Chapter 2 Direct Approach for Discrete Systems
Basic steps for FEM:
Step 1. Preprocessing
Step 2. Element formulation
Step 3. Assembly
Step 4. Solving equations
Step 5. Postprocessing
Bar element
Step 2. Describe the behavior of an element
Content: relate 'nodal internal forces' to 'nodal displacements'
nodal internal forces
\(\downarrow\) ------- Linearity
stress
\(\downarrow\) ------- Hook's Law
strain
\(\downarrow\) ------- Linearity
nodal displacements
\(\downarrow\) ------- Equilibrium of the element
element formulation
P13 sign convention for 'internal axial force' and 'nodal internal forces'
Step 3. Assembly
Essence: equilibrium equations for each node + compatibility (local and global)
P15 At each node, either the external forces or the nodal dsiplacements are known, but not both
P16 The sum of internal element forces is equal to that of the external forces and reactions
Basic idea: matrix scatter and add (argument,add)
Program implentation: Directly assembly (direct add based on global id)
Equations for assembly:
nodal displacements of each element: \(\pmb{d}^e=\pmb{L}^e\pmb{d}\)
where \(\pmb{L}^e\) gather matrices, \(\pmb{d}\) global displacement matrix
\(\pmb{K}^e\pmb{d}^e=\pmb{K}^e\pmb{L}^e\pmb{d}=\pmb{F}^e\)
Note that \((\pmb{L}^e)^T\) scatters the nodal forces into global matrix
\(\sum_{e=1}^{n_e}(\pmb{L}^e)^T\pmb{F}^e=\pmb{f}+\pmb{r}\)
where \(\pmb{f}\) prescribed external force, \(\pmb{r}\) reactions at essential boundary...
\((\pmb{L}^e)^T\pmb{K}^e\pmb{L}^e\pmb{d}=(\pmb{L}^e)^T\pmb{F}^e\), \(e=1,...,n^e\)
adding above element equations
\(\pmb{K}\pmb{d}=\pmb{f}+\pmb{r}\)
where \(\pmb{K}=\sum_{e=1}^{n_e}(\pmb{L}^e)^T\pmb{K}^e\pmb{L}^e\) is the global stiffness matrix
Above equations are equivalent to directly assembly and matrix scatter and add
Impose boundary conditions
- Global system partition
\(\left[\begin{matrix} \pmb{K}_{E} & \pmb{K}_{EF}\\ \pmb{K}_{EF}^T & \pmb{K}_{F} \end{matrix}\right] \left[\begin{matrix} \overline{\pmb{d}}_{E}\\ \pmb{d}_{F} \end{matrix}\right]= \left[\begin{matrix} \pmb{r}_{E}\\ \pmb{f}_{F} \end{matrix}\right]\)
where E-nodes for displacements known (essential), F-nodes for displacements unknown (free)
Solve \(\pmb{d}_{F}\) from second row then taken into first row to get \(\pmb{r}_{E}\) + Draw 0 and set 1
\(\left[\begin{matrix} 1 & 0 & 0\\ 0 & {K}_{22} & {K}_{23} \\ 0 & {K}_{32} & {K}_{33} \end{matrix}\right] \left[\begin{matrix} u_{1}\\ u_{2}\\ u_{3} \end{matrix}\right]= \left[\begin{matrix} \overline{u}_1\\ f_{2}-\overline{u}_1K_{21}\\ f_3-\overline{u}_1K_{31} \end{matrix}\right]\)
where \(\overline{u}_1\) is the prescribed displacement
Solve \(u_i\) then back substitution to get \(r_1\) + Penalty method
\(\left[\begin{matrix} \beta & {K}_{12} & {K}_{13}\\ {K}_{21} & {K}_{22} & {K}_{23} \\ {K}_{31} & {K}_{32} & {K}_{33} \end{matrix}\right] \left[\begin{matrix} u_{1}\\ u_{2}\\ u_{3} \end{matrix}\right]= \left[\begin{matrix} \beta\overline{u}_1\\ f_2\\ f_3 \end{matrix}\right]\)
where \(\beta\) is a very large number
Almost the same as 'Draw 0 and set 1'
For its tendency to decrease the conditioning of the equations，only applied for matrices of moderate size (up to about 10000 unknowns)
Application to other linear systems
Above theory applicable for system characterized by:
1 A balance or conservation law for the flux
2 A linear law relating the flux to the potential
3 A continuous potential (i.e. a compatible potential)
e.g.
mechanical bar (pot: displacement, flux: stress);
steady-state electrical flow in a circuit (pot: voltages, flux: current);
fluid flow in a hydraulic piping system (pot: pressure, flux: flow rate)...
Transformation law
For rotation from one coordinate to another or scatter operation
Objective: Get the expression of \(\overline{\pmb{K}^e}\)
Known:
transformation matrix \(\pmb{T}^e\)
\(\widehat{\pmb{d}^e}=\pmb{T}^e\overline{\pmb{d}^e}\)
\(\widehat{\pmb{F}^e}=\widehat{\pmb{K}^e}\widehat{\pmb{d}^e}\)
\(\delta\widehat{\pmb{d}^e}\) is an infinite small element displacement (\(\widehat{\pmb{F}^e}\) remains constant)
\(\delta W_{int}=\delta\widehat{\pmb{d}^e}^T\widehat{\pmb{F}^e}=\delta\widehat{u_1^e}\widehat{F_1^e}+\delta\widehat{u_2^e}\widehat{F_2^e}\) is the work done by the internal forces
The work should be the same in two coordinates system
\(\delta\widehat{\pmb{d}^e}^T\widehat{\pmb{F}^e}=\delta\overline{\pmb{d}^e}^T\overline{\pmb{F}^e}=\delta{\overline{\pmb{d}^e}}^T{\pmb{T}^e}^T\widehat{\pmb{F}^e}\)
So \(\delta\overline{\pmb{d}^e}^T({\pmb{T}^e}^T\widehat{\pmb{F}^e}-\overline{\pmb{F}^e})=0\)
For arbitary \(\delta\overline{\pmb{d}^e}\), above equations holds, according to the vector scalar product theorem
\(\overline{\pmb{F}^e}={\pmb{T}^e}^T\widehat{\pmb{F}^e}={\pmb{T}^e}^T\widehat{\pmb{K}^e}\widehat{\pmb{d}^e}={\pmb{T}^e}^T\widehat{\pmb{K}^e}\pmb{T}^e\overline{\pmb{d}^e}\)
So \(\overline{\pmb{K}^e}={\pmb{T}^e}^T\widehat{\pmb{K}^e}\pmb{T}^e\)
Energy is invariant with respect to the frame of reference -----> **Principle of virtual work** & **Theorem of minimum potential energy**