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
Discussion:
13 comments Page 1 of 2.
BSD said:
1 decade ago
= (6300/1400)*20 = 90 kg/sq cm.
Anu said:
1 decade ago
Please tell me formula.
Dfgh said:
10 years ago
Length in meter but per Stress in steel in cm2 how it correct?
Maggie said:
9 years ago
What is the formula? Please explain the solution.
Harshad said:
9 years ago
Stress = load/area.
So, area = load/stress.
So, area = load/stress.
Babar said:
8 years ago
At = ( load/stress)*length.
At= (6300/1400)*20.
At = 90 cm2.
At= (6300/1400)*20.
At = 90 cm2.
(2)
KVD said:
8 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?
But ans asked in kg/sq.m.
How it is correct?
Viplav said:
8 years ago
I think Unit of the answer is incorrect.
(3)
Sabbir said:
8 years ago
P=permissible stress in steel * Area of steel.
Kit said:
7 years ago
According to me, it is 90msq.Cm.
(1)
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