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Algebra questions

  1. Mar 1, 2013 #1
    HI all

    I have Two questions regarding algebra
    ,they are example in book, but the book simplify them quickly,and i dont understand how
    I just write them and upload the picture
    see below
    http://img268.imageshack.us/img268/5458/img051lg.jpg [Broken]

    Thank you
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 1, 2013 #2

    SammyS

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    The first one, is not so much an algebra problem, as it is a limit problem.

    Division by zero is undefined, so W is undefined for r = 0 .

    What they have called Wmax is actually
    lim r→0 W(r),​
    where W(r) is a function of r.


    Why integrate the second one?

    Take the indicated derivatives and solve for [itex]\ \alpha\ .[/itex]

    Added in Edit:

    After looking at these expressions more thoroughly, I suspect that there is some additional information that you haven't shared with us.

    For the first expression, [itex]\displaystyle \ \ W=\frac{Fr^2}{16\pi D}\left(2\ln\left(\frac{r}{R}\right)-1+\left(\frac{R}{r}\right)^2\, \right)[/itex]

    If we take W as being in cylindrical, or spherical coordinates, then W has a removable discontinuity at r = 0. If we assign W(r=0) a value equal to [itex]\displaystyle \ \ \lim_{r\to 0^+}W(r)\,,\ [/itex] then W is continuous at r=0. Furthermore, with this extension, W has a relative maximum at r = 0.

    I also suspect that this expression for W is only valid for r ≤ R .


    For your second question:
    I suspect that some integration is involved, but we are missing too much of the context of the problem to be of much -- if any -- help.
     
    Last edited by a moderator: May 6, 2017
  4. Mar 2, 2013 #3
    Thank you very much.
    I uploaded two pictures the first one is regarding the first question and the second one is regarding the second.
    You can zoom them to see more clearly.

    Appreciate your help ,to better understanding.

    Thank you again
     

    Attached Files:

  5. Mar 2, 2013 #4

    SammyS

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    \frac{}{}
    The two pages you posted were helpful, and I may be able to answer your questions to some extent, but some information is still lacking.

    I see by the chapter title in the page corner, that these questions refer to the bending of a thin circular plate. Therefore, I presume in these expressions, [itex]\ r\,,\ [/itex] is the radial coordinate in a cylindrical coordinate system. ... etc. ...
    Anyone helping you with such a problem, shouldn't be expected to be digging such information out, just to help you. You should be supplying such information.​

    I can only hazard a guess as to what [itex]\ \alpha\ [/itex] represents. I see that [itex]\ W\ [/itex] is a quantity referred to as deflection and is related to [itex]\ \alpha\ [/itex] by [itex]\displaystyle \ \alpha=\frac{dW}{dr}\ .[/itex]

    At any rate, my earlier post in this thread was pretty much on the mark regarding Question 1 . [itex]\ W\ [/itex] is defined only on the disc, r ≤ R .

    The quantity, [itex]\displaystyle \ \ W=\frac{Fr^2}{16\pi D}\left(2\ln\left(\frac{r}{R}\right)-1+\left(\frac{R}{r}\right)^2\, \right)\,, \ [/itex] has a removable discontinuity at r = 0 (the origin). It may be helpful to graph W as a function of r. To aid you in doing that, let u = r/R, then r = R∙u . This gives you
    [itex]\displaystyle \ \ W=\frac{FR^2}{16\pi D}u^2\left(2\ln\left(u\right)-1+\frac{1}{u^2}\, \right)\ . \ [/itex]​
    So you can actually graph the u dependence of this function. Where is its maximum? [itex]\ W\ [/itex] is defined only on the disc, r ≤ R , so in terms of u, it's only defined for |u| ≤ 1.

    What is [itex]\displaystyle \ \ \frac{dW}{dr}\ ?[/itex]
     
  6. Mar 2, 2013 #5

    SammyS

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    Regarding your Question 2:

    I had initially thought that [itex]\ \alpha\ [/itex] was simply a constant. That's why I wrote what I did in my first reply.

    This appears to be a more complicated situation than the situation for Question 1.

