$\tau_{max} = \dfrac{16T}{\pi d^3}$

$20(1000) = \dfrac{16T}{\pi (0.20)^3}$

$T = 10\pi \, \text{lb}\cdot\text{in}$

$L = \dfrac{T}{0.50 \, \text{lb}\cdot\text{in/in}}$

$L = \dfrac{10\pi \, \text{lb}\cdot\text{in}}{0.50 \, \text{lb}\cdot\text{in/in}}$

$L = 20\pi \, \text{in} = 62.83 \, \text{in}$

$\theta = \dfrac{TL}{JG}$

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?

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.