

Subject:
Beam span calculation
Category: Science > Instruments and Methods Asked by: jscott82ga List Price: $50.00 
Posted:
08 Feb 2006 21:03 PST
Expires: 10 Mar 2006 21:03 PST Question ID: 443448 
Ok.. I am building an arbor over my patio. I have checked my local building codes no permit or inspection is required. I still want to ensure I have a safe and structurally sound design. My problem lies in the wood I have chosen, Ipe. I cannot find a span table for this wood, it is very heavy (69lb/ft^3) and very strong (modulus of elasticity 3,140,000). The design is straightforward: two main beams spanning 14ft. Between the two beams will be 11 joists 10 ft long. No roof, no additional loads, but the structure itself will weigh nearly 3000 lbs. So the question: what size do I need for the main beams? Most importantly: What is the calc for this (be specific and detailed). Jeff  
 


Subject:
Re: Beam span calculation
Answered By: redhossga on 09 Feb 2006 10:28 PST Rated: 
Wow Jeff, that is some seriously bad boy lumber. Here are the calcs: First we need the section modulus and moment of inertia for a 2x12. (finished dimensions 1.5 x 11.5) S (section modulus) = bd^2 / 6 = 1.5 x 132.25 / 6 = 33.06 in^3 I (moment of inertia) = bd^3 / 12 = 1.5 x 1520.875 / 12 = 190.1 in^4 Next we need a reasonable weight to use. We will assume 2x12 beams and 2x6 joists: 2x12x14' weight = 2(1.5 x 11.5 x 14 x 69 / 144) = 232# 2x6x10' weight = 11(1.5 x 5.5 x 10 x 69 /144) = 435# TOTAL 667# PSF = 667 / 140 sq ft = 4.76 say 5 PSF The maximum bending moment is found by: M = (wl^2 / 8) x 144 sq in per sq ft Where w is applied load in # per ft = 5 x 5 = 25 M = 25 x 196 x 144 / 8 = 88,200 in# The bending stress is: s = M / S = 88,200 / 33.06 = 2,668 psi Compare this to the bending strength of 22,560 psi (from website) and you are very safe. Safety factor = 25,360 / 2,668 = 9.5 Now we will look at deflection: D = (5wl^4 / 384 E I) x 1728 cubic inches per cubic ft Where E is the modulus of elasticity = 3,140,000 psi D = (5 x 25 x 38,416 / 384 x 3,140,000 x 190.1) x 1728 = .036 inch Again this number is very safe. Just out of curiosity I ran the numbers for a 2x6 beam. To keep from boring you further (and me also) here are the results: S = 7.56 in^3 I = 20.8 in^4 s = 11,666 psi Now we have reduced the safety factor to 25,360 / 11,666 = 2.17 I would be a little nervous with this number. D = 0.33 inch Obviously your first inclination to use 2x12 beams is a very sound choice. You could certainly use 2x8 beams, but the question then becomes more about aesthetics. The 2x12 beams appear more massive and that may be the look you want. Hopefully you have been able to follow all of this. However, please ask for a clarification if there is something you don't understand. Good luck with your project, Redhoss  

jscott82ga
rated this answer:
redhoss, Thanks that is exacty what I am looking for. I have run into this same problem with other materials and will keep the math for future projects. This realy helps alot. Not part of my initial question but if you would be so kind. How would the bending stress equation change if it were a point load? Thanks Again. 

Subject:
Re: Beam span calculation
From: iainmcclatchiega on 04 Apr 2006 18:44 PDT 
Redhoss, >M = (wl^2 / 8) x 144 sq in per sq ft >Where w is applied load in # per ft = 5 x 5 = 25 >M = 25 x 196 x 144 / 8 = 88,200 in# Is there a unit error here? You've multiplied 25 pounds/ft by 196 ft^2, which gives you an answer in poundsft. To convert to poundsinches, multiply by 12, not 144. The resulting bending stress is quite a bit smaller; the "safety factor" would be positively enormous. Note that the usual bending stresses allowed (Fb) under code for joists made of stuff like Douglas Fir are on the order of 1000 psi  I'm not sure how to compare the Ipewood 25,400 psi number to this, but in any case the arbor will stay up just fine. The deflection calculation looks right (I checked against some beam tables). I get 972 psi bending stress for 2x6 main beams, which is within the code value for Douglas Fir, so I assume it would be for Ipe. And this makes intuitive sense as well. If I dropped a big load on a 14' 2x6, and the thing sagged 1/3 of an inch, I'd trust it to hold. I sure wouldn't let anyone walk on it, though. And so my reservation with all these calculations is that they're for a dead load only. Usually one has to design for a live load as well (folks standing on the thing). What does code say about arbors? Finally, and I'm much less sure of this, is modulus = bd^2 / 6 correct? If I integrate: integral(d/2, d/2) of b*x*dx I get b*d^2/4 If I try the integral(d/2, d/2) of b*x^2*dx, I get bd^3/12, just as you wrote, for moment of inertia. But perhaps there is a convention of which I am unaware. 
Subject:
Re: Beam span calculation
From: redhossga on 06 Apr 2006 07:40 PDT 
Very good iainmcclatchiega. I did use the wrong factor to convert to inlbs. I was lucky in this case that it made no difference in the outcome. In my defense, I used "w" (lbs/ft) rather than "W" (lbs) and didn't quite compensate. Thanks for keeping me honest. Yes, I would normally apply a live load. However, Jeff stated in his question, "no additional loads". I have never seen any code requirements for arbors. The value I used for section modulus is correct. What you have derived is the "polar section modulus" usually denoted as "J". 
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