Normal and Shearing Stresses for an Axial Force

Prequisite Knowledge
Stresses on an Oblique Plane Under Axial Loading

Problem
A body is subject to a downward vertical force, applied in the middle and bottom. The force is 800 N. The cross-sectional area is 20cm x 30cm. For a 60° section through the material as shown, find the normal and shearing stresses at the slice.


Solution
First determine the area of the body at a section of the body which is perpendicular to the force in question.

A = 0.2(0.3) = 0.06 m2

The equation for normal stress is the following:

σ = (P/A)*(cos θ)2

The equation for shearing stress is the following:

τ = (P/A)*(sin θ)*(cos θ)

The angle θ is NOT 60°, it is 90 – 60 = 30°.
This is because the necessary angle θ is the angle between the force P and the normal force at the surface of the cut section.



The normal stress is then:

σ = (800/0.06)*(cos30°)2

σ = 10 kPa

and the shear stress is:

τ = (800/0.06)*(sin30°)*(cos30°)

τ = 5.77 kPa