Analysis and Design of Steel Roof Trusses to EC3.....Ubani Obinna U (2017) PRACTICAL ANALYSIS AND DESIGN OF STEEL ROOF TRUSSES TO EUROCODE 3: A SAMPLE DESIGN Ubani Obinna Uzodimma Department of Civil Engineering, Nnamdi Azikiwe University PMB 5025, Awka, Anambra State, Nigeria E-mail: [email protected] Problem statement The skeletal structure of a roof system (18.0m long and 7.2m wide) is as shown in Figure 1 below. The truss is made up of Howe Truss configuration spaced at 3m intervals. It is desired to specify the appropriate angle sections that will safely carry the anticipated loading using Eurocode design code (Specified steel grade S 275). Figure 1: 3D skeletal representation of the roof to be designed Figure 2: Idealised 2D model of the roof truss system for manual analysis Downloaded from www.structville.blogspot.com (C) Ranks Michael Enterprises (2017) Page 1 blogspot..08 KN/m2 × 1..536 KN/m2 × 1. Ubani Obinna U (2017) 1..0m Nodal spacing of the trusses = 1.2m × 3m = 2.2 x 3) = 15 kg/m2 = 0.structville.0 LOAD ANALYSIS Span of roof truss = 7.9 = 1..9 EN 1991-1-1:2001) Imposed load on roof (Qk) = 0. the resultant pressure coefficient on the windward and leeward slopes with positive internal pressure (cpe) is taken as −0.2m × 3m = 1.35 kN/m2 Vertical component pev = qe cos θ = 1.2 KN/m2 Total deal load (Gk) = 0.2m Spacing of the truss = 3.1 KN/m2 Weight of purlin (assume CH 150 x 75 x 18 kg/m) = (18 x 3m)/(1.7 KN Wind Load Wind velocity pressure (dynamic) is assumed as = qp(z) = 1.5 × 0.9296 KN Variable (Imposed) Load Category of roof = Category H – Roof not accessible except for normal maintenance and repairs (Table 6.5 kN/m2 When the wind is blowing from right to left.019 KN/m2 Weight of ceiling (adopt 10mm insulation fibre board) = 0. Analysis and Design of Steel Roof Trusses to EC3.2m Permanent (dead) Loads Self weight of long span aluminium roofing sheet (0.2m × 3m = 3.536 KN/m2 Therefore the nodal permanent load (GK) = 0.869 = 1.75 KN/m2 Therefore the nodal variable load (QK) = 0.08 kN/m2 acting upwards ↑ Therefore the nodal wind load (WK) = 1.75 KN/m2 × 1.077 KN/m2 Weight of services = 0. qe = qpcpe = −1.147 KN/m2 Self weight of trusses (assume) = 0.35 × cos 36.888 KN Downloaded from www.55mm gauge thickness) = 0.com (C) Ranks Michael Enterprises (2017) Page 2 .9 Therefore the external wind pressure normal to the roof is. 93 × 4. F1-2 = si = −8.0 is now placed on the nodes of the trusses.com (C) Ranks Michael Enterprises (2017) Page 3 .965 × 7. Downloaded from www. then there are no horizontal reactive forces ANALYSIS OF INTERNAL FORCES In all cases note that Fi-j = Fj-i JOINT 1 .2/2.79 KN Since there are no horizontal loads.8) – (1. Analysis and Design of Steel Roof Trusses to EC3. R1 = .956 + F1-2 (sin 𝜃) = 0 − .1 