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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.
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?
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.
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].