From Illustrative Problem 591

τ = 100 MPa shear stress

σ_{b} = 220 MPa bearing stress for single shear rivet

σ_{b} = 280 MPa bearing stress for double shear rivet

By transfer formula for moment of inertia

$I = \bar{I} + Ad^2$

$I_{NA} = 2 \, [ \, (3.84 \times 10^6) + 7570(19.7^2) \, ]$

$I_{NA} = 13\,555\,682.6 \, \text{ mm}^4$

Rivet capacity in shear (single shear)

$R_s = 2(A_s \, \tau) = 2 \, [ \, \frac{1}{4}\pi (19^2)(100) \, ]$

$R_s = 18\,050\pi \, \text{ N} = 56.705 \, \text{ kN}$

Rivet capacity in bearing (use σ_{b} = 220 MPa)

$R_b = 2(A_b \, \sigma_b) = 2 \, [ \, 19(13.2)(220) \, ]$

$R_b = 110\,352 \, \text{ N} = 110.352 \, \text{ kN}$

Use *R* = 56.705 kN for safe value of *R*

From strength of rivets

$R = \dfrac{VQ_{NA}}{I}s$

$56.705 = \dfrac{V \, [ \, 7570(19.7) \, ]}{13\,555\,682.6}(200)$

$V = 25.77 \, \text{ kN}$ *answer*