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

1-D strain measures

• Engineering strain :

$\varepsilon _{E}:={\cfrac {l-L}{L}}={\cfrac {\Delta L}{L}}$

• Natural/Logarithmic/True strain:

$\varepsilon _{L}:=\int _{L}^{l}{\cfrac {dl}{l}}=\ln(l){\Big |}_{L}^{l}=\ln \left({\cfrac {l}{L}}\right)$

Relation between engineering and true strain:

$\varepsilon _{L}=\ln \left({\cfrac {L+\Delta L}{L}}\right)=\ln \left(1+{\cfrac {\Delta L}{L}}\right)=\ln \left(1+\varepsilon _{E}\right)$

• Green (Lagrangian) strain :

$\varepsilon _{G}:={\frac {1}{2}}\left({\cfrac {l^{2}-L^{2}}{L^{2}}}\right)={\frac {1}{2}}\left({\cfrac {l^{2}}{L^{2}}}-1\right)$

• Almansi-Hamel (Eulerian) strain :

$\varepsilon _{A}:={\frac {1}{2}}\left({\cfrac {l^{2}-L^{2}}{l^{2}}}\right)={\frac {1}{2}}\left(1-{\cfrac {L^{2}}{l^{2}}}\right)$

1-D stress measures

• Engineering/Nominal stress:

$P=\sigma _{E}:={\cfrac {T}{A}}$

• Cauchy/True stress:

$\sigma =\sigma _{T}:={\cfrac {T}{a}}$

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

$\sigma _{T}={\cfrac {T}{a}}={\cfrac {Tl}{AL}}=\sigma _{E}\left({\cfrac {L+\Delta L}{L}}\right)=\sigma _{E}\left(1+\varepsilon _{E}\right)$

1-D stress-strain relations

• True stress - Green strain:

$\sigma _{T}=E_{TG}\left({\cfrac {l^{2}-L^{2}}{2L^{2}}}\right)$

• True stress - True strain:

$\sigma _{T}=E_{TT}\ln \left({\cfrac {l}{L}}\right)$

Example

Stretching of a bar

$cock$

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

• Assume incompressible material.

$V=v~;~~AL=al$or :$V=al~;~~a=V/l$}

• Equilibrium.

$T(x)=F$:$\implies ~~R(x)=T(x)-F=0$

where

${\begin{aligned}T(x)&=\sigma (x)~a(x)~\sin \theta \\&={\cfrac {\sigma (x)~a(x)~x}{l(x)}}\\&={\cfrac {\sigma (x)~V~x}{l(x)^{2}}}\end{aligned}}$

Stress-strain relation 1

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

Then,

${\begin{aligned}T(x)&={\cfrac {E~a(x)~x}{l(x)}}\left({\cfrac {l(x)^{2}-L^{2}}{2L^{2}}}\right)\\&={\cfrac {E~V~x}{l^{2}}}\left({\cfrac {l^{2}-L^{2}}{2L^{2}}}\right)\end{aligned}}$

and

$R(x)={\cfrac {E~V~x}{l^{2}}}\left({\cfrac {l^{2}-L^{2}}{2L^{2}}}\right)-F$

Highly nonlinear in $x$.

Stiffness

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

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

Now,

$T(x)={\cfrac {\sigma (x)~V~x}{l(x)^{2}}}$

Therefore,

${\begin{aligned}{\cfrac {dT}{dx}}&=V\left[x~{\cfrac {d}{dx}}\left({\cfrac {\sigma }{l^{2}}}\right)+{\cfrac {\sigma }{l^{2}}}\right]\\&=V\left[x~{\cfrac {d}{d\sigma }}\left({\cfrac {\sigma }{l^{2}}}\right){\cfrac {d\sigma }{dx}}+x~{\cfrac {d}{dl}}\left({\cfrac {\sigma }{l^{2}}}\right){\cfrac {dl}{dx}}+{\cfrac {\sigma }{l^{2}}}\right]=V\left[{\cfrac {x}{l^{2}}}{\cfrac {d\sigma }{dl}}{\cfrac {dl}{dx}}-{\cfrac {2x\sigma }{l^{3}}}{\cfrac {dl}{dx}}+{\cfrac {\sigma }{l^{2}}}\right]\\&={\cfrac {Vx}{l^{2}}}\left({\cfrac {d\sigma }{dl}}-{\cfrac {2\sigma }{l}}\right){\cfrac {dl}{dx}}+{\cfrac {V\sigma }{l^{2}}}={\cfrac {Vx}{l^{2}}}\left({\cfrac {d\sigma }{dl}}-{\cfrac {2\sigma }{l}}\right){\cfrac {x}{l}}+{\cfrac {V\sigma }{l^{2}}}\\\implies K&=a\left({\cfrac {d\sigma }{dl}}-{\cfrac {2\sigma }{l}}\right){\cfrac {x^{2}}{l^{2}}}+{\cfrac {\sigma ~a}{l}}\end{aligned}}$