Analysis for dead load The nodal dead load from section 1..93 × 2..2R1 – (0.93 × 3. = 5. Figure 3: Truss carrying the nodal dead loads (GK) Support Reactions Let the summation of moment about joint 12 be zero. Ubani Obinna U (2017) 2.. The loading is shown in Figure 3 below.2) – (1.structville. Let ∑ 𝑭 = 𝟎 5. It should be realised that at the first and last nodes.79 KN Let ∑ 𝐹 = We can verify that R12 = R1 = 5.869 .blogspot.6) – (1.0568 KN (COMPRESSION) .93 × 6) – (1. 𝜃 = tan− = 36.0 STRUCTURAL ANALYSIS 2..93 × 1.2) = 0 . ∑ 𝑀 = 7.4) – (1.79 – 0. the loads are halved because the inter-nodal load distance will be 1. Ubani Obinna U (2017) Let ∑ 𝑭 = 𝟎 F1-2 (cos 𝜃) + F1-3 = 0 F1-3 = −(−8.4(cos 𝜃) + F2 .608 KN (COMPRESSION) JOINT 5 Let ∑ 𝑭 = 𝟎 F5 . F2-4 = − 6.0568(cos .4(cos .448 KN (COMPRESSION) F2-5 = −1. ) – 0 − F2 .com (C) Ranks Michael Enterprises (2017) Page 4 . = 6.. ) + F5 – 4 = 0 F5 – 4 = 0.93 + F2 .93 + F2 .F2 . Let ∑ 𝑭 = 𝟎 −1.904 -----------------.4(sin .5(sin 𝜃) .8 F2 – 5 = −6.5(cos .(2) Solving equations (1) and (2) simultaneously..F2 .445 KN (TENSION) JOINT 3 Let ∑ 𝑭 = 𝟎 F3 – 2 = 0 (NO FORCE) Let ∑ 𝑭 = 𝟎 F3 – 5 − F3 – 1 = 0 F3 – 1 = F3 – 5 = 6..0568 × cos .(1) Let ∑ 𝑭 = 𝟎 F2 .structville.2(sin ϕ) + F5 – 4 = 0 -1.5(cos 𝜃) .4(sin 𝜃) – F2 – 3 − F2 .blogspot. )) = 0 0. )) = 0 0. Analysis and Design of Steel Roof Trusses to EC3.8 F2 – 4 + 0.. ) – (−8.0568(sin . ) – (−8.445 KN (TENSION) JOINT 2 . 𝜙 = tan− = 36.9646 KN (TENSION) Downloaded from www.4455-----------------.869 = 𝜃 . ) + F2 .608 (sin .6 F2 – 4 − 0.5(sin .1(sin 𝜃) = 0 −1.6 F2 – 5 = −2.1(cos 𝜃) = 0 F2 . ) – (−6.8398(sin .93 – (−6.8398(sin .1586 KN (TENSION) JOINT 4 .309° .6(cos 𝜃) + F6 . )=0 − .5547 F4 – 7 + 0.319 KN (COMPRESSION) F4 .8398 KN (COMPRESSION) s . Let ∑ 𝑭 = 𝟎 −1.7 (sin .8 (cos ..6 = −4.9646 – F4 .structville.448(cos . F4 . ) + F6 .832 F4 – 7 + 0. )+ F4 – 6 (sin . )) – 0.com (C) Ranks Michael Enterprises (2017) Page 5 .93 – (−4. F6 .448(sin .608 (cos . )) = 0 −0.7(cos .8398 cos .6(cos 𝜃) – F4 .8398 KN (COMPRESSION) JOINT 6 Let ∑ 𝑭 = 𝟎 − F4 .8 (cos 𝜃) = 0 − (−4. Ubani Obinna U (2017) Let ∑ 𝑭 = 𝟎 − F5 – 3 − F5 – 2 (cos ϕ) + F5 – 7 = 0 − . )) = 0 0.8777 KN (TENSION) Downloaded from www..2(cos 𝜃) = 0 F4 .7 = − 2.93 – F4 – 2 (sin 𝜃) – F4 – 5 – F4 – 7 (sin 𝛼) + F4 – 6(sin 𝜃) = 0 −1. ) + F4 .9742 -----------------. − (−1. )) – F6 – 7 − (−4.8 F4 – 6 = − 5.(1) Let ∑ 𝑭 = 𝟎 F4 – 7 (cos 𝛼) + F4 .8 = c = − 4.blogspot. )) + F5 – 7 = 0 F5 – 7 = 5.6(cos .6 F4 – 6 = −0. Analysis and Design of Steel Roof Trusses to EC3. )) = 0 F6 – 7 = 3. Let ∑ 𝑭 = 𝟎 −1.(2) Solving equations (1) and (2) simultaneously.93 – F6 – 4 (sin 𝜃) – F6 – 7 − F6 – 8 (sin 𝜃) = 0 −1.. 𝛼 = tan− = 56.1584 -----------------.. 