# Why does the torsion in a circular bar result in shear stress along the axial direction?

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?