(*Image Copyright belongs to Albion Sections Limited, UK*)

Roof purlin are members used to directly support roof sheeting materials, and could be made of timber or steel. In timber construction, purlins are nailed to the rafter or supporting trusses, while in steel roof construction, they are welded or bolted to the rafters or trusses by the means of cleats. As structural members, they resist loads, and provide lateral restraints for truss members, therefore it is important to design them properly against forces such as bending, shear, torsion, buckling etc.

In their design life, purlins are subjected to dead load (e.g self weight of sheeting materials and accessories), live load (e.g. during maintenance services and repairs), and environmental loads (e.g. wind and snow load). Therefore, a purlin should be adequately strong to withstand the loads it will encounter during its design life, and should not sag in an obvious manner thereby giving the roof sheeting an undulating and/or unpleasant appearance. This post will be focusing on design of steel purlin using cold formed sections.

**Arrangement of Purlins**

By default, purlin sections assume the slope of the roof they are supporting. The spacing of purlins usually call for careful arrangement, in the sense that it should follow the nodal pattern of the supporting trusses. What I mean in this regard is that purlins should be placed at the nodes of trusses and not on the members themselves so as not to induce secondary bending and shear forces in the members of the truss. Furthermore, if manual analysis is employed to analyse a truss loaded in such manner, such secondary stresses cannot be captured since we normally assume pinned connections.

Cold formed Z (Zed) and C (channel) sections are normally specified for purlins in steel structures (see their form in image below).

As compared with thicker hot rolled shapes, they normally offer the advantages of lightness, high strength and stiffness, easy fabrication and installations, easy packaging and transportation etc. The connection of purlins can be sleeved or butted depending on the construction method adopted.

In terms of arrangement, we can have single spans with staggered sleeved/butt arrangement, single/double span with staggered sleeve arrangement, double span butt joint system, and single span butt joint system. The choice of the arrangement to be adopted can depend on the supply length of the sections as readily available in the market, the need to avoid wasteful offcuts, the loading and span of the roof, the arrangement of the rafters etc. Therefore the roof designer must plan from start to finish. However, single and double span butt joint system are the most popular in Nigeria, due to their simplicity, and the culture of adopting shorter roof spans in the country. However, they are less structurally efficient than sleeved connections.

**Design Example**

We are to provide a suitable cold formed channel section for the purlin of the roof arrangement shown below.

**Partial Factor for loads (BS EN 1990 NA 2.2.3.2 Table NA.A1.2(B)) **

Permanent action γ_{G} = 1.35 (unfavourable)

Variable action γ_{G} = 1.5

Combination factor for Roofs, ψ_{0} = 0.7

Wind loads, ψ_{0 }= 0.5

Reduction factor ξ = 0.925

**Initial Sizing of Sections**

The maximum spacing of the trusses is at 3000 mm c/c

Limiting the deflection to L/200;

For continuous purlins, minimum depth of section (preliminary guide) = L/45

3000/45 = 66.667

Minimum width = 0.5L/60 = (3000/2)/60 = 25

Try C120-15 section (C – purlin) (Section picked from Albion Technical Manual, 2010).

**Section Properties**

In the design of purlins using EN 1993-1-3:2006, we normally utilise the effective section properties. This calculation is actually very tedious and prone to error, hence it is very advisable to obtain information from manufacturer’s details or you can a write a program using Microsoft Excel or MATLAB for such calculations. However, I am going to make a sample calculation for the section that we are considering.

Thickness (t_{nom}) = 1.5mm

Depth = 120mm;

Flange width = 50mm;

Lips/Edge Fold = 15mm;

Steel core thickness (t) = 1.5 – 0.05 = 1.45mm (Note that EN 1993-1-3:2006 recommends a thickness of 0.04mm for zinc coated sections, but we are using 0.05mm here)

Unit weight = 2.8 kg/m^{2};

Web height h_{p} = h – t_{nom} = 120 – 1.5 = 118.5mm

Width of flange in tension = Width of flange in compression b_{p1} = b_{p2} = b – t_{nom} = 50 – 1.5 = 48.5mm

Width of edge fold c_{p} = C – t_{nom}/2 = 15 – 1.5/2 = 14.25mm

**Checking of geometrical proportions**

b/t ≤ 60 b_{1}/t = 48.5/1.45 = 33.448 < 60 – OK

c/t ≤ 50 c/t = 15/1.45 = 10.344 < 50 – OK

h/t ≤ 500 c/t = 120/1.45 = 82.758 < 500 – OK

0.2 ≤ c/b ≤ 0.6 c/b_{1} = 15/48.5 = 0.309 – OK

The effect of rounding corners due to root radius has been neglected in this design.

