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BASICS of Stress, Loads, Strength of Materials & how they relate 

The basics required to calculate the strength of a loaded beam.

Mike Waters, 2018

While all this information is available in text books and now on line … all my life as an engineer, I’ve heard the comment, ‘this is too complex for me’…. 

So, before I get too old and befuddled myself, let me have a shot at explaining this in a way that will hopefully prove more accessible to those who have not spent a life-time juggling numbers to calculate such stuff.       Here are the basic stresses we may commonly encounter.     This article will deal primarily with Bending.


We can all appreciate that when a load is applied to a beam or to a surface, the material is stressed.   It will also be deformed to various degrees .... sometimes very apparent but other times, almost invisibly.

If it’s deformed too far, it will not spring back.   We refer to this as ’exceeding the materials plastic limit’ or moving beyond its ‘Yield point’.  

OK, let’s get started:     Let’s learn what level of stress is caused by different loadings in a beam – which could also be a sample strip of a complete surface. Until we know that, we cannot calculate what material or size of beam is needed to survive the load in the long term.

Beams are loaded and bent in different ways. Taking the most common, there is first a beam that’s supported at both ends and carries a load between its supports.   This creates a maximum bending on the beam that can be expressed by P/2 x L/2, or PL/4.  If the same load is spread out equally along the beam, then the maximum bending moment is cut in half or PL/8.

The other common beam loading is when there’s a load on the end of a beam that’s only fixed or held at one end. We call this a cantilever beam.   Beams for trimaran amas (if without waterstay wires) are just one example and so is the upper tip of a mast that is unsupported by wires at the top.   If the load is on the end, the bending moment is simply the length times the load or PL, but if the same load is distributed over the beam, then the bending moment is cut in half.  In both cases, the maximum bending moment is at the fixed end, tapering to nothing at the free end.    See Beam Diagrams.

Note: So called ’beam diagrams & formulas’ as shown here are available for numerous alternative beam loadings … see the AISC Manual or elsewhere On Line.   Also note, for ease of comparison, I have used ‘P distributed’ instead of “w x L”, [which means L  x  weight per unit of L].

Note that if AKA beams DO have a waterstay holding them from bending up, this then releases them from the cantilever bending moment and converts the beam load to one of compression along the beam, which is easier to resist.  However, the beam may still be under some vertical load as it will still have to carry its own weight plus any crew member that could be standing on it.   If the waterstay (or an arm for a folding system) does not attach at the outboard end, then the part outboard of the attachment must still be considered as being a cantilever … just a shorter one.

[The estimate of any beam deflection can be given by formulae for each different load diagram and many are given in Ch.3 of the Manual published by the AISC - American Institute of Steel Construction.  Updated every 6 yrs, this Manual is expensive (~US$400) but content from the 14th Ed. is available here: 

https://www.slideshare.net/ingeangelbalseca/aisc-steel-construction-manual-14th with Beam Diagrams being viewable on pages 3-213 to  3-228, from where they can be future referenced].

The applicable Maximum Deflection Formula is shown for just three of the above examples, but as you will see, it consistently involves two additional factors.  One is the physical sectional shape, measured in this case by the Dimensional INERTIA (I) of the section under stress.  The other is the Modulus of Elasticity (E), a material-related value that reflects how ‘NON-elastic’ the beam material is.  The higher these two values (E and I) are, the lower the deflection will be.

So with the above formulas, we are now able to calculate the BENDING MT. for a beam  - be it centrally loaded, distributed or cantilevered.     So what now?    This bending load will need to be related to the physical dimension of the actual beam and the strength of the material that is to be used.

NOTE:  Now most younger viewers of this article, will be wondering ‘why all this, when you can just plug the numbers into a computer program and get a result?’   Well, ‘sorry guys’, but I like to understand the basics behind the calculation, so that when there’s some ‘non-standard’ layout or arrangement, I can adjust the inputs to match.   This gives the 'feel' that is needed in order to set the most applicable safety factor.  Computer programs work on the basis of a lot of refined assumptions and while they may be fine most of the time, they risk to leave the user ‘outside their inner workings’, not really knowing if the result is truly valid or not.  

