Acircular cross section, d = 25 mm, experiences a torque load, t = 25 n·m, and a shear force, v = 85 kn. calculate the shear stress at the center of the cross section. a. 231 mpa b. 239 mpa c. 8.15 mpa d. 95.6 mpa

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

where,

= shear stress at a distance 'r' from the center

T = is the applied torque

= polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

Applying values we get

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