# 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. 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

Using the total Lagrangian formulation, develop expressions for the internal nodal forces.

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

$\mathbf {u} (X,t)={\frac {1}{l_{0}}}[X_{2}-X\quad X-X_{1}]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}$ where $l_{0}=X_{2}-X_{1}$ . The strain measure is evaluated in terms of the nodal displacement,

${\boldsymbol {\varepsilon }}(X,t)=u_{,X}={\frac {1}{l_{0}}}[-1\quad 1]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}$ which defines the $\mathbf {B} _{0}$ matrix to be

$\mathbf {B} _{0}={\frac {1}{l_{0}}}[-1\quad 1].$ The internal nodal forces are then given by the usual relations.

$\mathbf {f} _{e}^{\mbox{int}}=\int _{X_{1}}^{X_{2}}\mathbf {B} _{0}^{T}(A_{0}P)dX={\frac {1}{l_{0}}}\int _{X_{1}}^{X_{2}}\left((1-\xi )A_{01}+\xi A_{02}\right)\left((1-\xi )P_{1}+\xi P_{2}\right){\begin{bmatrix}-1\\1\end{bmatrix}}dX$ Integrating the above integral with $\xi =(X-X_{1})/l_{0}$ to obtain

${\mathbf {f} _{e}^{\mbox{int}}={\frac {1}{6}}\left(2A_{02}P_{2}+A_{02}P_{1}+A_{01}P_{2}+2A_{01}P_{1}\right){\begin{bmatrix}-1\\1\end{bmatrix}}}$ #### Part 2

What are the internal nodal forces if the reference area and the nominal stress are constant over the element?

${\mathbf {f} _{e}^{\mbox{int}}=A_{0}P{\begin{bmatrix}-1\\1\end{bmatrix}}}$ #### Part 3

Assume that the body force is constant. Develop expressions for the external nodal forces for that case.

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

$\mathbf {f} _{e}^{\mbox{ext}}=\int _{X_{1}}^{X_{2}}\mathbf {N} ^{T}\rho _{0}A_{0}bdX={\frac {\rho _{0}b}{l_{0}}}\int _{X_{1}}^{X_{2}}A_{0}{\begin{bmatrix}X_{2}-X\\X-X_{1}\end{bmatrix}}dX$ ${\mathbf {f} _{e}^{\mbox{ext}}={\frac {\rho _{0}b}{6}}{\begin{bmatrix}x_{1}^{2}A_{02}+2A_{01}x_{1}^{2}-2x_{1}A_{02}x_{2}-4A_{01}x_{1}x_{2}+2A_{01}x_{2}^{2}+A_{02}x_{2}^{2}\\A_{01}x_{1}^{2}+2x_{1}^{2}A_{02}-4x_{1}A_{02}x_{2}-2A_{01}x_{1}x_{2}+A_{01}x_{2}^{2}+2A_{02}x_{2}^{2}\end{bmatrix}}}$ #### Part 4

What are the external nodal forces if the reference area and the nominal stress are constant over the element?

${\mathbf {f} _{e}^{\mbox{ext}}={\frac {bA_{0}\rho _{0}l_{0}^{2}}{2}}{\begin{bmatrix}1\\1\end{bmatrix}}}$ #### Part 5

Develop an expression for the consistent mass matrix for the element.

The element mass matrix is

$\mathbf {M} _{e}=\int _{X_{1}}^{X_{2}}\rho _{0}A_{0}\mathbf {N} ^{T}\mathbf {N} dX$ ${\mathbf {M} _{e}={\frac {\rho _{0}l_{0}}{12}}{\begin{bmatrix}3A_{01}+A_{02}&A_{01}+A_{02}\\A_{01}+A_{02}&A_{01}+3A_{02}\end{bmatrix}}}$ #### Part 6

Obtain the lumped (diagonal) mass matrix using the row-sum technique.

Lumped mass matrix is given by

$M_{ii}=\int _{\xi _{1}}^{\xi _{2}}\rho _{0}A_{0}N_{i}dX$ ${\mathbf {M} _{e}={\frac {\rho _{0}l_{0}}{6}}{\begin{bmatrix}2A_{01}+A_{02}&0\\0&A_{01}+2A_{02}\end{bmatrix}}}$ #### Part 7

Find the natural frequencies of a single element with consistent mass by solving the eigenvalue problem

$\mathbf {K} ~\mathbf {u} =\omega ^{2}~\mathbf {M} ~\mathbf {u}$ with

$\mathbf {K} ={\cfrac {E(A_{01}+A_{02})}{2l_{0}}}{\begin{bmatrix}1&-1\\-1&1\end{bmatrix}}$ where $E$ is the Young's modulus and $l_{0}$ is the initial length of the element.

