User:Eml4500.f08.delta 6.krueger/Notes Friday October 17th

When equations one and two, from the equilibrium of node two, are rewritten they become: $$f^{(1)}_{3} + f^{(2)}_{1}= 0$$ $$f^{(1)}_{4} + f^{(2)}_{2}= 0$$ The element forces can then be written in terms of the element stiffness and displacements. $$f_{3}^{(1)} = \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix}$$ $$f_{1}^{(2)} = \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}$$ The forces written in terms of the elements stiffness and displacements are then substituted into equation one and equation one becomes. $$ \begin{bmatrix} k_{31}^{(1)}d_{1}^{(1)} + k_{32}^{(1)}d_{2}^{(1)} + k_{33}^{(1)}d_{3}^{(1)} + k_{34}^{(1)}d_{4}^{(1)} \end{bmatrix} + \begin{bmatrix} k_{11}^{(2)}d_{1}^{(2)} + k_{12}^{(2)}d_{2}^{(2)} + k_{13}^{(2)}d_{3}^{(2)} + k_{14}^{(2)}d_{4}^{(2)} \end{bmatrix}=0$$ Now recall the relationship between the global degrees of freedom and the element degrees of freedom: $$d_{1}=d_{1}^{(1)},d_{2}=d_{2}^{(1)},d_{3}=d_{3}^{(1)}=d_{1}^{(2)},d_{4}=d_{4}^{(1)}=d_{2}^{(2)},d_{5}=d_{3}^{(2)},d_{6}=d_{4}^{(2)}$$ The global degrees of freedom can now be substituted into equation one for the element degrees of freedom and equation one becomes. $$ \begin{bmatrix} k_{31}^{(1)}d_{1} + k_{32}^{(1)}d_{2} + k_{33}^{(1)}d_{3} + k_{34}^{(1)}d_{4} \end{bmatrix} + \begin{bmatrix} k_{11}^{(2)}d_{3} + k_{12}^{(2)}d_{4} + k_{13}^{(2)}d_{5} + k_{14}^{(2)}d_{6} \end{bmatrix}=0$$ Now the common terms are grouped and factored (i.e. The terms that include $$d_{3}$$ and $$d_{4}$$). The same process is repeated for equation two. Equation one results in the third row of the global stiffness matrix and equation two results in the fourth row of the global stiffness matrix. The work is shown in the homework for this section. =Assembly of the global stiffness matrix= Before starting the assembly of the global stiffness matrix a quick review of the goals and the end that is being worked towards is necessary. The first goal is the use of The Principle of Virtual Work (PVW) which allows the formation of the global stiffness matrixK by eliminating rows that correspond to the boundary conditions. The ultimate goal is to use partial differential equations (PDE's) to derive the Finite Element Method (FEM). Recall now two equations that have already been derived in the process of the FEM. $$q^{(e)}_{2x1}$$ = $$T^{(e)}_{2x4}$$ $$d^{(e)}_{4x1}$$ $$K^{(e)}= T^{(e)^(T)} * \hat{K}^{(e)} * T^{(e)}$$

The assembly of the global stiffness matrix (K) is accomplished using the element stiffness matrices ($$k^{(e)}$$) and an assemble operator (A). Where e= 1,....,nel, and nel is the number of elements. The assembly operator is introduced because the sigma operator is not correct. The global stiffness matrix is assembled using a combination of element stiffness matrices. The element stiffness matrices are small when compared to the global stiffness matrix. The global stiffness matrix could be orders of magnitude larger than the element stiffness matrices. The sum of $$k_{nedxned}^{(e)}$$ doesn't equal to a $$K_{nxn}$$ matrix. Therefore the assembly operator (A) is necessary for the notation of the process to be correct. $$K=Ak^{(e)}$$ With this, new notation needs to be introduced. n is equal to the number of global degrees of freedom before eliminating the boundary conditions. ned is the number of element degrees of freedom. Where ned is very small when compared to n. Recall that the force displacement relationship for a bar or spring is kd=F. After some algebra: $$ kd-F=0$$ This will be called equation three and is equivalent to: w(kd-F)=0 This will be call equation four. This holds true for all values of w. Therefore it is trivial to imply that equation three is equal to equation four but is extremely important to understand that equation four is equal to equation three. This is true because equation four is valid for all w. So letting w=1, equation four is converted back into equation three.