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.
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)?
Ashwani bhandari said:
7 years ago
We know that e = (Z/A).
Section modulus for circular column Z = πx(D4-d4)÷(32D),
Area A= πx(D2-d2)÷4.
We know the formula of;
(D4-d4) is (D2+d2)( D2-d2),
Then e = Z ÷ A,
We get;
e = (D2+d2)/(8D).
Section modulus for circular column Z = πx(D4-d4)÷(32D),
Area A= πx(D2-d2)÷4.
We know the formula of;
(D4-d4) is (D2+d2)( D2-d2),
Then e = Z ÷ A,
We get;
e = (D2+d2)/(8D).
PRAMOD KUMAR said:
7 years ago
@AFRID:
Z=I/y I= (π*D^4)/64 y=D/2 so Z = (π*D^3)/32 .
But as D-d cannot cancel out by D so the term remains same and only 64 cancelled out by 2 hence it is 32.
Z=I/y I= (π*D^4)/64 y=D/2 so Z = (π*D^3)/32 .
But as D-d cannot cancel out by D so the term remains same and only 64 cancelled out by 2 hence it is 32.
Yempee said:
6 years ago
Thanks @Ashwani Bhandari.
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