---
title: "Single-room wall: brick strength and mortar"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:wall-single-room-8000kg"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-single-room-8000kg
site: BCT Engineering
---

# Single-room wall: brick strength and mortar

Concrete Technology and Masonry Structure · Chapter 8: Design of masonry walls for gravity loads

- Asked in: 2069 Chaitra, 10 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/wall-single-room-8000kg
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

A brick masonry wall of a single room building is 20 cm thick and is supported by 10 cm thick RCC slab at its top and bottom. The wall carries a vertical load (inclusive of its own weight) of 8000 kg/m at the base at an eccentricity ratio of 0.1. The length of the wall is 3 m between cross walls. The clear height of storey is 3 m. Determine the required crushing strength of bricks and the type of mortar to be used. Use modular bricks.

## Short answer

Required basic compressive stress 0.385 N/mm²; use bricks of crushing strength not less than 5 N/mm² with cement mortar M2 (1:6).

## Answer

#### Data and assumptions

- Wall 20 cm thick, modular bricks, 10 cm RCC slabs at top and bottom. Load inclusive of self weight $=8000\ \text{kg/m}\times9.81 = 78.48$ kN/m, $e/t=0.1$.
- Wall 3 m long between cross walls, clear height 3 m, so $H=3+0.10=3.1$ m.
- Design code: IS 1905:1987 (Tables 4 to 10); NBC 109 follows the same permissible-stress method.
- Brick masonry unit weight 19 kN/m$^3$ (IS 875 Part 1); modular brick, height/width $\le 0.75$, so $k_p=1$.

#### Step 1: Effective height, length and slenderness ratio (IS 1905 cl. 4.3, 4.4, 4.6)

Height between slab centres $H = 3.10$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.10 = 2.325$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 3.00$ m. The lesser value governs:

$$
SR = \frac{h_{ef}}{t} = \frac{2.325}{0.200} = 11.62\ (<27,\ \text{Table 7, cement mortar})
$$

#### Step 2: Design load per metre run at the base

Load at the base (self weight already included) $P = 78.48$ kN/m.

Stress on the wall:

$$
\sigma = \frac{P}{A} = \frac{78.48\times10^3}{200\times10^3\ \text{mm}^2/\text{m}} = 0.392\ \text{N/mm}^2
$$

#### Step 3: Reduction factors

- Stress reduction factor $k_s$ (Table 9, $SR=11.62$, $e/t=0.100$, by linear interpolation) $= 0.816$
- Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.20 m$^2$ is not less than 0.2 m$^2$, so $k_a = 1.0$
- Shape factor $k_p = 1.0$ (modular bricks, height/width $\le 0.75$, Table 10)
- Since $e/t > 1/24$, a 25 percent increase in permissible stress is allowed (cl. 5.4.1.4 a), factor 1.25

#### Step 4: Required basic compressive stress

$$
\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.392}{0.816\times1.0\times1.0\times1.25} = 0.385\ \text{N/mm}^2
$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.481$ N/mm$^2$.)

#### Step 5: Strength of brick and grade of mortar (IS 1905 Table 8)

Smallest standard unit strength giving $f_b \ge 0.385$ N/mm$^2$:

| Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) |
|---|---|---|
| M2 (cement:sand 1:6) | 5 | 0.44 |
| M1 (cement:sand 1:5) | 5 | 0.50 |
| H2 (cement:sand 1:4) | 5 | 0.50 |
| H1 (cement:sand 1:3) | 5 | 0.50 |

**Answer: use bricks of crushing strength not less than 5 N/mm$^2$ with mortar M2 (cement:sand 1:6)** ($f_b = 0.44$ N/mm$^2$ $\ge$ 0.385). Other combinations in the table are also acceptable; a stronger mortar allows a weaker brick.

## Check yourself

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

1. 8000 kg/m is converted to force by:
   A. x 9.81 = 78.48 kN/m
   B. x 10 = 80 N/m
   C. /9.81 = 815 kN/m
   D. used as 8000 kN/m
   Answer: A. 8000 x 9.81 N = 78.48 kN/m, already including self weight. (Key point: Load conversion)

2. H = 3 + 0.10 = 3.1 m, h_ef = 0.75H. SR for t = 0.20 m is:
   A. 11.62
   B. 15.5
   C. 8.7
   D. 23.25
   Answer: A. h_ef = 2.325 m, SR = 2.325/0.20 = 11.62; the 3 m length is greater so height governs. (Key point: Slenderness)

3. e/t = 0.1 exceeds 1/24 (0.0417), so f_b is:
   A. 0.392/0.816 = 0.481
   B. 0.392/(0.816 x 1.25) = 0.385 N/mm^2
   C. 0.392 x 0.816
   D. 0.392/1.25 = 0.314
   Answer: B. The 25 percent increase gives f_b = 0.385 N/mm^2. (Key point: 25 percent increase)

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