# Virtual work on rigid bodies

1. Jul 12, 2007

### hopelesskarma

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

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2. Jul 13, 2007

### hopelesskarma

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

3. Jul 13, 2007

### radou

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. Jul 13, 2007

### hopelesskarma

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. Jul 13, 2007

### radou

Your diagram is wrong. If you find it hard to picture the displacement of the system, you can always use Kennedy's theorem to draw the exact displacement diagram.

6. Jul 14, 2007

### hopelesskarma

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. Jul 14, 2007

### radou

Well, what I'm talking about is no hard way. First of all, do you know how to draw displacement diagrams for mechanisms?

8. Jul 14, 2007

### hopelesskarma

Yes, i think so. I know that the displacement diagrams can bend at pins but must pass through the roller/rocker supports etc, and remains zero at fixed supports. And initially it goes in whichever direction the force is acting.

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

9. Jul 14, 2007

### radou

No need to thank, that's why we're here!

Ok, http://usera.imagecave.com/polkijuhzu322/pf/displ.JPG" the displacement diagram of the system (for the bending moment at B). I named the disks "i" and "ii". Of course, you have to add the forces to the diagram and use the principle of virtual work now.

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

Last edited by a moderator: Apr 22, 2017
10. Jul 14, 2007

### hopelesskarma

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 :-)

11. Jul 15, 2007

### radou

The member DF isn't important, since there is no force to do any work on that member. (Formally, DF can't be seen in the diagram; it coincides with the projection of disc ii.)

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