6/3/2023 0 Comments The truss wizard program 3.0Otherwise, if (for example) the bridge is wide enough that the car may be closer to one truss than the other, then more of the live load will go to the near truss. If a car is driving along your bridge, will it be perfectly centered (or close enough)? Then sure, divide the live load evenly between both trusses. ![]() If the structure is transversally symmetrical (that is, there is nothing that'd draw more loads to one truss than the other), then that is a perfectly valid assumption for all the dead loads (paving, etc).įor the live loads, however, a bit more care may be required if discrete moving loads are considered. ![]() The division of the remaining loads among the trusses, however, is a bit more circumstantial.Ī reasonable first hypothesis would be to get the total load and divide it evenly between both trusses. Each of the trusses must obviously support its own self-weight. The first question is how to determine the loading applied to the truss. The real question, of course, is how to do this well. After all, how do you think engineers calculated complex structures before the advent of the computer? The answer to your question "whether or not I could analyse a 2D simplification and still come to a somewhat credible conclusion" is obviously yes.
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