---
title: "Wall with openings: pier loads and check"
subject: "Design of Timber and Masonry Structures"
chapter: 5
key: "semester_5/design_of_timber_and_masonry_structures:masonry-wall-with-openings-piers"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/masonry-wall-with-openings-piers
site: BCT Engineering
---

# Wall with openings: pier loads and check

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

- Asked in: year not given, 8 marks
- 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-wall-with-openings-piers
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-5

## Question

A 230 mm thick brick wall of clear height 3.0 m and length 6.0 m has two window openings. From left to right the lengths are: end pier 0.75 m, window 1.5 m, middle pier 1.5 m, window 1.5 m, end pier 0.75 m. Floor and roof above transfer a uniform working load of 85 kN/m along the top of the wall through an RC band, which provides lateral and rotational restraint at the top; the wall is also fixed at the base. Take $f_b = 1.0$ N/mm², masonry unit weight 19 kN/m³, and use $k_s$ from the IS 1905 table (SR 8 → 0.95, SR 10 → 0.89). Assume the load above each window is shared equally by the two adjacent piers. Check the piers.

## Short answer

H_eff=2.25 m, SR=9.78, k_s=0.897. End pier area 0.1725 m² gives k_a=0.959, permissible stress 0.860 N/mm²; middle pier k_a=1.0, 0.897 N/mm². Tributary loads plus self-weight give actual stress 0.796 N/mm² in both. Both safe (end pier 93% utilisation).

## Answer

#### Principle
The load over a window opening is carried by the lintel/band and shared by the piers on either side, so the piers carry their own length of wall load plus half of each adjacent opening. Area factor $k_a$ is applied to the smaller piers.

#### Step 1: Slenderness and $k_s$
$H_{eff} = 0.75 \times 3000 = 2250$ mm, $SR = 2250/230 = 9.78 < 27$.

$$
k_s = 0.89 + \frac{10 - 9.78}{2}\times 0.06 = 0.897
$$

#### Step 2: Load on piers
| Pier | Length (m) | Tributary length (m) | Load from top (kN) |
|---|---|---|---|
| End pier | 0.75 | $0.75 + 0.75 = 1.5$ | $1.5 \times 85 = 127.5$ |
| Middle pier | 1.5 | $0.75 + 1.5 + 0.75 = 3.0$ | $3.0 \times 85 = 255.0$ |

Check: $2 \times 127.5 + 255.0 = 510 = 6.0 \times 85$ kN. OK.

#### Step 3: Self-weight (to base)
- End pier: $19 \times 0.23 \times 0.75 \times 3.0 = 9.83$ kN, total $P = 137.3$ kN.
- Middle pier: $19 \times 0.23 \times 1.5 \times 3.0 = 19.67$ kN, total $P = 274.7$ kN.

#### Step 4: Area factor and permissible stress
- End pier: $A = 0.23 \times 0.75 = 0.1725$ m² $< 0.2$, so $k_a = 0.7 + 1.5 \times 0.1725 = 0.959$.
- Middle pier: $A = 0.345$ m² $> 0.2$, $k_a = 1.0$.

$$
f_{c,end} = 1.0 \times 0.897 \times 0.959 = 0.860\ \text{N/mm}^2, \qquad f_{c,mid} = 1.0 \times 0.897 = 0.897\ \text{N/mm}^2
$$

#### Step 5: Actual stress
$$
\begin{aligned}
f_{end} &= \frac{137\,330}{0.1725\times10^{6}} = 0.796\ \text{N/mm}^2 < 0.860 \\
f_{mid} &= \frac{274\,670}{0.345\times10^{6}} = 0.796\ \text{N/mm}^2 < 0.897
\end{aligned}
$$

| Pier | Actual (N/mm²) | Permissible (N/mm²) | Result |
|---|---|---|---|
| End | 0.796 | 0.860 | Safe |
| Middle | 0.796 | 0.897 | Safe |

The end pier has the smallest margin (93 percent utilisation); it should not be reduced in width further. Pier width must also satisfy seismic opening rules.

**Answer: Both piers are safe: stress 0.80 N/mm² against permissible 0.86 N/mm² (end pier) and 0.90 N/mm² (middle pier).**

## Check yourself

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

1. With 85 kN/m on the wall and load over windows shared equally by adjacent piers, what is the top load on the middle pier?
   A. 1.5 x 85 = 127.5 kN
   B. 6.0 x 85 = 510 kN
   C. 3.0 x 85 = 255 kN, from its length plus half of each adjacent window
   D. 0.75 x 85 = 63.8 kN
   Answer: C. The middle pier takes 0.75 + 1.5 + 0.75 = 3.0 m of wall load; each end pier takes 1.5 m, and the total checks to 510 kN. (Key point: Load sharing by piers)

2. For H = 3.0 m with top and base restraint and t = 230 mm, SR and k_s are:
   A. SR = 13, k_s = 0.78
   B. SR = 9.78, k_s = 0.897
   C. SR = 26, k_s = 0.46
   D. SR = 9.78, k_s = 1.0
   Answer: B. H_eff = 0.75 x 3000 = 2250 mm, so SR = 9.78, and interpolation between 8 (0.95) and 10 (0.89) gives 0.897. (Key point: Slenderness and k_s)

3. The end pier has area 0.23 x 0.75 = 0.1725 m^2. What is k_a?
   A. 1.0, because walls always get 1.0
   B. 0.7, the minimum
   C. 0.959, from 0.7 + 1.5A since A < 0.2 m^2
   D. 1.2
   Answer: C. Areas under 0.2 m^2 take k_a = 0.7 + 1.5A = 0.959; the middle pier (0.345 m^2) has k_a = 1.0. (Key point: Area factor for piers)

4. The end pier has actual stress 0.796 N/mm^2 and permissible stress 0.860 N/mm^2. What is the conclusion?
   A. Unsafe, since the actual stress exceeds the permissible
   B. Safe, with the smallest margin (about 93 percent utilisation), so it should not be narrowed
   C. Safe with a large reserve of 50 percent
   D. Cannot be checked without bending
   Answer: B. 0.796 < 0.860, so the pier is safe, but at about 93 percent utilisation it has the least margin of the piers. (Key point: Pier check result)

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