that is mated to another brackets that are

mounted on the floor slab of the base building.

Fig. 4.1-1 Dead Load Supports

4.1.2 Lateral Loads

Lateral load, due to wind load applied perpendicular to the face of the unit, is (1) transferred through the mullions

and to the brackets that are mounted on the floor slab of the base building (see Fig. 3). Part of the loads (2) are

transferred directly to the bracket and the other part goes through the sword and finally to the bracket on CW unit

below.

WIND LOAD

WIND LOAD

Fig. 4.1-2 Wind Load Supports

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4.2 Coupled Mullion

The Mullions consist of a coupled pair of two distinct aluminum profiles which are so called male and female parts

(see Figure below). A continuous part of the male profile penetrates a continuous groove part of the female to serve

as shear connector in coupling them together in resisting bending due to lateral loads.

Fig. 4.2-1 Couple Mullion Profiles

(a) Deflection. Given the fact that with the shear connector the pair of mullion work together in bending about the minor-axis, both parts will deflect along the major-axis, at the same rate

MALE FEMALEδ δ=

(b) Bending Stiffness. Since both parts have equal deflection, the stiffness of both parts are responsible for the load carrying capacity. Assuming that the longitudinal shear, due to friction of the shear connector contact with the

female groove, is negligible, the total stiffness, EI of the coupled mullion setup is given as:

TOTAL MALE FEMALEEI EI EI= +

(c) Load Sharing. Each part of the mullion setup share a certain amount of the internal moment in proportion to their stiffness:

MALE

MALE TOTAL

TOTAL

EIM M

EI

=

FEMALE

FEMALE TOTAL

TOTAL

EIM M

EI

=

(d) Effective Section Modulus. The effective section properties of the mullion profile is predicted by adopting the procedure outlined in AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing Systems.

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(e) Determining the Design Value of Gc.

The procedure below outlines the determination of the design shear modulus (Gc) of the thermal barrier being

utilized as a medium for composite action between connected extrusion parts.

1. A test beam is performed to determine the elasticity constant (c) of the combined profile assembly.

The elasticity constant is determined as the unit deformation of the assembly under an applied longitudinal shear

force. This deformation is the combined effect of slippage and shear deformation of the thermal break material.

The test is performed under different temperature conditions of -20°C, +20°C and +80°C.

Fcl l

∆=∆ ⋅

where: c = elasticity constant in N/mm2

= applied force in N

l = deformation in mm

l = length of the specimen in mm

Fig. 4.2-2 Shear Test Specimen

2. Test Results. Please refer to Appendix A for the complete test report.

Test Results

Temperature Shear

strength

Tensile strength Elasticity constant, c Factor of safety

for design strength1

-20°C 55.1 N/mm 28.9 N/mm 89 N/mm2 3.0

+20°C 50.4 N/mm - 69 N/mm2 3.0

+80°C 50.8 N/mm 61.3 N/mm 57 N/mm2 3.0

Note: 1 In checking the longitudinal shear (shear flow) in the thermal barrier, a factor of safety of 3.0 is considered

when calculated using a more precise FEM analysis. In cases where equation (28) of AAMA-TIR-A8-04 is used to

approximate the shear flow, a lower factor of safety is considered since the accuracy of this equation depends on

the degree of symmetry of the two faces being combined. In the case of the profiles used in the system, the two

faces are always very far from being symmetrical.

3. Design value Gc is calculated considering the relationship:

( )hGc c b= ⋅ where: Gc = shear modulus of the core in N/mm2 h = height of the thermal break in mm.

b = total thickness of the thermal break in mm.

Design Value of Gc. For h = 27mm and b = 4mm

Temperature Elasticity

constant, c

Shear Modulus,

Gc

-20°C 89 N/mm2 600.8 N/mm2

+20°C 69 N/mm2 465.8 N/mm2

+80°C 57 N/mm2 384.8 N/mm2

4. Predictions of effective moment of Inertia (I’e) are calculated for a range of Gc based on the shear test results.

The results are plotted on “predicted effective I” versus “span L” graph.

� Upper bound @ -20°C Gc = 600.8 N/mm2

� Lower bound @ +80°C Gc = 384.8 N/mm2

Using the conservative value from the test, the design value of Gc used in the analysis is when the thermal break is

softest @ elevated temperature of +80°C:

2384.8 /Gc N mm=

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4.3 Mullion Structural Systems

The sword inside the mullion has to transfer the shear reaction force of the CW unit above to the unit below. But

sometimes an amount of bending moment is transferred between mullions after a certain amount of rotation at the

ends of the mullion during service condition. Effectively at that stage, the mullion is acting as a continuous beam

system. Due to the gap between the extrusion dimensions and the sword dimensions a certain slip (free slip angle )

occurs and therefore the continuous system may only occur after a certain rotation at the beam ends.

(a) Section (b) Structural System - 1 (c) Structural System - 2

Fig. 4.3-1 Mullion Structural Systems

Applying the above theory to optimize the use of the mullion profiles two systems of structural behavior are

resolved with the following objective:

� The same male and female profile will be used throughout the entire job, and

� Only the swords will vary, in length and the grade of material.

4.3.1 Structural System 1 (): Sword is shorter and used as a shear connector only.

That is, no contact between the sword and the aluminum extrusion during service conditions, making it incapable of

transferring moment between mullions above or below. Therefore, the mullions act as simple beams. This is applied

where there is little load on the mullions.

