I think you are confusing a few things here. Slow day at work so let's try to break this down.
When saying that the torque transmitted through the torsion bar is unchanged, you seem to be referring to this diagram in your link:
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While I agree with the math throughout this example (of course I do, because I can't just disagree with physics), the
KEY CONSIDERATION here is that your tangential distance, the distance between your rotating axis (nut) and your force's line of action is always the same, in this case 120mm. All this example is really demonstrating is that your Fx and Fy components will always add to the same resultant moment
as long as your force continues moving through the same straight line.
In our trucks, the torsion bars acting through the control arm are more accurately represented by example 4, further down the page:
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This is just an upside-down version of control arms as they go through travel. In this case, instead of your tangential distance being a constant, it is now a function of "d" (or your control arm length) and it's angle as it rotates (Θ) around your pivot point (control arm mounting bolts), where your new tangential distance is calculated by the equation d*cos(Θ).
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Here is a quick sketch to show what I mean. The solid lines are our 18" control arm shown through 30 degree increments, and the dashed rectangles are our 33x10 tire going up and down through an arc. With the tire drooped down to the red position, just about to lift off the ground, lets say there are 200lbs on it. This is acting at an
effective radius of 9", so 200lb * 9" = 1800lb*in being resisted by the torsion bar.
In our blue position, we have put some weight back on the wheel, we are just cruising down the road with 1270lbs on that tire. That 1270lb * 15.59" = 19,800lb*in on the torsion bar.
In our green position, we are now trying to drive our lead tire up onto something. A curb, large rock, downed tree, ex-wife, whatever. With that extra compression on that corner, we now have 2000lb * 18" = 36000lb*in
While I am taking some liberties here with guesstimations and numbers to make the math easy, you can look at my resultant torques and compare them using a constant torsional spring rate of 600lb*in / degree. This torsional rate would then imply/require that we add force as we travel through our range of motion, contrary to your point.
Another scenario: First off, remove all sway bars. Now, let's say we are turning right. As we turn right, weight is transferred to the left side, and unloaded from the right. If the torsion bars were putting out a constant torque through their range of motion, the truck would just flop over to the left side because if your springs are always in equilibrium, they won't actually do anything to counteract your uneven roll force. In reality, when force is transferred to the left side, your left torsion bar compresses more, increasing force all while your right torsion bar unloads, decreasing force. These uneven forces push the body back to the right, allowing things to balance back out. The idea that "the “twist” on the torsion bar remains unchanged" would not allow for these uneven forces, and thus the suspension would be completely unable to keep the vehicle upright, minus the presence of sway bars (and the rear suspension being different of course).
Ok, 'tism rant over. Need to put all of those physics classes to work somehow. Hopefully that all made sense.