Civil Engineering - UPSC Civil Service Exam Questions - Discussion

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

[A]. 0.25
[B]. 0.30
[C]. 0.33
[D]. 0.35

Answer: Option B

Explanation:

No answer description available for this question.

Awad said: (Mar 18, 2016)  
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).

Dolly said: (Mar 4, 2018)  
How you take 0.0015?

Umesh Jit said: (Sep 17, 2018)  
0.0015 is a longitudinal strain.
0.03/20 = 0.0015.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.