Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 34)
34.
When a bar of length l and diameter d is rigidly fixed at the upper end and hanging freely, then the total elongation produced in the bar due to its own weight is (where w = Weight per unit volume of the bar)
Discussion:
6 comments Page 1 of 1.
Arjun said:
6 years ago
Elongation due to self weight equal to WL/2AE.
Given w is Weight is per unit volume so multiply L to convert area A in denominator into a volume to have total load W=w/v.
To do this we have to multiply on both numerator and denominator and hence the result
wL^2/2E.
Given w is Weight is per unit volume so multiply L to convert area A in denominator into a volume to have total load W=w/v.
To do this we have to multiply on both numerator and denominator and hence the result
wL^2/2E.
(1)
Mohan said:
8 years ago
No @Milindsardar
The answer is correct because total elongation due to own weight is WL/2EA.
Multiply by L on both up and down give u the same result as mentioned in the answer.
The answer is correct because total elongation due to own weight is WL/2EA.
Multiply by L on both up and down give u the same result as mentioned in the answer.
A Sibasankar Patro said:
3 years ago
I didn't understand.
Can you please elaborate the solution?
Can you please elaborate the solution?
MILINDSARDAR said:
9 years ago
I think it should be W^2l/2E.
Reena said:
7 years ago
Anyone, please explain this.
Numankhan said:
7 years ago
Explain it please.
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