Solid Mechanics: What is difference between torsional and direct shear stress?

I am not an engineer so I really am not qualified to give an answer, but  I would say that it really depends on the materials being used. A soft  substance may twist multiple times before it breaks and still have some  tensile strength, but have very poor direct sheer or load bearing strength. A hard substance may have superior load bearing and tensile properties, but have poor torque strength.

[The answer by provides the description as to how the "two different types" of shear stresses originate, so I'll be not going into that] I mean if given a value of simple shear stress, how to determine value of torsional shear stress? You can't. What you should be worrying about is the principal stresses, because that will determine your design. What is the physical difference between these two types of stresses? Physically, nothing. Except that they are both shear stresses, and hence can be 'rotated' to extract out the principal stresses which can then be used in failure theories to predict failure. And since shear stresses are frame dependent, it doesn't make sense to squabble about their relative magnitudes/nature of origin as they will ultimately result in the same end result when applied to failure theories. @ From the comment thread, I gather that the OP is trying to reconcile the idea of this torsional shear stress with the failure theories. Well, when you get the (max)  torsional shear stress, use that to calculate the principal stresses because failure theories deal with principal stresses. For a shaft under pure torsion, the stress tensor is [math]\large \left[\begin{array}{cc} 0 & \tau \\ \tau & 0\end{array}\right][/math] where [math]\tau = \tau(r) =Tr/J [/math] which would give the eigenvalues (hence the principal stresses)  as [math]\large \sigma_{1,2}=\pm \tau_{max}[/math]. Now that you have obtained the principal stresses, you can go about applying it to failure theories.