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Angled Beam Calculations | |||
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Posted by: mtp9302 ® 09/14/2006, 10:47:29 Author Profile eMail author Edit |
Hi, I am writing to ask about some angled beam reaction calculations I am doing. The beam is a piece of strut at an angle theta overhung on both ends with asymmetric supports. It experiences a uniformly distributed load and is simply supported (attached to perpendicular strut with gussets). I know that the front reaction will support more weight than the rear reaction due to the slope, but I don't know how to express this mathematically. I have attempted to do so, but i got a difference between the front and rear reactions of 19% for a 3 degree angle. I was expecting something more like 1-5%, so I am questioning what I have done. I have attached the free body diagram and my calculations, as well as an explanation of the variables. Any assistance you can give would be much appreciated. Thanks for your time. -Matt
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Re: Angled Beam Calculations | |||
Re: Angled Beam Calculations -- mtp9302 | Post Reply | Top of thread | Forum |
Posted by: swearingen ® 09/14/2006, 16:33:50 Author Profile eMail author Edit |
This one throws many of my students off. Some comments: 1. Your figure is not in static equilibrium because there is a force from the w acting along your beam that must be resisted by one or both of your reaction points. Your reactions should be pointing vertically. 2. For a zero-depth beam, the vertical reactions at the ends would always be w/2. The only reason why the lower support "feels" more load in the real world is because the thickness of the beam moves the centroid of the load over to the lower support as you raise it. To understand this a little better, try starting with just a central point load. Draw it to scale, preferably in CAD. Make the beam 20' long and 10" deep. Now, keeping the applied central load vertical, tilt the beam to 45 degrees. If you measure the distance from the center of the top of the beam to the bottom corners of the beam in the horizontal axis, you'll see that the load moved closer to the lower end. Note that if you measure the center of the bottom of the beam to the ends, it hasn't changed, hence the zero-depth beam equality above. What I'm driving at is that the amount of load difference felt between the supports varies with the depth of the beam. For 3 degrees, and using my example of a 10" deep, 20' long beam, the reactions would be 49.7% to the high end and 50.3% to the low end. For reference, at 45 degrees it goes to 32.4% to the high end and 67.6% to the low. These reactions are VERTICAL. |
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