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.

${\displaystyle \mathbf {u} (X,t)={\frac {1}{l_{0}}}[X_{2}-X\quad X-X_{1}]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}}$

where ${\displaystyle l_{0}=X_{2}-X_{1}}$. The strain measure is evaluated in terms of the nodal displacement,

${\displaystyle {\boldsymbol {\varepsilon }}(X,t)=u_{,X}={\frac {1}{l_{0}}}[-1\quad 1]{\begin{bmatrix}u_{1}(t)\\u_{2}(t)\end{bmatrix}}}$

which defines the ${\displaystyle \mathbf {B} _{0}}$ matrix to be

${\displaystyle \mathbf {B} _{0}={\frac {1}{l_{0}}}[-1\quad 1].}$

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

${\displaystyle \mathbf {f} _{e}^{\mbox{int}}=\int _{X_{1}}^{X_{2}}\mathbf {B} _{0}^{T}(A_{0}P)dX={\frac {1}{l_{0}}}\int _{X_{1}}^{X_{2}}\left((1-\xi )A_{01}+\xi A_{02}\right)\left((1-\xi )P_{1}+\xi P_{2}\right){\begin{bmatrix}-1\\1\end{bmatrix}}dX}$

Integrating the above integral with ${\displaystyle \xi =(X-X_{1})/l_{0}}$ to obtain

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

Part 2

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

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

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, ${\displaystyle b}$, are obtained by the usual procedure.

${\displaystyle \mathbf {f} _{e}^{\mbox{ext}}=\int _{X_{1}}^{X_{2}}\mathbf {N} ^{T}\rho _{0}A_{0}bdX={\frac {\rho _{0}b}{l_{0}}}\int _{X_{1}}^{X_{2}}A_{0}{\begin{bmatrix}X_{2}-X\\X-X_{1}\end{bmatrix}}dX}$

${\displaystyle {\mathbf {f} _{e}^{\mbox{ext}}={\frac {\rho _{0}b}{6}}{\begin{bmatrix}x_{1}^{2}A_{02}+2A_{01}x_{1}^{2}-2x_{1}A_{02}x_{2}-4A_{01}x_{1}x_{2}+2A_{01}x_{2}^{2}+A_{02}x_{2}^{2}\\A_{01}x_{1}^{2}+2x_{1}^{2}A_{02}-4x_{1}A_{02}x_{2}-2A_{01}x_{1}x_{2}+A_{01}x_{2}^{2}+2A_{02}x_{2}^{2}\end{bmatrix}}}}$

Part 4

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

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

Part 5

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

The element mass matrix is

${\displaystyle \mathbf {M} _{e}=\int _{X_{1}}^{X_{2}}\rho _{0}A_{0}\mathbf {N} ^{T}\mathbf {N} dX}$

${\displaystyle {\mathbf {M} _{e}={\frac {\rho _{0}l_{0}}{12}}{\begin{bmatrix}3A_{01}+A_{02}&A_{01}+A_{02}\\A_{01}+A_{02}&A_{01}+3A_{02}\end{bmatrix}}}}$

Part 6

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

Lumped mass matrix is given by

${\displaystyle M_{ii}=\int _{\xi _{1}}^{\xi _{2}}\rho _{0}A_{0}N_{i}dX}$

${\displaystyle {\mathbf {M} _{e}={\frac {\rho _{0}l_{0}}{6}}{\begin{bmatrix}2A_{01}+A_{02}&0\\0&A_{01}+2A_{02}\end{bmatrix}}}}$

Part 7

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

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

with

${\displaystyle \mathbf {K} ={\cfrac {E(A_{01}+A_{02})}{2l_{0}}}{\begin{bmatrix}1&-1\\-1&1\end{bmatrix}}}$

where ${\displaystyle E}$ is the Young's modulus and ${\displaystyle l_{0}}$ is the initial length of the element.

The above equation can be rewrite as

${\displaystyle (\mathbf {K} -\omega ^{2}\mathbf {M} )\mathbf {u} =0}$

which only has a solution if

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

Solving the above determinant for ${\displaystyle \omega }$, we have

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

Part 1

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

The velocity field is

${\displaystyle v(X,t)={\frac {1}{l_{0}}}\left[X_{2}-X\quad X-X_{1}\right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}}$

In term of element coordinates, the velocity field is

${\displaystyle v(\xi ,t)={\frac {1}{l_{0}}}\left[1-\xi \quad \xi \right]{\begin{bmatrix}v_{1}(t)\\v_{2}(t)\end{bmatrix}}\qquad \xi ={\frac {X-X_{1}}{l_{0}}}}$

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

${\displaystyle u(\xi ,t)=\mathbf {N} (\xi )\mathbf {u} _{e}(t)}$

Therefore, since ${\displaystyle x=X+u}$

${\displaystyle x(\xi ,t)=\mathbf {N} (\xi )\mathbf {x} _{e}(t)=\left[1-\xi \quad \xi \right]{\begin{bmatrix}x_{1}(t)\\x_{2}(t)\end{bmatrix}}\qquad \xi _{,\xi }=x_{2}-x_{1}=l}$

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

${\displaystyle \xi ={\frac {x-x_{1}}{x_{2}-x_{1}}}={\frac {x-x_{1}}{l}},\quad l=x_{2}-x_{1},\quad \xi _{,x}={\frac {1}{l}}}$

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

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

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

Using (146) in Handout 13, we have

${\displaystyle \mathbf {f} _{e}^{\mbox{int}}=\int _{x_{1}}^{x_{2}}\mathbf {B} ^{T}\sigma Adx=\int _{x_{1}}^{x_{2}}{\frac {\sigma A}{l}}{\begin{bmatrix}-1\\1\end{bmatrix}}dx}$

Integrating the above equation to obtain

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

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

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