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

Thread: Solving non linear Matrix

  1. #1
    Super Member sakonpure6's Avatar
    Joined
    Sep 2012
    From
    Canada
    Posts
    829
    Thanks
    83

    Solving non linear Matrix

    Hello,

    I am working on the derivation of ideal buckling of columns under fixed supports at both ends.
    The general equation that describes the deflection of the column is:

    y(x) = A \cos(kx) + B \sin (kx) + Cx + D ; k = \sqrt{ \frac{P}{EI}}

    Where A,B,C,D are constants, k is also a constant ( E = modulus of elasticity and I is second moment of inertia of the member)

    I have four boundary conditions for a fixed-fixed support that are:
    y(0)=0 , y(L)=0, y'(0)=0 , y'(L) = 0

    where L is the full length of the column. These boundary condition simply mean that the deflection at each support is 0 and the slope of the deflection curve is also 0.

    The purpose of this exercise is to determine the critical axial load P that will cause buckling.

    Ideally I want to solve for k, then isolate for P which is gives critical axial load. The solution to this is:

    P = \frac{4\pi^2 \cdot EI}{L} \cdot n where n is any natural number, which determines the shape of the deflected shape. For simplicity we will take n=1.

    I have generated this matrix to solve:

     \begin{bmatrix} 1 & 0 & 0 & 1 \\ \cos(kL) & \sin(kL) & L & 1\\ 0 & k & 1 & 0 \\ -k\sin(kL) & \cos(kL) & 1 & 0\\ \end{bmatrix} \cdot \begin{bmatrix} A \\B\\C\\D \end{bmatrix} = \{0\}

    I want the non trivial solution, that is the 4x4 matrix = 0.

    I have forgotten how to solve such matrices...

    However, in class the prof briefly mentioned that we use Cramer's method to solve
    \begin{bmatrix} 1 & 0 & 0 & 1 \\ \cos(kL) & \sin(kL) & L & 1\\ 0 & k & 1 & 0 \\ -k\sin(kL) & \cos(kL) & 1 & 0\\ \end{bmatrix} = \{0\}

    This included finding the determinant of the matrix until you get a numerical equation.

    I computed the determinant of the above and got:

    2\cos(kl) \cdot ( \cos(kl) - 1 ) + kl \sin (kl) = 0

    If the above is true, how would you explicitly solve for k?

    Any guidance is appreciated.
    Last edited by sakonpure6; Jul 20th 2017 at 07:45 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,345
    Thanks
    2860

    Re: Solving non linear Matrix

    If Ax= 0 then there are two possibilities:
    1) A is invertible. In that case, multiplying both sides by the inverse, x= A^(-1)0= 0. x= 0 is the only solution- there is no "non-trivial" solution.

    2) A is not invertible. In that case, the determinate of A must be 0. And there are an infinite number of solutions. In fact, the set of solutions form a subspace. The best way to find that subspace is to row reduce the matrix. You will wind up with quantity in the last row that must be set to 0 to be equal to the right hand side (which will always be the 0 vector) that will give the conditions on k.
    Thanks from sakonpure6
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member sakonpure6's Avatar
    Joined
    Sep 2012
    From
    Canada
    Posts
    829
    Thanks
    83

    Re: Solving non linear Matrix

    Thanks that makes more sense, yes there will be an infinite set of solutions! Can you recommend a numerical solution to solve for kL ?

    And I don't really see how I can row reduce when I have trig functions and constants, it would get really messy
    Last edited by sakonpure6; Jul 21st 2017 at 03:12 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member sakonpure6's Avatar
    Joined
    Sep 2012
    From
    Canada
    Posts
    829
    Thanks
    83

    Re: Solving non linear Matrix

    Hey Halls, I have solved the problem!

    Please ignore my final equation in post #1, it is incorrect.

    I re evaluated the determinant, simplified and got a final equation of:

     2 = x\sin x + 2\cos x

    Where x = kL

    The thing is that I had to use an online graphing calculator to determine that the solution was:

    x=2\pi \cdot n \to P = \frac{\pi^2\cdot 4 \cdot EI}{L^2}

    I was wondering, how would I determine this solution numerically?

    For exam/test situation do you recommend I use Newtons method?
    Last edited by sakonpure6; Jul 21st 2017 at 03:39 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving for a matrix?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Feb 25th 2014, 05:59 PM
  2. Linear Algebra Unitary Matrix and Triangular Matrix
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Mar 12th 2013, 08:30 AM
  3. Solving for a Matrix
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: Sep 7th 2011, 08:19 PM
  4. Help in solving a following matrix
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Feb 26th 2011, 07:58 AM
  5. Replies: 2
    Last Post: Nov 27th 2010, 03:07 PM

/mathhelpforum @mathhelpforum