Advances in low-cost 3D Printing and CNC Machining have made it easy for a growing number of hobbyists to design and manufacture complex parts in the privacy of their homes. But the technology itself is not always enough: functional prototypes made out of PLA or machined out of HDPE often prove to have surprisingly little rigidity and strength.

The intuitive reaction to this problem is to uniformly increase the size or bulk of the manufactured components, often at the expense of printing time, material cost, or the overall utility and aesthetics of the finished part. When such trade-offs are completely unacceptable, some DIYers will instead try to have the parts made out of more expensive or challenging materials, such as aluminum or PEEK.

As it happens, in most cases, such frustrations can be completely avoided. The manufacturing industry, ever-obsessed with saving time and money, has developed quite a few simple tricks that allow them to routinely tame even the most finicky materials and get precisely the results they want. To figure it all out, let’s have a casual stroll through some of the most basic concepts in engineering – and see how they translate into surprisingly simple yet powerful part design tips.

### The marvelous world of bending

Everyday engineering materials fail in a myriad ways, depending on whether they are being crushed, pulled apart, sheared, or abused in some other fashion. That said, in small-scale mechanical prototyping, there is one characteristic that almost always matters more than the rest: the ability for your parts to survive being bent.

To get a sense of why parts break when flexed, let’s consider a very simple yet common scenario – a rectangular beam anchored at one end, and subjected to a bending force (F) on the other:

Common sense dictates that the force applied to the beam will pull atoms near the upper surface apart from each other, and compress the atoms on the other side. As the load increases, the stress that these molecules are subjected to will increase – up to a point where it overcomes the electromagnetic forces that hold the whole thing together. This critical stress threshold tends to be constant for a particular type of material – but depending on how many molecules are there and what shape are they arranged into, the load required to reach this point will be different.

So, how does it play out in practice? For starters, we know that the outer surfaces of the beam are subjected to the highest levels of tension and compression, and that’s where the stress will be most pronounced. Simplifying a bit, there is a linear transition between these regions, producing a stress-free *neutral axis* that runs through the center of the cross-section:

*Distribution of bending stress, looking at a cross-section of a rectangular beam.*

Looking at the two-dimensional cross-section, the neutral axis is always perpendicular to the direction of the applied force. For rectangles and other symmetric shapes, it crosses through the middle of the beam; for other geometries, it goes through the *area centroid* – that is, the geometric “center of mass”. This location can be automatically identified by any competent CAD program; symbolic formulas and calculators for common cross-sections are easily found online, too.

The existence of a low-stress region in the center of the beam is interesting on its own: perhaps it is possible to intentionally remove some of the material from there and use it to reinforce the high-stress outer surfaces, so that we end up with a tougher part without increasing its weight? The concept is not as crazy as it may seem – that’s precisely how I-beams work:

*In construction, I-beams greatly outperform rectangular beams made with the same amount of steel. On the flip side, they are more vulnerable to torsion, buckling, and shear – especially if the center section is too thin.*

But let’s not get ahead of ourselves. When it comes to stress distribution, there is one more thing to keep in mind: as long as their cross-section is constant, beams effectively work as simple levers, with the bending moment proportional to the applied load and the effective distance. Therefore, the overall stress experienced within a cantilever beam also varies along its business length, starting at zero and ramping up linearly up until reaching the anchor point:

*Distribution of bending stress along the length of a cantilever beam.*

When we combine the two observations we have made so far, we can infer that the highest stress will be concentrated in a very specific location in our beam: the outermost surface just next to where the beam is attached to another structure. With all this in mind, let’s try to figure out the conditions under which this stress exceeds the limits of what the material can cope with.

### Calculating the bending stress

The basic experiment-backed formula to calculate the maximum stress (σ_{max}) present in our setup goes like this:

σ_{max} = F * L * c_{x} / I_{x}

Let’s try to make some sense out of it: it looks like the stress at the surface near the anchor point is linearly proportional to the applied force (F) times the effective length of the beam (L). On top of it, it’s proportional to the distance between the outermost bits of the cross-section and the part’s neutral axis (c_{x}) – which is equal to h / 2 in rectangular beams. All of this should be fairly clear.

