Designing a Shaft for Specific Torsional Loads

In summary, the conversation is about designing a shaft to resist a specific torsional load. The speaker has a safety factor of 2 and a table of material properties, but is struggling to find the diameter of the shaft needed. They mention using the formula \taumax=Tc/J and the angle of twist formula, but it is not useful without a specified length. The speaker also discusses the maximum shear stress and its relation to yield stress. They mention a proof for shear stress and obtaining a yield shear stress from the distortion energy theorem. The other person in the conversation suggests looking up the maximum shear stress in a table of material properties and mentions that for steel, it is about 0.6 Fy. They also mention using
  • #1
firebird90
3
0
I need to design a shaft that have a specific torsional load acting on it. I have the safety factor of 2 and have a table of material properties (yield stress, ultimate stress, shear modulus, modulus of elasticity etc.).

I need to find the diameter of the shaft that resist the specifies torque. I have the formulas of [itex]\tau[/itex]max=Tc/J and angle of twist formula. As the angle of twist formula is related to the length its useless because I don't have any specified. I need to get a [itex]\tau[/itex]max by using material properties but I couldn't find anything to relate them up to now. I have searched many shaft design documents but no results and also I am confused right now.
 
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  • #3
Now I can make a proof for shear stress with max distortion energy theorem. So I am going to get a yield shear stress from the theorem and using it with factor of safety I will get my desired [itex]\tau[/itex]max to use it in torsion formula, with using the yield stress to find diameter, I will get a diameter free of plastic deformation. Am I right with that?
 
  • #4
I don't know why you need a proof for shear stress when it's max value can be looked up in a table of material properties. For steel, it's about 0.6 Fy, where Fy is the tensile yield stress. So with a SF of 2, allowable shear stress would be about 0.3Fy and you needn't worry about plastic deformation.
 
  • #5
The proof is my biggest problem with my work, Because it must be a project style showing how can I derived this 0.6Fy. I need a source to show it.
 
  • #6
The Steel Code I use calculates the ultimate shear stress as the tensile yield stress divided by square root of 3, which rounds to 0.6 Fy. I do not know if that value comes from the distortion energy theorem to which you refer.
 

1. What is a torsional load?

A torsional load is a type of force that acts on a shaft, causing it to twist or rotate. This can be caused by various factors such as motor torque, weight distribution, or external forces.

2. How do you determine the specific torsional load for a shaft?

The specific torsional load for a shaft can be determined by calculating the torque, or twisting force, that will be applied to the shaft. This can be done by considering the type of load, its magnitude, and the distance from the shaft's axis of rotation.

3. What materials are commonly used for designing shafts for specific torsional loads?

Some common materials used for shafts designed to withstand torsional loads include steel, aluminum, and titanium. The material chosen will depend on factors such as the required strength, weight, and cost.

4. What factors should be considered when designing a shaft for specific torsional loads?

When designing a shaft for specific torsional loads, factors such as the type and magnitude of the load, the material properties, the shaft diameter, and the length of the shaft should all be taken into consideration. It is also important to consider any potential stress concentrations or points of failure.

5. How can I ensure the shaft is strong enough to withstand the torsional load?

To ensure that the shaft is strong enough to withstand the torsional load, it is important to perform thorough calculations and simulations using appropriate engineering principles. It may also be helpful to consult with experts or conduct physical testing to validate the design.

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