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

# Wind pressure on pitched roof on hill

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

- Asked in: 2081 Chaitra, 8 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-pitched-roof-eaves-10m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8

## Question

Determine wind pressure on pitched roof of a building having height of eaves 10 m, rise of pitched roof 3.5 m, length of building 30 m and width of building 10 m. Building is located in dense forest of a hill having average hill slope 1:4, height of hill is 450 m and building is located at mid height of hill. Consider normal opening of building.

## Short answer

Design wind pressure is 1.26 kN/m² from a speed of 45.8 m/s. Net design pressure on the roof reaches 1.26 kN/m² suction, giving a force of 230.2 kN per slope.

## Answer

**Assumptions (not all data are given).** $V_b$ is not stated; **47 m/s** is assumed. A dense forest with closely spaced trees is **terrain Category 3**. Greatest dimension $= 30$ m, so **Class B**. Design life is taken as 50 years ($k_1 = 1.0$). Roof: width $w = 10$ m, rise 3.5 m, so $\tan\alpha = 3.5/5 = 0.7$, $\alpha = 35.0^\circ$; $h/w = 10/10 = 1.0$. Height for $k_2$ = mean roof height $= 10 + 3.5/2 = 11.75$ m. Normal openings: $C_{pi} = \pm0.2$.

**Topography factor $k_3$ (Appendix C).** Hill slope 1:4, so $\theta = \tan^{-1}(1/4) = 14.0^\circ$, which lies between 3° and 17°. Height of hill $Z = 450$ m, so the upwind slope length is $L = Z/\tan\theta = 1800$ m and $L_e = L = 1800$ m. $C = 1.2Z/L = 1.2\times450/1800 = 0.30$. The building is at mid-height on the upwind slope, i.e. $X = -L/2 = -900$ m (upwind of the crest), so $X/L_e = -0.5$, and $H/L_e \approx 12/1800 \approx 0.007$. From Fig. 15 (hill and ridge, upwind section), $s \approx 0.27$ (read from the chart, approximate). So $k_3 = 1 + Cs = 1 + 0.30\times0.27 = 1.081$.

#### 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 3, Class B, at height 11.75 m, interpolating between 10 m (0.88) and 15 m (0.94): $k_2 = 0.901$ (Table 2).
- Topography factor $k_3 = 1.081$ (hill, Appendix C).

$$
\begin{aligned} V_z &= 47\times1.00\times0.901\times1.081 = 45.78\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times45.78^2 = 1257\ \text{N/m}^2 = 1.257\ \text{kN/m}^2 \end{aligned}
$$

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

Height ratio $h/w = 1.00$ (band $1/2 < h/w \le 3/2$), roof angle $\alpha = 35.0^\circ$. Values are interpolated between $\alpha = 30^\circ$ and $45^\circ$.

Internal pressure coefficient $C_{pi} = \pm0.2$ (normal openings), 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°) | -0.07 | -0.27 | +0.13 | -335 | +168 |
| Leeward slope GH (wind 0°) | -0.50 | -0.70 | -0.30 | -880 | -377 |
| Slope EG (wind 90°) | -0.80 | -1.00 | -0.60 | -1257 | -754 |
| Slope FH (wind 90°) | -0.80 | -1.00 | -0.60 | -1257 | -754 |

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 $p_d = -1257$ N/m² $= -1.257$ kN/m².

#### Step 3: Wind force on the roof

$F = (C_{pe} - C_{pi})\,A\,p_z$, acting normal to the roof surface. The most severe net coefficient, $-1.00$, is used for the maximum force.

- One roof slope (slope length $\sqrt{5^2 + 3.5^2} = 6.10$ m × 30 m): $A = 183.1$ m², $F = 1.00\times183.1\times1.257 = 230.2$ kN (suction, i.e. acting away from the roof).

#### Answer

- Design wind speed $V_z = 45.8$ m/s; design wind pressure $p_z = 1257$ N/m² (1.26 kN/m²).
- Net design pressure on the roof: up to $1257$ N/m² (1.26 kN/m²), suction.
- Force on one roof slope (6.10 m × 30 m): 230.2 kN.

## Check yourself

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

1. The building stands in a dense forest with closely spaced trees. Which terrain category is used for k2?
   A. Category 3
   B. Category 1
   C. Category 2
   D. Category 4
   Answer: A. Terrain with numerous closely spaced obstructions (dense forest) is Category 3. (Key point: Terrain category)

2. At what height is k2 read, and what value results?
   A. Eaves height 10 m; k2 = 0.88
   B. Mean roof height 11.75 m (eaves + half the rise); k2 = 0.901 by interpolation
   C. Ridge height 13.5 m; k2 = 0.92
   D. Hill height 450 m; k2 = 1.0
   Answer: B. k2 is taken at the mean roof height 10 + 3.5/2 = 11.75 m, interpolating between 0.88 at 10 m and 0.94 at 15 m. (Key point: Height used for k2)

3. How is the topography factor k3 found for the hill site?
   A. k3 = 1 because the building is below the crest
   B. k3 = C + s with C = 0.36
   C. k3 = 1 + C s, with C = 1.2Z/L = 0.30 for a slope of 3 to 17 degrees and s read from the chart
   D. k3 = Z/L = 0.25 directly
   Answer: C. For slopes between 3 and 17 degrees C = 1.2Z/L, and k3 = 1 + Cs = 1 + 0.30 x 0.15 = 1.045. (Key point: Topography factor)

4. What are Vz and pz after combining the factors?
   A. Vz = 47 m/s; pz = 1325 N/m^2
   B. Vz = 44.25 m/s; pz = 26.5 N/m^2
   C. Vz = 42.3 m/s; pz = 1073 N/m^2
   D. Vz = 44.25 m/s; pz = 1175 N/m^2
   Answer: D. Vz = 47 x 1.00 x 0.901 x 1.045 = 44.25 m/s and pz = 0.6 x 44.25^2 = 1175 N/m^2. (Key point: Design pressure)

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