# Pitfalls of FE Computing

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15 Νοε 2013 (πριν από 4 χρόνια και 6 μήνες)

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Lect_5.ppt

M. Okrouhlík

Ústav termomechaniky, AV ČR
,
Praha

Plze
ň
, 2010

Pitfalls of FE Computing

Numerical methods in FEA

Equilibrium problems

Q
q
q
K

)
(

solution of algebraic equations

-
state vibration problems

0
q
M
K

2
Ω

generalized eigenvalue problem

Propagation prob
lems

V
V
d
,
T
int
ext
int

σ
B
F
F
F
q
M

step by step integration in time

In linear cases we have

)
(
t
P
Kq
q
C
q
M

Now a few examples

from

Statics

-
state vibration

Transient dynamics

using

Rod elements

Beam elements

Bilinear (L) and biquadratic (Q) plane elements

force acting at the free end.

Beam 4
-
dof
elements
(Euler
-
Bernoulli) are used.

1
2
3
4

2
2
2
3
2
sym.
3
6
3
2
3
6
3
6
2
l
l
l
l
l
l
l
l
EI
k

P
Kq

EI
PL
v
3
3
tip

1
1.90476190476190
5e
-
003 1
.90476190476190
4e
-
003

10
1.90476190476
043e
-
0
03
1.90476190476
1904e
-
003

1
2
3
kmax
P

‘Exact’ formula for thin beams

A thin cantilever beam,

vertical point load at the free end,

four
-
node plane stress elements

FE technology

V
V
d
T
EB
B
k

full integration

1 … anal, 2 … Gauss q.,

different types of underintegration

0
1
2
3
4
5
6
7
8
9
10
0
1
2
3
4
5
6
x 10
-3
maximum displacements for kmax = 10
case
"EXACT"
KHGS
0
1
2
3
4
5
6
L

Default in ANSYS

Reduced integration in Marc

Exact integration

Exact integration

does not yield

‘exact’ results

Data obtained with

10 elements

underintegrated elements

Summary_2

Isoparametric approach

Four
-
node bilinear element, plane stress, f
ull integration

Four
-
node bilinear element, plane stress, f
ull integration

Comparison of beam and L elements

of a cantilever beam

Beam
… even one element gives a
negligible error

L

… too stiff in bending, tricks have to be
employed to get “correct” results

1
2
3
4
5
6
7
8
0
5
10
0
1
2
3
x 10
4
GEIVP
consistent
1
2
3
4
5
6
7
8
0
5
10
0
5000
10000
15000
GEIVP
diagonal

A single four
-
node plane stress element

Generalized eigenvalueproblem

Full integration

0
q
M
K

)
(

1
2
3
kmax
0
q
M
K

)
(

Free transverse vibration of a thin elastic cantilever beam

FE vs. continuum approach (Bernoulli_Euler theory)

Generalized eigenvalue problem

Two cases will be studied

-

L (bilinear, 4
-
node, plane stress elements)

-

Beam elements

0
1
2
3
4
5
6
7
8
9
10
0
2000
4000
6000
8000
10000
axial, bending and FE frequencies [Hz]
diagonal

1, 100.506
2, 586.4532
3, 1299.7804
4, 1501.8821
5, 2651.6171
6, 3862.638
7, 3931.881
8, 5258.5331
9, 6313.8768
10, 6567.2502
11, 7802.0473
12, 8571.0469
13, 8907.7934
14, 9804.6267
15, 10394.9981
16, 10520.9827
Natural frequencies and modes of a cantilever beam … diagonal mass m.

Four
-
node plane stress elements, full integration

Sixth FE frequency is the second axial

Seventh FE frequency is the fifth bending

Higher frequencies are useless

due to discretization errors

Analytical axial

Analytical bending

FE frequencies

This we cannot say without

looking at eigenmodes

Frequencies in [Hz] versus counter

L

0
1
2
3
4
5
6
7
8
9
10
0
2000
4000
6000
8000
10000
axial, bending and FE frequencies [Hz]
consistent

1, 101.2675
2, 615.4501
3, 1302.6992
4, 1663.1767
5, 3135.596
6, 3941.6886
7, 4996.7233
8, 6682.2631
9, 7215.1641
10, 9593.4882
11, 9739.1512
12, 12430.9913
13, 12740.3253
14, 14990.9545
15, 16164.1614
16, 16950.4893
Natural frequencies and modes of a cantilever beam … consistent mass m.

