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 Subject: Roof load calculation Category: Science Asked by: gvjeep-ga 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.``` Request for Question Clarification by redhoss-ga on 26 Apr 2005 15:22 PDT ```I think that I understand what you have described. The 2x4's that run in the 24' direction support only the dead load of the ceiling (assuming I understand correctly). The 40 PSF that you call "roof loading" seems to me would actually be your ceiling load. You say the rafters slope up from all four sides and I assume that they would carry the roof load (transfering it to the outer walls). Do you plan on inhabiting the attic area or storing items there. If not, the 40 PSF is very conservative. Let me know if you agree with what I have said and I can help you size a beam.``` Clarification of Question by gvjeep-ga on 26 Apr 2005 16:40 PDT ```I?m really just looking for a little help understanding the problem. I thought that seeing someone else doing the problem would be a good way to understand. I?m not going to use the result to get a permit. I really don?t even need an answer just a method. Like I said in the original question I?ll probably hire an engineer once I decide exactly what I want to do. To clarify the original question, the crawl space is only about 4 feet tall at the highest point and won?t be used for storage. The house is in Sacramento, CA so I doubt that there is a snow load requirement. The rafters do slope up from all four sides.```
 Subject: Re: Roof load calculation Answered By: redhoss-ga 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``` Request for Answer Clarification by gvjeep-ga on 27 Apr 2005 07:32 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.``` Clarification of Answer by redhoss-ga on 27 Apr 2005 11:31 PDT ```First, I noticed a typo in the first paragraph of my answer "supporter" should read "supported". The two end walls each carry half the distance from the end wall to the center or 6 foot. This leaves the center support (wall or beam) to carry the remaining 12 foot (2 x 6 FT.). If you draw yourself a picture of what is happening, I think you can see that the center support does carry 12 foot of the load. In my question to you about whether your attic space was to be used for storage or living area I was questioning the 40 PSF loading. If you are not using it for storage, as you say, the 40 PSF sounds excessive to me. I found a website with some info that gives some numbers for different UBC live load requirements: http://www.awc.org/technical/spantables/tutorial.htm Looking at this it seems that a more reasonable number for live load might be 10 PSF (no storage) plus a dead load of 10 PSF for a total of 20 PSF. Of course when you consult with an engineer in your area I am sure that they will be able to tell you better than me what is normally done locally. Here is a website with a good explanation and drawing of what I tried to explain on how I determined the 12 foot number (see bottom of the page): http://64.78.42.182/sweethaven/BldgConst/Carpentry01/lessonMain.asp?iNum=030106 Let me know if you need more explanation or information and I will try to help.```
 gvjeep-ga rated this answer: `Thanks!! just what I was looking for`

 ```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```
 ```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.```