Solving for location of centroid of rivets:
$A X_G = \Sigma ax$
Where
$A = \frac{1}{2}(80 + 160)(120) = 14\,400 \, \text{mm}^2$
$a_1 = a_2 = a_3 = \frac{1}{2}(80)(120) = 4800 \, \text{mm}^2$
$x_1 = x_3 = \frac{1}{3}(120) = 40 \, \text{mm}$
$x_2 = \frac{2}{3}(120) = 80 \, \text{mm}$
$14\,400X_G = 4800(40) + 4800(80) + 4800(40)$
$X_G = \frac{160}{3} \, \text{mm}$
$r_1 = \sqrt{(\frac{160}{3})^2 + 80^2} = 96.148 \, \text{mm}$
$r_2 = \sqrt{(120 - \frac{160}{3})^2 + 40^2} = 77.746 \, \text{mm}$
$J = \Sigma A\rho^2 = \frac{1}{4}\pi (10^2)(2r_1^2 + 2r_2^2 + X_G^2)$
$J = \frac{1}{4}\pi (10^2)\, [ \, 2(96.148^2) + 2(77.746^2) + (\frac{160}{3})^2 \, ]$
$J = 2\,624\,973.55 \, \text{mm}^4$
$T = 120P + 100P = 220P$
The critical rivets are at distance r1 from centroid:
$\tau = \dfrac{T \rho}{J}$
$\tau_{max} = \dfrac{Tr_1}{J}$
$70 = \dfrac{220P(96.148)}{2\,624\,973.55}$
$P = 8686.8 \, \text{N}$ answer
