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Highway Engineering

Q1: Interior thickness of concrete road slab for design wheel load 6300 kg and permissible flexural stress 21 kg/cm2, is

A 17.0 cm

B 25.5 cm

C 34.0 cm

D 42.5 cm

E 50.0 cm

ANS:B - 25.5 cm

depth of concrete = d = √(3w/stress),
d= √(3*6300/21)=30,

Now,
Thickness of Concrete= 0.85*d,
t=0.85*30,
t=25.5cm.



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