---
title: "Brick and mortar for 23 cm wall (165 kN/m)"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:wall-eccentric-23cm-165kn"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-eccentric-23cm-165kn
site: BCT Engineering
---

# Brick and mortar for 23 cm wall (165 kN/m)

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

- Asked in: 2074 Ashwin, 12 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-eccentric-23cm-165kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

A brick wall 23 cm thick using modular brick carries an eccentric load of 165 kN/m at base (eccentricity ratio at 1/12). The wall is 4.5 m long between cross walls. The clear height of the wall is 3.1 m between RCC slabs of 10 cm thick at top and bottom. What should be the strength of brick and grade of mortar? Assume that joints are not raked.

## Short answer

Required basic compressive stress 0.67 N/mm². Use bricks of crushing strength not less than 10 N/mm² with M2 mortar (cement:sand 1:6); permissible stress 0.81 N/mm².

## Answer

#### Data and assumptions

- Wall 23 cm thick, modular bricks, joints not raked. Load 165 kN/m at base **inclusive of self weight**, $e/t=1/12$.
- Wall 4.5 m long between cross walls, clear height 3.1 m between 10 cm RCC slabs, so $H=3.1+0.10=3.2$ m (centre to centre).
- 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 $H = 3.20$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.20 = 2.400$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 4.50$ m. The lesser value governs:

$$
SR = \frac{h_{ef}}{t} = \frac{2.400}{0.230} = 10.43\ (<27,\ \text{Table 7, cement mortar})
$$

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

Load at the base (self weight already included) $P = 165.00$ kN/m.

Stress on the wall:

$$
\sigma = \frac{P}{A} = \frac{165.00\times10^3}{230\times10^3\ \text{mm}^2/\text{m}} = 0.717\ \text{N/mm}^2
$$

#### Step 3: Reduction factors

- Stress reduction factor $k_s$ (Table 9, $SR=10.43$, $e/t=0.083$, by linear interpolation) $= 0.857$
- Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.23 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

$$
\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{0.717}{0.857\times1.0\times1.0\times1.25} = 0.670\ \text{N/mm}^2
$$

(Without the 25 percent increase the requirement would be $f_b \ge 0.837$ N/mm$^2$.)

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

Smallest standard unit strength giving $f_b \ge 0.670$ N/mm$^2$:

| Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) |
|---|---|---|
| M2 (cement:sand 1:6) | 10 | 0.81 |
| M1 (cement:sand 1:5) | 7.5 | 0.74 |
| H2 (cement:sand 1:4) | 7.5 | 0.74 |
| H1 (cement:sand 1:3) | 7.5 | 0.75 |

**Answer: use bricks of crushing strength not less than 10 N/mm$^2$ with mortar M2 (cement:sand 1:6)** ($f_b = 0.81$ N/mm$^2$ $\ge$ 0.670). 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. With a clear height 3.1 m between 10 cm RCC slabs, what H is used for h_ef?
   A. 3.1 m
   B. 3.2 m (centre to centre of slabs)
   C. 3.0 m
   D. 3.3 m
   Answer: B. H is measured centre to centre of slabs: 3.1 + 0.10 = 3.2 m, so h_ef = 0.75 x 3.2 = 2.4 m. (Key point: Height H)

2. SR of the wall is:
   A. 10.43
   B. 13.9
   C. 7.8
   D. 4.5
   Answer: A. SR = h_ef/t = 2.400/0.230 = 10.43, below 27. (Key point: Slenderness ratio)

3. The 165 kN/m load is already 'at base'. How is self weight treated?
   A. Added again to 165 kN/m
   B. Already included, so P = 165 kN/m
   C. Subtracted from 165 kN/m
   D. Replaced by 19 kN/m^3 x height
   Answer: B. The data states the load is inclusive of self weight, so sigma = 165,000/230,000 = 0.717 N/mm^2. (Key point: Self weight)

4. f_b = 0.717/(0.857 x 1.25) gives:
   A. 0.670 N/mm^2
   B. 0.837 N/mm^2
   C. 0.717 N/mm^2
   D. 0.35 N/mm^2
   Answer: A. With the 25 percent increase f_b = 0.670 N/mm^2; without it, 0.837. (Key point: Required f_b)

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