---
title: "Wind pressure on sloping roof, Madras"
subject: "Design of Steel and Timber Structure"
chapter: 8
key: "semester_6/design_of_steel_and_timber_structure:wind-sloping-roof-madras"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-sloping-roof-madras
site: BCT Engineering
---

# Wind pressure on sloping roof, Madras

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

- Asked in: 2071 Bhadra, 6 marks
- Video: [Calculation of Wind load  | Design of steel structures and timber | IOE III/II PU MU |](https://www.youtube.com/watch?v=32ze2SwlHXk) — Parash Joshi - Civil Construction and Tutor
- 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/wind-sloping-roof-madras
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8

## Question

Find the design wind pressure on a sloping roof of span 10 m and pitch ¼. The height of the eves is 6 m above ground. The building is situated in Madras and its permeability is normal.

## Short answer

Using IS 875, Madras Vb = 50 m/s; with Category 2, Class A and height below 10 m, Vz = 50 m/s and pz = 0.6Vz^2 = 1.50 kN/m^2. For alpha = 26.6 degrees, h/w = 0.6 and Cpi = +/-0.2, the worst net coefficient is -1.00, giving 1.50 kN/m^2 suction.

## Answer

**Data and assumptions.** Span 10 m, pitch $1/4$: rise 2.5 m, $\alpha = 26.57^\circ$. Eaves height $h = 6$ m, $w = 10$ m, $h/w = 0.6$. Madras: $V_b = 50$ m/s (IS 875 Part 3, Appendix A). Terrain Category 2, Class A (building below 20 m); mean roof height $= 7.25$ m, below 10 m. $k_1 = 1$, $k_3 = 1$ (not given). Normal permeability: $C_{pi} = \pm0.2$.

#### Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3)

$$
V_z = V_b\,k_1k_2k_3
$$

- Basic wind speed $V_b = 50$ m/s.
- Risk coefficient $k_1 = 1.00$ (50-year life; Table 1).
- Terrain, height and size factor $k_2$: Category 2, Class A, at height 7.25 m: $k_2 = 1.00$ (constant up to 10 m) (Table 2).
- Topography factor $k_3 = 1.000$ (flat site).

$$
\begin{aligned} V_z &= 50\times1.00\times1.000\times1.000 = 50.00\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times50.00^2 = 1500\ \text{N/m}^2 = 1.500\ \text{kN/m}^2 \end{aligned}
$$

#### Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building)

Height ratio $h/w = 0.60$, roof angle $\alpha = 26.6^\circ$. Values are interpolated between $\alpha = 20^\circ$ and $30^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3.

| Surface | $C_{pe}$ | $C_{pe} - C_{pi}$ ($C_{pi}=+0.2$) | $C_{pe} - C_{pi}$ ($C_{pi}=-0.2$) | $p_d$ (N/m²), $C_{pi}=+$ | $p_d$ (N/m²), $C_{pi}=-$ |
|---|---|---|---|---|---|
| Windward slope EF (wind 0°) | -0.37 | -0.57 | -0.17 | -858 | -258 |
| Leeward slope GH (wind 0°) | -0.50 | -0.70 | -0.30 | -1050 | -450 |
| Slope EG (wind 90°) | -0.80 | -1.00 | -0.60 | -1500 | -900 |
| Slope FH (wind 90°) | -0.73 | -0.93 | -0.53 | -1397 | -797 |

Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is Slope EG (wind 90°), with $C_{pe} - C_{pi} = -1.00$, so $p_d = -1500$ N/m² $= -1.500$ kN/m².

#### Answer

- Design wind speed $V_z = 50.0$ m/s; design wind pressure $p_z = 1500$ N/m² (1.50 kN/m²).
- Net design pressure on the roof: up to $1500$ N/m² (1.50 kN/m²), suction.

## Check yourself

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

1. What basic wind speed is taken for Madras?
   A. 50 m/s
   B. 47 m/s
   C. 39 m/s
   D. 60 m/s
   Answer: A. IS 875 Part 3 Appendix A gives Vb = 50 m/s for Madras. (Key point: Basic wind speed)

2. With mean roof height 7.25 m (Category 2, Class A), what is k2?
   A. 0.93
   B. 1.00, constant up to 10 m
   C. 1.05
   D. 0.725 by proportion to height
   Answer: B. k2 does not change below 10 m, so the 10 m value 1.00 is used. (Key point: k2 for low heights)

3. What are Vz and pz?
   A. Vz = 50 m/s; pz = 30 N/m^2
   B. Vz = 47 m/s; pz = 1325 N/m^2
   C. Vz = 50 m/s; pz = 1500 N/m^2
   D. Vz = 50 m/s; pz = 2500 N/m^2
   Answer: C. With all factors 1.00, Vz = 50 m/s and pz = 0.6 x 50^2 = 1500 N/m^2. (Key point: Pressure calculation)

4. What is the maximum net suction on the roof?
   A. 858 N/m^2 on slope EF
   B. 1050 N/m^2 on slope GH
   C. 900 N/m^2 on slope EG
   D. 1500 N/m^2 on slope EG (wind at 90 degrees), net coefficient -1.00
   Answer: D. Cpe = -0.80 and Cpi = +0.2 give -1.00, so pd = -1.00 x 1500 = -1500 N/m^2. (Key point: Net design pressure)

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