

Subject:
Roof load calculation
Category: Science Asked by: gvjeepga List Price: $40.00 
Posted:
26 Apr 2005 11:06 PDT
Expires: 26 May 2005 11:06 PDT Question ID: 514493 
I was wondering if someone can help me analyze my need for a structural beam in my remodel project. I will probably hire an engineer later I just want to see if I can get an idea about what I will need. First this room is the entire second floor of a 2 story house built in the 1920?s. The 2nd floor is currently 24?x15? split down the middle with a wall. The plan is to remove the wall and open up the room. The ceiling joists are full dimension 2?x4? spliced in the middle, overlapped about 4? running the 24? direction. The rafters slope up from all four sides creating a short peak in the center. What I would like to do is run an engineered beam across the top and perpendicular to the ceiling joists and use hangers to hang the 2?x4? for support. Or worst case a beam below the ceiling joists. I would like to know how to calculate the load the beam must carry, and what kind of beam I would need to carry that load. I?m hoping I can use two 2?x8? nailed together since I already have the lumber. Information: Room dimension  24?x15? Span of Beam ? 15? For this, let?s assume 40 lb/sf for roof loading. I?m still trying to figure out the UBC but as far as I can tell that is the requirement.  
 


Subject:
Re: Roof load calculation
Answered By: redhossga on 26 Apr 2005 20:54 PDT Rated: 
Hello gvjeep, I have worked for years in the metal building industry and am very familiar with this type of problem. What you describe is a simply supporter beam with a uniformly distributed load. The length is 15 feet and it supports 12 foot of the 40 PSF load. The formula for the maximum bending moment, which occurs at the center of the span, is given by: M = (w x L^2)/8 where w = 40 PSF x 12 FT. = 480 LB./FT. M = (480 x 225)/8 = 13,500 FT.LB. = 162,000 IN.LB. You don't say what type of wood you plan to use, so we will assume "Western white pine" which has an allowable bending stress of 800 PSI. We will try two 2x8's as you suggest. The section modulus for a single 2x8 is 13.14 IN.^3 or doubled we have 26.28 IN.^3. Bending stress = Max. moment / section modulus = 162,000 / 26.28 = 6,164 PSI Comparing 6,164 to our allowable bending stress of 800 PSI it is obvious that we will have to use a larger beam. Solving the above formula for section modulus we get: Section modulus = Max. moment / Allowable bending stress = 162,000 / 800 = 202.5 IN.^3 The formula for section modulus is: S = (b x d^2)/6 Assuming a 6 inch wide beam and solving for d we get: d = sqrt(s) = 14.23 IN. So you can see that we are talking about a fairly large beam. There are many other factors to be considered when dealing with a wood beam. However, this will get you in the ballpark. One thing you might want to consider is using a glulam beam. Also, since the beam will be hidden (if you hang the 2x4's underneath) you could use a steel beam. I hope that this hasn't confused you too much and you are able to follow each step. If you have further questions, please ask for a clarification and I will try and answer. Good luck, Redhoss  
 

gvjeepga
rated this answer:
Thanks!! just what I was looking for 

Subject:
Re: Roof load calculation
From: toufarooga on 26 Apr 2005 14:10 PDT 
Most engineers will be reluctant to answer this question for a few reasons. First of all, the Professional Engineers act of many states makes it unlawful to give out such advice unless you are a PE. Next, there are quite a few liabilities associated with this matter. I suggest you consult a local registered professional civil engineer for information on this project. The scope seems simple enough, but there are way too many other factors that need to be taken into account, such as local conditions, live loads, and so on that change from region to region. For such project, a professional engineer probably wouldn't charge much more than a couple hundred dollars, which isn't that much more than the $40 you are offering here. You'd be getting peace of mind knowing the work were done correctly. If you are insistent on doing the calculations yourself, then you should go to the local library and reference a copy of the current UBC along with a textbook on Timber Design, along with any applicable local building codes. Just so you know, the 40 psf number you gave is a live service load number only. You also need to add a number of other loads, such as the roof dead load, snow load, seismic load, among others 
Subject:
Re: Roof load calculation
From: gvjeepga on 27 Apr 2005 07:31 PDT 
I?m able to follow your calculation for the most part. One question I have is about you first moment calculation. How do you decide how much of the 40 lb/sf load the beam caries? You say w = 40 psf X 12ft = 480 ft/lb. Where do you come up with the 12 ft? The way I was looking at it was: 40 lb/sf x (15 ft x 24ft) = 14,400 lb total load from the roof. Then trying to figure out how much of that load each wall carries. Maybe I?m missing something. Thanks for the answer it really helped out. 
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