# Aluminum weight saving

"Aluminum saves weight! Its 3 times lighter than steel!"

Also its 3 times weaker and 3 times softer. When comparing stiffness and strenght to weight ratio, aluminum isnt any better than steel of comparable grade.

Yes, 7000 grade expensive aluminum is as strong as cheap steel, but top of the line high performance maraging steel is 3 times stronger than strongest of aluminum alloys.

So where does the idea that making stuff out of aluminum saves weight come from? What I am missing?

• What you are missing is the entire engineering design process. The two materials compete head to head in a lot of different products from wheels to pipe wrenches. 20 pounds, Sep 10, 2021 at 2:18
• 17.4 pounds Sep 10, 2021 at 2:19
• NASCAR is switching wheels - buildingspeed.org/2020/03/05/nascar-nextgen-wheels Sep 10, 2021 at 2:27
• I don't know much about metals, but I know one difference between aluminum and steel is the ratio between the yield strength and the ultimate strength. When that ratio is small, as is the case for many "steel" alloys, then you have a big margin of safety built-in to whatever it is you are making: If you build it strong enough that it's unlikely to get bent, then it's WAY unlikely to rupture. If you don't need that margin of safety, then aluminum may be a less expensive choice. Sep 10, 2021 at 19:50
• @SolomonSlow And the very special fatigue limit of steel Sep 10, 2021 at 21:01

I believe it has less to do with strength and more to do with stiffness. A rod of aluminum of the same length and weight as a a steel rod will be just as strong (force required to break) but have three times the cross-sectional area which hugely increases the second moment of inertia by nine times (it scales to the fourth power of distance/length which is the second power of area). That more than makes up for the fact that the aluminum's elastic modulus is 1/3 that of steel. So for fixed volume, the steel can be made stiffer but for fixed weight the the aluminum can be made stiffer.

This also helps with buckling if I am not mistaken.

ELABORATION: When bending something, a sheet, bar, whatever. One side is under tension and the other side is under compression. The material in the middle doesn't contribute much to strength or stiffness except to keep the two sides apart and stop the side under compression from buckling.

Tape a slack rubber band to a piece of paper and try to stretch the paper. The rubber band doesn't contribute any strength because it isn't being deformed enough. The middle material is the same in that it is not being deformed enough compared to the material on the outside to take up any internal stress so as to contribute to the strength and stiffness of the entire part.

This is why you can put a bunch of weak foam in between two thin carbon fiber skins. The carbon takes all the stress because it is deforming the most because it on the outside while the foam takes almost none. It's only job is to keep the two carbon surfaces apart from each other (and to stop the side under compression from buckling). Dry wall works this way too. Weak plaster in between two sheets of paper, but you wouldn't want to hit your head against it.

Second moment of inertia is a weird term for the measure of this. It is a measure of how much material is located away from some bending axis. It measures how an I-beam is almost as stiff as a solid beam or a tube is almost as stiff as a solid rod (as long as it doesn't buckle) but weighs a lot less. As much material as possible in the middle has been removed and the only material that remains there is just enough to keep apart the two opposite faces which take all the bending stress. Looked at another way, the material has been positioned to be on opposite ends as far away from each other as possible.

For an object of the same general shape and weight, using a material that has the same strength per weight, but lower strength per volume will inherently position more material away from a bending axis.

• What is that second moment of inertia? So you are saying thicker parts are better even when made from weaker material, how does that work? Does this effect work even for non-tube geometry, like flat sheet for example? Sep 9, 2021 at 23:58
• @Soliton See edits. Sep 10, 2021 at 1:02
• This was the correct answer, i csme here to say the same. The rigidity isnt so much about mass of metal, or volume of metal, but the placement of metal. (And ensuring it's kept in its designated placement, to ensure no buckling, which is easy and lightweight, as this answer correctly says) Rigidity scales hugely with distances. So a weaker metal, sliiiiightly further apart and sliiiiiightly thicker, can easily be as (or even more) rigid than the original steel proposal. Rigid enough that you can end up keeping most/all of the entire weight saving that comes from swapping steel to aluminium Sep 10, 2021 at 21:08

## TL;DR: Materials perform differently under different loading conditions. Some applications are more suited for steel others for aluminium

I will try to give another more general perspective/approach, using the concept of material indices (mainly other users gave perfectly adequate explanations of the differences between bending and tension and stiffness vs strength and also provided enough examples. ).

