Question

A uniform rectangular bar of area of cross-section $A$ is fixed at one end and on other end forces $F$ is applied as shown in figure. Find the shear stress at a plane through the bar making an angle $\theta$ with the vertical as shown in figure.

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

Answer 1

Answer: (C)

Consider $A^{\prime}$ is the area of cross-section of plane shown in figure

$\therefore A'=\frac{A}{\cos \theta}$
Now the tangential force on plane is $F \sin \theta$
Shear stress $=\frac{F \sin \theta}{\frac{A}{\cos \theta}}$
$=\frac{F}{A} \sin \theta \cos \theta=\frac{F}{2 A} 2 \sin \theta \cos \theta$
$=\frac{F}{2 A}(\sin 2 \theta)$