Solution to Problem 333 | Flanged bolt couplings

$\tau = \dfrac{T\rho}{J}$
 

Where:
$T = 14(1000)(240) = 3\,360\,000 \, \text{N}\cdot\text{mm}$

$J = \Sigma A\rho^2 = \frac{1}{4}\pi(20)^2 \, [ \, 2(40^2) + 2(120^2) \, ]$

$J = 3\,200\,000\pi \, \text{mm}^4$

 

 

Maximum shearing stress (ρ = 120 mm):
$\tau_{max} = \dfrac{3\,360\,000(120)}{3\,200\,000\pi}$

$\tau_{max} = 40.11 \, \text{ MPa}$           answer
 

Minimum shearing stress (ρ = 40 mm):
$\tau_{min} = \dfrac{3\,360\,000(40)}{3\,200\,000\pi}$

$\tau_{min} = 13.37 \, \text{ MPa}$           answer