Topic 7
Two- and Three­
Dimensional Solid
Elements; Plane
Stress, Plane
Strain, and
Axisymmetric
Conditions
Contents:
• Isoparametric interpolations of coordinates and
displacements
• Consistency between coordinate and displacement
interpolations
• Meaning of these interpolations in large displacement
analysis, motion of a material particle
• Evaluation of required derivatives
• The Jacobian transformations
• Details of strain-displacement matrices for total and
updated Lagrangian formulations
• Example of 4-node two-dimensional element, details of
matrices used
Textbook:
Sections 6.3.2, 6.3.3
Example:
6.17
Topic Seven 7-3
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. TI4~~ E ELf HE:NT~
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Markerboard
7-1
7-4 Two- and Three-Dimensional Solid Elements
Transparency
7-1
TWO- AND THREE-DIMENSIONAL
SOLID ELEMENTS
• Two-dimensional elements comprise
- plane stress and plane strain elements
- axisymmetric elements
• The derivations used for the two­
dimensional elements can be easily
extended to the derivation of three­
dimensional elements.
Hence we concentrate our discussion
now first on the two-dimensional
elements.
Transparency
7-2
TWO-DIMENSIONAL
AXISYMMETRIC, PLANE
STRAIN AND PLANE STRESS
ELEMENTS
3
+----------X1
Topic Seven 7-5
Because the elements are
isoparametric,
Transparency
7-3
0
N
X1 =
L
hk °x~ , °X2 =
N
hk tx~ , tX2 =
k=1
N
L
k=1
hk °x~
and
t
X1 =
L
k=1
N
L
k=1
hk tx~
where the hk's are the isoparametric
interpolation functions.
Example: A four-node element
s
4
tXi
= L hk tx~
k=1
4
°Xi
where
1
h 1 = 4: (1
=
2:
k=1
+ r)(1 + s)
1
4
h2 = - (1 - r)(1
time 0
hk °x~
+ s)
1
h3 = - (1 - r)(1 - s)
4
h4
1
= 4 (1 + r)(1
- s)
Transparency
7-4
7-6 1\vo- and Three-Dimensional Solid Elements
Transparency
7-5
x
tx; =
t hkl
k-1
Transparency
7-6
r
r;0.5
5=0.5
A major advantage of the isoparametric
finite element discretization is that we
may directly write
tU1
k=1
tU2 = N hk tu~
~
k=1
N
N
N
~ hk tu~
and
U1
~ hk u~
k=1
U2
= ~ hk u~
k=1
tx~
Topic Seven 7-7
This is easily shown: for example,
Transparency
N
t
Xi =
oXi =
~
£J
k=1
N
~
£J
7-7
hk t Xik
hk
0 k
Xi
k=1
Subtracting the second equation from
the first equation gives
tXi
N
-
0
,
Xi =
.
~
£J
k=1
hk (tXik
'
-
0 k)
,
Xi
.
The element matrices require the following
derivatives:
Transparency
7-8
7-8 'I\vo- and Three-Dimensional Solid Elements
Transparency
7·9
These derivatives are evaluated using a
Jacobian transformation (the chain rule):
ahk _ ahk aOx1 + ahk
ar - aOx1 ar
aOx2
ahk ahk aOx1
as - aOx 1 as +
aOx2
ar
ahk aOx2
as
aox;.
~REaUIRED
In matrix form,
~ DERIVATIVES
r
ahk
as
ilxl aOx2
as
as
A
,
ahk
aOX2
°4
Transparency
7-10
The required derivatives are computed
using a matrix inversion:
ah k
O
a x1
ahk
ar
-
= 0J-1
ah k
aOx2
ahk
as
-
The entries in oJ are computed using
the interpolation functions. For example,
a°X1
ar
=
LN ahk 0x~
k=1 ar
Topic Seven 7-9
The derivatives taken with respect to
the configuration at time t can also be
evaluated using a Jacobian transformation.
