Dealing with Hollow and Solid shafts....

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SUMMARY

When analyzing a bar fixed at one end with a hollow section transitioning to a solid section, the condition of static equilibrium can indeed be applied to determine the reaction torque. This method parallels the analysis of a bar with two different diameters, utilizing pure torsion formulas to assess shear stress and angle of twist. By segmenting the bar and employing free-body diagrams, one can accurately calculate the torque on the uniform part of the composite beam and derive the shear stress in the thin wall.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with torsion formulas in mechanics
  • Knowledge of shear stress calculations
  • Ability to create and interpret free-body diagrams
NEXT STEPS
  • Study the application of torsion formulas in composite beams
  • Learn about calculating shear stress in hollow and solid shafts
  • Explore advanced free-body diagram techniques for complex structures
  • Investigate the effects of material properties on torque and shear stress
USEFUL FOR

Mechanical engineers, structural analysts, and students studying mechanics of materials will benefit from this discussion, particularly those focused on torsion and shear stress in composite shafts.

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When dealing with a bar (fixed on one end), if it is hollow for a certain length from the edge and then turns solid up until the fixed point, can the condition of static equilibrium be used to find the reaction torque, just like you would of a bar with 2 different diameters.
(Assuming all this is of the same material and a torque is provided only on the non fixed end right at the edge)

To me it seems like the same method can be used for this, dividing the bar into segments to use pure torsion formulas and find the variation of shear stress and angle of twist.
 
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Sure, you can split up any object and make free-body diagrams of the components. From the free-body diagrams, you can find (for example) the torque on a uniform part of the composite beam and thus the shear stress in the thin wall.
 

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