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## 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.

## Homework Equations

## 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.

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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