Designing a Shaft for Specific Torsional Loads

  • Thread starter Thread starter firebird90
  • Start date Start date
  • Tags Tags
    Design Shaft
Click For Summary

Discussion Overview

The discussion centers on the design of a shaft to withstand specific torsional loads, focusing on determining the appropriate diameter based on material properties and safety factors. Participants explore theoretical approaches, material yield stress, and shear stress calculations relevant to shaft design.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks to determine the diameter of a shaft under a specific torsional load, expressing confusion about relating material properties to maximum shear stress.
  • Another participant states that maximum shear stress can be derived from yield stress, referencing external material.
  • A participant proposes using the maximum distortion energy theorem to derive yield shear stress and calculate the diameter while ensuring no plastic deformation occurs.
  • Another participant questions the need for a proof of shear stress, suggesting that it can be referenced from material property tables, specifically noting that for steel, the allowable shear stress is approximately 0.3 times the tensile yield stress with a safety factor of 2.
  • A participant expresses concern about needing a proof for their project, seeking a source to substantiate the relationship of shear stress to yield stress.
  • One participant mentions that the Steel Code provides a method for calculating ultimate shear stress as the tensile yield stress divided by the square root of 3, which aligns with the earlier claim of 0.6 times the yield stress.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of deriving shear stress through proof versus referencing established values from material property tables. The discussion reflects uncertainty regarding the application of the distortion energy theorem and its relevance to the calculations.

Contextual Notes

Participants note limitations in their understanding of the relationship between material properties and shear stress, as well as the lack of specified length for the shaft affecting the angle of twist calculations.

Who May Find This Useful

Individuals involved in mechanical engineering design, particularly those focused on shaft design and torsional load analysis, may find this discussion relevant.

firebird90
Messages
3
Reaction score
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 \taumax=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 \taumax 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.
 
Last edited:
Physics news on Phys.org
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 \taumax 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?
 
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.
 
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.
 
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.
 

Similar threads

Relative angle of twist between ends of the shaft
  • · Replies 1 ·
Replies
1
Views
2K
Mechanics of Materials — Torsional Stress on a Spinning Shaft
  • · Replies 5 ·
Replies
5
Views
4K
Engineering Loads on a shaft with V-Belts
  • · Replies 2 ·
Replies
2
Views
3K
Engineering Give examples of mechanical power transmission failure and solutions for a car
  • · Replies 1 ·
Replies
1
Views
3K
Torsion stress and twist criterion
  • · Replies 1 ·
Replies
1
Views
2K
Find Shear Stress in Shaft Design | Modulus of Rigidity 80kN/mm2
  • · Replies 1 ·
Replies
1
Views
2K
Finding diameter of torsion shaft
  • · Replies 6 ·
Replies
6
Views
3K
Solid Hollow Shaft Torsion Resistance: How Will It Change?
  • · Replies 10 ·
Replies
10
Views
3K
Designing Shafts to Withstand Shear Stress
  • · Replies 8 ·
Replies
8
Views
3K
Finding the shear stress and angle of twist in hollow shaft
  • · Replies 3 ·
Replies
3
Views
3K