$\sigma =E\left({\cfrac {l^{2}-L^{2}}{2L^{2}}}\right)\implies {\cfrac {d\sigma }{dl}}={\cfrac {E~l}{L^{2}}}$

${\begin{aligned}K&=a\left({\cfrac {d\sigma }{dl}}-{\cfrac {2\sigma }{l}}\right){\cfrac {x^{2}}{l^{2}}}+{\cfrac {\sigma ~a}{l}}\\&=a\left({\cfrac {E~l}{L^{2}}}-{\cfrac {2\sigma }{l}}\right){\cfrac {x^{2}}{l^{2}}}+{\cfrac {\sigma ~a}{l}}\\&={\cfrac {A}{L}}\left(E-2S\right){\cfrac {x^{2}}{l^{2}}}+{\cfrac {S~A}{L}}\end{aligned}}$

Initial stress/Geometric stiffness

${\cfrac {S~A}{L}}~;~~S=\sigma {\cfrac {L^{2}}{l^{2}}}=P{\cfrac {L}{l}}$

Stress-strain relation 2

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

Then,

${\begin{aligned}T(x)&={\cfrac {E~a(x)~x}{l(x)}}\ln \left({\cfrac {l(x)}{L}}\right)\\&={\cfrac {E~V~x}{l^{2}}}\ln \left({\cfrac {l}{L}}\right)\end{aligned}}$

and

$R(x)={\cfrac {E~V~x}{l^{2}}}\ln \left({\cfrac {l}{L}}\right)-F$

Highly nonlinear in $x$.

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

Stiffness

${\begin{aligned}K&=a\left({\cfrac {d\sigma }{dl}}-{\cfrac {2\sigma }{l}}\right){\cfrac {x^{2}}{l^{2}}}+{\cfrac {\sigma ~a}{l}}\\&=a\left({\cfrac {E}{l}}-{\cfrac {2\sigma }{l}}\right){\cfrac {x^{2}}{l^{2}}}+{\cfrac {\sigma ~a}{l}}\\&={\cfrac {a}{l}}\left(E-2\sigma \right){\cfrac {x^{2}}{l^{2}}}+{\cfrac {\sigma ~a}{l}}\end{aligned}}$

Initial stress/Geometric stiffness:

${\cfrac {\sigma ~a}{l}}$

Strain measures in two dimensions

Strains in two dimensions

Small strains

${\begin{aligned}\varepsilon _{xx}&={\frac {\partial u_{x}}{\partial x}}\\\varepsilon _{yy}&={\frac {\partial u_{y}}{\partial y}}\\\varepsilon _{xy}&={\frac {1}{2}}\left({\frac {\partial u_{x}}{\partial y}}+{\frac {\partial u_{y}}{\partial x}}\right)\end{aligned}}$

For 90$^{o}$ rotation,

$u_{x}=-Y-X~;~~u_{y}=X-Y$

Then strains are:

${\begin{aligned}\varepsilon _{xx}&=-1\\\varepsilon _{yy}&=-1\\\varepsilon _{xy}&=0\end{aligned}}$

Rotation should not lead to non-zero strains!

Finite strains

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)

$\varepsilon _{G}={\cfrac {l^{2}-L^{2}}{2L^{2}}}$

In 2-D:

$E_{xx}={\cfrac {ds^{2}-dX^{2}}{2dX^{2}}}$

Now,

$ds^{2}=dX^{2}\left(1+{\frac {\partial u_{x}}{\partial X}}\right)^{2}+dX^{2}\left({\frac {\partial u_{y}}{\partial X}}\right)^{2}$

Therefore,

${\begin{aligned}E_{xx}&={\cfrac {dX^{2}}{2dX^{2}}}\left[\left(1+{\frac {\partial u_{x}}{\partial X}}\right)^{2}+\left({\frac {\partial u_{y}}{\partial X}}\right)^{2}-1\right]\\&={\frac {\partial u_{x}}{\partial X}}+{\frac {1}{2}}\left[\left({\frac {\partial u_{x}}{\partial X}}\right)^{2}+\left({\frac {\partial u_{y}}{\partial X}}\right)^{2}\right]\end{aligned}}$

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

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