Skip to main content

Stair design, 1.8 m by 3.9 m

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.

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.

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=2.4H = 2.4 m; stair for a residential building, RCC; waist slab 150 mm (given).
  • Assumed riser R=150R = 150 mm and tread T=280T = 280 mm; flight width b=0.9b = 0.9 m.

Design

  1. Number of risers =H/R=2.4/0.15=16= H/R = 2.4/0.15 = 16, adopt 16 risers; exact riser =2.4/16=150= 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 =7×0.28=1.96= 7 \times 0.28 = 1.96 m.
  4. Width: two flights of 0.9 m side by side need 2b=1.82b = 1.8 m =1.8= 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 =8×0.15=1.2= 8 \times 0.15 = 1.2 m.
  6. Length check: going + landing =1.96+0.9=2.86= 1.96 + 0.9 = 2.86 m ≤3.9\le 3.9 m, so it fits; the spare 1.04 m is kept as entry/exit space.
  7. Check: 2R+T=2×150+280=5802R + T = 2 \times 150 + 280 = 580 mm (within 550-650 mm); pitch =tan⁡−1(R/T)=28.2∘= \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= 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= 143 mm ≤150\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

 +------------------------+---------+
 | 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)
 SECTION (not to scale)
                          +---- upper floor 2.4 m
                    _____|
               ____|   flight 2
        ______|  landing +1.2 m
   ____|  flight 1
 _|
 ground floor +0.00

Summary

ItemAdopted
Riser150 mm (16 nos.)
Tread280 mm
Flights2 flights x 8 risers
Width of flight0.9 m
Waist slab150 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Building Technology chapter 5 questions →