---
title: "Dog-legged stair, 4.8 m by 2.25 m"
subject: "Building Technology"
chapter: 5
key: "semester_6/building_technology:dog-legged-stair-4p8m-room"
source: https://www.bctengineer.com/q/semester_6/building_technology/dog-legged-stair-4p8m-room
site: BCT Engineering
---

# Dog-legged stair, 4.8 m by 2.25 m

Building Technology · Chapter 5: Stair, lifts and escalators

- Asked in: 2067 Asar (old course), 8 marks
- Video: [Design of RCC Dog-Legged Stair | Staircase | Design of Stair | Building design | Structural Design](https://www.youtube.com/watch?v=ZSurkcbC7GA) — Visual Civil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/building_technology/dog-legged-stair-4p8m-room
- Whole chapter: https://www.bctengineer.com/md/semester_6/building_technology/chapter-5

## Question

Define staircase. For a given space of 4.8m × 2.25m with floor height of 3.6m, design a RCC dog-legged staircase.

## Short answer

A staircase is a series of steps linking floors. With 3.6 m height in 4.8×2.25 m: 24 risers at 150 mm, tread 280 mm; two flights of 12 risers, 11 treads, flight width 1.05 m, waist slab 225 mm.

## Answer

A **staircase** is a set of steps (a series of risers and treads) that connects floors at different levels, giving access from one storey to another, usually with landings and a balustrade.

#### Given data and assumptions
- Stairwell (internal): 4.8 m x 2.25 m; floor-to-floor height $H = 3.6$ m; stair for a residential building, RCC.
- Assumed riser $R = 150$ mm and tread $T = 280$ mm; flight width $b = 1.05$ m. Gap between the flights = 2.25 - 2 x 1.05 = 0.15 m.

#### Design
1. **Number of risers** $= H/R = 3.6/0.15 = 24$, adopt **24 risers**; exact riser $= 3.6/24 = 150$ mm.
2. **Flights:** a dog-legged stair has two equal flights, so each flight has 24/2 = 12 risers and 12-1 = 11 treads (the landing or upper floor acts as the last tread).
3. **Going of one flight** $= 11 \times 0.28 = 3.08$ m.
4. **Width:** two flights of 1.05 m side by side need $2b = 2.1$ m; the remaining $W - 2b = 2.25 - 2.1 = 0.15$ m is the gap between the flights.
5. **Landing:** a half-space landing 1.05 m x 2.25 m (full stairwell width) at height $= 12 \times 0.15 = 1.8$ m.
6. **Length check:** going + landing $= 3.08 + 1.05 = 4.13$ m $\le 4.8$ m, so it fits; the spare 0.67 m is kept as entry/exit space.
7. **Check:** $2R + T = 2 \times 150 + 280 = 580$ mm (within 550-650 mm); pitch $= \tan^{-1}(R/T) = 28.2^\circ$ (within 25-40 degrees).
8. **Waist slab:** effective span about 4.13 m, so depth $\approx$ span/20 $= 206$ mm; adopt a 225 mm thick waist slab (reinforcement to be designed separately). Provide a handrail 0.9 m high, 20-25 mm nosing and a non-slip finish.

#### Plan and section
```text
 +------------------------+---------+
 | FLIGHT 2  <<<<<<  UP   |         |
 | 11 treads              | LANDING |
 +------------------------+ (b x W) |
 | FLIGHT 1  >>>>>>  UP   |         |
 | 11 treads              |         |
 +------------------------+---------+
 <-------- going -------> <-- b -->
 (entry at left end of flight 1; arrows point up the stair)
```
```text
 SECTION (not to scale)
                          +---- upper floor 3.6 m
                    _____|
               ____|   flight 2
        ______|  landing
   ____|  flight 1
 _|
 ground floor +0.00
```

#### Summary
| Item | Adopted |
|---|---|
| Riser | 150 mm (24 nos.) |
| Tread | 280 mm |
| Flights | 2 flights x 12 risers |
| Width of flight | 1.05 m |
| Waist slab | 225 mm |

**Answer: 24 risers of 150 mm and treads of 280 mm; 2 flights x 12 risers; flight width 1.05 m.**

## Check yourself

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

1. H = 3.6 m, R = 150 mm gives
   A. 20
   B. 24
   C. 30
   D. 12
   Answer: B. 3.6/0.15 = 24 risers. (Key point: Risers)

2. Treads in each flight of 12 risers
   A. 12
   B. 11
   C. 10
   D. 24
   Answer: B. Upper floor acts as last tread: 12 - 1 = 11. (Key point: Flights)

3. 11 treads of 280 mm give
   A. 3.08 m
   B. 2.8 m
   C. 3.3 m
   D. 2.25 m
   Answer: A. 11 x 0.28 = 3.08 m. (Key point: Going)

4. W = 2.25 m with 1.05 m flights gives gap
   A. 0.15 m
   B. 0.1 m
   C. 1.2 m
   D. 0.25 m
   Answer: A. 2.25 - 2.1 = 0.15 m. (Key point: Well gap)

## Related in Building Technology

- [Dog-legged RCC stair design for 3 m floor height](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Astair-dog-3m-assume) — Building Technology, chapter 5, asked 3×
- [RCC dog-legged stair for office building](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Arcc-stair-office-building) — Building Technology, chapter 5
- [Dog-legged RCC stair, 4.6 m by 2.6 m](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Adog-legged-stair-residential-4p6m) — Building Technology, chapter 5, asked 3×
- [Dog-legged stair design with elevators and ramp](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Astair-078c) — Building Technology, chapter 5
- [Vertical circulation elements; RCC dog-legged stair](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Astair-circulation-068ba) — Building Technology, chapter 5
- [Dog-legged stair, 5.0 m by 4.0 m](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Adog-legged-stair-public-5m) — Building Technology, chapter 5