SAMPLE PROJECT IN THE MIDDLE EAST DOCUMENT NO. STR-CALC UNITISED CURTAIN WALL 117 ENGINEER PROJECT. Pages REVISION TITLE

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1 PROJECT ENGINEER DOCUMENT NO. STR-CALC REVISION TITLE Pages UNITISED CURTAIN WALL 117

2 UNITISED CURTAIN WALL 2 of 117 Table of Contents 1 Summary 3 2 Basic Data Standards and References Materials Performance Criteria Programs used for the structural analysis 5 3 Layout and Dimensions Key Location: Wall Type Façade Elevation: Wall Type Design Narrative Load Path Coupled Mullion Mullion Structural Systems Critical Panel Evaluation CW In-plane Performance 14 5 Loading Dead Load, QD Wind Load, QW 16 6 Section Properties Structural System Mullion Profiles Stack Joint Transom Profiles Transom 1 Profile Transom 2 Profile Sword Profile 42 7 Section Properties Structural System Mullion Profiles Stack Joint Transom Profiles Transom 1 Profile Transom 2 Profile Sword Profile 56 8 Analysis & Code Check Structural System Mullion Check Check Stack Joint Header and Sill Transom Transom 1 (Type 1) Transom 2 (Type 1) Check Stainless Steel Spandrel Panel 74 9 Analysis & Code Check Structural System Mullion Check Check Stack Joint Header and Sill Transom Transom 1 (Type 2) Transom 2 (Type 2) Check Stainless Steel Spandrel Panel Joints & Connections Mullion Shear Connectors Header Transom Connection Check Sill Transom Connection Check Transom - 1 Connection Check Transom - 2 (System 1) Connection Check Transom - 2 (System 2) Connection Check Sword Connection Check 113

3 UNITISED CURTAIN WALL 3 of Back Pan Check 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: Element Material Table 1 Results Summary Stress Utilization Deflection Utilization Critical Male Mullion 6063-T6 98 % 98 % Biaxial bending 8.1 Female Mullion 6063-T6 43 % 62 % Bending 8.1 Section Referenc e Stack Joint - Header 6063-T6 65 % 38 % Biaxial bending Stack Joint - Sill 6063-T6 35 % 38 % Bending Transom 1 (Type1) 6063-T6 43 % 45 % Biaxial bending Transom 1 (Type2) 6063-T6 52 % 62 % Biaxial bending Transom 2 (Type1) 6063-T6 44 % 41 % Biaxial bending Transom 2 (Type2) 6063-T6 59 % 71 % Biaxial bending Sword (Type 1) 6063-T6 8 % - No bending (only shear) Sword (Type 2) 7022-T % - Biaxial bending Thermal Break Polyamide 95 % - Shear Connections - Screw ST4.8 A % - Shear

4 UNITISED CURTAIN WALL 4 of BASIC DATA 2.1 Standards and References Codes and Standards Table 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-A Metal Curtain Wall Fasteners 6 EN ISO 1478:1999 Tapping screw thread 7 ISO :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 Document reference Table Project Specification by Skidmore, Owings & Merrill LLP (SOM) Nr. Document Title 1 SOM Section Structural Steel - Part 1 - General 2 SOM Section Steel deck - Part 1 - General 3 SOM Section Strongback metal framing - Part 1 - General 4 SOM Section Exterior wall - Part 1 - General 5 SOM Section Glazing - Part 1 - General 6 SOM Section Window wall - Part 1 - General 7 SOM Section Wall louvers - Part 1 - General Drawing reference Table 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 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 3 SD-113 Wall type 1-5: SD-113 Rev 00 4 SD-150 Wall type 1-5: SD-150 Rev 00 5 SD-151 Wall type 1-5: SD-151 Rev 00 6 SD-170 Wall type 1-5: SD-170 Rev 00 7 SD-171 Wall type 1-5: SD-171 Rev 00

5 UNITISED CURTAIN WALL 5 of Materials Material Properties Material Extrusion 6063-T6 Sheets 5005-H14 Sword T6 Sword T651 Modulus of Elasticity, E (N/mm 2 ) Table Properties of Aluminum Min. Tensile Yield Strength, Modulus of Elasticity for deflection, E (N/mm 2 ) Fty (N/mm 2 ) Min. Tensile Ultimate Strength, Ftu (N/mm 2 ) Reference ADM 2005 Table 3.3-1M ADM 2005 Table 3.3-1M ADM 2005 Table 3.3-1M EN - AW7022 Table Properties of Stainless Steel Material Modulus Modulus Min. Tensile Yield Min. Tensile Reference of Elasticity, E (N/mm 2 ) of Elasticity for deflection, E (N/mm 2 ) Strength, Fty (N/mm 2 ) Ultimate Strength, Ftu (N/mm 2 ) Gr ASTM A240 Material A2/A (S31603) Modulus of Elasticity, E (N/mm 2 ) Table Properties of Screws Modulus Min. Tensile Yield of Elasticity for Strength, deflection, E (N/mm 2 ) Fty (N/mm 2 ) Min. Tensile Ultimate Strength, Ftu (N/mm 2 ) Reference EN ISO (ASTM A240) Note: 1 A4 austenitic steel is equivalent to S31603 (low carbon-316l) based on similar chemical properties in ASTM A240 and EN ISO , 2.3 Performance Criteria The deflection limits are based on the performance requirements stated in SOM Project Specification on Window Wall. Table 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 Programs used for the structural analysis Table Deflection Limits [Others] Criteria Member In-plane deflection 75% clearance Table 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 UP , ANSYS (R) Mechanical Toolbar Release 9.OA1 UP , Copyright 2004 SAS IP, Inc.

6 UNITISED CURTAIN WALL 6 of 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 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. 3.2 Façade Elevation: Wall Type 3 Fig Wall Type 3 Key Location Stack Jojnt Header Transom Transom 1 (Type 1) Stack Jojnt Header Transom Transom 1 (Type 2) Mullion Male & Female Mullion Male & Female Transom 2 (Type 1) Stack Joint Sill Transom Transom 2 (Type 2) Stack Joint Sill Transom (a) Structural System - 1 (b) Structural System - 2 Fig Wall Type 3 Layout

7 UNITISED CURTAIN WALL 7 of DESIGN NARRATIVE 4.1 Load Path 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 Lateral Loads Fig Dead Load Supports 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 Wind Load Supports

8 UNITISED CURTAIN WALL 8 of 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 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 δ MALE = δ FEMALE (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: EITOTAL = EIMALE + EI FEMALE (c) Load Sharing. Each part of the mullion setup share a certain amount of the internal moment in proportion to their stiffness: M MALE EI MALE = MTOTAL EITOTAL M FEMALE EI FEMALE = M TOTAL EITOTAL (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.

9 UNITISED CURTAIN WALL 9 of 117 (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. c = F l l where: c = elasticity constant in N/mm 2 = applied force in N l = deformation in mm l = length of the specimen in mm Fig 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 strength 1-20 C 55.1 N/mm 28.9 N/mm 89 N/mm C 50.4 N/mm - 69 N/mm C 50.8 N/mm 61.3 N/mm 57 N/mm 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: h ( ) Gc = c where: Gc = shear modulus of the core in N/mm2 b 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/mm N/mm C 69 N/mm N/mm C 57 N/mm N/mm 2 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 -20 C Gc = N/mm 2 Lower +80 C Gc = N/mm 2 Using the conservative value from the test, the design value of Gc used in the analysis is when the thermal break is elevated temperature of +80 C: Gc = N / mm 2

10 UNITISED CURTAIN WALL 10 of 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. q q (a) Section (b) Structural System - 1 (c) Structural System - 2 Fig 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 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. ρ, end rotation h SW Sword (short) gap/2 gap/2 q L SW EI' L M MAX δ MAX Deflected Mullion Mullion α, free slip angle Load Moment Deflection Fig Structural System - 1

11 UNITISED CURTAIN WALL 11 of 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 1 (ρ<α): Simple Beam ρ, end rotation h SW Sword (short) gap/2 gap/2 q 1 L SW EI' L M 1 δ 1 Deflected Mullion Mullion α, free slip angle Load Moment Deflection STAGE 2(ρ<α): Continuous Beam ρ, end rotation M SW Sword (longer) Sword q 2 EI L M 2 w 2 F couple,sw Sword F couple,sw Deflected Mullion Moment Deflection Load Coupling forces Fig Structural System - 2

12 UNITISED CURTAIN WALL 12 of 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. (3 CW UNITS BOTH SIDES) WING - A SYSTEM 2 (3 CW UNITS BOTH SIDES) WING - A SYSTEM 1 (ALL THE REST) SYSTEM 1 (ALL THE REST) SYSTEM 1 SYSTEM 1 (ALL THE REST) (ALL THE REST) WING - C WING - B WING - C WING - B SYSTEM 1 SYSTEM 2 (ALL THE REST) SYSTEM 2 (3 CW UNITS BOTH SIDES) LEVEL 7&8 TIER 0 (3 CW UNITS BOTH SIDES) SYSTEM 1 (ALL THE REST) SYSTEM 2 SYSTEM 2 (3 CW UNITS BOTH SIDES) (5 CW UNITS BOTH SIDES) LEVEL 25 TIER 1 SYSTEM 2 (3 CW UNITS BOTH SIDES) WING - A SYSTEM 2 (3 CW UNITS BOTH SIDES) WING - A SYSTEM 1 (ALL THE REST) SYSTEM 1 (ALL THE REST) SYSTEM 1 (ALL THE REST) SYSTEM 1 (ALL THE REST) WING - C WING - B WING - C WING - B SYSTEM 2 (5 CW UNITS BOTH SIDES) SYSTEM 1 (ALL THE REST) LEVEL 33 TIER 2 SYSTEM 2 (3 CW UNITS BOTH SIDES) SYSTEM 2 (3 CW UNITS BOTH SIDES) SYSTEM 1 (ALL THE REST) LEVEL 37 TIER 3 SYSTEM 2 (3 CW UNITS BOTH SIDES) SYSTEM 2 (3 CW UNITS BOTH SIDES) WING - A SYSTEM 2 (3 CW UNITS BOTH SIDES) WING - A SYSTEM 1 (ALL THE REST) SYSTEM 1 (ALL THE REST) SYSTEM 1 (ALL THE REST) SYSTEM 1 (ALL THE REST) WING - C WING - B WING - C WING - B SYSTEM 1 SYSTEM 2 (ALL THE REST) SYSTEM 2 (3 CW UNITS BOTH SIDES) LEVEL 52 TIER 4 (5 CW UNITS BOTH SIDES) SYSTEM 2 Fig Structural System Layout SYSTEM 1 SYSTEM 2 (ALL THE REST) (14 CW UNITS BOTH SIDES) (3 CW UNITS BOTH SIDES) LEVEL 63 TIER 5

13 UNITISED CURTAIN WALL 13 of 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. Constant Member Analysis Parameters - Span (H) Stress - Cross-sectional properties. Mullion - Span (H) Deflection - Cross-sectional properties. - Tributary widths (bu& bl) Stress - Cross-sectional properties Transom Deflection - Tributary widths (bu & bl) - Cross-sectional properties Table Critical Panel Evaluation Critical Variable Parameters Index Uniform load - Tributary width (W) - Wind load, Qw - Panel width (W) - Wind load, Qw - Span (W) - Wind load, Qw - Span (W) - Wind load, Qw Uniform load Bending moment Deflection System - 1 System - 2 W Qw = 4.83 KN/m Wing B (Zone C) QW = ±3.5 KPa W = m Qw W 2 = 7.58 KN Qw W 4 = 19.2 KN m 2 Wing B (Zone C) QW = ±3.0 KPa W = 1.59 m W Qw = 8.7 KN/m Qw W 2 = 16.9 KN Qw W 4 = 63.7 KN m 2 Wing B (Zone D) QW = ±4.5 KPa W = 1.94 m Wind load, ±Q w [KN/m 2 ] WING - A Table Panel Parameters WING - B WING - C Structural System System 1 System 2 System 1 System 2 System 1 System 2 Loading Zone A B C D 1 A B C D 1 A B C D 1 Level 7 & 8 (Tier 0) Level 25 (Tier 1) Level 33 (Tier 2) Level 37 (Tier 3) Level 52 (Tier 4) Level 63 (Tier 5) Panel width (up to ), m W = Mullion, Stress Index W Q w = Mullion, Deflection Index W Q w = Transom, Stress Index Q w W 2 = Transom, Deflection Index Q w W 4 = Note: 1 Structural system 2 extends to the shaded cell in the table. Fig Key Plan for Wind Load Distribution

14 UNITISED CURTAIN WALL 14 of 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. Wind force normal Wind force on protrusions (fins) Fig 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 Racking of CW Units

15 UNITISED CURTAIN WALL 15 of LOADING 5.1 Dead Load, QD Table Weight of Structural System - 1, Total weight = 3.07 kn (313 kg.) Element Refer. Specific Wt. Real constant 1 Unit Wt. Real constant 2 Total wt. k1 = k1 k2 W = k2 Aluminum Extrusion 0.53 kn Male mullion 6.1.1(a) 27.0 kn/m 3 A = 2414 mm kn/m L = 3.70m 0.24 kn Female mullion 6.1.2(a) 27.0 kn/m 3 A = 883 mm kn/m L = 3.70m 0.09 kn Header transom 6.2.1(a) 27.0 kn/m 3 A = 946 mm kn/m L = 1.39m 0.04 kn Sill transom 6.2.2(a) 27.0 kn/m 3 A = 1282 mm kn/m L = 1.39m 0.05 kn Transom (a) 27.0 kn/m 3 A = 1067 mm kn/m L = 1.39m 0.04 kn Transom (a) 27.0 kn/m 3 A = 984 mm kn/m L = 1.39m 0.04 kn Sword kn/m 3 A = 2100 mm kn/m L = 0.50m 0.03 kn Upper spandrel panel 0.41 kn/m kn Stainless Steel Sheet M kn/ m 3 t = 1.5mm 0.12 kn/m 2 A = 0.65 m kn Steel Sheet M kn/m 3 t = 1.5mm 0.12 kn/m 2 A = 0.65 m kn Insulation M9 0.7 kn/m 3 t = 70mm 0.05 kn/m 2 A = 0.65 m kn Steel Sheet M kn/m 3 t = 1.5mm 0.12 kn/m 2 A = 0.65 m kn Vision panel 0.45 kn/m kn Glass GL kn/m 3 t = 18mm 0.45 kn/m 2 A = 3.93 m kn Lower Spandrel panel 0.45 kn/m kn Glass GL kn/m 3 t = 18mm 0.45 kn/m 2 A = 0.54 m kn Accessories 0.05 kn/m kn Thermal break, setting block, gaskets, sealant, screws, etc. approx.10%w A = 5.12 m kn Table Weight of Structural System - 2, Total weight = 4.63 kn (472 kg.) Element Refer. Specific Wt. Real constant 1 Unit Wt. Real constant 2 Total wt. k1 = k1 k2 W = k2 Aluminum Extrusion 0.76 kn Male mullion 6.1.1(a) 27.0 kn/m 3 A = 2414 mm kn/m L = 3.70m 0.24 kn Female mullion 6.1.2(a) 27.0 kn/m 3 A = 883 mm kn/m L = 3.70m 0.09 kn Header transom 7.2.1(a) 27.0 kn/m 3 A = 946 mm kn/m L = 1.94m 0.05 kn Sill transom 7.2.2(a) 27.0 kn/m 3 A = 1282 mm kn/m L = 1.94m 0.07 kn Transom (a) 27.0 kn/m 3 A = 2181 mm kn/m L = 1.94m 0.11 kn Transom (a) 27.0 kn/m 3 A = 2050 mm kn/m L = 1.94m 0.11 kn Sword kn/m 3 A = 2100 mm kn/m L = 1.42m 0.09 kn Upper spandrel panel 0.41 kn/m kn Stainless Steel Sheet M kn/ m 3 t = 1.5mm 0.12 kn/m 2 A = 0.91 m kn Steel Sheet M kn/m 3 t = 1.5mm 0.12 kn/m 2 A = 0.91 m kn Insulation M9 0.7 kn/m 3 t = 70mm 0.05 kn/m 2 A = 0.91 m kn Steel Sheet M kn/m 3 t = 1.5mm 0.12 kn/m 2 A = 0.91 m kn Vision panel 0.50 kn/m kn Glass GL kn/m 3 t = mm 0.50 kn/m 2 A = 5.51 m kn Lower Spandrel panel 0.50 kn/m kn Glass GL kn/m 3 t = mm 0.50 kn/m 2 A = 0.76 m kn Accessories 0.05 kn/m kn Thermal break, setting block, gaskets, sealant, screws, etc. approx.10%w A = 7.18 m kn

16 UNITISED CURTAIN WALL 16 of Wind Load, QW Applied wind loads are in accordance to SOM Drawing Nr. AL-00-AR-502 Building Enclosure Wind Pressure Diagrams (Addendum 057) Wind Load on Wall Elements. A comprehensive summary of the wind load stated above is found on Table 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 Wind Load on Vertical Fins Vertical Fin 1 LOAD, KN/m 2 Zone E Level 7 & Level Level Level Level Level Note. 1 Values for the parapet walls are conservatively adopted for the vertical fins.

17 UNITISED CURTAIN WALL 17 of SECTION PROPERTIES STRUCTURAL SYSTEM Mullion Profiles Male Mullion (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig 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 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 A1 = 266.5mm2 c.g A2a = 623.7mm2 286 c.g. c.g.2 DIAGONALLY ORIENTED WEBS: A2b = 674.1mm A2 = A2a+ A2bxcos(13deg.) A2 = mm2 13 A = A1 + A2 A = mm2 SECTION PROPERTIES SHEAR AREA Fig Parameters in Determining the Effective Section Modulus DISTANCES

18 UNITISED CURTAIN WALL 18 of 117 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 = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 1.7 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 2.43E+05 mm 4 I o2 = 6.29E+06 mm 4 Moment of inertia c 11 = 30.4 mm c 22 = mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 1.09E+07 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 6.53E+06 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.74E+07 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 2.01E+06 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 4.46E-06 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G p 2 I) = 0.07 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = 0.00 Ditto D 2 = w I c /(2G p I) = 1.6E-07 /mm Ditto D 3 = w L/(12E I) = 2.6E-10 /mm 2 Ditto D 4 = -w/(24e I) = -3.5E-14 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -2.8E-05 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 3.91 for x = L/2 = 1850 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 2.57 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 1.72E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 1.38E+07 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 1.36E+07 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 1.34E+05 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 1.05E+05 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 2.20E-09 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 5.79 Shear flow per uniform unit load, w 2.00E+07 Effective I Curve Moment of Inertia [mm 4] 1.50E E E+07 I (fully composite) Ie for upper bound Gc Ie for lower bound Gc I'e for upper bound Gc I'e for lower bound Gc Io (non-composite) 5.00E Span, L [m]

19 UNITISED CURTAIN WALL 19 of 117 (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 L b1 = 2.84 m Unbraced length for face-1 under compression L b2 = 3.70 m Unbraced length for face-2 under compression I y = 6.4E+05 mm 4 Moment of inertia of profile J = 8.8E+05 mm 4 Torsional constant of profile S c1 = 1.3E+05 mm 3 Section modulus for face 1 under compression S c2 = 1.1E+05 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b1 S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b2 S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling Element 2 c.g. Element 3 Fig Elements Critical Under Local Buckling

20 UNITISED CURTAIN WALL 20 of 117 The governing allowable stress, under local buckling as calculated below, is summarized as follows: ace 1, considering elements 1, & 2 ace 2, considering element 3 c1 = MPa c2 = 131.0MPa Element 1 ADM 2005, : 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 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.35 Coefficient for determining slenderness limit B p = F cy [1+(F cy ) 1/3 /21.7] = Buckling formula intercept for comp. in flat element D p = (B p /10)(B p /E) 0.5 = 1.18 Buckilng formula slope for comp. in flat elements S 1 = (B p -F cy )/1.6D p = 23.0 Lower bound slender limit S 2 = k 1 B p /1.6D p = 39.5 Upper bound slender limit b/t = 16.9 Slenderness ratio Criteria = b/t < S1 Short compression element F c = F cy /n y = Mpa Allowable compressive stress Element 2 ADM 2005, : Compression in Elements of Beams - Flat Elements Supported on Both Edges Clause Action Notes b = 45.8 m Width of the flat element t = 1.8 mm Thickness of the flat element Cc = mm Distance from N.A. to heavily compressed edge Co = mm Distance from N.A. to other edge 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.50 Coefficient for determining slenderness limit Co/Cc = 0.66 m = 1.15+Co/(2Cc) = 1.48 for -1 < Co/Cc < B br = 1.3F cy [1+(F cy ) 1/3 /13.3] = Buckling formula intercept for comp. in flat element D br = (B br /20)(6B br /E) 0.5 = 2.57 Buckilng formula slope for comp. in flat elements S 1 = (B br -1.3F cy )/m D br = 24.2 Lower bound slender limit S 2 = k 1 B br /m D br = 41.1 Upper bound slender limit h/t = 25.4 Slenderness ratio Criteria = S1 < h/t < S2 Intermediate compression element F c = (1/n y )[B br -m D br (h/t)] = Mpa Allowable compressive stress

21 UNITISED CURTAIN WALL 21 of 117 Element 3 ADM 2005, : Compression in Elements of Beams - Flat Elements Supported on Both Edges Clause Action Notes b = 66.3 m Width of the flat element t = 2.5 mm Thickness of the flat element Cc = mm Distance from N.A. to heavily compressed edge Co = mm Distance from N.A. to other edge 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.50 Coefficient for determining slenderness limit Co/Cc = 0.54 m = 1.15+Co/(2Cc) = 1.42 for -1 < Co/Cc < B br = 1.3F cy [1+(F cy ) 1/3 /13.3] = Buckling formula intercept for comp. in flat element D br = (B br /20)(6B br /E) 0.5 = 2.57 Buckilng formula slope for comp. in flat elements S 1 = (B br -1.3F cy )/m D br = 25.2 Lower bound slender limit S 2 = k 1 B br /m D br = 42.9 Upper bound slender limit h/t = 26.5 Slenderness ratio Criteria = S1 < h/t < S2 Intermediate compression element F c = (1/n y )[B br -m D br (h/t)] = Mpa Allowable compressive stress Female Mullion (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig Section Properties (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 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.

22 UNITISED CURTAIN WALL 22 of 117 c.g.1 c.g. c.g.2 Fig Parameters in Determining the Effective Section Modulus

23 UNITISED CURTAIN WALL 23 of 117 Effective Section Modulus Calculation Female 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 = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 1.7 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 2.37E+05 mm 4 I o2 = 3.84E+04 mm 4 Moment of inertia c 11 = 29.8 mm c 22 = 20.7 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 69.0 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 9.67E+05 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 2.75E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.24E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 3.49E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 1.84E-05 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G p 2 I) = 0.12 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = Ditto D 2 = w I c /(2G p I) = 1.1E-06 /mm Ditto D 3 = w L/(12E I) = 3.6E-09 /mm 2 Ditto D 4 = -w/(24e I) = -4.9E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -1.6E-08 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 7.93 for x = L/2 = 1850 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 2.22E-05 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 1.10E+06 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 1.09E+06 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 2.30E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 1.80E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 3.12E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = Shear flow per uniform unit load, w Moment of Inertia [mm 4] 1.50E E E E E+06 Effective I Curve Ie I (fully for upper composite) bound I'e Gc for upper bound Gc Ie for lower bound I'e Gc for lower bound Gc Io (non-composite) Span, L [m]

24 UNITISED CURTAIN WALL 24 of 117 (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 L b = 2.84 m Unbraced length of the member for bending I y = 9.9E+04 mm 4 Moment of inertia of profile J = 6.7E+04 mm 4 Torsional constant of profile S c1 = 2.3E+04 mm 3 Section modulus for face 1 under compression S c2 = 1.8E+04 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling Element 1 Fig Elements Critical Under Local Buckling The governing allowable stress, under local buckling as calculated below, is summarized as follows: ace 1, considering element 1 c1 = 96.1MPa ace 2, no element is critical under local buckling.