The equations of performance of a material for a certain application (there are multiple) can be derived (e.g. an example of material indices derivation is in slide 17 in this here - or maybe later I'll give an example).

What you end up is an equation with the properties required to maximize stiffness or strength while minimising mass or cost.

So for minimising mass you end up with the following MI's:

Application Stiffness strength (to yield point)
Tension $$\frac{\rho}{E}$$ $$\frac{\rho}{\sigma_y}$$
bending beam $$\frac{\rho}{E^{1/2}}$$ $$\frac{\rho}{\sigma_y^{2/3}}$$
bending panel $$\frac{\rho}{E^{1/3}}$$ $$\frac{\rho}{\sigma_y^{1/2}}$$

Where:

• $$\rho$$ density
• $$E$$ modulus
• $$\sigma_y$$ yield strength/proof stress (you can use other properties also).

## Example for steel and aluminium comparison

Assuming the following properties are correct (for example I found that Al 6061 has better yield strength that S235, but for other steels that might not be the case) then:

property units Steel S235 aluminium 6061 Maraging steel
$$\rho$$ $$\frac{kg}{m^3}$$ 7800 2700 8100
$$E$$ $$GPa$$ 210 70 210
$$\sigma_y$$ $$MPa$$ 235 275 2000

Then you end up with the following MI's. In color and bold are the values that should be selected. (notice that you sometime authors use the inverted formula in that case the maximum values are preferable)

Max stiffness, for min mass Steel S235 Aluminimum 6061 Maraging steel
Tension (up to yield) $$\color{red}{37.1}$$ 38.6 38.6
bending beam 583 $$\color{red}{322.7}$$ 559
bending panel 1312 $$\color{red}{655}$$ 1362.7
Max strength, for min mass Steel S235 Aluminimum 6061 Maraging steel
Tension (up to yield) 33.2 9.8 $$\color{red}{4.1}$$
bending beam 204.8 63.8 $$\color{red}{51}$$
bending panel 508.8 $$\color{red}{162.8}$$ 181.1

From the above, when only the minimisation of mass is a criterion (cost can also be one), then for the grades I've selected for the example, aluminium outperforms steel S235 in all cases except for the stiffness in tension (and even then marginally). In most other cases the difference is significant.

On the other hand the maraging steel performs better than aluminium when strength is concerned, but even there when it comes to bending of panels Alumium has the edge

## Quantifying the weight saving.

Another benefit of the material indices is that they can directly estimate the weight saving. This is assuming the same basic cross-section. To do that you only need the ratio of the two mi's. So, in the above example for the maraging steel and aluminium.

Max stiffness, for min mass Aluminimum 6061 Maraging steel Weight saving
Tension (up to yield) 38.6 38.6 0
bending beam $$\color{red}{322.7}$$ 559 42%
bending panel $$\color{red}{655}$$ 1362.7 52%

the way to read the above, is that if you had a component that exhibited the same deformation/deflection, with similar rectangular cross-sections, the weight would be the same for tension, however for bending of a beam and panel there will be more that 40% and 50% weight saving respectively in the aluminium component.

The same can be done for strength, in which case:

Max strength, for min mass Aluminimum 6061 Maraging steel Weight saving
Tension (up to yield) 9.8 4.1 -142%
bending beam 63.8 51 -25%
bending panel 162.8 181 10%

The minus in this case signifies that the equivalent aluminium component would be heavier by that percent. So the maraging steel in this case outperforms the aluminium. Still, there is a 10% improvement in the bending of panel, even in strength.

## Ashby diagrams

You can visually do this material selection with an Ashby diagram with guide lines for material selection like the following.