241 COMPRESSIVE 2-4 − 8.1.445 TENSILE 5-7 5.445 TENSILE 3-5 6.2 Analysis for imposed load Figure 4: Truss carrying the nodal imposed loads (QK) Following the steps in section 2. Table 1..8398 COMPRESSIVE VERTICALS 2-3 0 NO FORCE 4-5 0. The summary of the result for analysis of dead load is given in Table 1.0 below.1586 TENSILE TOP CHORD 1-2 −8.998 COMPRESSIVE Downloaded from www..198 TENSILE TOP CHORD 1-2 −11.877 TENSILE DIAGONALS 2-5 −1.448 COMPRESSIVE 4-6 −4.0568 COMPRESSIVE 2-4 − 6.319 COMPRESSIVE 2.structville. what is happening at the left hand side is what is happening at the right hand side. Analysis and Design of Steel Roof Trusses to EC3. the following result as shown in Table 2.992 TENSILE 3-5 8. Table 2..0 can be obtained..blogspot.992 TENSILE 5-7 7. Ubani Obinna U (2017) By symmetry of load and structure.0: Summary of analysis results for imposed load (QK) MEMBER FORCE (KN) NATURE BOTTOM CHORD 1-3 8.608 COMPRESSIVE 4-7 − 2.9646 TENSILE 6-7 3.com (C) Ranks Michael Enterprises (2017) Page 6 .0: Summary of analysis results for dead load (GK) MEMBER FORCE (KN) NATURE BOTTOM CHORD 1-3 6. 228 TENSILE 4-7 4.1 also...0: Summary of analysis results for imposed load (QK) MEMBER FORCE (KN) NATURE BOTTOM CHORD 1-3 -12.242 COMPRESSIVE 4-7 − 3.3 Analysis for wind load Figure 5: Truss carrying the nodal wind loads (WK) Following the steps in section 2.957 TENSILE 4-6 9.1.948 COMPRESSIVE 5-7 -10. Analysis and Design of Steel Roof Trusses to EC3.187 TENSILE 2-4 12.717 TENSILE VERTICALS 2-3 0 NO FORCE 4-5 .structville.763 COMPRESSIVE DIAGONALS 2-5 3.346 TENSILE 6-7 5. the following result for the wind load as shown in Table 3.662 TENSILE Downloaded from www.0 can be obtained.938 COMPRESSIVE 6-7 -7. Table 3.948 COMPRESSIVE 3-5 -12..365 COMPRESSIVE TOP CHORD 1-2 16. Ubani Obinna U (2017) 4-6 −6.391 TENSILE DIAGONALS 2-5 −2.748 COMPRESSIVE VERTICALS 2-3 0 NO FORCE 4-5 1.238 COMPRESSIVE 2..blogspot.com (C) Ranks Michael Enterprises (2017) Page 7 . structville.5 × 0.896 KN (COMPRESSIVE) Therefore.445) + 1. Analysis and Design of Steel Roof Trusses to EC3.69 cm2 Radius of gyration (axis y-y) = 1.. NEd = 1.blogspot.0(6. Ubani Obinna U (2017) 3. refer to Table A1.com (C) Ranks Michael Enterprises (2017) Page 8 .Rd is the lesser of and 𝛾𝑀 𝛾𝑀 Downloaded from www. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1.992) = 22.35Gk + 1.6 = 0.5Qk + 0.5(12.445) + 1. 𝐴 𝑓 Nt. 2002b).5(8.189 KN (TENSILE) LOAD CASE 2: DEAD LOAD + IMPOSED LOAD + WIND LOAD acting simultaneously We use a partial factor for the accompanying variable actions of wind loads equal to γWkψ0 = 1.445) – 1.584 KN (TENSILE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1..894) = 10.35GK + 1.896 KN Length of longest bottom chord member = 1200mm Consider EQUAL ANGLES UA 50 X 50 X 6 Gross Area = 5.35(6. Fy (Yeild strength) = 275 N/mm2 Fu (ultimate tensile