From over 50 years back, I still remember the basic relationship of how section shape, material quality and stress are tied together, through a nickname we gave to this ….“MiFyEr”.     Here’s how it worked.      Each letter of this ‘Theory  of Bending’ meant something and reminds one of their important relationship.   Namely:  M/I = f/y = E/r

 Bending Moment(M) ÷ Mt. of Inertia (I) =  Stress (f) ÷ y  and this in turn was also equal to:   Mod. of Elasticity (E) ÷ Radius of Curvature (R)

with ‘y’ being the distance from the neutral axis of the beam section to the outer fiber farthest away.      This formula applies only to material being loaded and ‘bent’ within its Elastic Limit.

While the last part (E/R) can give you the curvature of the bending beam and therefore the deflection, it was the first part (MIFY) that we typically use the most.

The simple ‘M/I=F/Y=E/R’ formula can easily be shuffled around in order to solve the missing piece.   If we know the Bending Mt and the Allowable Stress for the material we are planning to use, then the formula will look like this:   M/f = I/y.    We just need to use the same ‘units’ for this to work.  ALL the aspects can be in Pounds and Inches for Imperial Units: so I/y will be in inch3 and the Stress in lbs/sq.in so the Bending Moment must also be in lb-ins (not lbs-ft).

[For working in Metric, also stay with constant units.    ie: suggest, kgs and cm’s for each item].

For the STRESS figure, we could use the limiting Rupture Stress to find the minimum dimensional section INERTIA at failure, but more useful will be final section INERTIA needed to keep the stress at its Allowable Stress Limit.

The ‘Allowable’ stress will be the stress at breaking point reduced (divided) by a Factor of Safety (FS).  This FS will vary,  depending typically on:

                                 -  how reliable and consistent the material will be relative to the tested sample

                                 -  how accurate the load could be calculated (are there unknowns?), and

                                 -  risks of failure, inconvenient or catastrophic, versus cost & weight etc.


From this, we can see that the FS is really made up of several factors.  Just to allow for a variance of tested material we could expect to add at least 10%.   Unknowns might add 30% and if there were severe shock loads expected, another 100%.  So this particular example would result in a FS of 1.1 x 1.3 x 2 = 2.9.

The final FS could be as low as 1.2 or might go over 5.    Typically, I might use 1.5-2 on a mast to not add excessive weight, but 3 elsewhere as a compromise.  However, 5 is recommended when the failure could be very serious, and possibly fatal.   

So deciding on what Allowable Stress to use, holds most of the secret behind correctly sizing the beam or structure.     For example, only the designer knows what force he has included in the load ‘P’.  Did he know the part would have shocks and therefore double the P value ?    If so, then he does not need such a large safety factor as he would if he had first ignored this aspect.     Are there some potential unknown loads on the part that are hard to estimate?     If so, did the designer include these with the original ‘P’ load, or does he need to add to the safety factor to cover for them?      So it’s a balancing act, between the value of P used in the calculation of the bending moment and that of the final safety factor, that will determine the allowable stress for the part being checked.   There will ALWAYS need to be some value to the safety factor, but what this factor is, will depend on how conservatively (over) estimated the original P load was.    Each designer will know where and how he covered the variables.

With the Bending Moment divided by the Allowable Stress, the target I/y or Section Modulus can be established.    So one just needs to calculate what this will be for the section you plan to use, to see if it’s up to the task or not.   If it’s too low, perhaps more depth will be required  ... so plan to adjust its dimensions to suit.

So the final part to address is how to calculate the I/y  or Section Modulus (SM) to see if it’s enough.

Common steel and alum. sections have such figures well tabulated for the construction business but for most boat parts, we may be left to figure this out as each shape is different.   Here is how.

One exercise I remember doing in college was to calculate the dimensional INERTIA of various I-beams.

(I think of this as ‘Dimensional Inertia’, just to clarify that there is nothing ‘dynamic’ about this value).

