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 75 KN/m2 Therefore the nodal variable load (QK) = 0.536 KN/m2 × 1..0m Nodal spacing of the trusses = 1.888 KN Downloaded from www.2m × 3m = 3.869 = 1.1 KN/m2 Weight of purlin (assume CH 150 x 75 x 18 kg/m) = (18 x 3m)/(1.2m × 3m = 2.5 × 0.2m Permanent (dead) Loads Self weight of long span aluminium roofing sheet (0.2 KN/m2 Total deal load (Gk) = 0.com (C) Ranks Michael Enterprises (2017) Page 2 .08 KN/m2 × 1. Ubani Obinna U (2017) 1.9 EN 1991-1-1:2001) Imposed load on roof (Qk) = 0.2m Spacing of the truss = 3.08 kN/m2 acting upwards ↑ Therefore the nodal wind load (WK) = 1. Analysis and Design of Steel Roof Trusses to EC3.7 KN Wind Load Wind velocity pressure (dynamic) is assumed as = qp(z) = 1.147 KN/m2 Self weight of trusses (assume) = 0.5 kN/m2 When the wind is blowing from right to left.019 KN/m2 Weight of ceiling (adopt 10mm insulation fibre board) = 0..75 KN/m2 × 1. qe = qpcpe = −1.blogspot.55mm gauge thickness) = 0.2 x 3) = 15 kg/m2 = 0.077 KN/m2 Weight of services = 0.35 × cos 36.536 KN/m2 Therefore the nodal permanent load (GK) = 0..0 LOAD ANALYSIS Span of roof truss = 7.9296 KN Variable (Imposed) Load Category of roof = Category H – Roof not accessible except for normal maintenance and repairs (Table 6..9 Therefore the external wind pressure normal to the roof is. the resultant pressure coefficient on the windward and leeward slopes with positive internal pressure (cpe) is taken as −0.structville.35 kN/m2 Vertical component pev = qe cos θ = 1.9 = 1.2m × 3m = 1. 0568 KN (COMPRESSION) . The loading is shown in Figure 3 below.0 STRUCTURAL ANALYSIS 2. Downloaded from www.1 Analysis for dead load The nodal dead load from section 1.965 × 7. the loads are halved because the inter-nodal load distance will be 1.0 is now placed on the nodes of the trusses.956 + F1-2 (sin ?) = 0 − . ? = tan− = 36..8) – (1..blogspot.2R1 – (0.structville. Let ∑ ? = ? 5.93 × 3. It should be realised that at the first and last nodes.2/2. Figure 3: Truss carrying the nodal dead loads (GK) Support Reactions Let the summation of moment about joint 12 be zero.2) = 0 .6) – (1.79 – 0. Analysis and Design of Steel Roof Trusses to EC3..869 . R1 = . = 5. ∑ ? = 7. then there are no horizontal reactive forces ANALYSIS OF INTERNAL FORCES In all cases note that Fi-j = Fj-i JOINT 1 .79 KN Since there are no horizontal loads.93 × 1.2) – (1.79 KN Let ∑ ? = We can verify that R12 = R1 = 5. Ubani Obinna U (2017) 2.com (C) Ranks Michael Enterprises (2017) Page 3 .93 × 4.93 × 6) – (1..4) – (1.93 × 2. F1-2 = si = −8. 4(cos .0568(sin .4(sin ?) – F2 – 3 − F2 .F2 .(1) Let ∑ ? = ? F2 .5(cos ?) .F2 . ? = tan− = 36.904 -----------------. Analysis and Design of Steel Roof Trusses to EC3. Ubani Obinna U (2017) Let ∑ ? = ? F1-2 (cos ?) + F1-3 = 0 F1-3 = −(−8.5(sin ?) . F2-4 = − 6.1(sin ?) = 0 −1.. Let ∑ ? = ? −1..1(cos ?) = 0 F2 . )) = 0 0.869 = ? .4(sin .6 F2 – 5 = −2.0568(cos .com (C) Ranks Michael Enterprises (2017) Page 4 . )) = 0 0.448 KN (COMPRESSION) F2-5 = −1.5(cos .8 F2 – 5 = −6.445 KN (TENSION) JOINT 3 Let ∑ ? = ? F3 – 2 = 0 (NO FORCE) Let ∑ ? = ? F3 – 5 − F3 – 1 = 0 F3 – 1 = F3 – 5 = 6. = 6.0568 × cos .9646 KN (TENSION) Downloaded from www.8 F2 – 4 + 0..4(cos ?) + F2 . ) + F2 .608 (sin . ) + F5 – 4 = 0 F5 – 4 = 0. ) – (−8..2(sin ϕ) + F5 – 4 = 0 -1.blogspot.