OK, let me try with some other assumptions.
Wiki gives 2 reactors at 140,000 shaft horsepower each, and 4 shafts. So let’s say 70,000 shaft horsepower per shaft. Now, let’s also take the assumption of Xema and say that the screws turn at 200 rpm (low-end of his range, meaning more torque on the shaft). And let’s take what Bad Samaritan and Rhubarb said, roughly, and assume a shaft of 36 inches in diameter, 3 inch wall thickness.
Now IIRC, to find the twist on a hollow shaft, we look at the outside and inside diameters.
K = ((pi)*(do^4 - di^4))/32
If do is 3 feet, and di is 2.5 feet (assuming 3 inch wall thickness), then
K = ((3.14159) * (3ft^4 - 2.5ft^4)) /32
K = 4.117 ft^4
And torque will be = (70,000 hp 5252)/200 rpm = 1,838,200 ftlbf
(note - this is much less torque than earlier assumed!)
So we have theta = (1,838,200 ft*lbf * 150ft) / (4.117 ft^4 * 1586.9 x 10^6 lb/ft^2)
(note I converted G to a foot-basis)
theta = (2.757x10^8 ft^2*lbf) / (6.534x10^9 ft^2 * lbf)
the units should cancel out, and I get
theta = 0.0422 rad (2.40 degrees of turn)
Am I the one who is doing something wrong here?