Torsional stiffness is calculated by taking the diameter to the fourth power. So diameter is VERY important. A hollow bar which has a slightly larger diameter can be stiffer than a solid bar that has a smaller diameter. Taking something to the fourth power really magnifies the differences. A hollow 7/16 pushrod is a lot stiffer than a solid 3/8 pushrod. I use hollow anti-sway bars in all of my cars. I think I made the first hollow anti-sway bar for a Mopar but I could be wrong. I had one fabricated for my Coronet by a chassis shop back in the early 80's. Nobody offered them for sale until fairly recently and that might have been because I kept bugging people to make them.

I went to the hollow Hotchkis. Can’t feel a difference, hits every parking bumper in the Western United States, and makes it difficult to remove the oil filter. The Zerks got knocked off in the first parking lot. What an improvement. Why don’t you guys work on this a bit more, maybe a few more pages of this thread will figure it all out.

I am pretty sure it's going to fix a number of problems I've had. Upper control arm adjustability is a DEFINITE benefit. Goes along with the pro touring/restomod track I'm taking the car in too.

Well, I'm sure that given the choice, the lighter part would be preferred if the cost was close. I'm going to need a new sway bar for my project car "Jigsaw". I was thinking that whatever I buy, I'd install in the other car and let Jigsaw get the hand me down stuff.

Google ‘polar moment of inertia’ - there are some images that show different cross sections and how it’s calculated - then ‘torsional stiffness’ to see how it’s applied. Polar moment of inertia is the variable that no ones really mentioned. If you still have questions, lemme know.

For my F body's and 4th gen mustang's, the hollow bars were more expensive by almost $100. That was a while ago, though...

https://www.engineering.com/calculators/beams.htm look at the deflection of a hollow tube compared to a solid one at 1000 pounds of force and it might help explain it better than I can.

Play with the numbers, a 1-3/8 hollow bar at 1' vs a 1-1/8" solid at the same distance with 1000 force and 2000 etc and you will see how much a bar will flex

