---
title: "Brick and mortar for 250 mm wall (350 kN/m)"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:wall-eccentric-250mm-350kn"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-eccentric-250mm-350kn
site: BCT Engineering
---

# Brick and mortar for 250 mm wall (350 kN/m)

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

- Asked in: 2079 Baisakh, 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-250mm-350kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

A wall 250 mm thick, using modular bricks, carries at the top a load of 350 kN/m having resultant eccentricity ratio of 1/12. The wall is 5 m long between cross walls and is 3.5 m clear height between RCC slabs at the top and bottom. What shall be the strength of brick and the grade of mortar? Assume that joints are not raked.

## Short answer

Required basic compressive stress 1.37 N/mm². Use bricks of crushing strength not less than 17.5 N/mm² with H1 mortar (cement:sand 1:3); permissible stress 1.45 N/mm².

## Answer

#### Data and assumptions

- Wall 250 mm thick, modular bricks, joints not raked. Load 350 kN/m at top, $e/t=1/12$.
- Wall 5 m long between cross walls, clear height 3.5 m between RCC slabs (slab thickness not given, $H=3.5$ m).
- Note: a 350 kN/m load is high for brick masonry, so strong units are needed.
- 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 $H = 3.50$ m. Effective height $h_{ef} = 0.75 H = 0.75\times3.50 = 2.625$ m. Effective length (wall supported by a cross wall at each end, Table 5 Sl. 3) $L_{ef} = 1.0L = 5.00$ m. The lesser value governs:

$$
SR = \frac{h_{ef}}{t} = \frac{2.625}{0.250} = 10.50\ (<27,\ \text{Table 7, cement mortar})
$$

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

$$
\begin{aligned} \text{Self weight} &= t\,H_c\,\gamma = 0.250\times3.50\times19 = 16.62\ \text{kN/m}\\ P &= 350.00 + 16.62 = 366.62\ \text{kN/m} \end{aligned}
$$

Stress on the wall:

$$
\sigma = \frac{P}{A} = \frac{366.62\times10^3}{250\times10^3\ \text{mm}^2/\text{m}} = 1.466\ \text{N/mm}^2
$$

#### Step 3: Reduction factors

- Stress reduction factor $k_s$ (Table 9, $SR=10.50$, $e/t=0.083$, by linear interpolation) $= 0.855$
- Area reduction factor $k_a$ (cl. 5.4.1.2): area per metre 0.25 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

$$
\sigma \le f_b\,k_s\,k_a\,k_p\times1.25 \Rightarrow f_b \ge \frac{1.466}{0.855\times1.0\times1.0\times1.25} = 1.372\ \text{N/mm}^2
$$

(Without the 25 percent increase the requirement would be $f_b \ge 1.715$ N/mm$^2$.)

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

Smallest standard unit strength giving $f_b \ge 1.372$ N/mm$^2$:

| Mortar (IS 1905 Table 1) | Min. unit strength (N/mm²) | Table 8 $f_b$ (N/mm²) |
|---|---|---|
| M2 (cement:sand 1:6) | 30 | 1.51 |
| M1 (cement:sand 1:5) | 25 | 1.47 |
| H2 (cement:sand 1:4) | 20 | 1.41 |
| H1 (cement:sand 1:3) | 17.5 | 1.45 |

**Answer: use bricks of crushing strength not less than 17.5 N/mm$^2$ with mortar H1 (cement:sand 1:3)** ($f_b = 1.45$ N/mm$^2$ $\ge$ 1.372). 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. What is the design load at the base?
   A. 350 kN/m only
   B. 350 + 16.62 = 366.62 kN/m
   C. 350 - 16.62 kN/m
   D. 350 + 3.5 kN/m
   Answer: B. Load given at top, so self weight 0.25 x 3.5 x 19 = 16.62 kN/m is added. (Key point: Total load at base)

2. SR for h_ef = 0.75 x 3.5 = 2.625 m and t = 0.25 m is:
   A. 10.50
   B. 14.0
   C. 7.9
   D. 21.0
   Answer: A. SR = 2.625/0.25 = 10.5; the 5 m wall length does not govern. (Key point: Slenderness)

3. f_b = 1.466/(0.855 x 1.25) is about:
   A. 1.715 N/mm^2
   B. 1.372 N/mm^2
   C. 0.855 N/mm^2
   D. 2.0 N/mm^2
   Answer: B. With the 25 percent increase, f_b = 1.372 N/mm^2 (1.715 without it). (Key point: Required f_b)

4. Smallest unit strength is obtained with which mortar?
   A. M2 (needs 30 N/mm^2)
   B. H1 (1:3) mortar with 17.5 N/mm^2 units
   C. M1 with 7.5 N/mm^2 units
   D. Lime mortar with 5 N/mm^2 units
   Answer: B. Richer mortar lowers the unit strength needed; H1 with 17.5 N/mm^2 gives 1.45 >= 1.372. (Key point: Selection)

## Close variants set in other exams

- [Brick and mortar for 230 mm wall (100 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-230mm-100kn-5m) — Concrete Technology and Masonry Structure, chapter 8, asked 7×
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- [Brick and mortar for 23 cm wall (165 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-23cm-165kn) — Concrete Technology and Masonry Structure, chapter 8, asked 7×

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