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Consider a long steel bar under a tensile stress due to force F acting at the edges along the length of the bar (as shown in figure). Consider a plane making an angle 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 ?

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/think-661001" style="font-weight:bold;" target="_blank" title="Click to know more about THINK">THINK</a> below figure, <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XI_P2_C09_E04_024_S01.png" width="80%"/> <br/> Suppose A is the cross sectional area of the bar. Let for balance of pane aa. angle `pi/<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> - theta` between perpendicular ON to the pane and <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> F. The parallel component of force to the pane, <br/> `F_(P) = F sin ((pi)/(2) - theta) = F cos theta and ` <br/> Perpendicular component, <br/> `F_(N) = F cos ((pi)/(2) -theta)= F sin theta` <br/> Suppose A is area of pane (face) aa. <br/> `therefore (A)/(A.) = sin theta` <br/> `therefore A. = (A)/(sin theta )` <br/> Tensile stress `= ("perpendicular force")/("area") = (F _(N))/(A.)= (F sin thea)/(A//sin theta)` <br/> `therefore` Tensile stress `= F/A sin ^(2) theta ` <br/> (a) For <a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> stress `F/A sin ^(2) theta ` should be maximum. <br/> `sin ^(2) theta =1` <br/>`therefore sin theta =1` <br/> `therefore theta = (pi)/(2)` <br/> For this `theta = <a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a>^(@)` <br/> Shearing stress `= ("parallel force")/("area") = (F _(p ))/(A.)` <br/> ` = (F cos theta )/(A //sin theta) = F/A sin theta cos theta= (F)/(2A) xx sin 2 theta` <br/> `[ because 2 sin theta cos theta = sin 2 theta]` <br/> (b) For shearing stress to be maximum `sin 2 theta =1` <br/> `therefore 2 theta =90^(@)` <br/> `therefore theta = 45^(@)`</body></html>


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