---
title: "Design wind pressure, large openings, 100 yr"
subject: "Design of Steel and Timber Structure"
chapter: 8
key: "semester_6/design_of_steel_and_timber_structure:wind-large-openings-hill-160m"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-large-openings-hill-160m
site: BCT Engineering
---

# Design wind pressure, large openings, 100 yr

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

- Asked in: 2080 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
- Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-large-openings-hill-160m
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8

## Question

A building with large openings of size over 50 m and height 15 m is planned in a city to have a 100 years life. The building is planned to be located on a downward slope of 160 m high hill, the slope of which is 1 vertical and 3 horizontal. The building has to be constructed at a height of 100 m above the base of hill. The terrain is an open terrain with scattered obstruction, the heights of obstructions being in the range of 1.5 m to 10 m above the ground level. Determine the design wind pressure.

## Short answer

Design wind pressure is 2.17 kN/m², corresponding to a design wind speed of 60.2 m/s.

## Answer

**Assumptions (not all data are given).** Basic wind speed is not stated, so **$V_b = 47$ m/s** is assumed. Building: dimension over 50 m, so **Class C**; height 15 m. Terrain: open with scattered obstructions 1.5 to 10 m high, so **Category 2**. Design life 100 years. Large openings (more than 20% of wall area).

#### Step 1: Design wind speed
- $k_1$: for a life of 100 years and $V_b = 47$ m/s, Table 1 gives $k_1 = 1.07$.
- $k_2$: Category 2, Class C, at 15 m: interpolating between 10 m (0.93) and 15 m (0.97): $k_2 = 0.970$, i.e. $0.97$ (Table 2).
- $k_3$ (Appendix C): hill height $Z = 160$ m, slope 1 vertical to 3 horizontal, $\theta = \tan^{-1}(1/3) = 18.4^\circ > 17^\circ$, so $C = 0.36$ and $L_e = Z/0.3 = 533$ m. The building is on the **downward (leeward) slope** at 100 m above the base, i.e. 60 m below the crest, so the horizontal distance from the crest is $X = 60\times3 = 180$ m downwind: $X/L_e = 0.337$; $H/L_e = 15/533 = 0.028$. From Fig. 15, $s \approx 0.65$ (read from the chart; approximate). $k_3 = 1 + Cs = 1 + 0.36\times0.65 = 1.234$ (within the limit 1.36).

$$
\begin{aligned} V_z &= V_bk_1k_2k_3 = 47\times1.07\times0.97\times1.234 = 60.2\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times60.2^2 = 2174\ \text{N/m}^2 = 2.17\ \text{kN/m}^2 \end{aligned}
$$

#### Step 2: Net pressure with large openings
For openings above 20% of the wall area, $C_{pi} = \pm0.7$ (cl. 6.2.3.2). Taking the external coefficients of a rectangular clad building (Table 4, $h/w < 1/2$, wind 0°): windward wall $C_{pe} = +0.7$, leeward wall $-0.5$.

| Surface | $C_{pe}$ | $C_{pe} - C_{pi}$ with $C_{pi} = -0.7$ | with $C_{pi} = +0.7$ |
|---|---|---|---|
| Windward wall | +0.7 | +1.4 (3.04 kN/m²) | 0.0 |
| Leeward wall | -0.5 | +0.2 (0.43 kN/m²) | -1.2 (-2.61 kN/m²) |

#### Answer
Design wind speed $V_z = 60.2$ m/s; **design wind pressure $p_z \approx 2174$ N/m² ($\approx 2.17$ kN/m²)**. The largest net pressure on the walls is $1.4p_z \approx 3.04$ kN/m² (inward on the windward wall with the internal suction case) and the largest net suction is $1.2p_z \approx 2.61$ kN/m².

## Check yourself

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

1. Why is this building classed as Class C?
   A. Its greatest dimension exceeds 50 m
   B. Its height is 15 m
   C. It has a design life of 100 years
   D. It is built on a hill
   Answer: A. Class C applies to buildings whose greatest horizontal or vertical dimension exceeds 50 m. (Key point: Building class)

2. How does the 100-year design life affect the calculation?
   A. It lowers k1 below 1.0
   B. It raises the risk coefficient to k1 = 1.07
   C. It changes the terrain category to 3
   D. It increases Cpi to 0.7
   Answer: B. A longer design life than 50 years gives a higher risk coefficient, here k1 = 1.07 from Table 1. (Key point: Risk factor k1)

3. The slope is 1 in 3 (18.4 degrees, over 17 degrees). What constants are used for k3?
   A. C = 1.2Z/L and Le = L
   B. C = 0 because the building is on the downward slope
   C. C = 0.36 and Le = Z/0.3 = 533 m, giving k3 = 1 + Cs = 1.234
   D. C = 0.36 and Le = Z = 160 m
   Answer: C. For slopes above 17 degrees the code uses C = 0.36 and effective length Le = Z/0.3; with s = 0.65, k3 = 1.234. (Key point: Topography on a steep hill)

4. What internal pressure coefficient applies for large openings (over 20% of wall area)?
   A. Cpi = +/- 0.2
   B. Cpi = +/- 0.5
   C. Cpi = 0
   D. Cpi = +/- 0.7
   Answer: D. Openings above 20% of wall area give Cpi = +/-0.7 (cl. 6.2.3.2), much larger than the normal +/-0.2. (Key point: Internal coefficient)

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