Question

A 135-N force is applied at a 50-degree angle to the surface of the end of a square bar. The surface is 2.50 cm on a side. What is the compressional stress?

Answer 1

Hmm, I think I remember how to do this from a previous materials analysis course.

Stress can be defined as:
\[\huge \sigma = \frac{\text{F}}{\text{A}}\]The issue here is that we want compressional stress, which is the force that compresses the square bar. I think what we have to do is take the component that is parallel to the "area vector" AKA perpendicular to the cross sectional area surface. Visually, we want: (assuming that I drew the right angle.. The component that is perpendicular to the cross sectional area would be \(\text{135} \cos(50)\). So
\[\huge \sigma =\frac{135 \cos(50)}{\text{side}^2}\]

Answer 2

Let's see what @IrishBoy123 @rvc @Michele_Laino says X)

Answer 3

Pretty sure this makes sense though. The component of the force that is parallel to the "area vector" causes compression/tensile stress and the component of the force that is perpendicular to the "area vector" will cause shear stress.

Answer 4

And an area vector is not a true vector, just one that helps us see if an area is perpendicular/parallel/etc to another measurable quantity, such as force

Answer 5

Well I guess I shouldn't say it's not a true vector. It's just not a normal vector that we would think of.

Answer 6

you are right! @CShrix