(2) Solving equations (1) and (2) simultaneously.4455-----------------.608 KN (COMPRESSION) JOINT 5 Let ∑ ? = ? F5 .6 F2 – 4 − 0.structville. ) – 0 − F2 .93 + F2 .445 KN (TENSION) JOINT 2 .93 + F2 . ) – (−8.5(sin . 6 = −4. )) – F6 – 7 − (−4.blogspot. ) + F4 .5547 F4 – 7 + 0. )=0 − .8398 KN (COMPRESSION) s . F4 . )) + F5 – 7 = 0 F5 – 7 = 5. )) = 0 F6 – 7 = 3.structville. Analysis and Design of Steel Roof Trusses to EC3.1586 KN (TENSION) JOINT 4 .93 – (−4. Let ∑ ? = ? −1.. ? = tan− = 56.. ) – (−6.2(cos ?) = 0 F4 .8398(sin .9742 -----------------.448(sin . Ubani Obinna U (2017) Let ∑ ? = ? − F5 – 3 − F5 – 2 (cos ϕ) + F5 – 7 = 0 − . )+ F4 – 6 (sin .608 (cos .832 F4 – 7 + 0.7(cos . − (−1.6(cos ?) – F4 .93 – F6 – 4 (sin ?) – F6 – 7 − F6 – 8 (sin ?) = 0 −1.6(cos .8 F4 – 6 = − 5.93 – (−6.309° . )) – 0..8 (cos .8 = c = − 4..(2) Solving equations (1) and (2) simultaneously.448(cos . )) = 0 −0. Let ∑ ? = ? −1.8398 cos .8398 KN (COMPRESSION) JOINT 6 Let ∑ ? = ? − F4 . )) = 0 0.8398(sin .6(cos ?) + F6 .319 KN (COMPRESSION) F4 .93 – F4 – 2 (sin ?) – F4 – 5 – F4 – 7 (sin ?) + F4 – 6(sin ?) = 0 −1.(1) Let ∑ ? = ? F4 – 7 (cos ?) + F4 . F6 .1584 -----------------.7 (sin .6 F4 – 6 = −0.8 (cos ?) = 0 − (−4.7 = − 2.9646 – F4 .8777 KN (TENSION) Downloaded from www.com (C) Ranks Michael Enterprises (2017) Page 5 . ) + F6 . 198 TENSILE TOP CHORD 1-2 −11.608 COMPRESSIVE 4-7 − 2.structville. the following result as shown in Table 2.877 TENSILE DIAGONALS 2-5 −1. Table 2.998 COMPRESSIVE Downloaded from www. Table 1..0 below.319 COMPRESSIVE 2.1.8398 COMPRESSIVE VERTICALS 2-3 0 NO FORCE 4-5 0.448 COMPRESSIVE 4-6 −4.. 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 8.992 TENSILE 3-5 8.0: Summary of analysis results for dead load (GK) MEMBER FORCE (KN) NATURE BOTTOM CHORD 1-3 6.0 can be obtained. what is happening at the left hand side is what is happening at the right hand side. The summary of the result for analysis of dead load is given in Table 1..1586 TENSILE TOP CHORD 1-2 −8.241 COMPRESSIVE 2-4 − 8.2 Analysis for imposed load Figure 4: Truss carrying the nodal imposed loads (QK) Following the steps in section 2.992 TENSILE 5-7 7.blogspot.9646 TENSILE 6-7 3..com (C) Ranks Michael Enterprises (2017) Page 6 .445 TENSILE 3-5 6.0568 COMPRESSIVE 2-4 − 6. Analysis and Design of Steel Roof Trusses to EC3.445 TENSILE 5-7 5. blogspot.346 TENSILE 6-7 5.1 also.0: Summary of analysis results for imposed load (QK) MEMBER FORCE (KN) NATURE BOTTOM CHORD 1-3 -12.763 COMPRESSIVE DIAGONALS 2-5 3.structville.748 COMPRESSIVE VERTICALS 2-3 0 NO FORCE 4-5 1..0 can be obtained.242 COMPRESSIVE 4-7 − 3.938 COMPRESSIVE 6-7 -7.391 TENSILE DIAGONALS 2-5 −2..662 TENSILE