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A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be?
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A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a...
Modulus of Rigidity
Modulus of Rigidity or Shear Modulus is the ratio of shear stress to the corresponding shear strain within the elastic limit. It is denoted by G.
Formula for Modulus of Rigidity
G = (E/2(1+v))
Where,
E = Modulus of Elasticity
v = Poisson's ratio
Given
Diameter of cylindrical bar, D = 40 mm
Length of cylindrical bar, L = 0.5 m
Longitudinal strain, εl = 4lateral strain
Modulus of Elasticity, E = 2 x 10^6 N/mm^2
Solution
Calculating Lateral Strain
Longitudinal strain, εl = 4lateral strain
4lateral strain = εl
lateral strain = εl/4
We know that,
εl = ΔL/L
where,
ΔL = Change in Length
L = Original Length
Lateral strain, εt = ΔD/D
where,
ΔD = Change in Diameter
D = Original Diameter
As the diameter of the cylindrical bar remains constant during the tensile test, ΔD = 0
Therefore, lateral strain, εt = 0
Now, we have,
4lateral strain = εl
4 x 0 = εl/4
εl = 0
This means that there is no change in diameter of the cylindrical bar during the tensile test.
Therefore, the ratio of shear strain to lateral strain, γ/εt = 0
Calculating Modulus of Rigidity
Now, we can use the formula for Modulus of Rigidity,
G = (E/2(1+v))
We need to calculate Poisson's ratio, v
We know that,
v = - lateral strain/longitudinal strain
As lateral strain, εt = 0 and longitudinal strain, εl = 4 x εt = 0
v = 0/0 = undefined
Therefore, we cannot calculate Poisson's ratio and hence, the Modulus of Rigidity cannot be calculated.
Conclusion
As there is no change in diameter of the cylindrical bar during the tensile test, the ratio of shear strain to lateral strain is zero. Hence, Poisson's ratio is undefined and the Modulus of Rigidity cannot be calculated.
Modulus of Rigidity or Shear Modulus is the ratio of shear stress to the corresponding shear strain within the elastic limit. It is denoted by G.
Formula for Modulus of Rigidity
G = (E/2(1+v))
Where,
E = Modulus of Elasticity
v = Poisson's ratio
Given
Diameter of cylindrical bar, D = 40 mm
Length of cylindrical bar, L = 0.5 m
Longitudinal strain, εl = 4lateral strain
Modulus of Elasticity, E = 2 x 10^6 N/mm^2
Solution
Calculating Lateral Strain
Longitudinal strain, εl = 4lateral strain
4lateral strain = εl
lateral strain = εl/4
We know that,
εl = ΔL/L
where,
ΔL = Change in Length
L = Original Length
Lateral strain, εt = ΔD/D
where,
ΔD = Change in Diameter
D = Original Diameter
As the diameter of the cylindrical bar remains constant during the tensile test, ΔD = 0
Therefore, lateral strain, εt = 0
Now, we have,
4lateral strain = εl
4 x 0 = εl/4
εl = 0
This means that there is no change in diameter of the cylindrical bar during the tensile test.
Therefore, the ratio of shear strain to lateral strain, γ/εt = 0
Calculating Modulus of Rigidity
Now, we can use the formula for Modulus of Rigidity,
G = (E/2(1+v))
We need to calculate Poisson's ratio, v
We know that,
v = - lateral strain/longitudinal strain
As lateral strain, εt = 0 and longitudinal strain, εl = 4 x εt = 0
v = 0/0 = undefined
Therefore, we cannot calculate Poisson's ratio and hence, the Modulus of Rigidity cannot be calculated.
Conclusion
As there is no change in diameter of the cylindrical bar during the tensile test, the ratio of shear strain to lateral strain is zero. Hence, Poisson's ratio is undefined and the Modulus of Rigidity cannot be calculated.
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A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be?
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A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be?.
A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical bar of 40 mm diameter and 0.5 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 10^6 N/mm^2, then its modulus of rigidity will be?.
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