    The textbook gives the differential equation for plate bending. This is the same as in your attachment in the OP.
    [itex]\displaystyle \frac{d}{dr} \left(\frac{1}{r}\frac{d}{dr}(\alpha r)\right)=-\,\frac{q}{D}\left(\frac{r^2-{r_1^2}}{2r}\right)[/itex]​
    The solution they give for this is:
    [itex]\displaystyle \alpha=C_1r+\frac{C_2}{r}-\frac{q}{16D}\left(\frac{r^4-{r_1^4}}{2r}-4r_1^2\,r\ln\left(\frac{r}{r_1}\right)\right)[/itex]​
    Your expression for [itex]\ \alpha\ [/itex] is the same if C1 and C2 are both set to zero. Why did you do that?

    It's pretty straight forward to integrate [itex]\displaystyle \ \ -\,\frac{q}{D}\left(\frac{r^2-{r_1^2}}{2r}\ \ \right)[/itex] with respect to r.

    Multiply the result by r, integrate again, then divide by r .

    Don't forget your constant of integration -- each time !
     
  7. Mar 5, 2013 #6
    thank you very much ,but i don't know why the integration is different from the book.
    see my solution please
     

    Attached Files:

  8. Mar 5, 2013 #7

    SammyS

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    When you integrate [itex]\displaystyle \ \ -\,\frac{q}{D}\left(\frac{r^2-{r_1^2}}{2r}\right)\ \ [/itex] there should be a constant of integration.

    [itex]\displaystyle \int\frac{-q}{D}\left(\frac{r}{2}-\frac{{r_1^2}}{2r}\right)dr= \frac{-q}{D}\left(\frac{r^2}{4}-\frac{{r_1^2}}{2}\ln(r)\ \right)+C_1[/itex]
     
  9. Mar 9, 2013 #8
    Thank you very much
    I show more detailed solution in the link below
    could you tell me where i did the mistake please
    http://img842.imageshack.us/img842/6430/pleaseyh.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  10. Mar 9, 2013 #9

    SammyS

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    What is the purpose of expanding [itex]\displaystyle \ \ \frac{d}{dr} \left(\frac{1}{r}\frac{d}{dr}(\alpha r)\right)\ \ [/itex] out to [itex]\displaystyle \ \ \frac{d^2\alpha}{dr^2}+ \frac{1}{r}\frac{d\alpha}{dr}-\frac{\alpha}{r^2}\ ?\ [/itex]

    Starting with [itex]\displaystyle \ \ \frac{d}{dr} \left(\frac{1}{r}\frac{d}{dr}(\alpha r)\right)=-\,\frac{q}{D}\left(\frac{r^2-{r_1^2}}{2r}\right)\,, \ [/itex]

    Integrate both sides with respect to r.

    [itex]\displaystyle \int \frac{d}{dr}\left(\frac{1}{r}\frac{d}{dr}(\alpha r)\right)\,dr=-\,\frac{q}{D}\int \left(\frac{r^2-{r_1^2}}{2r}\right)\,dr\ \ [/itex]

    The left hand side is simply [itex]\displaystyle \ \ \frac{1}{r}\frac{d}{dr}(\alpha r)+\text{constant}\ .\ [/itex]

    The right hand side has been done in another post in this thread, also in your attachment.

    Take the resulting equation and multiply both sides by r, and integrate again.
     
    Last edited by a moderator: May 6, 2017
  11. Mar 9, 2013 #10
    i dont know if you expand it or not ,it will be the same, but i didnt get the answer right.
    Anyway thank you
     
  12. Mar 9, 2013 #11
    the integration on the right is definitely not correct, but the problem i cant find the mistake
    idont knowhow he came up with r1^4 or lnr1, since r1 is constant??
     
  13. Mar 9, 2013 #12
    i try this in matlab ,i get exactly like my answer
    r1 is square not to the power 4 and there is no ln r1 only ln r
     
  14. Mar 10, 2013 #13
    i try using mat lab and i get exactly the answer with manual solution, which is differ from the book solution.

    Any help please
     
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