Solve Simple Torsion Problem: Calculate Maximum Torque on 25mm Shaft

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Discussion Overview

The discussion revolves around calculating the maximum torque that can be applied to a hollow shaft with specified dimensions and shear stress limits. The focus is on the application of torsion theory and shear stress calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a formula for calculating torque based on shear stress and polar moment of inertia, but reports an incorrect result.
  • Another participant suggests that the problem may be approached as a shear flow problem, providing an alternative torque calculation that yields a different value.
  • A participant points out a resource that may contain the correct answer but does not provide a detailed derivation.
  • There is a discussion about the shear distribution assumed in the calculations, with one participant stating it is assumed to be linear.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method for calculating the maximum torque, as different approaches and assumptions are presented, leading to varying results.

Contextual Notes

The discussion includes unresolved mathematical steps and assumptions about shear distribution, which may affect the accuracy of the torque calculations.

Iclaudius
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Hiya friends,

so i have a little problem,

A shaft is made from tube 25 mm outer diameter and 20 mm inner diameter. The shear stress must not exceed 150 Mpa. Calculate the maximum torque that should be placed on it

ok so we have T/J = tau/r
so T = (tau/r)*J - this should give me my answer right I'm using tau = 150Mpa, r= 20mm, and J = (pi/32)(25^(4)-20^(4)), this however does not give me the answer which is 271.69 Nm

ps i have tried using both 25 and 20 respectively for the r value without any sucess, any help advice is appreciated, thanks in advance

Claudius
 
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Nevermind people, a foolish oversight on my part :redface:
 
Is this not a shear flow problem

T=2aSflow

T = \frac{\pi }{4}*{\left( {22.5} \right)^2}*5*150*{10^{ - 3}}*{10^{ - 3}}*{10^{ - 3}}*{10^6} = 298.2
 
Sorry the I can't open the link.

However you have to admit that if your link provides a more accurate solution the shear flow approximation gets pretty close, pretty easily.

What shear distribution are you assuming?
 
The link unfortunately just gives a straight answer without working, to answer your question however the shear distribution is assumed to be linear
 

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