Question

A circular 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

Answer 1

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

Step-by-step explanation:

using the equation of torsion we Have

`(T)/(I_(p))=(\tau )/(r)\\\\\therefore \tau =(T)/(I_(p))* r`

where,

`\tau`= shear stress at a distance 'r' from the center

T = is the applied torque

`I_(p)` = 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

`\tau _(max)=(4)/(3)* (V)/(A)`

Applying values we get

`\tau _(max)=(4)/(3)* (85* 10^(3))/(0.25* \pi * (25* 10^(-3))^(2))\\\\\therefore \tau _(max)=230.88Mpa\approx 231Mpa`