Practical Analysis AND Design OF Steel ROOF Trusses TO Eurocode 3 PDF

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PRACTICAL ANALYSIS AND DESIGN OF STEEL ROOFTRUSSES TO EUROCODE 3: A SAMPLE DESIGNUbani Obinna UzodimmaDepartment of Civil Engineering, Nnamdi Azikiwe University PMB 5025, Awka, Anambra State, NigeriaE-mail: rankiesubani@gmailProblem statement The skeletal structure of a roof system (18 long and 7 wi...

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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

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Page 1

Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) 1.0 LOAD ANALYSIS Span of roof truss = 7.2m Spacing of the truss = 3.0m Nodal spacing of the trusses = 1.2m

Permanent (dead) Loads Self weight of long span aluminium roofing sheet (0.55mm gauge thickness) = 0.019 KN/m2 Weight of ceiling (adopt 10mm insulation fibre board) = 0.077 KN/m2 Weight of services = 0.1 KN/m2 Weight of purlin (assume CH 150 x 75 x 18 kg/m) = (18 x 3m)/(1.2 x 3) = 15 kg/m2 = 0.147 KN/m2 Self weight of trusses (assume) = 0.2 KN/m2 Total deal load (Gk) = 0.536 KN/m2 Therefore the nodal permanent load (GK) = 0.536 KN/m2 × 1.2m × 3m = 1.9296 KN

Variable (Imposed) Load Category of roof = Category H – Roof not accessible except for normal maintenance and repairs (Table 6.9 EN 1991-1-1:2001) Imposed load on roof (Qk) = 0.75 KN/m2 Therefore the nodal variable load (QK) = 0.75 KN/m2 × 1.2m × 3m = 2.7 KN

Wind Load Wind velocity pressure (dynamic) is assumed as = qp(z) = 1.5 kN/m2 When the wind is blowing from right to left, the resultant pressure coefficient on the windward and leeward slopes with positive internal pressure (cpe) is taken as −0.9 Therefore the external wind pressure normal to the roof is; qe = qpcpe = −1.5 × 0.9 = 1.35 kN/m2 Vertical component pev = qe cos θ = 1.35 × cos 36.869 = 1.08 kN/m2 acting upwards ↑ Therefore the nodal wind load (WK) = 1.08 KN/m2 × 1.2m × 3m = 3.888 KN

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Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) 2.0 STRUCTURAL ANALYSIS

2.1 Analysis for dead load The nodal dead load from section 1.0 is now placed on the nodes of the trusses. It should be realised that at the first and last nodes, the loads are halved because the inter-nodal load distance will be 1.2/2. The loading is shown in Figure 3 below.

Figure 3: Truss carrying the nodal dead loads (GK)

Support Reactions Let the summation of moment about joint 12 be zero; ∑ 𝑀 = 

7.2R1 – (0.965 × 7.2) – (1.93 × 6) – (1.93 × 4.8) – (1.93 × 3.6) – (1.93 × 2.4) – (1.93 × 1.2) = 0 R1 =

. .

= 5.79 KN

Let ∑ 𝐹 =  We can verify that R12 = R1 = 5.79 KN Since there are no horizontal loads, then there are no horizontal reactive forces ANALYSIS OF INTERNAL FORCES

In all cases note that Fi-j = Fj-i JOINT 1 .

𝜃 = tan− 󰇡 .󰇢 = 36.869

Let ∑ 𝑭 = 𝟎

5.79 – 0.956 + F1-2 (sin 𝜃) = 0 −.

F1-2 = si . = −8.0568 KN (COMPRESSION)

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Page 3

Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) Let ∑ 𝑭 = 𝟎

F1-2 (cos 𝜃) + F1-3 = 0 F1-3 = −(−8.0568 × 󰇛cos .󰇜󰇜 = 6.445 KN (TENSION) JOINT 3 Let ∑ 𝑭 = 𝟎 F3 – 2 = 0 (NO FORCE) Let ∑ 𝑭 = 𝟎

F3 – 5 − F3 – 1 = 0 F3 – 1 = F3 – 5 = 6.445 KN (TENSION) JOINT 2 .

