---
title: "Seismic base shear of a masonry building"
subject: "Design of Timber and Masonry Structures"
chapter: 7
key: "semester_5/design_of_timber_and_masonry_structures:base-shear-two-storey-masonry-building"
source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/base-shear-two-storey-masonry-building
site: BCT Engineering
---

# Seismic base shear of a masonry building

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

- Asked in: year not given, 8 marks
- Video: [Seismic base shear of a special reinforced masonry bearing shear wall of a building](https://www.youtube.com/watch?v=3HKFKJ5TkfM) — Olivier Ouedraogo
- 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/base-shear-two-storey-masonry-building
- Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-7

## Question

A two-storey load-bearing masonry building (plan 8 m × 6 m, height 6 m, each storey 3 m) is in Zone IV ($Z = 0.24$) on medium soil. Seismic weights are 450 kN at the roof and 600 kN at the first floor. Importance factor $I = 1.0$, response reduction factor $R = 1.5$. Use the seismic coefficient method of IS 1893: $A_h = \dfrac{Z}{2}\dfrac{I}{R}\dfrac{S_a}{g}$, with $T = 0.09h/\sqrt{d}$ taking $d$ the plan dimension in the direction of force (8 m). Take $S_a/g = 2.5$ for the period found. Find the design base shear and its distribution to floors.

## Short answer

Period T=0.09×6/√8=0.19 s; Ah=(0.24/2)×(1/1.5)×2.5=0.20; base shear Vb=0.20×1050=210 kN. Distribution: roof 157.5 kN, first floor 52.5 kN.

## Answer

#### Step 1: Fundamental period
$$
T = \frac{0.09\,h}{\sqrt{d}} = \frac{0.09 \times 6}{\sqrt{8}} = 0.19\ \text{s}
$$

For medium soil with $0.1 \le T \le 0.55$ s, the spectral value $S_a/g = 2.5$ (plateau).

#### Step 2: Design horizontal coefficient
$$
A_h = \frac{0.24}{2}\times\frac{1.0}{1.5}\times 2.5 = 0.12 \times 0.667 \times 2.5 = 0.20
$$

#### Step 3: Base shear
Seismic weight $W = 450 + 600 = 1050$ kN.

$$
V_B = A_h W = 0.20 \times 1050 = 210\ \text{kN}
$$

#### Step 4: Distribution of base shear
$$
Q_i = V_B \frac{W_i h_i^2}{\sum W_j h_j^2}
$$

| Level | $W_i$ (kN) | $h_i$ (m) | $W_i h_i^2$ | $Q_i$ (kN) | Storey shear (kN) |
|---|---|---|---|---|---|
| Roof | 450 | 6 | 16 200 | $210 \times 16200/21600 = 157.5$ | 157.5 |
| First floor | 600 | 3 | 5 400 | $210 \times 5400/21600 = 52.5$ | 210.0 |
| **Sum** | 1050 | | 21 600 | 210.0 | |

```text
        Q2 = 157.5 kN
   --->  =================  roof (6 m)
   |
        Q1 = 52.5 kN
   --->  =================  floor (3 m)
   |
   ^ base shear VB = 210 kN
```

#### Remarks
- The design force is 20 percent of the weight. NBC 105 uses a similar approach: base shear = $C_d(T)\,W$, where $C_d(T)$ depends on zone factor, soil type, importance factor, ductility and overstrength factors.
- The shear is then shared among the walls in the direction of force in proportion to their stiffness (rigid diaphragm) or their tributary area (flexible diaphragm), and each wall is checked for shear and overturning.

**Answer: $T = 0.19$ s, $A_h = 0.20$, base shear $V_B = 210$ kN; lateral forces: 157.5 kN at roof and 52.5 kN at first floor.**

## Check yourself

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

1. What is T for h = 6 m and d = 8 m?
   A. T = 0.09 x 6 x sqrt(8) = 1.53 s
   B. T = 0.09 x 6 / sqrt(8) = 0.19 s
   C. T = 0.09 x 8 / sqrt(6) = 0.29 s
   D. T = 0.09 x 6 / 8 = 0.07 s
   Answer: B. T = 0.09h/sqrt(d) = 0.54/2.83 = 0.19 s, within the plateau 0.1-0.55 s. (Key point: Fundamental period)

2. What is A_h?
   A. 0.24 x 2.5 = 0.60
   B. 0.24/2 x 1.5 x 2.5 = 0.45
   C. (0.24/2)(1.0/1.5)(2.5) = 0.20
   D. 0.24/1.5 = 0.16
   Answer: C. A_h = (Z/2)(I/R)(Sa/g) = 0.12 x 0.667 x 2.5 = 0.20. (Key point: Design coefficient)

3. What is the design base shear?
   A. V_B = 0.20 x 450 = 90 kN
   B. V_B = 0.20 x 600 = 120 kN
   C. V_B = 0.20 x (450 + 600) = 210 kN
   D. V_B = 0.24 x 1050 = 252 kN
   Answer: C. The seismic weight is 1050 kN and V_B = A_h W = 210 kN. (Key point: Base shear)

4. How is base shear distributed to levels?
   A. Q_i = V_B W_i h_i^2 / sum(W_j h_j^2), giving 157.5 kN at roof and 52.5 kN at first floor
   B. In proportion to W_i only, giving 90 and 120 kN
   C. Equally, 105 kN each
   D. In proportion to h_i only
   Answer: A. The weight times height squared is 16 200 and 5 400, so the roof takes 75 percent of 210 kN. (Key point: Distribution to floors)

## Related in Design of Timber and Masonry Structures

- [Multi-storey wall: stresses at each level](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Athree-storey-wall-base-stress-check) — Design of Timber and Masonry Structures, chapter 5
- [Distribution of lateral force among masonry walls](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Alateral-load-sharing-among-piers) — Design of Timber and Masonry Structures, chapter 6
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