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## Nonlinear Axial Bar

We describe the deformation of a nonlinear elastic bar. The non-linearity to be discussed can be classified as material non-linearity.

In our previous discussion we had assumed that the bar had a constant Young's modulus of ${\displaystyle E}$, that is, the constitutive model of the bar was linear elastic.

Let us now assume that the stress-strain relationship is nonlinear, and of the form

${\displaystyle {\sigma =E(u)~\varepsilon ,~\qquad \varepsilon :={\cfrac {du}{dx}},~\qquad ~E(u):=E_{0}\left(1-{\cfrac {du}{dx}}\right)}}$

where ${\displaystyle E_{0}}$ is a material constant which can be interpreted as the initial Young's modulus. Figure 2 shows how the stress-strain relationship for this material looks.

 Figure 2. Stress-strain relationship for the nonlinear bar.

The governing differential equation for the bar is

${\displaystyle {A~{\cfrac {d\sigma }{dx}}+ax=0~.}}$

If we plug in the stress-strain relation into the governing equation, we get

${\displaystyle {A~{\cfrac {d}{dx}}\left[E(u){\cfrac {du}{dx}}\right]+ax=0~.}}$

Notice that we have not included the expression for ${\displaystyle E(u)}$ in the above equation. The reason is that we want to maintain the symmetry of the stiffness matrix.

### Effect of including expression for ${\displaystyle E(u)}$

You can see what happens when we include the expression for ${\displaystyle E(u)}$ in the steps below. We get,

${\displaystyle A~{\cfrac {d}{dx}}\left[E_{0}\left(1-{\cfrac {du}{dx}}\right){\cfrac {du}{dx}}\right]+ax=0~.}$

or,

${\displaystyle AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}-2AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}~{\cfrac {du}{dx}}+ax=0~.}$

Now, when we try to derive the weak form, we get

${\displaystyle \int _{\Omega }AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}~w~dx-\int _{\Omega }2AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}~{\cfrac {du}{dx}}~w~dx+\int _{\Omega }ax~w~dx=0~.}$

The first integral is the same as before, and we get

${\displaystyle \int _{\Omega }AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}~w~dx=\left.AE_{0}{\cfrac {du}{dx}}~w\right|_{\Gamma _{t}}-\int _{\Omega }AE_{0}~{\cfrac {du}{dx}}~{\cfrac {dw}{dx}}~dx~.}$

The third integral remains the same. However, the second integral becomes

{\displaystyle {\begin{aligned}\int _{\Omega }2AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}~{\cfrac {du}{dx}}~w~dx&=\left.2AE_{0}{\cfrac {du}{dx}}~w~{\cfrac {du}{dx}}\right|_{\Gamma _{t}}-\int _{\Omega }2AE_{0}~{\cfrac {du}{dx}}~{\cfrac {d}{dx}}\left(w~{\cfrac {du}{dx}}\right)~dx\\&=\left.2AE_{0}\left({\cfrac {du}{dx}}\right)^{2}w\right|_{\Gamma _{t}}-\int _{\Omega }2AE_{0}~{\cfrac {du}{dx}}~\left({\cfrac {dw}{dx}}{\cfrac {du}{dx}}+w~{\cfrac {d^{2}u}{dx^{2}}}\right)~dx\\&=\left.2AE_{0}\left({\cfrac {du}{dx}}\right)^{2}w\right|_{\Gamma _{t}}-\int _{\Omega }2AE_{0}~\left({\cfrac {du}{dx}}\right)^{2}{\cfrac {dw}{dx}}~dx-\int _{\Omega }2AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}~{\cfrac {du}{dx}}~w~dx~.\end{aligned}}}

Rearranging, we get

${\displaystyle \int _{\Omega }2AE_{0}~{\cfrac {d^{2}u}{dx^{2}}}~{\cfrac {du}{dx}}~w~dx=\left.AE_{0}\left({\cfrac {du}{dx}}\right)^{2}w\right|_{\Gamma _{t}}-\int _{\Omega }AE_{0}~\left({\cfrac {du}{dx}}\right)^{2}{\cfrac {dw}{dx}}~dx~.}$

After collecting all the terms, the weak form becomes

${\displaystyle \left.AE_{0}{\cfrac {du}{dx}}~w\right|_{\Gamma _{t}}-\int _{\Omega }AE_{0}~{\cfrac {du}{dx}}~{\cfrac {dw}{dx}}~dx-\left.AE_{0}\left({\cfrac {du}{dx}}\right)^{2}w\right|_{\Gamma _{t}}+\int _{\Omega }AE_{0}~\left({\cfrac {du}{dx}}\right)^{2}{\cfrac {dw}{dx}}~dx+\int _{\Omega }ax~w~dx=0~.}$

Separating the terms corresponding to the internal and external forces, we get

${\displaystyle \int _{\Omega }AE_{0}~{\cfrac {du}{dx}}~{\cfrac {dw}{dx}}~dx-\int _{\Omega }AE_{0}~\left({\cfrac {du}{dx}}\right)^{2}{\cfrac {dw}{dx}}~dx=\int _{\Omega }ax~w~dx+\left.AE_{0}\left[{\cfrac {du}{dx}}-\left({\cfrac {du}{dx}}\right)^{2}\right]w\right|_{\Gamma _{t}}~.}$

If we choose trial and weighting functions

${\displaystyle u=\sum _{j}u_{j}N_{j}~;~\qquad w=N_{i}}$

and plug them into the weak form, we get

${\displaystyle \int _{\Omega }AE_{0}\left(\sum _{j}u_{j}{\cfrac {dN_{j}}{dx}}\right){\cfrac {dN_{i}}{dx}}~dx-\int _{\Omega }AE_{0}~\left(\sum _{j}u_{j}{\cfrac {dN_{j}}{dx}}\right)^{2}{\cfrac {dN_{i}}{dx}}~dx=\int _{\Omega }ax~N_{i}~dx+\left.AE_{0}\left[{\cfrac {du}{dx}}-\left({\cfrac {du}{dx}}\right)^{2}\right]w\right|_{\Gamma _{t}}~.}$

Taking the constants out of the integrals, we get (regarding this step, read the Discuss page of this article)

${\displaystyle AE_{0}\sum _{j}\left[\int _{\Omega }\left({\cfrac {dN_{j}}{dx}}{\cfrac {dN_{i}}{dx}}-u_{j}\left({\cfrac {dN_{j}}{dx}}\right)^{2}{\cfrac {dN_{i}}{dx}}\right)~dx\right]u_{j}=\int _{\Omega }ax~N_{i}~dx+\left.AE_{0}\left[{\cfrac {du}{dx}}-\left({\cfrac {du}{dx}}\right)^{2}\right]w\right|_{\Gamma _{t}}~.}$

Therefore, the coefficients of the stiffness matrix are given by

${\displaystyle K_{ij}(u_{j})=AE_{0}\int _{\Omega }\left[{\cfrac {dN_{i}}{dx}}{\cfrac {dN_{j}}{dx}}-u_{j}{\cfrac {dN_{i}}{dx}}\left({\cfrac {dN_{j}}{dx}}\right)^{2}\right]~dx~.}$

This stiffness matrix is not symmetric because of the second term inside the integral. That is, ${\displaystyle K_{ij}\neq K_{ji}}$. Also, we have not been able to remove the dependence of the stiffness matrix on the displacement.