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design of stair case
The document provides steps to design a dog-legged staircase. It specifies dimensions for the room and staircase including a rise of 150mm and tread of 300mm. It then calculates the total rise, number of flights, and load on each flight. The maximum bending moment of 37.75 kN-m is calculated. The required depth of 116mm is less than the permitted 180mm. The main steel reinforcement is calculated to be 626.38mm2 using 12mm diameter bars spaced at 180mm. The transverse reinforcement is calculated to be 225mm2 using 8mm diameter bars at 200mm spacing.
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design of stair case
- 1. 6.4 DESIGN OFSTAIRCASE 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 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.