Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 30)
30.
Interior thickness of concrete road slab for design wheel load 6300 kg and permissible flexural stress 21 kg/cm2, is
Discussion:
17 comments Page 1 of 2.
Dev said:
4 years ago
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.
d= √(3*6300/21)=30,
Now,
Thickness of Concrete= 0.85*d,
t=0.85*30,
t=25.5cm.
(6)
Qasim said:
8 years ago
depth of concrete=d= sqrt(3w/stress),
d=sqrt(3*6300/21)=30,
Now,
Thickness of Concrete= 0.85*d,
t=0.85*30,
t=25.5cm.
d=sqrt(3*6300/21)=30,
Now,
Thickness of Concrete= 0.85*d,
t=0.85*30,
t=25.5cm.
(2)
Ammy said:
7 years ago
Thank you @Qasim.
(1)
Arya said:
9 years ago
Stress = load/area,
Thickness at interior is 0.85 d.
Thickness at interior is 0.85 d.
Jehangir khan said:
9 years ago
Please provide the explanation.
SAM said:
9 years ago
d = (3W/stress)^0.5,
= (3 * 6300/21)^0.5,
= 30.
= (3 * 6300/21)^0.5,
= 30.
Praveen said:
9 years ago
Thickness will be 0.85 * 30 = 25.5.
Roy said:
8 years ago
Thanks @Praveen and @Sam.
Anbu star lion said:
8 years ago
Thanks @Praveen and @Sam.
Arnab Chanda said:
9 years ago
Please give me the explanation.
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