**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;

**2.0 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/m^{2}

Weight of ceiling (adopt 10mm insulation fibre board) = 0.077 KN/m^{2}

Weight of services = 0.1 KN/m^{2}

Weight of purlin (assume CH 150 x 75 x 18 kg/m) = (18 x 3m)/(1.2 x 3) = 15 kg/m^{2} = 0.147 KN/m^{2}

Self weight of trusses (assume) = 0.2 KN/m^{2}

Total deal load (Gk) = 0.536 KN/m^{2}

Therefore the nodal permanent load (GK) = 0.536 KN/m^{2} × 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/m^{2}

Therefore the nodal variable load (QK) = 0.75 KN/m^{2} × 1.2m × 3m = 2.7 KN

**Wind Load**

Wind velocity pressure (dynamic) is assumed as = q_{p(z)} = 1.5 kN/m^{2}

When the wind is blowing from right to left, the resultant pressure coefficient on the windward and leeward slopes with positive internal pressure (c_{pe}) is taken as −0.9

Therefore the external wind pressure normal to the roof is;

q_{e} = q_{p}c_{pe} = −1.5 × 0.9 = 1.35 kN/m^{2}

Vertical component p_{ev} = q_{e} cos θ = 1.35 × cos 36.869 = 1.08 kN/m2 acting upwards ↑

Therefore the nodal wind load (WK) = 1.08 KN/m^{2} × 1.2m × 3m = 3.888 KN

To see how wind load is analysed using Eurocode, click HERE

3**.0 ANALYSIS OF THE TRUSS FOR INTERNAL FORCES**

N/B: Please note that the internal forces in the members are denoted by F_{i-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

**ANALYSIS FOR DEAD LOAD**

**JOINT 1**

θ = tan^{-1}(2.7/3.6) = 36.869

Let ∑Fy = 0

5.79 – 0.956 + F_{1-2} (sin θ) = 0

F_{1-2} = (-4.834)/sin36.869 = -8.0568 KN (COMPRESSION)

Let ∑Fx = 0

F_{1-2} (cos θ) + F_{1-3} = 0

F_{1-3} = -(-8.0568 ×(cos36.869)) = 6.445 KN (TENSION)

**JOINT 3**

Let ∑Fy = 0

F_{3 – 2} = 0 (NO FORCE)

Let ∑Fx = 0

F_{3–5} – F_{3–1} = 0

F_{3–1} = F_{3–5} = 6.445 KN (TENSION)

**JOINT 2**

ϕ = tan^{-1}(0.9/1.2) = 36.869 = θ

Let ∑Fy = 0

-1.93 + F_{2-4}(sin θ) – F_{2–3} – F_{2-5}(sin θ) – F_{2-1}(sin θ) = 0

-1.93 + F_{2-4}(sin 36.869) – 0 – F_{2-5}(sin 36.869) – (-8.0568(sin 36.869)) = 0

0.6 F_{2–4} – 0.6F_{2–5} = -2.904 —————— (1)

Let ∑Fx = 0

F_{2-4}(cos θ) + F_{2-5}(cos θ) – F_{2-1}(cos θ) = 0

F_{2-4}(cos 36.869) + F_{2-5}(cos 36.869) – (-8.0568(cos36.869)) = 0

0.8 F_{2–4} + 0.8 F_{2–5} = -6.4455—————— (2)

Solving equations (1) and (2) simultaneously;

F_{2-4} = – 6.448 KN (COMPRESSION)

F_{2-5} = -1.608 KN (COMPRESSION)

**JOINT 5**

Let ∑Fy = 0

F_{5-2}(sin ϕ) + F_{5–4} = 0

-1.608 (sin 36.869) + F_{5–4} = 0

F_{5–4} = 0.9646 KN (TENSION)

Let ∑Fx = 0

–F_{5–3} – F_{5–2} (cos ϕ) + F_{5–7} = 0

-6.445– (–1.608 (cos 36.869)) + F_{5–7} = 0

F_{5–7} = 5.1586 KN (TENSION)

**JOINT 4**

α = tan^{-1}(1.8/1.2) = 56.309°

Let ∑Fy = 0

-1.93 – F_{4–2} (sin θ) – F_{4–5} – F_{4–7} (sin α) + F_{4–6}(sin θ) = 0

-1.93 – (-6.448(sin 36.869)) – 0.9646 – F_{4-7} (sin 56.309) + F_{4–6} (sin 36.869)) = 0

-0.832 F_{4–7} + 0.6 F_{4–6} = -0.9742 —————— (3)

Let ∑Fx = 0

F_{4–7} (cos α) + F_{4-6}(cos θ) – F_{4-2}(cos θ) = 0

F_{4-7}(cos 56.309) + F_{4-6}(cos 36.869) – (-6.448(cos36.869)) = 0

0.5547 F_{4–7} + 0.8 F_{4–6} = – 5.1584 —————— (4)

Solving equations (3) and (4) simultaneously;

F_{4-7} = – 2.319 KN (COMPRESSION)

F_{4-6} = -4.8398 KN (COMPRESSION)

**JOINT 6**

Let ∑Fx = 0

– F_{4-6}(cos θ) + F_{6-8} (cos θ) = 0

– (-4.8398 cos 36.869) + F_{6-8} (cos 36.869) = 0

F_{6-8} = (-3.87184)/cos 36.869 = – 4.8398 KN (COMPRESSION)

