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A bar of cross-sectional area A is subjected to equal and opposite tensile forces at its ends. Consider a plane section (PS) of the bar, whose normal makes an angle $ϕ$ with the axis (axis is along the length) of the bar.
Column-I Column-II
i. shearing stress on PS p. $\frac{F}{A} \cos^{2} ϕ$
ii. tensile stress on PS q. 0°
iii. the tensile stress is maximum for $ϕ$ = r. $\frac{F}{A} sinϕ cosϕ$
iv. the shearing stress is maximum for $ϕ$ = s. 60°
t. 45°
Match the given columns and select the correct option from the codes given below.
Codes

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i-p, ii-r, iii--q, iv-t

i-q, ii-r, iii-s, iv-p

i-r, ii-p, iii-q, iv-t

i-p, ii-q, iii-r, iv-s

answer is A.

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Detailed Solution

Tangential force is $F_{t} = F \sin ϕ$

Area of section considered is

$A^{'} = L (\frac{L}{\cos ϕ}) = \frac{L^{2}}{\cos ϕ} = \frac{A}{\cos ϕ}$

Shearing stress = $\frac{F_{t}}{A^{'}} = \frac{F \sin ϕ}{\frac{A}{\cos ϕ}} = \frac{F}{A} sinϕcosϕ$ --------------(i)

Normal force on plane section considered as
.'. Tensile stress = $\frac{F_{N}}{A^{'}} = \frac{F \cos ϕ}{\frac{A}{\cos ϕ}} = \frac{F}{A} \cos^{2} ϕ$ --------------(ii)

Shearing stress is maximum when

$ϕ = 45 °$

Tensile stress is maximum when

$ϕ = 0 ° [From equation (ii)]$

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