Downloaded from www.3 Analysis for wind load Figure 5: Truss carrying the nodal wind loads (WK) Following the steps in section 2.com (C) Ranks Michael Enterprises (2017) Page 7 . Table 3.948 COMPRESSIVE 3-5 -12. Ubani Obinna U (2017) 4-6 −6.957 TENSILE 4-6 9.365 COMPRESSIVE TOP CHORD 1-2 16.1..238 COMPRESSIVE 2..717 TENSILE VERTICALS 2-3 0 NO FORCE 4-5 .187 TENSILE 2-4 12.948 COMPRESSIVE 5-7 -10. Analysis and Design of Steel Roof Trusses to EC3.228 TENSILE 4-7 4. the following result for the wind load as shown in Table 3. 992) − 0. Analysis and Design of Steel Roof Trusses to EC3.189 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 = 1.5 × 0. 2002b).9(12..5 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.5.896 KN Length of longest bottom chord member = 1200mm Consider EQUAL ANGLES UA 50 X 50 X 6 Gross Area = 5.35(6.5QK NEd = 1.0 (favourable).5Wk.69 cm2 Radius of gyration (axis y-y) = 1.structville.9 (for the value of ψ0.72 cm2 Equivalent tension area for welded connection = 4.35GK + 1.0 STRUCTURAL DESIGN TO EC3 All structural steel employed has the following properties.584 KN (TENSILE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1. Ubani Obinna U (2017) 3.6 = 0.. Fu = γGjGk + γQkQk + γWkψ0Wk = 1.189KN and a possible reversal of stresses with a compressive load of 12.1 Design of the bottom chord (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only Fu = γGjGk + γQkQk ULTIMATE DESIGN FORCE (NEd) = 1.894) = 10. Partial factor for leading variable actions (WK) = γWk = 1.. NEd = 1..894) = −12.0(6.88cm2 ? ?? .445) – 1.com (C) Ranks Michael Enterprises (2017) Page 8 .992) = 22.5(12. refer to Table A1.35(6.5(8.896 KN (COMPRESSIVE) Therefore.445) + 1. NEd = 1. Fy (Yeild strength) = 275 N/mm2 Fu (ultimate tensile strength = 430 N/mm2) 3.445) + 1.35Gk + 1. Therefore the ultimate design force in the member is.blogspot.5(8.Rd is the lesser of and ?? ?? Downloaded from www.9Wk. all bottom chord members should be able to resist an axial tensile load of 22. ? ? Nt.5Qk + 0. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1.1 of BS EN 1990: 2002(E) (Eurocode. Analysis and Design of Steel Roof Trusses to EC3. ) + ? ] Downloaded from www.0824 < 1 Therefore section is ok for uniform compression. Buckling resistance of member (clause 5.2 (sheet 3) of Eurocode 3. Design yield strength Fy = 275 N/mm2 (Table 3.92 = 13. = = 0. ? .475 KN ?? .. Since t < 16mm.3) OK (h + b)/2t = 8.9244 ? h/t = 50/6 = 8. h/t ≤ 15ε and (h + b)/2t ≤ 11.92) OK Thus.. × = 102.1 EC3) Section classification ?=√? =√ = 0. = = 0.5 × 0. Compression and buckling resistance Thickness of section t = 6 mm.8 (11.blogspot.3 KN .2 of Eurocode 3 Φ = 0. Resistance of the member to uniform compression ??? . .9216 × .. × × − Also check. In our case. Fy = 275 N/mm2 .9 × 0. = 156475 N = 156. × = 115. Referring to Table 5.Rd = . ?. Buckling curve b is appropriate for all angle sections according to Table 6.33.801 In the planar axis ?