I'll throw out some of the equations:
T(orque) = JG*theta/l
where
theta = angle of twist for a given amount of body roll
J = pi/2*r^4 (solid)
J= p/2*(r_o^4 - r_i^4) (hollow)
G is modulus of elasticity (dependent on the kind of steel used)
l is the torsional length (twist length)
So you have a resistive torque and for a given vertical displacement of the suspension from 'rest'/equilibrium, if you know the moment arm (lets just assume the moment arm is perpendicular to the torsional arm) you can convert that displacement into an angle pretty easily with good ole' tan.
Also, you can convert the torque to an effective force at the endlink F=T/l_moment_arm.
Using that you have k = F/disp ie spring rate at the link.
And you mentioned the moment arm squared and that's true but 1/l^2 as follows:
F=T/l_moment_arm
displacement of bar end = l_moment_arm*tan(theta)
k=F/displacemnt_bar = T/(l_moment_arm^2*tan(theta))
Anyway, the important equation is:
k_bar_end*MR^2 = wheel rate
This is the equation important to the STI (getting back on topic

) since it has front strut suspension (say .99 MR) and rear double a arm (say .6 MR). So to get good balance of wheel rates, the rear needs more spring rate to get there. That is the nice thing about the suspension updates on Aarons STI SE since they went with a bigger RSB AND stiffer rear spring (compared to the front).