, free slip angle

Deflection

, end rotation

Load Moment

Deflected Mullion

q

EI'

L

M δ

ρ

Sword (short)

α

hSW

LSW

gap/2 gap/2

Mullion

MAX MAX

Fig. 4.3-2 Structural System - 1

q q

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4.3.2 Structural System 2 (�≥�): Sword is longer and is utilized to transfer moment.

That is, there is contact between the sword and the aluminum extrusion during service condition, allowing transfer

of bending moment between the mullion profiles. Therefore, the system acts as a continuous beam. This system is

used for cases where system 1 is not sufficient to cater the loads.

(a) Stage 1 (��): The system is initially behaving as a simple beam.

That is, no contact between the sword and the aluminum extrusion, making it incapable of transferring moment

between mullion profiles.

(b) Stage 2 (��): The system finally behaves as a continuous beam.

That is, there is contact between the sword and the aluminum extrusion, allowing transfer of bending moment

between the mullion profiles.

STAGE 2(ρ

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4.3.3 Structural System Layout

The figure below shows the locations where structural systems 1 and 2 are applied. Note that corner panels are not

included in this set of structural calculations but nevertheless have been indicated as system 2.

(ALL THE REST)

SYSTEM 1

SYSTEM 2(3 CW UNITS BOTH SIDES)

WING - A

SYSTEM 1(ALL THE REST)

WING - BWING - C

SYSTEM 2(14 CW UNITS BOTH SIDES) (3 CW UNITS BOTH SIDES)

SYSTEM 2

LEVEL 63TIER 5


(3 CW UNITS BOTH SIDES)SYSTEM 2

WING - A

(ALL THE REST)

SYSTEM 1

WING - B


WING - C

(3 CW UNITS BOTH SIDES)SYSTEM 2 SYSTEM 2

(5 CW UNITS BOTH SIDES)

(ALL THE REST)SYSTEM 1

TIER 4

LEVEL 52

(ALL THE REST)

SYSTEM 1

WING - C


LEVEL 37TIER 3



WING - B


WING - A


TIER 2

LEVEL 33

(ALL THE REST)SYSTEM 1

WING - CWING - B

(ALL THE REST)

SYSTEM 1SYSTEM 1(ALL THE REST)

WING - A

SYSTEM 2(3 CW UNITS BOTH SIDES)(5 CW UNITS BOTH SIDES)

SYSTEM 2



SYSTEM 2(3 CW UNITS BOTH SIDES) (5 CW UNITS BOTH SIDES)

SYSTEM 2(3 CW UNITS BOTH SIDES)SYSTEM 2 SYSTEM 2

(3 CW UNITS BOTH SIDES)

WING - A

(ALL THE REST)

SYSTEM 1 SYSTEM 1(ALL THE REST)

WING - BWING - C


LEVEL 25TIER 1TIER 0

LEVEL 7&8


(ALL THE REST)

(ALL THE REST)

WING - CWING - B

WING - A(3 CW UNITS BOTH SIDES)

SYSTEM 1

SYSTEM 1

Fig. 4.3-4 Structural System Layout

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4.4 Critical Panel Evaluation

To evaluate the most critical panel two criteria are to be checked, the stress index and the deflection index. The

indices of each panel type are determined from the variable parameters in calculating the bending moment and

deflections. The criteria for this evaluation are summarized in the tables below.

The panel under the most critical circumstances will be the subject of the structural system analysis in the

succeeding sections of this report.

Table 4.4-1 Critical Panel Evaluation

Member Analysis Constant

Parameters Variable Parameters

Critical

Index System - 1 System - 2

Mullion

Stress

- Span (H)

- Cross-sectional

properties.

- Tributary width (W)

- Wind load, Qw

Uniform

load

W∙Qw = 4.83 KN/m

Wing B (Zone C)

QW = ±3.5 KPa

W = 1.385 m W∙Qw = 8.7 KN/m

Qw∙W2 = 16.9 KN

Qw∙W4 = 63.7

KN∙m2

Wing B (Zone D)

QW = ±4.5 KPa

W = 1.94 m

Deflection

- Span (H)

- Cross-sectional

properties.

- Panel width (W)

- Wind load, Qw

Uniform

load

Transom

Stress

- Tributary widths

(bu & bl)

- Cross-sectional

properties

- Span (W)

- Wind load, Qw

Bending

moment

Qw∙W2 = 7.58 KN

Qw∙W4 = 19.2

KN∙m2

Wing B (Zone C)

QW = ±3.0 KPa

W = 1.59 m Deflection

- Tributary widths

(bu & bl)

- Cross-sectional

properties

- Span (W)

- Wind load, Qw

Deflection

Table 4.4-2 Panel Parameters

A B A B A B

Level 7 & 8 (Tier 0) 2.0 3.0 3.0 3.0 3.5 3.5 2.0 2.5 3.0 3.0 4.5 4.5 2.0 3.0 3.0 3.0 3.5 3.5

Level 25 (Tier 1) 2.0 2.5 3.0 3.0 3.5 3.5 2.0 3.5 3.5 3.5 3.5 3.5 2.0 2.5 3.0 3.0 3.5 3.5

Level 33 (Tier 2) 2.0 2.5 3.0 3.0 3.5 3.5 2.5 2.5 3.0 3.0 3.5 3.5 2.0 3.5 3.5 3.5 3.5 4.0