= = 0.8 > h/t (8.216 < 1.0 (Section is ok for tension resistance) . Lcr = 1200mm ? ? Slenderness ratio ? = ?? ? ? = 93.5 [ + ?(? − .com (C) Ranks Michael Enterprises (2017) Page 9 . × . critical buckling length is the same for all axis.structville.5ε.Rd = ? = .17 KN . Part 1-1. 15ε = 15 × 0.33 < 10. × × − Nt. ??.9? = 93.9244 = 86. × × NC. Ubani Obinna U (2017) Fu = 430 N/mm2.. for class 3 classification. the section satisfies both of the conditions.5 ENV 1993-1-1:1992) Since the member is pinned at both ends. 2002b). .187) = +16.241) + 0.5 × 0.5 −11.Rd = = = 101286.5(−11. +√ .2 Design of the top chord (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only ULTIMATE DESIGN FORCE (NEd) = 1.. all the top chord members should be able to resist an axial tensile load of 16.0473 ?= Φ+ √Φ − ? ? = = 0. Therefore. ?? ??? .0 (favourable).com (C) Ranks Michael Enterprises (2017) Page 10 .5Qk + 0.0(−8.structville...0568) + 1. Fu = γGjGk + γQkQk + γWkψ0Wk = 1. Analysis and Design of Steel Roof Trusses to EC3.9Wk. × .5QK NEd = 1. 3.35 −8. − .9 (for the value of ψ0.2237 KN (TENSILE) Therefore. × × Therefore Nb.35GK + 1. − .5.35Gk + 1. = = 0.5(16.6 = 0.5Wk.6473 < 1 .0568) + 1. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1..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 = 1.35(−8.241) = -27.169 KN (COMPRESSIVE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.738 KN Length of longest TOP chord member = 1500mm Let us also consider EQUAL ANGLES UA50 X 50 X 6 Downloaded from www. . Partial factor for leading variable actions (WK) = γWk = 1.127 < 1 Therefore section is ok for buckling ? . Ubani Obinna U (2017) Φ = 0.2237KN and a compressive load of 27.286 KN ? ?? .blogspot.187) = -13. refer to Table A1. NEd = 1.9(16. Therefore the ultimate design force in the member is. + .2675 N = 101.1 of BS EN 1990: 2002(E) (Eurocode. NEd = 1. ] = 1.5 [ + . the section is ok to resist all axial loads on it.0568) + 1. × × NC. Since t < 16mm..92 = 13.158 < 1. for class 3 classification.3 KN .475 KN ?? . 15ε = 15 × 0.8 (11.Rd = . In our case.structville. Design yield strength Fy = 275 N/mm2 (Table 3.9244 = 86.3) OK (h + b)/2t = 8.8 > h/t (8.69 cm2 Radius of gyration (axis y-y) = 1. = 0. ?.5 ENV 1993-1-1:1992) Since the member is pinned at both ends. × = 102. h/t ≤ 15ε and (h + b)/2t ≤ 11.blogspot.com (C) Ranks Michael Enterprises (2017) Page 11 .88cm2 ? ?? . . Analysis and Design of Steel Roof Trusses to EC3.5 × 0.2 (sheet 3) of Eurocode 3.1 EC3) Section classification ?=√? =√ = 0.. the section satisfies both of the conditions. ? ? Nt. Resistance of the member to uniform compression ??? .33 < 10. × × − Nt..9244 ? h/t = 50/6 = 8. = 156475 N = 156.5ε. = = 0.Rd = ? = .801 Downloaded from www.5 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.Rd is the lesser of ?? and ?? Fu = 430 N/mm .9? = 93. Ubani Obinna U (2017) Gross Area = 5. Part 1-1.177 < 1 Therefore section is ok for uniform compression. × .17 KN . ??. critical buckling length is the same for all axis Lcr = 1500mm ? Slenderness ratio ? = ? ?? ? ? = 93. = . Referring to Table 5.0 (Section is ok for tension resistance) Compression and buckling resistance Thickness of section t = 6 mm.92) OK Thus.72 cm2 Equivalent tension area for welded connection = 4.9 × 0. × = 115. Buckling resistance of member (clause 5.33.. ? . × × − Also check. Fy = 275 N/mm2 2 . − .391) − 0.325 ?