mechanics of materials 2 ,GAtech,Coursera

In the case of pure torsion, how does a differential area on cross-section of the cylinder with dx length undergo a shear force that is perpendicular to the cross-section ? I can understand that a shear force parallel to cross-section generates because if we divide the cylinder into many such disks of dx length then each disk rotates relative to each other, thus giving rise to such shear stress. But why the other one ?

If I haven't been able to make myself clear here's another way. On the 2D stress block, why is there a horizontal shear stress?


1 Answer 1


The two shear force components perpendicular to the cross section arise to eliminate rotational acceleration. Applying only a single couple would cause regions to begin spinning. This is true for any shear stress state on a nonaccelerating element: Fully four forces are necessary to maintain both translational and rotational equilibrium:

  • $\begingroup$ How do I add animations of this kind ? Do I create it elsewhere and then add it as a gif or is this an added functionality of the forum ? $\endgroup$
    – raconteur
    Mar 17 at 12:09
  • $\begingroup$ You can copy and paste them right into your question/answer, and the site will usually take care of interpreting them. This generally means they’ll be stored in imgur. $\endgroup$ Mar 17 at 15:04
  • $\begingroup$ These are from my site; I wrote them in Python. The code for this particular animation is here. $\endgroup$ Mar 19 at 23:27
  • $\begingroup$ Thanks @Chemomechanics. $\endgroup$
    – raconteur

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