$T = 0.05P = 0.05(5)$
$T = 0.25 ~ \text{kN}\cdot\text{m}$
Shear stress due to direct force
$\tau_1 = \dfrac{P}{A} = \dfrac{5(1000)}{\pi (5^2)}$
$\tau_1 = 63.66 ~ \text{MPa}$
Shear stress due to torque
$\tau_2 = \dfrac{Tr}{J} = \dfrac{0.25(1000^2)(5)}{\frac{1}{2}\pi (5^4)}$
$\tau_2 = 1273.24 ~ \text{MPa}$
Part 1 - Shear stress at A (maximum shear stress)
$\tau_A = \tau_1 + \tau_2$
$\tau_A = 63.66 + 1273.24$
$\tau_A = 1336.90 ~ \text{MPa}$ answer
Or you can use the formula for helical spring
$\tau = \dfrac{16PR}{\pi d^3} \left( 1 + \dfrac{d}{4R} \right)$
$\tau_\text{max} = \dfrac{16(5000)(50)}{\pi (10^3)} \left[ 1 + \dfrac{10}{4(50)} \right]$
$\tau_\text{max} = 1336.90~ \text{MPa}$ (okay!)
Part 2 - Shear stress at B
$\tau_B = \tau_2 - \tau_1$
$\tau_B = 1273.24 - 63.66$
$\tau_B = 1209.58 ~ \text{MPa}$ answer
Part 3 - At the point of zero shear, τtorque = τdirect-shear
$\dfrac{Tx}{J} = \tau_1$
$\dfrac{0.25(1000^2)x}{\frac{1}{2}\pi (5^4)} = 63.66$
$x = 0.25 ~ \text{mm from point } C$ answer
Another way to solve for x is by ratio and proportion
$\dfrac{x}{\tau_1} = \dfrac{r}{\tau_2}$
$\dfrac{x}{63.66} = \dfrac{5}{1273.24}$
$x = 0.25 ~ \text{mm from point } C$ (okay!)
