Nonlinear finite elements/Homework 9/Solutions

Problem 1: Total Lagrangian

Given:

Consider the tapered two-node element shown in Figure 1. The displacement field in the element is linear.

File:NFE HW9Prob1.png Figure 1. Tapered two-node element.

The reference (initial) cross-sectional area is

$A_0=(1-\xi )A_{01}+\xi A_{02}.$

Assume that the nominal (engineering) stress is also linear in the element, i.e.,

$P=(1-\xi )P_1+\xi P_2.$

Solution

Part 1

The displacement field is given by the linear Lagrange interpolation expressed in terms of the material coordinate.

$𝐮(X,t)=\frac{1}{l_0}[X_2-XX-X_1]\left[\begin{array}{c}{u}_1(t)\\ u_2(t)\end{array}\right]$

where $l_0=X_2-X_1$. The strain measure is evaluated in terms of the nodal displacement,

$\mathit{\epsilon }(X,t)=u_{,X}=\frac{1}{l_0}[-11]\left[\begin{array}{c}{u}_1(t)\\ u_2(t)\end{array}\right]$

which defines the ${𝐁}_0$ matrix to be

${𝐁}_0=\frac{1}{l_0}[-11].$

The internal nodal forces are then given by the usual relations.

${𝐟}_e^{\text{int}}=\underset{X_1}{\overset{X_2}{\int }}{𝐁}_0^T(A_{0}P)dX=\frac{1}{l_0}\underset{X_1}{\overset{X_2}{\int }}((1-\xi )A_{01}+\xi A_{02})((1-\xi )P_1+\xi P_2)\left[\begin{array}{c}-1\\ 1\end{array}\right]dX$

Integrating the above integral with $\xi =(X-X_1)/l_0$ to obtain

${𝐟}_e^{\text{int}}=\frac{1}{6}(2A_{02}{P}_2+A_{02}{P}_1+A_{01}{P}_2+2A_{01}{P}_1)\left[\begin{array}{c}-1\\ 1\end{array}\right]$

Part 2

${𝐟}_e^{\text{int}}=A_{0}P\left[\begin{array}{c}-1\\ 1\end{array}\right]$

Part 3

The external body forces arising from the body force, $b$, are obtained by the usual procedure.

${𝐟}_e^{\text{ext}}=\underset{X_1}{\overset{X_2}{\int }}{𝐍}^T{\rho }_0{A}_{0}bdX=\frac{{\rho }_{0}b}{l_0}\underset{X_1}{\overset{X_2}{\int }}{A}_0\left[\begin{array}{c}{X}_2-X\\ X-X_1\end{array}\right]dX$

${𝐟}_e^{\text{ext}}=\frac{{\rho }_{0}b}{6}\left[\begin{array}{c}{x}_1^2{A}_{02}+2A_{01}{x}_1^2-2x_1{A}_{02}{x}_2-4A_{01}{x}_1{x}_2+2A_{01}{x}_2^2+A_{02}{x}_2^2\\ A_{01}{x}_1^2+2x_1^2{A}_{02}-4x_1{A}_{02}{x}_2-2A_{01}{x}_1{x}_2+A_{01}{x}_2^2+2A_{02}{x}_2^2\end{array}\right]$

Part 4

${𝐟}_e^{\text{ext}}=\frac{b{A}_0{\rho }_0{l}_0^2}{2}\left[\begin{array}{c}1\\ 1\end{array}\right]$

Part 5

The element mass matrix is

${𝐌}_e=\underset{X_1}{\overset{X_2}{\int }}{\rho }_0{A}_0{𝐍}^T𝐍dX$

${𝐌}_e=\frac{{\rho }_0{l}_0}{12}\left[\begin{array}{cc}3A_{01}+A_{02}& A_{01}+A_{02}\\ A_{01}+A_{02}& A_{01}+3A_{02}\end{array}\right]$

Part 6

Lumped mass matrix is given by

$M_{ii}=\underset{{\xi }_1}{\overset{{\xi }_2}{\int }}{\rho }_0{A}_0{N}_{i}dX$

${𝐌}_e=\frac{{\rho }_0{l}_0}{6}\left[\begin{array}{cc}2A_{01}+A_{02}& 0\\ 0& A_{01}+2A_{02}\end{array}\right]$

Part 7

$𝐊𝐮={\omega }^2𝐌𝐮$

with

$𝐊=\frac{E(A_{01}+A_{02})}{2l_0}\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]$

where $E$ is the Young's modulus and $l_0$ is the initial length of the element.