Level 37 (Tier 3) 2.0 2.5 3.0 3.0 3.5 3.5 2.0 2.5 3.0 3.0 3.5 3.5 2.0 2.5 3.0 3.0 3.5 3.5

Level 52 (Tier 4) 2.0 2.5 3.0 3.0 3.5 3.5 2.0 3.5 3.5 3.5 3.5 3.5 2.0 2.5 3.0 3.0 3.5 3.5

Level 63 (Tier 5) 2.5 3.0 3.0 3.0 3.5 3.5 2.5 3.0 3.0 3.0 3.5 3.5 2.5 3.0 4.0 4.0 4.5 4.5

Panel width (up to…), m W = 1.80 1.34 1.38 1.59 1.94 2.25 1.80 1.34 1.38 1.59 1.94 2.25 1.80 1.34 1.38 1.59 1.94 2.25

Mullion, Stress Index W∙Qw = - 4.02 4.14 4.77 6.79 - - 4.69 4.83 4.77 8.73 - - 4.69 4.83 4.77 8.73 -Mullion, Deflection Index W∙Qw = - 4.02 4.14 4.77 6.79 - - 4.69 4.83 4.77 8.73 - - 4.69 4.83 4.77 8.73 -

Transom, Stress Index Qw∙W2 = - 5.39 5.71 7.58 13.2 - - 6.28 6.67 7.58 16.9 - - 6.28 6.67 7.58 16.9 -

Transom, Deflection Index Qw∙W4 = - 9.67 10.9 19.2 49.6 - - 11.3 12.7 19.2 63.7 - - 11.3 12.7 19.2 63.7 -

C D1

C D1

C D1

WING - B WING - C

System 1 System 2 System 1 System 2 System 1 System 2

Wind load, ±Qw [KN/m2]

Structural System

Loading Zone

WING - A

Note: 1Structural system 2 extends to the shaded cell in the table.

Fig. 4.4-1 Key Plan for Wind Load Distribution

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4.5 CW In-plane Performance

The in-plane behavior of the CW structural members is dealt due to the following situations:

(a) In-plane Wind load. Wind load applied to the protruding surface of the vertical fins produces torsion as well as bending on the weak axis of the mullion profiles.

This wind load is calculated as per the 1997 UBC code as a parapet wall with pressure coefficient of 1.3 inward and

outward. See §5.2 on Wind loads.

Fig. 4.4-1 Wind Load on Fins

(b) Slab deflection. Due to slab deflection two loading scenarios occur on the system which are addressed in this set of calculation:

• Where the slab has uneven deflection at the bracket locations of a given CW unit, the bracket that is located on

the lower point of the deflected slab will become a “dead bracket” which means it is carrying no weight because

the slab is hanging below. This results to a one-side supported unit wherein the other bracket carries the whole

weight and causes the CW unit to rotate due to the eccentric application of its weight (dead load). Couple force

is then produced to counteract the rotation. The couple is a set of horizontal forces acting on the “live bracket”

and on the sword.

• At the apex of the deflected slab, “live brackets” occur at the same point for two adjacent CW units. Thus, the

bracket mounted on the slab shall be designed to cater for this double loading.

Error! Objects cannot be created from editing field codes.

Fig. 4.4-2 Racking of CW Units

Wind force on

protrusions

(fins)

Wind force

normal

to CW unit

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5 LOADING

5.1 Dead Load, QD

Table 5.1-1 Weight of Structural System - 1, Total weight = 3.07 kN (313 kg.)

Element Refer. Specific Wt.

Real constant 1

k1

Unit Wt.

= ∙ k1

Real constant 2

k2

Total wt.

W = ∙k2

Aluminum Extrusion 0.53 kN

Male mullion §6.1.1(a) 27.0 kN/m3 A = 2414 mm2 0.06 kN/m L = 3.70m 0.24 kN

Female mullion §6.1.2(a) 27.0 kN/m3 A = 883 mm2 0.02 kN/m L = 3.70m 0.09 kN

Header transom §6.2.1(a) 27.0 kN/m3 A = 946 mm2 0.03 kN/m L = 1.39m 0.04 kN

Sill transom §6.2.2(a) 27.0 kN/m3 A = 1282 mm2 0.03 kN/m L = 1.39m 0.05 kN

Transom 1 §6.3.1(a) 27.0 kN/m3 A = 1067 mm2 0.03 kN/m L = 1.39m 0.04 kN


Sword §6.5.1 27.0 kN/m3 A = 2100 mm2 0.06 kN/m L = 0.50m 0.03 kN

Upper spandrel panel 0.41 kN/m2 0.27 kN

Stainless Steel Sheet M14 78.0 kN/ m3 t = 1.5mm 0.12 kN/m2 A = 0.65 m2 0.08 kN

Steel Sheet M38 78.0 kN/m3 t = 1.5mm 0.12 kN/m2 A = 0.65 m2 0.08 kN

Insulation M9 0.7 kN/m3 t = 70mm 0.05 kN/m2 A = 0.65 m2 0.03 kN


Vision panel 0.45 kN/m2 1.77 kN

Glass GL-1.3 25.0 kN/m3 t = 18mm 0.45 kN/m2 A = 3.93 m2 1.77 kN

Lower Spandrel panel 0.45 kN/m2 0.24 kN

Glass GL-2.3 25.0 kN/m3 t = 18mm 0.45 kN/m2 A = 0.54 m2 0.24 kN

Accessories 0.05 kN/m2 0.26 kN

Thermal break, setting block, gaskets, sealant, screws, etc. approx.10%W A = 5.12 m2 0.26 kN

Table 5.1-2 Weight of Structural System - 2, Total weight = 4.63 kN (472 kg.)

