design of one way slab
Design of one way slab, design procedure and a problem
PROCEDURE:
Find that the given question is one way slab or two way slab
=> to find whether one way or two way slab use the formulae: Ly/Lx>2 one way slab
Find that the given question is one way slab or two way slab
=> to find whether one way or two way slab use the formulae: Ly/Lx>2 one way slab
=> if Ly/Lx ≤2 then two way slab
In one way slab the moment action is predominant along shorter span dimension therefore the design of one way slab is normally based on moments along shorter span.
Determination of thickness of slab
Caluclation of load
In one way slab the moment action is predominant along shorter span dimension therefore the design of one way slab is normally based on moments along shorter span.
NOTE: The examples followed are only simply supported one way slab but not any other continuous slab.
..........as per IS-456-2000 (cl 23.2.1, pg 37)
i)Find d value and assume as 125 for further purpose
write all the loads given in the question and find if any thing necessary
calculate ultimate load = total load x F.O.S(factor of safety(1.5))
Effective span le= live load+b
le=live load+d
the lesser value should be taken
check effective depth of the slab
check effective depth of the slab
Mud=wl2/8
Design moment value
Mud=0.36 x fck x Xumax/d x (1 - 0.42 x Xumax/d ) x bd2
find d value from the above formulae
if d<d provided(126)
so safe
Caluclation of reinforcement:
Mud=0.87 x fy x Ast x (1-(Ast X fy)/(bd X fck)) x d
----as per Is 456-2000(Anexure G)
find Ast from above formulae
Minimum reinforcement
-------as per Is 456-2000(cl 26.5.2.1)
Astmin=0.12% x b x d (here d is overall depth)
Astprovided >Astmin
Provide spacing for the Ast and find number of bars required
(Spacing) S= (b X π/( 4) X Bars(10 or 8 or 16 or 20 etc))/Ast
Reinforcement along long span direction
For Astmin we have to calculate spacing
(Spacing) S= (b X π/( 4) X Bars(10 or 8 or 16 or 20 etc)/Astmin
The maximum spacing is 450mm as per Is 456-2000 (cl 26.3.3(b) pg:46
Check for shear
Usually shear effect in slab is very small the slab it self takes care of shear effect.
first calculate Vu=(w X l)/2
Nominal shear stress (Ï„v)=Vu/(b X d)
(Percentage of steel)Pt=(100 X Ast)/(b X d)
find the Ï„c value by using interpolation concept
as per IS 456-2000(cl 21.1.1) pg:72
For slab design shear =K x Ï„c
if design shear strength > Ï„v
Then no shear shear reinforcement is needed
Check for deflection:
Pt=(100 X Ast)/(b X d)
L/d=20 x M.F(Modification factor)—as per fig 4 pg:38
then find d value
Check for development length:
Ld=(0.87 X fy X d)/(4 X Ï„ bd)
Detailing: Detailed view of the slab.
=> Below is an example problem of One way slab and you can follow the above procedure to solve this problem, and also the solution of the below problem is given and refer that problem and solve it.
=> If you have any query you can free to comment in the bottom comment bar, we are here to help you.
Problem :
Τc= 0.45 N/MM2
à Design a simply supported one way slab for a room with clear dimensions of 3.5mx8m.The support having width of 250mm.let us take a live load of 3kN/m2 and floor finish of 1.5KN/m2 ,use M20 mix and Fe415 grade steel?
Sol:
Given data
Live load =3.5 KN/M2
floor finish=1.5KN/M2
use M20 mix & Fe415 steel
Edge support =250mm width
à Find whether the given question is one way or two way slab
Here ly=8m & lx=3.5m
Ly/Lx>2 ------ 8/(3.5)>2
so given problem is one way slab
3.5m=3500mm & 8m=8000mm
Note:
=> If Ly/Lx > 2 then the given problem is one way slab.
=> If Ly/Lx < 2 then the given problem is Two way slab.
