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 Subject: rectangular steel beam sizing Category: Science > Instruments and Methods Asked by: akmudman-ga List Price: \$60.00 Posted: 23 Feb 2006 22:33 PST Expires: 25 Mar 2006 22:33 PST Question ID: 700258
 ```I currently have a project to frame in and finish the inside of my 24'x30' "pole barn" construction shop. I need some help with sizing a rectangular steel tube to span the 23' 1/4" (inside dimension) width & help support an insulated sheetrocked ceiling. The beam in question would fit in the 3" void between two wooden trusses notched into either side of the 6 in square support poles (spaced on 10' centers along the length of the building), and would have to be tied into the poles & through bolted to the existing trusses. The wooden trusses have 2x6 top & horizontal members with a 2x4 "W" bracing. Perpendicular to the trusses would be 2x6 headers space 24" on center and 5/8 sheetrocked. There would be no storage above this ceiling, so a 10# live load & a 5# dead load would only be required. With a 19' 2" length, 23'1/4" width, and the beam centered in the width at 8' 7", I calculated the total load of the beam to be just over 3300". If possible include information on bolt size, grade, and spacing to attach to the trusses & poles.``` Request for Question Clarification by redhoss-ga on 24 Feb 2006 07:20 PST ```I have read your question several times and drawn a sketch of what I think your pole barn looks like. I guess that you have 8 wooden trusses and that the rectangular steel tube would be 3 inches x some height (to be determined) and would be a straight section with connection plates welded on both ends for through bolting to the columns. There are 4 columns down each 30' side so there would be 4 of these wood truss/steel tube beams. If this is correct, then I would say that the 2 inner beams would support 10 x 24 x 15 = 3600#. The 2 end beams would support 5 x 24 x 15 = 1800#. The end beams could be different (less depth) than the center beams if you wish. Does this accurately describe your situation. If so, I can have you an answer back soon after your reply.``` Clarification of Question by akmudman-ga on 24 Feb 2006 12:04 PST ```This is definitely hard to describe having no drawing, so I can see your difficulty. There are two trusses spanning across on each of the two inside poles and a single truss on each end wall. Each of the end walls are framed with the proposed headers resting on these. There is also a framed room 10'x11.5'in the back corner that has an end wall that is supporting half the distance of one of the two sets of the double trusses. Across the other half the distance there are 2-2x10's sandwiched inside the trusses (nailed & 1/2" grade 5 bolts every 2') and supported by a steel bracket on the outside end. The 2-2x10's by a beam chart would support 2850# across a 12' span. If this is adequate, I would leave these in place. The steel beam in question would fit inside the the other set of double trusses with 2x6's perpendicular, and attached with hangers to the reinforce or supported double trusses on one end and the framed outside wall on the opposite side. I hope this clarifies this for you.``` Clarification of Question by akmudman-ga on 24 Feb 2006 13:38 PST ```After reading my reply there appears to be some of your questions that were not answered. Yes there would be attaching plates to welded on to both through the columns for any 3" wide steel beams used. Available height for the beams is 14". The reinforced double truss not supported by the room wall would have to support roughly 1800# or 1/2 the area of 12'x 20' and the side supported by the room is 1/2 of 10'x12'. There are 4 poles along the lenth of the side as you assumed with double trusses on each of the two interior poles (square 6"x6"). There is a 4" concrete rebar reinforced floor throughout the building```
