Direction of Shear Stress on the periphery of a circular section beam

Question

The Question is from shear stresses in beams

Consider a beam made of circular cross section, in which at any cross section the shear force is V.

The textbook I'm following states that -

When a beam has a circular cross section, we can no longer assume that the shear stresses act parallel to the y axis. For instance, we can easily prove that at point m (on the boundary of the cross section) the shear stress $\tau$ must act tangent to the boundary. This observation follows from the fact that the outer surface of the beam is free of stress, and therefore the shear stress acting on the cross section can have no component in the radial direction.

I'm having trouble with understanding why the shear stress at m, should be tangent to the boundary.

Say I enlarge the element,

and assume that the shear stress acts along y axis. this shear stress can be resolved into tangential and radial components to get $\tau_t$ and $\tau_r$. The book says this $\tau_r$ can't be balanced by any other stress in the radial direction and hence it must be zero. But it could happen that a normal stress is developed on the surface as shown in (b), which balances $\tau_r$

Answer 1

Let's take a small element of the circular shaft and apply the shear stress $\tau_1$ in an arbitrary direction on the surface of the cut. The shear stress can be resolved into a radial shear stress $\tau_r$ and tangential shear stress $\tau_t$ as shown below.

We recall the state of shear stresses on a cube element - when shear stress presents on one plane, there must exist a shear stress on the face normal to that plane. In this case, there must be a shear stress $\tau_z$ corresponding to $\tau_r$, however, we know that the stress on the outermost surface of an element is zero, from this realization we can conclude that $\tau_r$ must be zero as well, so only $\tau_t$ remains effective.

Note that this phenomenon is valid for shear stress on the perimeter only.