𝜙 = tan− 󰇡.󰇢 = 36.869 = 𝜃 Let ∑ 𝑭 = 𝟎

−1.93 + F2 - 4(sin 𝜃) – F2 – 3 − F2 - 5(sin 𝜃) - F2 - 1(sin 𝜃) = 0 −1.93 + F2 - 4(sin .) – 0 − F2 - 5(sin .) – (−8.0568(sin .)) = 0 Let ∑ 𝑭 = 𝟎

0.6 F2 – 4 − 0.6 F2 – 5 = −2.904 ------------------ (1)

F2 - 4(cos 𝜃) + F2 - 5(cos 𝜃) - F2 - 1(cos 𝜃) = 0 F2 - 4(cos .) + F2 - 5(cos .) – (−8.0568(cos.)) = 0 0.8 F2 – 4 + 0.8 F2 – 5 = −6.4455------------------ (2) Solving equations (1) and (2) simultaneously; F2-4 = − 6.448 KN (COMPRESSION) F2-5 = −1.608 KN (COMPRESSION) JOINT 5 Let ∑ 𝑭 = 𝟎

F5 - 2(sin ϕ) + F5 – 4 = 0

-1.608 (sin .) + F5 – 4 = 0 F5 – 4 = 0.9646 KN (TENSION)

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Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) Let ∑ 𝑭 = 𝟎

− F5 – 3 − F5 – 2 (cos ϕ) + F5 – 7 = 0 −. − (−1.608 (cos .)) + F5 – 7 = 0 F5 – 7 = 5.1586 KN (TENSION) JOINT 4 𝛼 = tan− 󰇡

.

.

󰇢 = 56.309°

Let ∑ 𝑭 = 𝟎

−1.93 – F4 – 2 (sin 𝜃) – F4 – 5 – F4 – 7 (sin 𝛼) + F4 – 6(sin 𝜃) = 0 −1.93 – (−6.448(sin .)) – 0.9646 – F4 - 7 (sin .) + F4 – 6 (sin .)) = 0 −0.832 F4 – 7 + 0.6 F4 – 6 = −0.9742 ------------------ (1) Let ∑ 𝑭 = 𝟎

F4 – 7 (cos 𝛼) + F4 - 6(cos 𝜃) – F4 - 2(cos 𝜃) = 0 F4 - 7(cos .) + F4 - 6(cos .) – (−6.448(cos.)) = 0 0.5547 F4 – 7 + 0.8 F4 – 6 = − 5.1584 ------------------ (2) Solving equations (1) and (2) simultaneously; F4 - 7 = − 2.319 KN (COMPRESSION) F4 - 6 = −4.8398 KN (COMPRESSION) JOINT 6 Let ∑ 𝑭 = 𝟎

− F4 - 6(cos 𝜃) + F6 - 8 (cos 𝜃) = 0 − (−4.8398 cos .) + F6 - 8 (cos .) = 0 −.

F6 - 8 = cs . = − 4.8398 KN (COMPRESSION) Let ∑ 𝑭 = 𝟎

−1.93 – F6 – 4 (sin 𝜃) – F6 – 7 − F6 – 8 (sin 𝜃) = 0 −1.93 – (−4.8398(sin .)) – F6 – 7 − (−4.8398(sin .)) = 0 F6 – 7 = 3.8777 KN (TENSION) Downloaded from www.structville.blogspot.com (C) Ranks Michael Enterprises (2017)

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Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) By symmetry of load and structure, 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 1.0 below. Table 1.0: Summary of analysis results for dead load (GK)

MEMBER 1-3 3-5 5-7 1-2 2-4 4-6 2-3 4-5 6-7 2-5 4-7

FORCE (KN) BOTTOM CHORD 6.445 6.445 5.1586 TOP CHORD −8.0568 − 6.448 −4.8398 VERTICALS 0 0.9646 3.877 DIAGONALS −1.608 − 2.319

NATURE TENSILE TENSILE TENSILE COMPRESSIVE COMPRESSIVE COMPRESSIVE NO FORCE TENSILE TENSILE COMPRESSIVE COMPRESSIVE

2.2 Analysis for imposed load

Figure 4: Truss carrying the nodal imposed loads (QK)

Following the steps in section 2.1, the following result as shown in Table 2.0 can be obtained; Table 2.0: Summary of analysis results for imposed load (QK)

MEMBER 1-3 3-5 5-7 1-2 2-4

FORCE (KN) NATURE BOTTOM CHORD 8.992 TENSILE 8.992 TENSILE 7.198 TENSILE TOP CHORD COMPRESSIVE −11.241 − 8.998 COMPRESSIVE

