Homework #7 Solutions
Reminder - you're going to need these webpages:
http://www.continuummechanics.org/techforms/index.html
to do this homework.... unless
you prefer to use Matlab, Mathematica, etc.
-
Figure out the deformation gradient at the center of the element below.
The node displacements are
\[
\begin{eqnarray}
u_1 = 3 \qquad u_2 = 4 \qquad u_3 = 5 \qquad u_4 = 4 \\
v_1 = 2 \qquad v_2 = 2 \qquad v_3 = 2 \qquad v_4 = 2
\end{eqnarray}
\]
The deformation gradient components are
\[
\begin{eqnarray}
F_{11} & = & {1 \over 4} [-(1-Y)u_1 + (1-Y)u_2 + (1+Y)u_3 - (1+Y)u_4] + 1 \\
\\
& = & {1 \over 4} [-u_1 + u_2 + u_3 - u_4] + 1 \\
\\
\\
& = & 1.5
\\
\\
F_{12} & = & {1 \over 4} [-(1-X)u_1 - (1+X)u_2 + (1+X)u_3 + (1-X)u_4] \\
\\
& = & {1 \over 4} [-u_1 - u_2 + u_3 + u_4] \\
\\
\\
& = & 0.5
\\
\\
F_{21} & = & {1 \over 4} [-(1-Y)v_1 + (1-Y)v_2 + (1+Y)v_3 - (1+Y)v_4] \\
\\
& = & {1 \over 4} [-v_1 + v_2 + v_3 - v_4] \\
\\
\\
& = & 0.0
\\
\\
F_{22} & = & {1 \over 4} [-(1-X)v_1 - (1+X)v_2 + (1+X)v_3 + (1-X)v_4] + 1 \\
\\
& = & {1 \over 4} [-v_1 - v_2 + v_3 + v_4] + 1 \\
\\
\\
& = & 1.0
\end{eqnarray}
\]
\[
{\bf F} =
\left[ \matrix{
1.5 & 0.5 \\
0.0 & 1.0
} \right]
\]
-
Figure out the deformation gradient in this element.
The undeformed node coordinates are
\[
\begin{eqnarray}
X_1 = -1 & \qquad X_2 = 1 & \qquad X_3 = -1
\\
\\
Y_1 = -1 & \qquad Y_2 = 0 & \qquad Y_3 = 2
\end{eqnarray}
\]
The node displacements are
\[
\begin{eqnarray}
u_1 = 3 & \qquad u_2 = 3 & \qquad u_3 = 3
\\
\\
v_1 = 2 & \qquad v_2 = 3 & \qquad v_3 = 1
\end{eqnarray}
\]
The partial derivatives are
\[
\begin{eqnarray}
\left[ \matrix{
{\partial u \over \partial X} & {\partial u \over \partial Y} \\
\\
{\partial v \over \partial X} & {\partial v \over \partial Y}
} \right]
& = &
\left[ \matrix{
u_1 - u_3 & u_2 - u_3 \\
\\
v_1 - v_3 & v_2 - v_3 \\
} \right]
\left[ \matrix{
X_1 - X_3 & X_2 - X_3 \\
\\
Y_1 - Y_3 & Y_2 - Y_3 \\
} \right]^{-1}
\\
\\
\\
\\
& = &
\left[ \matrix{
0 & 0 \\
\\
1 & 2 \\
} \right]
\left[ \matrix{
\;\;\;0 & \;\;\;2\\
\\
-3 & -2 \\
} \right]^{-1}
\\
\\
\\
\\
\\
\\
& = &
\left[ \matrix{
0 & 0 \\
\\
1 & 2 \\
} \right]
\left[ \matrix{
-0.333 & -0.333 \\
\\
\;\;\;0.500 & \;\;\;0.000 \\
} \right]
\\
\\
\\
\\
\\
\\
& = &
\left[ \matrix{
0.000 & \;\;\;0.000 \\
\\
0.667 & -0.333 \\
} \right]
\end{eqnarray}
\]
And add \( {\bf I}\) to get \({\bf F}\).
\[
{\bf F}
=
\left[ \matrix{
1.000 & 0.000 \\
\\
0.667 & 0.667 \\
} \right]
\]
-
How many degrees of rigid body rotation is this element experiencing?
The polar decomposition can be applied to get
\[
{\bf R} =
\left[ \matrix{
0.928 & -0.371 \\
\\
0.371 & \;\;\;0.928
} \right]
\qquad \qquad
{\bf U} =
\left[ \matrix{
1.176 & 0.248 \\
\\
0.248 & 0.619
} \right]
\]
So this element rotates
\[
\theta \quad = \quad \text{Sin}^{-1}(0.371) \quad = \quad 21.8^\circ
\]