Find Max Torque Using Smaller Rad of Shaft

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SUMMARY

The discussion centers on determining the maximum torque in a shaft by analyzing the relationship between shear stress and the radius of the shaft. The correct approach involves using the radius of the smaller part of the shaft, as the torque is applied at that section, despite the larger radius being more intuitive. The formula for shear stress, given as shear stress = (max torque)(maximum radius)/J, where J = π/2c^4, highlights the importance of the polar moment of inertia in torque calculations. Ultimately, the smaller radius must transmit the same torque as the larger radius section.

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  • Understanding of shear stress and torque relationships
  • Familiarity with polar moment of inertia (J) calculations
  • Knowledge of basic mechanics of materials
  • Ability to interpret mechanical diagrams and stress distributions
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Mechanical engineers, students studying mechanics of materials, and professionals involved in shaft design and analysis will benefit from this discussion.

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Homework Statement


The question wants me to find the maximum torque.
I do not understand why the solution uses the radius of the smaller part of the shaft rather than the radius of the thicker part of the shaft. Please see the attached image.

Homework Equations



shear stress = (max torque)(maximum radius)/J where J = polar moment of inertia.

J = π/2c^4 where c = max radius of the rod

The Attempt at a Solution



I want to use 37.5mm as the radius (radius of larger part of the shaft), but it is not correct. The formula above directly relates torque T and c (radius), which is why I would assume it would be correct to use the largest c value possible. I would appreciate any insight. Thanks.
 

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I do not understand why the solution uses the radius of the smaller part of the shaft rather than the radius of the thicker part of the shaft.
Probably because the diagram shows the torque being applied to the thin part of the shaft.

It's shear stress you are calculating - the bigger part of the shaft may get a torsional (corkscrew) and a centrifugal stress - but to get a shear stress, something has to be pulling on the surface.
I suspect you are just taking your equation too literally - go back to the definition of a shear force and see how that applies to a torque. i.e. what if you applied the torque to the inner surface of a hollow shaft?
 
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Simon Bridge said:
Probably because the diagram shows the torque being applied to the thin part of the shaft.

Should have seen that, haha. Thank you.
 
The smaller radius shaft must transmit the same torque as the larger radius part.
 

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