Virtual Work. What is it and how does it work? - Straight Dope

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08-23-2002, 09:04 PM

#1

Cubsfan

Join Date: Jul 2002 Location: Ohio Posts: 5,190



Guest Virtual Work. What is it and how does it work?

I am in a Statics and Dynamics course and we are covering Virtual Work right now. Actually, I Am independant studying the course so I am trying to teach myself. I am having an extremely hard time figuring out how to use virtual work to solve statics problems. Is there anyone out there that can explain this concept to me in easy to understand terms? Why do we use virtual work? How do we know how to set up a statics problem using the method of virtual work? Why does virtual work work when regualr equations of equillibrium don't? I havent even been able to set up a problem using this concept because I am totally lost where to begin. I do have a diagram of the problem that I am working on that I could post to my site if anyone wants to get a look at it. Any help would be greatly appreciated. Reply With Quote 08-24-2002, 05:08 AM

#2

Jabba

Join Date: Mar 2002 Posts: 886



Guest

The principal of virtual work states: suppose that a body is in equilibrium under a system of forces F1,...,Fn and the particle undergoes a displacement during which the applied forces remain constant in magnitude and direction. Then the sum of the works done will be zero. An example may help to clarify. Suppose we have a particle of weight W at rest on a plane of slope q. The forces acting on the particle are its weight vertically downwards, a normal reaction N perpendicular to the plane and a frictional force F acting up the plane. Now imagine that the particle moves a distance x up the plane ( we call this a virtual displacement). If the particle were actually to do this the frictional force would change direction and act up the plane; also we would need an additional force to bring about the motion. We ignore both of these facts and consider only the original forces ( this is why it's virtual). The reaction N, being perpendicular to the direction of motion, does no work. The frictional force does work Fx. The component of the weight acting down the plane does work -Wx sin q. Thus, by the principal of virtual work, Fx - Wx sin q = 0 i.e. F = W sin q Now consider a displacement y along the normal to the plane. F does no work, N does work Ny and W does work -Wy cos q and so as above N = W cos q Of course this example is trivial without using virtual work but it illustrates the principal. If you have a specific example in mind then yes, a link would help. Reply With Quote 08-24-2002, 05:27 AM

sailor Registered User

#3 Join Date: Mar 2000 Location: Washington dc Posts: 16,441



>> The principal of virtual work states >> but it illustrates the principal The principle. The principle. The principle (yes, I *have* been diagnosed as anal-obsesive-compulsive-repulsive) Reply With Quote 08-24-2002, 05:46 AM

Jabba Guest

#4 Join Date: Mar 2002 Posts: 886



Damn. So am I usually. That'll teach me to get up so early. Reply With Quote

08-24-2002, 06:37 AM

#5

Cubsfan

Join Date: Jul 2002 Location: Ohio Posts: 5,190



Guest

This link will take you to a problem I am trying to work out of the book. Unfortunately I do not have an answer key availible to me to check my work. My library doesn't stock it. http://www.beefchips.com/Statics2 I must use virtual work to solve this problem. The details of it are on the web page. When you use virtual work do you pic an arbitrary point and riection to use as the origin of the x and y axis? This is so confusing. Reply With Quote 08-24-2002, 07:17 AM

Jabba Guest

#6 Join Date: Mar 2002 Posts: 886



I am not an engineer and so do not have much experience with problems like this. I assume the beams are light. What about forces at B? Do we assume they are zero? If so I think I've done it. I get Q = 100/3Ö3 ( consider what happens to the point D if q is increased slightly). Reply With Quote 08-24-2002, 03:17 PM

Achernar Guest

#7 Join Date: Aug 1999 Location: 23 male, Boston Posts: 5,791



Quote:

Originally posted by Jabba a body is in equilibrium under a system of forces F1,...,Fn and the particle undergoes a displacement during which the applied forces remain constant in magnitude and direction.

Isn't the body being in equilibrium an equivalent condition to the F's summing to 0? Reply With Quote 08-24-2002, 04:01 PM

#8

David Simmons

Join Date: Nov 2001 Posts: 12,684



Charter Member Quote:

Originally posted by Stinkpalm This link will take you to a problem I am trying to work out of the book. Unfortunately I do not have an answer key availible to me to check my work. My library doesn't stock it. http://www.beefchips.com/Statics2 I must use virtual work to solve this problem. The details of it are on the web page. When you use virtual work do you pic an arbitrary point and riection to use as the origin of the x and y axis? This is so confusing.

