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Mast Design Question

QUESTION: "How do I calculate how much load a column (like a mast) can take before it collapses?"

Well, I guess that's a million dollar question! The typical answer might be that 'it depends on SO many variables that it's not possible to calculate accurately'.
However, if the theorists and rocket-scientists out there will turn their backs for a moment and try not to either scream or cry, I will share a few thoughts from the 'practical' engineering aspect. Be warned though, that in order to simplify this complex issue, I need to make some general assumptions and ignore some lesser details.

Two areas where 'compression on a column' is of particular concern to boat designers are in the design of spars (masts, booms etc) and for the compression on struts, such as for the cross beams of multihulls—particularly when waterstays are used on trimarans.

If a column is very short, one can reasonably consider the load to be purely compression. Most materials available today have been tested for their capacity to accept compression and these figures are quoted as 'a load related to an area', such as lbs/in², kg/cm² etc, or as 'kips/in²' which simply means units of 1000 lbs per sq-in.

A note on Metric Units:
The swing from the British Imperial system to SI (metric standard), has brought new units to our attention and many of these units can be used in different ways for science, fluids and structural engineering etc, so this can really confuse the issue. But for the case being looked at here, we are strictly considering the structural use.
First we have the Newton: a SI unit of force, (mass × acceleration in scientific terms) but roughly equal to 0.225 lbs or 0.102 kg for our purpose.
Then we come to the Pascal: a scientific SI unit of pressure, where: 1 Pa = 0.000145 lbs/in² for structural comparison with old units.
With the spread of the metric system, we now also see the pascal expanded with hPa, KPa, MPa or GPa, (hecto, kilo, mega or giga pascals) meaning; 100, 1,000, 1,000,000 or 1,000,000,000 pascals respectively.
So 1 MPa = 145 lbs/in² and 1 in² = 6.45 cm² (approx).
(With all these conversions and calculations, it's just not necessary to use multiple decimal places, as not only are there still too many unknowns in the equations to justify that sort of accuracy, but we also need to add a safety factor into the end result anyway.)

OK, now we have that cleared away a bit, let's look at the two specific areas identified earlier.

Calculating the compression load per in² (or cm²) of any particular material is a straightforward matter, but the particular issue here, is the application of a load on a slender column. First of all, the permissible load will drop off very quickly if one exceeds a certain slenderness.

So what is this 'slenderness' and how is it measured?
Well, the engineering standard is to compare 'the effective length' with something called 'the radius of gyration' or 'r' as it is commonly termed. The effective length is the distance between supports, so that is not complicated to find. (For a deck-stepped mast, the ends are typically considered more as free ball-joints, so the effective length is the total length between these nominally 'fixed' points. For a strut that is held firmly at each end, this effective length' will be somewhat less.)
But to understand this 'r' value, we need to look at a cross section of our strut or mast, so let's imagine you've sliced through a mast and are looking down on the cross section.
The term 'radius of gyration' is quite misleading, as for such a structural application there IS no 'gyration' involved! However true that may be, there IS a certain distance out from the center (or centroid) of the cross section that geometrically represents the effective center of the material, as far as carrying compression is concerned.
Let's look at a few geometric shapes and see where this 'r' is located. For a full circle (a shaft, rather than a tube), r = d/4. In other words, if you draw a circle at ½ the full radius, this line will represent the 'effective center' of load carrying capacity. If you now remove material from the inside, the 'r' value will naturally move outwards—so for two pipes of identical outside diameter but of different wall thickness, the 'r' value will be slightly greater for the thinner pipe. Now if we change the shape to an elliptical one, the 'r' value will also lie on an elliptical line, being greater in line with the longer axis but closer to the center, where it's narrower. Naturally, if you now load such a section vertically, it will bend more easily in the direction with less section width, where the 'r' value is also much less.
This is all very logical and is the reason that spreaders and diamonds are used transversally, to oppose the bend in that direction. But our goal here is to see if we can determine how to estimate some reasonable figures of bend and/or load. So how do we figure this 'r' figure anyway?
Here's a little table that will help—keep in mind that I'm only giving here the 'r' value for the narrowest width W , the transverse thickness for a mast.

Type of Section Ratio of r to W  
*Thin wall rectangle 2:1 0.41  
*Thin wall elliptical 2:1 0.36 (guestimated)
*Thin wall circle 0.33 W = outside dia. in this case
Solid circle 0.25 ditto
(*Thin wall sections assumed as ~5% of W for these values but minor wall thickness variations will have little effect on values given.)