The above equation can be rewrite as

$(𝐊-{\omega }^2𝐌)𝐮=0$

which only has a solution if

$\text{det}(𝐊-{\omega }^2𝐌)=0$

Solving the above determinant for $\omega$, we have

$\omega =0,\pm \sqrt{\frac{18E(A_{01}^2+2A_{01}{A}_{02}+A_{02}^2)}{{\rho }_0{l}_0^2(A_{01}^2+4A_{01}{A}_{02}+A_{02}^2)}}$

Problem 2: Updated Lagrangian

Given:

Consider the tapered two-node element shown in Figure 1.

The current cross-sectional area is

$A=(1-\xi )A_1+\xi A_2.$

Assume that the Cauchy stress is also linear in the element, i.e.,

$\sigma =(1-\xi ){\sigma }_1+\xi {\sigma }_2.$

Solution

Part 1

The velocity field is

$v(X,t)=\frac{1}{l_0}[X_2-XX-X_1]\left[\begin{array}{c}{v}_1(t)\\ v_2(t)\end{array}\right]$

In term of element coordinates, the velocity field is

$v(\xi ,t)=\frac{1}{l_0}[1-\xi \xi ]\left[\begin{array}{c}{v}_1(t)\\ v_2(t)\end{array}\right]\xi =\frac{X-X_1}{l_0}$

The displacement is the time integrals of the velocity, and since $\xi$ is independent of time

$u(\xi ,t)=𝐍(\xi ){𝐮}_e(t)$

Therefore, since $x=X+u$

$x(\xi ,t)=𝐍(\xi ){𝐱}_e(t)=[1-\xi \xi ]\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\end{array}\right]{\xi }_{,\xi }=x_2-x_1=l$

where $l$ is the current length of the element. For this element, we can express $\xi$ in terms of the Eulerian coordinates by

$\xi =\frac{x-x_1}{x_2-x_1}=\frac{x-x_1}{l},l=x_2-x_1,{\xi }_{,x}=\frac{1}{l}$

So ${\xi }_{,x}$ can be obtained directly, instead of through the inverse of $x_{,\xi }$.

The $𝐁$ matrix is obtained by the chain rule

$𝐁={𝐍}_{,x}={𝐍}_{,\xi }{\xi }_{,x}=\frac{1}{l}[-11]$

Using (146) in Handout 13, we have

${𝐟}_e^{\text{int}}=\underset{x_1}{\overset{x_2}{\int }}{𝐁}^T\sigma Adx=\underset{x_1}{\overset{x_2}{\int }}\frac{\sigma A}{l}\left[\begin{array}{c}-1\\ 1\end{array}\right]dx$

Integrating the above equation to obtain

${𝐟}_e^{\text{int}}=\frac{1}{6}(A_1{\sigma }_2+2(A_2{\sigma }_2+A_1{\sigma }_1)+A_2{\sigma }_1)\left[\begin{array}{c}-1\\ 1\end{array}\right]$

Part 2

The external forces are given by

${𝐟}_e^{\text{ext}}=\frac{\rho b}{6}\left[\begin{array}{c}2x_2{A}_1+x_2{A}_2-2x_1{A}_1-x_1{A}_2\\ 2x_2{A}_2+x_2{A}_1-2x_1{A}_2-x_1{A}_1\end{array}\right]$

Problem 3: Modal Analysis

Given: Consider the axially loaded bar in problem VM 59 of the ANSYS Verification manual. Assume that the bar is made of Tungsten carbide.

The input file for ANSYS is as shown in VM 59 except the following material properties are used: $E=100$ Msi and $\rho =0.567/386$ lb$\cdot$s$2^{}/$in$4^{}$.

Solution

$f_1=36.963\text{ Hz}$

$f_1=33.174\text{ Hz}{f}_2=144.107\text{ Hz}{f}_3=328.443\text{ Hz}$

Given: Consider the stretched circular membrane in problem VM 55 of the ANSYS Verification manual. Assume that the membrane is made of OFHC (Oxygen-free High Conductivity) copper.

The input file for ANSYS is as shown in VM 55 except the following material properties are used: $E=17$ Msi and $\rho =0.322/386$ lb$\cdot$s$2^{}/$in$4^{}$, and the modes are expanded to 5.

Solution

$f_1=1.509\text{ Hz}$

$f_1=88.329\text{ Hz}{f}_2=203.114\text{ Hz}{f}_3=320.174\text{ Hz}{f}_4=441.655\text{ Hz}{f}_5=571.350\text{ Hz}$