= Φ+ √Φ − ? ? = = 0. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1..structville.877) + 1. × × Therefore Nb.5Wk. ) + ? ] Φ = 0. ] = 1.351< 1 Therefore section is ok for buckling ? .5QK NEd = 1.9(7.0355 KN ? ?? . NEd = 1.5(5.5.0 (favourable). refer to Table A1.334 KN (TENSILE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.35Gk + 1.2 of Eurocode 3 Φ = 0.763) = 6.9Wk..877) + 1.152 × . Analysis and Design of Steel Roof Trusses to EC3. + . Fu = γGjGk + γQkQk + γWkψ0Wk = 1.9 (for the value of ψ0.6 = 0.5 × 0.com (C) Ranks Michael Enterprises (2017) Page 12 .5051 < 1 .5Qk + 0. Downloaded from www.35(3.3 Design of the vertical members (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only Fu = γGjGk + γQkQk ULTIMATE DESIGN FORCE (NEd) = 1.. ?? ??? .5 [ + . = = 0.35(3.5225 N = 79.5 [ + ?(? − .5(5. Buckling curve b is appropriate for all angle sections according to Table 6. . Ubani Obinna U (2017) In the both planar axis ?= = 1.391) = 13. – .35GK + 1. × . 3. +√ . 2002b).. the section is ok to resist all axial load on it. Therefore. Partial factor for leading variable actions (WK) = γWk = 1.32 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 = 1.1 of BS EN 1990: 2002(E) (Eurocode. Therefore the ultimate design force in the member is.blogspot. .Rd = = = 79035. 27 cm2 Radius of gyration (axis y-y) = 0.5(7.3906 < 1.5 × 0.425 KN .Rd = . the section satisfies both of the conditions. × .2 (sheet 3) of Eurocode 3. × = 34. × × − Nt. ??.892 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 1.763) = −7. Part 1-1.Rd = ? = = 62425 N = 62.9244 ? h/t = 30/4 = 7. ? ? Nt. . ?? .8 (11.5 Referring to Table 5.7675 KN (COMPRESSIVE) Length of longest bottom chord member = 2700mm Consider EQUAL ANGLES 30 X 30 X 4 Gross Area = 2.5 < 10.. Resistance of the member to uniform compression ??? .. × = 38.5) OK (h + b)/2t = 7. Analysis and Design of Steel Roof Trusses to EC3.5ε.Rd is the lesser of ?? and ?? Fu = 430 N/mm2.0(3.1 EC3) Section classification ?=√? =√ = 0. ?. Design yield strength Fy = 275 N/mm2 (Table 3.5 ENV 1993-1-1:1992) Downloaded from www. × × NC.. In our case. Fy = 275 N/mm2 .877) – 1. .95 cm2 ? ?? .92 = 13. for class 3 classification. h/t ≤ 15ε and (h + b)/2t ≤ 11.1244 < 1 Therefore section is ok for uniform compression..8 > h/t (7.0 (Section is ok for tension resistance) Compression resistance Thickness of section t = 6 mm.3904 KN ? . 15ε = 15 × 0. = 0. = = 0.24 cm2 Equivalent tension area for welded connection = 1.1 KN . × × − Also check. Since t < 16mm.blogspot.structville.92) OK Thus.com (C) Ranks Michael Enterprises (2017) Page 13 . Buckling resistance of member (clause 5. Ubani Obinna U (2017) NEd = 1. = . 