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

# 3 m wall under wind: free top, fixed bottom

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

- Asked in: 2080 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-free-top
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-9

## Question

Design a wall 3.0 m high to resist a wind load. The wall is free at top and fixed at bottom supports. Assume uniformly distributed wind pressure of 12 N/m$^2$.

## Short answer

By calculation, a 230 mm thick wall is adequate (σmax=0.063 N/mm², stability factor 28). Use bricks of at least 3.5 N/mm² in cement mortar M1 (1:5). Without calculation, adopt 345 mm per IS 1905 Table 11 (H/t ≤ 10).

## Answer

#### Data and assumptions

- Wall 3.0 m high, free at the top and fixed at the base: a cantilever (free-standing) wall. Uniform wind pressure taken as stated, 12 N/m$^2$ (0.012 kN/m$^2$) [reading: given value is very small, so stability is easily satisfied; if the wind pressure is larger, repeat with the same steps].
- Design by IS 1905:1987 cl. 5.5.2.1 and Table 11, Table 4 Sl. 4. Per metre run; masonry 19 kN/m$^3$; trial thickness 230 mm.

```text
        ^ free top
        |<-- w = 12 N/m2
        |  wall H = 3.0 m
        |  M = wH^2/2 at base
  ======+======  fixed base
```

#### Step 1: Slenderness (cl. 4.3, Table 4 Sl. 4)

For a free top and a fixed base $h_{ef}=1.5H=4.5$ m, so $SR = 4.5/0.23 = 19.6<27$ (cement mortar), but the slenderness check alone would give $t\ge4.5/27=0.167$ m.

#### Step 2: Bending and stresses

$$
\begin{aligned} M &= \frac{wH^2}{2} = \frac{0.012\times3^2}{2} = 0.054\ \text{kNm/m}\\ \text{Self weight} &= 0.23\times3\times19 = 13.11\ \text{kN/m}\\ \sigma_a &= \frac{13.11\times10^3}{230\times10^3} = 0.0570\ \text{N/mm}^2\\ \sigma_b &= \frac{6M}{t^2} = \frac{6\times0.054\times10^6}{1000\times230^2} = 0.0061\ \text{N/mm}^2\\ \sigma_{max} &= 0.0631,\ \ \sigma_{min} = 0.0509\ \text{N/mm}^2 \end{aligned}
$$

No tension at the base ($\sigma_{min}>0$), and the compressive stress is far below the allowable value for the lowest brick grade.

#### Step 3: Stability (cl. 4.2.2.4 a)

Free-standing wall: stability moment must be at least 1.5 times the overturning moment:

$$
M_{stab}=W\frac{t}{2} = 13.11\times0.115 = 1.51\ \text{kNm/m}\ \gg\ 1.5\times0.054 = 0.081\ \text{kNm/m}
$$

#### Step 4: Table 11 check

IS 1905 Table 11 allows a free-standing wall without calculation (seismic zones I and II, mortar not leaner than M1) when $H/t \le 10$ for pressures up to 285 N/m$^2$: $t\ge3.0/10=0.30$ m. Our wall satisfies stress and stability by calculation; if the simplified table is preferred, adopt 345 mm (one and a half brick).

**Answer: the wall is safe with 230 mm thickness by calculation ($\sigma_{max}=0.063$ N/mm$^2$, stability factor 28); bricks of at least 3.5 N/mm$^2$ in cement mortar M1 (1:5), practical 5 N/mm$^2$.** If the wall is built without calculation use $t=345$ mm (Table 11, $H/t\le10$). Provide a damp-proof course and a coping to protect the top.

## Check yourself

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

1. A wall free at top and fixed at the base acts as what, and what is M at the base?
   A. Propped cantilever, wH^2/8
   B. Cantilever, M = wH^2/2
   C. Fixed-fixed, wH^2/12
   D. Simply supported, wH^2/8
   Answer: B. A free top and fixed base is a cantilever; M = 0.012 x 9/2 = 0.054 kNm/m. (Key point: Wall model)

2. What h_ef applies to the free-top wall?
   A. 0.75H
   B. 0.85H
   C. 1.0H
   D. 1.5H = 4.5 m
   Answer: D. Table 4 Sl. 4 gives h_ef = 1.5H = 4.5 m, so SR = 4.5/0.23 = 19.6 < 27. (Key point: Effective height)

3. What does the stability check require for a free-standing wall?
   A. Resisting moment at least 1.5 times overturning moment
   B. Resisting moment equal to overturning moment
   C. Resisting moment at least 3 times overturning moment
   D. No check is needed
   Answer: A. Cl. 4.2.2.4 a requires M_stab >= 1.5 x overturning; W t/2 = 1.51 kNm/m far exceeds 0.081. (Key point: Stability check)

4. If the wall is designed without calculation using Table 11 (H/t <= 10), what thickness results?
   A. 230 mm
   B. 345 mm (0.30 m rounded up to 1.5 brick)
   C. 115 mm
   D. 460 mm
   Answer: B. t >= 3.0/10 = 0.30 m, so the next brick size, 345 mm, is adopted. (Key point: Table 11 route)

## Close variants set in other exams

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

## Related in Concrete Technology and Masonry Structure

- [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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