$T = 0.05P = 0.05(5)$

$T = 0.25 ~ \text{kN}\cdot\text{m}$

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, τ_{torque} = τ_{direct-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!*)