reaction force?
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Posted by: hccedward ® 12/21/2009, 22:42:16 Author Profile eMail author Edit

Dear professional, Please see attached sketch and advise the reaction force at the question mark on sketch. P.S. assume all the member, bracket and connects are OK to transfer from pushing force 1.5KN to the reaction point mark on sketch. Thanks in advance. Edward

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: reaction force?
: reaction force? -- hccedward	Post Reply	Top of thread	Engineering Forum

Posted by: zekeman ® 12/22/2009, 15:14:52 Author Profile eMail author Edit

School question????

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Posted by: hccedward ® 12/22/2009, 23:10:07 Author Profile eMail author Edit

I am a designer of a curtain wall company. The engineer manager of my company make me crazy. She said the pivot point should be the connection point between the circular hollow and the bracket and told me the reaction force should be 1.5KN x (300mm/100mm)=4.5kn which I can no way to agree with. Could anyone provide a correct answer? Or tell me she is correct. Thanks.

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Posted by: zekeman ® 12/23/2009, 00:52:55 Author Profile eMail author Edit

Not enough information. You need to show constraints on teh right cylinder or forces there.

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Posted by: hccedward ® 12/23/2009, 01:55:51 Author Profile eMail author Edit

The drawing ly attached is the horizontal section of a sunshade blade which will be rotated about the center (the shaft at top and bottom) of the diameter 200mm cylinder when subjects to pushing force (1.5KN - converted from wind load). Assume the sunshade blade (two cylinders and plates) is rigid and free to rotate (nothing is holding the right cylinder.) I need to know the braking force at the point and direction as marked on the attachment. Thanks.

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Posted by: jboggs ® 12/23/2009, 08:40:00 Author Profile eMail author Edit

Its a simple force analysis. The 1.5Kn force at a perpendicular distance of 200 mm creates a moment: 1.5 x 200 = 300 Kn-mm. Your reaction force must create the same moment. Since the connection point is also 200 mm from the center, then the force at that point it is also 1.5Kn. But the direction of that 1.5Kn force is PERPENDICULAR to the line from the center of rotation to the connection point. Your line of action is not perpendicular to that line. The force along your line of action will be greater. Your sketch is missing some information. We know what the distance to the connection point is, 200 mm. But the line of action is not perpendicular to that distance, and we do not know the angle, so we do not know the perpendicular distance to the line of action. You must know that number to solve your problem. Then it is simple: Perpendicular distance 1 Force 2 ------------------------ = ------- Perpendicular distance 2 Force 1 You know what I mean by perpendicular distance, right?

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Posted by: hccedward ® 12/23/2009, 21:07:46 Author Profile eMail author Edit

understand. Thank you.

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Posted by: zekeman ® 12/23/2009, 10:03:17 Author Profile eMail author Edit

Your direction of the braking force should be opposite to that drawn. I scaled the brake force distance to the shaft centerline at 145mm, so assuming that the braking force is in the same direction as the 1.5KN force (as drawn) balancing moments, I get ?braking force =200/145*1.5KN= 2.06 KN Both forces are in the same direction so for equilibrium the reaction force at the shaft is the sum or 3.56 KN in the opposite direction of the other forces.

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Posted by: hccedward ® 12/23/2009, 21:11:17 Author Profile eMail author Edit

That's exactly the answer i looking for. Thank you very much!

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