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(A) Let the cross-sectional area of the rod be $a$. The area of the inclined section $AB$ is $A' = \frac{a}{\sin 	heta}$.The force $F$ acting on the

Vedclass · Accepted Answer

$\frac{F \sin 	heta \cos 	heta}{a}$

Vedclass · Answer

(A) Let the cross-sectional area of the rod be $a$. The area of the inclined section $AB$ is $A' = \frac{a}{\sin 	heta}$.The force $F$ acting on the rod can be resolved into two components along the section $AB$:$1$. Normal component: $F_n = F \sin 	heta$$2$. Tangential (shearing) component: $F_s = F \cos 	heta$The shearing stress $	au$ is defined as the tangential force divided by the area of the section:$	au = \frac{F_s}{A'} = \frac{F \cos 	heta}{a / \sin 	heta} = \frac{F \sin 	heta \cos 	heta}{a}$

Two equal and opposite forces each $F$ act on a rod of uniform cross-sectional area $a$ as shown in the figure. The shearing stress on the section $AB$ will be

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