---
title: "Masonry pier under axial load and in-plane moment"
subject: "Design of Timber and Masonry Structures"
chapter: 7
key: "semester_5/design_of_timber_and_masonry_structures:masonry-pier-axial-moment-check"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/masonry-pier-axial-moment-check
site: BCT Engineering
---

# Masonry pier under axial load and in-plane moment

Design of Timber and Masonry Structures · Chapter 7: Seismic Design and Strengthening of Masonry Buildings

- Asked in: year not given, 6 marks
- Video: [M3.1 Masonry Pier Design](https://www.youtube.com/watch?v=ikz_UbsKnGs) — Struct2009
- 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/masonry-pier-axial-moment-check
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-7

## Question

A brick masonry pier 230 mm thick and 1.5 m long, 3.0 m high, has lateral and rotational restraint at top and bottom. It carries an axial load of 150 kN and an in-plane seismic moment of 35 kN·m at the base. The basic compressive stress is $f_b = 1.2$ N/mm². Check the pier using the working stress method allowing a 33 percent increase in permissible stresses for earthquake. Use $k_s$ (SR 8 → 0.95, SR 10 → 0.89). State the permissible tension is zero.

## Short answer

Stresses: σmax=0.841 N/mm², σmin=0.029 N/mm² (no tension). Permissible compressive stress increased by 33% for earthquake is 1.43 N/mm²; pier is safe. Zero-tension moment limit is 37.5 kN·m.

## Answer

#### Step 1: Section properties
$$
\begin{aligned}
A &= 230 \times 1500 = 3.45\times10^5\ \text{mm}^2 \\
Z &= \frac{230 \times 1500^2}{6} = 8.625\times10^{7}\ \text{mm}^3
\end{aligned}
$$

#### Step 2: Stresses
$$
\begin{aligned}
\frac{N}{A} &= \frac{150\times10^3}{3.45\times10^5} = 0.435\ \text{N/mm}^2 \\
\frac{M}{Z} &= \frac{35\times10^6}{8.625\times10^7} = 0.406\ \text{N/mm}^2
\end{aligned}
$$

$$
\begin{aligned}
\sigma_{max} &= 0.435 + 0.406 = 0.841\ \text{N/mm}^2 \ (\text{compression}) \\
\sigma_{min} &= 0.435 - 0.406 = +0.029\ \text{N/mm}^2 \ (\text{compression})
\end{aligned}
$$

There is no tension (the resultant lies within the middle third).

#### Step 3: Permissible stress
$H_{eff} = 0.75 \times 3000 = 2250$ mm, $SR = 2250/230 = 9.78$ and $k_s = 0.89 + 0.22/2 \times 0.06 = 0.897$.

$A = 0.345$ m² $> 0.2$, so $k_a = 1.0$.

$$
f_c = 1.2 \times 0.897 = 1.076\ \text{N/mm}^2, \qquad f_{c,seismic} = 1.33 \times 1.076 = 1.43\ \text{N/mm}^2
$$

#### Step 4: Checks
- Maximum compression: $0.841 < 1.43$ N/mm². Safe.
- Gravity alone: $0.435 < 1.076$ N/mm². Safe.
- Tension: none.

The moment that just causes zero tension is $M_0 = \dfrac{N Z}{A} = N\dfrac{L}{6} = 150 \times 0.25 = 37.5$ kN·m, so the applied 35 kN·m is 93 percent of this limit. If the moment is larger, the pier will crack and tension must be taken by vertical bars (reinforced masonry), or the axial load must be increased or the pier lengthened.

**Answer: $\sigma_{max} = 0.841$ N/mm² < 1.43 N/mm² and $\sigma_{min} = +0.029$ N/mm² (no tension); the pier is safe. Zero-tension limit moment = 37.5 kN·m.**

## Check yourself

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

1. How are pier stresses found?
   A. N/A only
   B. M/Z only
   C. N/A +/- M/Z, giving 0.435 +/- 0.406 N/mm^2
   D. N/A x M/Z
   Answer: C. Axial stress plus bending stress gives sigma_max 0.841 and sigma_min +0.029 N/mm^2. (Key point: Combined stresses)

2. Why is there no tension in the pier?
   A. The resultant lies within the middle third, since sigma_min is positive
   B. Tension is permitted to be zero by ignoring it
   C. The moment is zero
   D. The pier is short
   Answer: A. N/A exceeds M/Z, so the whole section is in compression. (Key point: No tension)

3. How is permissible stress for earthquake obtained?
   A. f_b x k_s x 0.67
   B. f_b x k_s x k_a = 1.076 N/mm^2 increased by 33 percent to 1.43 N/mm^2
   C. Directly f_b = 1.2
   D. f_b x k_s decreased by 33 percent
   Answer: B. k_s = 0.897 (SR 9.78), k_a = 1, so f_c = 1.076, times 1.33 for earthquake = 1.43. (Key point: Permissible stress)

4. What moment just gives zero tension, and what does it show?
   A. N L/3 = 75 kN·m
   B. N L/6 = 37.5 kN·m; applied 35 kN·m is about 93 percent of it
   C. N L/2 = 112.5 kN·m
   D. N t/6 = 5.75 kN·m
   Answer: B. M0 = N Z/A = N L/6 = 150 x 0.25 = 37.5 kN·m, so the pier is close to cracking. (Key point: Zero-tension moment)

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