As a side reference, good engineering practice for steel structures typically states that for a pillar to be considered safe to take vertical load without buckling, the ratio of its effective length to its 'r' value, should be 120 or less. Naturally, the lower the value the less likelihood there is of buckling, but over that value, the permissible load drops off very quickly. For a slenderness ratio of 200 for example, the allowable load would only be about 1/3 of what it could be at L/r = 120. Now, I only mention that in passing as for our applications, to go over 120 would be asking for trouble. In fact, for any boat application, I would consider 100 as a maximum for L/r.

One of the reasons we cannot work with the 120 value is, that many times, our strut is not vertical like a pillar. If you imagine a strut that is horizontal, then its very weight will cause some deflection and that initial deflection could be greatly augmented if the strut is accessible for someone to walk on—as it would if used for a trimaran cross beam.

Now here is how I choose to look at this. Looking at the sketch, I find it reasonable (and arguably logical), to limit the deflection so that the 'r' stays within the limits of the outer mast skin fibers, or the mast risks to buckle and collapse.
One may then ask, well what about the masts we see with a foot or more of bend? Well, they are typically not under the same compression at that stage and therefore the buckling load is much reduced. For example, a free standing mast can (like a fishing rod) take an enormous bend and although this can well affect the form of the sail and its aero dynamic effectiveness, such a mast will not fail in compression due to stays—only from the nearly equal compression and tension forces in its outer fibers, due to the bending stress. [Still, ultimately a compression failure though, as most non-metallic materials are weaker in compression than tension.]
Somewhat strangely (to me anyway), mast suppliers rarely give an 'r' value for their sections. But this can be calculated by r = (I/A)½.

For a typical, stayed mast, compression is the main enemy. Dividing up the mast into shorter lengths by the addition of spreaders and/or diamonds is the typically way to handle this. Spreading the shrouds farther out also helps, as it reduces the compression load. In the case of cross beams for trimarans, we simply have to have enough material and 'r' value, to not reach a critical loading. In practice, the section typically used for such beams, often works out to be very similar to that of the mast on the same boat, unless the amas are small and the overall beam limited, when the beams may be lighter. (But also see further comment below.)

It's clear from the way 'r' is affected by the width or diameter of a section, that a round thin-wall tube is the most weight-effective column. However, unless you are considering a mast with a sleeve sail, there are other factors to consider that will generally move the section away from the perfect circle.

For a mast, the after edge needs a narrower area for sail attachment and a more aerodynamic 'pear' shape mast will give better flow to the important lee (downwind) side of the mainsail. A wing mast will be significantly narrower than the chord is long, though I do not recommend to exceed 3:1 for this ratio. Fortunately, we can add diamonds to such a mast and that will divide up the length and lower the L/r to something acceptable, without making the rig impractical to use. Spreaders for this must not be too short though, as diamond wires add further to the compression in their local area.

For a cross beam, although round tubes are often used here, there's a good argument to again use a mast section as these beams are often driven through some pretty solid water at times and then a wing section will offer less resistance and create less spray. A compromise has to be found though, as they should still be able to accept a mid-span vertical load of at least a 100kg (220lb) person and if too streamlined in section when combined with a high end loading, could collapse, unless one wants to add a diamond stay under each beam—as catamarans typically add dolphin strikers under their forward beams.

The mast ABOVE the hounds (the attachment point of the stays) is typically left unsupported and free to bend. While this does relieve this upper part from compression due to stays, some of the bend will transfer to the mast below the hounds and will need to be taken into consideration when calculating the L/r in that area, that in turn, will decide the number and location of diamond stays.

With this in mind, at least we can have some value in mind when we need to answer the question of "how much deflection will this mast take before it folds?" It also supports my answer when testing out say a new plywood wing mast, as I would typically suggest to not exceed a side bend of (w − r), where w = ½ width of mast. This is a tighter limit than some might accept and to keep within that limit, one will either have to reduce sail or add reinforcement to limit the bend. (A metal mast with higher compression resistance can normally exceed this deflection without collapse—say, up to w/2.) But assuming your wood wing mast is now already built, about the only practical reinforcement is the addition of some UNI carbon-fiber tape on each side. This leads to the justified argument that "if you're going to use carbon fiber, then you should use enough of it to take ALL the stress or else it will fail first" I certainly agree with this from a technical aspect and for a one-off CF mast, would only use a very minimal shell of some light material, purely to establish the shape. However, the fact is, that adding even only one layer of CF on the side of a too-flexible mast has proven to keep it straighter and if there's enough of the basic material (often wood) to take all the compression, the combination does seem to work within most practical limits. After all, at this point, there are really not a lot of options left ;-)

I had actually planned to close the article at this point but I was chatting recently with noted mast expert Eric Sponberg and he suggested that even if writing briefly about the subject, I should at least include the Euler formula typically used for mast calculation, being as it relates the vertical load to the actual section used for the material selected. Good idea, so here it is for those more technically oriented.

