A bar of cross-section A is subjected to 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 tensile stress is maximum?

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Solution

The correct option is A 0
$\begin{matrix} Tensile stress & = \frac{F_{Normal to plane}}{A plane} = \frac{F cos θ}{A sec θ} = \frac{F}{A} {cos}^{2} θ \end{matrix}$

$For tensile stress to naxe be maximum, θ = 0^{\circ}$

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A bar of cross-section A is subjected to 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 tensile stress is maximum?

The correct option is A 0 Tensile stress =FNormal to plane A plane =FcosθAsecθ=FAcos2θ For tensile stress to naxe be maximum, θ=0∘

A bar of cross-section A is subjected to 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 tensile stress is maximum?

The correct option is A 0
$\begin{matrix} Tensile stress & = \frac{F_{Normal to plane}}{A plane} = \frac{F cos θ}{A sec θ} = \frac{F}{A} {cos}^{2} θ \end{matrix}$

$For tensile stress to naxe be maximum, θ = 0^{\circ}$