 ```Okay akmudman, I understand exactly what you are building. Here are the calcs for your beam: First we will calculate the maximum bending moment and required section modulus: M = wl^2/8 Where w = 15 PSF x 10 ft = 150 #/ft M = 150 x 24^2 / 8 = 10,888 ft# = 129,600 in# S (section modulus) = M /s Where s is the allowable bending stress = 36,000 psi x .55 = 19,800 psi S = 129,600 / 19,800 = 6.54 in^3 Next we will calculate the moment of inertia required to limit the deflection to (beam length)/360 = 24 x 12 / 360 = 0.8 inch I (req'd moment of inertia) = (5wl^4/384 ED) x 1728 Where E is a constant for steel = 30,000,000 psi D is deflection 1728 is a constant to keep the units in inches I = (5 x 150 x 331,776 / 384 x 30,000,000 x 0.8) x 1728 = 47 in^4 Here is a chart of rectangular tubing properties: http://www.chicagotube.com/stockbook/structure/elements_rect.html Looking for a tube that has the I and S we need and also has a width of 3 inches we find: 9x3x1/4 S= 11.4 I= 51.1 8x3x3/8 S= 12.7 I= 51 Either of these would work. The 9x3x1/4 would be the better choice since it weighs less per foot. Now for the through bolts: The shear load per end is 3,600# / 2 = 1,800# A 1/2 inch bolt has a shear area of 0.196 sq in The allowable shear for an ASTM A307 bolt is 10 ksi. The allowable shear load for one bolt is 10 x .196 = 1,960# So, one bolt would actually be enough. However, two would make for a better connection and we need one top and bottom to prevent the beam from twisting. The plate itself could be made from 1/4 in flat bar or plate. I think that this answers your question, but if you need any further explanation or info please ask for a clarification. Good luck with your building, Redhoss``` Request for Answer Clarification by akmudman-ga on 25 Feb 2006 13:00 PST ```This looks to be just the information that I was looking for, but there are a couple of details that need some clarification. Are the ASTM A307 bolts a grade that can be requested at the local fastener store or does this cross over to something in the range of a grade 5 or 8 bolt? Also, are the stress & bending calculation inclusive of the weight of the beam itself? Thank you for getting back so quickly and appreciate your help in this matter.``` Clarification of Answer by redhoss-ga on 26 Feb 2006 07:38 PST ```You caught me in a little bit of laziness, which I will admit that I am prone to. No, I didn't include the weight of the beam in the calculations. Normally when a designer calculates roof members they would use a nominal dead load that includes an allowance for the members themselves. In your case I should have probably added more because your 5 PSF dead load only includes the actual ceiling and framing materials. However, the weight of the beam itself is somewhat insignificant if you consider that the 150 #/ft load that I used would become about 170 #/ft if we included the beam. The percentage difference being 150/170 = 88%. I hate to destroy your faith in the entire engineering profession, but most of the standard formulas used are not 100% accurate. That is the reason that the term "safety factor" exists. The safety factor in these calculations is that we only used 55% of the published yield strength for the beam material (.55 x 36,000 psi = 19,800 psi). To be more exact you can use what is called an iterative solution. Since we had no idea what beam we would be using when we started we could have assumed a value, solved the equations, and then plugged the beam weight we calculated back into the equations again until the answer came out more exact. I hope all this hasn't bored you too much, but I just wanted to explain why I wasn't too worried about the actual beam weight. However, I will solve the equations using 170 #/ft just to see how much difference it makes. First we will calculate the maximum bending moment and required section modulus: M = wl^2/8 Where w = 15 PSF x 10 ft = 170 #/ft M = 170 x 24^2 / 8 = 12,240 ft# = 146,880 in# S (section modulus) = M /s Where s is the allowable bending stress = 36,000 psi x .55 = 19,800 psi S = 146,880 / 19,800 = 7.41 in^3 NOTE: This would make no difference since both possible beam choices have an S value greater than 7.41 Next we will calculate the moment of inertia required to limit the deflection to (beam length)/360 = 24 x 12 / 360 = 0.8 inch I (req'd moment of inertia) = (5wl^4/384 ED) x 1728 Where E is a constant for steel = 30,000,000 psi D is deflection 1728 is a constant to keep the units in inches I = (5 x 170 x 331,776 / 384 x 30,000,000 x 0.8) x 1728 = 53 in^4 NOTE: We would still choose the same two beams. So, now I feel better and hope that you do also. Now on to the bolt question. ASTM A307 is equivalent to Grade 2. Very likely the bolts you buy at the hardware store will be either Grade 2 or 5. Thanks for keeping me honest, Redhoss```
 akmudman-ga rated this answer: ```This was just the information that I needed. Redhoss answered in a timely manner and clarified details of the answer. I decided to use this forum primarily after viewing some of his answers.```