Figure : source Wikipedia

Sometimes the strength isn't required, so size for size, aluminium is lighter.

e.g. bicycles often use aluminium screws to hold the bottle cage to the frame. The size of the screw is set to be comfortable to handle, and to accept the same size Allen key as other screws on the bike. But, it doesn't need to be super strong as it's not taking a critical or high load. Hence, using aluminium screws saves weight.

• The example is possible but is now pretty rare - real weight weenies use Ti screws, while the rest of us use steel. It's quite easy to strip the heads off the aluminium screws when the threads bind, unless well-protected with something like copper-loaded grease. But the general statement is very true; parts designed to mate with others or to be handled often have a minimum size independent of strength Sep 10, 2021 at 8:47
• Real weight weenies don't have bottle cages at all 😅 Sep 10, 2021 at 21:12
• I suppose dehydration also saves weight (I'm not quick, but do long distances - on a steel bike, and pack rat would be a more realistic description than weight weenie) Sep 12, 2021 at 8:37

When Zeppelins were designed, only non-sparking materials were allowed for them so the hydrogen gas that invariably seeps from the gas bags would not find an ignition source. Steel did not qualify.

But this is admittedly a rare example. The other answers which mention the second moment of inertia are spot-on, though. If you build a load-carrying shell you can either pick aluminium or steel, but the steel sheet thickness would require a much narrower stiffener spacing in order to avoid buckling of skins loaded in compression or shear, so the manufacturing effort would become insanely high. With the thicker aluminium skin, this is much more manageable. See the picture of a semi-finished fuselage of an RV-10 (below) as an example (picture source):

Now imagine you have to rivet three times as many stringers and more bulkheads, too. It can be done but is no fun.

where does the idea that making stuff out of aluminum saves weight come from?

Maybe from ignorance? Ed Swearingen once said that it doesn't matter much whether you use wood, aluminium, titanium or steel - if well designed, the airplane structure will turnout to be of comparable weight. The all-steel Junkers J 1 was only heavier than later Junkers aluminium airplanes because it used a smooth skin over a corrugated inner structure (where corrugation worked like stringers). The later designs simply omitted the outer skin.

Only supersonic heating forces you to use titanium or steel. Aluminium does have its disadvantages (fatigue and tricky to weld) but is ideal for lightly loaded shells.

• Of course, there are other ways to avoid buckling. For a hull of suitably cylindrical shape under enough pressure, thin steel may work even without any stringers. This is certainly one of the reasons why SpaceX decided to build Starship out of steel. Sep 11, 2021 at 15:36
• @leftaroundabout ... and needs to fill the stages with water for horizontal transportation. They learned this the hard way when they shipped the first one to the launch site. Of course, this is not a feasible possibility for most structures. Sep 11, 2021 at 16:32
• RV-10 use metal? I thought they used wood. But I also assume every plane that looks like a bushplane is made of wood. That's also why model airplanes are built out of wood rather than plastic or metal. It's not that the wood is necessarily stronger, it's just that by the time you make the plastic or metal thick enough to be stiff enough, it is way too heavy. Sep 11, 2021 at 18:01

And, in addition for sheet metal application like aircraft skin the aluminum could be 0.040 " ( 1 mm) thick ; the equal steel would be 0.013" thick . The steel would be so thin it would cause handling problems and , I expect other problems in the real world. PS ; maraging steels are good for NASA but not much use in the real world. Regular Q&T low alloys will handle most any critical application like aircraft landing gear.

• Other problems in the real world: yeah, corrosion! Carbon steel structures have between 1mm and 3mm corrosion allowance depending on application. Aluminium needs no corrosion allowance and can be used unpainted (often is used unpainted in aero applications.) Stainless steel doesn't corrode but has no cost advantage over aluminium. Sep 10, 2021 at 20:30

Some engineering aspects, related to designing end products (my experience is in prototypes and small-volume production):

• Machining costs (accounting for time and tool wear when milling/turning) are so much less on Al alloys that the costs of extra material can be tiny in comparison - if you're starting from billets.