**Gross section properties**

A_{br} = t(2c_{p} + b_{p1} + b_{p2} + h_{p}) = 1.45[(2 × 14.25) + 48.5 + 48.5 + 118.5)] = 353.8 mm^{2}

Position of the neutral axis with respect to the flange in compression;

Z_{b1} = 1.45[14.25(118.5 – 7.125) + (48.5 × 118.5) + 118.5^{2}/2 + 14.25^{2}/2] / 353.8 = 59.253mm

Effective section properties of the flange and lip in compression (clause 3.7.2)

Effective width of the compressed flange;

The stress ratio ψ = 1.0 (uniform compression)

k_{σ} = 4 for internal compression element (clause 3.7.2 Table 3.5)

ε = √(235/f_{yb}) = √(235/350) = 0.819

The effective width is;

b_{eff} = ρ_{b}b_{1} = 0.966 × 48.5 = 46.851mm

b_{e1} = b_{e2} = 0.5b_{eff} = 0.5 × 46.851 = 23.4255mm

Effective Width of the edge fold (lip) Clause 3.7.3.2.2 Equation 3.47

The buckling factor is;

c_{p }⁄ b_{p1 }= 14.25/48.5 = 0.2938 < 0.35

So k_{σ} = 0.5

Therefore since 1.149 < 1.0, take reduction factor as 1.0

The effective width is therefore c_{eff} = ρ_{cp} = 1.0 × 14.25 = 14.25mm

The effective area of the edge stiffener;

A_{s} = t(b_{e2} + c_{eff}) = 1.45 (23.4255 + 14.25) = 54.629 mm^{2}

We now have to use the initial effective cross-section of the stiffener to determine the reduction factor, allowing for the effects of the continuous spring restraint.

The elastic critical buckling stress for the edge stiffener is;

Where K is the spring stiffness per unit length;

b_{1} is the distance of the web to the centre of the effective area of the stiffener in compression flange (upper flange);

k_{f} = 0 for bending about the y-y axis

Therefore;

= 175882.211 × 0.00000491305 = 0.8641 N/mm

So the elastic critical buckling stress for the edge stiffener is;

σ_{cr,s} = (2√(0.8641 × 210000 × 1007.801))/(54.629 ) = 495.901 N/mm^{2}

The relative slenderness factor for the edge stiffener;

In our own case;

χ_{d} = 1.47 – 0.723(0.840)= 0.862

As the reduction factor for buckling is less than 1.0, we can optionally iterate to refine the value of the reduction factor for buckling of the stiffeners according to clause 5.5.3.2(3). But we are not iterating in this post.

Therefore;

χ_{d} = 0.862

b_{e2} = 23.4255 mm

c_{eff} = 14.25mm

The buckling factor k_{σ} = 7.81 – 6.29ψ + 9.78ψ^{2}

k_{σ} = 7.81 – 6.29(-0.9346) + 9.78(-0.9346)^{2} = 22.231

Therefore, take *ρ *as 1.0 since 1.137 > 1.0.

Therefore, the effective width of the zone in compression of the web is;

h_{eff} = ρh_{c} = 1.0 × 61.252 = 61.252mm

**Near the flange in compression;**

h_{e1} = 0.4h_{eff} = 0.4(61.252) = 24.5mm

h_{e2 } = 0.6h_{eff} = 0.6(61.252) = 36.7512mm

**The effective width of the web is;***Near the flange in compression;*

h_{1} = h_{e2} = 24.5mm

*Near the flange in tension;*

h_{2} = h_{p} – (h_{c} – h_{e2}) = 118.5 – (61.252 – 36.7512) = 93.992mm

**Effective section properties;**

A_{eff} = 1.45 × [14.25 + 48.5 + 24.5 + 93.992 + 23.4255 + (23.4255 + 14.25) 0.862]

A_{eff} = 343.858 mm^{2}

Position of the neutral axis with regard to the flange in compression;

zc = 1.45[(14.25 × 111.375) + (48.5 × 118.5) + (93.992 × 71.504) + 300.125 + 87.5199] / 343.858 = 60.903 mm

Position of the neutral axis with regard to the flange in tension;

Z_{t} = h_{p} – Z_{c}

**Second moment of area:**

**Effective section modulus;***With regard to the flange in compression;***W _{eff,y,c} = I_{eff,y}/z_{c} = (777557.517) / 60.903 = 12767 mm^{3}**