strength = 430 N/mm2) 3..5 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.5.1 of BS EN 1990: 2002(E) (Eurocode.894) = −12.9(12.0 (favourable). Therefore the ultimate design force in the member is..5(8.9 (for the value of ψ0.0 STRUCTURAL DESIGN TO EC3 All structural steel employed has the following properties.35(6.5Wk.88cm2 𝐴 𝐹𝑦 . NEd = 1. Partial factor for leading variable actions (WK) = γWk = 1. all bottom chord members should be able to resist an axial tensile load of 22.9Wk.72 cm2 Equivalent tension area for welded connection = 4.189KN and a possible reversal of stresses with a compressive load of 12.5QK NEd = 1.1 Design of the bottom chord (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only Fu = γGjGk + γQkQk ULTIMATE DESIGN FORCE (NEd) = 1.992) − 0. Fu = γGjGk + γQkQk + γWkψ0Wk = 1. .9244 = 86.5 [ + 𝛼(𝜆 − .1 EC3) Section classification 𝜀=√𝐹 =√ = 0. Buckling resistance of member (clause 5.8 (11.0824 < 1 Therefore section is ok for uniform compression. Part 1-1. × = 115. Analysis and Design of Steel Roof Trusses to EC3. = 156475 N = 156. ) + 𝜆 ] Downloaded from www.475 KN 𝑁𝐸 .blogspot. . × × − Nt.9216 × . 𝑁 . 𝑁𝐶. Fy = 275 N/mm2 . Since t < 16mm.9244 𝑦 h/t = 50/6 = 8. for class 3 classification.216 < 1.Rd = . × = 102.33.33 < 10.. critical buckling length is the same for all axis.3) OK (h + b)/2t = 8.17 KN . Buckling curve b is appropriate for all angle sections according to Table 6. × × − Also check. = = 0. Resistance of the member to uniform compression 𝐴𝐹𝑦 . 𝑁.2 (sheet 3) of Eurocode 3. Design yield strength Fy = 275 N/mm2 (Table 3.0 (Section is ok for tension resistance) .Rd = 𝛾 = .2 of Eurocode 3 Φ = 0.8 > h/t (8..structville. × . the section satisfies both of the conditions.92) OK Thus. h/t ≤ 15ε and (h + b)/2t ≤ 11.5 ENV 1993-1-1:1992) Since the member is pinned at both ends. × × NC.801 In the planar axis 𝜆= = 0.92 = 13. Ubani Obinna U (2017) Fu = 430 N/mm2.9𝜀 = 93.5ε.5 × 0. = = 0.9 × 0. 15ε = 15 × 0. Referring to Table 5. In our case.com (C) Ranks Michael Enterprises (2017) Page 9 . Compression and buckling resistance Thickness of section t = 6 mm. Lcr = 1200mm 𝐿 𝑟 Slenderness ratio 𝜆 = 𝑟𝑖 𝜆 𝜆 = 93.3 KN .. 35GK + 1.Rd = = = 101286. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1.1 of BS EN 1990: 2002(E) (Eurocode. .5Wk. the section is ok to resist all axial loads on it.5QK NEd = 1. 3.0 (favourable). Therefore.6473 < 1 .5 × 0.5.35(−8. × × Therefore Nb. NEd = 1. × .169 KN (COMPRESSIVE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.blogspot.241) + 0. Therefore the ultimate design force in the member is. Fu = γGjGk + γQkQk + γWkψ0Wk = 1. = = 0.0473 𝒳= Φ+ √Φ − 𝜆 𝒳 = = 0..35 −8. .structville.5Qk + 0.5(−11.5 −11.187) = -13..9(16. − .0(−8. Ubani Obinna U (2017) Φ = 0.6 = 0.127 < 1 Therefore section