Element Refer. Specific Wt.

Real constant 1

k1

Unit Wt.

= ∙ k1

Real constant 2

k2

Total wt.

W = ∙k2

Aluminum Extrusion 0.76 kN

Male mullion §6.1.1(a) 27.0 kN/m3 A = 2414 mm2 0.06 kN/m L = 3.70m 0.24 kN

Female mullion §6.1.2(a) 27.0 kN/m3 A = 883 mm2 0.02 kN/m L = 3.70m 0.09 kN

Header transom §7.2.1(a) 27.0 kN/m3 A = 946 mm2 0.03 kN/m L = 1.94m 0.05 kN

Sill transom §7.2.2(a) 27.0 kN/m3 A = 1282 mm2 0.03 kN/m L = 1.94m 0.07 kN



Sword §7.5.2 27.0 kN/m3 A = 2100 mm2 0.06 kN/m L = 1.42m 0.09 kN

Upper spandrel panel 0.41 kN/m2 0.37 kN

Stainless Steel Sheet M14 78.0 kN/ m3 t = 1.5mm 0.12 kN/m2 A = 0.91 m2 0.11 kN


Insulation M9 0.7 kN/m3 t = 70mm 0.05 kN/m2 A = 0.91 m2 0.04 kN


Vision panel 0.50 kN/m2 2.76 kN

Glass GL-1.3 25.0 kN/m3 t = 10 + 10mm 0.50 kN/m2 A = 5.51 m2 2.76 kN

Lower Spandrel panel 0.50 kN/m2 0.38 kN

Glass GL-2.3 25.0 kN/m3 t = 10 + 10mm 0.50 kN/m2 A = 0.76 m2 0.38 kN

Accessories 0.05 kN/m2 0.36 kN

Thermal break, setting block, gaskets, sealant, screws, etc. approx.10%W A = 7.18 m2 0.36 kN

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5.2 Wind Load, QW

Applied wind loads are in accordance to SOM Drawing Nr. AL-00-AR-502 “ Building Enclosure Wind Pressure

Diagrams (Addendum 057)”

5.2.1 Wind Load on Wall Elements.

A comprehensive summary of the wind load stated above is found on Table 4.4-2.

5.2.2 Wind loads on Vertical Fins.

The positive values from wind load zone `E´ on parapet walls are considered for the wind Loads on the vertical fins.

The negative values on parapet walls are not considered because this includes the trapped air at return corners

which does not happen on the vertical fins.

Table 5.2-4 Wind Load on Vertical Fins

LOAD, KN/m2 Vertical Fin1

Zone E

Level 7 & 8 +3.0

Level 25 +3.0

Level 33 +3.0

Level 37 +3.0

Level 52 +3.0

Level 63 +3.0

Note. 1Values for the parapet walls are conservatively adopted for the vertical fins.

PROJECT NAME DATE




6 SECTION PROPERTIES – STRUCTURAL SYSTEM 1

6.1 Mullion Profiles

6.1.1 Male Mullion

(a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS’ beam plot section command.

Fig. 6.1-1 Section Properties

(b) Effective Section Modulus. The effective section properties of the mullion profile are predicted by adopting the procedure outlined in AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing Systems.

§7.5.5 of AAMA TIR-A8-04 defines the effective shear area, A as the sum of the areas of the web elements. In this

case, all elements that are oriented parallel to the shear direction are considered effective in resisting shear.

c.g.1

c.g.2

SECTION PROPERTIES

13°

A2 = 1280.5mm2

A2 = A2a+ A2bxcos(13deg.)

DIAGONALLY ORIENTED

WEBS:

A2b = 674.1mm2

A2a = 623.7mm2

A1 = 266.5mm2

SHEAR AREA

A = 1547.0mm2

A = A1 + A2

c.g.

137

14

9

34.2 40.9

DISTANCES

10

7.2

17.9 2

27

30.4

148

.5

28

6

2

Fig. 6.1-2 Parameters in Determining the Effective Section Modulus

PROJECT NAME DATE




• Effective Section Modulus Calculation – Male Mullion (System 1)

AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System

Clause Action Notes

Extrusion Parameters L = 3.70 m. Unsupported span of the member

E = 68950 Mpa Young's modulus of aluminum faces

Gc = 384.8 Mpa Design shear modulus of thermal break

w = 1.0 kN/m Considered uniform unit load

M = w∙L2/8 = 1.7 KN∙m Maximum bending moment due to unit loadA = 1547.0 m Shear area of aluminum

h = 286.0 mm Overall depth of extrusion

g = 17.9 mm Gap (clearance) between faces

Dc = 27.0 mm Maximum cavity depth

b = 4.0 mm Average width of thermal break core

a1 = 686.9 mm2 a2 = 1750.0 mm

2 Cross sectional area

Io1 = 2.43E+05 mm4 Io2 = 6.29E+06 mm

4 Moment of inertia

c11 = 30.4 mm c22 = 107.2 mm Extreme fiber dist. to c.g.of face

D = h - (c11+c22) = 148.4 mm Distance between centroidal axes of both faces

7.5.4(1) Ic = a1∙a2∙D2/(a1+a2) = 1.09E+07 mm

4 For the case where both faces are same material

7.5.4(2) Io = Io1+Io2 = 6.53E+06 mm4 Lower bound on stiffness I'e (no composite action)

