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 Author: deajohn5556 Time: 12/14/06 10:26 PST This is a reply to message no. 18259 by timwpa Reply | Original Message | New Topic | List Topics | List Messages on This Topic
 Current Topic:Beam Loading Question Tim It’s been awhile (30 years) since I used the steel codes, but try this:  If a beam is to carry a load over a 28 foot span, or any span, its defection is  Y = 5wL^4/384EI       where L is span in inches, w is load in pounds per inch, E is modulus and I is beam inertia  Its stress is   S = Mc/I = wL^2 (depth/2)/8I  This is worst case and assumes pinned end.  Total load for floors is 60 pounds per sq ft typically  Now assume the beam carries half the floor load and the walls the other half.  The beam carries   28ft  x 30ft  x 60lb/ft^2 / 2   or 25,200 pounds.  This is 25,200/12/28 = 75 lb/in  I don’t have tables for inertia of your S beam, but it is roughly 30-50 in^4  Substitute into the equations above and find, using I = 30 and E = 30e06,  Y = 13.8 inchesS = 141,000 psi  Code allows S = 24,000 and Y= 1/360 span = .90 inch so both fail and the house falls down.  That is why you need posts.  When you add posts, the beam equation is indeterminate but you can use tables.  For worst case you can assume the longest span unsupported and use the equations above.  For two posts, nine feet apart, the max span is 10 feet and now you have  Y = .22 inch S = 18,000 psi so both values meet the code.  Now move the span to 15.5 feet and   Y = 1.3 inch S =  43000 psi and both criteria fail!  You may be asking for trouble, although the calculation is conservative.  If there is more than one S beam, things improve of course.  But I would not fool with the span without a review by a professional engineer who is independent of the building inspector