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Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) 4-6 2-3 4-5 6-7 2-5 4-7

−6.748 VERTICALS 0 1.346 5.391 DIAGONALS −2.242 − 3.238

COMPRESSIVE NO FORCE TENSILE TENSILE COMPRESSIVE COMPRESSIVE

2.3 Analysis for wind load

Figure 5: Truss carrying the nodal wind loads (WK)

Following the steps in section 2.1 also, the following result for the wind load as shown in Table 3.0 can be obtained;

Table 3.0: Summary of analysis results for imposed load (QK)

MEMBER 1-3 3-5 5-7 1-2 2-4 4-6 2-3 4-5 6-7 2-5 4-7

FORCE (KN) BOTTOM CHORD -12.948 -12.948 -10.365 TOP CHORD 16.187 12.957 9.717 VERTICALS 0 - 1.938 -7.763 DIAGONALS 3.228 4.662

NATURE COMPRESSIVE COMPRESSIVE COMPRESSIVE TENSILE TENSILE TENSILE NO FORCE COMPRESSIVE COMPRESSIVE TENSILE TENSILE

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Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) 3.0 STRUCTURAL DESIGN TO EC3 All structural steel employed has the following properties; Fy (Yeild strength) = 275 N/mm2 Fu (ultimate tensile strength = 430 N/mm2)

3.1 Design of the bottom chord (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only

Fu = γGjGk + γQk Qk ULTIMATE DESIGN FORCE (NEd) = 1.35GK + 1.5QK NEd = 1.35(6.445) + 1.5(8.992) = 22.189 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 × 0.6 = 0.9 (for the value of ψ0, refer to Table A1.1 of BS EN 1990: 2002(E) (Eurocode, 2002b). Therefore the ultimate design force in the member is;

Fu = γGjGk + γQk Qk + γWkψ0Wk = 1.35Gk + 1.5Qk + 0.9Wk. NEd = 1.35(6.445) + 1.5(8.992) − 0.9(12.894) = 10.584 KN (TENSILE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.0 (favourable). Partial factor for leading variable actions (WK) = γWk = 1.5. Therefore ultimate design force in the member = Fu = γGjGk + γWkWk = Gk + 1.5Wk.

NEd = 1.0(6.445) – 1.5(12.894) = −12.896 KN (COMPRESSIVE)

Therefore, all bottom chord members should be able to resist an axial tensile load of 22.189KN and a possible reversal of stresses with a compressive load of 12.896 KN

Length of longest bottom chord member = 1200mm Consider EQUAL ANGLES UA 50 X 50 X 6 Gross Area = 5.69 cm2 Radius of gyration (axis y-y) = 1.5 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.72 cm2 Equivalent tension area for welded connection = 4.88cm2 Nt,Rd is the lesser of

𝐴 𝐹𝑦 𝛾𝑀

and

.𝐴𝑓 𝛾𝑀

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Page 8

Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) Fu = 430 N/mm2; Fy = 275 N/mm2

Nt,Rd =

. ×  ×  .

. × . ×  ×  .

Also check; 𝑁 𝑁,

=

× − = 102.3 KN

.

× − = 115.17 KN

= 0.216 < 1.0 (Section is ok for tension resistance)

.

Compression and buckling resistance Thickness of section t = 6 mm. Since t < 16mm, Design yield strength Fy = 275 N/mm2 (Table 3.1 EC3) Section classification 



𝜀=√𝐹 =√ 𝑦



= 0.9244

h/t = 50/6 = 8.33. Referring to Table 5.2 (sheet 3) of Eurocode 3, Part 1-1, for class 3 classification, h/t ≤ 15ε and (h + b)/2t ≤ 11.5ε. In our case, 15ε = 15 × 0.92 = 13.8 > h/t (8.3) OK (h + b)/2t = 8.33 < 10.8 (11.5 × 0.92) OK Thus, the section satisfies both of the conditions. Resistance of the member to uniform compression NC,Rd =

𝐴𝐹𝑦

𝑁𝐸 𝑁𝐶,

. .

=

𝛾

=

. ×  ×  = .

156475 N = 156.475 KN

= 0.0824 < 1 Therefore section is ok for uniform compression.