7.5.4(3) I = Ic + Io = 1.74E+07 mm4 Upper bound on stiffness I'e (full composite action)

7.5.4(5) Composite Analysis Gp = I∙b∙D2Gc/(Ic∙Dc) = 2.01E+06 N Geometric and core material parameter

7.5.4(6) c = Gp/(E∙Io) = 4.46E-06 /mm2 Buckilng formula slope for comp. in beam flanges7.5.4(Table3) D0 = w∙E∙Io∙Ic/(Gp

2∙I) = 0.07 mm Constant for the elastic curve formulaD1 = -w∙L∙Ic/(2Gp∙I)-wL

3/(24E∙I) = 0.00 Ditto

D2 = w∙Ic/(2Gp∙I) = 1.6E-07 /mm DittoD3 = w∙L/(12E∙I) = 2.6E-10 /mm2 DittoD4 = -w/(24E∙I) = -3.5E-14 /mm3 D5 = 0.00

r = (L/2)(c)0.5 = 3.91

7.5.4(Table5) F1 = -w∙Ic∙e-r/[c∙Gp∙I(e

r+e-r)] = -2.8E-05 mm Complementary constants

F2 = F1∙e2r = -0.07 mm Ditto

7.5.4(8a) p = x(c)0.5 = 3.91 for x = L/2 = 1850 mm

7.5.4(8) y =D5x5+D4x

4+D3x3+D2x2+D1x+D0+F1e

p+F2/ep = 2.57 mm Calculated effective maximum deflection, D5 =0

7.5.4(21) y'' =20D5x3+12D4x

2+6D3x+2D2+c(F1e

p+F2/e

p) = 1.72E-06 /mm

7.5.4(10) Ie = w∙L4/(76.8E∙y) = 1.38E+07 mm4 Effective moment of inertia w/o shear deformation

7.5.4(17) I'e = Ie/{ 1+[25.6Ie/(L2∙A)} = 1.36E+07 mm4 Effective moment of inertia considering shear def.

7.5.4(19) Se1 = M/[(M-E∙Io∙y'')/(a1∙D)+E∙c11∙y''] = 1.34E+05 mm3 Effective section modulus @ face 17.5.4(20) Se2 = M/[(M-E∙Io∙y'')/(a2∙D)+E∙c22∙y''] = 1.05E+05 mm3 Effective section modulus @ face 27.5.4(25a) Shear Flow Data y''' =6D3+c

1.5(F1-F2) = 2.20E-09 /mm2

7.5.4(24) Vc/w = w∙L/2 -E∙Io∙y''' = 859.91 mm Shear resisted by thermal break per unif. unit load

7.5.4(28) Vc/(D∙w) = 5.79 Shear flow per uniform unit load, w

Effective I Curve

I (fully composite)

Ie for upper bound Gc

Io (non-composite)

Ie for lower bound Gc

I'e for lower bound Gc

I'e for upper bound Gc

1.36E+07

5.00E+06

1.00E+07

1.50E+07

2.00E+07

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0Span, L [m]

Mom

ent

of Iner

tia

[mm

4]

PROJECT NAME DATE




(c) Allowable Stress Under Lateral Buckling

The moment of inertia about the minor axis (Iy) and the torsional constant (J) are conservatively calculated as the

sum of the individual properties of face 1 (Izz,1, J1) and face 2 (Izz,2,J2) profiles.

ADM 2005: Specification for Aluminum Structures - Allowable Stress Design

Clause Action Notes

Extrusion Parameters Lb1 = 2.84 m Unbraced length for face-1 under compression

Lb2 = 3.70 m Unbraced length for face-2 under compression

Iy = 6.4E+05 mm4 Moment of inertia of profile

J = 8.8E+05 mm4 Torsional constant of profile

Sc1 = 1.3E+05 mm3 Section modulus for face 1 under compression


3.3-1M Fcy = 170 Mpa compressive yield strength

E = 69600 Mpa Young's Modulus

3.4-1 Buckling Analysis ny = 1.65 Factor of safety on yield strength

3.3-4 Bc = Fcy[1+(Fcy/15510)0.5] = 187.798 Buckling formula intercept for comp. in beam flanges

Dc = (Bc/10)(Bc/E)0.5 = 0.98 Buckilng formula slope for comp. in beam flanges

Cc = 0.41(Bc/Dc) = 78.93 Buckling formula intersection for comp. in beam flanges

3.4.14-4 S1 = [(Bc-Fcy)/1.6Dc]2 = 130.0 Lower bound slender limit

3.4.14-5 S2 = (Cc/1.6)2 = 2433.6 Upper bound slender limit

3.4.14 Face-1: S = Lb1∙Sc1/[0.5(Iy∙J)0.5] = 1009.9 Slenderness ratio for face-1 under compression

Criteria = S1 < S < S2 Intermediate beam-compression member

3.4.14-2 Fc1 = 1/ny∙(Bc-1.6Dc∙S0.5) = 83.75 Mpa Allowable compressive stress for LTB

3.4.14 Face-2: S = Lb2∙Sc2/[0.5(Iy∙J)0.5] = 1031.0 Slenderness ratio for face-2 under compression



6063-T6 Alloy

(d) Allowable Stress Under Local Buckling

Element 3

Element 2

c.g.

