# PRACTICAL ANALYSIS AND DESIGN OF STEEL ROOF TRUSSES TO EUROCODE 3: A SAMPLE DESIGN 1.0 INTRODUCTION
The most widespread alternative for roof construction in Nigeria is the use of trusses, of which timber and steel are the primary choice of materials. Careful attention must be paid to design of structural members and connection details of trusses, since their failure can be catastrophic both in terms of loss of life and economy. The roof of a church building in Uyo, Akwa Ibom state Nigeria collapsed on the 10th of December 2016, and left more than 60 people dead, and many injured. This is to show how important, and why engineers must pay careful attention to such design situations. The aim of this post is to show in the clearest manner, how steel roof design can be carried out using Eurocode 3 design code.

To illustrate this, a simple design example has been presented. The skeletal structure of a roof system (18.0m long and 7.2m wide) is as shown in Figure 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).

The idealised 2D model of the roof truss typical loading configuration is as shown below;

Span of roof truss = 7.2m
Spacing of the truss = 3.0m
Nodal spacing of the trusses = 1.2m

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

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 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

3.0 ANALYSIS OF THE TRUSS FOR INTERNAL FORCES
N/B: Please note that the internal forces in the members are denoted by Fi-j which is also equal to F j-i e.g. F 2-3 = F 3-2 ; so kindly distinguish this from other numeric elements

JOINT 1

θ = tan-1⁡(2.7/3.6) = 36.869
Let ∑Fy = 0
5.79 – 0.956 + F1-2 (sin θ) = 0
F1-2 = (-4.834)/sin⁡36.869 = -8.0568 KN (COMPRESSION)
Let ∑Fx = 0
F1-2 (cos θ) + F1-3 = 0
F1-3 = -(-8.0568 ×(cos⁡36.869)) = 6.445 KN (TENSION)

JOINT 3

Let ∑Fy = 0
F3 – 2 = 0 (NO FORCE)
Let ∑Fx = 0
F3–5 – F3–1 = 0
F3–1 = F3–5 = 6.445 KN (TENSION)

JOINT 2

ϕ = tan-1⁡(0.9/1.2) = 36.869 = θ
Let ∑Fy = 0
-1.93 + F2-4(sin θ) – F2–3 – F2-5(sin θ) – F2-1(sin θ) = 0
-1.93 + F2-4(sin 36.869) – 0 – F2-5(sin 36.869) – (-8.0568(sin 36.869)) = 0
0.6 F2–4 – 0.6F2–5 = -2.904 —————— (1)

Let ∑Fx = 0
F2-4(cos θ) + F2-5(cos θ) – F2-1(cos θ) = 0
F2-4(cos 36.869) + F2-5(cos 36.869) – (-8.0568(cos36.869)) = 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 ∑Fy = 0
F5-2(sin ϕ) + F5–4 = 0
-1.608 (sin 36.869) + F5–4 = 0
F5–4 = 0.9646 KN (TENSION)
Let ∑Fx = 0
–F5–3 – F5–2 (cos ϕ) + F5–7 = 0
-6.445– (–1.608 (cos 36.869)) + F5–7 = 0
F5–7 = 5.1586 KN (TENSION)

JOINT 4

α = tan-1(1.8/1.2) = 56.309°
Let ∑Fy = 0
-1.93 – F4–2 (sin θ) – F4–5 – F4–7 (sin α) + F4–6(sin θ) = 0
-1.93 – (-6.448(sin 36.869)) – 0.9646 – F4-7 (sin 56.309) + F4–6 (sin 36.869)) = 0
-0.832 F4–7 + 0.6 F4–6 = -0.9742 —————— (3)
Let ∑Fx = 0
F4–7 (cos α) + F4-6(cos θ) – F4-2(cos θ) = 0
F4-7(cos 56.309) + F4-6(cos 36.869) – (-6.448(cos36.869)) = 0
0.5547 F4–7 + 0.8 F4–6 = – 5.1584 —————— (4)
Solving equations (3) and (4) simultaneously;
F4-7 = – 2.319 KN (COMPRESSION)
F4-6 = -4.8398 KN (COMPRESSION)

JOINT 6

Let ∑Fx = 0
– F4-6(cos θ) + F6-8 (cos θ) = 0
– (-4.8398 cos 36.869) + F6-8 (cos 36.869) = 0
F6-8 = (-3.87184)/cos ⁡36.869 = – 4.8398 KN (COMPRESSION)
Let ∑Fy = 0
-1.93 – F6–4 (sin θ) – F6–7 – F6–8 (sin θ) = 0
-1.93 – (-4.8398(sin 36.869)) – F6–7 – (–4.8398(sin 36.869)) = 0
F6–7 = 3.8777 KN (TENSION)

BOTTOM CHORD
F1-3 = 6.445 KN (T)
F3-5 = 6.445KN (T)
F5-7 = 5.158 KN (T)

TOP CHORD
F1-2 = -8.0568 KN (C)
F2-4 = -6.448 KN (C)
F4-6 = -4.839 KN (C)

VERTICALS
F2-3 = 0 (NO FORCE)
F4-5 = 0.9646 KN (T)
F6-7 = 3.877 KN (T)

DIAGONALS
F2-5 = -1.608 KN (C)
F4-7 = – 2.319 KN (C)

Similarly,
SUMMARY OF RESULTS FOR IMPOSED LOAD (QK)

BOTTOM CHORD
F1 – 3 = 8.992KN (T)
F3 – 5 = 8.992KN (T)
F5 – 7 = 7.198KN (T)