à Deflection criteria
as per cl 23.2.1 pg:137
L/d=20 for simply supported beam
here we are using Fe415 steel
from pg:38, fig 4 find modification factor
l/d=20 X 0.8
d=3500/(20 X 0.8) = 218.75
How ever d=126mm (effective depth)
Let us use 8mm diameter bars and 20mm clear span
Overall depth=125+8/2+20 =149mm ~ 150mm
d=126mm
D=150mm
à Caluclation of one way slab load
self weight of the slab= D X ɤ D=150mm=0.15m
=0.15 x 25 ɤ=25 (assume)
=3.75 KN/M2
floor finish = 1.5 KN/M2
Live load =3.5 KN/M2
Total load =3.75+1.5+3.5=8.75 KN/M2
Dead load = 3.75+1.5= 5.25 KN/M2
Ultimate load (Wu)= Total Load x F.O.S = 8.75 x1.5= 13.125 KN/M
Effective span(le)= live load + d = 3.5+0.15= 3.65m
(le)= live load + b = 3.5+0.25= 3.75 m
le=lesser value
à Check for effective depth of the one way slab
Mud=wl2/8 =(13.125 X (3.65)2)/8 =21.85 KN-M
Design moment value
Mud=0.36 x fck x Xumax/d x (1 - 0.42 x Xumax/d ) x bd2
21.85 X106=0.36 x 20 x 0.48x (1 - 0.42 x 0.48) x 1000 x d2
d=89mm < dprovided (126mm)
so safe
Note: When d is greater than dprovided then the one way slab design criteria fails, so you have to re assume the depth of the slab once again and re calculate the effective depth of the slab and check up to then, d is less than dprovided.
à Caluclation of reinforcement
Mud=0.87 x fy x Ast x (1-(Ast X fy)/(bd X fck)) x d
21.85=0.87 x 415 x Ast x (1-(Ast X 415)/(1000 X 126 X 20)) x 126
Ast=525.82 MM2
Minimum reinforcement:- as per cl 26.5.2.1 pg:48
Astmin=0.12/100 x 1000 x 150
Astmin=180 mm2
Ast provided > Astmin
so safe
Spacing of 10mm dia bars
S= (1000X π/( 4) X 102 )/525.5= 149.36mm
so provide 10mm dia bars @145mm c/c
number of bars= 1000/145= 6.69 ~ 7 bars
à Reinforcement along long span direction
To arrest the cracks due to temperature and shrinkage effects minimum reinforcement is to be provided along longer span direction.
Astmin=180 mm2
S= S= (1000 X π/( 4) X102 )/180 =436.33 mm
The max spacing is 450 mm or 5 X125 =625, so provide 10mm bars @400 c/c
à Check for shear
Note: Usually shear effect in slab is very small the slab it self takes care of shear effect.
first calculate
Vu=(w X l)/2 =(13.125 X 3.65)/2= 23.95 KN
Nominal shear stress (Ï„v)=
=Vu/(b x d)=(23.95 X 103)/(1000 X 125) =0.19N/MM2
Pt=(100 X 549.7)/(1000 X 125) =0.439%
as per table 19 page :73 …,as M20
By interpolation concept pt= 0.25 Ï„c= 0.36
pt= 0.5 Ï„c= 0.48
Ï„c=0.36+ (0.48-0.36)/(0.5-0.25) x(0.439-0.25)
For slab design strength = K x Ï„c = 1.3 x 0.45 =0.58 > Ï„v
so no need of shear reinforcement
à Check for deflection
Pt=(100 X Ast)/(bxd) = (100 X 549.7 )/(1000 X 125)=0.439%
L/d=20 x M.F M.F=1.2
d= 3650/(20 X 1.2) =140.38
le=3.65m=3650mm
à Check for development length of the slab
Ld=(0.87 X fy X d)/(4 X Ï„ bd)= (0.87X 415 X 10)/(4 X1.2 X1.6)
=470 mm
Detailing:
=> Detailing should be done in two spans that is, in Shorter span and Longer span.
Detailing:
=> Detailing should be done in two spans that is, in Shorter span and Longer span.
=> First of all you should draw the detailing for shorter span.
=> Second you have to draw the detailing for longer span.
=> Detailing should be drawn carefully by using the values what you got in the above problem.
Conclusion:
=> The procedure should be read first before solving the problem
=> Detailing of the problem should be drawn carefully.
