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	$\left(\frac{F}{2A}\right)\sin 2\theta$
D	The shearing stress will be $\mathrm{maximum}$ 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 $△PQR$
$\because \frac{A}{A^{\prime }}=\sin \left(90^{\circ }-\theta \right)=\cos \theta$
$\therefore$ 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 $=\frac{\text{ Force }}{\text{ Area }}=\frac{F\sin \theta }{A\sec \theta }$
$=\frac{F}{A}\sin \theta \cos \theta =\frac{F}{2A}\sin 2\theta$
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, $A\to 2,B\to 3,C\to 1$ and $D\to 4$