Stress and Strain

As we are given the Modulus of Elasticity, we can transpose the following equation to make strain the subject, and solve:

Remember GIGA means trillion, or 10 to the ninth power (nine zeros)

Strain has no units.

This is one of my favorite questions, it combines stress with some interesting geometry. First, we will draw a picture:

The stress equation is easy enough, its the force divided by the area:

The interesting part of the question is the area over which the force is exerted. The area is the cross-hatched side of the cylinder in the diagram above. This is the area that used to stick the round that we just punched out, into the plate. We need to calculate the curved surface area of the cylinder.

Now we know our area, we can calculate our stress. This can be given in N/m^2 or N/mm^2 whichever is more appropriate:

First lets draw a diagram

Remember the Stress formula! In order to use it, we need to know the surface area. The surface area here is the curved surface area on the outside of the round. This curved surface area used to be in contact with the metal plate, and then we punched it out. First, we need to calculate the curved surface area:

Now we know how much surface area we are dealing with, and we know the ultimate shear strength of the steel (given in the question) we can calculate the force required to punch that hole:

We need the cross-sectional area to calculate the stress. This question looks hard at first, but we can see that we can split that complicated shape into 2 very simple shapes. It is made from a circle, well two halves and the straight section is just a rectangle.

Lets work on each section, and the sum the areas.

Next lets find the total:

Using the above equation for stress, all we need now is the downwards force, created by the mass of the object under gravity.

As we can see the stress is below the stress limit, we can see that the column can withstand the load.