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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 2 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 Calculation:



Given data:
- Diameter of the cylindrical bar (d) = 40 mm
- Length of the cylindrical bar (L) = 0.5 m
- Longitudinal strain (εl) = 2 x lateral strain (εt)
- Modulus of elasticity (E) = 2 x 10^6 N/mm^2



Step 1: Calculate the Longitudinal and Lateral Strains
- Longitudinal strain (εl) = Change in length/Original length = ΔL/L
- Lateral strain (εt) = Change in diameter/Original diameter = Δd/d

Since εl = 2 x εt, we can write: ΔL/L = 2 x Δd/d



Step 2: Calculate the Poisson's Ratio (ν)
- Poisson's ratio (ν) = εt/εl = 1/2



Step 3: Calculate the Modulus of Rigidity (G)
- Modulus of rigidity (G) = E/(2 x (1 + ν)) = E/(2 x (1 + 1/2)) = E/3

Substitute the given value of E into the equation:
G = 2 x 10^6 N/mm^2 / 3
G = 6.67 x 10^5 N/mm^2

Therefore, the modulus of rigidity of the cylindrical bar is 6.67 x 10^5 N/mm^2.
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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 2 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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