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 60M
N/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
