huangyhg

identificaiton of beam requirements
i am trying to calculate the size of a universal beam that i will need to replace an old wooden beam.
it is a simply supported beam that needs to hold up a uniformly distributed load of around 1.5 tonnes, over a length of 14.5ft. the beam is supported at either end by an upright column.
i have calculated this weight using the following:
surface area of slate & wood roof = 103 sq ft @ 16lbs / sq ft
= 1624 lbs
surface area of wood panal and partially glazed wall = 100sq ft @ 6lbs / sq ft =609lbs
total weight = 2233 lbs
safety factor = 2233*1.5
= 3349lbs
i have tried using bm = wl^2/8 and d= 5/384 x wl^3/ei , but i am not having much luck as i am getting figures that say a 203 x133 will be sufficient but this seems light.
any help would be greatly appreciated. if you could show me the calculations it would be fantastic so i can see where i am going wrong.
rob

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even the dead loads seem a bit light, but i don't see you using any live loads. be it interior or exterior application, you use some live load. since you name slate one would think it is a roof and hence most likely would settle the live load more or less equals at the worst case live load, this with a significant snow load. so look what th snow load needs be at the location, and if too weak, put alternatively anyway some live load for repairs. almost any code on loads has provision for this.
i'd be carefull about the mixed units ... 3349lbs is a running load of 231 lbs/ft = 3.37n/mm. so the moment is calculated to be 3.37*(14.5*12*25.4)^2/8 = 8.2e6nmm ... is that the capacity of your section ?
i assume that a universal beam is similar to a wide flange or i beam. you have not told us the properties of the beam, so it is difficult to respond intelligently to your question.
if w represents the uniform load per foot, then:
m = wl2/8
d = 5/384 * w*l4/ei (contrary to your formula)
if w represents the total uniform load on the beam, namely wl then:
m = wl/8
d = 5/384 *w * l3/ei
the bending strength of a steel beam depends on the lateral support of the compression (top) flange. but for such light loading as you are assuming, a section of 200 x 133 would likely be adequate.
the factored maximum moment is 3349 * 14.5/8 = 6070'# (8.22kn-m)
deflection need not be checked based on experience, but if you are going to check it, keep the units consistent.
according to my steel handbook, w200x19 has a resisting moment of 58.1 kn-m when the top flange is continuously supported and 16.4 kn-m when supported at 5m centers.
w200x19 is 200mm deep, has a flange width of 102mm and weighs 19 kg/m. its imperial equivalent is w8x13 which means it has a depth of 8" and weighs 13 pounds per foot.
ba
hi everyone, thaks for your input. looking back i should have included a little more information.
in my original calculations i applied a 40psi load for snow and wind, but i forgot the live load of individuals in the room itself.
i also used d = 5/384 *w * l3/ei instead of d = 5/384 *w * l4/ei, which would explain why i got a lighter value for the inertion.
the i-beam is to replace an old wooden beam, so i am trying to define what size of beam i would require, hence why i did not incldue properties of the beam.
i have carried out my calculations in imperial purely becasue i have an old structural steelwork handbook. i only stated the 1.5 tonne uniform lode in metric as i assumed that is what everyone would recognise.
if the factored moment is 8.22kn-m without the live loads, if i add in my live loads to get a new bm, then this should give me a more realistic value. should i then take that total bm value and select a beam smaller than the w200x19 depending on ym asnwer?
thanks again, and apologies for my ignorance, i studied a small amount of structural at uni, but 15 years on it seems to elude me !
rob
apologies, "40 psi" should read "40lbs sq ft".
thanks
gems:fff"> what are the cross-sectional dimensions of the existing wood beam?

ralph
structures consulting
northeast usa
hi rhtpe
the existing beam is 50mm x 150mm, however ti used to be supported at 7ft centres, and we want to be able to remove the beam, hence the new 14ft span.
thanks
rob

existing wood beam is only one 2x6? your safety factor? are you sure about tributary areas and loads? why not laminate wood, and make a built up wood beam to avoid trying to get a steel beam in there in the first place.
looks like a residential diy. d is ??
gems:fff"> do you really want to try to muscle in a 200+ pound steel beam in amongst the temporary supports you will have to install to support the floor joists (or rafters) until the replacement is complete?
i'm with beton1fff"> - install some temporary supports to take the load off the existing beam and laminate some additional lumber to it. perhaps build a flitch beam in-place.
the difficulty with these kinds of questions is that the folks you are asking advice from cannot see what you have seen, and we only get a small piece of the puzzle. which leads to additional questions, and by the time we converge on a possible solution, the complexion of the challenge has changed dramatically.
a picture is worth a thousand words ....
ralph
structures consulting
northeast usa
another idea is to use a channel each side of the existing beam and bolt it through, then remove the central column. before doing that, however it would be wise to ensure that the two remaining columns and foundations are capable of safely sustaining the entire load.
ba

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