---
title: "Wind pressure on sloping roof, Kathmandu"
subject: "Design of Steel and Timber Structure"
chapter: 8
key: "semester_6/design_of_steel_and_timber_structure:wind-sloping-roof-kathmandu-10m-9m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-sloping-roof-kathmandu-10m-9m
site: BCT Engineering
---

# Wind pressure on sloping roof, Kathmandu

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

- Asked in: 2076 Baisakh, 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-kathmandu-10m-9m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8

## Question

Find the wind pressure for design of a sloping roof of span 10m and pitch ¼. The height of eaves is 9m above the ground. The building is situated in Kathmandu, where the basic wind speed is found to be 47 m/sec. and permeability is normal.

## Short answer

Using IS 875 with Category 2, Class A, Vz = 47 x 1.0 x 1.002 x 1.0 = 47.1 m/s and pz = 0.6Vz^2 = 1.33 kN/m^2. For alpha = 26.6 degrees, h/w = 0.9 and Cpi = +/-0.2, the worst net coefficient is -1.00, so design pressure is 1.33 kN/m^2 suction.

## Answer

**Assumptions.** Span 10 m, pitch $1/4$, so rise 2.5 m, $\tan\alpha = 0.5$, $\alpha = 26.57^\circ$. Eaves at 9 m: $h/w = 9/10 = 0.9$; mean roof height $9 + 1.25 = 10.25$ m. Kathmandu: $V_b = 47$ m/s (given). Terrain is not stated: **Category 2** (open terrain), **Class A** (span below 20 m); $k_1 = 1$, $k_3 = 1$. Permeability normal: $C_{pi} = \pm0.2$.

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

$$
V_z = V_b\,k_1k_2k_3
$$

- Basic wind speed $V_b = 47$ 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 10.25 m: interpolating between 10 m (1.00) and 15 m (1.05): $k_2 = 1.002$ (Table 2).
- Topography factor $k_3 = 1.000$ (flat site).

$$
\begin{aligned} V_z &= 47\times1.00\times1.002\times1.000 = 47.12\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times47.12^2 = 1332\ \text{N/m}^2 = 1.332\ \text{kN/m}^2 \end{aligned}
$$

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

Height ratio $h/w = 0.90$, 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 | -762 | -229 |
| Leeward slope GH (wind 0°) | -0.50 | -0.70 | -0.30 | -932 | -400 |
| Slope EG (wind 90°) | -0.80 | -1.00 | -0.60 | -1332 | -799 |
| Slope FH (wind 90°) | -0.73 | -0.93 | -0.53 | -1241 | -708 |

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 $p_d = -1332$ N/m² $= -1.332$ kN/m².

#### Answer

- Design wind speed $V_z = 47.1$ m/s; design wind pressure $p_z = 1332$ N/m² (1.33 kN/m²).
- Net design pressure on the roof: up to $1332$ N/m² (1.33 kN/m²), suction.

## Check yourself

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

1. Why is the building Class A?
   A. Its greatest dimension (span below 20 m) is under 20 m
   B. It is in Kathmandu
   C. The eaves are 9 m high
   D. The terrain is Category 2
   Answer: A. Class A covers buildings whose greatest dimension is less than 20 m. (Key point: Class A)

2. The mean roof height is 10.25 m, slightly above 10 m. What is k2?
   A. 1.00 exactly because the roof is below 15 m
   B. About 1.002, interpolated between 1.00 at 10 m and 1.05 at 15 m
   C. 1.05 from the 15 m row
   D. 0.93 from Class B
   Answer: B. k2 = 1.00 + 0.05 x (0.25/5) = 1.002. (Key point: k2 at mean roof height)

3. What is the design wind pressure pz?
   A. 1325 N/m^2
   B. 1109 N/m^2
   C. 1332 N/m^2
   D. 47 N/m^2
   Answer: C. Vz = 47.12 m/s so pz = 0.6 x 47.12^2 = 1332 N/m^2. (Key point: Pressure)

4. Which surface gives the largest roof pressure and what is the net coefficient?
   A. Windward slope EF, -0.57
   B. Leeward slope GH, -0.70
   C. Slope FH, -0.53
   D. Slope EG for wind at 90 degrees, net coefficient -1.00 (suction of 1332 N/m^2)
   Answer: D. Cpe = -0.80 on EG with Cpi = +0.2 gives -1.00, so pd = -1.00 x 1.332 = -1.332 kN/m^2. (Key point: Severe net pressure)

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