Perhaps you meant to say: where I p = π D 4/32 is the polar moment of inertia of a circular cross section. Thus, the torque required for unit twist, i.e., T (θ) is called the torsional stiffness. The quantity is known as torsional rigidity and is represented ... Torsion equation or torsion constant is defined as the geometrical property of a bar's cross-section that is involved in the axis of the bar that has a relationship between the angle of twist and applied torque whose SI unit is m4. The torsion equation is given as follows: \frac{T}{J}=\frac{\tau}{r}=\frac{G\Theta}{L} And to furthur complicate things.... Design Methods to Improve Torsional Rigidity Products that are subject to a torsional load often require analysis similar to what we use for bending stiffness. The main differences are the specific material property to be used (Shear Modulus of Rigidity) and the polar moment of inertia (which is very similar to the area moment of inertia used in bending). If we define torsion as the act of twisting along a structural component from an applied load, one example of this is the twist induced in a drive shaft when throttle is applied in a race car. This article will walk product designers and engineers through the methods used to analyze and improve torsional rigidity in design. Problem Statement and Target Objectives The key problem we’ll solve in this article is that of enhancing the torsional rigidity of our design. The definition of rigid, in this case, is the resistance to twisting as a torque is applied to the component. Going back to our drive shaft example, if we allow too much twist, we may end up with unstable performance. The images below show two shafts that are being torqued by 100 ft-lbs, but they have different torsional rigidities. There is a small “flag” modeled on the outside of the shaft to give a visual of the feature twisting as the torque is applied. The design on the bottom (Steel) is much more rigid than the design on top (HDPE). The next section will walk through the properties that make this possible. Key Properties and Formulae If we want to improve the rigidity of a design, we have a few material properties with which we can play. The list below details the key properties: Shear Modulus, G (modulus of rigidity): Polar Moment of InertiaFurthermore, we can relate the shear modulus to the modulus of elasticity (E) for isotropic materials using the following formula: Where: E = Modulus of Elasticity v = Poisson’s RatioTo give a quick example, if we take the modulus of elasticity of steel at approximately 30e6 psi at room temperature, we can convert it to shear modulus with the following math: Now that we understand the material property of interest, let’s look at the formulas that actually allow us to determine how rigid a component is. We will start with the overall formula for angle of twist (θ) in radians: Where: T = Torque applied to the component L = Length of the Beam J = Polar Moment of Inertia (Torsional Constant) G = Shear Modulus (Modulus of Rigidity)The image below illustrates what exactly we mean when we say angle of twist: The polar moment of inertia, J, is the same thing as the area moment of inertia about the long axis. That means the formula for determining J will depend on the shape of your component. If we’re dealing with a solid shaft, the formula is: It’s important to remember to use the correct value of J for all problems you solve, as it can have a big impact on design decisions. We’ll end on a formula that uses some of the variables above and is known simply as torsional stiffness (k). We could use this value to compare the rigidity of components directly, without ever solving for the angle of twist. Now that we know the math, we’ll step into some examples in the next section. Applied Example In this section, we’ll solve for polar moment of inertia, shear modulus, stiffness, and angle of twist. We will start with a simple solid circular shaft, for simplicity’s sake. At the end of this section, we’ll look at how to use CAD programs to help us solve for more complex geometry. First, let’s start with a shaft that has the following geometry: To start out, we’ll calculate the polar moment of inertia: We’ll analyze this part using the properties of aluminum. This means that with an elastic modulus of about 10e6 psi and a Poisson’s ratio of 0.33, we can calculate the shear modulus to be: Solving for stiffness, we can see that we get a value of: Finally, let’s look at the angle of twist if we apply 200 in-lb:> If we go from aluminum to steel, we’ll increase the stiffness three times. This means that we’ll also have one-third of the deflection, or 0.07 rad. If you’re in a situation where the design is very sensitive to positional accuracy, going with a stiff material will be beneficial. If a greater increase in stiffness is needed than can be obtained from a material change alone, increasing the polar moment of inertia will be the best option. As an example, going from 0.5” to 0.75” brings the polar moment of inertia from 0.00614 in4 all the way up to 0.0311 in4. That means we’re making the part five times as rigid, with a 50% increase in diameter. In many instances, we’re not just dealing with simple geometries. While the equations for angle of twist, stiffness, and shear modulus apply, we can’t use the same formula for polar moment of inertia. As a matter of fact, there may not be a textbook-designed formula for your given geometry. In that event, it’s best to use a CAD system to calculate your polar moment of inertia. This value can be substituted in for J in the remaining equations. The picture below from my previous article on designing for stiffness illustrates how to do this. While the image was captured from CREO, Solidworks can do the same type of analysis. (Cross Section Properties in CREO 2.0) The highlighted quantities are Area MOI and Polar MOI (J), and as we’ve discussed, the former is the property of choice for this article. Using a CAD system makes this analysis much easier than hand-deriving a value. Design Guidelines Based on these formulas and the overall analytical approach, there are a few important guidelines to remember: Increases in stiffness, as a result of increasing diameters in circular or near circular sections, can often greatly exceed a material change.The angle of twist increases linearly with length, so making a member shorter can also reduce the total twist.You can compare Young’s modulus alone for a relative difference in rigidity if all other properties are held constant. There’s no need to calculate shear modulus until you’re looking to solve for stiffness or angle of twist.These formulas don’t solve for stress. If the loads are expected to induce stresses near the material strength, a more thorough stress analysis needs to be completed.This isn’t an exhaustive list, but it should get you on your way to improving the rigidity of your components.Main Takeaways Designing for torsional rigidity may sound daunting, but after breaking down the steps, it hopefully feels more manageable. These principles apply to products spanning drive shafts, quadcopters, 3D printers, and much more. I encourage anyone involved in design to explore the multiple variables here for help figuring out how to accomplish your design objectives. From Google torsional stress analysis....As noted.....the solution is elegantly simple....IF...you happen to be a student of Mechanical Engineering .... and if not...there is always Google BOB RENTON

Well all I can say is that the solid 1.125 bar I added to my 64 front suspension greatly improved the handling and feel of the car.

Not at all. Copy and pasting all of that will make a reader roll their eyes and move on to the next post...and doesn’t address the original question directly. Also, the polar moment of inertia is calculated differently for a purely circular cross section and whats effectively a tube (hollow...circular sectional area minus an internal circular sectional area) as well. Different equations. Introducing theta/deflection also gets away from the original question and complicates things. When speaking of bending or torsional loads, most of the load is carried by the material furthest away from its center, whether a beam in bending or torsion bar. The further away the material is from the center, the higher its load carrying capacity. This is why i beams w/ a tall web can carry more load than a shorter web even if the flange dimensions are identical, or why a 2x8 carries more than a 2x6. Load at the center is effectively ‘0’ for a sway bar, so any material in the region does nothing but add weight. This is why you see hollow bars/shafts or even in the case of shear through an I beam web lightening holes w/ no ill effects. As you move away from the center, there is some load carried, so when that material is removed the now hollow bar must increase in diameter relative to the solid bar. Overall the weight of the larger diameter sway bar will be less yet maintain the same torsional rigidity as the smaller solid bar. Ultimately it’s more effective to put the material where the load is acting, which is not at the center. Hope this makes sense and helps.

I thought I sensed the polar vortex Venturi influence while driving over Carson Pass. THATS what it was! Hollow tube changes everything.