Fig. 6.1-3 Elements Critical Under Local Buckling

PROJECT NAME DATE




The governing allowable stress, under local buckling as calculated below, is summarized as follows:

� ace 1, considering elements 1, & 2 c1 = 103.03MPa

� ace 2, considering element 3 c2 = 131.0MPa

• Element 1

ADM 2005, §3.4.16: Uniform Compression in Elements of Beams - Flat Elements Supported on Both Edges

Clause Action Notes

b = 30.5 m Width of the flat element

t = 1.8 mm Thickness of the flat element



3.4-1 ny = 1.65 Factor of safety on yield strength

3.3-4 k1 = 0.35 Coefficient for determining slenderness limit

3.3-4 Bp = Fcy[1+(Fcy)1/3/21.7] = 213.398 Buckling formula intercept for comp. in flat element

Dp = (Bp/10)(Bp/E)0.5

= 1.18 Buckilng formula slope for comp. in flat elements

3.4.16-4 S1 = (Bp-Fcy)/1.6Dp = 23.0 Lower bound slender limit

3.4.16-5 S2 = k1∙Bp/1.6Dp = 39.5 Upper bound slender limitb/t = 16.9 Slenderness ratio

Criteria = b/t < S1 Short compression element

3.4.16-1 Fc = Fcy/ny = 103.03 Mpa Allowable compressive stress

6063-T6 Alloy

• Element 2

ADM 2005, §3.4.18: Compression in Elements of Beams - Flat Elements Supported on Both Edges

Clause Action Notes



3.4.48 Cc = -135.2 mm Distance from N.A. to heavily compressed edge

Co = -89.4 mm Distance from N.A. to other edge





Co/Cc = 0.66

m = 1.15+Co/(2Cc) = 1.48 for -1 < Co/Cc < 1

3.3-4 Bbr = 1.3Fcy[1+(Fcy)1/3/13.3] = 313.0 Buckling formula intercept for comp. in flat element

Dbr = (Bbr/20)(6Bbr/E)0.5 = 2.57 Buckilng formula slope for comp. in flat elements

3.4.18-4 S1 = (Bbr-1.3Fcy)/m∙Dbr = 24.2 Lower bound slender limit3.4.18-5 S2 = k1∙Bbr/m∙Dbr = 41.1 Upper bound slender limit

h/t = 25.4 Slenderness ratio

Criteria = S1 < h/t < S2 Intermediate compression element

3.4.18-2 Fc = (1/ny)[Bbr-m∙Dbr(h/t)] = 131.02 Mpa Allowable compressive stress

6063-T6 Alloy

PROJECT NAME DATE




• Element 3

ADM 2005, §3.4.18: Compression in Elements of Beams - Flat Elements Supported on Both Edges

Clause Action Notes



3.4.48 Cc = -140.8 mm Distance from N.A. to heavily compressed edge

Co = -76.2 mm Distance from N.A. to other edge





Co/Cc = 0.54

m = 1.15+Co/(2Cc) = 1.42 for -1 < Co/Cc < 1

3.3-4 Bbr = 1.3Fcy[1+(Fcy)1/3/13.3] = 313.0 Buckling formula intercept for comp. in flat element

Dbr = (Bbr/20)(6Bbr/E)0.5 = 2.57 Buckilng formula slope for comp. in flat elements

3.4.18-4 S1 = (Bbr-1.3Fcy)/m∙Dbr = 25.2 Lower bound slender limit3.4.18-5 S2 = k1∙Bbr/m∙Dbr = 42.9 Upper bound slender limit

h/t = 26.5 Slenderness ratio

Criteria = S1 < h/t < S2 Intermediate compression element

3.4.18-2 Fc = (1/ny)[Bbr-m∙Dbr(h/t)] = 131.02 Mpa Allowable compressive stress

6063-T6 Alloy

6.1.2 Female Mullion



(b) Effective Section Modulus. The effective section properties of the mullion profile is predicted by adopting the procedure outlined in AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing Systems.



PROJECT NAME DATE




c.g.1

c.g.2

c.g.


PROJECT NAME DATE




• Effective Section Modulus Calculation – Female Mullion (System 1)


Clause Action Notes










a1 = 652.0 mm2 a2 = 294.9 mm


Io1 = 2.37E+05 mm4 Io2 = 3.84E+04 mm

4 Moment of inertia



7.5.4(1) Ic = a1∙a2∙D2/(a1+a2) = 9.67E+05 mm







3/(24E∙I) = -0.03 DittoD2 = w∙Ic/(2Gp∙I) = 1.1E-06 /mm DittoD3 = w∙L/(12E∙I) = 3.6E-09 /mm2 DittoD4 = -w/(24E∙I) = -4.9E-13 /mm3 D5 = 0.00

r = (L/2)(c)0.5 = 7.93



F2 = F1∙e2r = -0.12 mm Ditto

7.5.4(8a) p = x(c)0.5 = 7.93 for x = L/2 = 1850 mm

7.5.4(8) y =D5x5+D4x



7.5.4(21) y'' =20D5x3+12D4x

2+6D3x+2D2+c(F1ep+F2/e

p) = 2.22E-05 /mm




1.5(F1-F2) = 3.12E-08 /mm2



Effective I CurveI (fully composite)Ie for upper bound

Gc

Io (non-composite)

Ie for lower bound

GcI'e for lower bound

Gc

I'e for upper bound

Gc1.09E+06

0.00E+00

5.00E+05

1.00E+06

1.50E+06

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Span, L [m]

Mom

ent

of Iner

tia

[mm

4]

PROJECT NAME DATE






sum of the individual properties of face 1 (Izz,1, J1) and face 2 (Izz,2,J2) profiles.


