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Mechanical Properties of Solids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Solids5 Marks
Question
Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar. Consider a plane making an angle $\Delta$ with the length. What are the tensile and shearing stresses on this plane?
  1. For what angle is the tensile stress a maximum?
  2. For what angle is the shearing stress a maximum?
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Answer

According to the problem force F is applied along horizontal, so we resolve it in two perpendicular components - one is parallel to the inclined plane and other one is perpendicular to the inclined plane as shown in the diagram. Now, we can easily calculate the tensile and shearing stress. Here,$\text{F}_\bot=\text{F}\sin\theta,\text{F}_\parallel=\text{F}\cos\theta$
Let the cross - sectional area of the bar be A. Consider the equilibrium of the plane aa'. Here, $\text{F}_\bot$ produces tensile stress and $\text{F}_\parallel$ produces shear stress, on the plane aa'. Let the area of the face aa' be A, then$\therefore\ \sin\theta=\frac{\text{A}}{\text{A}'}\Rightarrow\text{A}'=\frac{\text{A}}{\sin\theta}$
Tensile stress on the plane aa'$\text{aa}'=\frac{\text{F}_\bot}{\text{A}'}=\frac{\text{F}\sin\theta}{\text{A}/\sin\theta}=\frac{\text{F}}{\text{A}}\sin^2\theta$
Shearing stress on the plane aa', Shearing stress $=\frac{\text{Parallel force}}{\text{Area}}$$=\frac{\text{F}_\parallel}{\text{A}'}=\frac{\text{F}\cos\theta}{\text{A}/\sin\theta}=\frac{\text{F}\sin\theta\cos\theta}{\text{A}}=\frac{\text{F}(2\sin\theta\cos\theta)}{2\text{A}}$
$=\frac{\text{F}\sin2\theta}{2\text{A}}$
  1. For tensile stress to be maximum,
$\sin^2\theta=0\Rightarrow\sin\theta=0\Rightarrow\theta=\frac{\pi}{2}\text{ or }\theta=90^0$
  1. For shearing stress to be maximum,
$\sin2\theta=1\Rightarrow2\theta=\frac{\pi}{2}\Rightarrow\theta=\frac{\pi}{4}\text{ or }\theta=45^0$

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