25 UNITISED CURTAIN WALL 25 of 117 Element 1 ADM 2005, : Uniform Compression in Elements of Beams - Flat Elements Supported on One Edge Clause Action Notes b = 27.3 m Width of the flat element t = 3.0 mm Thickness of the flat element 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.35 Coefficient for determining slenderness limit B p = F cy [1+(F cy ) 1/3 /21.7] = Buckling formula intercept for comp. in flat element D p = (B p /10)(B p /E) 0.5 = 1.18 Buckilng formula slope for comp. in flat elements S 1 = (B p -F cy )/5.1D p = 7.2 Lower bound slender limit S 2 = k 1 B p /5.1D p = 12.4 Upper bound slender limit b/t = 9.1 Slenderness ratio Criteria = S1 < b/t < S2 Intermediate compression element F c = (1/n y )[B p -5.1D p (b/t)] = Mpa Allowable compressive stress

26 UNITISED CURTAIN WALL 26 of Stack Joint Transom Profiles Header Transom (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS finite element analysis. Fig Section Properties (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 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. A1 = 225.5mm2 A2 = 121.1mm2 FACE-2 c.g.2 FACE-1 c.g. c.g.1 A = A1 + A2 A = 346.6mm2 Fig Parameters in Determining the Effective Section Modulus

27 UNITISED CURTAIN WALL 27 of 117 Effective Section Modulus Calculation Header Transom (System 1) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.59 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.3 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 9.41E+04 mm 4 I o2 = 2.36E+04 mm 4 Moment of inertia c 11 = 23.4 mm c 22 = 16.5 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 64.5 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 9.27E+05 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 1.18E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.05E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 2.67E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 3.29E-05 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G p 2 I) = 0.10 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = 0.00 Ditto D 2 = w I c /(2G p I) = 1.7E-06 /mm Ditto D 3 = w L/(12E I) = 1.8E-09 /mm 2 Ditto D 4 = -w/(24e I) = -5.8E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -1.1E-05 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 4.56 for x = L/2 = 795 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 2.21 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 7.64E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 5.47E+05 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 5.38E+05 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 1.66E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 1.61E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 3.01E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 8.54 Shear flow per uniform unit load, w Effective I Curve Moment of Inertia [mm 4] 1.00E E E+05 I (fully composite) Ie for upper bound Gc I'e for upper bound Gc Ie for lower bound Gc I'e for lower bound Gc Io (non-composite) 0.00E Span, L [m]

28 UNITISED CURTAIN WALL 28 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005: Specification for Aluminum Structures - Allowable Stress Design Clause Action Notes Extrusion Parameters L b = 1.59 m Unbraced length of the member for bending I y = 7.4E+05 mm 4 Moment of inertia of profile J = 8.4E+04 mm 4 Torsional constant of profile S c1 = 1.7E+04 mm 3 Section modulus for face 1 under compression S c2 = 1.6E+04 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling Element 2 c.g. Element 1 FACE-2 FACE-1 Fig Elements Critical Under Local Buckling The governing allowable stress, under local buckling as calculated below, is summarized as follows: ace 1, considering element 1 c1 = 88.2MPa ace 2, considering element 2 c2 = 34.8MPa

29 UNITISED CURTAIN WALL 29 of 117 Element 1 ADM 2005, : Uniform Compression in Elements of Beams - Flat Elements Supported on Both Edges Clause Action Notes b = 64.7 m Width of the flat element t = 1.8 mm Thickness of the flat element 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.35 Coefficient for determining slenderness limit B p = F cy [1+(F cy ) 1/3 /21.7] = Buckling formula intercept for comp. in flat element D p = (B p /10)(B p /E) 0.5 = 1.18 Buckilng formula slope for comp. in flat elements S 1 = (B p -F cy )/1.6D p = 23.0 Lower bound slender limit S 2 = k 1 B p /1.6D p = 39.5 Upper bound slender limit b/t = 35.9 Slenderness ratio Criteria = S1 < b/t < S2 Intermediate compression element F c = (1/n y )[B p -1.6D p (b/t)] = Mpa Allowable compressive stress Element 2 ADM 2005, : Uniform Compression in Elements of Beams - Flat Elements Supported on One Edge Clause Action Notes b = 59.8 m Width of the flat element t = 2.0 mm Thickness of the flat element 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.35 Coefficient for determining slenderness limit K 2 = 2.27 Ditto B p = F cy [1+(F cy ) 1/3 /21.7] = Buckling formula intercept for comp. in flat element D p = (B p /10)(B p /E) 0.5 = 1.18 Buckilng formula slope for comp. in flat elements S 1 = (B p -F cy )/5.1D p = 7.2 Lower bound slender limit S 2 = k 1 B p /5.1D p = 12.4 Upper bound slender limit b/t = 29.9 Slenderness ratio Criteria = S2 < b/t Slender compression element F c = k 2 /n y (B p E) 0.5 /(5.1b/t)= Mpa Allowable compressive stress

30 UNITISED CURTAIN WALL 30 of Sill Transom (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig Section Properties (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 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. A1 = 280.8mm2 A2 = 115.2mm2 2 x P 37.3 Z c.g. FACE-2 c.g.2 c.g.1 FACE-1 A = A1 + A2 A = 396.0mm2 Fig Parameters in Determining the Effective Section Modulus

31 UNITISED CURTAIN WALL 31 of 117 Effective Section Modulus Calculation Sill Transom (System 1) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.59 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.3 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 4.20E+05 mm 4 I o2 = 1.20E+04 mm 4 Moment of inertia c 11 = 32.8 mm c 22 = 18.8 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 59.4 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 8.12E+05 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 4.32E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.24E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 3.08E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 1.03E-05 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G 2 p I) = 0.20 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = 0.00 Ditto D 2 = w I c /(2G p I) = 1.1E-06 /mm Ditto D 3 = w L/(12E I) = 1.5E-09 /mm 2 Ditto D 4 = -w/(24e I) = -4.9E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -1.2E-03 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 2.56 for x = L/2 = 795 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 1.67 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 5.48E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 7.22E+05 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 7.09E+05 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 2.11E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 2.02E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 1.60E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 5.36 Shear flow per uniform unit load, w 1.50E+06 Effective I Curve Moment of Inertia [mm 4] 1.00E E E+05 I (fully composite) Ie for upper bound Gc I'e for upper bound Gc Ie for lower bound Gc I'e for lower bound Gc Io (non-composite) 0.00E Span, L [m]

32 UNITISED CURTAIN WALL 32 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005: Specification for Aluminum Structures - Allowable Stress Design Clause Action Notes Extrusion Parameters L b = 1.59 m Unbraced length of the member for bending I y = 9.9E+05 mm 4 Moment of inertia of profile J = 8.2E+04 mm 4 Torsional constant of profile S c1 = 2.1E+04 mm 3 Section modulus for face 1 under compression S c2 = 2.0E+04 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling FACE-1 FACE-2 c.g. Element 1 Fig Elements Critical Under Local Buckling The governing allowable stress, under local buckling as calculated below, is summarized as follows: ace 1, considering element 1 c1 = 69.9MPa ace 2, no element is critical under local buckling.

33 UNITISED CURTAIN WALL 33 of 117 Element 1 ADM 2005, : Uniform Compression in Elements of Beams - Flat Elements Supported on One Edge Clause Action Notes b = 55.0 m Width of the flat element t = 3.7 mm Thickness of the flat element 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.35 Coefficient for determining slenderness limit K 2 = 2.27 Ditto B p = F cy [1+(F cy ) 1/3 /21.7] = Buckling formula intercept for comp. in flat element D p = (B p /10)(B p /E) 0.5 = 1.18 Buckilng formula slope for comp. in flat elements S 1 = (B p -F cy )/5.1D p = 7.2 Lower bound slender limit S 2 = k 1 B p /5.1D p = 12.4 Upper bound slender limit b/t = 14.9 Slenderness ratio Criteria = S2 < b/t Slender compression element F c = k 2 /n y (B p E) 0.5 /(5.1b/t)= Mpa Allowable compressive stress

34 UNITISED CURTAIN WALL 34 of Transom 1 Profile Transom -1 (Type 1) (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig Section Properties (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 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. A1 = 242.2mm2 A2 = 154.7mm2 FACE-2 c.g.2 c.g.1 FACE-1 c.g. A = A1 + A2 A = 396.9mm2 Fig Parameters in Determining the Effective Section Modulus

35 UNITISED CURTAIN WALL 35 of 117 Effective Section Modulus Calculation Transom 1 (Type 1: System 1) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.59 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.3 KN m Maximum bending moment due to unit load A = 96.9 m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 3.15E+05 mm 4 I o2 = 3.52E+04 mm 4 Moment of inertia c 11 = 30.7 mm c 22 = 14.2 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 61.1 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 9.47E+05 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 3.50E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.30E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 2.92E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 1.21E-05 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G p 2 I) = 0.21 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = 0.00 Ditto D 2 = w I c /(2G p I) = 1.3E-06 /mm Ditto D 3 = w L/(12E I) = 1.5E-09 /mm 2 Ditto D 4 = -w/(24e I) = -4.7E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -8.2E-04 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 2.76 for x = L/2 = 795 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 1.75 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 5.72E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 6.91E+05 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 6.44E+05 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 1.91E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 2.51E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 1.75E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 6.09 Shear flow per uniform unit load, w 1.50E+06 Effective I Curve Moment of Inertia [mm 4] 1.00E E E+05 I (fully composite) Ie for upper bound Gc I'e for upper bound Gc Ie for lower bound Gc I'e for lower bound Gc Io (non-composite) 0.00E Span, L [m]

36 UNITISED CURTAIN WALL 36 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005, : Compression in Beams, Extreme Fiber, Gross Section - Tubular Shapes Clause Action Notes Extrusion Parameters L b = 1.39 m Unbraced length of the member for bending I y = 6.4E+05 mm 4 Moment of inertia of profile J = 5.3E+05 mm 4 Torsional constant of profile S c1 = 1.9E+04 mm 3 Section modulus for face 1 under compression S c2 = 2.5E+04 mm 4 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = 90.7 Slenderness ratio for face-1 under compression Criteria = S S1 Short beam-compression meber F c1 = F cy /n y = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S S1 Short beam-compression meber F c2 = F cy /n y = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling c.g. Element 1 Fig Elements Critical Under Local Buckling The governing allowable stress, under local buckling as calculated below, is summarized as follows: ace 1, considering element 1 c1 = 92.4MPa ace 2, no element is critical under local buckling.

37 UNITISED CURTAIN WALL 37 of 117 Element 1 ADM 2005, : Uniform Compression in Elements of Beams - Flat Elements Supported on Both Edges Clause Action Notes b = 64.4 m Width of the flat element t = 2.0 mm Thickness of the flat element 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.35 Coefficient for determining slenderness limit B p = F cy [1+(F cy ) 1/3 /21.7] = Buckling formula intercept for comp. in flat element D p = (B p /10)(B p /E) 0.5 = 1.18 Buckilng formula slope for comp. in flat elements S 1 = (B p -F cy )/1.6D p = 23.0 Lower bound slender limit S 2 = k 1 B p /1.6D p = 39.5 Upper bound slender limit b/t = 32.2 Slenderness ratio Criteria = S1 < b/t < S2 Intermediate compression element F c = (1/n y )[B p -1.6D p (b/t)] = Mpa Allowable compressive stress 6.4 Transom 2 Profile Transom 2 (Type 1) (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig Section Properties (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 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.

38 UNITISED CURTAIN WALL 38 of 117 A1 = 263.2mm2 A2 = 151.4mm x P DL FACE-2 c.g.2 c.g.1 FACE-1 c.g. A = A1 + A2 A = 414.6mm2 Fig Parameters in Determining the Effective Section Modulus

39 UNITISED CURTAIN WALL 39 of 117 Effective Section Modulus Calculation Transom 2 (Type 1: System 1) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.59 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.3 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 2.96E+05 mm 4 I o2 = 2.91E+04 mm 4 Moment of inertia c 11 = 31.0 mm c 22 = 12.7 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 62.3 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 8.52E+05 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 3.25E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.18E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 3.06E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 1.36E-05 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G 2 p I) = 0.17 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = 0.00 Ditto D 2 = w I c /(2G p I) = 1.2E-06 /mm Ditto D 3 = w L/(12E I) = 1.6E-09 /mm 2 Ditto D 4 = -w/(24e I) = -5.1E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -4.9E-04 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 2.94 for x = L/2 = 795 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 1.79 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 6.01E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 6.74E+05 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 6.63E+05 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 1.83E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 2.25E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 1.85E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 6.11 Shear flow per uniform unit load, w 1.50E+06 Effective I Curve Moment of Inertia [mm 4] 1.00E E E+05 I (fully composite) Ie for upper bound Gc I'e for upper bound Gc Ie for lower bound Gc I'e for lower bound Gc Io (non-composite) 0.00E Span, L [m]

40 UNITISED CURTAIN WALL 40 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005, : Compression in Beams, Extreme Fiber, Gross Section - Tubular Shapes Clause Action Notes Extrusion Parameters L b = 1.39 m Unbraced length of the member for bending I y = 5.5E+05 mm 4 Moment of inertia of profile J = 4.9E+05 mm 4 Torsional constant of profile S c1 = 1.8E+04 mm 3 Section modulus for face 1 under compression S c2 = 2.3E+04 mm 4 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = 97.4 Slenderness ratio for face-1 under compression Criteria = S S1 Short beam-compression meber F c1 = F cy /n y = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S S1 Short beam-compression meber F c2 = F cy /n y = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling c.g. Element 1 Fig Elements Critical Under Local Buckling The governing allowable stress, under local buckling as calculated below, is summarized as follows: ace 1, considering element 1 c1 = 89.0MPa ace 2 No element is critical under lateral buckling.

41 UNITISED CURTAIN WALL 41 of 117 Element 1 ADM 2005, : Uniform Compression in Elements of Beams - Flat Elements Supported on Both Edges Clause Action Notes b = 70.4 m Width of the flat element t = 2.0 mm Thickness of the flat element 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus n y = 1.65 Factor of safety on yield strength k 1 = 0.35 Coefficient for determining slenderness limit B p = F cy [1+(F cy ) 1/3 /21.7] = Buckling formula intercept for comp. in flat element D p = (B p /10)(B p /E) 0.5 = 1.18 Buckilng formula slope for comp. in flat elements S 1 = (B p -F cy )/1.6D p = 23.0 Lower bound slender limit S 2 = k 1 B p /1.6D p = 39.5 Upper bound slender limit b/t = 35.2 Slenderness ratio Criteria = S1 < b/t < S2 Intermediate compression element F c = (1/n y )[B p -1.6D p (b/t)] = Mpa Allowable compressive stress

42 UNITISED CURTAIN WALL 42 of Sword Profile Sword (Type-1) (a) Cross-sectional Properties. Fig Sword Length 140 Fig Sword Cross-Section

43 UNITISED CURTAIN WALL 43 of SECTION PROPERTIES STRUCTURAL SYSTEM Mullion Profiles Male Mullion Section properties of male mullion for Structural system 1 and 2 are the same. Please refer to Female Mullion Section properties of female mullion for Structural system 1 and 2 are the same. Please refer to 6.1.2

44 UNITISED CURTAIN WALL 44 of Stack Joint Transom Profiles Header Transom (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS finite element analysis. Fig Section Properties (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 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. A1 = 225.5mm2 A2 = 121.1mm2 FACE-2 c.g.2 FACE-1 c.g. c.g.1 A = A1 + A2 A = 346.6mm2 Fig Parameters in Determining the Effective Section Modulus

45 UNITISED CURTAIN WALL 45 of 117 Effective Section Modulus Calculation Header Transom (System 2) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.94 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.5 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 9.41E+04 mm 4 I o2 = 2.36E+04 mm 4 Moment of inertia c 11 = 23.4 mm c 22 = 16.5 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 64.5 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 9.27E+05 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 1.18E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.05E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 2.67E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 3.29E-05 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G 2 p I) = 0.10 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = Ditto D 2 = w I c /(2G p I) = 1.7E-06 /mm Ditto D 3 = w L/(12E I) = 2.2E-09 /mm 2 Ditto D 4 = -w/(24e I) = -5.8E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -1.5E-06 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 5.57 for x = L/2 = 970 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 4.12 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 9.82E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 6.49E+05 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 6.41E+05 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 1.80E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 1.68E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 3.25E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = Shear flow per uniform unit load, w Effective I Curve Moment of Inertia [mm 4] 1.00E E E+05 I (fully composite) Ie for upper bound Gc I'e for upper bound Gc Ie for lower bound Gc I'e for lower bound Gc Io (non-composite) 0.00E Span, L [m]

46 UNITISED CURTAIN WALL 46 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005: Specification for Aluminum Structures - Allowable Stress Design Clause Action Notes Extrusion Parameters L b = 1.94 m Unbraced length of the member for bending I y = 7.4E+05 mm 4 Moment of inertia of profile J = 8.4E+04 mm 4 Torsional constant of profile S c1 = 1.8E+04 mm 3 Section modulus for face 1 under compression S c2 = 1.7E+04 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling The governing allowable stress, under local buckling as calculated in 6.2.1(d), is summarized as follows: ace 1, considering element 1 ace 2, considering element 2 c1 = 88.2MPa c2 = 34.8MPa

47 UNITISED CURTAIN WALL 47 of Sill Transom (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig Section Properties (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 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. A1 = 280.8mm2 A2 = 115.2mm2 2 x P Z 37.3 c.g. FACE-2 c.g.2 c.g.1 FACE-1 A = A1 + A2 A = 396.0mm2 Fig Parameters in Determining the Effective Section Modulus

48 UNITISED CURTAIN WALL 48 of 117 Effective Section Modulus Calculation Sill Transom (System 2) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.94 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.5 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 4.20E+05 mm 4 I o2 = 1.20E+04 mm 4 Moment of inertia c 11 = 32.8 mm c 22 = 18.8 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 59.4 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 8.12E+05 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 4.32E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 1.24E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 3.08E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 1.03E-05 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G 2 p I) = 0.20 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = Ditto D 2 = w I c /(2G p I) = 1.1E-06 /mm Ditto D 3 = w L/(12E I) = 1.9E-09 /mm 2 Ditto D 4 = -w/(24e I) = -4.9E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -4.0E-04 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 3.12 for x = L/2 = 970 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 3.17 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 7.42E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 8.45E+05 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 8.33E+05 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 2.23E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 1.99E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 1.81E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 7.26 Shear flow per uniform unit load, w 1.50E+06 Effective I Curve Moment of Inertia [mm 4] 1.00E E E+05 I (fully composite) Ie for upper bound Gc I'e for upper bound Gc Ie for lower bound Gc I'e for lower bound Gc Io (non-composite) 0.00E Span, L [m]

49 UNITISED CURTAIN WALL 49 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005: Specification for Aluminum Structures - Allowable Stress Design Clause Action Notes Extrusion Parameters L b = 1.94 m Unbraced length of the member for bending I y = 9.9E+05 mm 4 Moment of inertia of profile J = 8.2E+04 mm 4 Torsional constant of profile S c1 = 2.2E+04 mm 3 Section modulus for face 1 under compression S c2 = 2.0E+04 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling The governing allowable stress, under local buckling as calculated in 6.2.2(d), is summarized as follows: ace 1, considering element 1 c1 = 69.9MPa ace 2, no element is critical under local buckling.

50 UNITISED CURTAIN WALL 50 of Transom 1 Profile Transom 1 (Type 2) (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig Section Properties (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 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. A1 = 342.1mm2 A2 = 161.0mm2 2 x P Z 38.7 c.g.2 c.g.1 c.g. FACE-2 FACE-1 A = A1 + A2 A = 503.1mm2 Fig Parameters in Determining the Effective Section Modulus

51 UNITISED CURTAIN WALL 51 of 117 Effective Section Modulus Calculation Transom 1 (Type 2: System 2) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.94 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.5 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 7.30E+05 mm 4 I o2 = 3.72E+04 mm 4 Moment of inertia c 11 = 23.6 mm c 22 = 15.9 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 66.5 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 1.81E+06 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 7.67E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 2.58E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 3.59E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 6.78E-06 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G 2 p I) = 0.29 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = 0.00 Ditto D 2 = w I c /(2G p I) = 9.8E-07 /mm Ditto D 3 = w L/(12E I) = 9.1E-10 /mm 2 Ditto D 4 = -w/(24e I) = -2.3E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -1.8E-03 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 2.53 for x = L/2 = 970 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 2.00 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 4.29E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 1.33E+06 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 1.31E+06 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 5.10E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 4.13E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 1.05E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 6.24 Shear flow per uniform unit load, w 3.00E+06 Effective I Curve Moment of Inertia [mm 4] 2.50E E E E E+06 I (fully composite) Ie for upper bound Gc Ie I'e for for lower upper bound bound Gc Gc I'e for lower bound Gc Io (non-composite) 5.00E Span, L [m]

52 UNITISED CURTAIN WALL 52 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005: Specification for Aluminum Structures - Allowable Stress Design Clause Action Notes Extrusion Parameters L b = 1.94 m Unbraced length of the member for bending I y = 1.0E+06 mm 4 Moment of inertia of profile J = 9.9E+05 mm 4 Torsional constant of profile S c1 = 4.5E+04 mm 3 Section modulus for face 1 under compression S c2 = 4.3E+04 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling No element is critical under local buckling.

