A circular bar is subjected to an axial pull of 100kN. if the maximum intensity of shear stress on any plane is not to exceed 60MN/m^2 determine the diameter of the bar.

I know the answer to this is 32.6mm, what I need to know is how to arrive at this number so that I know how to do it.

Use a Mohr circle to find the maximum shear stress relative to the axial stress.

Here we assume the axial stress is sigma, the transverse axial stress is zero.

So we have a Mohr circle with (0,0) and (0, sigma) as a diameter.

The centre of the circle is therefore (0, sigma/2), and the radius is sigma/2.

From the circle, we determine that the maximum stress is the maximum y-axis values, namely + / - sigma/2, at locations (sigma/2, sigma/2), and (sigma/2, - sigma/2).

Given that the maximum shear stress is 60 MPa, we have

sigma/2=60 MPa, or sigma=120 MPa.

(note: 1 MPa = 1N/mm^2)

Therefore

100 kN / (pi*d^2/4) = 100,000 N / (pi*d^2/4) = 120 MPa where d is in mm.

Solve for d

d=sqrt (100,000*4 / (120*pi))

=32.5735 mm