---
title: "I-section purlin for industrial building"
subject: "Design of Steel and Timber Structure"
chapter: 8
key: "semester_6/design_of_steel_and_timber_structure:purlin-i-section-industrial"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/purlin-i-section-industrial
site: BCT Engineering
---

# I-section purlin for industrial building

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

- Asked in: 2079 Chaitra, 8 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-i-section-industrial
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8

## Question

Design an I-section purlin for an industrial building to support a galvanized corrugated iron sheet roof from the following data:
Spacing of the trusses = 5 m
Spacing of purlins = 1.5 m
Inclination of main rafter to horizontal = 30°
Weight of galvanized sheets taking into account laps and connecting bolts = 130 N/m²
Wind load = 1 kN/m²

## Short answer

Provide an ISMB 125 purlin with a sag rod at mid-span. The section satisfies biaxial bending, shear, and deflection; use cleat connections with two M12 bolts at each end.

## Answer

**Approach.** The purlin is a simply supported beam bent about both axes, because the loads are inclined to its principal axes. Design to IS 800:2007 (cl. 8.2 and 9.3), with loads from IS 875.

#### Data

- Steel Fe 410 (E250): $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$. Purlin span $L = 5.0$ m (truss spacing); inclination of rafter $\alpha = 30.00^\circ$ ($\cos\alpha = 0.866$, $\sin\alpha = 0.500$).
- Sag rods are provided at mid-span to restrain the purlin in the plane of the roof. The 1 kN/m² wind load is taken as the net design pressure normal to the roof (acting either towards or away from the roof). Spacing of purlins along the slope = 1.5 m; trusses at 5 m.

#### Loads per metre run of purlin

- Sheeting: $0.130\times1.5 = 0.195$ kN/m (dead, vertical), plus the self-weight of the purlin (trial section below).
- Live load (IS 875 Part 2, roof slope above 10°): $0.75 - 0.02(\alpha - 10) = 0.35$ kN/m², but not less than 0.40 kN/m² on plan, so use $0.40$ kN/m². Per metre of purlin: $0.40\times1.5\cos\alpha = 0.520$ kN/m (vertical).
- Wind: given wind pressure $1.0$ kN/m² $\times 1.5$ m $= 1.500$ kN/m, acting **normal** to the roof.

#### Trial section and capacities

Try **ISMB 125**: $D = 125$, $b_f = 70$, $t_w = 5.0$, $t_f = 8.0$ mm, $I_{zz} = 445$ cm⁴, $I_{yy} = 38.5$ cm⁴, $Z_{ez} = 71.2$ cm³, mass 13.3 kg/m (self-weight $0.130$ kN/m).

$$
\begin{aligned} Z_{pz} &= b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 80.4\ \text{cm}^3,\quad M_{dz} = \frac{Z_{pz}f_y}{\gamma_{m0}} = 18.27\ \text{kN·m} \\ Z_{py} &= 2\cdot\frac{t_fb_f^2}{4} + \frac{(D - 2t_f)t_w^2}{4} = 20.3\ \text{cm}^3,\quad M_{dy} = \min(Z_{py}, 1.2Z_{ey})\frac{f_y}{\gamma_{m0}} = 3.00\ \text{kN·m} \end{aligned}
$$

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 | $M_z/M_{dz} + M_y/M_{dy}$ |
|---|---|---|---|---|---|---|
| 1.5(DL+LL) | +1.10 | 0.63 | 3.43 | 0.50 | 18.27 | 0.35 |
| 1.2(DL+LL+WL) | +2.68 | 0.51 | 8.37 | 0.40 | 18.27 | 0.59 |
| 1.5(DL+WL) | +2.67 | 0.24 | 8.35 | 0.19 | 18.27 | 0.52 |
| 0.9DL+1.5WL (uplift) | -2.00 | 0.15 | 6.24 | 0.11 | 11.36 | 0.59 |

For the uplift case the bottom flange is in compression and unrestrained between sag rods ($L_{LT} = L/2$), so the lateral-torsional buckling capacity is used for $M_{dz}$. The greatest ratio is 0.59 $\le 1.0$. **Safe in biaxial bending.**

**Shear:** $V_{max} = 6.70$ kN $\ll V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 82.0$ kN. OK.

**Deflection** (unfactored wind load of 1.5 kN/m normal to the roof, limit span/150 for elastic sheeting, Table 6): $\delta = 5wL^4/(384EI) = 13.72$ mm $< L/150 = 33.3$ mm. OK.

#### Result
Provide an **ISMB 125** 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. Why is a live load of 0.40 kN/m^2 used instead of the calculated 0.35 kN/m^2?
   A. The code gives a minimum roof live load of 0.40 kN/m^2
   B. The roof slope is under 10 degrees
   C. Sheeting weight is added to it
   D. Wind load replaces the calculated value
   Answer: A. 0.75 - 0.02(30 - 10) = 0.35 kN/m^2 is below the minimum, so 0.40 kN/m^2 governs. (Key point: Live load on the purlin)

2. How is the 1 kN/m^2 wind pressure applied to the purlin?
   A. As a vertical load resolved with cos(alpha) and sin(alpha)
   B. As 1.0 x 1.5 = 1.5 kN/m acting normal to the roof
   C. As a horizontal load of 1.5 kN/m
   D. As a load along the roof slope only
   Answer: B. Wind acts normal to the roof surface, so it adds only to the normal component w_n. (Key point: Wind load direction)

3. How is the minor-axis moment My calculated?
   A. My = w_t L^2/8 with L = 5 m
   B. My = w_n L^2/8
   C. My = w_t (L')^2/8 with L' = L/2 because of the mid-span sag rod
   D. My = w_t L^2/2
   Answer: C. The sag rod at mid-span halves the effective span in the roof plane, so L' = 2.5 m. (Key point: Minor-axis moment)

4. Why is the uplift case checked with a reduced Mdz (11.36 instead of 18.27 kN.m)?
   A. Wind factors reduce the yield stress
   B. The section becomes slender in uplift
   C. Shear and bending interact in uplift
   D. The bottom flange is in compression and unrestrained, so lateral-torsional buckling reduces capacity
   Answer: D. With the free bottom flange in compression over L/2, LTB governs and the design moment is lower. (Key point: Governing uplift check)

## Related in Design of Steel and Timber Structure

- [Purlin for roof truss, 10 m span](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Apurlin-roof-truss-10m-span) — Design of Steel and Timber Structure, chapter 8
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