Clause Action Notes

Extrusion Parameters Lb = 2.84 m Unbraced length of the member for bending









Dc = (Bc/10)(Bc/E)0.5

= 0.98 Buckilng formula slope for comp. in beam flanges




3.4.14 Face-1: S = Lb∙Sc1/[0.5(Iy∙J)0.5] = 1601.4 Slenderness ratio for face-1 under compression



3.4.14 Face-2: S = Lb∙Sc2/[0.5(Iy∙J)0.5

] = 1253.3 Slenderness ratio for face-2 under compression


3.4.14-2 Fc2 = 1/ny∙(Bc-1.6Dc∙S0.5

) = 80.33 Mpa Allowable compressive stress for LTB

6063-T6 Alloy


Element 1




� ace 2, no element is critical under local buckling.

PROJECT NAME DATE




• Element 1

ADM 2005, §3.4.15: Uniform Compression in Elements of Beams - Flat Elements Supported on One Edge

Clause Action Notes








Dp = (Bp/10)(Bp/E)0.5 = 1.18 Buckilng formula slope for comp. in flat elements



Criteria = S1 < b/t < S2 Intermediate compression element

3.4.15-2 Fc = (1/ny)[Bp-5.1Dp(b/t)] = 96.10 Mpa Allowable compressive stress

6063-T6 Alloy

PROJECT NAME DATE




6.2 Stack Joint Transom Profiles

6.2.1 Header Transom

(a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS finite element analysis.





A = A1 + A2

A = 346.6mm2

A1 = 225.5mm2

A2 = 121.1mm2

FACE-2 FACE-1

c.g.1

c.g.2 c.g.


PROJECT NAME DATE




• Effective Section Modulus Calculation – Header Transom (System 1)


Clause Action Notes










a1 = 587.0 mm2 a2 = 359.4 mm


Io1 = 9.41E+04 mm4 Io2 = 2.36E+04 mm

4 Moment of inertia



7.5.4(1) Ic = a1∙a2∙D2/(a1+a2) = 9.27E+05 mm







3/(24E∙I) = 0.00 DittoD2 = w∙Ic/(2Gp∙I) = 1.7E-06 /mm DittoD3 = w∙L/(12E∙I) = 1.8E-09 /mm2 DittoD4 = -w/(24E∙I) = -5.8E-13 /mm3 D5 = 0.00

r = (L/2)(c)0.5 = 4.56



F2 = F1∙e2r = -0.10 mm Ditto

7.5.4(8a) p = x(c)0.5 = 4.56 for x = L/2 = 795 mm

7.5.4(8) y =D5x5+D4x



7.5.4(21) y'' =20D5x3+12D4x


p) = 7.64E-06 /mm




1.5(F1-F2) = 3.01E-08 /mm2



Effective I Curve

I (fully composite)Ie for upper bound Gc

Io (non-composite)




5.38E+05

0.00E+00

5.00E+05

1.00E+06

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0Span, L [m]

Mom

ent

of Iner

tia

[mm

4]

PROJECT NAME DATE






sum of the individual properties of face 1 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles.


Clause Action Notes




















6063-T6 Alloy


c.g.Element 1Element 2

FACE-2 FACE-1





PROJECT NAME DATE




• Element 1


Clause Action Notes













6063-T6 Alloy

• Element 2


Clause Action Notes







K2 = 2.27 Ditto





Criteria = S2 < b/t Slender compression element

3.4.15-3 Fc = k2/ny (Bp∙E)0.5/(5.1b/t)= 34.77 Mpa Allowable compressive stress

6063-T6 Alloy

PROJECT NAME DATE




6.2.2 Sill Transom






FACE-2 FACE-1

A = 396.0mm2

A = A1 + A2

A2 = 115.2mm2

A1 = 280.8mm2

c.g.

c.g.1c.g.2

2 x P

37.3

Z


PROJECT NAME DATE




• Effective Section Modulus Calculation – Sill Transom (System 1)


Clause Action Notes










a1 = 981.6 mm2 a2 = 300.5 mm


Io1 = 4.20E+05 mm4 Io2 = 1.20E+04 mm

4 Moment of inertia



7.5.4(1) Ic = a1∙a2∙D2/(a1+a2) = 8.12E+05 mm







3/(24E∙I) = 0.00 DittoD2 = w∙Ic/(2Gp∙I) = 1.1E-06 /mm DittoD3 = w∙L/(12E∙I) = 1.5E-09 /mm2 DittoD4 = -w/(24E∙I) = -4.9E-13 /mm3 D5 = 0.00

r = (L/2)(c)0.5 = 2.56



F2 = F1∙e2r = -0.20 mm Ditto

7.5.4(8a) p = x(c)0.5 = 2.56 for x = L/2 = 795 mm

7.5.4(8) y =D5x5+D4x



7.5.4(21) y'' =20D5x3+12D4x


p) = 5.48E-06 /mm




1.5(F1-F2) = 1.60E-08 /mm2



Effective I Curve


Io (non-composite)




7.09E+05

0.00E+00

5.00E+05

1.00E+06

1.50E+06

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0Span, L [m]

Mom

ent

of

Iner

tia

[mm

4]

PROJECT NAME DATE








Clause Action Notes




















6063-T6 Alloy


c.g.

