---
title: "Warehouse exterior wall, wind 750 N/m2"
subject: "Concrete Technology and Masonry Structure"
chapter: 9
key: "semester_5/concrete_technology_and_masonry_structure:wind-wall-warehouse-750"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wind-wall-warehouse-750
site: BCT Engineering
---

# Warehouse exterior wall, wind 750 N/m2

Concrete Technology and Masonry Structure · Chapter 9: Masonry structures under lateral loads

- Asked in: 2082 Baisakh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wind-wall-warehouse-750
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-9

## Question

Design an exterior wall of a single storey warehouse of 3.5 m height. The loading on the wall consists of a vertical load of 30 kN/m from the roof and wind pressure of 750 N/m$^2$. The wall is tied with metal anchor at the floor and roof levels.

## Short answer

230 mm thick wall, bricks of minimum crushing strength 3.5 N/mm² in cement mortar M1 (1:5) or richer. Wind moment 1.148 kNm/m, maximum compressive stress 0.294 N/mm², no tension. Slenderness ratio 15.2 <27. Tie wall to roof and floor with metal anchors.

## Answer

#### Data and assumptions

- Exterior wall of a single-storey warehouse, height 3.5 m, tied by metal anchors at floor and roof levels, so it spans vertically as simply supported. Vertical roof load $P=30$ kN/m (assumed axial); wind pressure 750 N/m$^2$.
- Anchors give lateral restraint only (no rotational restraint), so the effective height is $h_{ef}=1.0H$ (IS 1905 Table 4 Sl. 3).
- Design by IS 1905:1987 cl. 5.4.1, 5.4.2, 5.5.2, 5.5.3 (NBC 109 follows the same method). Unit weight of masonry 19 kN/m$^3$. Per metre run of wall.

```text
     roof anchor (support)
        |<-- wind w
  ------+-----------   load P (roof)
        |   |
        | wall, H=3.5 m
        |   |  M = wH^2/8 at mid-height
  ------+-----------
        floor anchor (support)
```

#### Step 1: Wind moment

$$
\begin{aligned} w &= 0.750\ \text{kN/m (per m of wall)}\\ M &= 0.750\times3.5^2/8 = 1.148\ \text{kNm/m} \end{aligned}
$$

#### Step 2: Trial thickness 230 mm, load and slenderness

$$
\begin{aligned} \text{Self weight of wall} &= 0.230\times3.50\times19 = 15.30\ \text{kN/m}\\ N\ (\text{at mid-height}) &= 37.65\ \text{kN/m}\\ h_{ef} &= 1.0\times3.5=3.5\ \text{m} \\ SR &= \frac{3.500}{0.230} = 15.22\ (<27) \end{aligned}
$$

#### Step 3: Stresses at the mid-height section

$$
\begin{aligned} \sigma_a &= \frac{N}{A}=\frac{37.65\times10^3}{230\times10^3} = 0.1637\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times1.148\times10^6}{1000\times230^2} = 0.1303\ \text{N/mm}^2\\ \sigma_{max} &= \sigma_a+\sigma_b = 0.2939,\qquad \sigma_{min} = \sigma_a-\sigma_b = 0.0334\ \text{N/mm}^2 \end{aligned}
$$

$\sigma_{min}>0$: the whole section is in compression, no tension (the resultant eccentricity $e=M/N=0.0305$ m, $e/t=0.133<1/6$).

#### Step 4: Required basic compressive stress

Resultant eccentricity ratio $e/t = M/(Nt) = 0.133$. From Table 9: $k_s=0.676$ ($SR=15.22$). $k_a=1$, $k_p=1$. Since $1/24<e/t\le1/6$, a 25 percent increase is allowed (cl. 5.4.1.4 a).

$$
f_b \ge \frac{\sigma_a}{k_sk_ak_p\times1.25} = \frac{0.1637}{0.676\times1.25} = 0.194\ \text{N/mm}^2
$$

#### Step 5: Brick and mortar (IS 1905 Table 8)

| Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) |
|---|---|---|
| M2 (cement:sand 1:6) | 3.5 | 0.35 |
| M1 (cement:sand 1:5) | 3.5 | 0.35 |
| H2 (cement:sand 1:4) | 3.5 | 0.35 |
| H1 (cement:sand 1:3) | 3.5 | 0.35 |

**Answer: 230 mm thick wall (one brick). Bricks of at least 3.5 N/mm$^2$ crushing strength in cement mortar M1 (1:5) or richer; practically use 5 N/mm$^2$ bricks with 1:5 mortar.** Maximum stress 0.294 N/mm$^2$ (compression) and minimum 0.033 N/mm$^2$; slenderness ratio 15.2 < 27. Provide metal anchors at roof and floor at close spacing to pass the wind load into the roof diaphragm and floor.

## Check yourself

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

1. The wall is tied by anchors at floor and roof. How is the wind moment found?
   A. M = wH^2/2, as a cantilever
   B. M = wH^2/8 at mid-height, as a simply supported span
   C. M = wH^2/12 at both ends
   D. M = wH^2/8 at the base, as propped cantilever
   Answer: B. Anchors give lateral support only, so the wall spans vertically as simply supported and M = wH^2/8 = 0.750 x 3.5^2/8 = 1.148 kNm/m. (Key point: Wind moment model)

2. Why is h_ef = 1.0H taken for this wall?
   A. Anchors give lateral restraint but no rotational restraint
   B. The base is fully fixed
   C. The top is free
   D. The wall is taller than 3 m
   Answer: A. With lateral support but no rotational restraint at both ends, IS 1905 Table 4 gives h_ef = 1.0H. (Key point: Effective height)

3. Which loads make up N = 37.65 kN/m at mid-height?
   A. Roof load 30 kN/m only
   B. Roof load 30 plus self weight of the full wall height 15.30 kN/m, giving 45.3
   C. Roof load 30 plus self weight of the upper half of the wall, 7.65 kN/m
   D. Self weight 15.30 plus wind load 22.35
   Answer: C. At mid-height the wall carries the roof load plus the upper half of its self weight: 30 + 0.5 x 15.30 = 37.65 kN/m. (Key point: Direct load at mid-height)

4. Since sigma_min = 0.033 N/mm^2 > 0 and e/t = 0.133, what follows?
   A. The section cracks in tension
   B. The whole section is in compression; as 1/24 < e/t <= 1/6 a 25 percent increase in stress is allowed
   C. The wall must be thickened to 345 mm
   D. The 25 percent increase is not allowed
   Answer: B. No tension occurs, and for e/t between 1/24 and 1/6 IS 1905 cl. 5.4.1.4 permits a 25 percent increase, giving f_b >= 0.194 N/mm^2. (Key point: Stress check and 25% increase)

## Close variants set in other exams

- [Warehouse exterior wall, wind 860 N/m2, anchored](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awind-wall-warehouse-860-floor-roof) — Concrete Technology and Masonry Structure, chapter 9, asked 3×
- [Warehouse exterior wall, wind 860 N/m2, fixed base](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awind-wall-warehouse-860-fixed-base) — Concrete Technology and Masonry Structure, chapter 9, asked 3×

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