53 UNITISED CURTAIN WALL 53 of Transom 2 Profile Transom 2 (Type 2) (a) Cross-sectional Properties. The properties of the profile cross-section are calculated using ANSYS beam plot section command. Fig Section Properties (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 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. A1 = 334.8mm2 A2 = 199.3mm x P Z c.g.1 c.g. FACE-2 c.g.2 FACE-1 A = A1 + A2 A = 534.1mm2 Fig Parameters in Determining the Effective Section Modulus

54 UNITISED CURTAIN WALL 54 of 117 Effective Section Modulus Calculation Transom 2 (Type 2: System 2) AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes Extrusion Parameters L = 1.94 m. Unsupported span of the member E = Mpa Young's modulus of aluminum faces G c = Mpa Design shear modulus of thermal break w = 1.0 kn/m Considered uniform unit load M = w L 2 /8 = 0.5 KN m Maximum bending moment due to unit load A = m Shear area of aluminum h = mm Overall depth of extrusion g = 17.9 mm Gap (clearance) between faces D c = 27.0 mm Maximum cavity depth b = 4.0 mm Average width of thermal break core a 1 = mm 2 a 2 = mm 2 Cross sectional area I o1 = 7.30E+05 mm 4 I o2 = 2.84E+04 mm 4 Moment of inertia c 11 = 23.8 mm c 22 = 13.9 mm Extreme fiber dist. to c.g.of f D = h - (c 11 +c 22 ) = 68.3 mm Distance between centroidal axes of both faces 7.5.4(1) I c = a 1 a 2 D 2 /(a 1 +a 2 ) = 1.37E+06 mm 4 For the case where both faces are same material 7.5.4(2) I o = I o1 +I o2 = 7.58E+05 mm 4 Lower bound on stiffness I'e (no composite action) 7.5.4(3) I = I c + I o = 2.13E+06 mm 4 Upper bound on stiffness I'e (full composite action) 7.5.4(5) Composite Analysis G p = I b D 2 G c /(I c D c ) = 4.13E+05 N Geometric and core material parameter 7.5.4(6) c = G p /(E I o ) = 7.91E-06 /mm 2 Buckilng formula slope for comp. in beam flanges 7.5.4(Table3) D 0 = w E I o I c /(G p 2 I) = 0.20 mm Constant for the elastic curve formula D 1 = -w L I c /(2G p I)-wL 3 /(24E I) = 0.00 Ditto D 2 = w I c /(2G p I) = 7.8E-07 /mm Ditto D 3 = w L/(12E I) = 1.1E-09 /mm 2 Ditto D 4 = -w/(24e I) = -2.8E-13 /mm 3 D 5 = 0.00 r = (L/2)(c) 0.5 = (Table5) F 1 = -w I c e -r /[c G p I(e r +e -r )] = -8.4E-04 mm Complementary constants F 2 = F 1 e 2r = mm Ditto 7.5.4(8a) p = x(c) 0.5 = 2.73 for x = L/2 = 970 mm 7.5.4(8) y =D 5 x 5 +D 4 x 4 +D 3 x 3 +D 2 x2+d 1 x+d 0 +F 1 e p +F 2 /e p = 2.02 mm Calculated effective maximum deflection, D5 = (21) y'' =20D 5 x 3 +12D 4 x 2 +6D 3 x+2d 2 +c(f 1 e p +F 2 /e p ) = 4.56E-06 /mm 7.5.4(10) I e = w L 4 /(76.8E y) = 1.33E+06 mm 4 Effective moment of inertia w/o shear deformation 7.5.4(17) I' e = I e /{ 1+[25.6I e /(L 2 A)} = 1.30E+06 mm 4 Effective moment of inertia considering shear def (19) S e1 = M/[(M-E I o y'')/(a 1 D)+E c 11 y''] = 4.96E+04 mm 3 Effective section face (20) S e2 = M/[(M-E I o y'')/(a 2 D)+E c 22 y''] = 3.37E+04 mm 3 Effective section face (25a) Shear Flow Data y''' =6D 3 +c 1.5 (F 1 -F 2 ) = 1.10E-08 /mm (24) V c /w = w L/2 -E I o y''' = mm Shear resisted by thermal break per unif. unit load 7.5.4(28) V c /(D w) = 5.81 Shear flow per uniform unit load, w 2.50E+06 Effective I Curve Moment of Inertia [mm 4] 2.00E E E E+06 I (fully composite) Ie for upper bound Gc I'e for upper bound Gc Ie for lower bound Gc I'e for lower bound Gc Io (non-composite) 5.00E Span, L [m]

55 UNITISED CURTAIN WALL 55 of 117 (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 (Iyy,1, J1) and face 2 (Iyy,2,J2) profiles. ADM 2005: Specification for Aluminum Structures - Allowable Stress Design Clause Action Notes Extrusion Parameters L b = 1.94 m Unbraced length of the member for bending I y = 9.7E+05 mm 4 Moment of inertia of profile J = 9.4E+05 mm 4 Torsional constant of profile S c1 = 5.0E+04 mm 3 Section modulus for face 1 under compression S c2 = 3.4E+04 mm 3 Section modulus for face 2 under compression 3.3-1M F cy = 170 Mpa 6063-T6 Alloy compressive yield strength E = Mpa Young's Modulus Buckling Analysis n y = 1.65 Factor of safety on yield strength B c = F cy [1+(F cy /15510) 0.5 ] = Buckling formula intercept for comp. in beam flanges D c = (B c /10)(B c /E) 0.5 = 0.98 Buckilng formula slope for comp. in beam flanges C c = 0.41(B c /D c ) = Buckling formula intersection for comp. in beam flanges S 1 = [(B c -F cy )/1.6D c ] 2 = Lower bound slender limit S 2 = (C c /1.6) 2 = Upper bound slender limit Face-1: S = L b S c1 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-1 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c1 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB Face-2: S = L b S c2 /[0.5(I y J) 0.5 ] = Slenderness ratio for face-2 under compression Criteria = S1 < S < S2 Intermediate beam-compression member F c2 = 1/n y (B c -1.6D c S 0.5 ) = Mpa Allowable compressive stress for LTB (d) Allowable Stress Under Local Buckling No element is critical under local buckling.

56 UNITISED CURTAIN WALL 56 of Sword Profile Sword (Type-2) (a) Cross-sectional Properties. Fig Sword Length 140 Fig Sword Cross-Section

57 UNITISED CURTAIN WALL 57 of ANALYSIS & CODE CHECK STRUCTURAL SYSTEM 1 Refer to 4.3 for discussions on structural system 1. Summary of Section Properties for Structural System 1 I' e S e1 S e2 (V c /D)/w F c1 F c2 Local, F c1 Local, F Section Properties c2 mm 4 mm 3 mm 3 N/mm/N/mm Mpa Mpa Mpa Mpa Reference (b) (c) (d) Structural System - 1 Male Mullion E E E Female Mullion E E E Header Transom E E E Sill Transom E E E Transom E E E Transom E E E Mullion Check The mullions are subjected to biaxial bending stresses due to (i) wind load normal to the plane of the unit, and (ii) wind load parallel to the plane of the unit. The latter is being caused by the wind load on the protruding vertical fin on the façade unit. For discussion on this modes of bending refer to Biaxial Bending Analysis of Mullion ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes Male Mullion Bending Stress ratio, f yy-,m /F c1,m = 0.74 Major axis face 1 under wind suction Stress ratio, f yy+,m /F c2,m = 0.95 Major axis face 2 under wind pressure Stress ratio, f zz1,m /F c,m = 0.13 Minor axis face 1 under lateral wind load Stress ratio, f zz2,m /F c,m = 0.07 Minor axis face 2 under lateral wind load Biaxial bending, f yy-,m /F c1,m f zz1,m /F c,m = % Suction + 50% lateral wind: comp. on face f yy-,m /F c1,m + f zz1,m /F c,m = % Suction + 100% lateral wind: comp. on face 1 Biaxial bending, f yy+,m /F c2,m f zz2,m /F c,m = % Pressure + 50% lateral wind: comp. on face f yy+,m /F c2,m + f zz2,m /F c,m = % Pressure + 100% lateral wind: comp. on face 3 Maximum stress ratio = 0.98 <1.00 O.K! Female Mullion Bending Stress ratio, f yy-,f /F c1,f = 0.36 Major axis face 1 under wind suction Stress ratio, f yy+,f /F c2,f = 0.43 Major axis face 2 under wind pressure Stress ratio, f zz1,f /F c,f = 0 Minor axis face 1 under lateral wind load Stress ratio, f zz2,f /F c,f = 0 Minor axis face 2 under lateral wind load Biaxial bending, f yy-,f /F c1,f f zz1,f /F c,f = % Suction + 50% lateral wind: comp. on face f yy-,f /F c1,f + f zz1,f /F c,f = % Suction + 100% lateral wind: comp. on face 1 Biaxial bending, f yy+,f /F c2,f f zz2,f /F c,f = % Pressure + 50% lateral wind: comp. on face f yy+,f /F c2,f + f zz2,f /F c,f = % Pressure + 100% lateral wind: comp. on face 3 Maximum stress ratio = 0.43 <1.00 O.K! Sword Bending Stress ratio, f yy-,sw /F c,sw = 0 Major axis bending under wind suction Stress ratio, f yy+,sw /F c,sw = 0 Major axis bending under wind pressure Stress ratio, f zz,sw /F c,sw = 0 Minor axis bending under lateral wind load Biaxial bending, f yy-,sw /F c,sw f zz,sw /F c,sw = 0.00 Suction + 50% lateral wind load Biaxial bending, 0.5 f yy-,sw /F c,sw + f zz,sw /F c,sw = % Suction + lateral wind load Biaxial bending, f yy+,sw /F c,sw f zz,sw /F c,sw = 0.00 Pressure + 50% lateral wind load Biaxial bending, 0.5 f yy+,sw /F c,sw + f zz,sw /F c,sw = % Pressure + lateral wind load Maximum stress ratio = 0.00 <1.00 O.K!

58 UNITISED CURTAIN WALL 58 of Superposition Under Biaxial Bending The table above summarizes the biaxial bending analyses of the mullion in the succeeding sections as follows: (i) Mullion Analysis under Wind Load Normal to Plane of the Unit mullion check under wind suction, refer to 81.2, mullion check under wind pressure, refer to 8.1.3, bf bm STACK JOINT - HEADER WIND LOAD TRIBUTARY AREA ON FEMALE MULLION TRANSOM - 1 WIND LOAD TRIBUTARY AREA ON MALE MULLION H FEMALE MULLION MALE MULLION WIND LOAD M yy,1 δ z,1 WIND LOAD M yy,1 δ z,1 TRANSOM - 2 STACK JOINT - SILL W Mom - yy Def - z Mom - yy Def - z ELEVATION WIND SUCTION SYSTEM - 1 WIND PRESSURE SYSTEM - 1 Fig Wind Load Normal to the Plane of the Unit (ii) Mullion Analysis under Wind Load Parallel to Plane of the Unit mullion check under lateral wind load, refer to In checking the minor axis bending of the mullion, it is assumed that the upper spandrel panel is a rigid diaphragm in the axis on mullion bending, this by virtue of the double skin 1.5mm infill metal sheets. With this assumption, the mode of mullion bending is similar to a propped cantilever beam with the spandrel panel as its rigid support. Error! Objects cannot be created from editing field codes. Fig Mullion under Lateral Wind Load on Vertical Fin

59 UNITISED CURTAIN WALL 59 of Mullion Check for Wind Suction ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action 4.4 CW Unit Parameters H = 3.70 m Unsupported span of mullion W = 1.39 m Tributary width for wind load Wind load on wall element, Q z = Kpa Suction normal to the plane of CW unit Mullion Alloying compressive yield strength, F cy = 170 Mpa 6063-T6 E = Mpa 6.1.1(b) Male Profile: Effective moment of Inertia, I' e,m = 1.36E+07 mm 4 About y-y axis, composite male profile Effective section modulus, S e1,m = 1.34E+05 mm 3 About y-y axis, on face 1 (for wind suction) 6.1.2(b) Female Profile: Effective moment of Inertia, I' e,f = 1.09E+06 mm 4 About y-y axis, composite female profile Effective section modulus, S e1,f = 2.23E+04 mm 3 About y-y axis, on face 1 (for wind suction) Sword Alloying compressive yield strength, F cy,sw = 170 Mpa 6063-T6 E Sw = Mpa Sword length, L Sw = 120 mm Portion of sword penetrating the mullion profile Sword cross-section, d Sw x t Sw = 140 x 15 mm Sword slack on its depth, gap z = 2.0 mm Total gaps on each side of the sword Bending Analysis Uniform load, q z = W Q z = KN/m Parallel to the major axis of mullion 4.3 Slip angle, α yy = tan -1 (gap z /L Sw ) = rad Sword's freedom of rotation before it carries moment Stage 1: Initial load, q z,i = 24α yy E I' e /H 3 = 7.99 KN/m Amount of load to produce rotation mullion ends Applied load for simple beam bending, q z,1 = 4.85 KN/m qzi > qz : Structural system 1 (Stage 2 is null) Midspan moment, M yy,1 = q z,1 H 2 /8 = 8.30 KN m Bending moment in the mullion assembly - Stage 2: Net load, q z,2 = q z - q z,1 = 0.00 KN/m Balance load after undergoing rotation, ends Midspan moment, M yy,2 = q z,2 H 2 /24 = 0.00 KN m Bending moment in the mullion assembly Support moment, M yy,sw = q z,2 H 2 /12 = 0.00 KN m Bending moment in the sword Couple force on sword, F z,sw = M yy,sw /L Sw = 0.00 KN Developed couple on sword Total midspan moment, M yy = M yy,1 + M yy,2 = 8.30 KN m Total bending moment in the mullion assembly 4.2(c) Check Mullion Profile I' e = I 'e,m + I' e,f = 1.5E+07 mm 4 About y-y axis, composite profile 4.2(d) Male Mullion: Moment, M yy,m = M yy (I' e,m /I' e ) = 7.68 KN m Shared moment by male mullion Bending stress, f yy,m = M yy,m /S e1,m = 61.9 Mpa Maximum bending stress 6.1.1(c) Allowable bending comp. stress, F c1,m = 83.8 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,m = Mpa Under local buckling of beam elements on face 1 Stress ratio, f yy,m /F c1,m = 0.74 <1.00 O.K! 4.2(d) Female Mullion: Moment, M yy,f = M yy (I' e,f /I' e ) = 0.62 KN m Shared moment by male mullion Bending stress, f yy,f = M yy,f /S e2,f = 27.6 Mpa Maximum bending stress 6.1.2(c) Allowable bending comp. stress, F c1,f = 76.0 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,f = 96.1 Mpa Under local buckling of beam elements on face 1 Stress ratio, f yy,f /F c1,f = 0.36 <1.00 O.K! Check Sword S yy,sw = t Sw d 2 Sw /6 = 4.9E+04 mm 4 Section modulus Bending: Bending stress, f yy,sw = M yy,sw /S yy,sw = 0.0 Mpa Considering plastic modulus Allowable stress, F c,sw = 1.3 F cy,sw /n y = Mpa Allowable compressive stress under bending Stress ratio, f yy,sw /F c,sw = 0.00 <1.00 O.K! Shear: Shear on sword, V z,sw = q z H/2 = 8.97 KN Lateral shear force Shear stress, f sz,sw = V z,sw /(0.9d Sw t Sw ) = 4.7 Mpa Allowable shear stress, F s = F ty,sw /[(3) 0.5 n y ] = 59.5 Mpa F ty = F cy, n y = 1.65 Stress ratio, f sz,sw /F s = 0.08 <1.00 O.K! Check Deflection δ z,1 = 5/384 q z,1 H 4 /E I' e = mm Stage 1: simple beam deflection - δ z,2 = 1/384 q z,2 H 4 /E I' e = 0.00 mm Stage 2: continuous beam deflection Total deflection, δ z = d z,1 + d z,2 = mm Calculated maximum deflection 2.3 Allowable deflection, δ allow = mm Criteria: Span/175 or 19mm Strain ratio, δ z /δ allow = 0.62 <1.00 O.K! Shear Flow Check Parameters q z-,m = q z- (I' e,m /I' e ) = KN/m Uniform load shared by the male mullion q z-,f = q z- (I' e,f /I' e ) = KN/m Uniform load shared by the female mullion Notes

60 UNITISED CURTAIN WALL 60 of Mullion Check for Wind Pressure ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action 4.4 CW Unit Parameters H = 3.70 m Unsupported span of mullion W = 1.39 m Tributary width for wind load Wind load on wall element, Q z = 3.50 Kpa Pressure normal to the plane of CW unit Mullion Alloying compressive yield strength, F cy = 170 Mpa 6063-T6 E = Mpa 6.1.1(b) Male Profile: Effective moment of Inertia, I' e,m = 1.36E+07 mm 4 About y-y axis, composite male profile Effective section modulus, S e2,m = 1.05E+05 mm 3 About y-y axis, on face 2 (for wind pressure) 6.2.1(b) Female Profile: Effective moment of Inertia, I' e,f = 1.09E+06 mm 4 About y-y axis, composite female profile Effective section modulus, S e2,f = 1.80E+04 mm 3 About y-y axis, on face 2 (for wind pressure) Sword Alloying compressive yield strength, F cy,sw = 170 Mpa 6063-T6 E Sw = Mpa Sword length, L Sw = 120 mm Portion of sword penetrating the mullion profile Sword cross section, d Sw x t Sw = 140 x 15 mm Sword slack on its depth, gap z = 2.0 mm Total gaps on each side of the sword Bending Analysis Uniform load, q z = W Q z = 4.85 KN/m Parallel to the major axis of mullion 4.3 Slip angle, α yy = tan -1 (gap z /L Sw ) = rad Sword's freedom of rotation before it carries moment Stage 1: Initial load, q z,i = 24α yy E I' e /H 3 = 7.99 KN/m Amount of load to produce rotation mullion ends Applied load for simple beam bending, q z,1 = 4.85 KN/m qzi > qz : Structural system 1 (Stage 2 is null) Midspan moment, M yy,1 = q z,1 H 2 /8 = 8.30 KN m Bending moment in the mullion assembly - Stage 2: Net load, q z,2 = q z - q z,1 = 0.00 KN/m Balance load after undergoing rotation, ends Midspan moment, M yy,2 = q z,2 H 2 /24 = 0.00 KN m Bending moment in the mullion assembly Support moment, M yy,sw = q z,2 H 2 /12 = 0.00 KN m Bending moment in the sword Couple force on sword, F z,sw = M yy,sw /L Sw = 0.00 KN Developed couple on sword Total midspan moment, M yy = M yy,1 + M yy,2 = 8.30 KN m Total bending moment in the mullion assembly 4.2(c) Check Mullion Profile I' e = I 'e,m + I' e,f = 1.5E+07 mm 4 About y-y axis, composite profile 4.2(d) Male Mullion: Moment, M yy,m = M yy (I' e,m /I' e ) = 7.68 KN m Shared moment by male mullion Bending stress, f yy,m = M yy,m /S e2,m = 79.0 Mpa Maximum bending stress 6.1.1(c) Allowable bending comp. stress, F c2,m = 83.5 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2,m = Mpa Under local buckling of beam elements on face 2 Stress ratio, f yy,m /F c2,m = 0.95 <1.00 O.K! 4.2(d) Female Mullion: Moment, M yy,f = M yy (I' e,f /I' e ) = 0.62 KN m Shared moment by male mullion Bending stress, f yy,f = M yy,f /S e2,f = 34.2 Mpa Maximum bending stress 6.1.2(c) Allowable bending comp. stress, F c2,f = 79.1 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2,f = - Mpa Under local buckling of beam elements on face 2 Stress ratio, f yy,f /F c2,f = 0.43 <1.00 O.K! Check Sword S yy,sw = t Sw d 2 Sw /6 = 4.9E+04 mm 4 Section modulus Bending: Bending stress, f yy,sw = M yy,sw /S yy,sw = 0.0 Mpa Considering plastic modulus Allowable stress, F c,sw = 1.3 F cy,sw /n y = Mpa Allowable compressive stress under bending Stress ratio, f yy,sw /F c,sw = 0.00 <1.00 O.K! Shear: Shear on sword, V z,sw = q z H/2 = 8.97 KN Lateral shear force Shear stress, f sz,sw = V z,sw /(0.9d Sw t Sw ) = 4.7 Mpa Allowable shear stress, F s = F ty,sw /[(3) 0.5 n y ] = 59.5 Mpa F ty = F cy, n y = 1.65 Stress ratio, f sz,sw /F s = 0.08 <1.00 O.K! Check Deflection δ z,1 = 5/384 q z,1 H 4 /E I' e = mm Stage 1: simple beam deflection - δ z,2 = 1/384 q z,2 H 4 /E I' e = 0.00 mm Stage 2: continuous beam deflection Total deflection, δ z = d z,1 + d z,2 = mm Calculated maximum deflection 2.3 Allowable deflection, δ allow = mm Criteria: Span/175 or 19mm Strain ratio, δ z /δ allow = 0.62 <1.00 O.K! Shear Flow Check Parameters q z+,m = q z+ (I' e,m /I' e ) = 4.49 KN/m Uniform load shared by the male mullion q z+,f = q z+ (I' e,f /I' e ) = 0.36 KN/m Uniform load shared by the female mullion Notes

61 UNITISED CURTAIN WALL 61 of Mullion Check for Lateral Wind Load Element 1 - critical element under wind suction c.g. Element 4 - critical element under wind pressure Fig Values of cy for Critical Elements under Biaxial Bending ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Unitized Parameters H = 3.70 m Height of the unitized panel h = 3.23 m Height of the unsupported span in y axis b = 0.19 m Width of vertical fin projection Wind load on fin, Q y = 3.0 Kpa Normal to the axis of fin protrusion Mullion Alloying compressive yield strength, F cy = 170 Mpa 6063-T6 E = Mpa 6.1.1(b) Male Profile: Effective moment of Inertia, I zz,m = 6.50E+05 mm 4 About z-z axis, non-composite male profie Dist. to c.g. of fiber considered, c y1,m = 35.1 mm Along y axis, on face 1 Dist. to c.g. of fiber considered, c y2,m = 18.8 mm Along y axis, on face Sword Alloying compressive yield strength, F cy,sw = 160 Mpa 6063-T6 E Sw = Mpa Sword length, L Sw = 120 mm Portion of sword penetrating the mullion profile Sword depth, d Sw x t Sw = 140 x 15 mm Sword slack on its thickness, gap y = 2.0 mm Total gaps on each side of the sword Bending Analysis Uniform load, q y = b Q y = 0.56 KN/m Parallel to the major axis of mullion 4.3 Slip angle, α zz = tan -1 (gap y /L Sw ) = 0.02 rad Sword's freedom of rotation before it carries moment Stage 1: Initial load, q y,i = 48α zz E I zz,m /h 3 = 1.06 KN/m Amount of load to produce rotation mullion ends Applied load for propped beam bending, q y,1 = 0.56 KN/m qzi > qz : Structural system 1 (Stage 2 is null) Midspan moment, M zz,1 = 3q y,1 h(h/2)/8 + q y (H/2) 2 /2 = 0.30 KN m Bending moment in the male mullion Stage 2: Net load, q y,2 = q y - q y,1 = 0.00 KN/m Balance load after undergoing rotation, ends Midspan moment, M zz,2 = q y,2 h 2 /24 = 0.00 KN m Bending moment in the male mullion Support moment, M zz,sw = q y,2 h 2 /12 = 0.00 KN m Bending moment in the sword Couple force on sword, F y,sw = M zz,sw /L Sw = 0.00 KN Developed couple on sword Total midspan moment, M zz = M zz,1 + M zz,2 = 0.30 KN m Total minor axis bending moment in the male mullion Check Male Mullion Profile f zz1,m = M zz /(I zz,m /c y1,m ) = 16.0 Mpa Minor axis bending stress on critical face Face 1: Allow. comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about minor axis Stress ratio, f zz1,m /F c = 0.13 <1.00 O.K! Face 2: f zz1,m = M zz /(I zz,m /c y1,m ) = 8.6 Mpa Minor axis bending stress on critical face 2 Allow. comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about minor axis Stress ratio, f zz1,m /F c = 0.07 <1.00 O.K! Check Sword Section modulus, S zz,sw = d Sw t 2 Sw /6 = 5.25E+03 mm 4 Bending: Bending stress, f zz,sw = M zz,sw /S zz,sw = 0.0 Mpa Considering plastic modulus Allow. bending comp. Stress, F c,sw = 1.3 F cy,sw /n y = Mpa Stress ratio, f zz,sw /F c,sw = 0.00 <1.00 O.K! Shear: Shear on sword, V y,sw = q y H/2 = 1.04 KN Lateral shear force Shear stress, f sy,sw = V y,sw /(0.9d Sw t Sw ) = 0.5 Mpa Allowable shear stress, F s = F ty,sw /[(3) 0.5 n y ] = 56.0 Mpa Fty = Fcy, ny = 1.65 Stress ratio, f sy,sw /F s = 0.01 <1.00 O.K! Check Deflection of Mullion δ z,1 = q z,1 /185 h 4 /E I zz.m = 7.36 mm Stage 1: propped cantilever beam deflection δ z,2 = 1/384 q z,2 H 4 /E I zz,m = 0.00 mm Stage 2: continuous beam deflection Total lateral deflection, δ z = d z,1 + d z,2 = 7.36 mm Calculated maximum deflection 2.3 Allowable lateral deflection, δ allow = 7.50 mm Criteria: 75% of 10mm clearance Strain ratio, δ z /δ allow = 0.98 <1.00 O.K!

