Finding diameter of torsion shaft

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SUMMARY

The discussion focuses on calculating the required diameter of a brass torsion shaft subjected to a torque of 25 kNm, with an allowable shear stress of 100 MPa and a shear modulus of elasticity of 39 GPa. The length of the shaft is 2 m, and the allowable angle of twist is 0.10 rad. The correct formula to use is J = TL / Gθ, where J is the polar moment of inertia. The initial calculation of a 7mm diameter was incorrect due to algebraic errors in applying the torsion formula.

PREREQUISITES
  • Understanding of torsion in solid shafts
  • Familiarity with shear stress and shear modulus concepts
  • Knowledge of polar moment of inertia (J) calculation
  • Basic algebra for manipulating equations
NEXT STEPS
  • Review the derivation of the polar moment of inertia for circular shafts
  • Study the relationship between torque, shear stress, and angle of twist in torsion
  • Learn about the properties of materials, specifically brass, in mechanical applications
  • Practice solving torsion problems using different values for torque and material properties
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Mechanical engineers, students studying mechanics of materials, and professionals involved in shaft design and analysis will benefit from this discussion.

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
 

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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

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    picture022.jpg
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  • picture023.jpg
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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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