- #1

bimbambaby

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## Homework Statement

A hollow pipe has an inner diameter of 80 mm and an outer diameter of 100 mm. If its end is tightened using a torque wrench using 80 N forces, determine the maximum and minimum shear stress in the material. Where are they located?

Note: In the diagram of the picture, the left hand applies an 80 N force upward on the pipe, 200 mm from the shaft, and 80N downward with the right hand, 300 mm from the axis of the pipe. Had trouble getting the picture.

## Homework Equations

$$

\tau_{max} = \frac{T*radius}{I_p}\\

Torque = r X F\\

I_p = \frac{\pi}{32}((d_2)^4-(d_1)^4)

$$

## The Attempt at a Solution

So I know how to calculate the maximum shear stress in the pipe:

$$

Torque = r X F = (.200 m)*(80 N) + (.300 m)*(80 N) = 40 N*m\\

\\

I_p = \frac{\pi}{32}((d_2)^4-(d_1)^4) = \frac{\pi}{32}((.100 m)^4-(.080 m)^4) = 5.796e-6 m^4\\

\\

\tau_{max} = \frac{T*radius}{I_p} = \frac{(40 N*m)*(.050 m)}{5.796e-6 m^4} = 345051.4 N/m^2

$$

Therefore, tau_max takes place at the outer surface of the shaft.

**For tau_min, would I evaluate my expression for tau max at the inner diameter of the pipe?**It makes sense to me, but I was hoping someone could verify this