---
title: "Wall subjected to an eccentric load"
subject: "Design of Timber and Masonry Structures"
chapter: 5
key: "semester_5/design_of_timber_and_masonry_structures:wall-eccentric-load-stress-check"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/wall-eccentric-load-stress-check
site: BCT Engineering
---

# Wall subjected to an eccentric load

Design of Timber and Masonry Structures · Chapter 5: Design of Masonry Walls for Gravity Loads

- Asked in: year not given, 6 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 Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/wall-eccentric-load-stress-check
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-5

## Question

A 230 mm thick brick wall carries a vertical load of 60 kN per metre length from a beam bearing on it, with an eccentricity of 45 mm from the centre line of the wall. The permissible compressive stress of the masonry is 0.90 N/mm² and the permissible tensile stress (flexural) is 0.07 N/mm². Find the maximum and minimum stresses at the base of the wall section and check them. State the middle-third rule.

## Short answer

Middle-third rule: if e ≤ t/6 (38.3 mm) no tension. Here e=45 mm, so tension appears. Using σ = (P/A)(1±6e/t), σ_max=0.567 N/mm² (<0.90), σ_min=-0.045 N/mm² (<0.07 permissible tension). Wall safe.

## Answer

#### Middle-third rule
If the resultant load acts within the middle third of the thickness ($e \le t/6$), the whole section is in compression. If $e > t/6$, tension appears on the far face, and masonry (weak in tension) may crack.

#### Data
$P = 60$ kN/m, $t = 230$ mm, $e = 45$ mm, $b = 1000$ mm (per metre).

$$
\frac{t}{6} = \frac{230}{6} = 38.3\ \text{mm} < e = 45\ \text{mm}
$$

The load is outside the middle third, so some tension is expected.

#### Stresses
$$
\begin{aligned}
A &= 230 \times 1000 = 2.3\times10^5\ \text{mm}^2 \\
Z &= \frac{1000 \times 230^2}{6} = 8.817\times10^{6}\ \text{mm}^3 \\
M &= Pe = 60\,000 \times 45 = 2.7\times10^{6}\ \text{N·mm} \\
\frac{P}{A} &= \frac{60\,000}{2.3\times10^5} = 0.261\ \text{N/mm}^2 \\
\frac{M}{Z} &= \frac{2.7\times10^6}{8.817\times10^6} = 0.306\ \text{N/mm}^2
\end{aligned}
$$

Or, using the formula $\sigma = \dfrac{P}{A}\left(1 \pm \dfrac{6e}{t}\right)$ with $6e/t = 1.174$:

$$
\begin{aligned}
\sigma_{max} &= 0.261\,(1 + 1.174) = 0.567\ \text{N/mm}^2 \ (\text{compression}) \\
\sigma_{min} &= 0.261\,(1 - 1.174) = -0.045\ \text{N/mm}^2 \ (\text{tension})
\end{aligned}
$$

#### Checks
- Compression: $0.567 < 0.90$ N/mm². Safe.
- Tension: $0.045 < 0.07$ N/mm². Within the permissible flexural tension.

Both limits are satisfied. If the tension were more than allowed, the eccentricity would have to be reduced (for example, by centring the beam bearing with a bearing plate), or the wall thickness increased, or the wall reinforced.

**Answer: $\sigma_{max} = 0.567$ N/mm² (compression) and $\sigma_{min} = -0.045$ N/mm² (tension); both within permissible values, so the wall is safe.**

## Check yourself

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

1. What does the middle-third rule say about a wall section?
   A. If e <= t/6 the whole section is in compression; if e > t/6 tension appears on the far face
   B. If e <= t/3 the whole section is in compression
   C. If e > t/6 the wall is always safe because stress is reduced
   D. Tension appears only when e exceeds t/2
   Answer: A. With the resultant inside the middle third (e <= t/6) there is no tension; beyond it tension develops on the far face. (Key point: Middle-third rule)

2. For t = 230 mm and e = 45 mm, what is the conclusion about the middle third?
   A. Load is within the middle third since 45 < 230/3
   B. Load is outside the middle third since t/6 = 38.3 mm < 45 mm, so some tension is expected
   C. Load is exactly at the kern
   D. No conclusion is possible without the load value
   Answer: B. t/6 = 38.3 mm is less than e = 45 mm, so the resultant lies outside the middle third. (Key point: Eccentricity check)

3. Which formula gives the maximum and minimum stresses at the wall base?
   A. sigma = P/A x (1 +/- e/t)
   B. sigma = P/A x (1 +/- 3e/t)
   C. sigma = P/A x (1 +/- 6e/t)
   D. sigma = P/A x (1 +/- 12e/t)
   Answer: C. Since Z = bt^2/6, M/Z = (P/A)(6e/t), so sigma = (P/A)(1 +/- 6e/t). (Key point: Stress formula)

4. With sigma_max = 0.567 and sigma_min = -0.045 N/mm^2, what is the conclusion?
   A. Unsafe, because tension exists
   B. Unsafe, because compression exceeds 0.90 N/mm^2
   C. Safe: compression 0.567 < 0.90 and tension 0.045 < 0.07 permissible
   D. Safe only if the tension is ignored
   Answer: C. Both stresses lie within the permissible compressive (0.90) and flexural tensile (0.07) limits. (Key point: Final check)

## Related in Design of Timber and Masonry Structures

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