62 UNITISED CURTAIN WALL 62 of Check Sword Under Slab Deflection. Refer to 4.3 for discussions on slab deflection and in-plane performance of the curtain wall units. W R h R v W t H W/2 R h Fig Reaction Forces under Deflected Slab ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes I. Claculated shear on sword in the preceeding section Suction Wind on Mullion: Shear, V z-,m = 8.30 kn Calculated maximum shear on male mullion Shear, V z-,f = 0.70 kn Calculated maximum shear on female mullion Total sword shear, V z-,sw = V z-,m + V z-,f = 9.00 kn Total major axis shear on sword under wind suction Pressure Wind on Mullion: Shear, V z+,m = 8.30 kn Calculated maximum shear on male mullion Shear, V z+,f = 0.70 kn Calculated maximum shear on female mullion Total sword shear, V z+,sw = V z+,m + V z+,f = 9.00 kn Total major axis shear on sword under wind pressure Lateral Wind on Mullion: Sword shear, V y,sw = 1.41 kn Minor axis shear on sword under lateral wind II. Shear on sword due to slab deflection 5.2 CW Unit Parameters: Width of unit, W = 1.39 m Clear span between dead load support brackets Height of unit, H = 3.70 m Approx. vertical dist. bet. dead load bracket and sword j 5.1 Total weight of the unit, W t = 3.07 kn Calculated weight of all components 3.3-1M Sword Parameters: Sword, F ty = 170 Mpa Shear yield strength Depth, d Sw = 140 mm Sword gross depth Sword thickness, t Sw = 15 mm Sword gross thickness Reaction, R h = W t (W/2H) = 0.6 kn Lateral reaction force due to one lamed bracket support Sword shear, V yδ,sw = R h = 0.6 kn Minor axis shear on sword due tor slab deflection III. Combined shear on sword Suction wind + Lateral wind + Slab defleftion: Shear, V = [V 2 z-,sw +(V y,sw +V yδ,sw ) 2 ) 0.5 = 9.2 kn Combined major and minor axis shear in sword Shear stress, f s = V/(0.9d Sw t Sw ) = 4.9 Mpa Total shear stress in sword Allowable shear stress, F s = F ty /1.732n y = 59.5 Mpa where n y = 1.65 for yield failure Stress ratio = 0.08 <1.00 O.K! Pressure wind + Lateral wind + Slab defleftion: Shear, V = [V 2 z-,sw +(V y,sw +V yδ,sw ) 2 ) 0.5 = 9.2 kn Combined major and minor axis shear in sword Shear stress, f s = V/(0.9d Sw t Sw ) = 4.9 Mpa Total shear stress in sword Allowable shear stress, F s = F ty /1.732n y = 59.5 Mpa where n y = 1.65 for yield failure Stress ratio = 0.08 <1.00 O.K! Note that the calculated shear in the mullion is equal to the shear in the sword: V = V + V where: Vz,Sw = Max. shear on sword in the major axis z, Sw z, m z, f Vz,m = Max. shear on the male mullion in the major axis. Vz,f = Max. shear on the female mullion in the major axis.

63 UNITISED CURTAIN WALL 63 of Check Thermal Break Shear (a) Check Longitudinal Shear in Male Mullion The FEM Model Equation (28.a) of AAMA-TIR-A8-04 is an approximation for shear flow in a thermal barrier. The accuracy of this equation depends on the degree of symmetry of the two faces being combined. In the case of the male mullion profile, the two faces are very far from being symmetrical. Therefore a more precise FEM analysis is performed to calculate the actual shear flow in the thermal barrier. The assembly is modeled to include both faces of the male mullion profile as custom beam sections and the pair of thermal break strips as shell elements. Stiff link elements at 100mm spacing along the entire span join both faces with the edges of the thermal strips. Boundary conditions are defined at end spans and at intermediate points at the location of the transoms. The amount of uniform wind load shared by the male mullion (with its female partner) is applied as a point load on every node that are spaced 100mm along the whole span. Face-1 Link 100mm on centre Thermal break, 2 nos, 2mm thick Lateral and torsional restraints by the transoms on face-1 Link 100mm on centre Face-1 Face-2 Face-2 Wind load: 4.5N/mmm Roller support on face-2 Hinge support on face-2 Face-1 and face-2 are coupled together at this end (a) Male mullion Cross-section Fig Male mullion Thermal break Model (b) Male mullion - Boundary conditions

64 UNITISED CURTAIN WALL 64 of 117 Clause Action Notes Parameters L = 3.70 m Unsupported span of beam assembly Beam Assembly: h = mm Height of thermal break strip b = 2.00 mm Thickness of thermal break strip q z = 4.50 kn/m Applied uniform wind load on the assembly 6.2.1(b) Aluminium: E Al = Mpa Young's modulus of aluminium extrusion µ Al = 0.33 Poisson's ratio of aluminium extrusion Thermal Break: E Pm = 2000 Mpa Young's modulus of thermal break (polyamide) µ Pm = 0.49 Poisson's ratio of thermal break (polyamide) 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Link element: E Lk, I Lk, A Lk = 1.00E+09 Mpa Link element properties (very large number) ANSYS FEM Analysis Results Notes Check Shear Flow Shear stress, f v = 7.27 Mpa Result maximum shear stress in YZ plane Max. shear flow, q c = f v b = N/mm Calculated shear flow in the thermal break Allowable shear flow, Q'c = Qc/F.S. = N/mm Considerig a factor of safety, F.S. of 3.0 Stress ratio = 0.87 <1.00 O.K!

65 UNITISED CURTAIN WALL 65 of 117 Analysis results - Shear stress Diagram Fig Shear stress on YZ Plane Result verification Deflection Diagram The result of the analysis is verified by checking if the resulting deflection is close to the manually calculated deflection in Manually calculated deflection in : max = 11.78mm ANSYS FEM model deflection = 11.73mm = max Therefore, the result is reliable. Fig Male mullion - Deflection

66 UNITISED CURTAIN WALL 66 of 117 (b) Check Longitudinal Shear in emale Mullion AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 6.1.2(b) Extrusion Parameters V c /(D w) = 5.75 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 2.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.08 <1.00 O.K! Pressure Wind: q z+ = 0.36 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 2.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.08 <1.00 O.K!

67 UNITISED CURTAIN WALL 67 of Check Stack Joint Header and Sill Transom Check Header & Sill Transom under Wind Suction ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Unitized Parameters Unsupported span, W = 1.59 m 3.2 Upper panel height, h u = 0.39 m Lower panel height, h l = 0.47 m Cross-sectional area of header profile, A g,h = mm 2 P z 6.2.1(a) Cross-sectional area of sill profile, A g,s = mm 2 a 5.1 Weight of infill, if any, Q z = 0.00 Kpa Weight of glass above the transom, Q DL = 0.45 Kpa 4.4 Wind load on wall element, Q y = Kpa Setting block location, a = 190 mm 6.2.1(b) Eccentricity of setting block location, e = 37.3 mm Reference point for torsional deflection, r = 66.0 mm Transom Alloying material = Alloying compressive yield strength, F cy = 6063-T6 170 Mpa 6.2.1(b) Header Transom: Effective moment of Inertia, I' e,h = 5.38E+05 mm 4 Effective section modulus (face 1), S e1,h = 1.66E+04 mm (a) Moment of inertia, I yy,h = 7.40E+05 mm (b) Dist. of top extreme fibre on face 1, c z1,h = 53.4 mm 6.2.2(b) Sill Transom: Effective moment of inertia, I' e,s = Effective section modulus (face 1), S e1,s = 7.09E+05 mm E+04 mm (c) Moment of inertia, I yy,s = 9.90E+05 mm 4 Torsional constant, J s = 8.20E+04 mm 4 TRANSOM - 1 Wind load tributary area TRANSOM - 2 STACK JOINT - SILL 6.2.2(b) Dist. of top extreme fibre on face 1, c z1,s = 42.3 mm z P z e Analysis Glass weight, P z = Q DL h u W/2 = 0.14 KN q Upper tributary width, b u = (h u /2)[1-h u /(2W)] = 0.17 m y r Lower tributary width, b l = (h l /2)[1-h l /(2W)] = 0.20 m rotation c.g. Moment, M zz = Q y (b u +b l ) W 2 /8 = KN/m Check Stack Joint Profiles I' e = I 'e,h + I' e,s = 1.2E+06 mm 4 About y-y axis, composite profile Header Transom M yy,h = (1.25 A g,h γ)w 2 /8 = 0.01 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy,h = M yy,h /(I yy,h /c z1,h ) = 0.7 Mpa Maximum bending stress Allow. bending comp. stress, F c,h = 1.17 F cy /n y = Mpa Under bending about weak axis 4.2(d) Suction Wind: Moment, M zz,h = M zz (I' e,h /I' e ) = KN m Shared moment by header transom Bending stress, f zz,h = M zz,h /S e1,h = 9.1 Mpa Maximum bending stress on face (c) Allowable bending comp. stress, F c1,h = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,h = 88.2 Mpa Under local buckling of beam elements on face 1 Dead Load + Suction Wind: f yy,h /F c,h + f zz,h /F c1,h = 0.10 <1.00 O.K! Sill Transom M yy,s = P z a + (1.25 A g,s γ +Q DL h u )W 2 /8 = 0.04 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy,s = M yy,s /(I yy,s /c z1,s ) = 1.7 Mpa Maximum bending stress Allow. bending comp. stress, F c,s = 1.17 F cy /n y = Mpa Under bending about weak axis 4.2(d) Suction Wind: Moment, M zz,s = M zz (I' e,s /I' e ) = kn-m Shared moment by sill transom Bending stress, f zz,s = M zz,s /S e1,s = 9.5 Mpa Maximum bending stress on face (c) Allowable bending comp. stress, F c1,s = 99.3 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,s = 69.9 Mpa Under local buckling of beam elements on face 1 Dead Load + Suction Wind: f yy,s /F c,s + f zz,s /F c1,s = 0.15 <1.00 O.K! Check Deflection δ z,h = (5/384)(1.25 A g,h γ)w 4 ]//E I yy,h = 0.05 mm Deflection under dead load 2.3 Header Transom Allowable deflection, δ z,allow = mm Criteria: 75% of 20mm clearance Dead Load: Strain ratio, δ z /δ z,allow = 0.00 <1.00 O.K! δ z,s1 = [(P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g,s γ +Q DL h u )W 4 ]//E I yy,s = 0.17 mm Deflection under dead load Sill Transom δ z,2 = (P z e) W r/(j s G) = 0.26 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 0.43 mm Total deflection at the fibre closest to glass edge 2.3 Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.14 <1.00 O.K! Stack Joint Profiles δ y = (5/384) Q y (b u +b l )W 4 /[E I' e ] = 1.08 mm Calculated maximum deflection 2.3 Suction Wind: Allowable deflection, δ y,allow = 9.09 mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.12 <1.00 O.K! Shear Flow Check Parameters q y,h = Q y (b u +b l ) (I' e,h /I' e ) = KN/m Uniform load shared by the header transom q y,s = Q y (b u +b l ) (I' e,s /I' e ) = KN/m Uniform load shared by the sill transom End Connection Check Parameters V y,h = q y,h W/2 = 0.4 kn Horizontal reaction force on header profile Header Transom: V z,h = (1.25 A g,s γ)w/2 = 0.0 kn Vertical reaction force due to dead load on header profil Sill Trasnom: V y,s = q y,s W/2 = 0.5 kn Horizontal reaction force on sill profile V z,h = P z + (1.25 A g,s γ +Q z h u )W/2 = 0.2 kn Vertical reaction force due to dead load on sill profile FEMALE MULLION W TRANSOM - 2 SILL HEADER ELEVATION Pz a MALE MULLION Wind load tributary area b u b l Equiv. rectangular tributary width Weight of sheets 2 x P WIND LOAD SECTION y z e hu h l H

68 UNITISED CURTAIN WALL 68 of Stack Joint Header and Sill Transom (under Wind Pressure) ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Unitized Parameters Unsupported span, W = 1.59 m 3.2 Upper panel height, h u = 0.39 m Lower panel height, h l = 0.47 m Cross-sectional area of header profile, A g,h = mm 2 P z 6.2.1(a) Cross-sectional area of sill profile, A g,s = mm 2 a 5.1 Weight of infill, if any, Q z = 0.00 Kpa Weight of glass above the transom, Q DL = 0.45 Kpa 4.4 Wind load on wall element, Q y = 3.00 Kpa Setting block location, a = 190 mm 6.2.1(b) Eccentricity of setting block location, e = 37.3 mm Reference point for torsional deflection, r = 66.0 mm Transom Alloying material = Alloying compressive yield strength, F cy = 6063-T6 170 Mpa Header Transom: Effective moment of Inertia, I' e,h = 5.38E+05 mm 4 Effective section modulus (face 2), S e2,h = 1.61E+04 mm 3 Moment of inertia, I yy,h = 7.40E+05 mm (b) Dist. of top extreme fibre on face 2, c z2,h = 53.4 mm Sill Transom: Effective moment of inertia, I' e,s = 7.09E+05 mm (a) Effective section modulus (face 2), S e2,s = 2.02E+04 mm (b) Moment of inertia, I yy,s = 9.90E+05 mm (b) Torsional constant, J s = 8.20E+04 mm 4 Dist. of top extreme fibre on face 2, c z2,s = 42.3 mm 6.2.2(c) Glass weight, P z = Q DL h u W/2 = 0.14 KN Upper tributary width, b u = (h u /2)[1-h u /(2W)] = 0.17 m 6.2.2(b) Lower tributary width, b l = (h l /2)[1-h l /(2W)] = 0.20 m Analysis Moment, M zz = Q y (b u +b l ) W 2 /8 = 0.35 KN/m I' e = I 'e,h + I' e,s = 1.2E+06 mm 4 About y-y axis, composite profile M yy,h = (1.25 A g,h γ)w 2 /8 = 0.01 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy,h = M yy,h /(I yy,h /c z2,h ) = 0.7 Mpa Maximum bending stress Check Stack Joint Profiles Allow. bending comp. stress, F c,h = 1.17 F cy /n y = Mpa Under bending about weak axis Header TrSuction Wind: Moment, M zz,h = M zz (I' e,h /I' e ) = 0.15 KN m Shared moment by header transom Bending stress, f zz,h = M zz,h /S e2,h = 9.4 Mpa Maximum bending stress on face Allowable bending comp. stress, F c2,h = Mpa Under lateral buckling of beam section on face 2 4.2(d) Allowable bending comp. stress, F c2,h = 34.8 Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy,h /F c,h + f zz,h /F c2,h = 0.27 <1.00 O.K! 6.2.1(c) M yy,s = P z a + (1.25 A g,s γ +Q DL h u )W 2 /8 = 0.04 kn-m Accessories assumed 25% add'l to transom weight 6.2.1(d) Dead Load: Bending stress, f yy,s = M yy,s /(I yy,s /c z2,s ) = 1.7 Mpa Maximum bending stress Allow. bending comp. stress, F c,s = 1.17 F cy /n y = Mpa Under bending about weak axis Sill TransoSuction Wind: Moment, M zz,s = M zz (I' e,s /I' e ) = 0.20 kn-m Shared moment by sill transom Bending stress, f zz,s = M zz,s /S e2,s = 9.9 Mpa Maximum bending stress on face Allowable bending comp. stress, F c2,s = 99.6 Mpa Under lateral buckling of beam section on face 2 4.2(d) Allowable bending comp. stress, F c2,s = - Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy,s /F c,s + f zz,s /F c2,s = 0.11 <1.00 O.K! 6.2.2(c) δ z,h = (5/384)(1.25 A g,h γ)w 4 ]//E I yy,h = 0.05 mm Deflection under dead load 6.2.2(d) Header Transom Allowable deflection, δ z,allow = mm Criteria: 75% of 20mm clearance Dead Load: Strain ratio, δ z /δ z,allow = 0.00 <1.00 O.K! Check Deflection P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g,s γ +Q DL h u )W 4 ]//E I yy,s = 0.17 mm Deflection under dead load 2.3 Sill Transom δ z,2 = (P z e) W r/(j s G) = 0.26 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 0.43 mm Total deflection at the fibre closest to glass edge Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.14 <1.00 O.K! Stack Joint Profiles δ y = (5/384) Q y (b u +b l )W 4 /[E I' e ] = 1.08 mm Calculated maximum deflection 2.3 Suction Wind: Allowable deflection, δ y,allow = 9.09 mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.12 <1.00 O.K! q y,h = Q y (b u +b l ) (I' e,h /I' e ) = 0.48 KN/m Uniform load shared by the header transom 2.3 q y,s = Q y (b u +b l ) (I' e,s /I' e ) = 0.63 KN/m Uniform load shared by the sill transom V y,h = q y,h W/2 = 0.4 kn Horizontal reaction force on header profile Shear FlowHeader Transom: V z,h = (1.25 A g,s γ)w/2 = 0.0 kn Vertical reaction force due to dead load on header profi Sill Trasnom: V y,s = q y,s W/2 = 0.5 kn Horizontal reaction force on sill profile End Connection Check Parameters V z,h = P z + (1.25 A g,s γ +Q z h u )W/2 = 0.2 kn Vertical reaction force due to dead load on sill profile FEMALE MULLION W TRANSOM - 2 SILL HEADER TRANSOM - 1 Wind load tributary area TRANSOM - 2 STACK JOINT - SILL ELEVATION q y rotation P z Pz a MALE MULLION e r Wind load tributary area b u b l Equiv. rectangular tributary width Weight of sheets z c.g. WIND LOAD 2 x P SECTION z y e hu h l H

69 UNITISED CURTAIN WALL 69 of Check Thermal Break Shear (a) Check Longitudinal Shear in Header Transom AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 6.2.1(b) Extrusion Parameters V c /(D w) = 8.54 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 4.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.16 <1.00 O.K! Pressure Wind: q z+ = 0.48 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 4.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.16 <1.00 O.K! (b) Check Longitudinal Shear in Sill Transom AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 6.2.2(b) Extrusion Parameters V c /(D w) = 7.13 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 4.5 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.18 <1.00 O.K! Pressure Wind: q z+ = 0.63 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 4.5 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.18 <1.00 O.K!

70 UNITISED CURTAIN WALL 70 of Transom 1 (Type 1) Check Profile under Biaxial Bending ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Unitized Parameters Unsupported span, W = 1.59 m 3.2 Upper panel height, h u = 0.47 m Lower panel height, h l = 2.84 m 6.3.1(a) Cross-sectional area of profile, A g = mm Weight of infill, if any, Q z = 0.46 Kpa Weight of glass above the transom, Q DL = 0.00 Kpa 4.4 Design wind suction, Q y- = Kpa Design wind pressure, Q y+ = 3.00 Kpa Setting block location, a = 190 mm 6.3.1(b) Eccentricity of setting block location, e = 24.5 mm Reference point for torsional deflection, r = 55.4 mm Transom Alloying material = Alloying compressive yield strength, F cy = 6063-T6 170 Mpa Modulus of Elasticity, E = Mpa Shear Modulus of alloy, G = Mpa Unit weight of alloy, γ = 27 kn/m (b) Effective moment of inertia, I' e = 6.4E+05 mm 4 Effective section modulus (face 1), S e1 = (face 2), S e2 = 1.9E+04 mm 3 2.5E+04 mm (c) Moment of inertia, I yy = 6.4E+05 mm 4 Torsional constant, J = 5.3E+05 mm (b) Dist. from c.g. to extreme top fibre on face 1, c z1 = 34.6 mm Dist. from c.g. to extreme top fibre on face 2, c z2 = 34.6 mm Analysis Glass weight, P z = Q DL h u W/2 = 0.00 KN Applied as pair loads of glass chair Upper tributary width, b u = (h u /2)[1-h u /(2W)] = 0.20 m Width of the equivalent rectangular tributary area Lower tributary width, b l = W/3 = 0.53 m Width of the equivalent rectangular tributary area Uniform wind suction load, q y- = Q y- (b u +b l ) = Also used for checking of thermal break shear flow Uniform wind pressure load, q y+ = Q y+ (b u +b l ) = 2.19 Also used for checking of thermal break shear flow Check Transom M yy = P z a + (1.25 A g γ +Q z h u )W 2 /8 = 0.08 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy = M yy /(I yy /c z ) = 4.3 Mpa Maximum bending stress Allow. bending comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about weak axis Suction Wind: Moment, M zz- = q y- W 2 /8 = kn-m Bending moment about z-z axis Bending stress, f zz- = M zz- /S e1 = 36.2 Mpa Maximum bending stress 6.3.1(c) Allowable bending comp. stress, F c1 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1 = 92.4 Mpa Under local buckling of beam elements on face 1 Pressure Wind: Moment, M zz+ = q y+ W 2 /8 = 0.69 kn-m Bending moment about z-z axis Bending stress, f zz+ = M zz+ /S e2 = 27.6 Mpa Maximum bending stress 6.3.1(c) Allowable bending comp. stress, F c2 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2 = - Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy /F c + f zz- /F c1 = 0.43 <1.00 O.K! Pressure Wind + Dead Load: f yy /F c + f zz+ /F c2 = 0.30 <1.00 O.K! STACK JOINT - HEADER Wind load tributary area Weight of sheets Weight of infill (insulation) Equiv. rectangular MALE MULLION tributary width δ z,1 = [(P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g γ +Q DL h u )W 4 ]//E I yy = 0.48 mm Deflection under dead load Check Deflection δ z,2 = (P z e) W r/(j G) = 0.00 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 0.48 mm Total deflection at the fibre closest to glass edge 2.3 Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.16 <1.00 O.K! Suction Wind: δ y- = (5/384) Q y- (b u +b l )W 4 /[E I' e ] = mm Maximum deflection under suction wind Pressure Wind: δ y+ = (5/384) Q y+ (b u +b l )W 4 /[E I' e ] = 4.11 mm Maximum deflection under pressure wind 2.3 Allowable deflection, δ y,allow = 9.09 mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.45 <1.00 O.K! End Connection Check Parameters V y- = q y- W/2 = 1.7 kn End shear under wind suction V y+ = q y+ W/2 = 1.7 kn End shear under wind pressure V z = P z + (1.25 A g,s γ +Q z h u )W/2 = 0.2 kn End shear under dead load q z TRANSOM - 1 FEMALE MULLION TRANSOM - 2 STACK JOINT - SILL ELEVATION q y rotation z e r b u bl WIND LOAD 2 x P z c.g. e h u h l SECTION y H

71 UNITISED CURTAIN WALL 71 of Check Thermal Break Shear (a) Check Longitudinal Shear in Transom 1 AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 6.3.1(b) Extrusion Parameters V c /(D w) = 6.09 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 13.3 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.53 <1.00 O.K! Pressure Wind: q z+ = 2.19 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 13.3 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.53 <1.00 O.K!

72 UNITISED CURTAIN WALL 72 of Transom 2 (Type 1) Check Profile under Biaxial Bending ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action 4.4 Unitized Parameters Unsupported span, W = 1.59 m 3.2 Upper panel height, h u = 2.84 m Lower panel height, h l = 0.39 m 6.4.1(a) Cross-sectional area of profile, A g = mm Weight of infill, if any, Q z = 0.00 Kpa Weight of glass above the transom, Q DL = 0.45 Kpa 4.4 Design wind suction, Q y- = Kpa Design wind pressure, Q y+ = 3.00 Kpa Setting block location, a = 190 mm 6.4.1(b) Eccentricity of setting block location, e = 27.1 mm Reference point for torsional deflection, r = 56.8 mm Transom Alloying material = Alloying compressive yield strength, F cy = 6063-T6 170 Mpa Modulus of Elasticity, E = Mpa Shear Modulus of alloy, G = Mpa Unit weight of alloy, γ = 27 kn/m (b) Effective moment of inertia, I' e = 6.6E+05 mm 4 Effective section modulus (face 1), S e1 = (face 2), S e2 = 1.8E+04 mm 3 2.3E+04 mm (c) Moment of inertia, I yy = 5.5E+05 mm 4 Torsional constant, J = 4.9E+05 mm (b) Dist. from c.g. to extreme top fibre on face 1, c z1 = 37.5 mm Dist. from c.g. to extreme top fibre on face 2, c z2 = 17.5 mm Analysis Glass weight, P z = Q DL h u W/2 = 1.02 KN Applied as pair loads of glass chair Upper tributary width, b u = W/3 = 0.53 m Width of the equivalent rectangular tributary area Lower tributary width, b l = (h l /2)[1-h l /(2W)] = 0.17 m Width of the equivalent rectangular tributary area Uniform wind suction load, q y- = Q y- (b u +b l ) = Also used for checking of thermal break shear flow Uniform wind pressure load, q y+ = Q y+ (b u +b l ) = 2.10 Also used for checking of thermal break shear flow Check Transom M yy = P z a + (1.25 A g γ +Q z h u )W 2 /8 = 0.20 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy = M yy /(I yy /c z ) = 13.9 Mpa Maximum bending stress Allow. bending comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about weak axis Suction Wind: Moment, M zz- = q y- W 2 /8 = kn-m Bending moment about z-z axis Bending stress, f zz- = M zz- /S e1 = 36.3 Mpa Maximum bending stress 6.4.1(c) Allowable bending comp. stress, F c1 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1 = 89.0 Mpa Under local buckling of beam elements on face 1 Pressure Wind: Moment, M zz+ = q y+ W 2 /8 = 0.66 kn-m Bending moment about z-z axis Bending stress, f zz+ = M zz+ /S e2 = 29.5 Mpa Maximum bending stress 6.4.1(c) Allowable bending comp. stress, F c2 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2 = - Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy /F c + f zz- /F c1 = 0.52 <1.00 O.K! Pressure Wind + Dead Load: f yy /F c + f zz+ /F c2 = 0.40 <1.00 O.K! δ z,1 = [(P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g γ +Q DL h u )W 4 ]//E I yy = 1.65 mm Deflection under dead load Check Deflection δ z,2 = (P z e) W r/(j G) = 0.20 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 1.85 mm Total deflection at the fibre closest to glass edge 2.3 Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.62 <1.00 O.K! Suction Wind: δ y- = (5/384) Q y- (b u +b l )W 4 /[E I' e ] = mm Maximum deflection under suction wind Pressure Wind: δ y+ = (5/384) Q y+ (b u +b l )W 4 /[E I' e ] = 3.83 mm Maximum deflection under pressure wind 2.3 Allowable deflection, δ y,allow = 9.09 mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.42 <1.00 O.K! End Connection Check Parameters V y- = q y- W/2 = 1.7 kn End shear under wind suction V y+ = q y+ W/2 = 1.7 kn End shear under wind pressure V z = P z + (1.25 A g,s γ +Q z h u )W/2 = 1.0 kn End shear under dead load Notes a P z W STACK JOINT - HEADER TRANSOM - 1 FEMALE MULLION Wind load tributary area TRANSOM - 2 STACK JOINT - SILL Pz MALE MULLION a Equivalent rectangular tributary width ELEVATION q y rotation z e r bu bl c.g. WIND LOAD 2 x P z e h H u hl SECTION y

73 UNITISED CURTAIN WALL 73 of Check Thermal Break Shear (a) Check Longitudinal Shear in Transom 2 AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 6.4.1(b) Extrusion Parameters V c /(D w) = 6.11 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 12.8 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.51 <1.00 O.K! Pressure Wind: q z+ = 2.10 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 12.8 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.51 <1.00 O.K!

