Question

Two equal and opposite forces each $F$ act on a rod of uniform cross-sectional area $‘a'$ 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
$=\frac{F}{a} \sin \theta \cos \theta$