A bar of cross-section A is subjected two equal and opposite tensile forces at its ends consider a plane section of the bar whose normal makes an angle $θ$ with the axis of the bar.
For what value of $θ$ is the shearing stress maximum?

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Solution

The correct option is A 45

$\begin{matrix} Shear stress & = \frac{F along plane}{A plane} = \frac{F sin θ}{A sec θ} = \frac{F}{2 A} \cdot sin 2 θ \end{matrix}$

$For shear stress to maximise, θ = 45 d e g r e e$

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A bar of cross-section A is subjected two equal and opposite tensile forces at its ends consider a plane section of the bar whose normal makes an angle θ with the axis of the bar.For what value of θ is the shearing stress maximum?

The correct option is A 45 Shear stress =F along plane A plane =FsinθAsecθ=F2A⋅sin2θ For shear stress to maximise, θ=45degree

A bar of cross-section A is subjected two equal and opposite tensile forces at its ends consider a plane section of the bar whose normal makes an angle $θ$ with the axis of the bar.
For what value of $θ$ is the shearing stress maximum?

The correct option is A 45

$\begin{matrix} Shear stress & = \frac{F along plane}{A plane} = \frac{F sin θ}{A sec θ} = \frac{F}{2 A} \cdot sin 2 θ \end{matrix}$

$For shear stress to maximise, θ = 45 d e g r e e$