74 UNITISED CURTAIN WALL 74 of Check Stainless Steel Spandrel Panel Details For structural system 1, the 1.5mm thick stainless steel spandrel panel is reinforced with a continuous bent aluminum sheet bonded at the back of the sheet via structural silicone. Fig Spandrel Panel Details Where: K = 341.5mm mm Thick Aluminium Reinforcement The metal sheet and its reinforcement are considered as non-composite or layered elements. Which means the structural sealant bonding is not considered to transfer shear between the metals. Therefore, every adjacent nodes that are co-incident on the plane of the metal sheet are coupled in the in-out direction (as shown below). 1.5mm thk sheet 4mm thk reinforce ment In-out Coupling Fig Coupled Nodes

75 UNITISED CURTAIN WALL 75 of Check Sheet for Full Wind Load Shell Element Model with Boundary conditions Perimeter Restraint in Z - dir. Wind load: - Coupled Restraint in Fig Spandrel Sheet Applied Load & Boundary Conditions (a) Calculations Summary ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes Metal Sheet Parameters Width, W = 1385 mm Net dimension considering width of mullions Height, h = 400 mm Net height considering grip of glazing bead Wind load, Q = -3.5 KN/m 2 Design wind load in suction Alloy, F ty = 170 Mpa Tensile yield strength of Stainless steel S31203 ANSYS FEM Anlysis Results Notes Check Stress VM =0.707[(S min -S min ) 2 +S 2 max +S 2 min ) 0.5 = Mpa Von mises stress results (from ANSYS FEM anlaysis) Allowable stress, F c = 1.3F ty /n y = Mpa For rectangular solid plates Stress ratio, VM/F c = 0.84 <1.00 O.K! Check Deflection δ max = 6.82 mm Maximum deflection (from ANSYS FEM analysis) Allowable deflection, δ allow = 8.00 mm Architectural requirement (span/60 = 400mm/80) Strain ratio, δ z /δ allow = 0.85 <1.00 O.K!

76 UNITISED CURTAIN WALL 76 of 117 (b) Analysis Results (i.) Von Mises Stress Diagram (ii.) Deflection Diagram Fig Spandrel Sheet Analysis Results

77 UNITISED CURTAIN WALL 77 of ANALYSIS & CODE CHECK STRUCTURAL SYSTEM 2 Refer to 4.3 for discussions on structural system 2. Summary of Section Properties for Structural System 2 I' e S e1 S e2 (V c /D)/w F c1 F c2 Local, F c1 Local, F Section Properties c2 mm 4 mm 3 mm 3 N/mm/N/mm Mpa Mpa Mpa Mpa Reference (b) (c) (d) Structural System - 2 Male Mullion E E E Female Mullion E E E Header Transom E E E Sill Transom E E E Transom E E E Transom E E E Mullion Check The mullions are subjected to biaxial bending stresses due to (i) wind load normal to the plane of the unit, and (ii) wind load parallel to the plane of the unit. The latter is being caused by the wind load on the protruding vertical fin on the façade unit. For discussion on this modes of bending refer to Biaxial Bending Analysis of Mullion ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes Male Mullion Bending Stress ratio, f yy-,m /F c1,m = 0.61 Major axis face 1 under wind suction Stress ratio, f yy+,m /F c1,m = 0.78 Major axis face 2 under wind pressure Stress ratio, f zz1,m /F c,m = 0.20 Minor axis face 1 under lateral wind load Stress ratio, f zz2,m /F c,m = 0.11 Minor axis face 2 under lateral wind load Biaxial bending, f yy-,m /F c1,m f zz1,m /F c,m = % Suction + 50% lateral wind: comp. on face f yy-,m /F c1,m + f zz1,m /F c,m = % Suction + 100% lateral wind: comp. on face 1 Biaxial bending, f yy+,m /F c2,m f zz2,m /F c,m = % Pressure + 50% lateral wind: comp. on face f yy+,m /F c2,m + f zz2,m /F c,m = % Pressure + 100% lateral wind: comp. on face 3 Maximum stress ratio = 0.84 <1.00 O.K! Female Mullion Bending Stress ratio, f yy-,f /F c1,f = 0.30 Major axis face 1 under wind suction Stress ratio, f yy+,f /F c1,f = 0.35 Major axis face 2 under wind pressure Stress ratio, f zz1,f /F c,f = 0 Minor axis face 1 under lateral wind load Stress ratio, f zz2,f /F c,f = 0 Minor axis face 2 under lateral wind load Biaxial bending, f yy-,f /F c1,f f zz1,f /F c,f = % Suction + 50% lateral wind: comp. on face f yy-,f /F c1,f + f zz1,f /F c,f = % Suction + 100% lateral wind: comp. on face 1 Biaxial bending, f yy+,f /F c2,f f zz2,f /F c,f = % Pressure + 50% lateral wind: comp. on face f yy+,f /F c2,f + f zz2,f /F c,f = % Pressure + 100% lateral wind: comp. on face 3 Maximum stress ratio = 0.35 <1.00 O.K! Sword Bending Stress ratio, f yy-,sw /F c,sw = 0.43 Major axis bending under wind suction Stress ratio, f yy+,sw /F c,sw = 0.43 Major axis bending under wind pressure Stress ratio, f zz,sw /F c,sw = 0.15 Minor axis bending under lateral wind load Biaxial bending, f yy-,sw /F c,sw f zz,sw /F c,sw = 0.51 Suction + 50% lateral wind load Biaxial bending, 0.5 f yy-,sw /F c,sw + f zz,sw /F c,sw = % Suction + lateral wind load Biaxial bending, f yy+,sw /F c,sw f zz,sw /F c,sw = 0.51 Pressure + 50% lateral wind load Biaxial bending, 0.5 f yy+,sw /F c,sw + f zz,sw /F c,sw = % Pressure + lateral wind load Maximum stress ratio = 0.51 <1.00 O.K!

78 UNITISED CURTAIN WALL 78 of Superposition Under Biaxial Bending The table above summarizes the biaxial bending analyses of the mullion in the succeeding sections as follows: (i) Mullion Analysis under Wind Load Normal to Plane of the Unit mullion check under wind suction, refer to 9.1.2, mullion check under wind pressure, refer to 9.1.3, and b f b m Stage 1: Simple beam Stage 2: Continuous beam Stage 1: Simple beam Stage 2: Continuous beam STACK JOINT - HEADER M yy,sw M yy,sw TRANSOM - 1 H FEMALE MULLION MALE MULLION M yy,1 δ z,1 M yy,2 δ z,2 Myy,1 δ z,1 M yy,2 δz,2 TRANSOM - 2 STACK JOINT - SILL M yy,sw M yy,sw W Mom - yy Def - z Mom - yy Def - z Mom - yy Def - z Mom - yy Def - z ELEVATION WIND SUCTION: SYSTEM - 2 WIND PRESSURE: SYSTEM - 2 Fig Wind Load Normal to the Unit Plane (ii) Mullion Analysis under Wind Load Parallel to Plane of the Unit mullion check under lateral wind load, refer to In checking the minor axis bending of the mullion, it is assumed that the upper spandrel panel is a rigid diaphragm in the axis on mullion bending, this by virtue of the double skin 1.5mm infill metal sheets. With this assumption, the mode of mullion bending is similar to a propped cantilever beam with the spandrel panel as its rigid support. Error! Objects cannot be created from editing field codes. Fig Mullion under Lateral Wind Load on Vertical Fin

79 UNITISED CURTAIN WALL 79 of Mullion Check for Wind Suction ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action 4.4 CW Unit Parameters H = 3.70 m Unsupported span of mullion W = 1.94 m Tributary width for wind load Wind load on wall element, Q z = Kpa Suction normal to the plane of CW unit Mullion Alloying compressive yield strength, F cy = 170 Mpa 6063-T6 E = Mpa 7.1.1(b) Male Profile: Effective moment of Inertia, I' e,m = 1.36E+07 mm 4 About y-y axis, composite male profile Effective section modulus, S e1,m = 1.34E+05 mm 3 About y-y axis, on face 1 (for wind suction) 7.1.2(b) Female Profile: Effective moment of Inertia, I' e,f = 1.09E+06 mm 4 About y-y axis, composite female profile Effective section modulus, S e1,f = 2.23E+04 mm 3 About y-y axis, on face 1 (for wind suction) Sword Alloying compressive yield strength, F cy,sw = 490 Mpa 7022-T651 E Sw = Mpa Sword length, L Sw = 600 mm Portion of sword penetrating the mullion profile Sword cross-section, d Sw x t Sw = 140 x 15 mm Sword slack on its depth, gap z = 2.0 mm Total gaps on each side of the sword Bending Analysis Uniform load, q z = W Q z = KN/m Parallel to the major axis of mullion 4.3 Slip angle, α yy = tan -1 (gap z /L Sw ) = rad Sword's freedom of rotation before it carries moment 4.3.2(a) Stage 1: Initial load, q z,i = 24α yy E I' e /H 3 = 1.60 KN/m Amount of load to produce rotation mullion ends Applied load for simple beam bending, q z,1 = 1.60 KN/m qzi < qz : Structural system 2 Midspan moment, M yy,1 = q z,1 H 2 /8 = 2.74 KN m Bending moment in the mullion assembly 4.3.2(b) Stage 2: Net load, q z,2 = q z - q z,1 = 7.13 KN/m Balance load after undergoing rotation, ends Midspan moment, M yy,2 = q z,2 H 2 /24 = 4.07 KN m Bending moment in the mullion assembly Support moment, M yy,sw = q z,2 H 2 /12 = 8.14 KN m Bending moment in the sword Couple force on sword, F z,sw = M yy,sw /L Sw = KN Developed couple on sword Total midspan moment, M yy = M yy,1 + M yy,2 = 6.80 KN m Total bending moment in the mullion assembly 4.2(c) Check Mullion Profile I' e = I 'e,m + I' e,f = 1.5E+07 mm 4 About y-y axis, composite profile 4.2(d) Male Mullion: Moment, M yy,m = M yy (I' e,m /I' e ) = 6.30 KN m Shared moment by male mullion Bending stress, f yy,m = M yy,m /S e1,m = 50.8 Mpa Maximum bending stress 7.1.1(c) Allowable bending comp. stress, F c1,m = 83.8 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,m = Mpa Under local buckling of beam elements on face 1 Stress ratio, f yy,m /F c1,m = 0.61 <1.00 O.K! 4.2(d) Female Mullion: Moment, M yy,f = M yy (I' e,f /I' e ) = 0.50 KN m Shared moment by male mullion Bending stress, f yy,f = M yy,f /S e2,f = 22.6 Mpa Maximum bending stress 7.1.2(c) Allowable bending comp. stress, F c1,f = 76.0 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,f = 96.1 Mpa Under local buckling of beam elements on face 1 Stress ratio, f yy,f /F c1,f = 0.30 <1.00 O.K! Check Sword S yy,sw = t Sw d 2 Sw /6 = 4.9E+04 mm 4 Section modulus Bending: Bending stress, f yy,sw = M yy,sw /S yy,sw = Mpa Considering plastic modulus Allowable stress, F c,sw = 1.3 F cy,sw /n y = Mpa Allowable compressive stress under bending Stress ratio, f yy,sw /F c,sw = 0.43 <1.00 O.K! Shear: Shear on sword, V z,sw = q z H/2 = KN Lateral shear force Shear stress, f sz,sw = V z,sw /(0.9d Sw t Sw ) = 8.5 Mpa Allowable shear stress, F s = F ty,sw /[(3) 0.5 n y ] = Mpa F ty = F cy, n y = 1.65 Stress ratio, f sz,sw /F s = 0.05 <1.00 O.K! 4.3.2(a) Check Deflection δ z,1 = 5/384 q z,1 H 4 /E I' e = 3.85 mm Stage 1: simple beam deflection 4.3.2(b) δ z,2 = 1/384 q z,2 H 4 /E I' e = 3.44 mm Stage 2: continuous beam deflection Total deflection, δ z = d z,1 + d z,2 = 7.29 mm Calculated maximum deflection 2.3 Allowable deflection, δ allow = mm Criteria: Span/175 or 19mm Strain ratio, δ z /δ allow = 0.38 <1.00 O.K! Shear Flow Check Parameters q z-,m = q z- (I' e,m /I' e ) = KN/m Uniform load shared by the male mullion q z-,f = q z- (I' e,f /I' e ) = KN/m Uniform load shared by the female mullion Notes

80 UNITISED CURTAIN WALL 80 of Mullion Check for Wind Pressure ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action 4.4 CW Unit Parameters H = 3.70 m Unsupported span of mullion W = 1.94 m Tributary width for wind load Wind load on wall element, Q z = 4.50 Kpa Pressure normal to the plane of CW unit Mullion Alloying compressive yield strength, F cy = 170 Mpa 6063-T6 E = Mpa 7.1.1(b) Male Profile: Effective moment of Inertia, I' e,m = 1.36E+07 mm 4 About y-y axis, composite male profile Effective section modulus, S e2,m = 1.05E+05 mm 3 About y-y axis, on face 2 (for wind pressure) 7.1.2(b) Female Profile: Effective moment of Inertia, I' e,f = 1.09E+06 mm 4 About y-y axis, composite female profile Effective section modulus, S e2,f = 1.80E+04 mm 3 About y-y axis, on face 2 (for wind pressure) Sword Alloying compressive yield strength, F cy,sw = 490 Mpa 7022-T651 E Sw = Mpa Sword length, L Sw = 600 mm Portion of sword penetrating the mullion profile Sword cross section, d Sw x t Sw = 140 x 15 mm Sword slack on its depth, gap z = 2.0 mm Total gaps on each side of the sword Bending Analysis Uniform load, q z = W Q z = 8.73 KN/m Parallel to the major axis of mullion 4.3 Slip angle, α yy = tan -1 (gap z /L Sw ) = rad Sword's freedom of rotation before it carries moment 4.3.2(a) Stage 1: Initial load, q z,i = 24α yy E I' e /H 3 = 1.60 KN/m Amount of load to produce rotation mullion ends Applied load for simple beam bending, q z,1 = 1.60 KN/m qzi < qz : Structural system 2 Midspan moment, M yy,1 = q z,1 H 2 /8 = 2.74 KN m Bending moment in the mullion assembly 4.3.2(b) Stage 2: Net load, q z,2 = q z - q z,1 = 7.13 KN/m Balance load after undergoing rotation, ends Midspan moment, M yy,2 = q z,2 H 2 /24 = 4.07 KN m Bending moment in the mullion assembly Support moment, M yy,sw = q z,2 H 2 /12 = 8.14 KN m Bending moment in the sword Couple force on sword, F z,sw = M yy,sw /L Sw = KN Developed couple on sword Total midspan moment, M yy = M yy,1 + M yy,2 = 6.80 KN m Total bending moment in the mullion assembly 4.2(c) Check Mullion Profile I' e = I 'e,m + I' e,f = 1.5E+07 mm 4 About y-y axis, composite profile 4.2(d) Male Mullion: Moment, M yy,m = M yy (I' e,m /I' e ) = 6.30 KN m Shared moment by male mullion Bending stress, f yy,m = M yy,m /S e2,m = 64.8 Mpa Maximum bending stress 7.1.1(c) Allowable bending comp. stress, F c2,m = 83.4 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2,m = Mpa Under local buckling of beam elements on face 2 Stress ratio, f yy,m /F c2,m = 0.78 <1.00 O.K! 4.2(d) Female Mullion: Moment, M yy,f = M yy (I' e,f /I' e ) = 0.50 KN m Shared moment by male mullion Bending stress, f yy,f = M yy,f /S e2,f = 28.0 Mpa Maximum bending stress 7.1.2(c) Allowable bending comp. stress, F c2,f = 80.3 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2,f = - Mpa Under local buckling of beam elements on face 2 Stress ratio, f yy,f /F c2,f = 0.35 <1.00 O.K! Check Sword S yy,sw = t Sw d 2 Sw /6 = 4.9E+04 mm 4 Section modulus Bending: Bending stress, f yy,sw = M yy,sw /S yy,sw = Mpa Considering plastic modulus Allowable stress, F c,sw = 1.3 F cy,sw /n y = Mpa Allowable compressive stress under bending Stress ratio, f yy,sw /F c,sw = 0.43 <1.00 O.K! Shear: Shear on sword, V z,sw = q z H/2 = KN Lateral shear force Shear stress, f sz,sw = V z,sw /(0.9d Sw t Sw ) = 8.5 Mpa Allowable shear stress, F s = F ty,sw /[(3) 0.5 n y ] = Mpa F ty = F cy, n y = 1.65 Stress ratio, f sz,sw /F s = 0.05 <1.00 O.K! 4.3.2(a) Check Deflection δ z,1 = 5/384 q z,1 H 4 /E I' e = 3.85 mm Stage 1: simple beam deflection 4.3.2(b) δ z,2 = 1/384 q z,2 H 4 /E I' e = 3.44 mm Stage 2: continuous beam deflection Total deflection, δ z = d z,1 + d z,2 = 7.29 mm Calculated maximum deflection 2.3 Allowable deflection, δ allow = mm Criteria: Span/175 or 19mm Strain ratio, δ z /δ allow = 0.38 <1.00 O.K! Shear Flow Check Parameters q z+,m = q z+ (I' e,m /I' e ) = 8.08 KN/m Uniform load shared by the male mullion q z+,f = q z+ (I' e,f /I' e ) = 0.65 KN/m Uniform load shared by the female mullion Notes

81 UNITISED CURTAIN WALL 81 of Mullion Check for Lateral Wind Load Element 1 - critical element under wind suction c.g. Element 4 - critical element under wind pressure Fig Values of cy for Critical Elements under Biaxial Bending ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action 4.4 Unitized Parameters H = 3.70 m Height of the unitized panel h = 3.23 m Height of the unsupported span in y axis b = 0.19 m Width of vertical fin projection Wind load on fin, Q y = 3.00 Kpa Normal to the axis of fin protrusion Mullion Alloying compressive yield strength, F cy = 170 Mpa 6063-T6 E = Mpa 7.1.1(b) Male Profile: Effective moment of Inertia, I zz,m = 6.50E+05 mm 4 About z-z axis, non-composite male profie Dist. to c.g. of fiber considered, c y1,m = 35.1 mm Along y axis, on face 1 Dist. to c.g. of fiber considered, c y2,m = 18.8 mm Along y axis, on face Sword Alloying compressive yield strength, F cy,sw = 490 Mpa 7022-T651 E Sw = Mpa Sword length, L Sw = 600 mm Portion of sword penetrating the mullion profile Sword depth, d Sw x t Sw = 140 x 15 mm Sword slack on its thickness, gap y = 2.0 mm Total gaps on each side of the sword Bending Analysis Uniform load, q y = b Q y = 0.57 KN/m Parallel to the major axis of mullion 4.3 Slip angle, α zz = tan -1 (gap y /L Sw ) = 0.00 rad Sword's freedom of rotation before it carries moment Stage 1: Initial load, q y,i = 48α zz E I zz,m /h 3 = 0.21 KN/m Amount of load to produce rotation mullion ends Applied load for propped beam bending, q y,1 = 0.21 KN/m qzi < qz : Structural system 2 Midspan moment, M zz,1 = 3q y,1 h(h/2)/8 + q y (H/2) 2 /2 = 0.30 KN m Bending moment in the male mullion Stage 2: Net load, q y,2 = q y - q y,1 = 0.36 KN/m Balance load after undergoing rotation, ends Midspan moment, M zz,2 = q y,2 h 2 /24 = 0.16 KN m Bending moment in the male mullion Support moment, M zz,sw = q y,2 h 2 /12 = 0.31 KN m Bending moment in the sword Couple force on sword, F y,sw = M zz,sw /L Sw = 0.52 KN Developed couple on sword Total midspan moment, M zz = M zz,1 + M zz,2 = 0.46 KN m Total minor axis bending moment in the male mullion Check Male Mullion Profile f zz1,m = M zz /(I zz,m /c y1,m ) = 24.7 Mpa Minor axis bending stress on critical face Face 1: Allow. comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about minor axis Stress ratio, f zz1,m /F c = 0.20 <1.00 O.K! Face 2: f zz1,m = M zz /(I zz,m /c y1,m ) = 13.2 Mpa Minor axis bending stress on critical face 2 Allow. comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about minor axis Stress ratio, f zz1,m /F c = 0.11 <1.00 O.K! Check Sword Section modulus, S zz,sw = d Sw t 2 Sw /6 = 5.25E+03 mm 4 Bending: Bending stress, f zz,sw = M zz,sw /S zz,sw = 59.2 Mpa Considering plastic modulus Allow. bending comp. Stress, F c,sw = 1.3 F cy,sw /n y = Mpa Stress ratio, f zz,sw /F c,sw = 0.15 <1.00 O.K! Shear: Shear on sword, V y,sw = q y H/2 = 1.05 KN Lateral shear force Shear stress, f sy,sw = V y,sw /(0.9d Sw t Sw ) = 0.6 Mpa Allowable shear stress, F s = F ty,sw /[(3) 0.5 n y ] = Mpa Fty = Fcy, ny = 1.65 Stress ratio, f sy,sw /F s = 0.00 <1.00 O.K! Check Deflection of Mullion δ z,1 = q z,1 /185 h 4 /E I zz.m = 2.79 mm Stage 1: propped cantilever beam deflection δ z,2 = 1/384 q z,2 H 4 /E I zz,m = 2.26 mm Stage 2: continuous beam deflection Total lateral deflection, δ z = d z,1 + d z,2 = 5.06 mm Calculated maximum deflection 2.3 Allowable lateral deflection, δ allow = 7.50 mm Criteria: 75% of 10mm clearance Strain ratio, δ z /δ allow = 0.67 <1.00 O.K! Notes

82 UNITISED CURTAIN WALL 82 of Check Sword Under Slab Deflection. Refer to 4.3 for discussions on slab deflection and in-plane performance of the curtain wall units. W R h R v W t H W/2 R h Fig Reaction Forces under Deflected Slab ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes I. Claculated shear on sword in the preceeding section Suction Wind on Mullion: Shear, V z-,m = kn Calculated maximum shear on male mullion Shear, V z-,f = 2.30 kn Calculated maximum shear on female mullion Total sword shear, V z-,sw = V z-,m + V z-,f = kn Total major axis shear on sword under wind suction Pressure Wind on Mullion: Shear, V z+,m = kn Calculated maximum shear on male mullion Shear, V z+,f = 2.30 kn Calculated maximum shear on female mullion Total sword shear, V z+,sw = V z+,m + V z+,f = kn Total major axis shear on sword under wind pressure Lateral Wind on Mullion: Sword shear, V y,sw = 1.41 kn Minor axis shear on sword under lateral wind II. Shear on sword due to slab deflection 5.2 CW Unit Parameters: Width of unit, W = 1.94 m Clear span between dead load support brackets Height of unit, H = 3.70 m Approx. vertical dist. bet. dead load bracket and sword j 5.1 Total weight of the unit, W t = 4.63 kn Calculated weight of all components 3.3-1M Sword Parameters: Sword, F ty = 490 Mpa Shear yield strength Depth, d Sw = 140 mm Sword gross depth Sword thickness, t Sw = 15 mm Sword gross thickness Reaction, R h = W t (W/2H) = 1.2 kn Lateral reaction force due to one lamed bracket support Sword shear, V yδ,sw = R h = 1.2 kn Minor axis shear on sword due tor slab deflection III. Combined shear on sword Suction wind + Lateral wind + Slab defleftion: Shear, V = [V 2 z-,sw +(V y,sw +V yδ,sw ) 2 ) 0.5 = 16.4 kn Combined major and minor axis shear in sword Shear stress, f s = V/(0.9d Sw t Sw ) = 8.7 Mpa Total shear stress in sword Allowable shear stress, F s = F ty /1.732n y = Mpa where n y = 1.65 for yield failure Stress ratio = 0.05 <1.00 O.K! Pressure wind + Lateral wind + Slab defleftion: Shear, V = [V 2 z-,sw +(V y,sw +V yδ,sw ) 2 ) 0.5 = 16.4 kn Combined major and minor axis shear in sword Shear stress, f s = V/(0.9d Sw t Sw ) = 8.7 Mpa Total shear stress in sword Allowable shear stress, F s = F ty /1.732n y = Mpa where n y = 1.65 for yield failure Stress ratio = 0.05 <1.00 O.K! Note that the calculated shear in the mullion is equal to the shear in the sword: V = V + V where: Vz,Sw = Max. shear on sword in the major axis z, Sw z, m z, f Vz,m = Max. shear on the male mullion in the major axis. Vz,f = Max. shear on the female mullion in the major axis.

