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The section properties of wide flange beams can be found here: Link If we limit the deflection under load to L/240 the steel stress should be OK. So for a 28 foot span the allowable center deflection is 1.4 inches. Using the beam bending equation: Deflection = W L^3 / (48*E*I) For steel use E = 300000 0 L = 28 * 12 or 336 inches Solve for W W = 1.4 * 48 * 300000 0 * I / 336 ^ 3 W = 53.1 * I Looking up Ixx in the table, We find a W12x14 beam has an Ixx of 88.6 and could support 4705 pounds at the mid point. This is the TOTAL, so subtract out the dead weight of the trolley for useable capacity A W12x22 beam has an Ixx of 156 and could support 8291 pounds at the mid point (You can look up the others on your own. Measure the thickness and width of the flanges to figure out what size beam you have). Very important: The classic beam bending equation ignores the possibility of compression flange buckling. This is when the beam twists or rotates out of plane under load. In structures this is usually addressed by laterally supporting the compression flange, typically because it is built into a floor. This bracing holds the beam in one vertical plane and enables it to develop its full load capacity. A beam which is NOT laterally supported on the compression flange (the top flange in your case) supports far less total load before compression flange or lateral/torsional buckling limits it. So don't use the method above if you can't laterally brace the beam at (say) the 1/4, 1/2, and 3/4 points.
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