Do you have to use virtual work to solve this because the course instructions require it? Resolve the force P = 100N into one component along the beam and one perpendicular to it. The one along the beam only stretches the beam and causes no rotation. The one perpendicular is P(sin 30o) = 50N. Resolve force Q into the same two components. The one perpendicular to the beam is opposed to the equivalent P component and equals Q(cos 30o). So Q(cos 30o) = 50N or Q = 50/.866 = 57.73N unless I am completely screwed up. At least this is a check for your answer by virtual work. Reply With Quote 08-24-2002, 04:43 PM

Achernar Guest

#9 Join Date: Aug 1999 Location: 23 male, Boston Posts: 5,791



I don't think that C is a pivot point, so you can't ignore the parallel component of P. Isn't C a slidey-pointy-bolt thing? (Sorry that I'm not used to Engineering Physics.) Reply With Quote

08-24-2002, 05:21 PM

#10

David Simmons

Join Date: Nov 2001 Posts: 12,684



Charter Member Quote:

Originally posted by Achernar I don't think that C is a pivot point, so you can't ignore the parallel component of P. Isn't C a slidey-pointy-bolt thing? (Sorry that I'm not used to Engineering Physics.)

I think you are right. After I posted I got to thinking that unless the point C could move the problem didn't make sense. So, everyone, just ignore my solution and go on without me, at least for the time being. Reply With Quote 08-24-2002, 08:32 PM

zut Charter Member

#11 Join Date: Apr 2000 Location: Detroit, MI Posts: 3,724



To use the principle of virtual work, apply the procedure that Jabba outlined. Let's apply it to your linked-to diagram. First step is to imagine the linkage moving just a tiny little bit. "Tiny" means "such a small distance that all the relative angles don't change." Now, the question is just exactly which direction does everything move? It's really just geometry, but it takes some thinking through. Let's step through the linkage: Point A doesn't move at all (of course). Point B rotates around point A. You can express this by saying that if point B moves a distance dx to the right, it moves a distance cos(60)dx downwards. Point C moves only horizontally; by symmetry it will move a distance 2dx to the right. Point D is a little harder to figure out: go through the geometry by imagining the entire beam BCD translating the same distance that B does, and then rotating about B whatever distance it takes to return C to the same horizontal line. When you go through that exercise, if I did the math right, you'll find that point C moves a distance 3dx to the right and a distance cos(60)dx upwards. Note that the value of dx is arbitrary, so you might have different numbers for the motion, but the relative motion of all the joints ought to be the same. OK, now that you know the geometry, apply the principle of virtual work. In a nutshell, this says: the sum of all forces multiplied by the virtual distance they move, in the direction of the force, is zero. Let's look at each point: Point A: Has some applied forces, but it doesn't move. Virtual work = 0 Point B: Moves, but all the forces are internal, so they cancel each other out. Virtual work = 0. Point C: Has a vertical force, and moves a distance 2dx horizontally. No motion in the direction of the force. Virtual work = 0. Point D: Moves 3dx in the direction of force P, and cos(60)dx opposite force Q. Virtual work = P*3dx - Qcos(60)dx. Add 'em up, and find that = P*3dx - Qcos(60)dx = 0. You can do the final math. Note that dx drops out. You can also solve this problem by summing forces and moments, and indeed I get the same answer (good sign!), but it's a little more involved. Reply With Quote 08-24-2002, 09:16 PM

Cubsfan Guest

#12 Join Date: Jul 2002 Location: Ohio Posts: 5,190



Great explaination. The part that was confusing me was the distance of dx, and how the geometry was used to set up the problem. One question though, where did you find that the distance the point B moves is cos(60) downwards? Is it not cos(30) downwards? I am just not seeing where the 60 came from. And are you also telling me that the given length of 2 for each section is completely unused when solving this problem? Thanks for the help. The book is very wordy and tought follow for me. Reply With Quote

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