ah k
-
ar
=
ah k
alX2
ar
-
a1x1
ar
-
alX1
a1x1
as
-
a1x2
as
ah k
-
-
as
IJ
ah
- k
ah k
1
a x1
alX2
ah k
a1x2
f
k=1
ah k IX~
as
ar
= IJ-1
ah k
ah k
-
as
We can now compute the required
element matrices for the total
Lagrangian formulation:
Element Matrix
Transparency
7-11
Matrices Required
dB L
ds , dB NL
oC ,
tAt
oS , oB L
Transparency
7-12
7-10 Two- and Three-Dimensional Solid Elements
We define oC so that
Transparency
7-13
08 11
08 22
08 12
08 33
analogous to
OSij. = oC~,s
For example, we may choose
(axisymmetric analysis),
1
oe,s
v
1- v
o
v
_
oQ - (1 + v)(1 - 2v)
Transparency
7-14
v
0
1- v
E(1 - v)
1- v
1 - 2v
2(1 - v)
0
0
v
1 - v
1- v
v
0
We note that, in two-dimensional
analysis,
t
Oe11 = OU1,1 +loU1,1
OU1,1
t
+ OU2,1
t
Oe22 = OU2,2 + IOU 1,2 oU 1,2
2 Oe12
t
t
+ OU2,2 OU2,2,
,(J U 1,1
= (OU1,2 + OU2,1) +
t
~e:~: ;~2:+(~~~1)2O:~" +
I
X1
X1
OU1,2
°f
t
X1
OU2,1,
U2
,2 OU2,1
I
""-INITIAL DISPLACEMENT
EFFECT
)
I
0
Topic Seven 7-11
and
Transparency
7-15
01")33
=
Derivation of
X2
(U1)2
1
2 Ox 1
0833, o~:
./'
axis of
reVolut~
time 0
r-\
L-:=1
/time t+dt
CI
Transparenc)
7-16
~
' " - - - - - - - - - - X1
We see that
7-12 'I\vo- and Three-Dimensional Solid Elements
Transparency
7-17
Transparency
7-18
H.1toE _ 1 [(H.1tdS)2
Hence
33 -
-
2
----0-:;-:-
ds
-
1
]
We construct riB l so that
~
J
oe11]
Oe22
t
t
2 Oe12 = o~ = (06l0 + 06L1) a
De..
is only included
for axisymmetric analysis
Oe33
I
JBL
~
contains initial
displacement effect
'Ibpic Seven 7-13
Entries in 6Sl0:
Transparency
node k
.
.....
ohk,1
0
ohk,2
hkfx1
...
-
.
..
I
I
I
I
I
I
I
0
ohk,2
ohk,1
0
included only for
axisymmetric analysis
7-19
-
...
t
u~
node k
u~
l
-
This is similar in form to the B matrix used in
linear analysis.
Entries in 6Bl1: node k
I
...
~
u~
6U1,1 ohk,1
6U1,20hk,2
6U1,1 Ohk,2
t
+ OU1,2
0hk,1
t
U1 hk
°X1 °X1
..
u~
.
I 6U2,1 ohk,1
I
I 6U2, 20hk,2
I
I 6U2,10hk,2
II + 6U2,2 ohk,1
I
I
I
I
Transparency
-
...
0
-
only
Th e ...
InltlaI d'ISPIacement effect t u forincluded
axisym~etric
is contained in the terms 6Ui,~, OX:. analysIs
7-20
7-14 1\vo- and Three-Dimensional Solid Elements
Transparency
7-21
We construct dSNL and ds so that
s:: ~ T tST ts ts
~
ts s::
u!! O_NL 0_ O_NL!! = i}Uolli}
°
Entries in ds:
-
r-
0
0
dS11 dS12 0
0
0
dS21 dS22 0
0
0 dS11 dS12 0
0
0 dS21 dS22 0
0
0
0 dS33
..... 0
-
included only
for axisymmetric
analysis
Entries in dSNL:
Transparency
node k
7-22
r-
...
u~ ,I u~
I
ohk,1 I 0
ohk,2 I 0
I
0 I Oh k1
0 I ohk,2
hk/OX1 I 0
-
...
