Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 24)
24.
The figure refers to a channel section of uniform thickness of 1 cm used as a beam with the cross-sectional dimensions shown in cm units. If the Moment of Inertia of channel Ixx be equal to 246 cm4 and C be the shear centre, then the value of e in cm is equal to
Discussion:
9 comments Page 1 of 1.
Pratap said:
7 years ago
1.67 is correct only if it is from C.G.
So, the correct one is 1.5.
So, the correct one is 1.5.
Aamir said:
7 years ago
1.67 is the correct answer.
Eiman arbab said:
7 years ago
It should be 1.67.
(1)
Fuad bhai said:
8 years ago
Please explain clearly.
RAHUL said:
9 years ago
Please clearly explain the answer.
Sahil garg said:
9 years ago
Shear center = t * b^2 * h^2/(4 * Ixx)
= 1 * 10^2 * 5^2/(4 * 246).
= 2.54 it is form 1cm thicknes i.e. (e+1).
So e = 2.54 - 1 = 1.54 cm --> Answer.
= 1 * 10^2 * 5^2/(4 * 246).
= 2.54 it is form 1cm thicknes i.e. (e+1).
So e = 2.54 - 1 = 1.54 cm --> Answer.
(2)
Mounir Matta said:
9 years ago
It is 1.67 from the web center.
So 1.67 - 0.5 = 1.17 from external edge.
Then e = 1.17 cm.
So 1.67 - 0.5 = 1.17 from external edge.
Then e = 1.17 cm.
MAYANK said:
10 years ago
It should be 1.67 from direct formula.
E = t*h^2*b^2/4Ixx;
Where h and b are c2c distance and t is thickness.
E = t*h^2*b^2/4Ixx;
Where h and b are c2c distance and t is thickness.
Shraddha said:
10 years ago
Explain it detail?
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