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 $θ$ with the axis of the bar.
For what value of $θ$ is the tensile stress maximum?

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Solution

The correct option is A 0
Tensile stress $= \frac{f o r c e}{a r e a} = \frac{F cos θ}{A sec θ}$
$= \frac{F}{A} {cos}^{2} θ$
Tensile strength will be maximum when ${cos}^{2} θ$ is maximum i.e., $cos θ = 1$ or $θ = 0 °$

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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 θ with the axis of the bar.For what value of θ is the tensile stress maximum?

The correct option is A 0Tensile stress=forcearea=FcosθAsecθ=FAcos2θTensile strength will be maximum when cos2θ is maximum i.e., cosθ=1 or θ=0°

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 $θ$ with the axis of the bar.
For what value of $θ$ is the tensile stress maximum?

The correct option is A 0
Tensile stress $= \frac{f o r c e}{a r e a} = \frac{F cos θ}{A sec θ}$
$= \frac{F}{A} {cos}^{2} θ$
Tensile strength will be maximum when ${cos}^{2} θ$ is maximum i.e., $cos θ = 1$ or $θ = 0 °$