$\tau_{max} = \dfrac{16T}{\pi d^3}$
$20(1000) = \dfrac{16T}{\pi (0.20)^3}$
$T = 10\pi \, \text{lb}\cdot\text{in}$
$L = \dfrac{T}{0.50 \, \text{lb}\cdot\text{in/in}}$
$L = \dfrac{10\pi \, \text{lb}\cdot\text{in}}{0.50 \, \text{lb}\cdot\text{in/in}}$
$L = 20\pi \, \text{in} = 62.83 \, \text{in}$
$\theta = \dfrac{TL}{JG}$
If θ = dθ, T = 0.5L and L = dL
$\displaystyle \int d\theta = \frac{1}{JG} \int_0^{20\pi} (0.5L) \, dL$
$\theta = \left[ \dfrac{0.5L^2}{2} \right]_0^{20\pi} = \dfrac{1}{JG} \, [ \, 0.25(20\pi)^2 - 0.25(0)^2 \, ]$
$\theta = \dfrac{100\pi^2}{\frac{1}{32}\pi (0.20^4)(12 \times 10^6)}$
$\theta = 0.5234 \, \text{rad} = 30^\circ$ answer
Why is the upper limit 20pi?
That is the total length of
In reply to Why is the upper limit 20pi? by BeforeWinter
That is the total length of the shaft L, the solution for $L = 20\pi ~\text{inches}$ is also shown above.