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:
 * $$ \sigma_{T} = E_{TG} \left(\cfrac{l^2 - L^2}{2 L^2}\right)$$


 * True stress - True strain:
 * $$ \sigma_{T} = E_{TT} \ln\left(\cfrac{l}{L}\right)$$

Example


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

Stress-strain relation 1


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

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

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



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



\begin{align} 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{align} $$

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{align} 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{align} $$ 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{align} 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{align} $$

Initial stress/Geometric stiffness:

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

Small strains


\begin{align} \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{align} $$ For 90$$^o$$ rotation,

u_x = -Y - X ~; u_y = X - Y $$ Then strains are:

\begin{align} \varepsilon_{xx} & = -1 \\ \varepsilon_{yy} & = -1 \\ \varepsilon_{xy} & = 0 \end{align} $$ Rotation should not lead to non-zero strains!

Finite strains
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{align} 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{align} $$ Similar for $$E_{yy}$$ and $$E_{xy}$$.