User:Eml4500.f08.team.guan/hw2v2

9.1-9.2

Global Stiffness Matrix
Going from the elemental stiffness matrices to the global stiffness matrix, the correspondence between displacement degrees of freedom, globally and locally, must be identified. Due to the sharing of nodes of different elements, some local displacement degrees of freedom will equal each other because the deflection of the local nodes are constrained by the fact that these local nodes share a same global node. Upon, identifying these similarities and relationships, the assembly of the global stiffness matrix can be written in terms of elemental stiffness values. This is the topology of $$\displaystyle K$$. First, it is necessary to just recall the element stiffness matrices. $$\displaystyle k^{(1)}_{matrix} $$ = $$\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} &k_{13}^{(1)} &k_{14}^{(1)} \\ k_{21}^{(1)} & k_{22}^{(1)} &k_{23}^{(1)} &k_{24}^{(1)} \\ k_{31}^{(1)} & k_{32}^{(1)} &k_{33}^{(1)} &k_{34}^{(1)} \\ k_{41}^{(1)} & k_{42}^{(1)} &k_{43}^{(1)} &k_{44}^{(1)} \\ \end{bmatrix} $$  , $$\displaystyle k^{(2)}_{matrix} $$ = $$\begin{bmatrix} k_{11}^{(2)} & k_{12}^{(2)} &k_{13}^{(2)} &k_{14}^{(2)} \\ k_{21}^{(2)} & k_{22}^{(2)} &k_{23}^{(2)} &k_{24}^{(2)} \\ k_{31}^{(2)} & k_{32}^{(2)} &k_{33}^{(2)} &k_{34}^{(2)} \\ k_{41}^{(2)} & k_{42}^{(2)} &k_{43}^{(2)} &k_{44}^{(2)} \\

\end{bmatrix} $$ Also, it is necessary to recall the relationships between the local displacement degrees of freedom and the global displacement degrees of freedom: $$\displaystyle d_1^{(1)}=d_1, d_2^{(1)}=d_2, d_3^{(1)}=d_3=d_1^{(2)}, d_4^{(1)}=d_4=d_2^{(2)}, d_5^{(2)}=d_5, d_4^{(2)}=d_6$$ $$\displaystyle d_{global} $$ = $$\begin{bmatrix} d_{1} \\ d_{2} \\ d_{3} \\ d_{4} \\ d_5 \\ d_6\\ \end{bmatrix}$$  , $$\displaystyle d^{(1)} $$ = $$\begin{bmatrix} d_{1}^{(1)} \\ d_{2}^{(1)} \\ d_{3}^{(1)} \\ d_{4}^{(1)} \\ \end{bmatrix}$$  , $$\displaystyle d^{(2)} $$ = $$\begin{bmatrix} d_{1}^{(2)} \\ d_{2}^{(2)} \\ d_{3}^{(2)} \\ d_{4}^{(2)} \\ \end{bmatrix}$$

 Now, it is possible to construct the global stiffness matrix: $$\displaystyle K_{global} $$ = $$\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0  & 0  \\ k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)} & 0  & 0  \\ k_{31}^{(1)} & k_{32}^{(1)}  & a  & b & k_{13}^{(2)} & k_{14}^{(2)} \\ k_{41}^{(1)} & k_{42}^{(1)}  & c  & d  & k_{23}^{(2)} & k_{24}^{(2)} \\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} & k_{34}^{(2)}  \\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)}  \\ \end{bmatrix} $$ = $$\begin{bmatrix} K_{11} & K_{12} & K_{13} & K_{14} & K_{15}  & K_{16}  \\ K_{21} & K_{22} & K_{23} & K_{24} & K_{25}  & K_{26}  \\ K_{31} & K_{32} & K_{33} & K_{34} & K_{35}  & K_{36}  \\ K_{41} & K_{42} & K_{43} & K_{44} & K_{45}  & K_{46}  \\ K_{51} & K_{52} & K_{53} & K_{54} & K_{55}  & K_{56}  \\ K_{61} & K_{62} & K_{63} & K_{64} & K_{65}  & K_{66}  \\ \end{bmatrix} $$

where $$ a=k_{33}^{(1)}+k_{11}^{(2)}, b=k_{34}^{(1)}+k_{12}^{(2)}, c=k_{43}^{(1)}+k_{21}^{(2)}, d=k_{44}^{(1)}+k_{22}^{(2)}$$ and $$K_{11}= k_{11}^{(1)}, K_{12}=k_{12}^{(1)}...K_{33}=a=k_{33}^{(1)}+k_{11}^{(2)},K_{34}=b=k_{34}^{(1)}+k_{12}^{(2)}...$$ The numeric values of the global stiffness matrix are as follows: $$K_{11}=\frac{9}{16},K_{12}=\frac{3\sqrt{3}}{16},K_{33}=\frac{9}{16}+\frac{5}{2},K_{34}=\frac{3\sqrt{3}}{16}-\frac{5}{2}, K_{44}=\frac{3}{16}+\frac{5}{2}$$ Applying the boundary conditions where the global nodes one and three are fixed due to the pinned joints ($$d_1=d_2=d_5=d_6=0$$). Thus, elimination of known displacement degrees of freedom it is possible to reduce the global force displacement relationship leaving $$d_3 and d_4$$ to be unknown. Removing $$d_1, d_2, d_5$$ and $$d_6$$ is like replacing columns 1,2,5,6 with zeros because of matrix algebra.