*With regard to the flange in tension;***W _{eff,y,t} = I_{eff,y}/z_{t} = (777557.517) / 57.697 = 13476 mm^{3}**

**LOAD ANALYSIS***Permanent loads*

Employing long span aluminium roofing sheet (gauge thickness = 0.55mm)

Load due to sheeting = 0.019 kN/m^{2}

Other permanent accessories and fittings = 0.15 kN/m^{2}

Total = 0.169 KN/m^{2}

At a spacing of 1.2m, = 0.169 KN/m2 × 1.2m = 0.2028 KN/m

Self weight of purlin = 2.8 kg/m = 0.0275 KN/m

Total Gk = 0.2028 KN/m + 0.0275 KN/m = 0.230 KN/m

*Live load*

For a roof with 20° slope and no access except for normal repairs and maintenance, let us adopt a live load of 0.75 KN/m^{2}

At a spacing of 1.2m, Qk = 0.75 KN/m^{2} × 1.2m = 0.9 KN/m

*Wind Load*

Taking a dynamic wind pressure of 1.5 KN/m^{2}

When the wind is blowing from right to left, the resultant pressure coefficient on a windward slope with positive internal pressure is;

c_{pe} = −0.90 upwards

Therefore the external wind pressure normal to the roof is;

p_{e} = q_{p}c_{pe} = 1.5 × − 0.90 = −1.35 kN/m^{2}

The vertical component of the wind pressure is;

p_{ev} = p_{e}cosθ = −1.35 × cos 20° = −1.268 kN/m^{2} acting upwards.

At a spacing of 1.2m;

W_{k} = −1.268 kN/m^{2} × 1.2m = 1.522 KN/m

**STATIC SYSTEMS**

We are adopting two possible systems that will offer us continuous and single span systems. The 6m span is based on supply length.** **

*Static Model 1*

*Static Model 2*

**Load Case 1**

When Dead load and live load are acting alone;

q = 1.35Gk + 1.5Qk = 1.35(0.230) + 1.5(0.9) = 1.6605 KN/m

*Model 1*

**Model 2****Load Case 2**

When Dead load, live load and wind load are acting alone;

q = 1.35Gk + 1.5Qk + 0.9Wk

Where the live load is the leading variable action

q = 1.35(0.230) + 1.5(0.9) – 0.9 (1.522) = 0.2907 KN/m

*Model 1 *

*Model 2 *

** ****Load Case 3**

When Dead load and wind load are acting alone;

q = 1.0Gk – 1.5Wk

Where dead load is favourable

q = 1.0(0.230) – 1.5(1.522) = -2.053 KN/m

* Model 1*

*Model 2*

Maximum span design moment M_{Ed} = 2.31 KNm

Maximum shear force V_{Ed }= 3.85 KN

**Verification of Bending**

Design moment resistance

M_{C,Rd} = W_{eff,y}.F_{y} / γ_{m0} [Clause 6.1.4.1(1) of EN 1993-1-3:2006]

From our calculations;

W_{eff,y} = min(W_{eff,y,c }, W_{eff,y,t}) = 12767 mm^{3}

Therefore;

M_{C,Rd} = W_{eff,y}.F_{y}/γ_{m0} = [(12767 mm^{3}) × (350) × 10^{-6}] / 1.0 = 4.46845 KNm

Check of span and single support;

M_{Ed}/M_{C,Rd} = 2.31/4.46845 = 0.5169 < 1.0 OK!

**Check of shear resistance at ULS**** **The design shear resistance is given by;

_{}_{Vb,Rd} = (h_{w}/sinφ.t.f_{bv}) / γ_{M0} (Clause 6.1.5)

The shear buckling strength (f_{bv}) which is based on the relative web slenderness can be obtained from the table below (Table 6.1 of EN 1993-1-3).

Where λ_{w} ̅ is the relative slenderness for webs without longitudinal stiffeners.

Since 1.154 < 0.83 but less than 1.40;

f_{bv} = 0.48f_{yb} = 0.48 × 350 = 168 N/mm^{2}

V_{b,Rd} = [(118.5 / sin90°) × 1.45 × 168] / 1.0 = 28866.6 N = 28.867 KN

Check for shear (using maximum shear force);

V_{Ed }/ V_{b,Rd} = (3.85)/(28.867) = 0.1333 < 1.0 Shear is ok

**Deflection Check**

Maximum deflection under SLS (1.0gk + 1.0qk) = 0.07mm

Limiting deflection = L/200 = 3000/200 = 15 mm

Since 0.07 < 15mm, deflection is OK!

Therefore, the channel Z120-15 section is adequate for the applied load.

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