Buckling resistance of member (clause 5.5 ENV 1993-1-1:1992) Since the member is pinned at both ends, critical buckling length is the same for all axis; Lcr = 1200mm Slenderness ratio 𝜆 =

𝐿𝑟

𝑟𝑖 𝜆

𝜆 = 93.9𝜀 = 93.9 × 0.9244 = 86.801 In the planar axis 𝜆=



 × .

= 0.9216

Buckling curve b is appropriate for all angle sections according to Table 6.2 of Eurocode 3 

Φ = 0.5 [ + 𝛼(𝜆 − .) + 𝜆 ] Downloaded from www.structville.blogspot.com (C) Ranks Michael Enterprises (2017)

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Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) Φ = 0.5 [ + .󰇛. − .󰇜 + . ] = 1.0473 𝒳=



Φ+ √Φ − 𝜆

𝒳 =





. + √. − .

Therefore Nb,Rd = 𝑁𝐸 𝑁,

=

. = .

𝒳𝑦 𝐴𝐹𝑦 𝛾

=

= 0.6473 < 1

. ×. ×  ×  = 

101286.2675 N = 101.286 KN

0.127 < 1 Therefore section is ok for buckling

Therefore, the section is ok to resist all axial loads on it.

3.2 Design of the top chord (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only ULTIMATE DESIGN FORCE (NEd) = 1.35GK + 1.5QK NEd = 1.35󰇛−8.0568) + 1.5󰇛−11.241) = -27.738 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 = 1.5 × 0.6 = 0.9 (for the value of ψ0, refer to Table A1.1 of BS EN 1990: 2002(E) (Eurocode, 2002b). Therefore the ultimate design force in the member is;

Fu = γGjGk + γQk Qk + γWkψ0Wk = 1.35Gk + 1.5Qk + 0.9Wk. NEd = 1.35(−8.0568) + 1.5(−11.241) + 0.9(16.187) = -13.169 KN (COMPRESSIVE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.0 (favourable). Partial factor for leading variable actions (WK) = γWk = 1.5. Therefore ultimate design force in the member = Fu = γGjGk + γWk Wk = Gk + 1.5Wk. NEd = 1.0(−8.0568) + 1.5(16.187) = +16.2237 KN (TENSILE) Therefore, all the top chord members should be able to resist an axial tensile load of 16.2237KN and a compressive load of 27.738 KN

Length of longest TOP chord member = 1500mm Let us also consider EQUAL ANGLES UA50 X 50 X 6 Downloaded from www.structville.blogspot.com (C) Ranks Michael Enterprises (2017)

Page 10

Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) Gross Area = 5.69 cm2 Radius of gyration (axis y-y) = 1.5 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.72 cm2 Equivalent tension area for welded connection = 4.88cm2 𝐴 𝐹𝑦

Nt,Rd is the lesser of

𝛾𝑀

and

.𝐴 𝑓 𝛾𝑀

Fu = 430 N/mm2; Fy = 275 N/mm2 Nt,Rd =

. ×  ×  .

. × . ×  × 

Also check; 𝑁

=

𝑁,

× − = 102.3 KN

. .

.

× − = 115.17 KN

= 0.158 < 1.0 (Section is ok for tension resistance)

Compression and buckling resistance Thickness of section t = 6 mm. Since t < 16mm, Design yield strength Fy = 275 N/mm2 (Table 3.1 EC3) Section classification 

𝜀=√

𝐹𝑦



= √  = 0.9244

h/t = 50/6 = 8.33. Referring to Table 5.2 (sheet 3) of Eurocode 3, Part 1-1, for class 3 classification, h/t ≤ 15ε and (h + b)/2t ≤ 11.5ε. In our case, 15ε = 15 × 0.92 = 13.8 > h/t (8.3) OK (h + b)/2t = 8.33 < 10.8 (11.5 × 0.92) OK Thus, the section satisfies both of the conditions. Resistance of the member to uniform compression 𝐴𝐹𝑦

NC,Rd = 𝛾



𝑁𝐸

𝑁𝐶,

=

=

.

. ×  × 

.

.

= 156475 N = 156.475 KN

= 0.177 < 1 Therefore section is ok for uniform compression.