TOP CHORD
F1 – 2 = -11.241 KN (C)
F2 – 4 = – 8.998 KN (C)
F4 – 6 = -6.748KN (C)

VERTICALS
F2 – 3 = 0 (NO FORCE)
F4 – 5 = 1.346 KN (T)
F6 – 7 = 5.391 KN (T)

DIAGONALS
F2 – 5 = -2.242 KN (C)
F4 – 7 = – 3.238KN (T)

Similarly,
SUMMARY OF RESULTS FOR WIND LOAD (WK)

BOTTOM CHORD
F1 – 3 = -12.948 KN (C)
F3 – 5 = -12.948 KN (C)
F5 – 7 = -10.365 KN (C)

TOP CHORD
F1 – 2 = 16.187 KN (T)
F2 – 4 = 12.957 KN (T)
F4 – 6 = 9.717 KN (T)
VERTICALS
F2 – 3 = 0 (NO FORCE)
F4 – 5 = – 1.938 KN (C)
F6 – 7 = -7.763 KN (C)

DIAGONALS
F2 – 5 = 3.228 KN (T)
F4 – 7 = 4.662 KN (T)

In all cases, (T) – Tensile force; (C) – Compressive force

4.0 STRUCTURAL DESIGN TO EUROCODE 3

All structural steel employed has the following properties;
Fy (Yeild strength) = 275 N/mm2
Fu (ultimate tensile strength = 430 N/mm2)

Design of the bottom chord (considering maximum effects)

Fu = γGjGk + γQkQk

ULTIMATE DESIGN FORCE (NEd) = 1.35GK + 1.5QK
NEd = 1.35(6.445) + 1.5(8.992) = 22.189 KN (TENSILE)

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 + γQkQk + γWkψ0Wk = 1.35Gk + 1.5Qk + 0.9Wk
NEd = 1.35(6.445) + 1.5(8.992) – 0.9(12.894) = 10.584 KN (TENSILE)

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) ri = 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;
(Anet × Fy)/γM0 and (0.9Anet × fu)/γM2
Fu = 430 N/mm2; Fy = 275 N/mm2
Nt,Rd = (3.72 × 102 × 275)/1.0 × 10-3) = 102.3 KN
Also check; (0.9 × 3.72 × 102 × 430)/1.25 × 10-3 = 115.17 KN

Therefore;

NSd/Nt,Rd = 22.189/102.3 = 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 = √(235/Fy ) = √(235/275) = 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,
5ε = 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 = (A × Fy)/γM0 = (5.69 × 102 × 275)/1.0 = 156475 N = 156.475 KN
NEd/NC,Rd = 12.896/156.475 = 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 = Lcr/(ri × λ1 )
λ1= 93.9ε = 93.9 × 0.9244 = 86.801
In the planar axis (z-z and y-y)
¯λ = 1200/(15 × 86.801) = 0.9216

Buckling curve b is appropriate for all angle sections according to Table 6.2 of Eurocode 3
α = 0.34 for buckling curve b
Φ = 0.5 [1+ α(¯λ – 0.2) + ¯λ2 ]
Φ = 0.5 [1+ 0.34(0.9216 – 0.2)+ 0.92162 ] = 1.0473
X = 1/[Φ+ √(Φ2 – ¯λ2)]
X = 1/[1.0473 + √(1.0472 – 0.92162)] = 0.6473 < 1

Therefore Nb,Rd = (X × A × Fy)/γm1 = (0.6473 × 5.69 × 102× 275)/1.0 = 101286.2675 N = 101.286 KN
NEd/Nb,Rd = 12.869/101.286 = 0.127 < 1 Therefore section is ok for buckling

Therefore, UA 50 x 50 x 6 is ok to resist all axial loads on the bottom chord of the truss.
Following the method shown above in section 4.0, other members of the truss can be efficiently designed.

Thank you for visiting

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3. Nice presentation. I have a question though: When designing for roof trusses in Nigeria, is 0.75 Kn/m2 a reasonable imposed load to adopt? What could possibly create such a load? Thanks

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12. This design is lacking in two main ways:
1. Eurocode for wind loading doesn't allow us to use a uniform distributed load over the whole roof. There are multiple uneven wind load scenarios to be calculated.
2. Buckling resistance has been calculated in the vertical planes but not the horizontal planes. Without cross-bracing between trusses, they could buckle sideways as you effectively have a critical buckling length of 7.2m on the bottom elements and 4.5m on the top. Perhaps you have assumed the purlins will prevent this, but this is insufficient without cross-bracing.

• Victor. O

1. The U.K national anex allow the use of a factor for calculating the overall wind load on a roof or building. Thus the wind load may be considered to be constant over the entire structure by multiplying with the overall cp’s values.

2. The trusse are so closely spaced for buckling in the horizontal plane to be a issue

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15. Roof pitches are typically measured by an equation called the "rise over the run" and is figured in the number of feet that the roof rises in a 12 foot horizontal run. A "12/12 pitched roof" means that the roof rises 12 feet in a horizontal run of 12 feet and results in a 45 degree angle. Roofing Companies Sudbury

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17. Victor. O

1. The U.K national anex allow the use of a factor for calculating the overall wind load on a roof or building. Thus the wind load may be considered to be constant over the entire structure by multiplying with the overall cp’s values.

2. The trusse are so closely spaced for buckling in the horizontal plane to be a issue

• Sivaguru

Couldn’t find the overall wind load factor in U.K national annex.

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