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Poisson's ratio and E from rod and G

Short answer

From axial stress 127.32 N/mm², lateral strain 3×10⁻⁴, and G=0.5×10⁵ N/mm², solve ν = 0.308 using ε_lat = νσ/[2G(1+ν)]. Then E = 2G(1+ν) = 1.308×10⁵ N/mm².

The modulus of rigidity for a material is 0.5×105 N/mm20.5\times10^5\ \text{N/mm}^2. A 10mm diameter rod of the material was subjected to an axial pull of 10kN and the change in diameter was observed to be 3×10−3 mm3\times10^{-3}\ \text{mm}. Calculate the Poisson's ratio and the modulus of elasticity.

Method: use the lateral strain from the change in diameter, the axial stress, and the relation E=2G(1+ν)E = 2G(1+\nu).

Stress and lateral strain

σ=PA=10 000π4(10)2=127.32 N/mm2,εlat=Δdd=3×10−310=3×10−4 (contraction)\sigma = \frac{P}{A} = \frac{10\,000}{\frac{\pi}{4}(10)^2} = 127.32\ \text{N/mm}^2, \qquad \varepsilon_{lat} = \frac{\Delta d}{d} = \frac{3\times10^{-3}}{10} = 3\times10^{-4}\ \text{(contraction)}

Poisson's ratio

Lateral strain =ν σE= \nu\,\dfrac{\sigma}{E} and E=2G(1+ν)E = 2G(1+\nu), so

εlat=νσ2G(1+ν)⇒3×10−4=ν(127.32)2(0.5×105)(1+ν)\varepsilon_{lat} = \frac{\nu\sigma}{2G(1+\nu)} \Rightarrow 3\times10^{-4} = \frac{\nu(127.32)}{2(0.5\times10^5)(1+\nu)} 127.32 ν=30(1+ν)⇒ν=30127.32−30=0.3082127.32\,\nu = 30(1+\nu) \Rightarrow \nu = \frac{30}{127.32-30} = 0.3082

Modulus of elasticity

E=2G(1+ν)=2(0.5×105)(1+0.3082)=130825 N/mm2E = 2G(1+\nu) = 2(0.5\times10^5)(1+0.3082) = 130825\ \text{N/mm}^2

Answer: ν=0.308\nu = 0.308 and E≈1.308×105 N/mm2E \approx 1.308\times10^5\ \text{N/mm}^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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