Design a rectangular beam continuous over 4 column supports of effective shap 6m. The beam is subjected to an imposed dead load of 10 KN/m and a live load of 15 KN/m. use M20 and Fe415 steel.
Given
Effective shap (L eff. ) = 6m
(f ck ) = 20 N/mm 2
(f y ) = 415 N/mm 2
Imposed dead load = 10 KN/m
Imposed live load = 15 KN/m
a. Effective depth:-
Assuming effective depth = = = 400 mm
Trying a total depth of D = 500 mm
And width (b) = to i.e,
Nearly 250mm
Trying a width (b) of 300mm
Assuming an effective cover of 30mm
⸫ effective depth (d) = 500 – 30 = 470mm
⸫ the dimensions are:-
Width(b) = 300mm
Effective depth (d) = 470mm
Total depth (D) = 500mm
Effective depth (d) = 470mm
Total depth (D) = 500mm
b. Load calculation:-
Self weight of beam = 0.3×0.5×25 = 3.75 KN/m
Imposed dead load = 10 KN/m
⸫ total dead load = 10 + 3.75 = 13.75 KN/m
⸫ factored dead load (W d )= 13.75 x 1.5 = 20.625 KN/m
⸫ factored live load (W L ) = 15 x 1.5 = 22.5 KN/m
c. B.M. & S.F. using coefficient:-
(ref. table- 12,13, IS 456:2000)
Now,
M = WL 2
the maximum negative moment occurs at support next to end support,
M = – (20.625 x 6 2 ) + (22.5 x 6 2 )
= -164.25 KNm
The maximum positive moment occurs near the middle of the end shap,
M = (20.625 x 6 2 ) + (22.5 x 6 2 )
= 142.875 KNm
And,
V = WL
Maximum shear force occure at outer side of the support next to end support,
V u = 0.6 ( 20.625×6 ) + 0.6 ( 22.5×6 )
= 155.25 KN
d. Depth check :-
( Ref. CL. 31.1.1, IS 456 : 2000 )
M u = 0.36 bd 2 f ck 1- 0.42
= 0.36 x 0.48 x 1- ( 0.42 x 0.48 )
= 0.138 f ck bd 2
d req. =
=
= 446 mm < 470 mm
e. Area of steel :-
(Ref. CL. 31.1.1 IS 456 : 2000 )
Calculating the limiting moment of resisitance,
We know,
M u = 0.138 f ck bd 2
= 0.138 x 20 x 300 x 470 2
= 182.9 KNm > M u (164.28KNm)
Hence, the singly reinforced section can be designed,
Ast = 0.5 1 –
Now,
Designing the section at intermediate support,
M u = 0.87 f y Ast d 1-
164.25 x 10 6 = 0.87 x 415 x Ast x 470 1 –
Solving we get,
Ast = 1014 mm 2
No o bars = 4
Using 4 – 20 mm ϕ bars (Ast Pro. = 1256 mm 2 ) at intermediate support as negative
Reinforcement.
At mid-span –
M u = 142.875 KNm
M u = 0.87 f y Ast d 1-
142.875 x 10 6 = 0.87 x 415 x Ast x 470 1-
Solving we get,
Ast = 876 mm 2
No. of bars = 3
Using 3 – 20 mm ϕ bars (Ast = 942 mm 2 ) at mid-span as positive Reinforcement.
Ast = (Ref. CL 26.5.1.1, IS 456:2000)
Ast = = 289mm 2 > Ast
Hence, Ast provided is OK.
f. Shear design :-
At support next to end support,
V u = 155.25KN
τ v = (Ref. CL 40.1 IS 456:2000)
τ v = = 1.1 N/mm 2
from table 20, IS 456:2000, for M20,
τ c = 2.8 N/mm 2 > τ v
At support,
P t = 100 x = = 0.89%
From table 19, IS 456:2000, For M20, concrete & P t = 0.89%
P t = 0.89 N/mm 2 < τ v
Hence, shear reinforcement is necessary
V us = V o – τ c bd (Ref. of CL 40.4)
V us = 155.25 x 10 3 – 0.02 x 300 x 470 = 67830 N
Spacing of 2 legged 8mm ϕ bars stirrups
S v = (Ref. CL 40.4(a) IS 456:2000)
S v = = 251 mm
Maximum spacing of 2 legged 8 mm dia stirrups should not excced
I. 0.75d = 0.75 x 470 = 352 mm
II. 300 mm
III. S v max =
= = 302 mm
Hence provide 2 legged 8mm ϕ @ 250 mm c/c at end support,
V u = ( 0.4 x 20.625 x 6) +( 0.45 x 22.5 x 6 ) = 110.25 KN
τ v = = = 0.78 N/mm 2
From table 20 , IS 456 : 2000;
τ c max = 2.8 N/mm 2 > τ v
P t at end support = 100 x = = 0.66 %
From table 19, IS 456 2000,For M20 concrete,
P t = 0.66%
τ c = 0.48 + (0.66 – 0.5 ) = 0.53 N/mm 2 < τ v
Hence, shear reinforcement is necessary
V us = V u – τ c bd (Ref. CL 40.4 IS 456:2000 )
V us = (110.25 x 10 3 ) – (0.53 x 300 x 470 )
V us = 35520 N
Spacing of 2 legged 8 mm ϕ stirrups
S v = (Ref. CL 40.4(a) IS 456:2000)
S v = = 480 mm > 300mm
Provide 2 legged 8 mm ϕ stirrups @ 300 mm c/c. hence, provide 2 legged 8 mm ϕ stirrups @
250 mm c/c at intermediate support and gradually increasing the spacing to 300 mm c/c near the
mid-span & at the end support.
