Twist a Shaft and the Shear Grows with Radius
A car's drive shaft, a screw you tighten with a driver, a generator's rotor shaft. These are not pulled; they are twisted. What happens inside a shaft when you twist it? The tension story you saw earlier repeats almost verbatim, with one decisive difference: in torsion the stress varies with position. The central axis barely moves, while the outer surface twists the hardest. The shear stress grows in direct proportion to the radius. τ = TρJ says exactly that. And the angle of twist is φ = TLGJ, a twin of B1's δ = PLAE with only the letters swapped. This one fact, the dependence on radius, even explains why you can hollow out a heavy shaft.
First, see what twisting is. Fix one end and apply a torque T at the other to twist it. The cross-sections then rotate relative to one another. The fixed end stays put while the free end turns the most. The difference in rotation between the ends is the angle of twist φ. Watch the straight line drawn along the surface: the more you twist, the more it slants into a spiral. The key is that the cross-section itself stays flat and keeps its round shape. A circular section simply rotates in place, without warping or distorting. This clean behavior is what lets torsion collapse into a very simple formula.
Now the key insight. A whole cross-section rotating means that the farther from the center, the more a point moves. For the same angle of rotation, a point at a larger radius traces a longer arc. So the shear strain γ is in direct proportion to the radius ρ: zero on the central axis, maximum at the surface. In the widget, increase the twist. The centerline stays lying flat, while lines farther out tilt more steeply. This is all that torsion is: the strain distributes linearly with radius. In tension the whole section stretched equally, but torsion varies with position. It is exactly this difference that sets the stress distribution next.
If the shear strain is proportional to radius, then by Hooke's law τ = G·γ the shear stress is proportional to radius too: zero at the center, maximum at the surface. Written as a formula, τ = TρJ. Here T is the applied torque, ρ is the radius of the point you care about, and J is a geometric quantity called the polar moment of inertia. J takes the place of the area A from tension, with one difference. A is just area, but J is area weighted by radius squared and integrated, so the outer material contributes far more. Drag T. The stress distribution glows blue in the middle and red at the surface, growing linearly like a triangle. That is why torsional failure always begins at the surface.
Now for how much the whole shaft rotates, the angle of twist φ. It is φ = TLGJ. Set it beside B1's δ = PLAE. The force P is replaced by the torque T, the length L stays, the area A becomes the polar moment of inertia J, and Young's modulus E becomes the shear modulus G. A perfect one-to-one match. The larger the torque and the longer the shaft, the more it twists; the larger GJ, the less. So GJ is called the torsional stiffness. Drag T and the free end turns by φ. Just as a bar in tension was a spring, a twisting shaft is a torsional spring, returning an angle for the torque you apply.
Finally, the most useful conclusion in practice. Both stress and strain were nearly zero near the center. The central material barely resists torsion while only adding weight. So what if you hollow it out? Drag the inner radius to carve out the core. Weight drops fast, as the radius squared, but the stiffness J drops slowly, as the radius to the fourth. So you can hollow out quite a lot of the center and still keep nearly all the torsional stiffness. The efficiency per unit weight shoots up. That is why bicycle frames, drive shafts, and golf clubs are all hollow tubes: as stiff for the same weight, or the same stiffness for less weight. It is a gift from torsion's dependence on radius.