Four
-
node plane stress elements, full integration

0
1
2
3
4
5
6
7
8
9
10
0
2000
4000
6000
8000
10000
frekvence analyticke axialni (x), ohybove (o) a MKP frekvence (*)
1
1.5
2
2.5
3
3.5
4
4.5
5
0
5
10
15
20
25
konsistentni formulace
relativni chyba v % pro axialni (x) a ohybove (o) frekvence
0
1
2
3
4
5
6
7
8
9
10
0
2000
4000
6000
8000
10000
frekvence analyticke axialni (x), ohybove (o) a MKP frekvence (*)
1
1.5
2
2.5
3
3.5
4
4.5
5
-20
-10
0
10
20
diagonalni formulace
relativni chyba v % pro axialni (x) a ohybove (o) frekvence
Relative errors for axial (x) and bending (o) freq.

Relative errors for axial (x) and bending (o) freq.

Eigenfrequencies of a cantilever beam

Four
-
node bilinear element, plane strain
Relative errors [%] of FE frequencies

x

… axial

continuum,
o

… bending

continuum,
*

… FE frequencies

1
2
3
kmax

1
2
3
4

2
2
2
2
4
sym.
22
156
3
13
4
13
54
22
156
420
l
l
l
l
l
l
l
lA

m

2
2
2
3
2
sym.
3
6
3
2
3
6
3
6
2
l
l
l
l
l
l
l
l
EI
k
0
q
M
K

)
(

Free transverse vibration of a thin elastic cantilever beam

FE (
beam element
) vs. continuum approach (
Bernoulli_Euler theory
)

Free transverse vibration of a thin elastic beam

ANALYTICAL APPROACH

The equation of motion of a long thin beam considered as
continuum

undergoing transverse vibration is derived under Bernoulli
-
Euler assumptions,
namely

-
there is an axis, say x, of the beam that undergoes no extension,

-
the x
-
axis is located along the neutral axis of the beam,

-
cross sections perpendicular to the neutral axis remain planar during the

deformation

transverse shear deformation is neglected,

-
material is linearly elastic and homogeneous,

-
the y
-
axis, perpendicular to the x
-
axis, together with x
-
axis form a principal plane of

the beam.

These

assumptions

are

acceptable

for

thin

beams

the

model

ignores

shear

deformations

of

a

beam

element

and

rotary

inertia

forces
.

For

more

details

see

Craig,

R
.
R
.:

Structural

Dynamics
.

John

Wiley,

New

York,

1981

or

Clough,

R
.
W
.

and

Penzien,

J
.:

Dynamics

of

Structures,

McGraw
-
Hill,

New

York,

1993
.

The equation is usually presented in the form

)
,
(
2
2
2
2
2
2
t
x
p
t
v
A
x
v
EI
x

where
x
is a longitudinal coordinate,
v
is a transversal displacement of the beam in

y

direction, which is perpendicular to
x
,
t
is time,
E
is the Young’s modulus,
I
is the planar
moment of inertia of the cross section,
A
is the cross sectional area and

is the density.
On the right hand side of the equation there
)
,
(
t
x
p

-
generally a function
of space and time
-
acting in the
xy
plane.
For
free
transverse
vibrations
we have zero on
the right
-
hand side of Eq. (1). If the bending stiffness
EI
is independent of time and
space coordinate
s we can write

0
2
2
4
4

t
v
EI
A
x
v

.

(4
a
)

Assuming the
in a
harmonic

form

)
cos(
)
(
)
,
(

t
x
V
t
x
v

we get

0
)
(
d
)
(
d
4
4
4

x
V
x
x
V

(4
b
)

where we have introduced an auxiliary variable by

)
/(
2
4
EI
A

(5)

The general solution of Eq. (4) can be assumed (see
Engineering Mathematics, John Wiley & Sons, New York, 1993
) in the form

x
C
x
C
x
C
x
C
x
V

cos
sin
cosh
sinh
)
(
4
3
2
1

(6)

where constants
4
1
to
C
C
depend on boun
dary conditions.

