---
title: "Wind pressure on sloping roof, Delhi"
subject: "Design of Steel and Timber Structure"
chapter: 8
key: "semester_6/design_of_steel_and_timber_structure:wind-sloping-roof-delhi-10m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-sloping-roof-delhi-10m
site: BCT Engineering
---

# Wind pressure on sloping roof, Delhi

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

- Asked in: 2070 Bhadra, 6 marks; 2069 Bhadra, 6 marks (asked 2 times)
- 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-delhi-10m
- 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 meter and pitch 1/4. The height of eaves is 5 meter above ground. The building is situated in Delhi and its permeability is normal. (Assume $K_1$ = 1; $K_2$ = 0.8; $K_3$ = 1.)

## Short answer

Vz = 47×1×0.8×1 = 37.6 m/s; pz = 0.6×37.6² = 848 N/m². For span 10 m, eaves 5 m, α=26.6°, h/w=0.5, normal permeability: net coefficient -0.90 (wind 90°, slope EG). Net pressure = 0.90×848 = 763 N/m² (0.763 kN/m²) suction.

## Answer

**Data.** Span 10 m, pitch $1/4$ (rise/span), so rise $= 2.5$ m and $\tan\alpha = 2.5/5 = 0.5$, $\alpha = 26.57^\circ$. Eaves height $h = 5$ m; take the lesser plan dimension $w = 10$ m, so $h/w = 0.5$. Delhi: $V_b = 47$ m/s (IS 875 Part 3, Appendix A). Given: $k_1 = 1$, $k_2 = 0.8$, $k_3 = 1$. Permeability normal.

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

$$
V_z = V_b\,k_1k_2k_3
$$

- Basic wind speed $V_b = 47$ m/s.
- Risk coefficient $k_1 = 1.00$ (given).
- Terrain, height and size factor $k_2 = 0.8$ (given).
- Topography factor $k_3 = 1.000$ (given).

$$
\begin{aligned} V_z &= 47\times1.00\times0.800\times1.000 = 37.60\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times37.60^2 = 848\ \text{N/m}^2 = 0.848\ \text{kN/m}^2 \end{aligned}
$$

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

Height ratio $h/w = 0.50$ (band $h/w \le 1/2$), 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°) | -0.14 | -0.34 | +0.06 | -286 | +53 |
| Leeward slope GH (wind 0°) | -0.40 | -0.60 | -0.20 | -509 | -170 |
| Slope EG (wind 90°) | -0.70 | -0.90 | -0.50 | -763 | -424 |
| Slope FH (wind 90°) | -0.60 | -0.80 | -0.40 | -679 | -339 |

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 $C_{pe} - C_{pi} = -0.90$, so $p_d = -763$ N/m² $= -0.763$ kN/m².

#### Answer (design wind pressure)

- Design wind speed $V_z = 37.6$ m/s; design wind pressure $p_z = 848$ N/m² (0.85 kN/m²).
- Net design pressure on the roof: up to $763$ N/m² (0.76 kN/m²), suction.

## Check yourself

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

1. For Delhi (Vb = 47 m/s) with k1 = 1, k2 = 0.8, k3 = 1, what is the design wind speed Vz?
   A. 37.6 m/s
   B. 47 m/s
   C. 58.8 m/s
   D. 30.1 m/s
   Answer: A. Vz = Vb k1 k2 k3 = 47 x 1 x 0.8 x 1 = 37.6 m/s. (Key point: Design wind speed)

2. How is the design wind pressure pz obtained from Vz, and what is its value here?
   A. pz = 0.6 Vz = 22.6 N/m^2
   B. pz = 0.6 Vz^2 = 848 N/m^2
   C. pz = 0.6 Vb^2 = 1325 N/m^2
   D. pz = Vz^2/0.6 = 2357 N/m^2
   Answer: B. pz = 0.6 x 37.6^2 = 848 N/m^2 (0.848 kN/m^2); the factor 0.6 multiplies the square of Vz. (Key point: Design wind pressure)

3. What internal pressure coefficient Cpi is used for a building of normal permeability?
   A. +/- 0.7
   B. +/- 0.5
   C. +/- 0.2
   D. 0 (no internal pressure)
   Answer: C. Normal permeability gives Cpi = +/-0.2, and both signs are checked to find the worst net pressure. (Key point: Internal pressure coefficient)

4. Which case gives the most severe net pressure on this roof, and what is its value?
   A. Windward slope EF, wind at 0 degrees: about 286 N/m^2 suction
   B. Leeward slope GH, wind at 0 degrees: about 509 N/m^2 suction
   C. Windward slope EF with Cpi = -0.2: about 53 N/m^2 pressure
   D. Slope EG, wind at 90 degrees: net coefficient -0.90, suction of about 763 N/m^2
   Answer: D. EG at 90 degrees has Cpe - Cpi = -0.90, so pd = -0.90 x 848 = -763 N/m^2, the largest suction. (Key point: Governing roof case)

## Close variants set in other exams

- [Wind pressure on sloping roof, Madras](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Awind-sloping-roof-madras) — Design of Steel and Timber Structure, chapter 8, asked 3×

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