83 UNITISED CURTAIN WALL 83 of Check Thermal Break Shear (a) Check Thermal Break Longitudinal Shear in Male Mullion The FEM Model Equation (28.a) of AAMA-TIR-A8-04 is an approximation for shear flow in a thermal barrier. The accuracy of this equation depends on the degree of symmetry of the two faces being combined. In the case of the male mullion profile, the two faces are very far from being symmetrical. Therefore a more precise FEM analysis is performed to calculate the actual shear flow in the thermal barrier. The assembly is modeled to include both faces of the male mullion profile as custom beam sections and the pair of thermal break strips as shell elements. Stiff link elements at 100mm spacing along the entire span join both faces with the edges of the thermal strips. Boundary conditions are defined at end spans and at intermediate points at the location of the transoms. The amount of uniform wind load shared by the male mullion (with its female partner) is applied as a point load on every node that are spaced 100mm along the whole span. Face-1 Face-2 Thermal break, 2 nos, 2mm thick Link 100mm on centre Lateral and torsional restraints by the transoms on face 1 Link 100mm on centre Face- 1 Face- 2 Sword Sword Wind load: 8.1N/mmm Coupled joint in vertical translation Sword Rigid constraint on sword Face-1 and face-2 are coupled together at this end Rigid constraint on sword (a) Male mullion Cross-section (b) Male mullion Elevation - Boundary conditions

84 UNITISED CURTAIN WALL 84 of 117 Fig Male mullion Thermal break Model Clause Action Notes Parameters L = 3.70 m Unsupported span of beam assembly Beam Assembly: h = mm Height of thermal break strip b = 2.00 mm Thickness of thermal break strip q z = 8.08 kn/m Applied uniform wind load on the assembly Aluminium Extrusion: E Al = Mpa Young's modulus of aluminium extrusion µ Al = 0.33 Poisson's ratio of aluminium extrusion Aluminium Sword: Eal, Sw = Mpa Young's modulus of sword (modified by inspection) µ Al,Sw = 0.33 Poisson's ratio of aluminium sword Thermal Break: E Pm = 2000 Mpa Young's modulus of thermal break (polyamide) µ Pm = 0.49 Poisson's ratio of thermal break (polyamide) 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Link element: E Lk, I Lk, A Lk = 1.00E+09 Mpa Link element properties (very large number) ANSYS FEM Analysis Results Notes Check Shear Flow Shear stress, f v = 5.24 Mpa Result maximum shear stress in YZ plane Max. shear flow, q c = f v b = N/mm Calculated shear flow in the thermal break Allowable shear flow, Q'c = Qc/F.S. = N/mm Considerig a factor of safety, F.S. of 3.0 Stress ratio = 0.62 <1.00 O.K! Analysis Results - Shear Stress Diagram Fig Shear stress on YZ Plane Investigate Shear Deformation Check that the there is no shear deformation of the thermal break at the right end. Whereas maximum shear deformation occurs at the far left end.

85 UNITISED CURTAIN WALL 85 of 117 (a) Longitudinal shear deformation (a) Left end Maximum shear deformation Fig Longitudinal Shear Deformation (b) Right end Zero shear deformation

86 UNITISED CURTAIN WALL 86 of 117 Result verification Deflection Diagram The result of the analysis is verified by checking if the resulting deflection is close to the manually calculated deflection in Manually calculated deflection in : max = 7.29mm ANSYS FEM model deflection = 9.37mm > max Therefore, the FEM analysis result is conservative. Fig Male mullion - Deflection (b) Check Thermal Break Shear on emale Mullion AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 7.1.2(b) Extrusion Parameters V c /(D w) = mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 11.8 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.47 <1.00 O.K! Pressure Wind: q z+ = 0.65 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 11.8 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.47 <1.00 O.K!

87 UNITISED CURTAIN WALL 87 of Check Stack Joint Header and Sill Transom Check Header & Sill Transom under Wind Suction ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Unitized Parameters Unsupported span, W = 1.94 m 3.2 Upper panel height, h u = 0.39 m Lower panel height, h l = 0.47 m Cross-sectional area of header profile, A g,h = mm 2 P z 7.2.1(a) Cross-sectional area of sill profile, A g,s = mm 2 a 5.1 Weight of infill, if any, Q z = 0.00 Kpa Weight of glass above the transom, Q DL = 0.45 Kpa 4.4 Wind load on wall element, Q y = Kpa Setting block location, a = 190 mm 7.2.1(b) Eccentricity of setting block location, e = 37 mm Reference point for torsional deflection, r = 66.0 mm Transom Alloying material = Alloying compressive yield strength, F cy = 6063-T6 170 Mpa 7.2.1(b) Header Transom: Effective moment of Inertia, I' e,h = 4.66E+05 mm 4 Effective section modulus (face 1), S e1,h = 1.55E+04 mm (a) Moment of inertia, I yy,h = 7.40E+05 mm (b) Dist. of top extreme fibre on face 1, c z1,h = 53.4 mm 7.2.2(b) Sill Transom: Effective moment of inertia, I' e,s = 6.13E+05 mm 4 Effective section modulus (face 1), S e1,s = 2.01E+04 mm (c) Moment of inertia, I yy,s = 9.90E+05 mm 4 Torsional constant, J s = 8.20E+04 mm 4 TRANSOM - 1 Wind load tributary area TRANSOM - 2 STACK JOINT - SILL 7.2.2(b) Dist. of top extreme fibre on face 1, c z1,s = 42.3 mm e Analysis Glass weight, P z = Q DL h u W/2 = 0.17 KN q y r Upper tributary width, b u = (h u /2)[1-h u /(2W)] = 0.18 m Lower tributary width, b l = (h l /2)[1-h l /(2W)] = 0.21 m rotation Moment, M zz = Q y (b u +b l ) W 2 /8 = KN/m Check Stack Joint Profiles I' e = I 'e,h + I' e,s = 1.1E+06 mm 4 About y-y axis, composite profile Header Transom M yy,h = (1.25 A g,h γ)w 2 /8 = 0.02 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy,h = M yy,h /(I yy,h /c z1,h ) = 1.1 Mpa Maximum bending stress Allow. bending comp. stress, F c,h = 1.17 F cy /n y = Mpa Under bending about weak axis 4.2(d) Suction Wind: Moment, M zz,h = M zz (I' e,h /I' e ) = KN m Shared moment by header transom Bending stress, f zz,h = M zz,h /S e1,h = 22.5 Mpa Maximum bending stress on face (c) Allowable bending comp. stress, F c1,h = 98.0 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,h = 88.2 Mpa Under local buckling of beam elements on face 1 Dead Load + Suction Wind: f yy,h /F c,h + f zz,h /F c1,h = 0.26 <1.00 O.K! Sill Transom M yy,s = P z a + (1.25 A g,s γ +Q DL h u )W 2 /8 = 0.05 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy,s = M yy,s /(I yy,s /c z1,s ) = 2.3 Mpa Maximum bending stress Allow. bending comp. stress, F c,s = 1.17 F cy /n y = Mpa Under bending about weak axis 4.2(d) Suction Wind: Moment, M zz,s = M zz (I' e,s /I' e ) = kn-m Shared moment by sill transom Bending stress, f zz,s = M zz,s /S e1,s = 22.9 Mpa Maximum bending stress on face (c) Allowable bending comp. stress, F c1,s = 97.4 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1,s = 69.9 Mpa Under local buckling of beam elements on face 1 Dead Load + Suction Wind: f yy,s /F c,s + f zz,s /F c1,s = 0.35 <1.00 O.K! Check Deflection δ z,h = (5/384)(1.25 A g,h γ)w 4 ]//E I yy,h = 0.12 mm Deflection under dead load 2.3 Header Transom Allowable deflection, δ z,allow = mm Criteria: 75% of 20mm clearance Dead Load: Strain ratio, δ z /δ z,allow = 0.01 <1.00 O.K! δ z,s1 = [(P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g,s γ +Q DL h u )W 4 ]//E I yy,s = 0.34 mm Deflection under dead load Sill Transom δ z,2 = (P z e) W r/(j s G) = 0.38 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 0.72 mm Total deflection at the fibre closest to glass edge 2.3 Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.24 <1.00 O.K! Stack Joint Profiles δ y = (5/384) Q y (b u +b l )W 4 /[E I' e ] = 4.26 mm Calculated maximum deflection 2.3 Suction Wind: Allowable deflection, δ y,allow = mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.38 <1.00 O.K! Shear Flow Check Parameters q y,h = Q y (b u +b l ) (I' e,h /I' e ) = KN/m Uniform load shared by the header transom q y,s = Q y (b u +b l ) (I' e,s /I' e ) = KN/m Uniform load shared by the sill transom End Connection Check Parameters V y,h = q y,h W/2 = 0.7 kn Horizontal reaction force on header profile Header Transom: V z,h = (1.25 A g,s γ)w/2 = 0.0 kn Vertical reaction force due to dead load on header profil Sill Trasnom: V y,s = q y,s W/2 = 0.9 kn Horizontal reaction force on sill profile V z,h = P z + (1.25 A g,s γ +Q z h u )W/2 = 0.2 kn Vertical reaction force due to dead load on sill profile FEMALE MULLION W TRANSOM - 2 SILL HEADER ELEVATION P z Pz a MALE MULLION Wind load tributary area b u b l Equiv. rectangular tributary width Weight of sheets z c.g. 2 x P WIND LOAD SECTION z y e hu h l H

88 UNITISED CURTAIN WALL 88 of Check Header & Sill Transom under Wind Pressure ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Unitized Parameters Unsupported span, W = 1.94 m 3.2 Upper panel height, h u = 0.39 m Lower panel height, h l = 0.47 m Cross-sectional area of header profile, A g,h = mm (a) Cross-sectional area of sill profile, A g,s = mm Weight of infill, if any, Q z = 0.00 Kpa Weight of glass above the transom, Q DL = 0.45 Kpa 4.4 Wind load on wall element, Q y = 4.50 Kpa Setting block location, a = 190 mm 7.2.1(b) Eccentricity of setting block location, e = 47 mm Reference point for torsional deflection, r = 66.0 mm Transom Alloying material = Alloying compressive yield strength, F cy = 6063-T6 170 Mpa 7.2.1(b) Header Transom: Effective moment of Inertia, I' e,h = 4.66E+05 mm 4 Effective section modulus (face 2), S e2,h = 1.55E+04 mm (a) Moment of inertia, I yy,h = 7.40E+05 mm (b) Dist. of top extreme fibre on face 2, c z2,h = 53.4 mm 7.2.2(b) Sill Transom: Effective moment of inertia, I' e,s = 6.13E+05 mm 4 Effective section modulus (face 2), S e2,s = 2.04E+04 mm (c) Moment of inertia, I yy,s = 9.90E+05 mm 4 Torsional constant, J s = 8.20E+04 mm (b) Dist. of top extreme fibre on face 2, c z2,s = 42.3 mm Analysis Glass weight, P z = Q DL h u W/2 = 0.17 KN Applied as pair loads of glass chair Upper tributary width, b u = (h u /2)[1-h u /(2W)] = 0.18 m Width of the equivalent rectangular tributary area Lower tributary width, b l = (h l /2)[1-h l /(2W)] = 0.21 m Width of the equivalent rectangular tributary area Moment, M zz = Q y (b u +b l ) W 2 /8 = 0.81 KN/m Suction normal to the plane of CW unit Check Stack Joint Profiles I' e = I 'e,h + I' e,s = 1.1E+06 mm 4 About y-y axis, composite profile Header Transom M yy,h = (1.25 A g,h γ)w 2 /8 = 0.02 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy,h = M yy,h /(I yy,h /c z2,h ) = 1.1 Mpa Maximum bending stress Allow. bending comp. stress, F c,h = 1.17 F cy /n y = Mpa Under bending about weak axis 4.2(d) Suction Wind: Moment, M zz,h = M zz (I' e,h /I' e ) = 0.35 KN m Shared moment by header transom Bending stress, f zz,h = M zz,h /S e2,h = 22.5 Mpa Maximum bending stress on face (c) Allowable bending comp. stress, F c2,h = 98.5 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2,h = 34.8 Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy,h /F c,h + f zz,h /F c2,h = 0.65 <1.00 O.K! TRANSOM - 1 Wind load tributary area TRANSOM - 2 P z STACK JOINT - SILL q y ELEVATION rotation Sill Transom M yy,s = P z a + (1.25 A g,s γ +Q DL h u )W 2 /8 = 0.05 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy,s = M yy,s /(I yy,s /c z2,s ) = 2.3 Mpa Maximum bending stress Allow. bending comp. stress, F c,s = 1.17 F cy /n y = Mpa Under bending about weak axis 4.2(d) Suction Wind: Moment, M zz,s = M zz (I' e,s /I' e ) = 0.46 kn-m Shared moment by sill transom Bending stress, f zz,s = M zz,s /S e2,s = 22.5 Mpa Maximum bending stress on face (c) Allowable bending comp. stress, F c2,s = 98.3 Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2,s = - Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy,s /F c,s + f zz,s /F c2,s = 0.25 <1.00 O.K! Check Deflection δ z,h = (5/384)(1.25 A g,h γ)w 4 ]//E I yy,h = 0.12 mm Deflection under dead load 2.3 Header Transom Allowable deflection, δ z,allow = mm Criteria: 75% of 20mm clearance Dead Load: Strain ratio, δ z /δ z,allow = 0.01 <1.00 O.K! δ z,s1 = [(P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g,s γ +Q DL h u )W 4 ]//E I yy,s = 0.34 mm Deflection under dead load Sill Transom δ z,2 = (P z e) W r/(j s G) = 0.48 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 0.82 mm Total deflection at the fibre closest to glass edge 2.3 Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.27 <1.00 O.K! Stack Joint Profiles δ y = (5/384) Q y (b u +b l )W 4 /[E I' e ] = 4.26 mm Calculated maximum deflection 2.3 Suction Wind: Allowable deflection, δ y,allow = mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.38 <1.00 O.K! Shear Flow Check Parameters q y,h = Q y (b u +b l ) (I' e,h /I' e ) = 0.74 KN/m Uniform load shared by the header transom q y,s = Q y (b u +b l ) (I' e,s /I' e ) = 0.98 KN/m Uniform load shared by the sill transom End Connection Check Parameters V y,h = q y,h W/2 = 0.7 kn Horizontal reaction force on header profile Header Transom: V z,h = (1.25 A g,s γ)w/2 = 0.0 kn Vertical reaction force due to dead load on header profil Sill Trasnom: V y,s = q y,s W/2 = 0.9 kn Horizontal reaction force on sill profile V z,h = P z + (1.25 A g,s γ +Q z h u )W/2 = 0.2 kn Vertical reaction force due to dead load on sill profile P z a FEMALE MULLION W TRANSOM - 2 SILL HEADER Pz a MALE MULLION e r Wind load tributary area b u b l Equiv. rectangular tributary width Weight of sheets z WIND LOAD 2 x P c.g. SECTION z y e hu h l H

89 UNITISED CURTAIN WALL 89 of Check Thermal Break Shear (a) Check Longitudinal Shear in Header Transom AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 7.2.1(b) Extrusion Parameters V c /(D w) = mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 8.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.32 <1.00 O.K! Pressure Wind: q z+ = 0.74 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 8.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.32 <1.00 O.K! (b) Check Longitudinal Shear in Sill Transom AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 7.2.2(b) Extrusion Parameters V c /(D w) = 7.26 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 7.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.28 <1.00 O.K! Pressure Wind: q z+ = 0.98 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 7.1 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.28 <1.00 O.K!

90 UNITISED CURTAIN WALL 90 of Transom 1 (Type 2) Check Profile under Biaxial Bending ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Unitized Parameters Unsupported span, W = 1.94 m 3.2 Upper panel height, h u = 0.47 m Lower panel height, h l = 2.84 m 7.3.1(a) Cross-sectional area of profile, A g = mm Weight of infill, if any, Q z = 0.46 Kpa Weight of glass above the transom, Q DL = 0.00 Kpa 4.4 Design wind suction, Q y- = Kpa Design wind pressure, Q y+ = 4.50 Kpa Setting block location, a = 190 mm 7.3.1(b) Eccentricity of setting block location, e = 38.7 mm Reference point for torsional deflection, r = Transom Alloying material = Alloying compressive yield strength, F cy = Modulus of Elasticity, E = Shear Modulus of alloy, G = 67.2 mm 6063-T6 170 Mpa Mpa Mpa Unit weight of alloy, γ = 27 kn/m (b) Effective moment of inertia, I' e = 1.3E+06 mm 4 Effective section modulus (face 1), S e1 = (face 2), S e2 = 5.1E+04 mm 3 4.1E+04 mm (c) Moment of inertia, I yy = 1.0E+06 mm 4 Torsional constant, J = 9.9E+05 mm (b) Dist. from c.g. to extreme top fibre on face 1, c z1 = 34.3 mm Dist. from c.g. to extreme top fibre on face 2, c z2 = 34.3 mm Analysis Glass weight, P z = Q DL h u W/2 = 0.00 KN Applied as pair loads of glass chair Upper tributary width, b u = (h u /2)[1-h u /(2W)] = 0.21 m Width of the equivalent rectangular tributary area Lower tributary width, b l = W/3 = 0.65 m Width of the equivalent rectangular tributary area Uniform wind suction load, q y- = Q y- (b u +b l ) = Also used for checking of thermal break shear flow Uniform wind pressure load, q y+ = Q y+ (b u +b l ) = 3.84 Also used for checking of thermal break shear flow Check Transom M yy = P z a + (1.25 A g γ +Q z h u )W 2 /8 = 0.14 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy = M yy /(I yy /c z ) = 4.7 Mpa Maximum bending stress Allow. bending comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about weak axis Suction Wind: Moment, M zz- = q y- W 2 /8 = kn-m Bending moment about z-z axis Bending stress, f zz- = M zz- /S e1 = 35.4 Mpa Maximum bending stress 7.3.1(c) Allowable bending comp. stress, F c1 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1 = - Mpa Under local buckling of beam elements on face 1 Pressure Wind: Moment, M zz+ = q y+ W 2 /8 = 1.81 kn-m Bending moment about z-z axis Bending stress, f zz+ = M zz+ /S e2 = 43.7 Mpa Maximum bending stress 7.3.1(c) Allowable bending comp. stress, F c2 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2 = - Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy /F c + f zz- /F c1 = 0.39 <1.00 O.K! Pressure Wind + Dead Load: f yy /F c + f zz+ /F c2 = 0.47 <1.00 O.K! STACK JOINT - HEADER δ z,1 = [(P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g γ +Q DL h u )W 4 ]//E I yy = 0.78 mm Deflection under dead load Check Deflection δ z,2 = (P z e) W r/(j G) = 0.00 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 0.78 mm Total deflection at the fibre closest to glass edge 2.3 Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.26 <1.00 O.K! Suction Wind: δ y- = (5/384) Q y- (b u +b l )W 4 /[E I' e ] = mm Maximum deflection under suction wind Pressure Wind: δ y+ = (5/384) Q y+ (b u +b l )W 4 /[E I' e ] = 7.85 mm Maximum deflection under pressure wind 2.3 Allowable deflection, δ y,allow = mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.71 <1.00 O.K! End Connection Check Parameters V y- = q y- W/2 = 3.7 kn End shear under wind suction V y+ = q y+ W/2 = 3.7 kn End shear under wind pressure V z = P z + (1.25 A g,s γ +Q z h u )W/2 = 0.3 kn End shear under dead load q FEMALE TRANSOM MULLION - 1 z Wind load tributary area TRANSOM - 2 STACK JOINT - SILL ELEVATION qy rotation z bl MALE MULLION e r Weight of sheets Weight of infill (insulation) b u Equiv. rectangular tributary width WIND LOAD 2 x P c.g. z e hu Hhl SECTION y

91 UNITISED CURTAIN WALL 91 of Check Thermal Break Shear (a) Check Longitudinal Shear in Transom-1 AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 7.3.1(b) Extrusion Parameters V c /(D w) = 6.24 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 24.0 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.95 <1.00 O.K! Pressure Wind: q z+ = 3.84 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 24.0 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.95 <1.00 O.K!

