Nonlinear finite elements/Stress and strain in one and two dimension

1-D strain measures

• Engineering strain :

${\epsilon }_E:=\frac{l-L}{L}=\frac{\mathrm{\Delta }L}{L}$

• Natural/Logarithmic/True strain:

${\epsilon }_L:=\underset{L}{\overset{l}{\int }}\frac{dl}{l}=\ln (l){|}_L^l=\ln \left(\frac{l}{L}\right)$

Relation between engineering and true strain:

${\epsilon }_L=\ln \left(\frac{L+\mathrm{\Delta }L}{L}\right)=\ln (1+\frac{\mathrm{\Delta }L}{L})=\ln (1+{\epsilon }_E)$

• Green (Lagrangian) strain :

${\epsilon }_G:=\frac{1}{2}\left(\frac{l^2-L^2}{L^2}\right)=\frac{1}{2}(\frac{l^2}{L^2}-1)$

• Almansi-Hamel (Eulerian) strain :

${\epsilon }_A:=\frac{1}{2}\left(\frac{l^2-L^2}{l^2}\right)=\frac{1}{2}(1-\frac{L^2}{l^2})$

1-D stress measures

• Engineering/Nominal stress:

$P={\sigma }_E:=\frac{T}{A}$

• Cauchy/True stress:

$\sigma ={\sigma }_T:=\frac{T}{a}$

Relation between engineering and true stress (no volume change):

${\sigma }_T=\frac{T}{a}=\frac{Tl}{AL}={\sigma }_E\left(\frac{L+\mathrm{\Delta }L}{L}\right)={\sigma }_E(1+{\epsilon }_E)$

1-D stress-strain relations

• True stress - Green strain:

${\sigma }_T=E_{TG}\left(\frac{l^2-L^2}{2L^2}\right)$

• True stress - True strain:

${\sigma }_T=E_{TT}\ln \left(\frac{l}{L}\right)$

Example

File:BarStretch.png Stretching of a bar

$cock$

$\sin \theta =\frac{x}{l(x)}$

• Assume incompressible material.

$V=v;AL=al$or :$V=al;a=V/l$}

• Equilibrium.

$T(x)=F$:$⟹R(x)=T(x)-F=0$

where

$\begin{array}{cc}T(x)& =\sigma (x)a(x)\sin \theta \\ & =\frac{\sigma (x)a(x)x}{l(x)}\\ & =\frac{\sigma (x)Vx}{l(x{)}^2}\end{array}$

Stress-strain relation 1

$\sigma (x)=E\left(\frac{l(x{)}^2-L^2}{2L^2}\right)$

Then,

$\begin{array}{cc}T(x)& =\frac{Ea(x)x}{l(x)}\left(\frac{l(x{)}^2-L^2}{2L^2}\right)\\ & =\frac{EVx}{l^2}\left(\frac{l^2-L^2}{2L^2}\right)\end{array}$

and

$R(x)=\frac{EVx}{l^2}\left(\frac{l^2-L^2}{2L^2}\right)-F$

Highly nonlinear in $x$.

Stiffness

Stiffness = change in equilibrium equation due to change in position.

$K(x)=\frac{dR(x)}{dx}=\frac{dT(x)}{dx}(\text{if}F\text{ is constant})$

Now,

$T(x)=\frac{\sigma (x)Vx}{l(x{)}^2}$

Therefore,

$\begin{array}{cc}\frac{dT}{dx}& =V[x\frac{d}{dx}\left(\frac{\sigma }{l^2}\right)+\frac{\sigma }{l^2}]\\ & =V[x\frac{d}{d\sigma }\left(\frac{\sigma }{l^2}\right)\frac{d\sigma }{dx}+x\frac{d}{dl}\left(\frac{\sigma }{l^2}\right)\frac{dl}{dx}+\frac{\sigma }{l^2}]=V[\frac{x}{l^2}\frac{d\sigma }{dl}\frac{dl}{dx}-\frac{2x\sigma }{l^3}\frac{dl}{dx}+\frac{\sigma }{l^2}]\\ & =\frac{Vx}{l^2}(\frac{d\sigma }{dl}-\frac{2\sigma }{l})\frac{dl}{dx}+\frac{V\sigma }{l^2}=\frac{Vx}{l^2}(\frac{d\sigma }{dl}-\frac{2\sigma }{l})\frac{x}{l}+\frac{V\sigma }{l^2}\\ ⟹K& =a(\frac{d\sigma }{dl}-\frac{2\sigma }{l})\frac{x^2}{l^2}+\frac{\sigma a}{l}\end{array}$

