MATH SOLVE

2 months ago

Q:
# 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.

Accepted Solution

A:

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

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