92 UNITISED CURTAIN WALL 92 of Transom 2 (Type 2) Check Profile under Biaxial Bending ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action 4.4 Unitized Parameters Unsupported span, W = 1.94 m 3.2 Upper panel height, h u = 2.84 m Lower panel height, h l = 0.39 m 7.4.1(a) Cross-sectional area of profile, A g = mm Weight of infill, if any, Q z = 0.00 Kpa Weight of glass above the transom, Q DL = 0.45 Kpa 4.4 Design wind suction, Q y- = Kpa Design wind pressure, Q y+ = 4.50 Kpa Setting block location, a = 190 mm 7.4.1(b) Eccentricity of setting block location, e = 43.4 mm Reference point for torsional deflection, r = 72.3 mm Transom Alloying material = Alloying compressive yield strength, F cy = 6063-T6 170 Mpa Modulus of Elasticity, E = Shear Modulus of alloy, G = Mpa Mpa Unit weight of alloy, γ = 27 kn/m (b) Effective moment of inertia, I' e = 1.3E+06 mm 4 Effective section modulus (face 1), S e1 = (face 2), S e2 = 5.0E+04 mm 3 3.4E+04 mm (c) Moment of inertia, I 9.7E+05 mm 4 yy = Torsional constant, J = 9.4E+05 mm (b) Dist. from c.g. to extreme top fibre on face 1, c z1 = 38.6 mm Dist. from c.g. to extreme top fibre on face 2, c z2 = 12.1 mm Analysis Glass weight, P z = Q DL h u W/2 = 1.24 KN Applied as pair loads of glass chair Upper tributary width, b u = W/3 = 0.65 m Width of the equivalent rectangular tributary area Lower tributary width, b l = (h l /2)[1-h l /(2W)] = 0.18 m Width of the equivalent rectangular tributary area Uniform wind suction load, q y- = Q y- (b u +b l ) = Also used for checking of thermal break shear flow Uniform wind pressure load, q y+ = Q y+ (b u +b l ) = 3.70 Also used for checking of thermal break shear flow Check Transom M yy = P z a + (1.25 A g γ +Q z h u )W 2 /8 = 0.27 kn-m Accessories assumed 25% add'l to transom weight Dead Load: Bending stress, f yy = M yy /(I yy /c z ) = 10.7 Mpa Maximum bending stress Allow. bending comp. stress, F c = 1.17 F cy /n y = Mpa Under bending about weak axis Suction Wind: Moment, M zz- = q y- W 2 /8 = kn-m Bending moment about z-z axis Bending stress, f zz- = M zz- /S e1 = 35.1 Mpa Maximum bending stress 6.4.1(c) Allowable bending comp. stress, F c1 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c1 = 89.0 Mpa Under local buckling of beam elements on face 1 Pressure Wind: Moment, M zz+ = q y+ W 2 /8 = 1.74 kn-m Bending moment about z-z axis Bending stress, f zz+ = M zz+ /S e2 = 51.6 Mpa Maximum bending stress 6.4.1(c) Allowable bending comp. stress, F c2 = Mpa Under lateral buckling of beam section on face (d) Allowable bending comp. stress, F c2 = - Mpa Under local buckling of beam elements on face 2 Dead Load + Suction Wind: f yy /F c + f zz- /F c1 = 0.48 <1.00 O.K! Pressure Wind + Dead Load: f yy /F c + f zz+ /F c2 = 0.59 <1.00 O.K! δ z,1 = [(P z a/24)(3w 2-4a 2 )+(5/384)(1.25 A g γ +Q DL h u )W 4 ]//E I yy = 1.83 mm Deflection under dead load Check Deflection δ z,2 = (P z e) W r/(j G) = 0.31 mm Add'l deflection due to torsion under eccentric load Dead Load: δ z = δ z,1 + δ z,2 = 2.14 mm Total deflection at the fibre closest to glass edge 2.3 Allowable deflection, δ z,allow = 3.00 mm Criteria: 75% of 4mm clearance Strain ratio, δ z /δ z,allow = 0.71 <1.00 O.K! Suction Wind: δ y- = (5/384) Q y- (b u +b l )W 4 /[E I' e ] = mm Maximum deflection under suction wind Pressure Wind: δ y+ = (5/384) Q y+ (b u +b l )W 4 /[E I' e ] = 7.62 mm Maximum deflection under pressure wind 2.3 Allowable deflection, δ y,allow = mm Criteria: Span/175 or 19mm Strain Ratio, δ y,max /δ y,allow = 0.69 <1.00 O.K! End Connection Check Parameters V y- = q y- W/2 = 3.6 kn End shear under wind suction V y+ = q y+ W/2 = 3.6 kn End shear under wind pressure V z = P z + (1.25 A g,s γ +Q z h u )W/2 = 1.3 kn End shear under dead load Notes P z a W STACK JOINT - HEADER TRANSOM - 1 FEMALE MULLION Wind load tributary area TRANSOM - 2 STACK JOINT - SILL P z Equivalent rectangular tributary width ELEVATION q y rotation z MALE MULLION a e r b u b l c.g. WIND LOAD2 x P z e h SECTION y h u H l

93 UNITISED CURTAIN WALL 93 of Check Thermal Break Shear (a) Check Longitudinal Shear in Transom-2 AAMA TIR-A8-04: Structural Performance of Composite Thermal Barrier Framing System Clause Action Notes 7.4.1(b) Extrusion Parameters V c /(D w) = 5.81 mm Shear flow per uniform unit load, w 4.2(f) Test shear strength, Q c = N/mm Taken from shear test of thermal break assembly Suction Wind: q z- = kn/m Applied uniform wind suctinon load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 21.5 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.85 <1.00 O.K! Pressure Wind: q z+ = 3.70 kn Applied uniform wind pressure load 7.5.7(28) Shear stress, q c = q z- V c /(D w) = 21.5 N/mm Max. longitudinal shear through the thermal break Design shear strength, Q' c = Q c /F.S. = 25.2 N/mm Considering a factor of safety, F.S. = 2.0 Stress ratio = 0.85 <1.00 O.K!

94 UNITISED CURTAIN WALL 94 of Check Stainless Steel Spandrel Panel Details For structural system 2, the 1.5mm thick stainless steel spandrel panel is reinforced with a continuous bent stainless steel sheet bonded at the back of the sheet via structural silicone. Fig Spandrel Panel Details Where: K = 341.5mm mm Thick Stainless Steel Reinforcement The metal sheet and its reinforcement are considered as non-composite or layered elements. Which means the structural sealant bonding is not considered as to transfer shear between the metals. Therefore, every adjacent nodes that are co-incident on the plane of the metal sheet are coupled in the in-out direction (as shown below). 1.5mm thk sheet 4mm thk reinforce ment In-out Coupling Fig Coupled Nodes

95 UNITISED CURTAIN WALL 95 of Check Sheet for Full Wind Load Shell Element Model with Boundary conditions Perimeter Restraint in Z - dir. Wind load: - Coupled Restraint in Fig Spandrel Sheet Applied Load & Boundary Conditions (a) Calculations Summary ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes Metal Sheet Parameters Width, W = 1940 mm Net dimension considering width of mullions Height, h = 400 mm Net height considering grip of glazing bead Wind load, Q = -4.5 KN/m 2 Design wind load in suction Alloy, F ty = 170 Mpa Tensile yield strength of Stainless steel S31203 ANSYS FEM Anlysis Results Notes Check Stress VM =0.707[(S min -S min ) 2 +S 2 max +S 2 min ) 0.5 = Mpa Von mises stress results (from ANSYS FEM anlaysis) Allowable stress, F c = 1.3F ty /n y = Mpa For rectangular solid plates Stress ratio, VM/F c = 0.76 <1.00 O.K! Check Deflection δ max = 5.46 mm Maximum deflection (from ANSYS FEM analysis) Allowable deflection, δ allow = 8.00 mm Architectural requirement (span/60 = 400mm/80) Strain ratio, δ z /δ allow = 0.68 <1.00 O.K!

96 UNITISED CURTAIN WALL 96 of 117 (b) Analysis Results (i.) Von Mises Stress Diagram (ii.) Deflection Diagram

97 3 PROJECT NAME DATE UNITISED CURTAIN WALL 97 of JOINTS & CONNECTIONS Fig Spandrel Sheet Analysis Results In all cases in this section, the most critical structural system-2 is checked for joint connections Mullion Shear Connectors qy-,m q q ± q y+,f q y Fig Male and Female Mullion Shear Connectors ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes Mullion Parameters Yield strength, F cy = 170 Mpa 6063-T6 Aluminum alloy Male Profile: Male leg length, l m = 34.0 mm From the toe of fillet to the farthest contact edge Male leg thickness, t m = 3.0 mm Thickness at the toe of fillet Female Profile: Female leg length, l f = 26.5 mm From the toe of fillet to the farthest contact edge Female leg thickness, t f = 3.0 mm Thickness at the toe of fillet Allowable compressive stress, F c = 1.3F cy /n y = Mpa Bending for rectangular solid plates Allowable shear stress, F s = F ty /(1.732n y ) = Mpa Considering F ty is equal to F cy Loads Uniform wind suction load, q z- = kn/m Total suction load carried by the mullion assembly Wind suction: Shared load by male mullion, q z-,m = kn/m In proportion of male-to-total bending stiffness Shared load by female mullion, q z-,f = kn/m In proportion of female-to-total bending stiffness Wind pressure: Uniform wind pressure load, q z+ = 8.73 kn/m Total suction load carried by the mullion assembly Shared load by male mullion, q z+,m = 8.08 kn/m In proportion of male-to-total bending stiffness Shared load by female mullion, q z+,f = 0.65 kn/m In proportion of female-to-total bending stiffness Check Male Shear Connetor Wind suction: Max. load by male, q zmax-,m = 0.5 q z- = kn/m Assuming male and female have equal bending stiffness Applied load, q -,m = q zmax-,m - q z-,m = 3.72 kn/m Uniform load on protrusion of male mullion Bending moment, M m- = q -,m (1mm)(l m ) = N mm Considering a typical 1mm strip Bending stress, f b-,m = 6M m- /[(1mm)t 2 m ] = Mpa Maximum compressive stress of extreme fiber Shear stress, f s-,m = q -,m /t m = 1.24 Mpa 4.4 Stress ratio, f b-,m /F c + (f s-,m /F s ) 2 = 0.63 <1.00 O.K! Wind pressure: Max. load by male, q zmax+,m = 0.5 q z+ = 4.37 kn/m Assuming male and female have equal bending stiffness Applied load, q +,m = q zmax+,m - q z+,m = kn/m Uniform load on protrusion of male mullion Bending moment, M m+ = q +,m (1mm)(l m ) = N mm Considering a typical 1mm strip Bending stress, f b+,m = 6M m+ /[(1mm)t 2 m ] = Mpa Maximum compressive stress of extreme fiber Shear stress, f s+,m = q +m /t m = 5.72 Mpa 4.4 Stress ratio, f b+,m /F c + (f s+,m /F s ) 2 = 0.64 <1.00 O.K! Check Male Shear Connetor Wind suction: Max. load by male, q zmax-,f = 0.5 q z- = kn/m Assuming male and female have equal bending stiffness Applied load, q -,f = q zmax-,f - q z-,f = kn/m Uniform load on protrusion of male mullion Bending moment, M m- = q -,f (1mm)(l f ) = N mm Considering a typical 1mm strip Bending stress, f b-,f = 6M f- /[(1mm)t 2 f ] = Mpa Maximum compressive stress of extreme fiber Shear stress, f s-,f = q -,f /t f = 1.24 Mpa 4.4 Stress ratio, f b-,f /F c + (f s-,f /F s ) 2 = 0.49 <1.00 O.K!

98 UNITISED CURTAIN WALL 98 of Glazing Bead Check W Stack joint - Header TRANSOM GLAZING BEAD Transom-1 b t h Female mullion Wind load tributary area Male mullion MULLION GLAZING BEAD Transom-2 b m Maximum tributary width tributary width Stack joint - Sill Fig CW unit Elevation MULLION GLAZING BEAD TRANSOM GLAZING BEAD MULLION GLAZING BEAD TRANSOM GLAZING BEAD Fig Glazing Bead Types

99 UNITISED CURTAIN WALL 99 of Check Glazing Bead on Transom (a) Solid Model with Boundary conditions Lateral Lateral Outward Pressure Fig Glazing Bead on Transom - Boundary Conditions (20mm long representation)

100 UNITISED CURTAIN WALL 100 of 117 (b) Calculations Summary ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Glass Prameters Glass width, W = 1.94 m Width of glass supported by the glazing bead Glass height, h = 2.84 m Height of glass supported by the glazing bead 4.4 Wind load, Q = -4.5 KN/m 2 Design wind load in suction 3.3-1M Glazing Bead Parameters Alloy, F cy = 170 Mpa Compressive yield strength of alloy 10.1 Bearing width, w= 3.0 pressure point to gasket at the tip of glazing bead Transom glazing bead, b t = W/2 = 0.97 m Maximum trbutary width Pressure load, q t = b t Q/w = 1.46 N/mm 2 the pressure point to gasket at tip of bead ANSYS FEM Anlysis Results Notes Check Stress VM =0.707[S min -S min ) 2 +S 2 max +S 2 min ) 0.5 = Mpa Von mises stress results (from ANSYS FEM anlaysis) Allowable stress, F c = 1.3F cy /n y = Mpa For rectangular solid plates Stress ratio, VM/F c = 0.87 <1.00 O.K! Check Deflection δ max = 0.48 mm Maximum deflection (from ANSYS FEM analysis) Allowable deflection, δ allow = 1.00 mm For gasket Strain ratio, δ z /δ allow = 0.48 <1.00 O.K! (i.) Von Mises Stress Diagram (ii.) Deflection Diagram Fig Glazing Bead on Transom - Analysis Results

101 UNITISED CURTAIN WALL 101 of Check Glazing Bead on Mullion (a) Solid Model with Boundary conditions Lateral Pressure Outward Lateral Fig Glazing Bead on Mullion - Boundary Conditions (20mm long representation)

102 UNITISED CURTAIN WALL 102 of 117 (b) Calculations Summary ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes 4.4 Glass Prameters Glass width, W = 1.94 m Width of glass supported by the glazing bead Glass height, h = 2.84 m Height of glass supported by the glazing bead 4.4 Wind load, Q = -4.5 KN/m 2 Design wind load in suction 3.3-1M Glazing Bead Parameters Alloy, F cy = 170 Mpa Compressive yield strength of alloy 10.1 Bearing width, w= 3.0 pressure point to gasket at the tip of glazing bead Mullion glazing bead, b m = W/2 = 0.97 m Maximum tributary width Pressure load, q m = b m Q/w = 1.46 N/mm 2 the pressure point to gasket at tip of bead ANSYS FEM Anlysis Results Notes Check Stress VM =0.707[S min -S min ) 2 +S 2 max +S 2 min ) 0.5 = Mpa Von mises stress results (from ANSYS FEM anlaysis) Allowable stress, F c = 1.3F cy /n y = Mpa For rectangular solid plates Stress ratio, VM/F c = 0.53 <1.00 O.K! Check Deflection δ max = 0.32 mm Maximum deflection (from ANSYS FEM analysis) Allowable deflection, δ allow = 1.00 mm For gasket Strain ratio, δ z /δ allow = 0.32 <1.00 O.K! (i.) Von Mises Stress Diagram (ii.) Deflection Diagram Fig Glazing Bead on Mullion - Analysis Results

103 UNITISED CURTAIN WALL 103 of Header Transom Connection Check Connection Detail P V y f1 f3 V y +Y f Z Analysis of Screw Group Analysis of Group of Fasteners Fig Header Joint Connection Reference Action Notes Applied Forces Horizontal shear force, V y = 0.70 kn Under most critical wind load Eccentricity of V y, e z = 0.00 mm Eccentric application of wind load Vertical shear force, V z = 0.00 kn Under dead load Eccentricity of V z, e y = 0.00 mm Eccentric location of setting block Fastener Group Tension force, T = 0.00 kn Parallel to the axis of the fasteners Eccentricity of T, e ty = 0.00 mm Eccentric Application of the tension force Total applied moment, M = V y e z + V z e y = 0.00 knm Sum of the torsion and all effects of eccentricity Number of fasteners, n = 3 nos. Fasteners in the group Calculated: Shear Force Tension Size A y z Q z I y I z F v,y1 F v,y2 F v,y F v,z1 F v,z2 F v,z Fv F t Fastener mm 2 mm mm mm 3 mm 4 mm 4 kn kn kn kn kn kn kn kn f1: ST f2: ST f3: ST Σ = Definition of Annotations A = Shear area of individual fasteners ey +z y, z = Coordinate location of individual fasteners from c.g. of fastener group Q z = A y+e ty = Statical moment of area about z-axis at center of T application I y = A z 2 = Inertia about y-axis of individual fasteners y y V 4 1 z f I z = A y 2 4 f1 = Inertia about z-axis of individual fasteners V F v,y1 = V y /n = Shear force in y-axis of individual fastener due to shear force, V y e y ty z 4 z 1 F v,y2 = M (A z/σi y ) = Shear force in y-axis of individual fastener due to moment, M T F v,y = F v,y1 + F v,y2 = Total shear force in y-axis of individual fastener z e z3 f 3 y 3 fasteners c.g. y 2 f 2 z2 +y F v,z1 = V z /n = Shear force in z-axis of individual fastener due to shear force, V z F v,z2 = M (A y/σi z ) = Shear force in z-axisof individual fastener due to moment, M F v,z = F v,z1 + F v,z2 = Total shear force in z-axis of individual fastener F v = (F v,z1 2 + F v,z2 2)0.5 = Total shear force in the individual fasteners F t = T(Q z /ΣQ z ) = Tension force in the individual fasteners

104 UNITISED CURTAIN WALL 104 of Check Most Critical Screw The most critical screw among the group is fastener f3. AAMA TIR-A9-1991: Metal Curtain Wall Fasteners Clause Action Notes Applied Forces Applied Tension, F t = 0.00 KN. Calculated tension force in the most critical screw Applied Shear, F v = 0.23 KN. Calculated shear force in the most critical screw Connected Parts 1st connected part, F u1 = 205 Mpa. Thickness, t p1 = 2nd connected part, F u2 = Thickness, t p2 = Number of shear plane(s), m = Screw Parameters Screw size, d b = 3.0 mm. L e1 = 6.4 mm. Dist. from centre of screw to the nearest free edge 205 Mpa. F y2 = 170 Mpa. Yield strength of connected material 3.0 mm. L e2 = 5.0 mm. Dist. from centre of screw to the nearest free edge LE = 20.7 mm. Length of fastener engagement 1 plane(s) #8 (ST4.2) Pan head screw A4-70 Stainless steel (BS EN ISO ) F y = 450 Mpa. Minimum tensile yield strength (BS EN ISO ) F u = 700 Mpa. Minimum ultimate tensile strength 9. Table 25 D = 4.16 mm. Nominal thread diameter 9. Table 25 K = 3.19 mm. Basic minor diameter 9. Table 25 Thread pitch, P = 1.41 mm. Giving a value of N = 18 threads per inch 9. Table 25 DSMIN = 3.99 mm. Minimum major diameter of external thread 9. Table 25 ENMAX = 3.76 mm. Maximum pitch diameter of internal thread 9. Table 25 A(S) = (K)² = 7.99 mm 2 Tensile stress area for spaced threads 9. Table 25 A(R) = (K)² = 7.99 mm 2 Thread root area for spaced threads TSA(I) = π[dsmin][(12.7/n) (dsmin-enmax)] = 6.45 mm 2 Thread stripping area per thread, internal 6. Check Tension P t = 0.75Fy[A(S)] = 2.70 KN. Yielding failure allowable tension P t = 0.40Fu[A(S)] = 2.24 KN. Fracture failure governs allowable tension F t /P t = 0.00 <1.00 O.K! 6. Check Shear P v = 0.75Fy[A(R)]/1.732 = 1.56 KN. Yielding failure allowable shear P v = 0.40Fu[A(R)]/1.732 = 1.29 KN. Fracture failure governs allowable shear F s /(m P v ) = 0.18 <1.00 O.K! Interaction, [F t /P t ] 2 + [F s /(m P v )] 2 = Check Bearing Fb1 = (2/1.95)*Fu1/1.65 = Mpa. Allowable bearing stress on 1st connected part t p1 '= 3.00 mm. Effective bearing thickness P b1 = F b D t p1 ' = 1.6 KN. Allowable bearing capacity P b1 /(m F s ) = 0.14 <1.00 O.K! Fb2 = (2/1.95)*Fu2/1.65 = Mpa. Allowable bearing stress on 2nd connected part P b2 = F b D t p2 = 1.4 KN. Allowable bearing capacity P b2 /F s = 0.16 <1.00 O.K! 6063-T6 Aluminium Standard round hole 6063-T6 Aluminium Standard round hole 9. Check Pullout Fv = 0.4*Fsu = 52 Mpa. Fsu = Shear ultimate strength, see ADM Table 3.3-1M P p = [LE/25.4-1/N] F v TSA(I) N = 4.6 KN. Pullout strength on thread of 2nd connected part F t /P p = 0.00 <1.00 O.K!

105 UNITISED CURTAIN WALL 105 of Sill Transom Connection Check Connection Detail V y f f X +Z +Y f2 5 V y 9.2 T Analysis of Screw Group Analysis of Group of Fasteners Fig Sill Joint Connection Reference Action Notes Applied Forces Horizontal shear force, V y = 0.90 kn Under most critical wind load Eccentricity of V y, e z = 0.00 mm Eccentric application of wind load Vertical shear force, V z = 0.20 kn Under dead load Eccentricity of V z, e y = mm Eccentric location of setting block Tension force, T = 0.20 kn Parallel to the axis of the fasteners Eccentricity of T, e ty = mm Eccentric Application of the tension force Total applied moment, M = V y e z + V z e y = knm Sum of the torsion and all effects of eccentricity Number of fasteners, n = 3 nos. Fasteners in the group Fastener Group Calculated: Shear Force Tension Size A y z Q z I y I z F v,y1 F v,y2 F v,y F v,z1 F v,z2 F v,z Fv F t Fastener mm 2 mm mm mm 3 mm 4 mm 4 kn kn kn kn kn kn kn kn f1: ST f2: ST f3: ST Σ = Definition of Annotations A = Shear area of individual fasteners ey +z y, z = Coordinate location of individual fasteners from c.g. of fastener group Q z = A y+e ty = Statical moment of area about z-axis at center of T application I y = A z 2 = Inertia about y-axis of individual fasteners y y V 4 1 z f I z = A y 2 4 f1 = Inertia about z-axis of individual fasteners V F v,y1 = V y /n = Shear force in y-axis of individual fastener due to shear force, V y e y ty z 4 z 1 F v,y2 = M (A z/σi y ) = Shear force in y-axis of individual fastener due to moment, M T fasteners c.g. F v,y = F v,y1 + F v,y2 = Total shear force in y-axis of individual fastener +y z 3 z 2 F v,z1 = V z /n = Shear force in z-axis of individual fastener due to shear force, V z F v,z2 = M (A y/σi z ) = Shear force in z-axisof individual fastener due to moment, M f 3 f 2 F v,z = F v,z1 + F v,z2 = Total shear force in z-axis of individual fastener z e y 3 y 2 F v = (F v,z1 2 + F v,z2 2)0.5 = Total shear force in the individual fasteners F t = T(Q z /ΣQ z ) = Tension force in the individual fasteners

106 UNITISED CURTAIN WALL 106 of Check Most Critical Screw The most critical screw among the group is fastener f2. AAMA TIR-A9-1991: Metal Curtain Wall Fasteners Clause Action Notes Applied Forces Applied Tension, F t = 0.09 KN. Calculated tension force in the most critical screw Applied Shear, F v = 0.69 KN. Calculated shear force in the most critical screw Connected Parts 1st connected part, F u1 = 205 Mpa. Thickness, t p1 = 2nd connected part, F u2 = Thickness, t p2 = Number of shear plane(s), m = Screw Parameters Screw size, d b = 3.0 mm. L e1 = 9.2 mm. Dist. from centre of screw to the nearest free edge 205 Mpa. F y2 = 170 Mpa. Yield strength of connected material 3.0 mm. L e2 = 5.0 mm. Dist. from centre of screw to the nearest free edge LE = 20.7 mm. Length of fastener engagement 1 plane(s) (BS EN ISO ) F y = 450 Mpa. Minimum tensile yield strength (BS EN ISO ) F u = 700 Mpa. Minimum ultimate tensile strength 9. Table 25 D = 4.83 mm. Nominal thread diameter 9. Table 25 K = 3.53 mm. Basic minor diameter 9. Table 25 Thread pitch, P = 1.59 mm. Giving a value of N = 16 threads per inch 9. Table 25 DSMIN = 4.62 mm. Minimum major diameter of external thread 9. Table 25 ENMAX = 4.24 mm. Maximum pitch diameter of internal thread 9. Table 25 A(S) = (K)² = 9.79 mm 2 Tensile stress area for spaced threads 9. Table 25 A(R) = (K)² = 9.79 mm 2 Thread root area for spaced threads TSA(I) = π[dsmin][(12.7/n) (dsmin-enmax)] = mm 2 Thread stripping area per thread, internal 6. Check Tension P t = 0.75Fy[A(S)] = 3.30 KN. Yielding failure allowable tension P t = 0.40Fu[A(S)] = 2.74 KN. Fracture failure governs allowable tension F t /P t = 0.03 <1.00 O.K! 6. Check Shear P v = 0.75Fy[A(R)]/1.732 = 1.91 KN. Yielding failure allowable shear P v = 0.40Fu[A(R)]/1.732 = 1.58 KN. Fracture failure governs allowable shear F s /(m P v ) = 0.44 <1.00 O.K! Interaction, [F t /P t ] 2 + [F s /(m P v )] 2 = 0.19 <1.00 O.K! 7. Check Bearing Fb1 = (2/1.95)*Fu1/1.65 = Mpa. Allowable bearing stress on 1st connected part t p1 '= 3.00 mm. Effective bearing thickness P b1 = F b D t p1 ' = 1.8 KN. Allowable bearing capacity P b1 /(m F s ) = 0.37 <1.00 O.K! Fb2 = (2/1.95)*Fu2/1.65 = Mpa. Allowable bearing stress on 2nd connected part P b2 = F b D t p2 = 1.6 KN. Allowable bearing capacity P b2 /F s = 0.43 <1.00 O.K! 6063-T6 Aluminium Standard round hole 6063-T6 Aluminium Standard round hole #10 (ST4.8) Pan head screw A4-70 Stainless steel 9. Check Pullout Fv = 0.4*Fsu = 52 Mpa. Fsu = Shear ultimate strength, see ADM Table 3.3-1M P p = [LE/25.4-1/N] F v TSA(I) N = 6.9 KN. Pullout strength on thread of 2nd connected part F t /P p = 0.01 <1.00 O.K! 12. Check Srew Engagement Torque, T = 6.0 N m. Applied tigthening torque on screw or bolt f = 0.47 Coefficient of friction for steel on aluminium R = D/2 = 2.42 mm. Major radius of screw thread r = D/ /N = 1.44 mm. Minor radius of screw thread r m = (R+r)/2 = 1.93 mm. Mean radius of screw thread a = 2cos -1 (r/r) = Angle defining limits of screw engagement A e = R 2 [π(a/180)-sin(a)] = 5.28 mm 2 Total area of screw thread engagement A t = π(r 2 -r 2 ) = mm 2 Thread area of fastener Re = Ae/At = 0.45 Ratio of engagement and total thread area sec(c) = { [24(R-r)] 2 +(8.5P) 2 } 0.5 /[24(R-r)] = 1.15 Secant of 1/2 the angle between faces of a thread SF =R e (T/r m )[P+2π r m f sec(c)]/[2π r m -P f sec(c)] = 1.01 kn Ultimate lateral frictional resistance of screw per thread P S = SF(LE/P)/2.34 = 5.63 kn Allowable design lateral frictional resistance of crew F t /P S = 0.02 <1.00 O.K!

