Lecture 23 More Stress Prisms Example Problem

Lecture 23 Example Problem

More Stress Prisms

Given:
a 1" diameter rod loaded with a 5k force as indicated.
Determine:
the stress prism.
Lecture 23

Solution:
The stress prism of the axial load would be evenly distributed over the cross-section. The magnitude of the stress (f) will equal the force divided by the area. In this case the cross section is circular so the area will equal:

A = ½r^2 = 3.14(.5in)^2 = .785 in^2
The stress will equal:

f = 5k / .785 in^2 = 6.37 ksi or 6370 psi
The stress prism will take the shape of a cylinder with the area or the cross section as its base and the magnitude of the stress as its height.