stair case design

1. 6.4 DESIGN OF STAIRCASE
STEP 1:- DIMENSIONS
Room dimensions = 3.5mx4.0m
Assuming Rise (R) = 150mm
Thread (T) = 300mm
Width of flight (Wf) =2m
Width of landing = 3m
Height of ech storey = 3.5m
Height of each flight (hf) =1.75m
Total No. of Rises = 3500/150
= 24 nos
Let us keep width of each flight = 1.45 m
STEP 2:- DESIGN OF DOG - LEGGED STAIRCASE
No. of Rises in each flight = 24 nos
No. of Threads required in each flight = 24-1 = 23 nos
Space occupied by Thread = 23x25 = 0.92m
Length of landing = 4- 0.92 = 3.08m
Effective length of flight (l) = 4.00mm
STEP 3:- DEAD LOAD OF FLIGHT
Assuming the thickness is slab = 150mm
Self weight of the slab = 0.15 x 25
= 3.75 KN/m

2. Live load = 4 KN/m
Floor finishes = 0.8 KN/m
Total load = 6.62 KN/m
Factored load = 6.62x1.5
= 9.93 x 2
=19.86 KN/m
Length of each flight = 13 x 0.3 = 3.9m
STEP 4:- BENDINGMOMENT
Maximum bending moment (Mu) = 





8
2
Wl
= 




 
8
9.386.19 2
= 37.75 kN-m
Vu = (19.86 x 3.9) / 2
= 38.727 kN
STEP 5:- CHECK FOR DEPTH
Bending moment (Mu) =0.138 fck b d²
Effective depth (d) =
100020138.0
1075.37 6


= 116mm<180mm
Hence safe
STEP 6:- AREAOF MAIN STEEL
Mx = 0.87 fy Ast d
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

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

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






bdf
fA
ck
yst
1

3. 37.75х106 =0.87х415х Astх180
 
  














180100020
415
1 stA
Ast = 626.38mm²
Using 12mm dia bars
Spacing = 





st
st
A
a
х1000 = ((π/4)х122)/626.38 =180mm
Therefore, provide 12mm dia bars @ 180mm C/C
STEP 7:- CALCULATION FOR TRANSVERS REINFORCMENT
Distribution of reinforcement @ 0.15% of gross c/s area
Asd = 0.15 x 150 x 1000/100
= 225mm2
Assume 8 mm # bars
Spacing (s) = ( 𝜋/4 x 82)/225 = 223 mm c/c
Hence provide 8mm dia bars @ 200mm c/c as distribution reinforcement.