Design of Riveted Joints Questions and Answers

Answer: A

The minimum plate thickness is \(\frac{F}{S_{f}(W - 2d)}\).

The weakest tensile section of each plate is the line through the rivet holes. Two rivets of diameter d lie in that row, so the net width there is W - 2d, and the whole load F passes through it:

\[ \begin{aligned} S_{f}\,t\,(W - 2d) &\ge F \\ t &\ge \frac{F}{S_{f}(W - 2d)} \end{aligned} \]

F/(S_fW) ignores the holes (the gross section). W - d removes only one hole, and 2F/(S_fW) doubles the load without removing any holes.

Answer: A

Each rivet needs at least 800 mm².

The rivets are one above the other, a distance \(a = 50\) mm from their centroid. The load gives each rivet a direct shear \(P/2 = 5\) kN downwards, and the moment \(Pe\) gives each a secondary shear at right angles to the line of rivets, that is horizontal:

\[ \begin{aligned} F_{s} &= \frac{Pe\,a}{2a^{2}} = \frac{P e}{2a} = \frac{10 \times 3\sqrt{7}}{2} \\ &= 15\sqrt{7} \approx 39.7\ \text{kN} \end{aligned} \]

The two components are perpendicular, so each rivet carries

\[ \begin{aligned} F &= \sqrt{5^{2} + (15\sqrt{7})^{2}} = \sqrt{25 + 1575} = 40\ \text{kN} \\ A &= \frac{40000}{50} = 800\ \text{mm}^{2} \end{aligned} \]

200 mm² ignores the moment, \(100\sqrt{7}\) and 25 mm² do not follow from combining the two shears. Adding the components directly would apply only if the rivets were side by side.