Acircular 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

Idrk but i think u should figure out the n dia n then the units it will make it way more easier