A singly supported rectangular beam is to be provided over a clear span 10 m to carry on load of 25 KN/m excluding its self-weight .design mid section using M20 grade concrete and fe 415 steel assume width of support as 300 mm
SOLUTION
Trial section
Clear span of the baam =10m
Assume effective depth of the beam d = span/12
= 10000 / 12
= 833 mm
Say 850 mm
Assume breadth of the beam b = 0.4d
= 0.4 x 850
= 340 mm
Say 350 mm
Assume eff cover = 50 mm
Over all depth D = 850 + 50 =900 mm
LOADS
Loads on the beam = 25 KN/M
Self weight of the beam = 0.35 x 0.9 x 1 x 25
Total characteristic load = 25 + 7.875
= 32.875 KN/m
Taking the partial safety factor = 1.5
Design load Wu = 1.5 x 32.875
Wu = 49.313 KN/m
EFFECTIVE SPAN
Effective span of the beam is least of
1. c/c distance of support = 10 + 0.3
= 10.30 m
2.clear span + eff.depth = 10 + 0.85
= 10.85m
So eff span lef = 10.3 m
Design bending moment
Design bending moment at mid span mu = WU l2/ 8
MU = (49.313 x 10.32)/8
= 653.95 KN.M
= 653.95 x 106 N.MM
Effective depth requirement
For Balanced section of M20 grade of concrete
Mu lim=2.76 bd2
D2 =mulim/2.76 b
D =Mulim2.76b
D =653.95 X 1062.76 X 350
D = 822 mm
We are take already eff.depth d =850 mm
So 822 mm < 850 mm
Provided depth is more than required depth Hence ok
AREA OF STEEL REQUIREMENT
Resisting moment of under reinforced section
Mu=0.87 fy Ast [d-(fy Ast/fck b)]
Fy= tensile strength of steel
Fck=compressive strength of concrete
Ast= Area of steel
B= breadth of beam
Mu=design bending moment
653.95 X 106 = 0.87 x 415 x Ast [850-(415 Ast/20 x 350)]
Ast = 2600 mm2
Provide 6 numbers of 25 mm dia bars (Area = 2940 mm2 )
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