Applying boundary conditions for a thin cantilever beam (clamped

free) we get a
frequency determinant

[0, 1, 0, 1 ]

[lam, 0, lam,
0 ]

[sinh(lam*L)*lam^2, cosh(lam*L)*lam^2,
-
sin(lam*L)*lam^2,
-
cos(lam*L)*lam^2]

[cosh(lam*L)*lam^3, sinh(lam*L)*lam^3,
-
cos(lam*L)*lam^3, sin(lam*L)*lam^3]

From the condition that the frequency determinant is
equal to zero we get the frequency equation in he
form

0
1
cos
cosh

L
L

Roots of this equation can only be found numerically, Denoting
L
x
i
i

we get the
natural frequencies in the
form

...
,
3
,
2
,
1
2

i
E
A
I
L
x
i
i

Comparison of analytical and FE results

counter continuum frequencies FE frequencies

1
5.
26650

4690912090e+002
5.26650
9194371887e+002

2
3.300
462151726965e+003
3.300
571391657554e+003

3
9.24
1389593048039e+003
9.24
3742518773286e+003

4
1.81
0943523875022e+004
1.81
2669270993247e+004

5
2.99
3619402962561e+004
3.0
0
1165614576545e+004

6
4.4
71949023233439e+004
4.4
96087393371327e+004

7
6
. 245945376065551e+004
6
. 308228786109306e+004

8
8
. 315607746908118e+004
8
. 451287572802173e+004

9
1.0
68093617279631e+005
1.0
92740977881639e+005

0
5
10
15
20
0
1
2
3
4
5
6
7
8
9
x 10
5
o - FEM, x - analytical
counter
angular frequencies
0
5
10
0
0.5
1
1.5
2
2.5
relative errors for FE frequencies [%]
counter
Are analytically computed

frequencies exact to be used

as an etalon for error analysis?

To answer this you have to recall

the assumptions used for the

thin beam theory

FE computation with

10 beam elements

Consistent mass matrix

Full integration

Comparison of Beam and Bilinear Elements

Used for Cantilever Beam Vibration

10 beam elements

… the ninth bending
frequency with 2.5% error

10 beam elements

… this element does not
yield axial frequencies

10 bilinear elements

… the first bending
frequency with 20% error, the errors goes
down with increasing frequency counter

10 bilinear elements

… the errors of axial
frequencies are positive for consistent mass
matrix, negative for diagonal mass matrix

Where is the truth?

Transient problems in linear dynamics, no damping

t
P
Kq
q
M

Modelling the 1D wave equation

2
2
2
0
2
2
x
u
c
t
u

0
0.5
1
-0.5
0
0.5
1
1.5
eps t = 1.6
0
0.5
1
0.4
0.6
0.8
1
1.2
1.4
dis
0
0.5
1
-0.5
0
0.5
1
1.5
vel
L1 cons 100 elem Houbolt (red)
0
0.5
1
-100
-50
0
50
100
acc
Newmark (green), h= 0.005, gamma=0.5
Rod elements used here, the results depend on the method of integration

Classical Lamb’s problem
B
A
C
1 m
1 m
axial
p
0
T
i
m
p
t
L or Q

Elements

L, Q, full int.

Consistent mass

axisymmetric

Mesh

Coarse

20x20

Medium

40x40

Fine

80x80

Newmark

a point force

equiv. pressure

Example of a transient problem

Coarse, L

Fine, Q

Fine L and medium Q

Medium L and coarse Q

0
1
2
x 10
-4
0
2000
4000
6000
8000
10000
12000
Point and pressure loading of the solid cylinder by a rectangular pulse in time
Time [s]
Total energy [J]
point Q fine
point Q medium
point L fine
point Q coarse
point L medium
point L coarse
For all FEM-models with distributed loads
P-L
coarse
P-L
medium
P-L
fine
P-Q
coarse
P-Q
medium
P-Q
fine
Total energy

Pollution
-
free energy production by a proper misuse of FE analysis