Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 3 (Q.No. 22)
22.
If the modulus of elasticity of a material is twice its modulus of rigidity, then the Poisson's ratio of the material is equal to zero.
Correct
Incorrect
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Danis said:   8 years ago
E = 2C ( 1+1/M)
E = 2C ( 1+ V ).
WHERE , V= POISSON RATIO
E = MODULUS OF ELASTICITY
C = MODULUS OF RIGIDITY
NOW,
C = E/ 2(1+V).
ACCORDING TO QUE.
E = 2C,
C = 2C / 2(1+V),
1= 1+ V,
V = 0.

HENCE PROVED.
(4)

Rathi said:   1 decade ago
G = E/(2(1+ v)).
Where G is the rigidity modulus.
E is the Young's modulus(modulus of elasticity) and
v is the Poisson's ratio.

Acc. to the question,
G = 2G/(2(1+v)).

(1+v) = 2G/2G.

1+v = 1.

v = 0.

Balram kumar said:   1 year ago
E = 2G(1+μ).
Given E = 2G.
2G = 2G(1+μ).
2G/2G = (1+μ).
1 = 1+μ.
1-1 = μ.
μ = 0.
Where E = modulus of elasticity.
G = modulus of rigidity.
μ = Poisson ratio.
(1)

Yogendra Singh Chouhan said:   6 years ago
No @Danis

And you are right @Mr. Krunal.

G = vE/(2(1+v)).
1+v = v2G/2G.
1+v = v.
(1/v)+1 = 1.
1/v = 0 or v = 0.
(1)

Krunal said:   1 decade ago
Here equation is:

G = vE/(2(1+v)).
1+v = v2G/2G.

1+v = v.
(1/v)+1 = 1.

1/v = 0 or v = 0.

Pradeep gk said:   8 years ago
E = 2C(1+u), u=poisson's ratio
E = 2C(1+0),
E = 2C.
(2)

Pradeep yadav said:   7 years ago
Thanks all for explaining the solution.
(1)

Ramen Baidya said:   6 years ago
In the question said 2E=C , NOT E=2C.
(1)

Rajesh said:   8 years ago
E = 2C(1+1/M),
E = 2C(1+0),
E = 2C.

Ishwar said:   8 years ago
Thank you for explaining it.


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