Question

A bar of cross-section $A$ is subjected to equal and opposite tensile forces at its ends. Consider a plane section of the bar, whose normal makes an angle $\theta$ with the axis of the bar.

Match the Column I (stress) with Column II (value) and select the correct answer from the codes given below.

Column I		Column II
A	The tensile stress on this plane will be	1	$0$
B	The shearing stress on this plane will be	2	$(F / A) \cos ^{2} \theta$
C	The tensile strength will be maximum at $\theta=$	3	$\left(\frac{F}{2 A}\right) \sin 2 \theta$
D	The shearing stress will be $\operatorname{maximum}$ at $\theta=$	4	$45^{\circ}$

• A. A - (3) , B - (2) , C - (4) , D - (1)
• B. A - (2) , B - (3) , C - (1) , D - (4)
• C. A - (2) , B - (3) , C - (4) , D - (1)
• D. A - (3) , B - (2) , C - (1) , D - (4)

Answer 1

Answer: (B)

(b) The resolved part of $F$ along the normal is the tensile stress on this plane and the resolved part parallel to the plane is the shearing stress on the plane as shown below.

Given, cross-section area of bar $=A$
Let cross-section area of plane $=A^{\prime}$
then, from $\triangle P Q R$
$\because \frac{A}{A^{\prime}}=\sin \left(90^{\circ}-\theta\right)=\cos \theta$
$\therefore$ Area, $A^{\prime}=\frac{A}{\cos \theta}=A \sec$
A. Tensile stress $=\frac{\text { Force }}{\text { Area }}=\frac{F \cos \theta}{A \sec \theta}=\frac{F}{A} \cos ^{2} \theta$
$($ area of plane section $=A \sec \theta)$
B. Shearing stress $=\frac{\text { Force }}{\text { Area }}=\frac{F \sin \theta}{A \sec \theta}$
$=\frac{F}{A} \sin \theta \cos \theta=\frac{F}{2 A} \sin 2 \theta$
C. Tensile stress or strength will be maximum when $\cos ^{2} \theta$ is maximum,
i.e. $\cos \theta=1$ or $\theta=0^{\circ}$.
D. Shearing stress will be maximum when $\sin 2 \theta$ is maximum,
i.e. or $2 \theta=90^{\circ}$, i.e. $\theta=45^{\circ}$.
Hence, $A \rightarrow 2, B \rightarrow 3, C \rightarrow 1$ and $D \rightarrow 4$