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Virtual work on rigid bodies 
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#1
Jul1207, 08:19 PM

P: 6

1. The problem statement, all variables and given/known data
Im having real trouble with drawing diagrams for virtual work. Can someone please, please help me! We were asked, in the diagram i posted, to find 1)vertical reaction at F, 2) all reactions at A, 3) bending moment and shear force at B using virtual work. i think i know how to do the actual working out but to do that i need the diagram, which i dont know how to draw. Any help/explanation would be very much appreciated! thanks in advance! 2. Relevant equations 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Jul1307, 08:12 AM

P: 6

the diagrams dont seem to be showing up.... how long normally does it take to be approved?



#3
Jul1307, 01:45 PM

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P: 3,220

To find the value of a statical quantity Q at a point x0, draw the displacement diagram of the system "caused" by a differential displacement at the point x0 "in the direction" of the quantity Q, and apply the principle of virtual work. (Sounds nice in theory, I know, but this is still a hint. )



#4
Jul1307, 05:59 PM

P: 6

Virtual work on rigid bodies
Dear Radou
Yes that is the general principal of virtual work as i understand it too but its very confusing to apply. Ive realised, since posting the diagrams up, that i made a few mistakes in my attempted solution. Since it takes so long for attachments to be approved maybe i can just describe in words what i think and hopefully u'll be able to follow... For the bending moment at B, Ive used the convention in my worked solution. However ive modified the diagram like this: The force acting on the beam will turn the beam clockwise, so it will slope down from the top of B to D, passing through C. From D to E it will be a straight constant line. The extension from D to F will be like it was in the original diagram. Thus sum bending moment at B=0 M(B) x d* + ()V(B) x 1.5d* + 10kN x 3d* = 0 M(B) x d* + ()(20kNm) x 1.5d* + 10kN x 3d* = 0 M(B) + 30 +30 = 0 M(B) = 60kNm NOTE: V(B) stands for the shear at B. i worked it out to be 20 d* is the angle between the virtual line and the real line However using statics i found the moment at B to be 30 not 60. Did i draw my diagram wrong? Or is it my working out? thanks in advance! 


#5
Jul1307, 06:27 PM

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#6
Jul1407, 03:44 AM

P: 6

We havnt learnt kennedy's theorem yet and googling it turned up nothing educational either....or at least nothing i could understand. Is there a simpler way about it?



#7
Jul1407, 05:22 AM

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#8
Jul1407, 05:53 AM

P: 6

Thanks a lot for going through this stuff with me! i know that it can be hard to explain....really appreicate it. 


#9
Jul1407, 07:33 AM

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Edit: your statics seems to be wrong too, since the moment you should obtain does not equal 30. 


#10
Jul1407, 07:10 PM

P: 6

So in the diagram, where did DF go? Or it doesnt move?
Yes using the diagram the moment at B should equal 15...my statics was wrong but i know wat went wrong now :) 


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