It’s purely a calculation that explains just WHERE the material is located relative to its effective center (or neutral axis).   An I-beam with thick flanges will have a higher I value than one with thin flanges, even if the weight is the same.    So the location of material within the section makes quite a difference in stiffness when the section is bent.      When making that calculation manually, we became aware of how little the connecting web added to the stiffness, relative to the effect of the exterior flanges.   So much so that one could justify making a quick estimate using only the INERTIA of the flanges (see below).

Take this example. Here is a fabricated I-beam (dimensioned here in mm’s for ease of calculation, but works the same in decimals of inches).

If you check the ‘Properties of Geometric Sections’ (also available on page 17-36 at the above link for the Steel Construction Manual), then you can see that the final Inertia (I) for this beam, can be calculated in two parts.   One for the central web and then for the 2 flanges – flanges being equal in this example.

For the central web, the formula for I = bd3 ÷ 12 (about the neutral axis shown), so in this case, that is:  (10 x 1683) ÷ 12,  or  3,951,360 mm4

For the two flanges, one can either use this related formula:

  I = b (dout3 – din3) ÷ 12  (with the ‘out’ and ‘in’ referring to which side of the flange the measurement is taken), OR more easily, use this one for the two flanges : A x h2… Total flange Area x (distance from NA)2.

In this case that is: (100 x 16) x 2 x 922   or  27,084,800 mm4.     The Total Beam Inertia will be the sum of the two values, but right away, you can see that the vertical web accounts for only a minor part of the total.        

So for most practical purposes, you can save some brainwork by just calculating the INERTIA using the simply Ah2. Then add 10%* and you’ll still be on the safe side. But remember to multiply the flange Area by two, for the top and bottom flanges.  (Of course, if you were working in ‘inches’, the figure would look a lot smaller as you’d have to divide by 25.44 or 416,231, bringing the imperial dimension to 65 in4.). 

*Whether one uses 10% or 12% has a very minor effect on the final result, as the choice of Factor of Safety is far more critical to the end result.

Now back to the ‘MIFy’ formula above and we see there’s only the Allowable Stress of the Material left to discuss.

This can get confusing as there are so many ‘stress values’ out there … so what do they all mean?

Well, some refer to pretty much the same thing.  If we make the assumption that the material we are using is homogeneous, then BENDING stress, FLEXURAL stress or RUPTURE modulus, can be taken as the same thing for all practical purposes.    So now, where does the TENSILE strength value fit in with this mix ?    It should/could also be the same, but think about this for a moment.

If the part you are designing is being bent, only the material fibers on the surface away from the load are in tension.  Such a test will define the Flexural or Bending strength.   But if the part were to be held at each end and stretched, the whole piece would be under tension and that will define its Tensile Strength.   So test results may show different figures, depending on the consistency of the sample under tension.    

Though the purists among you may not accept this easily, let’s try to keep this as simple yet safe as we can.   So I’ll add this practical note.   I’d not get too worried about which of the above values you are using in practice, as any differences will pale in relation to the Allowable Stress you need to use after you’ve applied a compiled Factor of Safety.   The FS will change the Allowable Stress by 100% or more, whereas any subtle differences in the other named stresses will likely be closer to 10%, except for some polymers (plastics) such as Kevlar, when you may need to pick a more conservative allowable stress value.    For a stressed beam of aluminum or stainless steel, you can be fairly confident of tested stress values.     For wood,  an extra 20% Factor of Safety is typically justified to cover for species variation and perhaps 15% added margin (over tests) for composite materials such as Carbon Fiber or Glass with epoxy.  

Guideline data and strength values for many materials will be posted in another article, but for a list of Elastic Modulus (E) for various materials, here is a list giving values in both SI (metric) and Imperial units.   The higher the value, the stiffer (less elastic) is the material.      (Although the listed figure for resin-reinforced carbon fiber starts at 18 x 10^6, this is typically for a professionally produced product for shop testing.    So I would suggest to derate that down to 12 to 15 x 10^6 for hand layups without vacuum-bagging, as the final result is hard to predict.  That is still higher than Al-alloy 6061 at 10 x 10^6).



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