Topic: Machine Design
A rectangular steel plate is joined to a vertical post using four identical rivets arranged as shown below in the figure. The shear load on the worst loaded rivet (in N) approximately is
- A. 1700
- B. 1500
- C. 1650
- D. 1750
Answer: C. 1650
Explanation: The rivets form a $60\times60$ square. Eccentric load $P = 840~N$ at eccentricity $e = 300~mm$. Primary shear force per rivet $P_{p} = P/4 = 840/4 = 210~N$. Distance from CG to each rivet $r = \sqrt{30^{2}+30^{2}} = 30\sqrt{2}~mm$. Secondary shear force $P_{s} = \frac{P \cdot e \cdot r}{4r^{2}} = \frac{P \cdot e}{4r} = \frac{840 \times 300}{4 \times 30\sqrt{2}} = \frac{252000}{169.7} \approx 1485~N$. The worst loaded rivet (top right) has the angle between primary and secondary shear forces as $\theta = 45^{\circ}$. Resultant force $R = \sqrt{P_{p}^{2} + P_{s}^{2} + 2P_{p}P_{s}\cos 45^{\circ}} = \sqrt{210^{2} + 1485^{2} + 2(210)(1485)(\frac{1}{\sqrt{2}})} = \sqrt{44100 + 2205225 + 441021} \approx \sqrt{2690346} \approx 1640.2~N$. The closest option is 1650 N.