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Am I correct that I need to use the sum weight of the steel rod and the weight of the uniform load to calculate the total load on beam (W)?

https://engineersedge.com/beam_bending/beam_bending1.htm

I ask because the calculations do not seem to be correct using a load of zero (so only the weight of the rod) or very small loads like 1 or 2 lbs for a large span.

The purpose is for a 1.5" diameter steel curtain rod that needs to span 232" supported only on both ends, not in the center. The curtains will weight 2 Lbs in total, and so the deflection in the center with this uniform load will be mainly with the weight of the steel rod.

Mark Main
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  • It seems to me both the weight of the rod itself and the curtain are distributed uniformly (or near enough to uniformly) between the two supports, so you should just add both together to get $W$ for those formulas. But I don't think you need to count the weight of any part of the rod that hangs past the supports on either end. – David K Aug 18 '19 at 19:46
  • Thanks, that is what I was thinking as well because the formula does not provide results that make sense to me otherwise. At 1.5" diameter with this wide span, the schedule 40 offer the least sag (deflection) under its own weight compared the schedule 80 and a solid bar; this is true up to 34 lbs. distributed load, and then the solid round bar takes over and becomes the better choice. – Mark Main Aug 18 '19 at 20:40

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