Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE
bx
F
L = 1m, A = 100mm2, E = 100 GPa, F = 10kN, bx = 10kN/m
Approximate solution
u( x ) = c1x + c2 x 2
Strain energy
U=
L
L
0 UL ( x )dx = 0
1
2
AE ex dx =
2
L
0
2
1 du
AE dx
2 dx
Potential energy of forces
L
2
1
1
4
2
2
U ( c1, c2 ) = AE
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
RAYLEIGHRITZ METHOD
PMPE is good for discrete system (exact solution)
RayleighRitz method approximates a continuous system as a
discrete system with finite number of DOFs
Approximate the displacements by a function containing finite
number of coeffic
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE cont.
Total potential energy is minimized with respect to the DOFs
P
P
P
P
= 0,
= 0,
= 0 or,
=0
u1
u 2
u3
cfw_Q
u1 F1
[ K ] u2 = F2
u F
3 3
Global FE equations
[K ]cfw_Q = cfw_F
Finite element equations
300 100 200 0 F1
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE cont.
3
U = U ( e )
Strain energy of the system
e=1
U=
1
u1 u2
2
k (1) + k (2)
u3 k (1)
k (2)
1
U = cfw_Q T [K ]cfw_Q
2
k (1)
k (1) + k (3)
k (3)
cfw_Q = cfw_u1, u2, u3 T
Potential energy of applied forces
V = ( F1u1 + F2u2 + F3u3 ) = 
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
PRINCIPLE OF MINIMUM POTENTIAL ENERGY
Strain energy density of 1D body
1
1 2
U 0 = s x ex = E ex
2
2
Variation in the strain energy density by du(x)
dU0 = E ex dex = s x dex
Variation of strain energy
L
dU =
L
L
dU0dAdx = sx dex dAdx = Pdex dx
0 A
dU
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
VARIATION OF A FUNCTION
Virtual displacements in the previous section can be considered as a
variation of real displacements
Perturbation of displ u(x) by arbitrary virtual displ du(x)
ut ( x ) = u ( x ) + tdu ( x )
Variation of displacement
dut ( x )
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
PMPE cont.
Potential energy of external forces
Force F is applied at x = L with corresponding virtual displ du(L)
Work done by the force = Fdu(L)
The potential is reduced by the amount of work
dV = d ( Fu (L ) )
dV = F du (L )
With distributed forc
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE: PMPE TO DISCRETE SYSTEMS
Express U and V in terms of
displacements, and then
differential P w.r.t displacements
k(1) = 100 N/mm, k(2) = 200 N/mm
k(3) = 150 N/mm, F2 = 1,000 N
F3 = 500 N
Strain energy of elements (springs)
U
(1)
1 (1)
2
= k ( u
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
PVW cont.
Integration by parts
L
L
P du  P
0
0
L
d ( du )
dx + bx du( x )dx = 0
dx
0
At x = 0, u(0) = 0. Thus, du(0) = 0
the virtual displacement should be consistent with the displacement
constraints of the body
At x = L, P(L) = F
Virtual strain
P
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
ENERGY METHOD
Powerful alternative method to obtain FE equations
Principle of virtual work for a particle
for a particle in equilibrium the virtual work is identically equal to zero
Virtual work: work done by the (real) external forces through the vir
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
PVW cont.
in equilibrium, the sum of external and internal virtual work is zero
for every virtual displacement field
3D PVW has the same form with different expressions
With distributed forces and concentrated forces
dWe =
( t x du + t y dv + tzdw ) d
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
PRINCIPLE OF VIRTUAL WORK
Deformable body (uniaxial bar under body force and tip force)
x
E, A(x)
F
Bx
L
Equilibrium equation:
PVW
L
dsx
+ Bx = 0
dx
dsx
dx + Bx du( x )dAdx = 0
0 A
Integrate over the area, axial force P(x) = As(x)
L
dP
+ bx du( x )
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE
Use three equallength elements
d 2u
+ x = 0,
2
dx
0 x 1
u(0) = 0, u(1) = 0
All elements have the same coefficient matrix
1 1 1 3 3
k(e )
= (e )
22 L 1 1 = 3 3 , (e = 1,2,3)
Change variable of p(x) = x to p(x):
RHS
cfw_f
(e )
(e )
=
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
FORMAL PROCEDURE cont.
