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

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SUMMARY

The discussion focuses on calculating the maximum torque for a hollow shaft with a 25 mm outer diameter and a 20 mm inner diameter, where the shear stress must not exceed 150 MPa. The formula used is T/J = τ/r, leading to the calculation of T = (τ/r) * J, where J is determined using the formula J = (π/32)(D^4 - d^4). The correct maximum torque calculated is 271.69 Nm, with an alternative approach using shear flow yielding a torque of 298.2 Nm. The conversation highlights the importance of shear distribution assumptions in torque calculations.

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  • Understanding of shear stress and torque calculations
  • Familiarity with the polar moment of inertia (J) for hollow shafts
  • Knowledge of shear flow concepts in mechanical engineering
  • Basic proficiency in using formulas for mechanical properties of materials
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  • Study the derivation and application of the polar moment of inertia for hollow shafts
  • Learn about shear flow calculations and their relevance in torsion problems
  • Research the effects of shear distribution assumptions on torque calculations
  • Explore advanced topics in torsion and shear stress in mechanical design
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Mechanical engineers, students studying materials science, and professionals involved in the design and analysis of rotating shafts will benefit from this discussion.

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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