Question

A horizontal plate has been joined to a vertical post using four rivets arranged as shown in the figure. the magnitude of the load on the worst loaded rivet (in N) is

• A. 1839.83

Answer 1

The correct option is A 1839.83
Primary force $\left(P_1\right)=100N$

$P_1"=P_2"=P_3"=P_4"=\frac{P_e\times r}{4r^2}$

$=\frac{400\times 500\times \left(40\sqrt{2}/2\right)}{4\times {\left(\frac{40\sqrt{2}}{2}\right)}^2}$

$P_1"=\frac{400\times 500\times 20\sqrt{2}}{4\times (20\sqrt{2}{)}^2}=\frac{400\times 500}{4\times 20\sqrt{2}}$
$P_1"=1767.766N$

(1 and 4) are worst loaded

$\therefore P_{net}^2=P_1+P_1{"}^2+2P_1{P}_1"\cos {45}^{\circ }$

$P_{net}=1839.83N$