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solve the structure in the diagram below |
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| Feb24-10, 02:44 PM | #1 |
| Feb24-10, 03:52 PM | #2 |
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Mentor
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| Feb25-10, 01:23 AM | #3 |
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sorry, should have added names for the points,
the bottom left hand and bottom right hand circles are section joints- like a ball and socket joint- which do not at all resist to circular motion (no moment) the top joint is a half joint meaning - a permanent connection to the top piece and a ball and socket joint to the vertical bar. i am not sure what the correct term is in english hope that it is clear now |
| Feb25-10, 10:37 AM | #4 |
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solve the structure in the diagram below
There is a triangle between the joints, two of its sides being r, and the third is [tex]\sqrt{2} r[/tex]
As [tex]\sqrt{2} < 2[/tex] The thing won't collapse to the ground. Moreover (setting mechanics aside for a moment) as none of the sides of the triangle would change geometry, it won't even move. Now with mechanics. As the parts stretch, contract and bend, the system will move a bit. I am also courious how to calculate tension in non-straight beams. I feel I will learn something from this thread. |
| Feb25-10, 11:06 AM | #5 |
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agwas
mim not 100% sure i sunderstand what you are saying but i think you misunderstood the diagram, the bottom left hand support is a fixed support, lets call it A, so i have reactions Ax and Ay the second support -lets call it B, is not fixed and i only have reaction Bx, i think the straight beam can fall causing the arc to fall with it. |
| Feb25-10, 11:23 AM | #6 |
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I see.
And I suppose we suppose there is no friction between B and ground. |
| Feb25-10, 11:41 AM | #7 |
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correct,
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| Feb25-10, 05:30 PM | #8 |
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It may be badly drawn. It looks like A and B are both supposed to be hinged supports. But at B, there can only be a By force. I think Magwas would learn more from this thread if P was horizontal, because then, Ax would have a value. Really this is just a simply supported bent beam, isn't it?
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