More cryptically, the stress is also inversely proportional to a value called the* area moment of inertia* (I_{x}), which quantifies the distribution of material in relation to the beam’s neutral axis – effectively describing how many molecules the applied force will have to fight, and how strongly they will be pulled apart. Deriving the formula for this parameter requires solving an integral, but CAD applications often automate the process of finding specific values for the shapes you have designed; in absence of that, online calculators offer values for common beam geometries for those on the go.

In any case, the formula to calculate the moment of inertia for a rectangular beam of given height (h) and width (w) is well-known:

I_{x} = w * h³ / 12

If we combine all the equations together, we get this equation for the maximum stress in a cantilevered rectangular beam:

σ_{max} = 6 * F * L / (w * h²)

In practice, the suppliers of common materials usually perform standardized bending tests to calculate σ_{max} for their product; this value is then advertised as* flexural strength* in the technical datasheet. Since we usually know the value of σ_{max}, let’s solve the equation for F instead; that would give us the maximum force at which the component is expected to fail, which sounds like a pretty useful thing. The general formula is:

F_{break} = σ_{max} * I_{x} / (L * c_{x})

If we plug in the equations for I_{x} and c_{x} in rectangular beams, we get this form:

F_{break} = ⅙ * w * h² * σ_{max} / L

Let’s try that out in practice: let’s say that you have a piece of acrylic that is 1 cm wide, 3 mm thick, and 10 cm long. A quick search reveals that the flexural strength of acrylic is in the vicinity of 110 MPa (that’s 110,000,000 N/m², or 110 N/mm²). So, let’s do the math:

F_{break} = ⅙ * 10 mm * (3 mm)² * 110 N/mm² / 100 mm = 16.5 N ≈ 1.68 kgf

Looks like a cantilever load applied to the end of this piece should not exceed approximately 1.6 kgf – and that’s without any safety margin to speak of. Were you to use polystyrene (flexural strength 40 MPa), as little as 0.6 kgf would be a grave concern. Gee, who would have guessed?

### Optimal geometries for load-bearing parts

Of course, solving such equations for every single functional component of your project is overkill; on top of that, not everything in life resembles a beam, even if you squint real hard. Nevertheless, the theory discussed so far highlights several surprisingly versatile design strategies that directly translate into tough, durable parts of almost any shape.

How so? Well, let’s use another simple example, and say that you are troubleshooting a mechanical linkage that uses a plastic rod, about 1 mm thick and 5 mm wide. The rod keeps breaking under loads applied along the X axis; after giving the problem some thought, you decide to increase its load-bearing capacity by – dice roll – a factor of three.

What would be the best way to achieve that goal? Well, a quick look at the formula for F_{break} in rectangular beams reveals that there is a linear relationship between width (w) and the resulting load limit, so one obvious choice would be to triple this dimension of the beam, going from 5 to 15 mm. On the flip side, the rod will be now pretty unwieldy; its weight and the amount of material needed to make it will increase by 200%, too.

*Tripling the load capacity of a beam, the naive way.*

Of course, we can do better than that. The aforementioned formula also tells us that the load limit increases in proportion to the square of part thickness (h). In other words, you need to multiply the height only by √3 ≈ 1.73 mm to get a threefold increase in load capacity. Such a change requires just 73% more material, and keeps the part close to its original shape:

*A better approach to the problem: increasing the thickness of the part.*

In fact, even this approach is fairly conservative: if there is no reason to maintain the original width of the beam, we could make the cross-section square, effectively equalizing its flexural performance in both X and Y. If so, the cross-section could be just 2.47 mm wide and 2.47 mm tall. This solution requires just 22% more material than the starting 5 x 1 mm beam, but can support three times its load.

But wait, there is more: because of what we know about the distribution of bending stress within the cross-section of the beam, we have a general suspicion that we could get rid of some of the material near the center of the shape without significantly compromising its overall flexural properties. So, let’s try to construct an I-beam to see what the practical impact would be.

