Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 7)
7.
A short column of external diameter D and internal diameter d, is subjected to a load W, with an eccentricity 'e', causing zero stress at an extreme fibre. Then the value of 'e' must be
Discussion:
12 comments Page 1 of 2.
Saqib khattak said:
6 years ago
As we know the formula for Z for ext and int dia is Z= 3.14(D^4-d^4)/32D and we also know that (D^4-d^4)= (D^2-d^2)(D^2+d^2) and we also know the area A= (3.14(D^2-d^2))/4
e=Z/A.
e=(3.14(D^4-d^4)/32D)/(3.14(D^2-d^2))/4.
Hence e will comes after solving.
e= (D^2+d^2)/8D.
e=Z/A.
e=(3.14(D^4-d^4)/32D)/(3.14(D^2-d^2))/4.
Hence e will comes after solving.
e= (D^2+d^2)/8D.
(3)
Vikas said:
6 years ago
Thanks, @Ashwani.
(2)
Yempee said:
6 years ago
Thanks @Ashwani Bhandari.
(1)
Gaurav said:
9 years ago
Please provide the explanation.
Swapnil said:
9 years ago
Can anyone please explain answer?
Vikas Niswade said:
9 years ago
Please provide the explanation if anyone can.
Asaithambi said:
9 years ago
σ(min) = 0 = (p/A) - (p.e/z),
e = z/A,
z = (3.14 * (D^2 - d^2))/(32 * D),
A = (3.14 * (D^2 - d^2),
σ(min)=(D^2 + d^2)/8D.
e = z/A,
z = (3.14 * (D^2 - d^2))/(32 * D),
A = (3.14 * (D^2 - d^2),
σ(min)=(D^2 + d^2)/8D.
Chhaya said:
8 years ago
Z= (3.14* (D^4 - d^4))/32 D.
A = (3.14 * (D^2 - d^2))/4.
e= (D^2 + d^2)/8 D.
A = (3.14 * (D^2 - d^2))/4.
e= (D^2 + d^2)/8 D.
Kavya said:
8 years ago
But here Z = (3.14*(D^4-d^4))/16D?
Afrid said:
7 years ago
How come (32*D)?
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