---
title: "Distribution of lateral force among masonry walls"
subject: "Design of Timber and Masonry Structures"
chapter: 6
key: "semester_5/design_of_timber_and_masonry_structures:lateral-load-sharing-among-piers"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/lateral-load-sharing-among-piers
site: BCT Engineering
---

# Distribution of lateral force among masonry walls

Design of Timber and Masonry Structures · Chapter 6: Masonry Structures Under Lateral Loads

- Asked in: year not given, 8 marks
- Video: [How do structures carry wind and seismic loads? An Intro to Lateral Force Resisting Systems](https://www.youtube.com/watch?v=yFX4JLe3JJM) — Structures with Prof. H
- 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/lateral-load-sharing-among-piers
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-6

## Question

A single-storey brick building has a rigid roof slab which transfers a total lateral seismic force of 240 kN to four parallel in-plane shear walls (piers), each 3.0 m high and 230 mm thick. The wall lengths are 1.0 m, 1.5 m, 2.0 m and 3.0 m. Treat each wall as fixed at the base and restrained against rotation at the top by the slab. Take $E = 2500$ N/mm² and $G = 0.4E$. Calculate the share of each wall (consider both flexural and shear deformations).

## Short answer

Considering flexural and shear deformations, wall stiffnesses are computed; force shares are proportional to stiffness. The 1.0 m, 1.5 m, 2.0 m, and 3.0 m walls take 14 kN, 36 kN, 64 kN, and 126 kN, respectively.

## Answer

#### Principle
A rigid diaphragm gives the same horizontal displacement $\Delta$ to all walls, so each wall takes force in proportion to its **stiffness** $k = V/\Delta$.

For a wall fixed at both ends (top rotation prevented):

$$
\Delta = \frac{Vh^3}{12EI} + \frac{1.2\,Vh}{GA}
\quad\Rightarrow\quad
k = \frac{1}{\dfrac{h^3}{12EI} + \dfrac{1.2h}{GA}}
$$

where $I = tL^3/12$ and $A = tL$ (the factor 1.2 is the shape factor for rectangular shear).

Data: $h = 3000$ mm, $t = 230$ mm, $E = 2500$ N/mm², $G = 1000$ N/mm².

#### Stiffness of each wall
| Wall $L$ (m) | $I$ (mm⁴) | $A$ (mm²) | Flexural term $\frac{h^3}{12EI}$ | Shear term $\frac{1.2h}{GA}$ | $k$ (N/mm) |
|---|---|---|---|---|---|
| 1.0 | $1.917\times10^{10}$ | $2.3\times10^5$ | $4.7\times10^{-5}$ | $1.565\times10^{-5}$ | 15 970 |
| 1.5 | $6.469\times10^{10}$ | $3.45\times10^5$ | $1.39\times10^{-5}$ | $1.043\times10^{-5}$ | 41 070 |
| 2.0 | $1.533\times10^{11}$ | $4.6\times10^5$ | $5.87\times10^{-6}$ | $7.83\times10^{-6}$ | 73 020 |
| 3.0 | $5.175\times10^{11}$ | $6.9\times10^5$ | $1.74\times10^{-6}$ | $5.22\times10^{-6}$ | 143 750 |

Sum $\Sigma k = 273\,810$ N/mm.

#### Shares of 240 kN
$$
V_i = V \frac{k_i}{\Sigma k}
$$

| Wall $L$ (m) | $k_i/\Sigma k$ | Force $V_i$ (kN) |
|---|---|---|
| 1.0 | 0.058 | 14.0 |
| 1.5 | 0.150 | 36.0 |
| 2.0 | 0.267 | 64.0 |
| 3.0 | 0.525 | 126.0 |
| **Total** | 1.000 | **240.0** |

Displacement of the slab: $\Delta = 240\,000/273\,810 = 0.88$ mm (same for all walls).

#### Remarks
- The longest wall (3.0 m) takes 52.5 percent of the force even though it is only 3 times as long as the shortest, because stiffness grows faster than length.
- Slender walls are weaker in resisting lateral force. For these properties the flexural and shear terms are equal at $L = h/\sqrt{3} \approx 1.73$ m, so shorter piers (about $L/h < 0.58$) are dominated by flexure. If the walls are treated as cantilevers (free at top), the shares become 8.8, 27.1, 57.2 and 147.0 kN, so the long wall attracts even more.
- If only shear deformation were considered, the shares would be in proportion to the wall lengths.

**Answer: Shares are 14 kN, 36 kN, 64 kN and 126 kN for the 1.0, 1.5, 2.0 and 3.0 m walls respectively.**

## Check yourself

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

1. With a rigid diaphragm, how is lateral force shared among walls?
   A. Equally among walls
   B. In proportion to wall height
   C. In proportion to the weight of each wall
   D. In proportion to stiffness, because all walls have the same displacement
   Answer: D. A rigid slab imposes the same displacement, so force = k x delta and walls share by stiffness. (Key point: Load sharing principle)

2. For a wall fixed at both ends, what is its deflection?
   A. Flexural term Vh^3/(12EI) plus shear term 1.2Vh/(GA)
   B. Only Vh^3/(3EI)
   C. Only Vh/(GA)
   D. Vh^3/(12EI) minus 1.2Vh/(GA)
   Answer: A. Both bending and shear deformation add, with 1.2 as the rectangular shear shape factor, and k is the inverse. (Key point: Stiffness formula)

3. Which share is obtained for the 3.0 m wall of 240 kN?
   A. 60 kN (equal share)
   B. 126 kN (about 52.5 percent)
   C. 180 kN
   D. 80 kN (proportional to length)
   Answer: B. Its stiffness is 143 750 of 273 810 N/mm, giving 0.525 x 240 = 126 kN. (Key point: Result of sharing)

4. Why does the longest wall take more than its length share?
   A. It is heavier
   B. Stiffness grows faster than length, since I varies with L^3
   C. It has a higher E
   D. The slab is thicker above it
   Answer: B. Flexural stiffness grows roughly with L^3, so long walls attract disproportionately more force; only pure shear would give length proportion. (Key point: Effect of length)

## Related in Design of Timber and Masonry Structures

- [Seismic base shear of a masonry building](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Abase-shear-two-storey-masonry-building) — Design of Timber and Masonry Structures, chapter 7
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