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Standard XII

Physics

Moment of Inertia of a Rod

A point in a ...

Question

A point in a strained material is subjected to the stresses as shown in figure.

The value of major principal stress is ____ $k N / m^{2}$
97.23

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Solution

The correct option is A 97.23

$σ_{1 / 2} = \frac{σ_{x} + σ_{y}}{2} \pm \sqrt{(\frac{σ_{x} - σ_{y}}{2}) + τ_{x y}^{2}}$

$σ_{x} = 80 sin 60^{\circ} = 69.28 k N / m^{2}$

$σ_{y} = 40 k N / m^{2}$

$τ_{x y} = 80 cos 60^{\circ} = 40 k N / m^{2}$

$Major principal stress (σ_{1})$

$= \frac{69.28 + 40}{2} \sqrt{{(\frac{69.28 - 40}{2})}^{2} + 40^{2}} = 97.23 k N / m^{2}$

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A point in a strained material is subjected to the stresses as shown in figure. The value of major principal stress is ____ kN/m297.23

The correct option is A 97.23 σ1/2=σx+σy2±√(σx−σy2)+τ2xy σx=80sin60∘=69.28 kN/m2 σy=40 kN/m2 τxy=80cos60∘=40 kN/m2 Major principal stress(σ1) =69.28+402√(69.28−402)2+402=97.23 kN/m2

A point in a strained material is subjected to the stresses as shown in figure.

The value of major principal stress is ____ $k N / m^{2}$
97.23