# Elastic properties of solids problem

• BrainMan

## Homework Statement

If the shear stress in steel exceeds about 4.0 x 10^8 N/m^2, the steel ruptures. Determine the shear force necessary to punch a 1-cm diameter hole in a steel plate that is 0.5 cm thick.

## The Attempt at a Solution

The way I tried to solve this problem was by doing

4.0 x 10^8 N/m^2 = x/((.005^2)*3.14*.005)

and solving for x, where x is the amount of Newtons necessary to punch the hole in the steel plate. I did this because the ratio of Newtons to m^2 must be 4.0 x 10^8 N/m^2. So I found the area of the plate punched out and solved for x. The solution to this problem is 6.28 x 10^4 N and I got 157 N.

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What is the surface area of the hole?

What is the surface area of the hole?
The surface area of the hole should be 7.85 x 10^-5 m^2.

The surface area of the hole should be 7.85 x 10^-5 m^2.
That's not what I get. How are you calculating it?
You are told the diameter and the depth, and you want the surface area inside the hole.

• BrainMan
That's not what I get. How are you calculating it?
You are told the diameter and the depth, and you want the surface area inside the hole.
Ok I realized my mistake. Thanks!