Civil Engineering - Highway Engineering - Discussion

Discussion Forum : Highway Engineering - Section 3 (Q.No. 9)
9.
Area of steel required per metre width of pavement for a length of 20 m for design wheel load 6300 kg and permissible stress in steel 1400 kg/cm2, is
70 kg/sq cm
80 kg/sq cm
90 kg/sq cm
100 kg/sq cm
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

GorkheeJr said:   5 months ago
Thanks for explaining.

Sabyasachi said:   4 years ago
We know that,
As=LfW/2s.
As=Area of steel.
L=Length.
f=Co-efficiant of friction, here f=2.
W=Weight.
s=Working stress in steel in kg/cm2.
As=(20X2X6300) /(2X1400).
= 90 Kg/cm2.

TANU said:   4 years ago
We know,

Area= (force or load)/stress.
=6300/1400.
= 4.5 cm2 , per metre width.
For 20 m length it will be = 4.5x20 = 90 kg/cm2.
(13)

Kit said:   6 years ago
According to me, it is 90msq.Cm.
(1)

Sabbir said:   6 years ago
P=permissible stress in steel * Area of steel.

Viplav said:   7 years ago
I think Unit of the answer is incorrect.
(2)

KVD said:   7 years ago
(6300/1400)*20= 90 is correct but i.e., in cm2-m.
But ans asked in kg/sq.m.

How it is correct?

Babar said:   7 years ago
At = ( load/stress)*length.
At= (6300/1400)*20.
At = 90 cm2.
(2)

Harshad said:   8 years ago
Stress = load/area.
So, area = load/stress.

Maggie said:   8 years ago
What is the formula? Please explain the solution.


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