Let ∑Fy = 0

-1.93 – F_{6–4} (sin θ) – F_{6–7} – F_{6–8} (sin θ) = 0

-1.93 – (-4.8398(sin 36.869)) – F_{6–7} – (–4.8398(sin 36.869)) = 0

F_{6–7} = 3.8777 KN (TENSION)

**SUMMARY OF RESULTS FOR DEAD LOAD** (GK)

**BOTTOM CHORD**

F_{1-3} = 6.445 KN (T)

F_{3-5} = 6.445KN (T)

F_{5-7} = 5.158 KN (T)

**TOP CHORD**

F_{1-2} = -8.0568 KN (C)

F_{2-4} = -6.448 KN (C)

F_{4-6} = -4.839 KN (C)

**VERTICALS**

F_{2-3} = 0 (NO FORCE)

F_{4-5} = 0.9646 KN (T)

F_{6-7} = 3.877 KN (T)

**DIAGONALS**

F_{2-5} = -1.608 KN (C)

F_{4-7} = – 2.319 KN (C)

Similarly,

**SUMMARY OF RESULTS FOR IMPOSED LOAD (QK**)

**BOTTOM CHORD**

F_{1 – 3} = 8.992KN (T)

F_{3 – 5} = 8.992KN (T)

F_{5 – 7} = 7.198KN (T)

**TOP CHORD**

F_{1 – 2} = -11.241 KN (C)

F_{2 – 4} = – 8.998 KN (C)

F_{4 – 6} = -6.748KN (C)

**VERTICALS**

F_{2 – 3} = 0 (NO FORCE)

F_{4 – 5} = 1.346 KN (T)

F_{6 – 7} = 5.391 KN (T)

**DIAGONALS**

F_{2 – 5} = -2.242 KN (C)

F_{4 – 7} = – 3.238KN (T)

Similarly,

**SUMMARY OF RESULTS FOR WIND LOAD (WK**)

**BOTTOM CHORD**

F_{1 – 3} = -12.948 KN (C)

F_{3 – 5} = -12.948 KN (C)

F_{5 – 7} = -10.365 KN (C)

**TOP CHORD**

F_{1 – 2} = 16.187 KN (T)

F_{2 – 4} = 12.957 KN (T)

F_{4 – 6} = 9.717 KN (T)

**VERTICALS**

F_{2 – 3} = 0 (NO FORCE)

F_{4 – 5} = – 1.938 KN (C)

F_{6 – 7} = -7.763 KN (C)

**DIAGONALS**

F_{2 – 5} = 3.228 KN (T)

F_{4 – 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/mm^{2}

Fu (ultimate tensile strength = 430 N/mm^{2})

Design of the bottom chord (considering maximum effects)

**LOAD CASE 1**: DEAD LOAD + IMPOSED LOAD only

Fu = γ_{Gj}Gk + γ_{Qk}Qk

ULTIMATE DESIGN FORCE (N_{Ed}) = 1.35GK + 1.5QK

N_{Ed} = 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 γ_{Wk}ψ_{0} = 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 = γ_{Gj}Gk + γ_{Qk}Qk + γ_{Wk}ψ_{0}Wk = 1.35Gk + 1.5Qk + 0.9Wk

N_{Ed} = 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) = γ_{Gj} = 1.0 (favourable)

Partial factor for leading variable actions (WK) = γ_{Wk} = 1.5

Therefore ultimate design force in the member = Fu = γ_{Gj}Gk + γ_{Wk}Wk = Gk + 1.5Wk.

N_{Ed} = 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 cm^{2}

Radius of gyration (axis y-y) ri = 1.5 cm

Considering one M12 bolt (14mm diameter allowance) – Equivalent tension area = 3.72 cm^{2}

Equivalent tension area for welded connection = 4.88cm^{2}

N_{t,Rd} is the lesser of;

(A_{net} × Fy)/γ_{M0} and (0.9A_{net} × fu)/γ_{M2}

Fu = 430 N/mm^{2}; Fy = 275 N/mm^{2}

N_{t,Rd} = (3.72 × 10^{2} × 275)/1.0 × 10^{-3}) = 102.3 KN

Also check; (0.9 × 3.72 × 10^{2} × 430)/1.25 × 10^{-3} = 115.17 KN

Therefore;

N_{Sd}/N_{t,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**

N_{C,Rd} = (A × Fy)/γ_{M0} = (5.69 × 10^{2} × 275)/1.0 = 156475 N = 156.475 KN

N_{Ed}/N_{C,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) + ¯λ^{2} ]

Φ = 0.5 [1+ 0.34(0.9216 – 0.2)+ 0.9216^{2} ] = 1.0473

X = 1/[Φ+ √(Φ^{2} – ¯λ^{2})]

X = 1/[1.0473 + √(1.047^{2} – 0.9216^{2})] = 0.6473 < 1

Therefore N_{b,Rd} = (X × A × Fy)/γ_{m1} = (0.6473 × 5.69 × 10^{2}× 275)/1.0 = 101286.2675 N = 101.286 KN

N_{Ed/Nb,Rd} = 12.869/101.286 = 0.127 < 1 Therefore section is ok for buckling

Therefore, **UA 50 x 50 x 6 i**s 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

TO DOWNLOAD THE FULL DESIGN PAPER IN PRINTABLE PDF FORMAT, CLICK HERE

Very educative!! Well done bro_!

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Thank you Afam

Thank you

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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

The code gives some specific guidelines, and range of possible imposed loads. I know in Nigeria it may be hard to imagine what will generate such load, but going to the roof for repairs and maintenance, or even the effect of rain can be seen as imposed load.

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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.

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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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