Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Discussion Forum : UPSC Civil Service Exam Questions - Section 22 (Q.No. 1)
1.
A bar 4 cm in diameter is subjected to an axial load of 41. The extension of the bar over a gauge length of 20 cm is 0.03 cm. The decrease in diameter is 0.0018 cm. The Poisson's ratio is
0.25
0.30
0.33
0.35
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
3 comments Page 1 of 1.

Umesh JIT said:   6 years ago
0.0015 is a longitudinal strain.
0.03/20 = 0.0015.

Dolly said:   6 years ago
How you take 0.0015?

AWAD said:   8 years ago
μ = - εt / εl= -(-0.00045) /0.0015= 0.3 ---- (1)

Where,

μ = Poisson's ratio.

εt = transverse strain.

εl = longitudinal or axial strain.

Strain can be expressed as,

εt = dl / L = 0.0018/4=0.00045 , εl=0.0015.
Where,

dl = change in length (m, ft).

L = initial length (m, ft).

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