Poisson's ratio from torsion and tension tests

Poisson's ratio follows from $E = 2G(1+\nu)$. Find $G$ from the torsion test and $E$ from the tension test. The torque is taken as 300 N·m ($= 3\times10^5$ N·mm).

Modulus of rigidity (torsion test)

$$J = \frac{\pi d^4}{32} = \frac{\pi (25)^4}{32} = 38349.5\ \text{mm}^4, \qquad \theta = 1.35^\circ = 0.02356\ \text{rad}$$ $$\frac{T}{J} = \frac{G\theta}{L} \Rightarrow G = \frac{TL}{J\theta} = \frac{3\times10^5 \times 250}{38349.5\times 0.02356} = 83002\ \text{N/mm}^2$$

Modulus of elasticity (tension test)

$A = \dfrac{\pi}{4}(25)^2 = 490.87\ \text{mm}^2$, $\delta = 0.01\ \text{cm} = 0.1\ \text{mm}$, $L = 150$ mm.

$$E = \frac{PL}{A\delta} = \frac{70\,000\times150}{490.87\times0.1} = 213904\ \text{N/mm}^2$$

Poisson's ratio

$$\nu = \frac{E}{2G} - 1 = \frac{213904}{2(83002)} - 1 = 0.289$$

Answer: $\nu \approx 0.289$.