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Stair case are generally help for the transportation and communication between the floors of the multi storey structure. Stair cases are most important in any building structure for the communication between the floors in building structures. In this article you can able to learn about how to design the reinforcement values of main and distribution in the stair slab according to the IS 456 2000 code standards.
For the designing of the stair cases we will follow the steps of the slab design as per IS 456. The complete procedure of RCC slab design as per IS 456 2000 is explained in my previous blog post read with below link.
Various steps used in two way slab design as per IS 456 2000 code
Stair case design
The above image shows the elevation view of the stair case which is used in this example. The dimensions of the stair slab is 4.87mX1.21m cross section.
The following steps are considered in the designing of stair cases
- Calculation of long span to short span ratio
- Calculation of effective depth
- Loads calculations
- Bending moment calculations
- Check for the Depth
- Calculation of Area of the steel
- Check for the deflection
4.87mX1.21m dimensions stair design as per IS 456 2000 code standards
Step 1: Calculation of long span to short span ratio
As per the dimensions of the stair case the long span is taken as 4.87m and short span is taken as 1.21m.
Ll/Ls = 4.87/1.21 = 4.02
If the ratio of long span to short span is less than 2 we will consider two way slab design condition and if the ratio of long span to short span is greater than 2 we will consider one way slab design condition.
Here the factor 4.02 is greater than 2 so we have to follow the one way slab design steps.
Step 2: Calculation of effective depth
Let us consider the thickness of the slab = 140mm
Cover = 15mm, Diameter of the bar = 10mm
Effective depth = Thickness of slab – (cover+dia/2)
= 140-(15+10/2)
So, Effective depth = 120mm.
Step 3: Loads calculations
The dead load of stair case slab is considered as sum of W1+W2
Where W1 = (tx√(tX(1+(R/T)2)) X25
Let us consider T=0.28m and R =0.17m
W1 = (0.14x√(1+(0.17/0.28)2) X25
= 4.09kN/m2
And W2 = (1/2)XRX25
= (1/2)X0.17X25
=2.125 kN/m2
Now the sum of W1+W2 = 4.09+2.125 = 6.125 kN/m2
Floor finishing load = 1.5 kN/m2
Assume live load = 3 kN/m2
Total load W = 6.125+1.5+3 = 10.715 kN/ m2
Now the factored load = Wu = 1.5XW = 1.5X10.715 = 16.0725 kN/ m2
Step 4: Bending moment calculations
One Side Continuous
The slab is simply supported
The maximum Bending Moment for simply supported structure =Wl2/8
=16.0725X1.212/8
= 2.941 KN-m
Step 5: Check for Depth
Mu = 0.138XfckXbXd2
=d2 = Mu/(0.138XfckXb)
d =√(Mu/(0.138XfckXb))
let us consider the grade of the concrete = 20Mpa and grade of the steel is 500Mpa.
=√(2.941X1000000/(0.138X20X1000))
d = 32.64
Assume effective depth 120mm > check for depth 32.64mm
Hence depth is safe according to the consideration
Step 6: Calculation of Area of steel (Ast):-
- a) Design of reinforcement along short span:-
Required Ast = 0.5X(fck/fy) X[1-(√1-(4.6XMu/fckXbXd2) XbXd
= 0.5X(20/500) X[1-(√1-(4.6X2.941X1000000/20X1000X1202) X1000X120
= 57.06 square mm
Spacing (s) =((Π/4 (dxd))/Ast)X1000
=((Π/4 (10×10))/57.06)X1000
= 1375.83 mm
Adopt (s) = 150mm
Provide Ast =((Π/4 (10×10))/150) x 1000
= 523.33m
“T10 @ 150 mm c/c ”
The main reinforcement for the 4.87mX1.21m cross section we will consider 10mm diameter bars which are placed at 150mm center to center distance.
b) Distribution steel:-
Minimum Area Of Steel = 0.12 % x b x d
=(12/100)x1000x 120
=144 mm2
Spacing (s) = ((Π/4 (dxd))/Ast)X1000
= ((Π/4 (10×10))/144)X1000
= 545.23 mm
Adopt (s) = 200mm
Provide Ast = ((Π/4 (10×10))/200)X1000
= 392.5m“ T10 @ 200 mm c/c ”
The distribution reinforcement for the 4.87mX1.21m cross section we will consider 10mm diameter bars which are placed at 200mm center to center distance.
8) Check for Deflection :-
l/d provided<l/d max
provided =1210/120 = 10.083
l/d max = Modification Factor x 32
Modification Factor value comes from graph with the help of &
in IS 456 : 2000 page no:38
fs= steel stress of services = 0.58 x fyx(Ast required/Ast Provided)
= 0.58 x 500 x (144/392.5)
= 106.39 N/mm2
Pt= Percentage of steel Ast/bd = (144/(120X1000))x 100
= 0.12%
max = 2 x 32 = 64
l/d provided = 10.083<l/d max = 64
Hence deflection is safe
For the detailed calculations and design of stair cases design see the video in my YouTube channel Civil engineering by shravan
Follow our previous blog posts here
Various steps used in the two way slab design according to IS 456 : 2000 code standard
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Various types of the wind design codes used in the building design all over the world
Different types of the seismic design codes used in the building design all over the world
Types of beams and how to calculate the volume of concrete required for the RCC beam
Conclusions of stair cases design as per IS 456 2000 code standards
The stairs are used for the communication between the floors in the buildings. As per the design considerations made with IS 456 2000 standards it is obtained the load which is 10.715 kN/ m2, bending moment is 2.941 KN-m, main reinforcement is 10mm diameter bars which are placed at 150mm center to center distance and distribution reinforcement of 10mm diameter bars with 200mm center to center distance.
For more interesting updates and civil engineering concepts follow my YouTube channel Civil engineering by shravan.
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