Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By romsek

Thread: Derivation formula for b(x)

  1. #1
    Member
    Joined
    Mar 2017
    From
    Netherlands
    Posts
    106

    Derivation formula for b(x)

    Hello,


    A non prismatic figure / beam I need to derive a formula for b (x).
    See the picture in the link below:

    https://myalbum.com/album/Qs5BcjTUrEdt


    The beam has a varying cross-section.

    It is given that the dimensions at the end of the beam is determined by the relationship:
    b (l) = αb (0), with α less than or equal to 1.

    asked:
    lead a formula for the dimension b (x) at a distance x from the left bearing.



    My book gives answers:

    b(x) = b(l) + ((l-x)/l)* (b(0)-b(l))


    I self must step by step show the derivation of the function.
    How do I do that ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2017
    From
    Netherlands
    Posts
    106

    Re: Derivation formula for b(x)

    Can someone help me?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,923
    Thanks
    2491

    Re: Derivation formula for b(x)

    so if I understand this you've got a beam that has a cross section that varies as the distance from the left bearing.

    at the left bearing it's width $b(0)$, at the furthest end, a distance $\ell$ from the left bearing it is width $b(\ell)$

    the cross section changes linearly with the distance, $x$, from the left bearing.

    We have values at two points, the left and right edges of the beam.

    At the left edge we have point $(0, b(0))$

    At the right edge we have point $(\ell, b(\ell))$

    The slope of the width change is

    $m = \dfrac{b(\ell)-b(0)}{\ell - 0} = \dfrac{b(\ell)-b(0)}{\ell }$

    and using the point-slope formula for a line we have

    $b(x) - b(0) = \dfrac{b(\ell)-b(0)}{\ell }\left(x - 0\right)$

    $b(x) = \dfrac{b(\ell)-b(0)}{\ell }x + b(0)$

    and I see for some reason your book reversed everything.

    Well if we look at it in reverse

    $m = \dfrac{b(0)-b(\ell)}{-\ell}$

    $b(x)-b(\ell) = \dfrac{b(0)-b(\ell)}{-\ell} (x - \ell)$

    $b(x) = \dfrac{b(0)-b(\ell)}{\ell}(\ell - x)+b(\ell)$

    and this matches the formula in your book.
    Thanks from skeeter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2017
    From
    Netherlands
    Posts
    106

    Re: Derivation formula for b(x)

    Thank you very much for your comment!

    If I understand correctly, the book has a function
    given b(x) for the right bearing? Not from the left bearing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. derivation formula of variance
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Feb 11th 2015, 07:43 PM
  2. Limit formula derivation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 25th 2012, 05:27 AM
  3. axisymmetric jet formula derivation
    Posted in the Advanced Applied Math Forum
    Replies: 11
    Last Post: Jun 19th 2010, 11:53 AM
  4. Derivation of Derangement formula
    Posted in the Advanced Statistics Forum
    Replies: 9
    Last Post: Jun 6th 2010, 02:42 PM

/mathhelpforum @mathhelpforum