is ok for buckling 𝑁 .286 KN 𝛾 𝑁𝐸 .738 KN (COMPRESSIVE) LOAD CASE 2: DEAD LOAD + IMPOSED LOAD + WIND LOAD acting simultaneously We use a partial factor for the accompanying variable actions of wind loads equal to γWkψ0 = 1.738 KN Length of longest TOP chord member = 1500mm Let us also consider EQUAL ANGLES UA50 X 50 X 6 Downloaded from www. + .2675 N = 101.2 Design of the top chord (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only ULTIMATE DESIGN FORCE (NEd) = 1.5(16. Analysis and Design of Steel Roof Trusses to EC3.0568) + 1.35Gk + 1.5 [ + . ] = 1. 𝒳𝑦 𝐴𝐹𝑦 .0568) + 1. − .187) = +16. all the top chord members should be able to resist an axial tensile load of 16. +√ .2237 KN (TENSILE) Therefore.2237KN and a compressive load of 27.9 (for the value of ψ0. Partial factor for leading variable actions (WK) = γWk = 1..0568) + 1.9Wk.com (C) Ranks Michael Enterprises (2017) Page 10 .. 2002b). NEd = 1. refer to Table A1.241) = -27. 92 = 13.2 (sheet 3) of Eurocode 3. for class 3 classification.blogspot..structville. h/t ≤ 15ε and (h + b)/2t ≤ 11.33 < 10. × × − Nt. = = 0. 𝑁𝐶.17 KN .3 KN .5 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.9244 = 86.com (C) Ranks Michael Enterprises (2017) Page 11 .1 EC3) Section classification 𝜀=√𝐹 =√ = 0.Rd = . × × NC. Part 1-1. = 0.3) OK (h + b)/2t = 8.5 ENV 1993-1-1:1992) Since the member is pinned at both ends. Since t < 16mm.. Ubani Obinna U (2017) Gross Area = 5.69 cm2 Radius of gyration (axis y-y) = 1.801 Downloaded from www. 𝐴 𝑓 Nt. Fy = 275 N/mm2 2 . Design yield strength Fy = 275 N/mm2 (Table 3.9𝜀 = 93. × × − Also check.0 (Section is ok for tension resistance) Compression and buckling resistance Thickness of section t = 6 mm.Rd is the lesser of 𝛾𝑀 and 𝛾𝑀 Fu = 430 N/mm . 𝑁..88cm2 𝐴 𝐹𝑦 .158 < 1. . Buckling resistance of member (clause 5.8 > h/t (8. × = 102. × = 115. Referring to Table 5.177 < 1 Therefore section is ok for uniform compression.Rd = 𝛾 = .9244 𝑦 h/t = 50/6 = 8. = .72 cm2 Equivalent tension area for welded connection = 4.5 × 0. critical buckling length is the same for all axis Lcr = 1500mm 𝐿 Slenderness ratio 𝜆 = 𝑟 𝜆𝑟 𝑖 𝜆 = 93.33. Analysis and Design of Steel Roof Trusses to EC3.475 KN 𝑁𝐸 .9 × 0.8 (11. Resistance of the member to uniform compression 𝐴𝐹𝑦 ..92) OK Thus. 15ε = 15 × 0. the section satisfies both of the conditions. × . In our case. 𝑁 .5ε. = 156475 N = 156. Buckling curve b is appropriate for all angle sections according to Table 6.9(7.Rd = = = 79035.5225 N = 79.5(5.35(3. = = 0. 2002b). 3.9 (for the value of ψ0.com (C) Ranks Michael Enterprises (2017) Page 12 . Downloaded from www.5051 < 1 .2 of Eurocode 3 Φ = 0. NEd = 1.9Wk. refer to Table A1. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1. × . the section is ok to resist all axial load on it.5 [ + 𝛼(𝜆 − .. × × Therefore Nb.5 × 0.. ] = 1. − . Therefore the ultimate design force in the member is.35(3.877) + 1.763) = 6. 