Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle θ with the length. What are the tensile and shearing stresses on this plane?

(a) For what angle is the tensile stress a maximum?

(b) For what angle is the shearing stress a maximum? 

This is a long answer type question as classified in NCERT Exemplar

Let the cross sectional area A . consider the equilibrium of the plane aa’. A force F must be acting on this plane making an angle of $\frac{\pi }{2}-\theta$ with the normal ON. Resolving F into components along the plane (FP) and normal to the plane.

By resolving into components

We get F_p= Fcos $\theta$

And F_N= Fsin $\theta$

Let area of the face aa’ be A’ then

A/A’=sin $\theta$  so A=A’sin $\theta$

The tensile stress = normal force/area=Fsin $\theta /A\text{'}$

= $\frac{Fsin\theta }{A/sin\theta }=\frac{F}{A}$ sin^{2 $\theta$}

Shearing stress = parallel foce/Area

= $\frac{Fcos\theta }{A/sin\theta }=\frac{F}{A}sin\theta cos\theta$

$\frac{F}{2A}(2sin\theta cos\theta =\frac{F}{2A}sin2\theta$

a) For stress to be maximum , sin^{2 $\theta$} =1

So $\theta$ = $\frac{\pi }{2}$

b) Shearing stress to be maximum

sin2 $\theta =1$

So $\theta$ = $\frac{\pi }{4}$

<div><div><p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>Let the cross sectional area A . consider the equilibrium of the plane aa’. A force F must be acting on this plane making an angle of <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mi>θ</mi> </math> </span>with the normal ON. Resolving F into components along the plane (FP) and normal to the plane.</p><p>By resolving into components</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745488871phprwKc99_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745488871phprwKc99.jpeg" alt="" width="233" height="166"></picture></div><div><p>We get F_p= Fcos<span contenteditable="false"> <math> <mi>θ</mi> </math> </span></p><p>And F_N= Fsin<span contenteditable="false"> <math> <mi>θ</mi> </math> </span></p><p>Let area of the face aa’ be A’ then</p><p>A/A’=sin<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>&nbsp;so A=A’sin<span contenteditable="false"> <math> <mi>θ</mi> </math> </span></p><p>The tensile stress = normal force/area=Fsin<span contenteditable="false"> <math> <mi>θ</mi> <mo>/</mo> <mi>A</mi> <mi>'</mi> </math> </span></p><p>= <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>F</mi> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> <mo>/</mo> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>θ</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </mfrac> </math> </span>sin^{2<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>}</p><p>Shearing stress = parallel foce/Area</p><p>=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>F</mi> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> <mo>/</mo> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>θ</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </mfrac> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>θ</mi> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>θ</mi> </math> </span></p><p><span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>A</mi> </mrow> </mrow> </mfrac> <mo>(</mo> <mn>2</mn> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>θ</mi> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>θ</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>A</mi> </mrow> </mrow> </mfrac> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mn>2</mn> <mi>θ</mi> </math> </span></p><p><strong>a)</strong>&nbsp;For stress to be maximum , sin^{2<span contenteditable="false"> <math> <mi>θ</mi> </math> </span>}=1</p><p>So <span contenteditable="false"> <math> <mi>θ</mi> </math> </span>=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span></p><p><strong>b)</strong> Shearing stress to be maximum</p><p>sin2<span contenteditable="false"> <math> <mi>θ</mi> <mo>=</mo> <mn>1</mn> </math> </span></p><p>So <span contenteditable="false"> <math> <mi>θ</mi> </math> </span>=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math> </span></p></div></div></div></div>

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