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In shear force in a rectangular bar,the relevant area is the cross sectional area parallel to the applied force.But in torsion which also undergo shearing we get shear stress from torsion equation.I like to know if we get a shear stress due to torsion in a rod,relevant area should be the cross section of the rod or is it irrelevant because rather than cross section its polar moment of area that influence shearing?

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  • $\begingroup$ or is it the outer surface area? $\endgroup$ – Navaneeth Jan 11 at 5:31
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Relevant area while applying torsional moment will one which is perpendicular to Moment direction. Shearing action will happen between surfaces which are perpendicular to the direction of the moment. Even though the polar moment of the area appears in the equation saying that area is irrelevant doesn't make sense because it's the polar moment of that cross-sectional area where shear action is happening.

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The terms shear stress and torsion stress are synonymous when it comes to stress due to torque versus stress due to shearing force. The torsion or shear stress in a circular member is given by

$$τ=\frac{Tr}{J}$$

Where $T$ is the applied torque, $r$ is the radius and $J$ is the polar moment of inertia. The equation applies to a solid circular shaft or a thick walled shaft where the wall thickness $t$ is given by $t>0.1r$.

The cross sectional area is taken into account by the polar moment of inertia. The polar moment of inertia of a solid shaft is given by

$$J=\frac{πa^4}{2}$$

Where $a$ is the radius.

For a circular disc, it is given by

$$J=\frac{π(a^4-b^4)}{2}$$

Where $a$ is the outside radius of the disc and $b$ the inside radius.

You can see that the cross sectional material area of the disc is less than the solid circular area, and therefore the torsional stress due to torque on the disc is greater than the solid circle as you would expect. The same would apply if you apply the same shearing force to a rectangular bar of less cross sectional area.

Hope this helps.

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