A cylindrical bar of 25 mm diameter and 1.20 m length is subjected to a longitudinal strain during a tensile test is four times the lateral strain, what is the value of shear modulus? (Take the modulus of elasticity as 1.2 × 10⁵ N/mm²)
- A5.8 kN/mm²
- B4.8 kN/mm²
- C58 kN/mm²
- D48 kN/mm²Correct
Solution
Concept
Shear modulus is found from E and Poisson's ratio using E = 2G(1 + μ).
Formula
μ = εlateral / εlongitudinal; G = E / (2(1 + μ))
Step-by-step solution
1. Longitudinal strain = 4 × lateral strain, so μ = 1/4 = 0.25. 2. Use E = 2G(1 + μ). 3. G = 1.2 × 105 / (2 × 1.25) = 1.2 × 105 / 2.5 = 48000 N/mm². 4. 48000 N/mm² = 48 kN/mm².
Explanation
Poisson's ratio is lateral strain divided by longitudinal strain, so μ = 0.25. Putting this in E = 2G(1 + μ) gives G = 48000 N/mm², which is 48 kN/mm². The diameter and length are not needed.
Why the other options are wrong
A: 5.8 kN/mm² has a wrong formula and a wrong decimal place. B: 4.8 kN/mm² is a unit conversion error (it is 10 times too small). C: 58 kN/mm² comes from a wrong relation or wrong μ.
Shortcut
G = E/2.5 when μ = 0.25, so G = 0.4E = 0.4 × 120 = 48 kN/mm².
Common mistake
Taking μ = 4 instead of 1/4, or dividing wrongly when converting N/mm² to kN/mm².
Correct answer: D) 48 kN/mm²