Question

Two equal and opposite forces each $F$ act on a rod of uniform cross-sectional area $‘a^{\prime }$ as shown in the figure. Shearing stress on the section $AB$ will be

• A. $\frac{F\sin \theta \cos \theta }{a}$
• B. $\frac{F\sin \theta }{a}$
• C. $\frac{F\cos \theta }{a}$
• D. $\frac{F{\sin }^2\theta }{a}$

Answer 1

Answer: (A)

Shearing stress
$=-\frac{F\sin \theta }{a_x}$
$=-\frac{F}{a}\sin \theta \cos \theta$
Magnitude of shearing stress
$=\frac{F}{a}\sin \theta \cos \theta$