---
title: "Max wind force on roof truss, 50 yr, hill crest"
subject: "Design of Steel and Timber Structure"
chapter: 8
key: "semester_6/design_of_steel_and_timber_structure:wind-max-force-roof-truss-kathmandu"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-max-force-roof-truss-kathmandu
site: BCT Engineering
---

# Max wind force on roof truss, 50 yr, hill crest

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

- Asked in: 2078 Chaitra, 8 marks
- 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-max-force-roof-truss-kathmandu
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8

## Question

A roof truss building (20m ×30m) has height 5m up to eaves level and rise of a truss is 2m and is designed for the highly-dense city (Kathmandu). The minimum lifetime of the structure is 50 years. The site should be located 15m above the mean ground level and 100m far. If the crest level of the ridge is 200m from mean ground level, determine the maximum wind force on the roof truss.

## Short answer

Design wind pressure is 1.20 kN/m² (speed 44.7 m/s). The maximum wind force on one roof truss at 5 m spacing is 158.5 kN suction.

## Answer

**Assumptions (data not fully given).** Kathmandu: $V_b = 47$ m/s. A "highly dense city" is **terrain Category 4**. Plan 20 m × 30 m, so the largest dimension is 30 m: **Class B**. Life 50 years: $k_1 = 1.0$. Roof: span 20 m, rise 2 m, so $\tan\alpha = 2/10$, $\alpha = 11.3^\circ$ and $h/w = 5/20 = 0.25$. Normal permeability ($C_{pi} = \pm0.2$). Height for $k_2$: eaves 5 m plus half the rise = 6 m, which is below 10 m, so $k_2$ is the value at 10 m. The site is 15 m above the mean ground level and 100 m from the crest of a ridge 200 m high. The slope of the ridge is not given and is **assumed steeper than 17°**, so (Appendix C) $C = 0.36$ and $L_e = Z/0.3 = 667$ m. The site is upwind of the crest: $X/L_e = -100/667 = -0.15$; $H/L_e = 15/667 = 0.022$; from Fig. 15, $s \approx 0.7$ (read from the chart, approximate). Truss spacing is not given; **5 m** is assumed (6 bays along 30 m).

#### 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 4, Class B, at height 6 m (use 10 m): $k_2 = 0.76$ (constant up to 10 m) (Table 2).
- Topography factor $k_3 = 1 + Cs = 1 + 0.36\times0.7 = 1.252$ (ridge).

$$
\begin{aligned} V_z &= 47\times1.00\times0.760\times1.252 = 44.72\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times44.72^2 = 1200\ \text{N/m}^2 = 1.200\ \text{kN/m}^2 \end{aligned}
$$

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

Height ratio $h/w = 0.25$, roof angle $\alpha = 11.3^\circ$. Values are interpolated between $\alpha = 10^\circ$ and $20^\circ$.

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°) | -1.10 | -1.30 | -0.90 | -1554 | -1074 |
| Leeward slope GH (wind 0°) | -0.40 | -0.60 | -0.20 | -720 | -240 |
| Slope EG (wind 90°) | -0.79 | -0.99 | -0.59 | -1184 | -704 |
| Slope FH (wind 90°) | -0.60 | -0.80 | -0.40 | -960 | -480 |

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 the windward slope EF (wind 0°), with $C_{pe} - C_{pi} = -1.295$ (shown as $-1.30$ in the table), so $p_d = -1554$ N/m² $= -1.554$ kN/m².

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

$F = (C_{pe} - C_{pi})\,A\,p_z$, acting normal to the roof surface. The most severe net coefficient, $-1.295$, is used for the maximum force.

- Whole roof, 20 m × 30 m plan (slope length 10.20 m, both slopes): $A = 611.9$ m², $F = 1.295\times611.9\times1.200 = 951.0$ kN (suction, i.e. acting away from the roof).

- One truss at 5 m spacing (both slopes): $A = 102$ m², $F = 1.295\times102\times1.200 = 158.5$ kN (suction, i.e. acting away from the roof).

#### Answer

- Design wind speed $V_z = 44.7$ m/s; design wind pressure $p_z = 1200$ N/m² (1.20 kN/m²).
- Net design pressure on the roof: up to $1554$ N/m² (1.55 kN/m²), suction.
- Whole roof, 20 m × 30 m plan (slope length 10.20 m, both slopes): 951.0 kN.
- One truss at 5 m spacing (both slopes): 158.5 kN.

## Check yourself

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

1. Why is Category 4 chosen for Kathmandu here?
   A. The site is a highly dense city with closely spaced tall structures
   B. It is open terrain with few obstructions
   C. It is a hill site
   D. The life is 50 years
   Answer: A. Category 4 covers dense urban areas with large, tall, closely spaced obstructions. (Key point: Terrain category)

2. Mean roof height is 6 m. How is k2 chosen?
   A. k2 is interpolated down to 0 at ground level
   B. The 10 m value is used (k2 = 0.76), as k2 is constant up to 10 m
   C. k2 = 1.0 for any height below 10 m
   D. k2 is read at the crest of the ridge
   Answer: B. Table 2 gives a constant k2 for heights up to 10 m, so the 10 m value applies. (Key point: k2 below 10 m)

3. For the site near the ridge crest, how is k3 evaluated?
   A. k3 = 1 because the site is on the plain
   B. k3 = 0.36 x 0.7 = 0.252
   C. k3 = 1 + C s = 1 + 0.36 x 0.7 = 1.252
   D. k3 = 1 - Cs = 0.748
   Answer: C. Slopes over 17 degrees use C = 0.36, and with s = 0.7 the speed is raised by the factor 1.252. (Key point: Ridge topography factor)

4. Which net coefficient is used for the maximum wind force on the roof, and what is the force formula?
   A. Slope EG, Cpe - Cpi = -0.99; F = Cp A pz
   B. Leeward slope GH, -0.60; F = Cp A pz
   C. Cpi only, +0.2; F = Cpi A pz
   D. Windward slope EF, Cpe - Cpi = -1.30; F = (Cpe - Cpi) A pz as suction
   Answer: D. Because the roof is low-pitched, the windward slope has the strongest suction (-1.10 external, -1.30 net), applied over the area with pz = 1.2 kN/m^2. (Key point: Maximum force on the roof)

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