Normal Stress question - Mechanics of Materials

[koab1mjr]

Homework Statement

Conceptual question - When considering beam connected to something say a wall by a cylindar shaped pin, why is it when the beam is in tension near the pin the stress is high as the load is applied over a smaller cross section (normal cross section - profile of the pin) but when in compression one ignores the pin and takes the full cross sectional area?

Homework Equations

stress is force/ area

The Attempt at a Solution

The book I am using ignores the pin when beams are in compression. When using that approach i get the write answer but without understanding it is meaningless

Thanks in advance

 

[Saladsamurai]

I can't say that I actually know the answer or that I recall this approach. But, if I had to rationalize it, I would say that it is because a lot of materials tend to fail in tension rather then compression.

When in compression, the pin almost acts as a type of 'filler' becoming one with the hole. When in tension, the circular hole becomes more elliptical about the longitudinal axis of the beam. This would probably be a good place for a crack to form and propagate.

Perhaps an expert will chime in here, but that is my guess.

 

[Mapes]

It has nothing to do with crack formation or failure in tension or compression; it's just a matter of whether there's any force applied the narrower cross section area on either side of the hole.

Think about which way the pin is pulling/pushing on the beam. If it's pulling on the beam (i.e., the beam is in tension), then the tensile stress is uniform along the beam except at the hole, where the cross section is smaller and therefore the stress is higher. If the pin is pushing on the beam, however, then no stress acts on that narrowed area because the pin is pushing on the other side of the hole. Thus, we use the uniform compression stress calculated by using the beam's full cross section. Does this answer your question?