## The Problem

Let’s say we need to design a structural section of a robot arm, as shown.

We’ve been given a few requirements:

- The mass of the component must be minimized at all costs.
- The length of the link must be 300 mm.
- The maximum allowed deflection (caused by the component) is 1 mm.
- The combined mass of the maximum payload and end-effector is 30 kg.
- The combined center of mass of the payload and end-effector is 50 mm from the joint.
- Both the base and end-effector joints can rotate in any axis.

We are allowed to make the following assumptions to make our job easier:

- We can assume that the arm does not move dynamically, so that our analysis can be static-only.
- We can assume that the end-effector itself does not deflect or deform.

So how do we optimally meet these requirements?

## Abstraction

Let’s start by simplifying the situation by only looking at the worst-case scenario, as shown. In this case, maximum bending moment will be acting along the length of the arm and maximum torsion will be acting on it in the axial direction.

Let’s redraw it as a cantilever beam.

Remember what we know from the requirements:

- $L_x = 300 \text{ mm}$
- $L_y = 50 \text{ mm}$
- $m = 30 \text{ kg}$

## Choice of Cross-Section

Choosing the cross-sectional shape of a cantilever beam is all about maximizing your second moment of area, a major determining factor in both stiffness and strength. On the other hand, you’ll also want to minimize cross-sectional area to ensure that the mass of the beam is as low as possible.

How do we maximize second moment of area while minimizing cross-sectional area? Well, second moment of area can be maximized by locating as much of the shape as far away from the neutral axis as possible, whereas area can be minimized by squeezing that shape into a thin band.

When bending is expected to occur in only one axis, the best solution is an I-beam. But because this section of the arm has powered rotation about its longitudinal axis, the beam must be able to support an equal load in any direction perpendicular to its length. As such, the cross-sectional shape of choice is a hollow circle, which has the best ratio of second moment of area to area when considering any neutral axis angle.

Area of a hollow circle: $$ \begin{equation} A = \pi(R^2 - r^2) \end{equation} $$

Second moment of area of a hollow circle: $$ \begin{equation} I = \frac{\pi}{4} (R^4 - r^4) \end{equation} $$

Second polar moment of area of a hollow circle: $$ \begin{equation} J = \frac{\pi}{2} (R^4 - r^4) \end{equation} $$

## Equations

Now that we’ve simplified the problem down to a cantilever beam with a hollow cylindrical cross-section, we can use several equations that will describe deflection and stress acting on the system.

### Shear Stress

Max shear stress for a beam with a thin-walled tube* $$ \begin{equation} \tau_{beam, max} = \frac{2 F}{A} = \frac{2 F}{\pi (R^2 - r^2)} \end{equation} $$

Shear stress due to torsion for a circular shaft* $$ \begin{equation} \tau_{torsion} = \frac{M_x R}{J} = \frac{F L_y R}{\frac{\pi}{2} (R^4 - r^4)} \end{equation} $$

Max shear stress for a thin-walled tube with downward force and torsion $$ \begin{equation} \tau_{max} = \tau_{torsion} + \tau_{beam, max} \end{equation} $$

### Tensile Stress

Max tensile stress for a beam due to bending* $$ \begin{equation} \sigma_{max} = \frac{M_y c}{I} = \frac{F L_x R}{\frac{\pi}{4} (R^4 - r^4)} \end{equation} $$

### Deflection

Deflection of a cantilever beam due to a force on the tip** $$ \begin{equation} \delta_{max} = \frac{F L_x^3}{3 E I} = \frac{F L_x^3}{3 E \frac{\pi}{4} (R^4 - r^4)} \end{equation} $$

*BasicStressEqns-DBWallace.pdf

**FE Reference Handbook, 9.4 Edition

## Optimization

To evaluate this problem holistically, I’ve characterized it as an optimization problem. Our objective is to minimize the cross-sectional area (which is proportional to $R^2 - r^2$) as a function of $R$ and $r$.

$$ \begin{align} \min_{R,r} \quad & R^2 - r^2 \\ \textrm{s.t.} \quad & \frac{F L_y R}{\frac{\pi}{2} (R^4 - r^4)} + \frac{2 F}{\pi (R^2 - r^2)} - \tau_{max} \leq 0 \\ & F L_x R - \frac{\pi}{4} (R^4 - r^4)\sigma_{max} \leq 0 \\ & F L_x^3 - 3 E \frac{\pi}{4} (R^4 - r^4)\delta_{limit} \leq 0 \\ & r - R + t_{min} \leq 0 \\ & 0.01 \leq R \leq 0.075 \\ & 0.01 \leq r \leq 0.075 \end{align} $$Where:

- $t_{min}$ is the minimum thickness chosen by the user.
- (10) is the shear stress constraint.
- (11) is the bending stress constraint.
- (12) is the deflection constraint.
- (13) limits the minimum wall thickness.
- (14) and (15) define the min/max dimensions.

But before we can see how this performs, we need to choose the material.

## Materials

Let’s look at three of the most popular materials when it comes to lightweight structural design: titanium, aluminum, and CFRP. There’s a reason these are the top three: strength-to-weight ratio, AKA specific strength.

