User:Eml4500.f08.team.guan/hw5

Derivation of Applying the P.V.W to Eliminate Rows Corresponding to the Boundary Conditions in the Stiffness Matrix
In this lecture, the justification of eliminating rows 1, 2, 5, and 6 in the global stiffness matrix of the 2-bar truss problem in order to obtain the 2 by 2 stiffness matrix $$\bar{k}_{2x2}$$ will be shown. The derivation is as follows. Recall the global force displacement relationship. $$\displaystyle K_{6x6}d_{6x1}=F_{6x1}$$ Subtracting the force column matrix to both sides will yield the following equation. $$\displaystyle K_{6x6}d_{6x1}-F_{6x1}=0_{6x1}...(1)$$ where 0 is a 6 by 6 column matrix of zeros. Now, the principle of virtual work is applied. The principle of virtual work states that "forces applied to a static system do no virtual work". Also, for rigid bodies, if "a rigid body that is in equilibrium is subject to virtual compatible displacements, the total virtual work of all external forces is zero; and conversely, if the total virtual work of all external forces acting on a rigid body is zero then the body is in equilibrium" (source: ). Applying it, we have the following equation. $$w_{6x1} \cdot (Kd-F)_{6x1}=0_{1x1}...(2)$$ Equation 1 is true for all matrices $$w_{6x1}$$. Here, a virtual weighting function ("$$w$$") or coefficient is dotted with KD-F. The weighting function or coefficient means relative weight between functions. In equation 2, 0 is a 1 by 1 matrix meaning it is a scalar. In order to multiply two column matrices and result in scalar, the dot product must be used. Now, the fact that equation 1 is equivalent to equation 2 is trivial, however, the fact that equation 2 is equivalent to equation 1 is not trivial. Here, we have $$w \cdot (Kd-F)=0$$ for all $$w$$. To prove this, let us select some choices for the weighting function matrix. Choice 1: We select the weighting function such that, $$\displaystyle w_1=1, w_2=0, w_3=0, w_4=0, w_5=0, w_6=0$$ $$w^T= \begin{bmatrix} 1 & 0& 0& 0& 0& 0\\ \end{bmatrix} $$

Plugging this column matrix into equation 2 will yield the following. $$w \cdot (Kd-F)= 1[\sum_{j=1}^6{K_{1j}d_j-F_1}]+0[\sum_{j=1}^6{K_{2j}d_j-F_2}]+0[\sum_{j=1}^6{K_{3j}d_j-F_3}]+...+0[\sum_{j=1}^6{K_{6j}d_j-F_6}]=0 $$ Multiplying it all out gives $$\sum_{j=1}^6{K_{1j}d_j}=F_1 $$

For choice 2, we set the second row of the weighting column matrix to be 1 and the reset of the values to be 0. $$\displaystyle w_1=0, w_2=1, w_3=0, w_4=0, w_5=0, w_6=0$$ $$w^T= \begin{bmatrix} 0 & 1& 0& 0& 0& 0\\ \end{bmatrix} $$

Plugging this column matrix into equation 2 will yield the following. $$w \cdot (Kd-F)= 0[\sum_{j=1}^6{K_{1j}d_j-F_1}]+1[\sum_{j=1}^6{K_{2j}d_j-F_2}]+0[\sum_{j=1}^6{K_{3j}d_j-F_3}]+...+0[\sum_{j=1}^6{K_{6j}d_j-F_6}]=0 $$ Multiplying it all out gives $$\sum_{j=1}^6{K_{2j}d_j}=F_2 $$

Considering all the choices of the weighting function column matrix, it is concluded that $$\displaystyle Kd=F$$ Thus, equation 1 is confirmed. However, this did not eliminate rows 1, 2, 5, and 6. The boundary conditions for the 2-bar truss problem must also be adhered to while considering the principle of virtual work. Here are the boundary conditions. $$\displaystyle d_1=d_2=d_5=d_6=0$$ The weighting column matrix (now called the virtual displacement) must be "kinematically admissible" meaning that it also must follow the boundary conditions. This means that $$\displaystyle w_1=w_2=w_5=w_6=0$$ From 10.1, ou can see that $$ Kd$$ can be reduced to a rectangular matrix multiplied by a 2 by 1 matrix and ultimately, after applying boundary conditions and the principle of virtual work with corresponding virtual displacements equaling zero, we have the following expression. $$ w \cdot (Kd-F)= \begin{bmatrix} w_3\\ w_4\\ \end{bmatrix}\cdot

(\bar{K} \bar{d}-\bar{F})=0...(3) $$ for all $$w_3$$ and $$ w_4$$ where $$ \bar{K}=\begin{bmatrix} k_{33} & k_{34}\\ k_{43} & k_{44}\\ \end{bmatrix} , \bar{d}= \begin{bmatrix} d_{3}\\ d_{4} \\ \end{bmatrix} , \bar{F}=\begin{bmatrix} F_{3}\\ F_{4} \\ \end{bmatrix} $$