Solution to Problem 329 | Flanged bolt couplings

$\dfrac{A \tau_1}{6} = \dfrac{A \tau_2}{4.5}$

$\tau_2 = 0.75\tau_1$
 

$T = P_1 R_1 n_1 + P_2 R_2 n_2$

$700(12) = \frac{1}{4}\pi (1/2)^2 \tau_1 (6)(8) + \frac{1}{4}\pi (1/2)^2 \tau_2 (4.5)(6)$

$8400 = 3\pi \tau_1 + 1.6875\pi (0.75\tau_1)$

$8400 = 13.4\tau_1$

$\tau_1 = 626.87 \, \text{psi}$   →   bolts in the outer circle       answer

$\tau_2 = 0.75(626.87) = 470.15 \, \text{psi}$   →   bolts in the inner circle       answer