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This is from Finite Element Methods. Part of potential energy approach. Can someone show me how to come to this symetric matrix: $$ [K]=EI/L^3\begin{bmatrix} 12&6L&-12&6L\\ &4L^2&-6L&2L^2\\ & & 12&-6L\\ &&&4L^2 \end{bmatrix} $$ EI - constant

Using this equation: $$ [K]=EI\int_0^L {[B]^T [B]}dx $$ Given: $[B]= \begin{bmatrix} \frac{12x-6L}{L^3} & \frac{6xL-4L^2}{L^3} & \frac{-12x+6L}{L^3} & \frac{6xL-2L^2}{L^3}\\ \end{bmatrix}$


I don't understand how I can have a 4x4 matrix when multiplying vector and its transpose gives me a scalar. Here is what I have so far: $$[B]^T[B]=\frac{144x^2-144xL+36L}{L^6}+\frac{36x^2L^2-48xL^3+16L^4}{L^6}+\frac{144x^2-144xl+36L}{L^6}+\frac{36x^2L^2-24xL^3+4L^4}{L^6}=$$ $$=\frac{1}{L^6}(288x^2+24x^2L^2-288xL-72xL^3+72xL^3+72L+20L^4)$$

$$ EI\int_0^L {[B]^T [B]}dx =$$ $$=\frac{EI}{L^6}(96x^3+24x^3L^2-144x^2L-36x^2L^3+72xL+20xL^4)_0^L=$$ $$=EI(\frac{-48}{L^3}+\frac{8}{L}+\frac{72}{L^4})$$ And I got scalar. Not a 4x4 matrix. Thank you in advance!

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1 Answer 1

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As is written, $[B]$ is a row vector. Then $[B]^T$ is a column vector. So, $[B]^T[B]$ is of the form $$\pmatrix{b_1\\ b_2\\ \vdots\\ b_n}\pmatrix{b_1&b_2&\dots&b_n}=\pmatrix{{b_1}^2&b_1b_2&\dots\\ \vdots & \ddots \\ b_nb_1&\dots & {b_n}^2}$$

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  • $\begingroup$ Oh, what a glorious face palm moment. Thank you. $\endgroup$
    – Jek Denys
    Commented Oct 14, 2018 at 22:12

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