huangyhg

moving point load
i have been checking some existing joists to see if can add a concentrated load. the size of the load is known, just not its exact location on the joist. i decided to put the point load in the center to produce the largest moment and then backed out the equivalent uniform load to compare to the joist tables. then i applied the load at the end of the joist to determine the maximum shear and backed out the uniform load again for comparison to the joist load tables. i realized after a couple of these calculations that i was coming up with the same equivalent uniform load for worst moment and worst shear for moving point load. here are the equations:
worst moment - point load in the center
w = (p*l/4)*8/l^2 = 2*p/l
worst shear - point near support
w = 2*p/l
i am missing something? it just doesn't seem correct. i can accept the math just having trouble with the concept. according to this all i need to do is multiply the point load by 2 and divide by the joist span and then add the resulting line load to the existing loads, compare to the joist load table and if the new load is less than the load listed in the table, the joist is adequate. i realize that some of the web members may be smaller in the center than the ends. lets assume the joist has a constant cross section and local bending of chord members is not an issue, like a w-shape beam. am i correct that doing this checks worst moment and worst shear of a moving point load in one calculation?
thanks for any input
jechols - refer to diagram 40 (simple beam - one concentrated moving load) of "beam diagrams and formulas" in the aisc manual of steel construction - page 2-310 in my copy of 9th edition (asd). since you are assuming that the joist is acting like a w section beam, the math presented here should apply. as you state, the joist is not a wf beam, but this diagram makes a good starting point.
this diagram, along with #41, #42, and general rules for moving concentrated loads are easily overlooked, but can be quite useful.
you should think in terms of shear and moment envelopes and not just look at the load at midspan and the load at the ends.
for example, suppose you move your p load out about 10% from the end. the shear diagram for this is a block where 90% of the shear is at one reaction and the other 10% is at the far end. the diagonal shear diagram formed by a uniform load will certainly cross below this point-load shear diagram and the equivalent w is higher than that derived from your equation above....hope this makes sense. just try graphing out a few shear diagrams for the p load at 10, 20 and 30 percent of the span and you'll see that when you try to draw a symmetrical "bowtie" shear diagram for a uniform load you'll get really really high shears as the concentrated load approaches the center of the span.
jechols,
can that point load be applied anywhere on the joist chord? or just on the panel points? if its off of a panel point, then i think there is some more checks to be made, like bending of the chord (bottom chord right?) on a length from panel point to panel point. also, consider the detail of the connection to the joist chord- it must be something that does not create a local bending problem in an angle leg, or a buckling problem.
regards,
chichuck
jechols - in order to check and see if an existing joist can take a point load at any location the following procedure should be used:
1. draw an applied shear envelope by adding the applied shear influence envelope (top line from +p to 0 on one end and lower line from 0 to -p on the other end) to the currently applied uniform load shear diagram.
2. superimpose the applied shear diagram onto the allowable shear diagram. the allowable shear diagram will range from vallow= wl/2 to -wl/2 where w equals the allowable uniform load. note however that joists are typically designed to have a minumum vallow of wl/8 even in the center where the allowable shear based on allowable uniform load would technically be zero.
any web member located where the applied envelope is greater than the allowable envelope will need to be reinforced. just checking the maximum shear of p at the end is not adequate as different shear diagrams result from different locations of applied p. these diagrams may be adequate when compared to the allowable on the end but not nearer the center.
3. in order to check the top and bottom chords the procedure is a bit easier as typically these
thanks for responding everyone,
i don't think i explained my question very well. lets consider the member is some undefined shape and is unloaded. i have a load called p that i want to be able to support anywhere along the
sorry what i meant to say is "these two equivalent loads are always equal to each other regardless of span length or size of p"
thanks
jechols, i don't think there is a direct flaw in your logic - your math and formulae are correct as far as they go.
what some of us are saying is that the critical shear for your undefined shape may not be at the end of the
only flaw that i can see is trying to determine a equivalent uniform load from the second case where p is applied near one support to obtain the maximum shear. moment will be zero, therefore equivalent uniform force will be zero.
two flaws;
1.) equation finds equivalent dist. load when point load is applied at mid span. not a moving case! you cannot assume this "equivalent" load will be accurate when the point load is somewhere other than the midspan. the long way to do what you are trying to do is you need an expression for moment,shear, and any other information for your design that relates moments to displacements, shears with displacements, and etc..then with excel, for example, run some interval to see which case is the worst.
2.) why do you want an equivalent load (w)? use an influence line, determine worst case scenario, do the checks, i.e shear, bending, deflections...etc.. check against allowable/lrfd/aci what ever the case may be. and size the member accordingly. are you havng trouble determining stress from loads?

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