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Posted by: John Hill ® 07/12/2010, 14:12:44 Author Profile eMail author Edit |
In all Deflection in Beam calculations, I find no case where there is one support in the center and either an evenly distributed load along the full length or equally concentrated loads at each far end, simulating a spreader bar. Can I use the cases where the beam is fixed at one end and the length and load are 1/2 of the totals, as if the remaing halves (length and load) are on the opposite side of the fixed point - as if the fixed point is the centered single lifting eye for the spreader bar and half the length and load is on each side of that lifting eye? Or is there another formula one must use? Thanks, John |
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Posted by: CFour ® 07/22/2010, 23:12:28 Author Profile eMail author Edit |
Spreader bars are NOT designed with only one "picking point." There should be at least 2 picking points on the spreader otherwise this is a safety hazard. Spreaders can have (and usually do unless designed for one specific reason) multiple attachments points where the load is connected in order to be used for various objects. Hope this helps. |
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Posted by: efendy ® 07/12/2010, 23:03:15 Author Profile eMail author Edit |
i think this is what you mean. i show you only the first case. the deflection you should be able to find it in engineering text book.
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Posted by: Kelly Bramble ® 07/12/2010, 19:24:52 Author Profile eMail author Edit |
You might post a picture of the spreader bar design you are refering to. We can then be sure about the loading configuration you are try to calculate The load configuration is defined by the static load points. The towing spreader bars I'm familiar with are best represented by a single point loading in the center and load point on the ends. |
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Posted by: Pinkerton ® 07/15/2010, 19:34:02 Author Profile eMail author Edit |
+1 for Kelly, same for my experience with lifting spreaders. Dave |
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