Element 1

FACE-2

FACE-1





PROJECT NAME DATE




• Element 1


Clause Action Notes







K2 = 2.27 Ditto





Criteria = S2 < b/t Slender compression element

3.4.15-3 Fc = k2/ny (Bp∙E)0.5/(5.1b/t)= 69.94 Mpa Allowable compressive stress

6063-T6 Alloy

PROJECT NAME DATE




6.3 Transom – 1 Profile

6.3.1 Transom -1 (Type 1)






A1 = 242.2mm2

A2 = 154.7mm2

A = A1 + A2

A = 396.9mm2

FACE-2 FACE-1c.g.1c.g.2 c.g.


PROJECT NAME DATE




• Effective Section Modulus Calculation – Transom – 1 (Type 1: System 1)


Clause Action Notes










a1 = 652.1 mm2 a2 = 415.1 mm


Io1 = 3.15E+05 mm4 Io2 = 3.52E+04 mm

4 Moment of inertia



7.5.4(1) Ic = a1∙a2∙D2/(a1+a2) = 9.47E+05 mm







3/(24E∙I) = 0.00 Ditto


r = (L/2)(c)0.5

= 2.76



F2 = F1∙e2r = -0.21 mm Ditto

7.5.4(8a) p = x(c)0.5 = 2.76 for x = L/2 = 795 mm

7.5.4(8) y =D5x5+D4x



7.5.4(21) y'' =20D5x3+12D4x


p) = 5.72E-06 /mm




1.5(F1-F2) = 1.75E-08 /mm

2



Effective I Curve


Io (non-composite)

Ie for lower bound GcI'e for lower bound Gc


6.44E+05

0.00E+00

5.00E+05

1.00E+06

1.50E+06

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0Span, L [m]

Mom

ent

of

Iner

tia

[mm

4]

PROJECT NAME DATE







ADM 2005, §3.4.14: Compression in Beams, Extreme Fiber, Gross Section - Tubular Shapes

Clause Action Notes









3.3-4 Bc = Fcy[1+(Fcy/15510)0.5

] = 187.798 Buckling formula intercept for comp. in beam flanges






Criteria = S ≤ S1 Short beam-compression meber

3.4.14-1 Fc1 = Fcy/ny = 103.03 Mpa Allowable compressive stress for LTB

3.4.14 Face-2: S = Lb∙Sc2/[0.5(Iy∙J)0.5

] = 119.1 Slenderness ratio for face-2 under compression



6063-T6 Alloy


c.g.

Element 1





PROJECT NAME DATE




• Element 1


Clause Action Notes













6063-T6 Alloy

6.4 Transom – 2 Profile

6.4.1 Transom – 2 (Type 1)






PROJECT NAME DATE




A2 = 151.4mm2

A1 = 263.2mm2

FACE-2 FACE-1

A = 414.6mm2

A = A1 + A2

c.g.1c.g.2 c.g.

DL2 x P

27.1


PROJECT NAME DATE




• Effective Section Modulus Calculation – Transom 2 (Type 1: System 1)


Clause Action Notes










a1 = 653.9 mm2 a2 = 330.7 mm


Io1 = 2.96E+05 mm4 Io2 = 2.91E+04 mm

4 Moment of inertia



7.5.4(1) Ic = a1∙a2∙D2/(a1+a2) = 8.52E+05 mm







3/(24E∙I) = 0.00 Ditto


r = (L/2)(c)0.5

= 2.94


r+e

-r)] = -4.9E-04 mm Complementary constants

F2 = F1∙e2r = -0.17 mm Ditto

7.5.4(8a) p = x(c)0.5 = 2.94 for x = L/2 = 795 mm

7.5.4(8) y =D5x5+D4x

4+D3x

3+D2x2+D1x+D0+F1e

p+F2/e

p = 1.79 mm Calculated effective maximum deflection, D5 =0

7.5.4(21) y'' =20D5x3+12D4x

2+6D3x+2D2+c(F1e

p+F2/e

p) = 6.01E-06 /mm




1.5(F1-F2) = 1.85E-08 /mm

2



Effective I Curve

I (fully composite)

Io (non-composite)

Ie for lower bound GcI'e for lower bound Gc

Ie for upper bound GcI'e for upper bound Gc

6.63E+05

0.00E+00

5.00E+05

1.00E+06

1.50E+06

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0Span, L [m]

Mom

ent

of Iner

tia

[mm

4]

PROJECT NAME DATE







ADM 2005, §3.4.14: Compression in Beams, Extreme Fiber, Gross Section - Tubular Shapes

Clause Action Notes




















6063-T6 Alloy


c.g.

Element 1




� ace 2 No element is critical under lateral buckling.

PROJECT NAME DATE




• Element 1


Clause Action Notes













6063-T6 Alloy

PROJECT NAME DATE




6.5 Sword Profile

Fig. 6.5-1 Sword Length

6.5.1 Sword (Type-1)

(a) Cross-sectional Properties.

140

Fig. 6.5-2 Sword Cross-Section

PROJECT NAME DATE




7 SECTION PROPERTIES – STRUCTURAL SYSTEM 2

7.1 Mullion Profiles

7.1.1 Male Mullion

Section properties of male mullion for Structural system 1 and 2 are the same.

Please refer to §6.1.1

7.1.2 Female Mullion

Section properties of female mullion for Structural system 1 and 2 are the same.

Please refer to §6.1.2

PROJECT NAME

ocument . r str-calc-548 0 · project amen date sample project in the middle east title revision...

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