𝒳𝑦 𝐴𝐹𝑦 . Partial factor for leading variable actions (WK) = γWk = 1.6 = 0.5Qk + 0.152 × .3 Design of the vertical members (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only Fu = γGjGk + γQkQk ULTIMATE DESIGN FORCE (NEd) = 1. Fu = γGjGk + γQkQk + γWkψ0Wk = 1. Ubani Obinna U (2017) In the both planar axis 𝜆= = 1.334 KN (TENSILE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1. + .5 [ + . . Therefore.877) + 1.0355 KN 𝛾 𝑁𝐸 .5QK NEd = 1.1 of BS EN 1990: 2002(E) (Eurocode..391) = 13.32 KN (TENSILE) LOAD CASE 2: DEAD LOAD + IMPOSED LOAD + WIND LOAD acting simultaneously We use a partial factor for the accompanying variable actions of wind loads equal to γWkψ0 = 1. +√ . Analysis and Design of Steel Roof Trusses to EC3.structville.391) − 0.0 (favourable).325 𝒳= Φ+ √Φ − 𝜆 𝒳 = = 0.5(5.blogspot. – .35Gk + 1.35GK + 1.5.351< 1 Therefore section is ok for buckling 𝑁 . .5Wk. ) + 𝜆 ] Φ = 0.. 2 (sheet 3) of Eurocode 3. 𝑁..5(7. Fy = 275 N/mm2 . = = 0.. 𝑁𝐶. Resistance of the member to uniform compression 𝐴𝐹𝑦 .8 (11. 𝑁𝐸 .95 cm2 𝐴 𝐹𝑦 .Rd is the lesser of 𝛾𝑀 and 𝛾𝑀 Fu = 430 N/mm2.. × = 34.892 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 1. Ubani Obinna U (2017) NEd = 1.com (C) Ranks Michael Enterprises (2017) Page 13 . .5ε..763) = −7. Buckling resistance of member (clause 5.9244 𝑦 h/t = 30/4 = 7.3906 < 1. 𝐴 𝑓 Nt.Rd = 𝛾 = = 62425 N = 62. Design yield strength Fy = 275 N/mm2 (Table 3. for class 3 classification.92) OK Thus. = .Rd = .877) – 1. × = 38.1244 < 1 Therefore section is ok for uniform compression.5 ENV 1993-1-1:1992) Downloaded from www. the section satisfies both of the conditions.7675 KN (COMPRESSIVE) Length of longest bottom chord member = 2700mm Consider EQUAL ANGLES 30 X 30 X 4 Gross Area = 2.blogspot. × × − Also check.5) OK (h + b)/2t = 7.27 cm2 Radius of gyration (axis y-y) = 0. × . Since t < 16mm. . × × − Nt. Analysis and Design of Steel Roof Trusses to EC3.8 > h/t (7.0(3.3904 KN 𝑁 .5 Referring to Table 5. × × NC.92 = 13.0 (Section is ok for tension resistance) Compression resistance Thickness of section t = 6 mm.structville. 15ε = 15 × 0. = 0.5 × 0.24 cm2 Equivalent tension area for welded connection = 1.1 KN . In our case.5 < 10.1 EC3) Section classification 𝜀=√𝐹 =√ = 0. h/t ≤ 15ε and (h + b)/2t ≤ 11. Part 1-1.425 KN . × × Therefore Nb. – . × .9372 𝒳= Φ+ √Φ − 𝜆 𝒳 = = 0. . ) + 𝜆 ] Φ = 0.13899 𝒳= Φ+ √Φ − 𝜆 𝒳 = = 0.9𝜀 = 93.Rd = = = 4669.48717 Buckling curve b is appropriate for all angle sections according to Table 6.801 In the both planar axis 𝜆= . = = 0.9 × 0.2 of Eurocode 3 Φ = 0. + .5 [ + . + . × × Therefore Nb. ) + 𝜆 ] Φ = 0. +√ .597 KN 𝛾 𝑁𝐸 .. ] = 7. − .9244 = 86. 