The above equation can be rewrite as

$(\mathbf {K} -\omega ^{2}\mathbf {M} )\mathbf {u} =0$ which only has a solution if

${\mbox{det}}(\mathbf {K} -\omega ^{2}\mathbf {M} )=0$ 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

Using the updated Lagrangian formulation, develop expressions for the internal nodal forces.

The velocity field is

$v(X,t)={\frac {1}{l_{0}}}\left[X_{2}-X\quad X-X_{1}\right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}$ In term of element coordinates, the velocity field is

$v(\xi ,t)={\frac {1}{l_{0}}}\left[1-\xi \quad \xi \right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}\qquad \xi ={\frac {X-X_{1}}{l_{0}}}$ The displacement is the time integrals of the velocity, and since $\xi$ is independent of time

$u(\xi ,t)=\mathbf {N} (\xi )\mathbf {u} _{e}(t)$ Therefore, since $x=X+u$ $x(\xi ,t)=\mathbf {N} (\xi )\mathbf {x} _{e}(t)=\left[1-\xi \quad \xi \right]{\begin{bmatrix}x_{1}(t)\\x_{2}(t)\end{bmatrix}}\qquad \xi _{,\xi }=x_{2}-x_{1}=l$ where $l$ is the current length of the element. For this element, we can express $\xi$ in terms of the Eulerian coordinates by

$\xi ={\frac {x-x_{1}}{x_{2}-x_{1}}}={\frac {x-x_{1}}{l}},\quad l=x_{2}-x_{1},\quad \xi _{,x}={\frac {1}{l}}$ So $\xi _{,x}$ can be obtained directly, instead of through the inverse of $x_{,\xi }$ .

The $\mathbf {B}$ matrix is obtained by the chain rule

$\mathbf {B} =\mathbf {N} _{,x}=\mathbf {N} _{,\xi }\xi _{,x}={\frac {1}{l}}[-1\quad 1]$ Using (146) in Handout 13, we have

$\mathbf {f} _{e}^{\mbox{int}}=\int _{x_{1}}^{x_{2}}\mathbf {B} ^{T}\sigma Adx=\int _{x_{1}}^{x_{2}}{\frac {\sigma A}{l}}{\begin{bmatrix}-1\\1\end{bmatrix}}dx$ Integrating the above equation to obtain

${\mathbf {f} _{e}^{\mbox{int}}={\frac {1}{6}}\left(A_{1}\sigma _{2}+2(A_{2}\sigma _{2}+A_{1}\sigma _{1})+A_{2}\sigma _{1}\right){\begin{bmatrix}-1\\1\end{bmatrix}}}$ #### Part 2

Assume that the body force is constant. Develop expressions for the external nodal forces for that case.

The external forces are given by

${\mathbf {f} _{e}^{\mbox{ext}}={\frac {\rho b}{6}}{\begin{bmatrix}2x_{2}A_{1}+x_{2}A_{2}-2x_{1}A_{1}-x_{1}A_{2}\\2x_{2}A_{2}+x_{2}A_{1}-2x_{1}A_{2}-x_{1}A_{1}\end{bmatrix}}}$ ## 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

Find the fundamental natural frequency of the bar.

${f_{1}=36.963{\mbox{ Hz}}}$ Find the first three modal frequencies for a load of 40,000 lbf.

${f_{1}=33.174{\mbox{ Hz}}\quad f_{2}=144.107{\mbox{ Hz}}\quad f_{3}=328.443{\mbox{ Hz}}}$ 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

Find the fundamental natural frequency of the bar.

${f_{1}=1.509{\mbox{ Hz}}}$ Find the first five modal frequencies for a load of 10,000 lbf.

${f_{1}=88.329{\mbox{ Hz}}\quad f_{2}=203.114{\mbox{ Hz}}\quad f_{3}=320.174{\mbox{ Hz}}\quad f_{4}=441.655{\mbox{ Hz}}\quad f_{5}=571.350{\mbox{ Hz}}}$ 