Minimum Shear Stress in Hollow Circular Tube

[bimbambaby]

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

 

[SteamKing]

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

Yes. Where else could you take it?

 

[bimbambaby]

I'm having trouble understanding the tone of your question. Are you saying there is another location?

 

[bimbambaby]

Yes. Where else could you take it?

Sorry, I'm having trouble understanding the tone of your response. Are you implying there are other locations?

 

[HalfLostAndSearching]

.

Yes, your approach to finding the maximum shear stress is correct. To find the minimum shear stress, you can use the same equation, but with the inner diameter of the pipe (d1) instead of the outer diameter (d2).

$$
\tau_{min} = \frac{T*radius}{I_p} = \frac{(40 N*m)*(.040 m)}{5.796e-6 m^4} = 276041.1 N/m^2
$$

Therefore, tau_min takes place at the inner surface of the shaft. This makes sense because the torque is applied at the outer surface, causing the maximum shear stress, and the minimum shear stress is located at the opposite surface.