Buckling resistance of member (clause 5.5 ENV 1993-1-1:1992) Since the member is pinned at both ends, critical buckling length is the same for all axis Lcr = 1500mm Slenderness ratio 𝜆 =

𝐿𝑟

𝑟𝑖 𝜆

𝜆 = 93.9𝜀 = 93.9 × 0.9244 = 86.801 Downloaded from www.structville.blogspot.com (C) Ranks Michael Enterprises (2017)

Page 11

Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) In the both planar axis 𝜆=



 × .

= 1.152

Buckling curve b is appropriate for all angle sections according to Table 6.2 of Eurocode 3 

Φ = 0.5 [ + 𝛼(𝜆 − .) + 𝜆 ] Φ = 0.5 [ + .󰇛. − .󰇜 + . ] = 1.325 𝒳=



Φ+ √Φ − 𝜆

𝒳 =





. + √. – .

Therefore Nb,Rd = 𝑁𝐸 𝑁,

=

. .

𝒳𝑦 𝐴𝐹𝑦 𝛾

=

= 0.5051 < 1

. ×. ×  ×  

= 79035.5225 N = 79.0355 KN

= 0.351< 1 Therefore section is ok for buckling

Therefore, the section is ok to resist all axial load on it.

3.3 Design of the vertical members (considering maximum effects) LOAD CASE 1: DEAD LOAD + IMPOSED LOAD only

Fu = γGjGk + γQk Qk ULTIMATE DESIGN FORCE (NEd) = 1.35GK + 1.5QK NEd = 1.35(3.877) + 1.5(5.391) = 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 = 1.5 × 0.6 = 0.9 (for the value of ψ0, refer to Table A1.1 of BS EN 1990: 2002(E) (Eurocode, 2002b). Therefore the ultimate design force in the member is;

Fu = γGjGk + γQk Qk + γWkψ0Wk = 1.35Gk + 1.5Qk + 0.9Wk. NEd = 1.35(3.877) + 1.5(5.391) − 0.9(7.763) = 6.334 KN (TENSILE) LOAD CASE 3: DEAD LOAD + WIND LOAD acting simultaneously Partial factor for permanent actions (DK) = γGj = 1.0 (favourable). Partial factor for leading variable actions (WK) = γWk = 1.5. Therefore ultimate design force in the member = Fu = γGjGk + γWk Wk = Gk + 1.5Wk. Downloaded from www.structville.blogspot.com (C) Ranks Michael Enterprises (2017)

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Analysis and Design of Steel Roof Trusses to EC3..... Ubani Obinna U (2017) NEd = 1.0(3.877) – 1.5(7.763) = −7.7675 KN (COMPRESSIVE) Length of longest bottom chord member = 2700mm Consider EQUAL ANGLES 30 X 30 X 4 Gross Area = 2.27 cm2 Radius of gyration (axis y-y) = 0.892 cm Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 1.24 cm2 Equivalent tension area for welded connection = 1.95 cm2 Nt,Rd is the lesser of

𝐴 𝐹𝑦 𝛾𝑀

and

.𝐴 𝑓 𝛾𝑀

Fu = 430 N/mm2; Fy = 275 N/mm2

Nt,Rd =

. ×  ×  .

. × . ×  × 

Also check; 𝑁

𝑁,

=

× − = 34.1 KN

.

.

× − = 38.3904 KN

= 0.3906 < 1.0 (Section is ok for tension resistance)

.

Compression resistance Thickness of section t = 6 mm. Since t < 16mm, Design yield strength Fy = 275 N/mm2 (Table 3.1 EC3) Section classification 



𝜀=√𝐹 =√ 𝑦



= 0.9244

h/t = 30/4 = 7.5 Referring to Table 5.2 (sheet 3) of Eurocode 3, Part 1-1, for class 3 classification, h/t ≤ 15ε and (h + b)/2t ≤ 11.5ε. In our case, 15ε = 15 × 0.92 = 13.8 > h/t (7.5) OK (h + b)/2t = 7.5 < 10.8 (11.5 × 0.92) OK Thus, the section satisfies both of the conditions. Resistance of the member to uniform compression NC,Rd = 𝑁𝐸 𝑁𝐶,

=

𝐴𝐹𝑦

𝛾

=

.

.

. ×  ×  = .

62425 N = 62.425 KN

= 0.1244 < 1 Therefore section is ok for uniform compression.

Buckling resistance of member (clause 5.5 ENV 1993-1-1:1992)

Downloaded from www.structville.blogspot.com (C) Ranks Michael Enterprises (2017)

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Analysis and Design ...