Can structural trusses in centrifuges be helpful at all?

Question

Imagine a rotating space station for the purpose of creating artificial gravity. Generally, such a space station is portrayed as a ring or a cylinder. Images of these space stations often have lines running through the center of the circle (examples 1, 2, 3). These could be interpreted to serve two different purposes:

1. simply providing access
2. aiding in the tensile strength required to hold the load in place against the centrifugal force

I would like to ask: From a structural standpoint, would the #2 reason make any sense at all? Would it always be favorable to have all structural material around the outside, or could a centrifuge be made more material efficient by running members across the structure, much like a bridge?

I can address the case where there are NO additional trusses through the center. Consider there is a load and some structural material holding it in place.

Use the equation for tension in a rotating circle, with linear mass density, radius, and acceleration.

$$ T = \lambda R a $$

$$ \lambda = \lambda_{load}+\lambda_{struct} $$

Introduce

$$ f = \frac{ \text{ mass of structural material} }{ \text{total mass} } = \frac{ \lambda_{struct} }{ \lambda_{struct} + \lambda_{load} } $$

$$ \lambda = \lambda_{struct} / f $$

$$ T = \lambda_{struct} R a / f $$

Divide by area of the cross section of the structural material.

$$ \sigma = \rho R a / f $$

$$ f = \frac{ R a }{ \left( \frac{ \sigma}{\rho} \right) } $$

With this equation we can identify the minimum fraction of the station's mass that must be structural material, given the specific strength of the material $\left( \frac{ \sigma}{\rho} \right)$, the radius, and desired magnitude of artificial gravity. So say we added trusses:

Assuming that the load is not moved, is there any possibility that adding the trusses (made of material with the same specific strength) will decrease the amount of structural material needed?

In one extreme, one could consider only a dumbbell structure as an example of the truss approach. I'm speaking about space stations, but I think the problem should generalize to any given centrifuge.

Answer 1

Due to specific function forms I encountered, I found that I had to abandon the metric I formalized in the question in favor of the slightly different metric which I'll arbitrarily call h.

$$ h = \frac{ \text{ mass of structural material} }{ \text{ mass of load } } $$

$$ h = \frac{ f }{ 1-f } $$

First, we can easily address the case where $h \ll 1$. Considering the dumbbell form, the cross section of the member follows directly from the magnitude of the load times acceleration, then the total mass of the member follows from the cross section times length times density.

$$ T = M_{load} a = \sigma A $$

$$ M_{struct} = \rho V = \rho A R $$

For brevity, I will introduce $s=\left( \frac{ \sigma }{ \rho} \right)$. Algebra leads to:

$$ M_{load} a R = \frac{ \sigma }{ \rho} \rho A R = s M_{struct} $$

$$ h = \frac{ R a }{ s }$$

This tells us that if structural materials are a small fraction of the total mass it doesn't matter what approach you use. In the case where the structural material is a non-negligible fraction of the total mass, we encounter a major complication where the cross sectional area of a member spanning the diameter of the centrifuge changes as a function of radius. That is to say, it will be thickest in the center, similar to the mathematics of a space elevator. We would reasonably assume that one would build this with the least mass possible, so that means that the linear mass density of the member will be proportional to the tension at that point in the member.

$$ \lambda(r) = \rho A(r) = \frac{ \sigma(r) A(r) }{ s } = \frac{ T(r) }{ s } $$

The tension in the member is the sum of the centrifugal force on the load plus the centrifugal force on the structural member integrated from the load to $r$. To put this into an equation we require a differential equation.

Initial condition:

$$ T(R) = M_{load} a $$

Develop expression for centripetal acceleration based on $a=\omega^2 r$:

$$ a(r) = a \frac{r}{R} $$

Differential equation accounts for mass contribution of the member itself. Then prior expressions are combined:

$$ \frac{ dT }{ dr } = - \lambda(r) a(r) = - r T(r) \frac{ a }{ R s } $$

For housekeeping, I will introduce a dimensionless parameter, $\alpha=R a/s$. With this I reformulate the differential equation.

$$ \frac{1}{T(r)} \frac{ dT }{ dr } = -r \frac{ \alpha }{ R^2 } $$

$$ T(R) = \alpha \frac{ M_{load} s }{ R} $$

The solution:

$$ T(r) = M_{load} a \exp{ \left( \frac{ \alpha }{2} \left( 1- \left( \frac{ r}{R} \right)^2 \right) \right) } $$

This is used to find the mass of the structural material.

$$ M_{struct} = \int_0^R \lambda(r) dr =\int_0^R \frac{ T(r) }{ s } dr $$

$$ M_{struct} = M_{load} \sqrt{ \frac{ \pi \alpha }{ 2 } } \text{erf} \left( \sqrt{ \frac{ \alpha }{ 2} } \right) \exp{ \left( \frac{ \alpha }{ 2 } \right) } $$

This effectively gives us the expression for $h$ in the case of the dumbbell approach. From the question, we know that $f=\alpha$ in the circumferential approach.

$$ h_{circumferential } = \frac{\alpha}{1-\alpha}$$

$$ h_{dumbbell} = \sqrt{ \frac{ \pi \alpha }{ 2 } } \text{erf} \left( \sqrt{ \frac{ \alpha }{ 2} } \right) \exp{ \left( \frac{ \alpha }{ 2 } \right) } $$

Here are plots, using Wolfram Alpha so that you can play with them yourself if desired. These represent the mass of structural material compared to the mass of the load for the two different constructions, given as a function of the dimensionless parameter alpha. Alpha is a proxy for the engineering difficulty of creating the needed acceleration with the available material.

Now I can draw conclusions. I have three main points.

1. If the structural material is not a major fraction of the total weight the two methods are roughly equally good.
2. Technically, the dumbbell method is more material efficient in all cases, and this advantage is larger when the mass of structural material is a larger fraction of the total.
3. Beyond the limit of $\alpha=1$ all structures are impossible with the circumferential method but there is no such limit for the dumbbell method. You could theoretically build an artificial gravity space station of any given radius with the dumbbell method, even if it meant the structural material mass would be many times more than the load.

The simple answer to the question I asked appears to be yes, dumbbell-type structural members as described would reduce the necessary materials for the same load, radius, and acceleration.