Material | Young’s Modulus (Pa) | Shear Str (Pa) | Yield Str (Pa) | Density (kg/m^3) |
---|---|---|---|---|

Annealed Grade 5 Titanium | 110e9 | 600e6 | 910e6 | 4430 |

Aluminum 7075-T6 | 70e9 | 330e6 | 480e6 | 2710 |

CFRP Tube, 0/90 Ply | 517e9 | 90e6 | 600e6 | 1522 |

Because CFRP is anisotropic, we should take our CFRP-based predictions with a larger grain of salt than with the metals. As such, while I’m using a safety factor of 2 for titanium and aluminum, I’m using a whopping safety factor of 4 for CFRP.

Finally, I found that deciding on minimum thicknesses $t_{min}$ is very tricky. In the end, I went with the pragmatic approach of checking the minimum tube thicknesses available online for each material. For titanium and aluminum, that’s 0.4064 mm (~1/64"), whereas for CFRP it’s 0.762 mm (~0.03").

## Results

I used NLOPT, a library for nonlinear optimization, to solve the aforementioned optimization problem. The code can be found here.

Material | R (mm) | r (mm) | Thickness (mm) | Mass (kg) |
---|---|---|---|---|

Annealed Grade 5 Titanium | 34.357 | 33.972 | 0.385 | 0.110 |

Aluminum 7075-T6 | 39.463 | 39.063 | 0.400 | 0.0798 |

CFRP Tube, 0/90 Ply | 20.834 | 20.072 | 0.762 | 0.0447 |

## Analysis

After looking at the equations from earlier, one’s first impression might be that the cylinder radius should jump to the maximum radius possible. This is not the case, because doing so would enlarge the cross-sectional area (and therefore mass) unnecessarily. Instead, it’s all about minimum wall thickness.

As previously mentioned, you want to squeeze your geometry into as thin of a band as possible to take advantage of the second moment of area without using up too much cross-sectional area.

Note that if minimum wall thickness were no object (i.e. you could have infinitesimal walls), then the optimal choice *would* be to maximize your radius! Sadly, we live in the real world, where we have to care about manufacturability or fabricatability.

With the given boundary conditions, the deflection limit is the next deciding factor after minimum wall thickness. As an example, using an end-effector mass of 80 kg:

Of course, if the deflection limit is high enough, it becomes irrelevant and the shear and tensile stresses become more important.

## Dethroning CFRP

Despite the doubled safety factor and higher minimum thickness, CFRP is the clear winner in our example scenario. But is it always? Let’s see if we can dethrone CFRP by hitting it where it hurts: the shear strength. As you may have noticed from the first table, 0/90 ply CFRP has a much lower shear strength than either aluminum or titanium. We can increase shear stress by increasing the axial offset distance $L_y$:

You can see that as $L_y$ (and therefore torsion) increases, the CFRP begins to lose its advantage over the other two options. The other two remain constant because their shear stress limits are not reached.

## Size vs. Strength

I have to admit that I was initially left scratching my head after looking at these plots. You may find yourself asking the same question:

“Why is the optimal titanium beam heavier than the optimal aluminum beam if titanium has a higher specific strength than aluminum?”

Recall Equations (5) through (8): the torsional stress, tensile stress, and deflection are all inversely proportional to the 4th power of the radius, whereas area (and therefore mass) is proportional to the 2nd power of the radius. This means that increasing the radius will reduce material stress at a greater rate than it will increase mass. As such, geometry can be more important than specific strength when it comes to saving weight!

“Okay, so we should 3D print everything out of foaming PLA instead of using metal?”

In most cases, absolutely not! First of all, keep in mind that in pure compression (no shear or bending stresses) this doesn’t apply, unless buckling comes into play.

Anyway, in most cases your design will have more stringent constraints on the bounding box of the design, and given a small enough maximum radius, the material with the higher specific strength will win. For example, say we were to reduce the outer radius of the aluminum beam down to that of the optimal titanium beam.

Let’s see how that looks: optimizing for inner radius $r$ only given a constant outer radius $R$, with an end-effector load of 80 kg:

Material | R (mm) | r (mm) | Thickness (mm) | Mass (kg) |
---|---|---|---|---|

Annealed Grade 5 Titanium | 30.0 | 26.362 | 3.639 | 0.856 |

Aluminum 7075-T6 | 30.0 | 22.215 | 7.785 | 1.038 |

CFRP Tube, 0/90 Ply | 30.0 | 28.619 | 1.381 | 0.116 |

This time the titanium beam is lighter!

As you can see, second moment of area is more important than specific strength–if you have the space to exploit it.

## Limitations

There are many other practical and theoretical factors that go into beam optimization and material choice that were out of the scope of this blog post. Here are some listed below:

- My optimization code is finicky and I have to do some playing around with the warm starting to prevent it from breaking. I should figure out how to make it a more robust tool.
- This analysis only looks at beams of constant radius. Beam taper actually makes a significant difference, and it’s something I may cover in a future blog post…
- CFRP can’t be machined whereas titanium and aluminum can. That means CFRP usually requires extra interfacing components, which increases its effective weight.
- Risk of impact to the sides of the tube is an important consideration which I completely ignored in this analysis (and it’s why I don’t think CFRP bikes are worth the cost).
- Be careful about where you order CFRP from. The actual carbon fiber content (vs epoxy) per unit volume varies wildly between manufacturers, so you can’t just look at the dimensions!
- Manufacturing cost: Aluminum is the cheapest by far.
- Fabrication: CFRP is difficult to work with and has health risks.
- Thermal properties: the metals win here.
- Corrosion: CFRP wins here, although it can cause corrosion in contact with metals.
- And more…