---
title: "3 m wall under wind: fixed top and bottom"
subject: "Concrete Technology and Masonry Structure"
chapter: 9
key: "semester_5/concrete_technology_and_masonry_structure:wind-wall-3m-fixed-top-bottom"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wind-wall-3m-fixed-top-bottom
site: BCT Engineering
---

# 3 m wall under wind: fixed top and bottom

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

- Asked in: 2081 Bhadra, 7 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-3m-fixed-top-bottom
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-9

## Question

Design a wall 3 m high to resist a wind load. The wall is fixed at top and bottom supports. Assume uniformly distributed wind pressure of 16 N/m$^2$.

## Short answer

A half-brick wall of 115 mm thickness satisfies stress requirements (slenderness ratio 19.6 < 27), but for practical stability use a 230 mm one-brick wall. Use bricks of 3.5 N/mm² with cement mortar M1 (1:5). Stresses are negligible: σmax=0.062 N/mm².

## Answer

#### Data and assumptions

- The wall is fixed at top and bottom, so $h_{ef}=0.75H=2.25$ m (full restraint, Table 4 Sl. 1). No vertical load other than self weight is given. The pressure is taken as stated, 16 N/m$^2$ (0.016 kN/m$^2$) [reading: a very small wind pressure, so stresses are tiny and slenderness controls the thickness].
- End moment of a wall fixed at both ends under uniform load: $M=wH^2/12$ at each end (sagging $wH^2/24$ at mid-height). Section checked at the base, where self weight is the largest.
- 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
  ======+===========  fixed (top)
        |<-- wind w
        |  wall H=3 m
        |  M = wH^2/12 at ends
  ======+===========  fixed (bottom)
```

#### Step 1: Wind moment

$$
\begin{aligned} w &= 0.016\ \text{kN/m (per m of wall)}\\ M &= 0.016\times3.0^2/12 = 0.012\ \text{kNm/m} \end{aligned}
$$

#### Step 2: Trial thickness 115 mm, load and slenderness

$$
\begin{aligned} \text{Self weight of wall} &= 0.115\times3.00\times19 = 6.56\ \text{kN/m}\\ N\ (\text{at base}) &= 6.56\ \text{kN/m}\\ h_{ef} &= 0.75\times3.0=2.25\ \text{m} \\ SR &= \frac{2.250}{0.115} = 19.57\ (<27) \end{aligned}
$$

#### Step 3: Stresses at the base section

$$
\begin{aligned} \sigma_a &= \frac{N}{A}=\frac{6.56\times10^3}{115\times10^3} = 0.0570\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times0.012\times10^6}{1000\times115^2} = 0.0054\ \text{N/mm}^2\\ \sigma_{max} &= \sigma_a+\sigma_b = 0.0624,\qquad \sigma_{min} = \sigma_a-\sigma_b = 0.0516\ \text{N/mm}^2 \end{aligned}
$$

$\sigma_{min}>0$: the whole section is in compression, no tension (the resultant eccentricity $e=M/N=0.0018$ m, $e/t=0.016<1/6$).

#### Step 4: Required basic compressive stress

Resultant eccentricity ratio $e/t = M/(Nt) = 0.016$. From Table 9: $k_s=0.619$ ($SR=19.57$). $k_a=1$, $k_p=1$.

$$
f_b \ge \frac{\sigma_a}{k_sk_ak_p\times1.00} = \frac{0.0570}{0.619\times1.00} = 0.092\ \text{N/mm}^2
$$

#### 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: a half-brick wall of 115 mm is sufficient for stress (SR = 19.6 < 27), but for stability and practical robustness adopt a 230 mm (one-brick) wall; bricks of 3.5 N/mm$^2$ with cement mortar M1 (1:5).** Stresses are very small: $\sigma_{max}=0.0624$, $\sigma_{min}=0.0516$ N/mm$^2$, far below the permissible values.

## Check yourself

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

1. What is the end moment of a wall fixed at both ends under uniform wind?
   A. wH^2/8
   B. wH^2/2
   C. wH^2/12
   D. wH^2/24 at the ends
   Answer: C. For fixed-fixed ends M = wH^2/12 at each end, with wH^2/24 sagging at mid-height. (Key point: Fixed-end moment)

2. What is h_ef for a wall fixed at top and bottom?
   A. 0.75H = 2.25 m
   B. 1.0H = 3.0 m
   C. 1.5H = 4.5 m
   D. 0.85H = 2.55 m
   Answer: A. Full restraint at both ends gives h_ef = 0.75H (Table 4 Sl. 1), so SR = 2.25/0.115 = 19.57. (Key point: Effective height)

## Close variants set in other exams

- [3 m wall under wind: free top, fixed bottom](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awind-wall-3m-free-top) — Concrete Technology and Masonry Structure, chapter 9, asked 2×

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