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06142014, 06:37 AM  #1 
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Beam Sizing. . .Section Modulus
I am designing a storage shed. I want a gable roof with with vaulted ceilings to gain a bit more headroom storage space.
This means I will need a load bearing ridge beam. I found some old papers laying around that explain how to calculate beam sizes and I just want to make sure I am using the correct section modulus numbers. When a beam is doubled do you just calculate the Section Modulus based on the outside dimensions of the doubled beam? Or is there loss involved when doubling a beam? For example: Section Modulus of a 2x10 is 21.4in^3 Is a doubel 2x10 42.8in^3? Thanks, Jake Advertisement 
06142014, 08:19 AM  #2 
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When you double a beam by sistering it, you double the section modulus, provided the sistered beam is exactly the same dimensions as the original. If you use a sistered element that is not exactly the same size, you sum the moments of inertia of the two beams, and use the total moment of inertia in your strength calculations.
When you sister two elements together for a structural ridge beam, you need to verify that you use an adequate number of appropriate fasteners to make the two elements act as one beam, rather than as two separate elements. Advertisement 
06142014, 08:28 AM  #3 
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Thanks for the clarification. I wanted to make sure there wasn't a known fractional multiplier used when doubling beams.
Just for some extra reassurance, is the following formula correct for beam sizing. It is what I used. S=[3*w*(L^2)] / [2(Fb)] S= section Modulus w= load per linear foot L= span in ft Fb= Max fiber stress in bending. (I was using 1200psi for a #1 2x12) 
06142014, 09:34 AM  #4 
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That formula is incorrect. Not sure where you found it, it does not compute the section modulus. Section modulus is defined as the moment of inertia I divided by the distance y from the centroid of the beam to the extreme fiber. So S = I / y. For a prismatic beam, the centroid is at the center of the beam, so the distance y = 1/2 the depth of the beam. For a nonprismatic beam, the computation is more difficult, and I will not bore you with it here.
For a prismatic beam, the moment of inertia I = bd^3/12, where b is the width of the beam, d is the depth of the beam, and d^3 means d cubed. In a prismatic beam, y = d/2, so S = bd^2/6 Make sure you use consistent units when using this equation, in the United States most tables use inches for b, d and y. 
06142014, 10:01 AM  #5 
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The equation came from some old literature from the Western Wood Products Association.
I think the formula has been adapted for unit conversions within the equation. I'm guessing so it's easier to work with on the fly. It basically gives you the required section modulus for a beam, when you know the material strength, span, and load. I hate to bore you with my measly plan here, but I attached the calculations that I did for my shed. 10'x16' Beam will run the 16' length. 20psf snow load. The formula seems to make sense to me, although it's been a long time since I took strength of materials in college. My only concern is if there is a safety factor built into this formula. The literature where I got it did not explain its derivation in the least. 
06142014, 10:50 AM  #6 
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That formula appears to compute the REQUIRED section modulus for the beam. It does NOT compute the actual section modulus of a beam. Not to get picky, but you asked if the formula computes the section modulus of a given beam, it does not, but it does appear to claim to compute the required section modulus for a given uniform load w on the beam, clear span length L, and allowable maximum fiber bending stress Fb. I have not verified this formula.
Unfortunately this is where I need to get off the train, since as a professional engineer I am liable for my designs, and you are effectively asking me to verify your design for a given load. I can't do that for anyone on this forum, since I have not seen the site, and I do not have a contract with you to perform work. I can tell you this. It is dangerous to pull a formula out of a book when you do not understand the derivation of the formula, the limitations, and the relationship of the formula to code requirements. I strongly urge you to contact your local building inspector to discuss this project. You need to fully understand all of the loads your beam needs to resist, and most of that information comes out of your local code. You also need to understand the required factor of safety for your project, and any load factors related to exposure and other site specific factors. 
06142014, 10:58 AM  #7 
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I respect your liability concerns. I appreciate your help and explanations.
I will have this plan reviewed by an engineer if I ever actually get around to building it. 
06142014, 04:08 PM  #8 
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@ j.robinson;
Your ridge beam will need to satisfy two main criteria  1. bending stress and 2. deflection. It is fairly simple to work out the section modulus required to keep within the limit quoted for bending stress. First you have to calculate your bending moment (M). For a uniformly loaded beam, it is W*L/8  the units will be lbs inch. You know your maximum fibre stress, so the section modulus comes from the equation M = fZ, where M = bending moment, f = permissible stress and Z = section modulus. As Daniel has pointed out, it is most important to keep to consistent units, otherwise your figures will be all over the place. What makes it a little more complex is that there are various factors which you need to apply to the figure for bending stress. I don't know what your Code will specify for these, but generally you can increase the allowable stress if it is a temporary load eg snow (our code allows 1.25 increase in allowable stress for that) and if two pieces are sistered (our allows a 1.1 increase). Deflection is a different matter, and there are also equations which enable you to check this. Most codes limit allowable deflection, expressed as a fraction of the span  for a beam in your case it is likely to be around span/150. 
06152014, 08:39 AM  #9 
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Just build a second floor on it
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