107 UNITISED CURTAIN WALL 107 of Transom - 1 Connection Check Connection Detail V z f f Z V y 6.3 +Y f f Analysis of Screw Group Analysis of Group of Fasteners Fig Transom-2 End Connection Reference Action Notes Applied Forces Horizontal shear force, V y = 3.70 kn Under most critical wind load Eccentricity of V y, e z = mm Eccentric application of wind load Vertical shear force, V z = 0.30 kn Under dead load Eccentricity of V z, e y = mm Eccentric location of setting block Fastener Group Tension force, T = 0.00 kn Parallel to the axis of the fasteners Eccentricity of T, e ty = 0.00 mm Eccentric Application of the tension force Total applied moment, M = V y e z + V z e y = knm Sum of the torsion and all effects of eccentricity Number of fasteners, n = 4 nos. Fasteners in the group Calculated: Shear Force Tension Fastener Size A y z Q z I y I z F v,y1 F v,y2 F v,y F v,z1 F v,z2 F v,z Fv F t mm 2 mm mm mm 3 mm 4 mm 4 kn kn kn kn kn kn kn kn f1: ST f2: ST f3: ST f4: ST Σ = Definition of Annotations A = Shear area of individual fasteners ey +z y, z = Coordinate location of individual fasteners from c.g. of fastener group Q z = A y+e ty = Statical moment of area about z-axis at center of T application V y z e V z f 4 y 1 z 4 z3 f 3 T y 4 y 3 e ty fasteners c.g. y 2 f 1 f 2 z 1 z2 +y I y = A z 2 = Inertia about y-axis of individual fasteners I z = A y 2 = Inertia about z-axis of individual fasteners F v,y1 = V y /n = Shear force in y-axis of individual fastener due to shear force, V y F v,y2 = M (A z/σi y ) = Shear force in y-axis of individual fastener due to moment, M F v,y = F v,y1 + F v,y2 = Total shear force in y-axis of individual fastener F v,z1 = V z /n = Shear force in z-axis of individual fastener due to shear force, V z F v,z2 = M (A y/σi z ) = Shear force in z-axisof individual fastener due to moment, M F v,z = F v,z1 + F v,z2 = Total shear force in z-axis of individual fastener F v = (F v,z1 2 + F v,z2 2 ) 0.5 = Total shear force in the individual fasteners F t = T(Q z /ΣQ z ) = Tension force in the individual fasteners

108 UNITISED CURTAIN WALL 108 of Check Most Critical Screw The most critical screw among the group is fastener f2. AAMA TIR-A9-1991: Metal Curtain Wall Fasteners Clause Action Notes Applied Forces Applied Tension, F t = 0.00 KN. Calculated tension force in the most critical screw Applied Shear, F v = 1.12 KN. Calculated shear force in the most critical screw Connected Parts 1st connected part, F u1 = 205 Mpa. Thickness, t p1 = 2nd connected part, F u2 = Thickness, t p2 = Number of shear plane(s), m = Screw Parameters Screw size, d b = 3.0 mm. L e1 = 16.5 mm. Dist. from centre of screw to the nearest free edge 205 Mpa. F y2 = 170 Mpa. Yield strength of connected material 3.0 mm. L e2 = 5.0 mm. Dist. from centre of screw to the nearest free edge LE = 20.7 mm. Length of fastener engagement 1 plane(s) #10 (ST4.8) Pan head screw A4-70 Stainless steel (BS EN ISO ) F y = 450 Mpa. Minimum tensile yield strength (BS EN ISO ) F u = 700 Mpa. Minimum ultimate tensile strength 9. Table 25 D = 4.83 mm. Nominal thread diameter 9. Table 25 K = 3.53 mm. Basic minor diameter 9. Table 25 Thread pitch, P = 1.59 mm. Giving a value of N = 16 threads per inch 9. Table 25 DSMIN = 4.62 mm. Minimum major diameter of external thread 9. Table 25 ENMAX = 4.24 mm. Maximum pitch diameter of internal thread 9. Table 25 A(S) = (K)² = 9.79 mm 2 Tensile stress area for spaced threads 9. Table 25 A(R) = (K)² = 9.79 mm 2 Thread root area for spaced threads TSA(I) = π[dsmin][(12.7/n) (dsmin-enmax)] = mm 2 Thread stripping area per thread, internal 6. Check Tension P t = 0.75Fy[A(S)] = 3.30 KN. Yielding failure allowable tension P t = 0.40Fu[A(S)] = 2.74 KN. Fracture failure governs allowable tension F t /P t = 0.00 <1.00 O.K! 6. Check Shear P v = 0.75Fy[A(R)]/1.732 = 1.91 KN. Yielding failure allowable shear P v = 0.40Fu[A(R)]/1.732 = 1.58 KN. Fracture failure governs allowable shear F s /(m P v ) = 0.71 <1.00 O.K! Interaction, [F t /P t ] 2 + [F s /(m P v )] 2 = Check Bearing Fb1 = (2/1.95)*Fu1/1.65 = Mpa. Allowable bearing stress on 1st connected part t p1 '= 3.00 mm. Effective bearing thickness P b1 = F b D t p1 ' = 1.8 KN. Allowable bearing capacity P b1 /(m F s ) = 0.61 <1.00 O.K! Fb2 = (2/1.95)*Fu2/1.65 = Mpa. Allowable bearing stress on 2nd connected part P b2 = F b D t p2 = 1.6 KN. Allowable bearing capacity P b2 /F s = 0.69 <1.00 O.K! 6063-T6 Aluminium Standard round hole 6063-T6 Aluminium Standard round hole 9. Check Pullout Fv = 0.4*Fsu = 52 Mpa. Fsu = Shear ultimate strength, see ADM Table 3.3-1M P p = [LE/25.4-1/N] F v TSA(I) N = 6.9 KN. Pullout strength on thread of 2nd connected part F t /P p = 0.00 <1.00 O.K!

109 UNITISED CURTAIN WALL 109 of Transom - 2 (System 1) Connection Check Connection Detail V y f4 f Vz f Z +Y 3 f Analysis of Screw Group Analysis of Group of Fasteners Fig Transom-2 End Connection Reference Action Notes Applied Forces Horizontal shear force, V y = 1.70 kn Under most critical wind load Eccentricity of V y, e z = 0.00 mm Eccentric application of wind load Vertical shear force, V z = 1.00 kn Under dead load Eccentricity of V z, e y = mm Eccentric location of setting block Fastener Group Tension force, T = 0.00 kn Parallel to the axis of the fasteners Eccentricity of T, e ty = 0.00 mm Eccentric Application of the tension force Total applied moment, M = V y e z + V z e y = knm Sum of the torsion and all effects of eccentricity Number of fasteners, n = 4 nos. Fasteners in the group Calculated: Shear Force Tension Size A y z Q z I y I z F v,y1 F v,y2 F v,y F v,z1 F v,z2 F v,z Fv F t Fastener mm 2 mm mm mm 3 mm 4 mm 4 kn kn kn kn kn kn kn kn f1: ST f2: ST f3: ST f4: ST Σ = Definition of Annotations A = Shear area of individual fasteners ey +z y, z = Coordinate location of individual fasteners from c.g. of fastener group Q z = A y+e ty = Statical moment of area about z-axis at center of T application I y = A z 2 = Inertia about y-axis of individual fasteners y y V 4 1 z f I z = A y 2 4 f1 = Inertia about z-axis of individual fasteners V F v,y1 = V y /n = Shear force in y-axis of individual fastener due to shear force, V y e y ty z 4 z 1 F v,y2 = M (A z/σi y ) = Shear force in y-axis of individual fastener due to moment, M T fasteners c.g. F v,y = F v,y1 + F v,y2 = Total shear force in y-axis of individual fastener +y z3 z2 F v,z1 = V z /n = Shear force in z-axis of individual fastener due to shear force, V z F v,z2 = M (A y/σi z ) = Shear force in z-axisof individual fastener due to moment, M f 3 f 2 F v,z = F v,z1 + F v,z2 = Total shear force in z-axis of individual fastener z e y 3 y 2 F v = (F v,z1 2 + F v,z2 2 ) 0.5 = Total shear force in the individual fasteners F t = T(Q z /ΣQ z ) = Tension force in the individual fasteners

110 UNITISED CURTAIN WALL 110 of Check Most Critical Screw The most critical screw among the group is fastener f1. AAMA TIR-A9-1991: Metal Curtain Wall Fasteners Clause Action Notes Applied Forces Applied Tension, F t = 0.00 KN. Calculated tension force in the most critical screw Applied Shear, F v = 0.63 KN. Calculated shear force in the most critical screw Connected Parts 1st connected part, F u1 = 205 Mpa. Thickness, t p1 = 2nd connected part, F u2 = Thickness, t p2 = Number of shear plane(s), m = Screw Parameters Screw size, d b = 3.0 mm. L e1 = 16.5 mm. Dist. from centre of screw to the nearest free edge 205 Mpa. F y2 = 170 Mpa. Yield strength of connected material 3.0 mm. L e2 = 5.0 mm. Dist. from centre of screw to the nearest free edge LE = 20.7 mm. Length of fastener engagement 1 plane(s) #10 (ST4.8) Pan head screw A4-70 Stainless steel (BS EN ISO ) F y = 450 Mpa. Minimum tensile yield strength (BS EN ISO ) F u = 700 Mpa. Minimum ultimate tensile strength 9. Table 25 D = 4.83 mm. Nominal thread diameter 9. Table 25 K = 3.53 mm. Basic minor diameter 9. Table 25 Thread pitch, P = 1.59 mm. Giving a value of N = 16 threads per inch 9. Table 25 DSMIN = 4.62 mm. Minimum major diameter of external thread 9. Table 25 ENMAX = 4.24 mm. Maximum pitch diameter of internal thread 9. Table 25 A(S) = (K)² = 9.79 mm 2 Tensile stress area for spaced threads 9. Table 25 A(R) = (K)² = 9.79 mm 2 Thread root area for spaced threads TSA(I) = π[dsmin][(12.7/n) (dsmin-enmax)] = mm 2 Thread stripping area per thread, internal 6. Check Tension P t = 0.75Fy[A(S)] = 3.30 KN. Yielding failure allowable tension P t = 0.40Fu[A(S)] = 2.74 KN. Fracture failure governs allowable tension F t /P t = 0.00 <1.00 O.K! 6. Check Shear P v = 0.75Fy[A(R)]/1.732 = 1.91 KN. Yielding failure allowable shear P v = 0.40Fu[A(R)]/1.732 = 1.58 KN. Fracture failure governs allowable shear F s /(m P v ) = 0.40 <1.00 O.K! Interaction, [F t /P t ] 2 + [F s /(m P v )] 2 = Check Bearing Fb1 = (2/1.95)*Fu1/1.65 = Mpa. Allowable bearing stress on 1st connected part t p1 '= 3.00 mm. Effective bearing thickness P b1 = F b D t p1 ' = 1.8 KN. Allowable bearing capacity P b1 /(m F s ) = 0.34 <1.00 O.K! Fb2 = (2/1.95)*Fu2/1.65 = Mpa. Allowable bearing stress on 2nd connected part P b2 = F b D t p2 = 1.6 KN. Allowable bearing capacity P b2 /F s = 0.39 <1.00 O.K! 6063-T6 Aluminium Standard round hole 6063-T6 Aluminium Standard round hole 9. Check Pullout Fv = 0.4*Fsu = 52 Mpa. Fsu = Shear ultimate strength, see ADM Table 3.3-1M P p = [LE/25.4-1/N] F v TSA(I) N = 6.9 KN. Pullout strength on thread of 2nd connected part F t /P p = 0.00 <1.00 O.K!

111 UNITISED CURTAIN WALL 111 of Transom - 2 (System 2) Connection Check Connection Detail V z 15.3 f1 20x20x2.5t (15mm long) Aluminium Chair Additional screw V y f3 f f2 +Z +Y Analysis of Screw Group Analysis of Group of Fasteners Fig Transom-2 End Connection Reference Action Notes Applied Forces Horizontal shear force, V y = 3.60 kn Under most critical wind load Eccentricity of V y, e z = 0.00 mm Eccentric application of wind load Vertical shear force, V z = 1.30 kn Under dead load Eccentricity of V z, e y = mm Eccentric location of setting block Tension force, T = 0.00 kn Parallel to the axis of the fasteners Eccentricity of T, e ty = 0.00 mm Eccentric Application of the tension force Total applied moment, M = V y e z + V z e y = knm Sum of the torsion and all effects of eccentricity Number of fasteners, n = 4 nos. Fasteners in the group Fastener Group Calculated: Shear Force Tension Size A y z Q z I y I z F v,y1 F v,y2 F v,y F v,z1 F v,z2 F v,z Fv F t Fastener mm 2 mm mm mm 3 mm 4 mm 4 kn kn kn kn kn kn kn kn f1: ST f2: ST f3: ST f4: ST V y z e V z Definition of Annotations e y 73 Σ = f 4 y 1 z 4 z 3 f 3 T y 4 y 3 e ty +z fasteners c.g. y 2 f 1 f 2 z 1 z 2 +y A = Shear area of individual fasteners y, z = Coordinate location of individual fasteners from c.g. of fastener group Q z = A y+e ty = Statical moment of area about z-axis at center of T application I y = A z 2 = Inertia about y-axis of individual fasteners I z = A y 2 = Inertia about z-axis of individual fasteners F v,y1 = V y /n = Shear force in y-axis of individual fastener due to shear force, V y F v,y2 = M (A z/σi y ) = Shear force in y-axis of individual fastener due to moment, M F v,y = F v,y1 + F v,y2 = Total shear force in y-axis of individual fastener F v,z1 = V z /n = Shear force in z-axis of individual fastener due to shear force, V z F v,z2 = M (A y/σi z ) = Shear force in z-axisof individual fastener due to moment, M F v,z = F v,z1 + F v,z2 = Total shear force in z-axis of individual fastener F v = (F v,z1 2 + F v,z2 2 ) 0.5 = Total shear force in the individual fasteners F t = T(Q z /ΣQ z ) = Tension force in the individual fasteners

112 UNITISED CURTAIN WALL 112 of Check Most Critical Screw The most critical screw among the group is fastener f2. AAMA TIR-A9-1991: Metal Curtain Wall Fasteners Clause Action Notes Applied Forces Applied Tension, F t = 0.00 KN. Calculated tension force in the most critical screw Applied Shear, F v = 1.37 KN. Calculated shear force in the most critical screw Connected Parts 1st connected part, F u1 = 205 Mpa. Thickness, t p1 = 2nd connected part, F u2 = Thickness, t p2 = Number of shear plane(s), m = Screw Parameters Screw size, d b = 3.0 mm. L e1 = 16.5 mm. Dist. from centre of screw to the nearest free edge 205 Mpa. F y2 = 170 Mpa. Yield strength of connected material 3.0 mm. L e2 = 5.0 mm. Dist. from centre of screw to the nearest free edge LE = 20.7 mm. Length of fastener engagement 1 plane(s) #10 (ST4.8) Pan head screw A4-70 Stainless steel (BS EN ISO ) F y = 450 Mpa. Minimum tensile yield strength (BS EN ISO ) F u = 700 Mpa. Minimum ultimate tensile strength 9. Table 25 D = 4.83 mm. Nominal thread diameter 9. Table 25 K = 3.53 mm. Basic minor diameter 9. Table 25 Thread pitch, P = 1.59 mm. Giving a value of N = 16 threads per inch 9. Table 25 DSMIN = 4.62 mm. Minimum major diameter of external thread 9. Table 25 ENMAX = 4.24 mm. Maximum pitch diameter of internal thread 9. Table 25 A(S) = (K)² = 9.79 mm 2 Tensile stress area for spaced threads 9. Table 25 A(R) = (K)² = 9.79 mm 2 Thread root area for spaced threads TSA(I) = π[dsmin][(12.7/n) (dsmin-enmax)] = mm 2 Thread stripping area per thread, internal 6. Check Tension P t = 0.75Fy[A(S)] = 3.30 KN. Yielding failure allowable tension P t = 0.40Fu[A(S)] = 2.74 KN. Fracture failure governs allowable tension F t /P t = 0.00 <1.00 O.K! 6. Check Shear P v = 0.75Fy[A(R)]/1.732 = 1.91 KN. Yielding failure allowable shear P v = 0.40Fu[A(R)]/1.732 = 1.58 KN. Fracture failure governs allowable shear F s /(m P v ) = 0.87 <1.00 O.K! Interaction, [F t /P t ] 2 + [F s /(m P v )] 2 = Check Bearing Fb1 = (2/1.95)*Fu1/1.65 = Mpa. Allowable bearing stress on 1st connected part t p1 '= 3.00 mm. Effective bearing thickness P b1 = F b D t p1 ' = 1.8 KN. Allowable bearing capacity P b1 /(m F s ) = 0.74 <1.00 O.K! Fb2 = (2/1.95)*Fu2/1.65 = Mpa. Allowable bearing stress on 2nd connected part P b2 = F b D t p2 = 1.6 KN. Allowable bearing capacity P b2 /F s = 0.85 <1.00 O.K! 6063-T6 Aluminium Standard round hole 6063-T6 Aluminium Standard round hole 9. Check Pullout Fv = 0.4*Fsu = 52 Mpa. Fsu = Shear ultimate strength, see ADM Table 3.3-1M P p = [LE/25.4-1/N] F v TSA(I) N = 6.9 KN. Pullout strength on thread of 2nd connected part F t /P p = 0.00 <1.00 O.K!

113 UNITISED CURTAIN WALL 113 of Sword Connection Check The sword is freely bearing on a stopper bolt underneath Connection Detail Check Stopper Bolt The stopper bolt carries only the weight of the sword. Therefore it is designed for shear due to the weight of the sword with a factor of safety of 2. Fv = 2(0.09 KN) 0.2KN Weight of sword = 0.09KN see 5.1 Fig Sword Connection AAMA TIR-A9-1991: Metal Curtain Wall Fasteners Clause Action Notes Applied Forces Applied Tension, F t = 0.00 KN. Calculated tension force in the most critical screw Applied Shear, F v = 0.20 KN. Calculated shear force in the most critical screw Connected Parts 1st connected part, F u1 = 205 Mpa. Thickness, t p1 = Screw Parameters Screw size, d b = 3.0 mm. L e1 = 50.0 mm. Dist. from centre of screw to the nearest free edge (BS EN ISO ) F y = 450 Mpa. Minimum tensile yield strength (BS EN ISO ) F u = 700 Mpa. Minimum ultimate tensile strength 9. Table 4 D = 8 mm. Nominal thread diameter 9. Table 4 Thread pitch, P = 1.00 mm. Giving a value of N = 25.4 threads per 25mm 9. Table 4 DSMIN = 7.80 mm. Minimum major diameter of external thread 9. Table 4 ENMAX = 7.10 mm. Maximum pitch diameter of internal thread 6. Eq. 1 A(S) = (π/4)[d-(24.75/n)] 2 = mm 2 Tensile stress area 6. Eq. 2 A(R) = (π/4)[d-(31.16/n)] 2 = mm 2 Thread root area TSA(I) = π[dsmin][(12.7/n) (dsmin-enmax)] = mm 2 Thread stripping area per thread, internal 6. Check Shear P v = 0.75Fy[A(R)]/1.732 = 7.02 KN. Yielding failure allowable shear P v = 0.40Fu[A(R)]/1.732 = 5.82 KN. Fracture failure governs allowable shear F s /(m P v ) = 0.03 <1.00 O.K! Interaction, [F t /P t ] 2 + [F s /(m P v )] 2 = T6 Aluminium Standard round hole M8 Hexagonal head screw A4-70 Stainless steel 7. Check Bearing Fb1 = (2/1.95)*Fu1/1.65 = Mpa. Allowable bearing stress on 1st connected part t p1 '= 3.00 mm. Effective bearing thickness P b1 = F b D t p1 ' = 3.1 KN. Allowable bearing capacity P b1 /(m F s ) = 0.07 <1.00 O.K!

114 UNITISED CURTAIN WALL 114 of Back Pan Check Back Pan Details Fig Back Pan Details Check Back pan for Full Wind Load (a) Shell Element Model with Boundary conditions Hinge Wind suction Fixed rotation Fig Back Pan Applied Load & Boundary Conditions

115 UNITISED CURTAIN WALL 115 of 117 (b) Calculations Summary ADM2005: Part I-A Specification for Aluminum Structures - Allowable Stress Design Reference Action Notes Metal Sheet Parameters Width, W = 1887 mm Net dimension considering width of mullions Height, h = mm Net height considering depth of transoms Wind load, Q = -4.5 KN/m 2 Design wind load in suction Alloy, F cy = 115 Mpa Compressive yield strength of H14 alloy sheet ANSYS FEM Anlysis Results Notes Check Stress VM =0.707[(S min -S min ) 2 +S 2 max +S 2 min ) 0.5 = Mpa Von mises stress results (from ANSYS FEM anlaysis) Allowable stress, F c = 1.3F cy /n y = Mpa For rectangular solid plates Stress ratio, VM/F c = 0.97 <1.00 O.K! Check Deflection δ max = 7.90 mm Maximum deflection (from ANSYS FEM analysis) Allowable deflection, δ allow = mm 75% of 30mm clearance Strain ratio, δ z /δ allow = 0.35 <1.00 O.K! (i.) Von Mises Stress Diagram (ii.) Deflection Diagram Fig Back Pan Analysis Results