---
title: "Purlin for roof truss, 10 m span"
subject: "Design of Steel and Timber Structure"
chapter: 8
key: "semester_6/design_of_steel_and_timber_structure:purlin-roof-truss-10m-span"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/purlin-roof-truss-10m-span
site: BCT Engineering
---

# Purlin for roof truss, 10 m span

Design of Steel and Timber Structure · Chapter 8: Design of Roof Trusses

- Asked in: 2068 Magh (old course), 10 marks
- Video: [Design of I section purlin | Design of purlin | Design of purlin in steel structure](https://www.youtube.com/watch?v=agOhq23leAk) — Civil learning online
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/purlin-roof-truss-10m-span
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8

## Question

Design a purlin using suitable section for a roof-truss. Span of roof = 10m, spacing of truss is 4m and pitch is 1/4. Assume vertical load of 500N/m (including self wt) and wind load 2 KN/m.

## Short answer

Using IS 800:2007, try ISMB 100 purlin on 4 m span with sag rod at mid-span. Biaxial bending from vertical and wind loads gives maximum interaction ratio 0.98, shear 7.34 kN is far below capacity, and deflection 18.2 mm is within L/150. Provide ISMB 100 with 12 mm sag rod and two M12 bolts at each end.

## Answer

**Approach.** Bi-axial bending of a simply supported purlin, IS 800:2007.

#### Data

- Steel Fe 410 (E250): $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$. Purlin span $L = 4.0$ m (truss spacing); inclination of rafter $\alpha = 26.57^\circ$ ($\cos\alpha = 0.894$, $\sin\alpha = 0.447$).
- Roof span 10 m, pitch 1/4 (rise/span), so $\tan\alpha = 2.5/5 = 0.5$. Truss spacing (purlin span) 4 m. The vertical load of $500$ N/m (including self-weight) and the wind load of $2$ kN/m are taken as loads per metre of purlin. A sag rod at mid-span is assumed.

#### Loads per metre run of purlin

- Vertical load (including self-weight) $= 0.50$ kN/m; no separate live load is given.
- Wind load $= 2.00$ kN/m acting **normal** to the roof (taken as pressure for the downward case and as suction for the uplift case).

#### Trial section and capacities

Try **ISMB 100** (IS 808:1989 values): $D = 100$, $b_f = 50$, $t_w = 4.7$, $t_f = 7.0$ mm, $I_{zz} = 183$ cm⁴, $I_{yy} = 12.9$ cm⁴, $Z_{ez} = 36.6$ cm³, mass 8.9 kg/m (self-weight is already included in the given load). Older steel tables (SP 6(1)) list ISMB 100 as a larger section (75 mm flange, 11.5 kg/m, $Z_{ez} = 51.5$ cm³); if those values are used, every check below is met with a bigger margin.

$$
\begin{aligned} Z_{pz} &= b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 41.2\ \text{cm}^3,\quad M_{dz} = \frac{Z_{pz}f_y}{\gamma_{m0}} = 9.37\ \text{kN·m} \\ Z_{py} &= 2\cdot\frac{t_fb_f^2}{4} + \frac{(D - 2t_f)t_w^2}{4} = 9.2\ \text{cm}^3,\quad M_{dy} = \min(Z_{py}, 1.2Z_{ey})\frac{f_y}{\gamma_{m0}} = 1.41\ \text{kN·m} \end{aligned}
$$

Section class: $b/t_f$ and $d/t_w$ are within the plastic limits ($9.4$ and $84$).

#### Load combinations and moments (IS 800 Table 4)

Normal component $w_n$ and tangential component $w_t$ (kN/m), with $M_z = w_nL^2/8$ and $M_y = w_t(L')^2/8$, where $L' = L/2$ (one sag rod at mid-span):

| Combination | $w_n$ | $w_t$ | $M_z$ (kN·m) | $M_y$ (kN·m) | $M_{dz}$ used | $M_z/M_{dz} + M_y/M_{dy}$ |
|---|---|---|---|---|---|---|
| 1.2(DL+LL+WL) | +2.94 | 0.27 | 5.87 | 0.13 | 9.37 | 0.72 |
| 1.5(DL+WL) | +3.67 | 0.34 | 7.34 | 0.17 | 9.37 | 0.90 |
| 0.9DL+1.5WL (uplift) | -2.60 | 0.20 | 5.20 | 0.10 | 5.74 | 0.98 |

For the uplift case the bottom flange is in compression and is unrestrained between sag rods ($L_{LT} = L/2$), so the lateral-torsional buckling capacity is used for $M_{dz}$. The greatest ratio is 0.98 $\le 1.0$. **Safe in bi-axial bending.**

**Shear:** $V_{max} = 7.34$ kN $\ll V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 61.7$ kN. OK.

**Deflection** (unfactored wind load, limit span/150 for elastic sheeting, Table 6): $\delta = 5wL^4/(384EI) = 18.21$ mm $< L/150 = 26.7$ mm. OK.

#### Result
Provide an **ISMB 100** purlin, simply supported on the rafters, with a sag rod (12 mm) at mid-span and cleat connections with two M12 bolts at each end.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the purlin span and rafter inclination?
   A. Span = truss spacing 4 m; tan(alpha) = 2.5/5 = 0.5, alpha = 26.57 degrees
   B. Span = 10 m; alpha = 14 degrees
   C. Span = 4 m; alpha = 14 degrees
   D. Span = 5 m; alpha = 26.57 degrees
   Answer: A. The purlin spans between trusses (4 m), and a pitch of 1/4 means rise 2.5 m over half-span 5 m. (Key point: Purlin span and slope)

2. Which load combination controls the design?
   A. 1.2(DL + LL + WL), ratio 0.72
   B. 0.9DL + 1.5WL (uplift), with ratio 0.98
   C. 1.5(DL + WL), ratio 0.90
   D. 1.5(DL + LL), ratio 0.35
   Answer: B. Uplift gives a lower Mdz (LTB) and the highest interaction ratio, 0.98 <= 1. (Key point: Governing combination)

3. How is the uplift normal load found?
   A. wn = 1.5 x 2.00 + 0.9 x 0.447 = 3.40 kN/m
   B. wn = 2.00 - 0.50 = 1.50 kN/m
   C. wn = 1.5 x 2.00 - 0.9 x 0.447 = 2.60 kN/m upward
   D. wn = 1.5 x 2.00 = 3.00 kN/m
   Answer: C. Dead load acts against uplift, so only 0.9 DL cos(alpha) = 0.40 offsets 1.5WL = 3.0. (Key point: Uplift normal load)

4. Why is Mdz reduced to 5.74 kN.m in the uplift case?
   A. The yield stress falls under wind
   B. The section is slender
   C. The plastic modulus is limited to 1.2 Ze
   D. The bottom flange is in compression and unbraced over L/2, so LTB governs
   Answer: D. With the free flange in compression between sag rods, lateral-torsional buckling gives a lower design moment than Zpz fy/gamma_m0. (Key point: Lateral-torsional buckling)

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