If an element is subjected to pure shearing stress τxy then the maximum principal stress is equal to-A-2τxyB-τxy/2C-τxyD- .root { background image: url(/_files/images/chemical engineering/common/root.gif); background position: top left; background repeat: no repeat; padding left: 10px; padding top: 1px; } 1 (τxy)2-C-τxy

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If an element is subjected to pure shearing stress &tau;xy then the maximum principal stress is equal to-A-2&tau;xyB-&tau;xy/2C-&tau;xyD- .root { background image: url(/_files/images/chemical engineering/common/root.gif); background position: top left; background repeat: no repeat; padding left: 10px; padding top: 1px; } 1 (τxy)2-C-&tau;xy

APTITUDE CRACK · Accepted Answer

C &tau;xy Since pure shearing stress subjected, then no normal stress so it becomes zero then now, root over Txy^2. Cancel root with square and the final answer Txy.

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If an element is subjected to pure shearing stress τ_xy then the maximum principal stress is equal to

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If an element is subjected to pure shearing stress τxy then the maximum principal stress is equal to

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If an element is subjected to pure shearing stress τ_xy then the maximum principal stress is equal to

For help Students Orientation
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