The standard Euler formula for buckling axial load (P) on a slender mast is:

P = C π² E I
E=(Young's) Modulus of Elasticity
I=Moment of Inertia of the section
L=effective length of strut, or mast
π=3.1416 (so π² = 9.87)
C=a constant depending on how the strut is end supported.
A deck mounted mast is free to bend at the ends, and as C would then equal only 1, can be forgotten. (More rigidly fixed ends could theoretically raise C significantly, but it would be unwise to use over 2 in practice.)
(Watch your units when solving these equations, to make sure you're not mixing feet and inches, or metres and cms, or Newtons and lbs etc.)

The Inertia 'I' relates to the EFFECTIVE SIZE of the section—its geometry and stiffness simply due to its form and distribution of material, while the E relates to the stiffness of the actual material itself, with fiberglass being very low, wood also generally low but wide ranging, but then a jump up to the harder aluminum-alloys and carbon fiber at the top end. [Steel would be even higher but seldom used for sailboat masts due to insufficient wall thickness, as would be required to reduce its weight.]
The so-called 'Modulus of Elasticity' is often confusing for those first learning of it. It relates Stress to Strain (or load to deformation), but is often easier to understand if you think of "the load it can take for a unit of extension, while still being considered elastic"—i.e. still with the ability to return to its original length or form. A material with a high M of E actually may have LOW elasticity or ductility (like carbon fiber), so actual elasticity is not related to the value of Modulus of Elasticity. [As Eric gently reminds us, although steel has a very high Modulus of Elasticity, the material is very ductile by comparison with carbon fiber, and WILL show signs of ductility before ultimate failure.]
While mentioning this, it's important to realize that the stiffer (less ductile) the material, the more suddenly it can fail and CF [with no plastic range] is notorious for just exploding once that limit has been reached, as there are few warning signs. For maximum strength to weight, UNI material is typically used but this does have low elasticity. (Perhaps there's a case for experimenting with cloth on a slight bias here—anybody tried that? What do you think Eric?

Eric replied: Additional 'carbon fiber material on a slight bias' would be the wrong thing to do and a waste of material. The strength of composite fibers in a laminate, drops off drastically with as little as 3° of off-axis angle, leaving only the resin to handle the loads. To increase shear strength of a structure with additional composite layer, one should lay up the fibers at +45/-45° to the axis, as that's the direction the shear stresses run.
Typically, masts and tubes are built with a mix of 3 fiber orientations, 0°, +/-45° and 90° with 80% UNI + 10% + 10% mix respectively, for a typical stayed mast. The UNI should be sandwiched between (inside) the other cloths. For a composite wing mast, one might use a [70%+15%+15%] or [60%+20%+20%] mix, depending on the wing design.

So back to Euler's Formula. It's useful to re-arrange this formula so that we can find the one thing we probably don't yet know—the required Moment of Inertia of the mast or strut section that will do the job.

So now:

I =  P × L² 
C π² E
 and if you want the I in in4, then P will be in lbs, the L in inches, and the E in lbs/in²

Keep in mind that this load P is for the buckling state, so in order to have some margin, the calculated I (or Mt of Inertia) should be increased by some factor; one that will depend on the reliability of the material E value, as well as how close one is willing to go to the limit in an effort to save weight aloft. This factor might be anywhere from 1.5 to say 3 for a cruising mast. More would likely make the section too large or too heavy and typically, a mast is shaved closer to its limit than other parts of the boat or rig, due to the undesirable effect of too much weight aloft.

For a wing mast, the fore and aft bend is usually very low and the mast of more than adequate strength in that direction. But for a regular mast, you will need to check the M of I in the longitudinal direction also and compare that to the loads induced by the stays based on their location. If the I is not enough, then the options are to: change the rigging location, increase the section size or change the material—or any combination of these. No one said this was 'going to be real easy' ;-)

…mike, march 2012

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