I'm studying Finite Element Procedures by K.J. Bathe and I have trouble understanding how the the total and updated lagrangian formulations should give same equations. Taking a simple truss element as an example we can calculate the internal nodal point forces and compare them in the two formulations. The truss is fixed at one end (pendulum). Here is what I got. First from simple kinematics of the truss we can write:

Kinematics of the truss in TL and UL formulations For TL:

$\frac{\partial{}^{t}_{0}u_{1}}{\partial{}^{0}X_1}=\frac{{}^{0}L+\Delta L}{{}^{0} L}\cos(\theta)-1$, and $\frac{\partial{}^{t}_{0}u_{2}}{\partial{}^{0}X_1}=\frac{{}^{0}L+\Delta L}{{}^{0} L}\sin(\theta)$

For UL:

$\frac{\partial{}^{t}_{0}u_{1}}{\partial{}^{t}x_1}=\cos(\theta)-\frac{{}^{0} L }{{}^{0}L+\Delta L}$, and $\frac{\partial{}^{t}_{0}u_{2}}{\partial{}^{t}x_1}=\sin(\theta)$

1. Total Lagrangian formulation

The internal forces in TL formulation are given by

$F_{int}=\int_{{}^{0} V}{}^{t}_{0}S_{ij}\delta{}_{0} e_{ij}dV$, where ${}^{t}_{0}S_{ij}$ is the second PK stress at time $t$ measured from configuration $\Omega_0$ and ${}_{0} e_{ij}$ is the incremental linear strain term.

In Bathe's lecturse it is assumed that the stresses are known, but very little examples are given on how these are practically calculated. I know that one can write the stresses using the constitutive relations, i.e. ${}^{t}_{0}S_{ij}={}_{0} C_{ijkl}{}^{t}_{0}E_{kl}$, where $E_{kl}$ is the Green-Lagrange strain. So this is what I did. Now for the 2d truss example we have only one stress componenent giving,

$F_{int}=\int_{{}^{0} V}{}^{t}_{0}S_{11}\delta{}_{0} e_{11}dV=\int_{{}^{0} V}{}_{0} C_{1111}{}^{t}_{0}E_{11}\delta{}_{0} e_{11}dV$,


$E_{11}=u_{1,1}+0.5(u_{1,1}u_{1,1}+u_{2,1}u_{2,1})$ and $\delta{}_{0} e_{11}=\frac{1}{{}^{0}L}[-\cos(\theta),-\sin(\theta),\cos(\theta),\sin(\theta)][\delta u^1_1,\delta u^1_2,\delta u^2_1,\delta u^2_2]^T$

and using the above kinematic relations we get,

$F_{int}= A{}_{0} C_{1111}\frac{({}^{0}L+\Delta L)(1-{}^{0}L^2)}{2{}^{0}L^3}\begin{bmatrix}-\cos(\theta)\\ -\sin(\theta)\\\cos(\theta)\\ \sin(\theta)\end{bmatrix}$

2. Updated Lagrangian Formulation

Now doing the same analysis for the UL formulation using the UL kinematic relations defined in the beginning. We have for the internal forces,

$\tilde{F}_{int}=\int_{{}^{t} V}{}^{t} \sigma_{ij}\delta{}_{t} \tilde{e}_{ij}dV$, where the tilde indicates that the quantity is in the local coordinate frame.

Now the Cauchy stress can be written in terms of the Euler-Almansi strain tensor. Here is where my confusion starts. Bathe mentions that it is computationally more efficient to calculate the Cauchy stress by transforming the PK2 stress to the Cauchy stress using the deformation gradient rather than using the Almansi strain. I guess my first question is Which method should you use to calculate the Cauchy stress? Or should you use the Jaumann rate instead? Either way I'll use in this example the Almansi strain. So we get, ${}^{t} \sigma_{11}={}_{t} C_{1111}{}^{t}_{t}\epsilon^A_{11}$, where,


Now solving for the internal forces and transforming the forces to the global coordinate system using $F_{int}=T^T\tilde{F}_{int}$, where $T=\begin{bmatrix}c & s & 0 & 0\\-s & c & 0 & 0\\0&0&c&s\\0&0&-s&c\end{bmatrix}$. We also now have $\delta{}_{t} \tilde{e}_{ij}=\frac{1}{{}^{1}L}[-1,0,1,0][\delta \tilde{u}^1_1,\delta \tilde{u}^1_2,\delta \tilde{u}^2_1,\delta \tilde{u}^2_2]^T=\frac{1}{{}^{0}L+\Delta L}[-1,0,1,0][\delta \tilde{u}^1_1,\delta \tilde{u}^1_2,\delta \tilde{u}^2_1,\delta \tilde{u}^2_2]^T$. Therefore, we get,

$F_{int}=A{}_{0}C_{1111}\frac{{}^{0} L+\Delta L}{2{}^{0} L^3}(2{}^{0} L{}^{1} L+{}^{1} L^2(\sin(\theta)^2-2\cos(\theta))+({}^{0} L+{}^{1} L\cos(\theta))^2)\begin{bmatrix}\cos(\theta)\\ \sin(\theta)\\-\cos(\theta)\\ -\sin(\theta)\end{bmatrix}$

The results are obviously different, but I'm not sure why. So my second question is why do the formulations give different results?


I realized that I used the approximation ${}_{0}C_{1111}={}_{1}C_{1111}$, but I don't understand why the material would have different properties in the different formulations. Either way I tried using the relationship given in Finite Element Procedures on page 549. So i set

${}_{1}C_{1111}=(\frac{{}^{0} L+\Delta L}{{}^{0} L})^3{}_{0}C_{1111}$,

but the forces are still different.



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