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A cylindrical bar of 20 mm diameter and 1 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 105 N/mm2, then its modulus of rigidity will be
  • a)
    8 x 106 N/mm2
  • b)
    8x105 N/mm2
  • c)
    0.8 x 104 N/mm2
  • d)
    0.8 x 105 N/mm2
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A cylindrical bar of 20 mm diameter and 1 m length is subjected to a t...

But, 
= 0.8 x 105 N/mm2
This question is part of UPSC exam. View all Mechanical Engineering courses
Most Upvoted Answer
A cylindrical bar of 20 mm diameter and 1 m length is subjected to a t...
Given:
Diameter of cylindrical bar = 20 mm
Length of cylindrical bar = 1 m
Longitudinal strain = 4 times that of lateral strain
Modulus of elasticity (E) = 2 x 105 N/mm2

To find: Modulus of rigidity (G)

Formula:
Modulus of elasticity (E) = stress/longitudinal strain
Modulus of rigidity (G) = stress/lateral strain

Calculation:

1. Calculation of longitudinal strain:
Given that, longitudinal strain is 4 times that of lateral strain
Let lateral strain be εl
Then, longitudinal strain εt = 4εl

2. Calculation of stress:
The bar is under tensile load. Let the load applied be P.
Area of cross-section of the bar = π/4 × (diameter)2 = π/4 × (20)2 = 314.16 mm2
Stress σ = P/A

3. Calculation of modulus of elasticity:
Modulus of elasticity (E) = stress/longitudinal strain
E = σ/εt = (P/A)/(4εl) = P/(4Aεl)

4. Calculation of lateral strain:
From the given data, we know that the ratio between longitudinal and lateral strains is 4:1.
Let the lateral strain be εl
Then, longitudinal strain εt = 4εl
εl = εt/4

5. Calculation of modulus of rigidity:
Modulus of rigidity (G) = stress/lateral strain
G = σ/εl = (P/A)/εl = P/(Aεl)

Substituting the values obtained in steps 3 and 4, we get
G = P/(Aεl) = E/4

Substituting the given value of E, we get
G = 2 x 105 N/mm2 / 4 = 0.5 x 105 N/mm2 = 0.8 x 105 N/mm2

Therefore, the modulus of rigidity of the cylindrical bar is 0.8 x 105 N/mm2. The correct option is (D).
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A cylindrical bar of 20 mm diameter and 1 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 105 N/mm2, then its modulus of rigidity will bea)8 x 106 N/mm2b)8x105 N/mm2c)0.8 x 104 N/mm2d)0.8 x 105 N/mm2Correct answer is option 'D'. Can you explain this answer?
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A cylindrical bar of 20 mm diameter and 1 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 105 N/mm2, then its modulus of rigidity will bea)8 x 106 N/mm2b)8x105 N/mm2c)0.8 x 104 N/mm2d)0.8 x 105 N/mm2Correct answer is option 'D'. Can you explain this answer? 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 20 mm diameter and 1 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 105 N/mm2, then its modulus of rigidity will bea)8 x 106 N/mm2b)8x105 N/mm2c)0.8 x 104 N/mm2d)0.8 x 105 N/mm2Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical bar of 20 mm diameter and 1 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 105 N/mm2, then its modulus of rigidity will bea)8 x 106 N/mm2b)8x105 N/mm2c)0.8 x 104 N/mm2d)0.8 x 105 N/mm2Correct answer is option 'D'. Can you explain this answer?.
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