• Al is also easy to melt, so casting is more viable for small production runs than with steel - and minimum wall thicknesses for casting would limit how light you can go with steel. This isn't something I've been involved in myself, but a cast and post-machined aluminium housing is a good option for a lot of lab equipment, being stable, rigid, of reasonably cost, and easier to lift than steel would be. The alternative is actually more likely to be aluminium plates screwed together.

• When a fixed or minimum size is needed for mating with other parts or hands, steel would be unnecessarily heavy, or would need a hollow structure. This includes the pipe wrenches in the comments under the question, but also a lot of bike parts.

• If you're using a tube or shaped beam anyway (as a structural member rather than as a pipe), you can often gain stiffness by increasing the diameter, but the limits often comes from have sufficient wall thickness to avoid minor dents leading to buckling - think of something like a bicycle frame, where aluminium ends up lighter than steel, but not by all that much, even when the steel tubes are more complicated (double-butted). Steel's better fatigue performance helps its case here, but still Al comes in lighter.

• Because of the lower melting point (actually it's not melted, but it is softened until it's malleable) aluminium can be extruded using steel parts, in complex shapes that can make for efficient beams (if stiffness is a major requirement), or provide optimal shapes for attaching other parts. These are easier to work into structures by end users, perhaps installing on-site, or by low-volume manufacturers

One example is a couple of lab periscopes I designed and assembled recently. I didn't do the machining. The taller is about 1.8 m long. They needed to have at least about 80mm internal width to fit the mirrors and mounts, square section, walls thick enough to tap a thread into (no access to add nuts on the inside, and threaded inserts would have interfered in places). For optical alignment, stiffness is important considering vibration rather than gross deformation. Finally they had to be light enough to be assembled, fitted onto optic tables, and adjusted into position by one person. In that case 4" square section aluminium extrusion was the obvious starting point. Comparable steel tube would have be several kg/m heavier despite thinner walls, have needed painting against rust, and needed more machining as the initial surface quality is far worse for mounting against. Some of that extra machining would be on the inside, meaning large cutouts for access, compromising the structure

Wider tubes and beams are stronger against bending and less susceptible to buckling. Look up second moment of area.

There is a limit to how thin you can make a tube before it becomes impossible (or very hard) to weld.

With aluminium you can make much wider tubes which can still be welded.

Just compare modern aluminium bicycle frames to older, high-end steel frames.

Aluminum has a couple of favorite properties.

• rust resistant.

• easy to extude

• decent strength

its ideal for uses where one needs decent strength combined with light weight but enough meet to paroved space for drilling bolt into to attach brackets and or other parts. Engine block is a good example.

Aluminum windows and shelving is another one. they strong enough for the job and have a clean quiet functionalty compared to steel.

In aviation it is used to be material of choice for wings and fuselage.

If we compare a birds wing bone and joint to that of a horse, we notice the bird bone is much lighter with foamy matrix but has large joints to accomodate large wing muscle tendons.

maybe thats whay they are doing research on foam alloys.

Simple and short answer is it depends where do you want to use Steel or Aluminum, and what is your target. If your target is to use a material so that it will not suffer from any plastic failure or fracture when subjected to extreme loading conditions, then you have to choose a material which will have higher yield and tensile strengths (that is why steel rods are preferred in civil related structures). Furthermore, if the target is reduce the deformations of the rod (i.e. increase the stiffness), then steel is again preferred since it has a higher Elastic Modulus. It should be noted that in the examples mentioned in this paragraph, it is assumed that the cross sectional area of the structure is fixed and cannot be changed.

However, steel costs more and is also heavy due to its density. Lets consider an aircraft's wing for example; if you want the maximum stresses within the wing to reach the value of the yield strength of the material (industries usually use a safety factor of 1.5 as well), and if you choose steel then the thickness of the wing's skin will be extremely low which is impractical to manufacture. This is why Aluminum (a lower strength material) is preferred so that the resulting thicknesses will be somewhat logical and practically possible to manufacture and will also result in lower mass overall.

In this case, the deformations resulting from the steel and aluminum both will be same since, although they have have differences in E, they both have different cross sectional areas now as well.