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 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 effective cover of 30mm

⸫ effective depth (d) = 500 – 30 = 470mm

⸫ the dimensions are:-
Width(b) = 300mm
Effective depth (d) = 470mm
Total depth (D) = 500mm

Self weight of beam = 0.3x0.5x25 = 3.75 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. &amp; 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 ( 20.625x6 ) + 0.6 ( 22.5x6 )
= 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 &lt; 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 &gt; 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 Ï• 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 Ï• bars (Ast = 942 mm 2 ) at mid-span as positive Reinforcement.
Ast = (Ref. CL 26.5.1.1, IS 456:2000)
Ast = = 289mm 2 &gt; Ast
Hence, Ast provided is OK.

f. Shear design :-
At support next to end support,
V u = 155.25KN
Ï„ v = (Ref. CL 40.1 IS 456:2000)
Ï„ v = = 1.1 N/mm 2
from table 20, IS 456:2000, for M20,
Ï„ c = 2.8 N/mm 2 &gt; Ï„ v
At support,
P t = 100 x = = 0.89%
From table 19, IS 456:2000, For M20, concrete &amp; P t = 0.89%

P t = 0.89 N/mm 2 &lt; Ï„ v
Hence, shear reinforcement is necessary
V us = V o - Ï„ 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 Ï• 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 Ï• @ 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
Ï„ v = = = 0.78 N/mm 2
From table 20 , IS 456 : 2000;
Ï„ c max = 2.8 N/mm 2 &gt; Ï„ v
P t at end support = 100 x = = 0.66 %
From table 19, IS 456 2000,For M20 concrete,

P t = 0.66%
Ï„ c = 0.48 + (0.66 – 0.5 ) = 0.53 N/mm 2 &lt; Ï„ v
Hence, shear reinforcement is necessary
V us = V u - Ï„ 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 Ï• stirrups
S v = (Ref. CL 40.4(a) IS 456:2000)
S v = = 480 mm &gt; 300mm

Provide 2 legged 8 mm Ï• stirrups @ 300 mm c/c. hence, provide 2 legged 8 mm Ï• stirrups @
250 mm c/c at intermediate support and gradually increasing the spacing to 300 mm c/c near the
mid-span &amp; at the end support.