9244 = 86.0737 Buckling curve b is appropriate for all angle sections according to Table 6. × × Therefore Nb.9? = 93.801 In the both planar axis ?= .6775 N = 13. .5 [ + ?(? − .0748 < 1 .663 > 1 Therefore section is NOT OK for buckling resistance ? . = 3.2 of Eurocode 3 Φ = 0. = = 1. provide UA 50 X 50 X 6 for all vertical members Downloaded from www. – . ) + ? ] Φ = 0. = = 0. × .. +√ . ] = 7. − . ?? ??? .5 [ + .5 [ + .39 N = 4. × . Therefore. critical buckling length is the same for all axis.Rd = = = 4669. Analysis and Design of Steel Roof Trusses to EC3. × × Therefore Nb.com (C) Ranks Michael Enterprises (2017) Page 14 .5 [ + ?(? − .structville.. ) + ? ] Φ = 0. ] = 5. Try another section UA 50 X 50 X 6 (see data from previous designs) ?= × . = 2.blogspot. + .9 × 0. ?? ??? . − . .48717 Buckling curve b is appropriate for all angle sections according to Table 6.0869 < 1 .2 of Eurocode 3 Φ = 0.13899 ?= Φ+ √Φ − ? ? = = 0. + . . Ubani Obinna U (2017) Since the member is pinned at both ends..597 KN ? ?? . +√ .9372 ?= Φ+ √Φ − ? ? = = 0.5712 < 1 Therefore section is ok for buckling ? .. .669 KN ? ?? . – . Lcr = 2700mm ? ? Slenderness ratio ? = ?? ? ? = 93.Rd = = = 13597. × . 4 Design of the diagonal members (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only Fu = γGjGk + γQkQk ULTIMATE DESIGN FORCE (NEd) = 1.662) = ?. Therefore deflection is ok 4..35Gk + 1.35(− 2.0(− 2.9Wk. Therefore the ultimate design force in the member is. one should always watch out for possible reversal of stresses during load combinations. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1.structville.. refer to Table A1.9 (for the value of ψ0.7918 KN (COMPRESSIVE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.6 = 0.5.5(− 3.98765 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. CHECK FOR DEFLECTION Employing UA 50 X 50 X 6 in all members of the roof truss. Ubani Obinna U (2017) 3.5Wk.1 of BS EN 1990: 2002(E) (Eurocode.blogspot.0 Conclusion The analysis and design for the section members have been successfully carried as shown in the calculations above. 2002b). NEd = 1. Fu = γGjGk + γQkQk + γWkψ0Wk = 1.662) = -3. the deflection due to unfactored imposed load calculated using STAADPRO V8i software is 1.319) + 1. Whenever the wind load effect is greater than the live load effect.com (C) Ranks Michael Enterprises (2017) Page 15 .35(− 2. The design has shown that the provision of UA 50 X 50 X 6 for the entire roof truss will be adequate.5Qk + 0.319) + 1. Partial factor for leading variable actions (WK) = γWk = 1.5QK NEd = 1. ? KN (TENSILE) With a maximum length of 2163mm.5(4.0 (favourable). NEd = 1.319) + 1.. Downloaded from www. a little consideration will show that UA 50 X 50 X 6 will satisfy all the necessary limit state requirements.5(− 3..8mm 1.5 × 0.028mm < 28.1 ENV 1993-1-1:1992) = L/250 = 7200/250 = 28.8mm.028mm Allowable deflection for roofs (Table 4.35GK + 1.238) = -7.238) + 0. Analysis and Design of Steel Roof Trusses to EC3.9(4.
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