Mechanics: Eccentric torsion + non circular tube

In summary, the objective is to find the torsion shear stress at the cross section of the dark grey tube, using the equation T = (J/2R) * (F1 - F2). This can be found by using the parallel axis theorem to transform the moment of inertia from the centroid to the torque center, and then using the torsion equation that contains J.
  • #1
Su Solberg
75
0
dear all, i have just received a question about torsion.

Homework Statement


*************** a graph has been attached as attachment at bottom*************

The objective is to find the torsion shear stress at the cross section of the "dark Grey tube".

Let's have a look:
1.There is a reaction torque acting on the motor M,
2.The force is then transferred to dark grey tube by a light grey carrier.
3. blue cross is the center of the motor
4. purple cross is the centroid of the dark grey tube
5. i have J / Ixx,Iyy of that tube
6. Only the dark grey tube is fixed at 2 ends.
7. carrier is mush shorter than the tube


Homework Equations



I think I am finding the suitable eqts. So I miss this section.
Or
You think T/J= shear/R ??

The Attempt at a Solution


So,

#1 i would like to ask whether the following steps are correct
and
#2 What is the torsion equations for irregular hollow section.
#3 how to find the R, outer radius of the shaft, for the case that equation T/J= shear/R is used.


My Methodology:
1. Shear force is acting at the 2 hand sides of the tube as resultant.
2. the Shear force acts like a torque on the hollow section
3. use parallel axis theorem to transform "moment of inertia" from the original centroid to the torque center
4. use a torsion equation that contain J to fix it (<------------#2 that's what i would like to find now)

Thank you very much for your help.
 

Attachments

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  • #2
The torsion equation to use for an irregular hollow section in this case is: T = (J/2R) * (F1 - F2), where F1 and F2 are the shear forces on the two ends of the tube, J is the polar moment of inertia of the cross-section, and R is the outer radius of the shaft. To find R, you can measure the outer diameter of the shaft and divide it by two.
 
  • #3


Hello,

Thank you for your question about eccentric torsion and non-circular tubes. It seems like you are working on finding the torsion shear stress at the cross section of the dark grey tube, which is fixed at two ends and has a reaction torque acting on it from a motor.

To answer your questions:

#1 Your steps seem to be on the right track. To find the torsion shear stress, you will need to use a torsion equation that involves the moment of inertia (J) of the tube. You can use the parallel axis theorem to transform the moment of inertia from the centroid to the torque center.

#2 The torsion equation for an irregular hollow section will depend on the specific geometry of the tube. Generally, the torsion equation for a circular tube can be used for non-circular tubes, but it may not provide an accurate result. It is best to use a torsion equation that is specifically designed for the geometry of the tube you are working with. You may be able to find this equation in a reference book or by conducting a literature search.

#3 To find the outer radius (R) of the shaft, you will need to know the inner radius and the thickness of the tube. You can then use the equation T/J = shear/R to solve for R.

I hope this helps. Good luck with your calculations!
 

1. What is eccentric torsion in mechanics?

Eccentric torsion refers to the twisting or rotational force that occurs when an object is subjected to an off-center or eccentric load. This can cause a bending or twisting motion in the object, which can affect its overall stability and strength.

2. How does eccentric torsion affect non-circular tubes?

Eccentric torsion can have a significant impact on non-circular tubes, as the off-center load can cause the tube to twist and buckle. This can lead to structural instability and reduced strength, which can be a concern in engineering and construction applications.

3. What factors influence the amount of eccentric torsion experienced by a non-circular tube?

The amount of eccentric torsion experienced by a non-circular tube depends on several factors, including the magnitude and direction of the load, the shape and size of the tube, and the material properties of the tube. The location of the load on the tube also plays a significant role.

4. How can eccentric torsion be calculated for non-circular tubes?

There are several methods for calculating eccentric torsion in non-circular tubes, including analytical methods, such as the Navier's method, and numerical methods, such as finite element analysis. These methods take into account the various factors that influence eccentric torsion and provide accurate predictions of the behavior of the tube under off-center loads.

5. What are some practical applications of eccentric torsion in non-circular tubes?

Eccentric torsion in non-circular tubes is a critical consideration in many engineering and construction applications. It is commonly seen in structures such as bridges, cranes, and buildings, where the tubes are subjected to off-center loads. Understanding and accounting for eccentric torsion is essential for ensuring the safety and stability of these structures.

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