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.
Discussion:
13 comments Page 1 of 2.
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.
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.
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.
G = vE/(2(1+v)).
1+v = v2G/2G.
1+v = v.
(1/v)+1 = 1.
1/v = 0 or v = 0.
Srikanth said:
9 years ago
Thank you @Rathi & @Krunal.
Ishwar said:
8 years ago
Thank you for explaining it.
Rajesh said:
8 years ago
E = 2C(1+1/M),
E = 2C(1+0),
E = 2C.
E = 2C(1+0),
E = 2C.
Pradeep gk said:
8 years ago
E = 2C(1+u), u=poisson's ratio
E = 2C(1+0),
E = 2C.
E = 2C(1+0),
E = 2C.
(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.
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)
IMARUL said:
8 years ago
Thank you @Danis.
Pradeep yadav said:
7 years ago
Thanks all for explaining the solution.
(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.
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)
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