If permissible compressive stress in concrete is 15 kg/cm^{2} , tensile stress in steel is 1400 kg/cm^{2} and modular ratio is 18, the depth of the beam is (1) d= \sqrt{{0.11765\timesB.M. }/ \breadth} (2) d= \sqrt{\frac{0.22765\timesB.M. }{ \breadth} (3) d= \sqrt{{0.33765\timesB.M.}/ {\breadth}} (4) d= \sqrt{{0.44765\timesB.M. }/ {\breadth} }

SSC JE Civil Engineering 2018 Set 1 Solved Paper

Question : 183 of 200

Marks: +1, -0

If permissible compressive stress in concrete is 15 kg/cm

^{2}

, tensile stress in steel is 1400 kg/cm

^{2}

and modular ratio is 18, the depth of the beam is
(1)

d= √{{0.11765×B.M. }/ \breadth}

(2)

d= √{{0.22765×B.M. }/{ \breadth}

(3)

d= √{{0.33765×B.M.}/ {\breadth}}

(4)

d= √{{0.44765×B.M. }/ {\breadth} }

Solution:

Compressive strength of concrete

σ_\cbc

= 50 kg/cm

^{2}

.
Tensile stress in steel = 1400 kg/cm

^{2 }

.
Modulus Ratio m = 18

K={mσ_\cbc}/{mσ_\cktσ_\st}={18×50}/{18×50+1400}

=0.3913

j=1-K/3={1-0.3913}/3=0.87

Q=1/2 σ_\cbc×jk

= 1/2× 50 ×0.87 ×0.3913= 8.510775

BM = Q bd

^2

√{{BM} /{Q b}}=d

[d= √{{0.1175×BM }/ \breadth}]

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