To understand what will happen, we can no longer look at the formula for beams with a rectangular cross-section; instead, we have to go back to the general formula for F_{break}:

F_{break} = σ_{max} * I_{x} / (L * c_{x})

Since I-beams are symmetric, the neutral axis is always in the middle, and therefore, c_{x} = h / 2. The only major unknown is I_{x}: the symbolic formula for the area moment of inertia can be easily looked up online, but is pretty messy for this particular beam shape; because of this, it’s best to use your CAD program or an online calculator to experimentally figure out how the value of I_{x} changes as you remove a sections in the middle – and then experiment with how much would we need to add to the outer flanges to get back to the desired value of F_{break}.

*Constructing an I-beam to match our previous best result. Material is removed in the red areas, and then the green sections are extruded until the calculated maximum load matches the starting shape.*

In the example shown above, we had to add only a minuscule amount of material at the outer surfaces to compensate for a huge undercut in the middle. In fact, the process yielded a beam that has three times the load capacity of the original 5 x 1 mm rod are trying to replace… yet uses 20% less material. Neat, huh?

In practice, I-beams are not a common sight in small-scale applications, in large part because integrating them into part geometry would complicate most of the staple manufacturing processes (say, injection molding or metal stamping.) But if you look closely enough, you will notice that their close relatives – T-beams and U-shaped channels – are ubiquitous in industrial design:

*Common methods of reinforcing thin-walled parts. The profile of the channel (right) can be rounded, too.*

Such features are commonly integrated with the design of the part itself: internal ribs, cleverly curved surfaces, flanged lids, rims, and many other elements of contemporary industrial aesthetics aren’t there just for the looks. Without them, our mobile phones, storage bins, and plastic cups would fall apart right away.

*Extensive use of reinforcing ribs and cleverly shaped walls in an otherwise paper-thin injection-molded enclosure for AVR ISP mkII, a popular programmer for ATmega MCUs.*

*Tune in next week for part two of “Prototypes That Last”. In the second installment, we’ll talk about the ways to predict – and tame – the rigidity and impact resistance of common materials used in 3D printing and CNC work.*

There’s a marvelous overview of all this and much, much more, over on Instructables:

http://www.instructables.com/id/How-to-Build-your-Everything-Really-Really-Fast/

It’s saved me a great deal of whiteboard time. :)

I think this is way too detailed for the average maker. I’m a mechanical engineer and I feel that this series, which I find a great asset, is diving way too deep into mathematical stresses and strain to simply talk about long twigs breaking easier than short twigs. It doesn’t really show “why” the last pictured prototype is the way it is, or how a previous prototype failed. With Autodesk’s new dedication to makers, how about just showing some heatmaps of stress and strain (or deflection) and how to use simple flanges to show how much better it is than a thicker wall? A really important topic and I’m glad you’re taking it on, but I think you might lose a lot of readers along the way.

Yup, that’s my fear – the second part has a lot less math and talks about practical consequences of it all; that said, with the article split into two pieces, the first chunk does seem to have more math than necessary.

Michal,

i am not good at Math, but i am not stressed looking at the formulars. This post contains a lot more usefull information i have expected. I also did not miss a heat map. I think its okay when people already know about this topic, they should skip it – but for me this will be the place to look at… Thanks !

Found this post interesting mostly because the maths you put on it, you should not give up. Rarely read this kind of article nowadays: either you find university books with boring stuff or vague rules for design as put radius here and reinforce there. Keep the great work.

@subreyes94:

I only wish I were a mechanical engineer; but as Mark Twain supposedly said, I’ve never let college stand in the way of my education. My personal preference is a mix of simple “how-to”, like you’re advocating, and the more granular detail of why one design works better than another. I, like Ayoub, was never a shining star in math class but I won’t shy away from a challenge if it helps me realize my designs in the physical world.

@Michal:

Thanks for the great article. I look forward to part 2!

you are awesome

Great Article . Eager to see part 2!