A staircase is an important component of a building that is planned and designed based on the type and orientation of the building. Therefore, it is impossible to recommend a definite dimension for the stair without a clear idea about the building's general design. The staircase designed must properly fit the given building plan.
Staircase can be designed based on the direction along which the stair slab spans. It is classified as:
- Stairs spanning horizontally
- Stairs spanning vertically
This article explains the common design steps involved in the design of R.C.C stairs spanning longitudinally with the help of an example. Indian standard codes, IS 456-2000 and SP 16-1980 is used to conduct the design.
Design of Stairs Spanning Longitudinally
The stairs spanning longitudinally are supported at the bottom and the top of the flights. These types of stairs are not supported at the sides by means of beams. Here, for the design purpose, the bending moment is calculated per unit width as wl2/10, where ‘w’ is the load per unit horizontal area and ‘l’ is the effective horizontal span.
Sometimes, stairs are cast along with the landings that are supported on the walls (Figure-1). In such cases, the effective span is taken as the horizontal distance between the centres of the bearings. Here, the maximum bending moment is calculated as wl2/8.
Example-1
The main stair of an office building should be located in an area measuring 3.5 m x 5.5 m. The vertical distance between the floors is 3.75 m. Design the stairs. Take live load (L.L) of 2000 N/m2, M20 grade concrete, and Fe 415 steel.
Solution
1. Preliminary Dimensioning of the Stairs
The height to be covered is 3.75 m. Hence, it is proposed to have two flights for the stairway.
The height of each flight = Total height to be covered / No. of flights = 3.75 /2
The height of each flight = 1.875 m
Assume the height of risers = 150 mm
No. of Risers = Height of flight / height of riser= 1875/150
No. of Risers = 12 nos
Therefore, actual height of riser ‘R’ = 1875 /12
Riser Height, r = 156 mm
No. of threads = No. of risers – 1= 12 – 1= 11 nos;
Let,
- The width of the stair = 1600 mm
- The width of thread ‘T’ = 270 mm
2. Design of Flight AB
As shown in Figure-2 above, the flights are spanning longitudinally by means of landing slabs at the top and bottom of the flight. Let the bearing of the flight be 150 mm.
2.1. Effective Horizontal Span (Leff)
Leff = 2.97 + 1.60 + (0.150 /2) = 4.645 m
2.2. Thickness of Waist Slab
The thickness of waist slab can be assumed as 40 mm to 50 mm per metre run of horizontal span. Here, a waist slab thickness is taken as 220 mm.
2.3. Load Calculation
Load on Waist Slab
a. Self-weight of waist slab = weight of concrete x thickness of waist slab
= 25 x 0.22 = 5.5 kN/m2
= 5500 N/m2
b. Ceiling Finish of thickness 12.5 mm = weight of P.C.C x thickness of finish
= 24 x 0.125 = 0.3 kN/m2
= 300 N/m2
Therefore, total load on waist slab 'ws'= 5500 + 300 = 5800 N/m2
The above load obtained is inclined. This must be converted into a horizontal load (w) by multiplying it with the factor. If the length of riser and thread is 'R' and 'T,' then,
c. Dead load of steps = (½) x Riser length x Concrete unit weight
= 0.5 x (156.2) x 25 = 1952 N/m2
d. Top finish of 12.5 mm= 12.5 x 24 = 300 N/m2
e. Live load on waist slab= 2000 N/m2
Total load on waist slab = 6700 + 1952 + 300 + 2000 = 10,952 N/m2
2.4. Moment on Waist Slab
Maximum bending moment per metre width of the stairs,
M = wl2/8;
= (10,952 x 4.6452)/8
M = 29538 Nm
Ultimate Moment, Mu = 1.5 x M = 1.5 x 29538 = 44307 Nm
2.5. Check for Depth of Waist Slab
From SP 16-1980, From Table C, Clause 2.3;
Mu,limit = 0.138fckbd2;
Therefore, dreq = sqrt [Mu,limit / (0.138 x fckb)
= sqrt [44307 x 1000 / (0.138 x 25 x 1000)]
dreq = 127 mm
If 10 mm diameter bars are provided, the effective cover = clear cover + half bar diameter = 15 mm + 5 mm = 20 mm
Then, overall depth required D = 127 + 20 = 147;
Provide an overall depth of 220 mm.
Effective depth = overall depth – effective cover = 220 – 20 = 200 mm;
2.6. Calculation of Reinforcement
Main Reinforcement for Stairs
From SP 16-1980, Table-3
Mu/bd2 = [44307 x 1000] / [1000 x 2002] = 1.11
By interpolation, percentage steel pt = 0.33%
Ast,req = (pt x bd)/100 = (0.33 x 1000 x 200)/100 = 660 mm2
Assume 10 mm diameter bars;
Area of single bar = (3.14 x r x r) = 3.14 x 5 x 5 = 78.5 mm2
Spacing of main reinforcement bars = 1000/ [Ast,req / Area of single bar]
= 1000/[660 / 78.5]
= 120 mm c/c
Provide main reinforcement of 10 mm diameter bars at 120 mm c/c
Distribution Reinforcement for Stairs
Provide 0.12% of cross-sectional area as distribution rebars as per IS:456-2000,
= 0.12% x (1000 x 220) = 264 mm2
Assume 8 mm diameter bars
Spacing of main reinforcement bars = 1000/ [Ast,req / Area of single bar]
= 1000/ [264 / 50.24]
= 190 mm c/c
Hence,
Provide distribution bars of 8mm diameter @ 180mm c/c
3. Reinforcement Details of Staircase
FAQs
A longitudinally spanning staircase spans horizontally and is supported at the bottom and top of the flights either by beams, walls, or landings.
Given the height of flight and the depth of riser of the steps, the number of steps = height of flight / riser
Read More
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Requirements of Staircases – General Guidelines about Heights, Headroom, Treads, and Risers