---
title: "Wall under axial and eccentric load"
subject: "Concrete Technology and Masonry Structure"
chapter: 8
key: "semester_5/concrete_technology_and_masonry_structure:wall-axial-and-eccentric-slenderness-16"
source: https://www.bctengineer.com/q/semester_5/concrete_technology_and_masonry_structure/wall-axial-and-eccentric-slenderness-16
site: BCT Engineering
---

# Wall under axial and eccentric load

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

- Asked in: 2078 Bhadra, 8 marks
- Video: [Masonry Wall Design - Part 4 || Design of Wall Subjected to Eccentric Loading || IS 1905:1987 ||](https://www.youtube.com/watch?v=7tXfjcYMwDc) — Sumit Paudel
- 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-axial-and-eccentric-slenderness-16
- Whole chapter: https://www.bctengineer.com/md/semester_5/concrete_technology_and_masonry_structure/chapter-8

## Question

A 23 cm thick brick masonry wall carries an axial load of 12 kN per meter length and an eccentric load of 27 kN per meter length acting at a distance of 7.33 cm from the axis of the wall. Design the masonry wall if its slenderness ratio is 16. Assume that joints are non-raked.

## Short answer

For slenderness ratio 16 and eccentricity 0.221t, required basic compressive stress is 0.519 N/mm². Use bricks of crushing strength not less than 7.5 N/mm² with cement mortar M2 (1:6); reduce eccentricity and provide an RCC band.

## Answer

#### Data and assumptions

- Wall 230 mm thick; the joints are not raked, so the full thickness is effective. $SR=16$ (given, $<27$). Axial load $12$ kN/m and eccentric load $27$ kN/m at $7.33$ cm from the axis.
- Per metre length of wall; $k_a=1$ (area $0.23$ m$^2$/m $\ge 0.2$) and $k_p=1$ (modular bricks).
- Design by IS 1905:1987 (Tables 8, 9; cl. 5.4.1.4, 5.4.4).

#### Step 1: Resultant load and eccentricity

$$
\begin{aligned} P &= 12+27 = 39\ \text{kN/m}\\ e &= \frac{12\times0+27\times7.33}{39} = 5.075\ \text{cm}\\ \frac{e}{t} &= \frac{5.075}{23} = 0.221 \end{aligned}
$$

Since $e/t>1/6$ (0.167), tension would develop. Masonry cannot carry tension, so the area under tension is disregarded (cl. 5.4.1.4 b, 5.4.4). The stress over the compressed part is triangular, with a contact (compression) width of

$$
3\left(\frac{t}{2}-e\right) = 3\times(11.5-5.075) = 19.3\ \text{cm}
$$

#### Step 2: Maximum stress and stress reduction factor

For a triangular stress block, the maximum stress at the compressed face is twice the average over the contact width:

$$
\sigma_{max} = \frac{2P}{3(t/2-e)} = \frac{2\times39}{0.1928} = 404.7\ \text{kN/m}^2 = 0.405\ \text{N/mm}^2
$$

$k_s$ from Table 9 at $SR=16$, $e/t=0.221$ (between $1/6$, 0.65, and $1/4$, 0.61): $0.65-\frac{(0.221-0.1667)}{0.0833}\times(0.65-0.61)=0.624$.

Since $e/t>1/6$, a 25 percent increase in permissible stress is allowed, together with neglecting the tension area (cl. 5.4.1.4 b).

#### Step 3: Required basic compressive stress

$$
f_b \ge \frac{\sigma_{max}}{k_s k_a k_p\times1.25} = \frac{0.405}{0.624\times1\times1\times1.25} = 0.519\ \text{N/mm}^2
$$

IS 1905 Table 8: 5 N/mm$^2$ units give at most $f_b = 0.50$ (M1, H2, H1), which is not enough. Units of 7.5 N/mm$^2$ with M2 mortar give $f_b=0.59 \ge 0.519$. (Without the 25 percent increase, $f_b \ge 0.649$ N/mm$^2$, which needs 7.5 N/mm$^2$ units with M1 mortar, $f_b = 0.74$.)

**Answer: use bricks of crushing strength not less than 7.5 N/mm$^2$ with cement mortar M2 (1:6), $f_b=0.59$ N/mm$^2$** (SR 16 < 27); use M1 (1:5) if the 25 percent increase is not taken. The eccentricity is large, so reduce it by centring the load, tie the wall to cross walls and provide an RCC band.

## Check yourself

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

1. With axial 12 kN/m at the axis and 27 kN/m at 7.33 cm, e of the resultant is:
   A. 7.33 cm
   B. 27 x 7.33/39 = 5.075 cm
   C. 12 x 7.33/39 = 2.25 cm
   D. 39 x 7.33 = 286 cm
   Answer: B. Moments about the axis: e = (12 x 0 + 27 x 7.33)/39 = 5.075 cm, e/t = 0.221. (Key point: Resultant eccentricity)

2. What does e/t = 0.221 > 1/6 imply?
   A. Tension would develop, which masonry cannot carry, so only the compression area is used
   B. The wall is safe against everything
   C. The 25 percent increase must be used
   D. The wall acts as axially loaded
   Answer: A. Masonry takes no tension; the stress is P/[3(t/2 - e)] over a contact width of 19.3 cm. (Key point: Large eccentricity)

3. The contact stress is P divided by:
   A. t x 1 m
   B. 3(t/2 - e) = 19.3 cm width, giving 0.202 N/mm^2
   C. t/2
   D. 6e
   Answer: B. sigma = 39/0.1928 = 202 kN/m^2 = 0.202 N/mm^2. (Key point: Stress)

4. With k_s = 0.598, f_b = 0.339. The choice is:
   A. 3.5 N/mm^2 brick with M2 mortar (0.35); 5 N/mm^2 in practice
   B. Lime mortar with 3.5 N/mm^2 brick
   C. 20 N/mm^2 brick
   D. No brick possible
   Answer: A. Table 8 gives 0.35 >= 0.339. Lime mortar is unsuitable (SR limit 20, f_b 0.25). (Key point: Selection)

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