Shear Stress/Torsion

Question

--see Attachment

satis · Answer

Outer diameter d = 2.5 cm = 0.025m
  
Inner diameter di = 1.95 cm = 0.0195m
  
Polar second moment of area J = π/2* (d4 – di4) / 32
  
 = 3.142/2 *(0.0254 – 0.01954)/32 m4   
=  12.08 x 10-9 m4
  
Maximum shear stress  τmax =  Tr/J   
= [40KN-m x (0.025/2) m] / (12.08 x 10-9 m4) MPa
  
= 41.39 x 103 MPa

KennTutor · Answer

Check the attached file below for the solution to your summer assignment.

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Shear Stress/Torsion

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