**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 Î¸ = 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**

Î¸ = tan

^{-1}(2.7/3.6) = 36.869

Let ∑Fy = 0

5.79 – 0.956 + F

_{1-2}(sin Î¸) = 0

F

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

Let ∑Fx = 0

F

_{1-2}(cos Î¸) + 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**

Ï• = tan

^{-1}(0.9/1.2) = 36.869 = Î¸

Let ∑Fy = 0

-1.93 + F

_{2-4}(sin Î¸) – F

_{2–3}– F

_{2-5}(sin Î¸) – F

_{2-1}(sin Î¸) = 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 Î¸) + F

_{2-5}(cos Î¸) - F

_{2-1}(cos Î¸) = 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 Ï•) + 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 Ï•) + F

_{5–7}= 0

-6.445– (–1.608 (cos 36.869)) + F

_{5–7}= 0

F

_{5–7}= 5.1586 KN (TENSION)

**JOINT 4**

Î± = tan

^{-1}(1.8/1.2) = 56.309°

Let ∑Fy = 0

-1.93 – F

_{4–2}(sin Î¸) – F

_{4–5}– F

_{4–7}(sin Î±) + F

_{4–6}(sin Î¸) = 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 Î±) + F

_{4-6}(cos Î¸) – F

_{4-2}(cos Î¸) = 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 Î¸) + F

_{6-8}(cos Î¸) = 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 Î¸) – F

_{6–7}– F

_{6–8}(sin Î¸) = 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 = Î³

_{Gj}Gk + Î³

_{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 Î³

_{Wk}Ïˆ

_{0}= 1.5 × 0.6 = 0.9 (for the value of Ïˆ0, refer to Table A1.1 of BS EN 1990: 2002(E) (Eurocode, 2002b). Therefore the ultimate design force in the member is;

Fu = Î³

_{Gj}Gk + Î³

_{Qk}Qk + Î³

_{Wk}Ïˆ

_{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) = Î³

_{Gj}= 1.0 (favourable)

Partial factor for leading variable actions (WK) = Î³

_{Wk}= 1.5

Therefore ultimate design force in the member = Fu = Î³

_{Gj}Gk + Î³

_{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)/Î³

_{M0}and (0.9A

_{net}× fu)/Î³

_{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**

Îµ = √(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 ≤ 15Îµ and (h + b)/2t ≤ 11.5Îµ. In our case,

5Îµ = 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)/Î³

_{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 ¯Î» = Lcr/(ri × Î»1 )

Î»1= 93.9Îµ = 93.9 × 0.9244 = 86.801

In the planar axis (z-z and y-y)

¯Î» = 1200/(15 × 86.801) = 0.9216

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

Î± = 0.34 for buckling curve b

Î¦ = 0.5 [1+ Î±(¯Î» - 0.2) + ¯Î»

^{2}]

Î¦ = 0.5 [1+ 0.34(0.9216 - 0.2)+ 0.9216

^{2}] = 1.0473

X = 1/[Î¦+ √(Î¦

^{2}- ¯Î»

^{2})]

X = 1/[1.0473 + √(1.047

^{2}- 0.9216

^{2})] = 0.6473 < 1

Therefore N

_{b,Rd}= (X × A × Fy)/Î³

_{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_!

ReplyDeleteThank you Afam

DeleteVery good

ReplyDeleteThank you

DeleteUba my man! Pls make a simple duplex design with staadpro... thanks

Deletewell written. thank you

ReplyDeleteA job well done

ReplyDeleteNice 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

ReplyDeleteThe 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.

DeleteThough these kits come with highly competent equipment, there are certain precautions stated in the product manuals of such products which need to be followed for user's safety. guttering repair services london

ReplyDeleteInstallation of an expert. Sure the art, such that the ratio is held firmly and is installed in the roof of his own affection.check my blog

ReplyDeleteRoof repairing is one of the hardest work of the home, for that you have to choose right servicer like Maciej Kiersnowski.

ReplyDeleteSlip-ups in rooftop repairs or establishments will prompt the need of supplanting the framework once more, in this way the need to spend more money. Extra costs can be evaded by procuring a solid roofing organization. Roofing ladders

ReplyDeleteVery nice presentation. Do you by any chance have design calculations for the connections? Thanks.

ReplyDeleteHowever, it is tricky business and you can never know how much a roof repair job is going to cost you and that is the very reason that you need to estimate the roof repair job before hiring a professional roofing company.roof consulting services Toronto

ReplyDelete• I will dÐµfinitelydigg it and personally suggest to my friends. I’m sure thÐµywilⅼ be benefited from thiÑ• web site.

ReplyDeleteBIM Project Management in India

BIM Implementation in India

Documentation Services in India

ANIMATION SERVICES in India

CAD to BIM conversion in India

More established homes frequently have lengths of 2" x 6" introduced rather than pressed wood or OSB board.Roofing Company Kansas City

ReplyDelete