---
title: "Stair design, 1.8 m by 3.9 m"
subject: "Building Technology"
chapter: 5
key: "semester_6/building_technology:stair-residential-1p8m-wide"
source: https://www.bctengineer.com/q/semester_6/building_technology/stair-residential-1p8m-wide
site: BCT Engineering
---

# Stair design, 1.8 m by 3.9 m

Building Technology · Chapter 5: Stair, lifts and escalators

- Asked in: 2073 Magh
- Video: [Design of Staircase](https://www.youtube.com/watch?v=vJCQEShrgis) — Engineering Motive
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/building_technology/stair-residential-1p8m-wide
- Whole chapter: https://www.bctengineer.com/md/semester_6/building_technology/chapter-5

## Question

The inside dimension of a stair case in a residential building are 1.8 m×3.9 m. The height of floor to floor is 2.4 m. The thickness of waist slab is 0.15 m. Design a proper layout with neat sketches.

## Short answer

Dog-legged stair with 16 risers of 150 mm, treads 280 mm, two flights of 8 risers each, flight width 0.9 m, half-space landing at 1.2 m.

## Answer

Data: stair room 3.9 m x 1.8 m, floor height 2.4 m, waist slab thickness 150 mm. A single straight flight would need 16 risers in one flight (more than the usual maximum of 12) and would fill the whole 3.9 m, leaving no landing for entry and exit. The 1.8 m width suits a **dog-legged stair** with two 0.9 m flights side by side, so this is adopted.

#### Given data and assumptions
- Stairwell (internal): 3.9 m x 1.8 m; floor-to-floor height $H = 2.4$ m; stair for a residential building, RCC; waist slab 150 mm (given).
- Assumed riser $R = 150$ mm and tread $T = 280$ mm; flight width $b = 0.9$ m.

#### Design
1. **Number of risers** $= H/R = 2.4/0.15 = 16$, adopt **16 risers**; exact riser $= 2.4/16 = 150$ mm.
2. **Flights:** a dog-legged stair has two equal flights, so each flight has 16/2 = 8 risers and 8-1 = 7 treads (the landing or upper floor acts as the last tread).
3. **Going of one flight** $= 7 \times 0.28 = 1.96$ m.
4. **Width:** two flights of 0.9 m side by side need $2b = 1.8$ m $= 1.8$ m, so there is no well between the flights (a true dog-legged stair).
5. **Landing:** a half-space landing 0.9 m x 1.8 m (full stairwell width) at height $= 8 \times 0.15 = 1.2$ m.
6. **Length check:** going + landing $= 1.96 + 0.9 = 2.86$ m $\le 3.9$ m, so it fits; the spare 1.04 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). Headroom over the first step of flight 1 (below the upper floor slab) $= 2.4 - 0.15 - 0.15 = 2.1$ m, which meets the usual minimum of 2.0-2.1 m.
8. **Waist slab:** effective span about 2.86 m, so depth $\approx$ span/20 $= 143$ mm $\le 150$ mm, so the given 150 mm waist slab is adequate (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   |         |
 | 7 treads               | LANDING |
 +------------------------+ (b x W) |
 | FLIGHT 1  >>>>>>  UP   |         |
 | 7 treads               |         |
 +------------------------+---------+
 <-------- going -------> <-- b -->
 (entry at left end of flight 1; arrows point up the stair)
```
```text
 SECTION (not to scale)
                          +---- upper floor 2.4 m
                    _____|
               ____|   flight 2
        ______|  landing +1.2 m
   ____|  flight 1
 _|
 ground floor +0.00
```

#### Summary
| Item | Adopted |
|---|---|
| Riser | 150 mm (16 nos.) |
| Tread | 280 mm |
| Flights | 2 flights x 8 risers |
| Width of flight | 0.9 m |
| Waist slab | 150 mm (given; span/20 = 143 mm, OK) |

**Answer: dog-legged stair with 16 risers of 150 mm and treads of 280 mm; 2 flights x 8 risers; flight width 0.9 m; half-space landing at 1.2 m.**

## Check yourself

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

1. Why is a single straight flight adopted?
   A. Going 15 x 260 = 3.9 m fits the 3.9 m length
   B. Floor is very tall
   C. Room is square
   D. Risers are 300 mm
   Answer: A. The full 3.9 m length is used by 15 treads of 260 mm. (Key point: Layout)

2. H = 2.4 m, R = 150 mm gives
   A. 12
   B. 14
   C. 16
   D. 18
   Answer: C. 2.4/0.15 = 16 risers, so 15 treads. (Key point: Risers)

3. What is added because the 1.8 m flight is wide?
   A. A middle handrail
   B. A lift
   C. An extra landing
   D. A ramp
   Answer: A. A middle handrail helps on a wide flight. (Key point: Wide flight)

4. 2R + T = 2 x 150 + 260 equals
   A. 560 mm, within 550-650 mm
   B. 460 mm
   C. 660 mm
   D. 300 mm
   Answer: A. 560 mm is within the comfort range. (Key point: Comfort)

## 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 stair for 6.0 m by 4.2 m lobby](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Arcc-stair-public-lobby-6m) — Building Technology, chapter 5
- [RCC stair, 2.1 m by 4.5 m room](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Arcc-stair-room-2p1m) — Building Technology, chapter 5
- [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
- [Expansion joint in a residential building](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Aexpansion-joint-350wall) — Building Technology, chapter 10
- [Passive solar design and building layout](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Apassive-solar-design) — Building Technology, chapter 1

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- [Slab panel 2 for BM and SF (floor plan)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-panel-2-for-bm-and) — Design of RCC Structure, chapter 9
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