𝒳𝑦 𝐴𝐹𝑦 . × .com (C) Ranks Michael Enterprises (2017) Page 14 .39 N = 4. .structville.6775 N = 13. . provide UA 50 X 50 X 6 for all vertical members Downloaded from www. Therefore. Try another section UA 50 X 50 X 6 (see data from previous designs) 𝜆= × .0869 < 1 .2 of Eurocode 3 Φ = 0.blogspot. = = 1.. ] = 5. +√ . Lcr = 2700mm 𝐿 𝑟 Slenderness ratio 𝜆 = 𝑟𝑖 𝜆 𝜆 = 93. critical buckling length is the same for all axis.0737 Buckling curve b is appropriate for all angle sections according to Table 6.663 > 1 Therefore section is NOT OK for buckling resistance 𝑁 .669 KN 𝛾 𝑁𝐸 . – .Rd = = = 13597.5 [ + 𝛼(𝜆 − . = 2.5712 < 1 Therefore section is ok for buckling 𝑁 . − . × .5 [ + .0748 < 1 . Ubani Obinna U (2017) Since the member is pinned at both ends. Analysis and Design of Steel Roof Trusses to EC3.5 [ + 𝛼(𝜆 − .. . = 3. 𝒳𝑦 𝐴𝐹𝑦 .. 238) + 0.35GK + 1. CHECK FOR DEFLECTION Employing UA 50 X 50 X 6 in all members of the roof truss. Analysis and Design of Steel Roof Trusses to EC3.5Qk + 0.blogspot. one should always watch out for possible reversal of stresses during load combinations.4 Design of the diagonal members (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only Fu = γGjGk + γQkQk ULTIMATE DESIGN FORCE (NEd) = 1.8mm. NEd = 1.662) = -3.35Gk + 1. the deflection due to unfactored imposed load calculated using STAADPRO V8i software is 1.6 = 0. refer to Table A1.9(4.5(− 3. NEd = 1. Downloaded from www.1 ENV 1993-1-1:1992) = L/250 = 7200/250 = 28.319) + 1.0(− 2. Therefore deflection is ok 4.7918 KN (COMPRESSIVE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.5.5(4.0 Conclusion The analysis and design for the section members have been successfully carried as shown in the calculations above. The design has shown that the provision of UA 50 X 50 X 6 for the entire roof truss will be adequate.35(− 2. Fu = γGjGk + γQkQk + γWkψ0Wk = 1.5 × 0. 2002b).319) + 1. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1.5QK NEd = 1. a little consideration will show that UA 50 X 50 X 6 will satisfy all the necessary limit state requirements..9 (for the value of ψ0.9Wk.5Wk.8mm 1.com (C) Ranks Michael Enterprises (2017) Page 15 .662) = 𝟒.0 (favourable). Whenever the wind load effect is greater than the live load effect. Therefore the ultimate design force in the member is.5(− 3.structville...238) = -7.1 of BS EN 1990: 2002(E) (Eurocode.319) + 1. Ubani Obinna U (2017) 3.35(− 2. 𝟒 KN (TENSILE) With a maximum length of 2163mm. Partial factor for leading variable actions (WK) = γWk = 1.028mm < 28..028mm Allowable deflection for roofs (Table 4.98765 KN (COMPRESSIVE) LOAD CASE 2: DEAD LOAD + IMPOSED LOAD + WIND LOAD acting simultaneously We use a partial factor for the accompanying variable actions of wind loads equal to γWkψ0 = 1.
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