Change variable from x to x
1 1 dNi
L( e ) 0 d x
+
dN
1
d x
1
u1
dN2
(e )
d x = L p( x )Ni (x )d x
0
d x
u2
du
du
( x j )Ni (1) ( x )N (0), i = 1,2
dx
dx i i
Do not use approximate solution for boundary terms
Elem
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
FORMAL PROCEDURE cont.
Need to derive the elementlevel equation for all elements
Consider Elements 1 and 2 (connected at Node 2)
k11
k21
k11
k21
du
 (x )
dx 1
k12 u1 f1
= +
du
k22 u2 f2
+ ( x )
dx 2
du
 (x )
(2)
(2)
2
dx
k12
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
FORMAL PROCEDURE
Galerkin method is still not general enough for computer code
Apply Galerkin method to one element (e) at a time
Introduce a local coordinate
x=
x = xi (1  x ) + x j x
x  xi
xx
= (e ) i
x j  xi
L
Approximate solution within the el
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE cont.
1.6
u(x)
1.2
0.8
uexact
0.4
uapprox.
0
0
0.2
0.4
x
0.6
0.8
1
2
1.5
du/dx
Solution comparison
Approx. solution has about 8%
error
Derivative shows a large
discrepancy
Approx. derivative is constant as
the solution is piecewise linear
1
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE cont.
Matrix equation
2 2 0 u1 F1
2 4 2 u2 = 0.5
0 2 2 u3 1.25
Consider it as unknown
Striking the 1st row and striking the 1st column (BC)
4 2 u2 0.5
=
2 2 u3 1.25
Solve for u2 = 0.875, u3 = 1.5
Approximate
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
FORMAL PROCEDURE cont.
Interpolation property
N1( xi ) = 1 N1( x j ) = 0
,
N2 ( xi ) = 0, N2 ( x j ) = 1
u( xi ) = ui
u( x j ) = u j
Derivative of approx. solution
du
dN1
dN2
= ui
+ uj
dx
dx
dx
1 dN
du dN1 dN2 u1
=
= (e ) 1
dx dx
dx u2 L d x
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE
Solve using two equallength elements
d 2u
+ 1 = 0,0 x 1
2
dx
u (0) = 0
Boundary conditions
du
(1) = 1
dx
Three nodes at x = 0, 0.5, 1.0; displ at nodes = u1, u2, u3
Approximate solution
u ( x ) = u1f1( x ) + u2f2 ( x ) + u3f3 ( x )
1  2 x,
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
EXAMPLE cont.
Derivatives of interpolation functions
0 x 0.5
d f1( x ) 2,
=
dx
0, 0.5 < x 1
d f3 ( x ) 0, 0 x 0.5
=
dx
2, 0.5 < x 1
d f2 ( x ) 2, 0 x 0.5
=
dx
2, 0.5 < x 1
Coefficient matrix
0.5
1
d f1 d f2
K12 =
dx =
(2)(2)dx + (0)(2)dx = 2
0
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
GALERKIN METHOD
Relation between interpolation functions and trial functions
1D problem with linear interpolation
0,
0 x xi 1
( i 1)
Ni ( x ) = x  xi 1 , xi 1 < x xi
ND
L( i 1)
fi ( x ) =
u( x ) =
ui fi ( x )
(i )
N ( x ) = xi +1  x , xi <
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
TRIAL SOLUTION cont.
Observations
Solution u(x) is interpolated using its nodal values ui and ui+1.
Ni(x) = 1 at node xi, and =0 at node xi+1.
Ni(x)
Ni+1(x)
xi
xi+1
The solution is approximated by piecewise linear polynomial and its
gradient is consta
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
FINITE ELEMENT METHOD cont.
Types of finite elements
1D
2D
3D
Variational equation is imposed on each element.
1
0
dx =
0.1
0
dx +
0.2
0.1
dx + +
1
0.9
dx
One element
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
TRIAL SOLUTION cont.
Substitute two nodal values
u( xi ) = ui = a0 + a1xi
u( xi +1) = ui +1 = a0 + a1xi +1
Express a0 and a1 in terms of ui and ui+1. Then, the solution is
approximated by
u( x ) =
xi +1  x
xx
ui + ( i ) i ui +1
L( i )
L
Ni ( x )
Korea Advanced Institute of Science and Technology
intro to FEM
MAE 321

Fall 2014
TRIAL SOLUTION
Solution within an element is approximated using simple polynomials.
1
1
2
2
n
n1
3
n1
xi
n
n+1
xi+1
i
ith element is composed of two nodes: xi and xi+1. Since two unknowns
are involved, linear polynomial can be used:
u ( x ) = a0 + a1x,