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RCC dog-legged stair for office building

Short answer

Riser 150 mm for easy climbing, tread 280 mm for safe footing (2R+T=580 mm). For 3.6 m height in 4.8×3.0 m: 24 risers at 150 mm, two flights of 12 risers, 11 treads; flight width 1.45 m, waist slab 250 mm.

Draw complete details of an R.C.C. dog-legged staircase for an office building leading from the ground floor to first floor. State reasons for size of the risers and treads adopted by you. The size of stair is 4.8m×3.0m×3.6m (height).

Reasons for the size of riser and tread

  • Riser 150 mm: an office building has heavy traffic of people of all ages, so the riser is kept at the lower limit (150 mm) for comfortable climbing; the floor height of 3.6 m divides exactly into 24 equal risers.
  • Tread 280 mm: a wide tread (250-300 mm) gives safe foot placing; 2R+T=2×150+280=5802R + T = 2 \times 150 + 280 = 580 mm, which is within the usual comfort range of 550-650 mm.
  • The riser and tread are kept equal throughout so that people do not trip.

Given data and assumptions

  • Stairwell (internal): 4.8 m x 3 m; floor-to-floor height H=3.6H = 3.6 m; stair for an office building, RCC.
  • Assumed riser R=150R = 150 mm and tread T=280T = 280 mm; flight width b=1.45b = 1.45 m. Stair size 4.8 m x 3.0 m x 3.6 m (length x width x height). Gap between flights = 3.0 - 2 x 1.45 = 0.1 m.

Design

  1. Number of risers =H/R=3.6/0.15=24= H/R = 3.6/0.15 = 24, adopt 24 risers; exact riser =3.6/24=150= 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 =11×0.28=3.08= 11 \times 0.28 = 3.08 m.
  4. Width: two flights of 1.45 m side by side need 2b=2.92b = 2.9 m; the remaining W−2b=3−2.9=0.1W - 2b = 3 - 2.9 = 0.1 m is the gap between the flights.
  5. Landing: a half-space landing 1.45 m x 3 m (full stairwell width) at height =12×0.15=1.8= 12 \times 0.15 = 1.8 m.
  6. Length check: going + landing =3.08+1.45=4.53= 3.08 + 1.45 = 4.53 m ≤4.8\le 4.8 m, so it fits; the spare 0.27 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).
  8. Waist slab: effective span about 4.53 m, so depth ≈\approx span/20 =226= 226 mm; adopt a 250 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

 +------------------------+---------+
 | 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)
 SECTION (not to scale)
                          +---- upper floor 3.6 m
                    _____|
               ____|   flight 2
        ______|  landing
   ____|  flight 1
 _|
 ground floor +0.00

Summary

ItemAdopted
Riser150 mm (24 nos.)
Tread280 mm
Flights2 flights x 12 risers
Width of flight1.45 m
Waist slab250 mm

Answer: 24 risers of 150 mm and treads of 280 mm; 2 flights x 12 risers; flight width 1.45 m.

Details to be drawn

  • Plans at ground floor and first floor with UP/DN arrows, flight width and landing.
  • Section showing the 150 mm riser, 280 mm tread, waist slab, landing slab, nosing (25 mm) and balustrade (0.9 m high).
  • Reinforcement: main bars along the slope (e.g. 12 mm dia at 150 c/c), distribution bars 8 mm at 200 c/c, concrete M20, steel Fe415 (to be designed separately).

✓ Answer checked by Claude Opus 5.5 (high effort)

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