How Do You Calculate Bending Moment Using Virtual Work Principles?

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laura001 said:
ok 1st one I've solved is Q5, i got that the reaction moment at A is

2.6133333 kNm... is that right?

Yes, seems correct. But do post all the results on one post. :smile:
 
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radou said:
http://usera.imagecave.com/polkijuhzu322/system1.bmp.jpg"

Ok, as said, the system is replaced with a mechanism, where a hinge is put in the place where you have to find the moment, and there is a couple of moments M added to that same place. Now, you have to construct an initial relative rotation between the two diskd connected by the hinge, so do it as is done in the displacement sketch. Now all you have to do is apply the principle of virtual work to get the moment M: [tex]F\cdot d_{F}+M(A_{1}+A_{2}) = 0[/tex], where A1 and A2 are the angles of rotation of the two disks.

I believe I am in the same course and am having difficulties with this problem. The imagecave server is down so i can't have a look at your diagram. If you would reupload the image if you have it I would be very grateful.

I want to ask certain questions that arose in my head from this incomplete understanding of your solution. I was wondering if you introduced a virtual displacement at x (the point where the hinge is added) and a virtual rotation is created between A and B, won't it also lead a displacement of the GIVEN hinge "s"? If this displacement were to exist won't we also have to add the virtual work carried out by vertical force acting on the hinge "s"?
 
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radou said:
M should equal -0.54 kNm, I checked on it. You obviously didn't do the trigonometry right. [tex]\frac{0.6}{4}=\frac{d_{F}}{4-2.2}[/tex], [tex]A_{1} = \tan A_{1} = \frac{3.4}{4}[/tex], and [tex]A_{2} = \tan A_{2} = \frac{0.6}{4}[/tex]. I forgot to point this out - A1 and A2 are differential rotations, so in the theory of small displacements you can use the identity [tex]\alpha \approx \tan(\alpha)[/tex].

Also I do not understand how one could possibly arrive at the above trigonometric relations. Maybe if I saw the diagram it would be clear, but to the extent of my understanding of your words, it does not make sense . I guess without the diagram I won't understand anything properly.
 
Ok, forget my first post and the question therein. I got how the problem works. However I still would like an answer to the question I posed in my second post, it would clarify my understanding of these things. I used some other trigonometric ratios. Thanks