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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