# Minimum Shear Stress in Hollow Circular Tube

• bimbambaby
In summary, the conversation discusses a hollow pipe with specific dimensions and an applied torque wrench force. The maximum and minimum shear stress in the material are calculated using the given equations, with the maximum stress occurring at the outer surface of the shaft and the minimum stress occurring at the inner diameter of the pipe.
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

bimbambaby said:

## 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?

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

SteamKing said:
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?

.

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.

## 1. What is minimum shear stress in a hollow circular tube?

Minimum shear stress in a hollow circular tube refers to the minimum amount of force per unit area required to cause shearing or sliding of the material in the tube. It is an important factor to consider in the design and analysis of structures, as it can affect the stability and strength of the tube.

## 2. How is minimum shear stress calculated in a hollow circular tube?

The minimum shear stress in a hollow circular tube can be calculated by dividing the applied shear force by the cross-sectional area of the tube. This can be represented by the equation τ = F/A, where τ is the minimum shear stress, F is the applied shear force, and A is the cross-sectional area of the tube.

## 3. What factors can affect the minimum shear stress in a hollow circular tube?

There are several factors that can affect the minimum shear stress in a hollow circular tube, including the dimensions and material properties of the tube, the applied load or force, and the type of support or boundary conditions at the ends of the tube.

## 4. How does the minimum shear stress in a hollow circular tube relate to its strength?

The minimum shear stress in a hollow circular tube is directly related to its strength. A higher minimum shear stress can lead to a higher likelihood of failure or instability in the tube, while a lower minimum shear stress can result in a stronger and more stable structure.

## 5. What are some common applications of minimum shear stress in hollow circular tubes?

Minimum shear stress in hollow circular tubes is commonly used in the design and analysis of various structures, including bridges, towers, and buildings. It is also important in the aerospace and automotive industries, where tubes are used in the construction of aircrafts, spacecrafts, and vehicles.

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