torsion of circular shafts
A car's driveshaft, a screw turned by a driver, a generator's rotor. These are not pulled. They are twisted. What happens inside a shaft when you twist it? The tension story from before repeats almost as it stands, with one decisive difference. In torsion the stress is not the same everywhere. The centerline hardly moves and the outer surface twists the hardest, so the shear stress grows in direct proportion to the radius. τ = TρJ says that. And the angle of twist is φ = TLGJ, a twin of B1's δ = PLAE with only the letters changed. That one dependence on radius even explains why a heavy shaft may be hollowed out. The radius is why the inside can be empty.
First, see what torsion is. Fix one end and give a torque T at the other, and the cross-sections rotate relative to each other. The fixed end stays still and the free end turns the most, and the difference of rotation from end to end is the angle of twist. A straight line drawn along the surface leans into a helix as the twist grows. What matters is that the section itself stays flat, and the round shape stays as it was. A circular section only turns in place. It does not bend or wrinkle. Because of this clean property, torsion settles into a very simple formula. A circular section only turns, and that is why the formula stays simple.
Now the central insight. That the section turns as a whole means a point farther from the center moves more. Even for the same angle, a point at a larger radius travels a longer arc. So the shear strain is directly proportional to the radius. At the center the strain is 0, and at the surface it is largest. Raise the twist and the centerline stays lying flat, while the outer vertical lines lean more steeply. That is the whole of torsion. The deformation is spread linearly along the radius. In tension the whole section lengthened alike, but torsion differs by position, and that difference sets the stress distribution that follows. Position, not a uniform stretch, sets what follows.
If the shear strain is proportional to the radius, then by Hooke's law τ = G·γ the shear stress is proportional to the radius too. 0 at the center, largest at the surface. Written as a formula, τ = TρJ. T is the twisting torque, there is the radius of the place you are looking at, and J is a geometric value called the polar moment of inertia of the section. J takes the place of the area A of tension, but there is a difference. A is just area, while J integrates area times the square of the radius, so material farther out contributes much more. Move T and the stress distribution stays low in the middle and high at the surface, growing linearly like a triangle. A torsion failure therefore always starts at the surface.
Now how far the shaft turns as a whole, the angle of twist. φ = TLGJ. Set it beside B1's δ = PLAE and the correspondence shows. Torque T takes the place of the force P, the length L stays, the polar moment J takes the place of the area A, and the shear modulus G takes the place of Young's modulus E. A perfect one-to-one match. A larger torque and a longer shaft twist more, and a larger GJ twists less. So GJ is called the torsional stiffness. Move T and the free end turns through that angle. Just as a bar in tension was a spring, a shaft in torsion is a torsion spring that answers a torque with an angle.
Last, the conclusion most useful in practice. Both stress and deformation were nearly 0 near the center. The material at the center hardly resists the twist and only adds weight. What if the inside is emptied? Move the inner radius and hollow it out. The weight falls quickly with the square of the radius, but the stiffness J falls slowly because it goes with the fourth power of the radius. So you can hollow the middle quite a lot and the torsional stiffness stays almost as it was, while the efficiency per weight jumps. Bicycle frames, driveshafts, and golf clubs are hollow tubes for this reason. Stiffer for the same weight, or lighter for the same stiffness. A gift of torsion's dependence on radius.