$\sigma =E\left(\frac{l^2-L^2}{2L^2}\right)⟹\frac{d\sigma }{dl}=\frac{El}{L^2}$

$\begin{array}{cc}K& =a(\frac{d\sigma }{dl}-\frac{2\sigma }{l})\frac{x^2}{l^2}+\frac{\sigma a}{l}\\ & =a(\frac{El}{L^2}-\frac{2\sigma }{l})\frac{x^2}{l^2}+\frac{\sigma a}{l}\\ & =\frac{A}{L}(E-2S)\frac{x^2}{l^2}+\frac{SA}{L}\end{array}$

Initial stress/Geometric stiffness

$\frac{SA}{L};S=\sigma \frac{L^2}{l^2}=P\frac{L}{l}$

Stress-strain relation 2

$\sigma (x)=E\ln \left(\frac{l(x)}{L}\right)$

Then,

$\begin{array}{cc}T(x)& =\frac{Ea(x)x}{l(x)}\ln \left(\frac{l(x)}{L}\right)\\ & =\frac{EVx}{l^2}\ln \left(\frac{l}{L}\right)\end{array}$

and

$R(x)=\frac{EVx}{l^2}\ln \left(\frac{l}{L}\right)-F$

Highly nonlinear in $x$.

$\sigma =E\ln \left(\frac{l}{L}\right)⟹\frac{d\sigma }{dl}=\frac{E}{l}$

Stiffness

$\begin{array}{cc}K& =a(\frac{d\sigma }{dl}-\frac{2\sigma }{l})\frac{x^2}{l^2}+\frac{\sigma a}{l}\\ & =a(\frac{E}{l}-\frac{2\sigma }{l})\frac{x^2}{l^2}+\frac{\sigma a}{l}\\ & =\frac{a}{l}(E-2\sigma )\frac{x^2}{l^2}+\frac{\sigma a}{l}\end{array}$

Initial stress/Geometric stiffness:

$\frac{\sigma a}{l}$

Strain measures in two dimensions

File:Rot2D.png Strains in two dimensions

Small strains

$\begin{array}{cc}{\epsilon }_{xx}& =\frac{\partial u_x}{\partial x}\\ {\epsilon }_{yy}& =\frac{\partial u_y}{\partial y}\\ {\epsilon }_{xy}& =\frac{1}{2}(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x})\end{array}$

For 90$o^{}$ rotation,

$u_x=-Y-X;u_y=X-Y$

Then strains are:

$\begin{array}{cc}{\epsilon }_{xx}& =-1\\ {\epsilon }_{yy}& =-1\\ {\epsilon }_{xy}& =0\end{array}$

Rotation should not lead to non-zero strains!

Finite strains

File:Green2D.png Green strain in two dimensions

For 90$o^{}$ rotation,

$u_x=-Y-X;u_y=X-Y$

Then,

$E_{xx}=0;E_{yy}=0;E_{xy}=0$

Green strain (1-D)

${\epsilon }_G=\frac{l^2-L^2}{2L^2}$

In 2-D:

$E_{xx}=\frac{d{s}^2-d{X}^2}{2d{X}^2}$

Now,

$d{s}^2=d{X}^2{(1+\frac{\partial u_x}{\partial X})}^2+d{X}^2{\left(\frac{\partial u_y}{\partial X}\right)}^2$

Therefore,

$\begin{array}{cc}{E}_{xx}& =\frac{d{X}^2}{2d{X}^2}[{(1+\frac{\partial u_x}{\partial X})}^2+{\left(\frac{\partial u_y}{\partial X}\right)}^2-1]\\ & =\frac{\partial u_x}{\partial X}+\frac{1}{2}[{\left(\frac{\partial u_x}{\partial X}\right)}^2+{\left(\frac{\partial u_y}{\partial X}\right)}^2]\end{array}$

Similar for $E_{yy}$ and $E_{xy}$.

Template:Subpage navbar