Finding diameter of torsion shaft

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Discussion Overview

The discussion revolves around calculating the required diameter of a solid brass torsion shaft based on given parameters such as allowable shear stress, shear modulus, length, angle of twist, and applied torque. It involves theoretical and mathematical reasoning related to mechanics of materials.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants discuss the relationship between shear strain, shear stress, and the angle of twist in the context of the torsion shaft.
  • One participant questions the variation of shear strain with axial and radial locations along the shaft.
  • Another participant emphasizes that the original poster must solve the problem independently, as the forum is not a homework service.
  • The original poster mentions arriving at a diameter of 7mm but believes it to be incorrect and seeks assistance in identifying errors.
  • A participant points out a potential algebraic error in the original poster's calculations, specifically in the equation for the polar moment of inertia (J).

Areas of Agreement / Disagreement

There is no consensus on the correctness of the original poster's calculations, as they believe their answer is incorrect while others have not confirmed or disputed this. The discussion remains unresolved regarding the correct diameter.

Contextual Notes

Participants have noted the importance of understanding the equations involved and have highlighted potential errors in the original poster's algebra. There may be missing assumptions or steps in the calculations that have not been fully explored.

Mechanical94
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Homework Statement



The solid shaft is made from brass that has an allowable shear stress
= 100 MPa and a shear modulus of elasticity G = 39 GPa. The length of the shaft is L = 2 m, and over the length of the shaft the allowable angle of twist is θ = 0.10 rad.the shaft is subjected to a torque of T = 25 kNm, what is the required diameter of the shaft

Homework Equations



T/J=t/r=Gθ/L
 
Last edited:
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Mechanical94 said:

Homework Statement



The solid shaft is made from brass that has an allowable shear stress
= 100 MPa and a shear modulus of elasticity G = 39 GPa. The length of the shaft is L = 2 m, and over the length of the shaft the allowable angle of twist is θ = 0.10 rad.the shaft is subjected to a torque of T = 25 kNm, what is the required diameter of the shaft

Homework Equations



T/J=t/r=Gθ/L

Hi Mechanical94. Welcome to Physics Forums.

In this situation, what is the equation for the local shear strain as a function of how θ is varying with distance along the shaft, and as a function of radial distance r from the axis of the shaft? Does the shear strain vary with axial location? Does the shear strain vary with radial location? What is the equation for the local shear stress?

Chet
 
this is all the information given to me i have attached a picture to this post hopefully it can answer your questions
 

Attachments

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    question torsion bar.jpg
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Those are questions which you, Mechanical94, have to answer. After all, it is your problem to solve. In case you missed it when you read the Terms & Conditions, PF is not a homework service.
 
As SteamKing said, you already have enough information to answer all my questions. I already know the answers to these questions. Now, it's up to you.

Chet
 
i understand about this not being a homework service however this is my workings out and i got a diameter of 7mm, apparently my answer is not correct can anyone spot my error or can anyone help me to get the correct answer??
 

Attachments

  • picture022.jpg
    picture022.jpg
    38.7 KB · Views: 679
  • picture023.jpg
    picture023.jpg
    27.8 KB · Views: 707
Last edited:
Mechanical94 said:
i understand about this not being a homework service however this is my workings out and i got a diameter of 7mm, apparently my answer is not correct can anyone spot my error or can anyone help me to get the correct answer??

In the right hand side of the equation in picture 1, the algebra is incorrect. It should be:

J=\frac{TL}{Gθ}

Chet
 

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