I
f
L-
I
included only for
axisymmetric
analysis
-
u~
---
u~
t
node k
1
Topic Seven 7-15
A
t
oS
is constructed so that
Entries in
Transparency
7-23
JS:
JS11
J8 22
J8 12
J833
~ included
only for
axisymmetric analysis
7-16 'I\vo- and Three-Dimensional Solid Elements
Exam~: Calculation of
Transparency
0.2
0.1
7-25
X2
Plane strain
conditions
0 .1
0.2
-l-.L"'"--time 0
t-+
Exam~: Calculation of
JB lI JBNl
0.2
0.1
Transparency
JB lI JB Nl
7-26
Plane strain
conditions
2
X2
0 .1
.2
r-t:======:i.J-J-.. . ~time 0
4
0.2
(0.1,0.1 )
·1
material fibers
have only translated
rigidly
Topic Seven 7-17
Example: Calculation of
dB L • dB NL
0.2
0.1
Transparency
7-27
Plane strain
conditions
0.2
0.2
x,
(0.1,0.1)
Exam~: Calculation of
dB Ll dB NL
0.2
0.1
Transparency
7-28
X2
Plane strain
conditions
0 .1
0.2
time 0
material fibers have
stretched and rotated
0.2
(0.1,0.1)
·1
x,
7-18 Two- and Three-Dimensional Solid Elements
Exam~: Calculation of JBL, JBNL
Transparency
0.1
7-29
X2
0 .1
0.2
Plane strain
condffions
2
s
1
0.2
---1----t~~rtime
·1
0.2
(0.1.0.1)
Transparency
7-30
0
At time 0,
We can now perform a Jacobian
transformation between the [, s)
coordinate system and the ( X1 ,OX2)
coordinate system:
By inspection,
a°arx1
=
aOx1
as
=
1
:;CC-a
X1 =
°-ar
1
X2 =
° 'as
aOx2
0~1]
°
a
a a
Hence oJ = [0.1
and
°., a°ar °
,
= 0.1
1°41 = 0.01
':;CC-a
X2 =
a
-as
°
1
'Ibpic Seven 7-19
Now we use the interpolation functions
to compute JU1,1 , JU1,2:
node
k
ahk
aOx1
ahk
aOx2
Transparency
7-31
ahk t k
ahk tU1k
-aOx2
U1
X1
2.5(1 + r) 0.1 0.25(1 + 5) 0.25(1 + r)
2.5(1 - r) 0.1 -0.25(1 + 5) 0.25(1 - r\
tu~
r
1 2.5(1 + 5)
2 -2.5(1 + 5)
3 -2.5(1 - 5) -2.5(1 - r) 0.0
4 2.5(1 - 5) -2.5(1 + r) 0.0
0
0
Sum:
0
0
0.0
"
4
0.5
,/
t
OU1,1
"
4
OU1,2
For this simple problem, we can
compute the displacement derivatives
by inspection:
From the given dimensions,
Jx =
-
Hence
[1.0 0.5]
0.0 1.5
JU1,1 = JX 11 -
1 =0
t
tx
t
tx
0.5
=0
1 = 0.5
OU1,2 = 0
12
OU2,1 = 0 21
JU2,2
= JX22 -
=
j
t
Transparency
7-32
7-20 Two- and Three-Dimensional Solid Elements
Transparency
7-33
We can now construct the columns in
riB L that correspond to node 3:
-2.5~1
!
I
-2.5(1 - 5) I
['0
o
I
-2.5(1 - r)
I
- r) ..
-2.5(1 - 5)
0
:
0
... -1.25(1 - r) : -1.25(1 - r)
[
-1.25(1-5)1- 1.25(1-5)
Transparency
7-34
J
ooJ
Similarly, we construct the columns in
riB NL that correspond to node 3:
I
-2.5(1 - 5) I
-2.5(1 - r) ,
o
o
0
0
I
I -2.5(1 - 5)
I -2.5(1 - r)
I
Thpic Seven 7-21
Transparency
7-35
Consider next the element matrices
required for the updated Lagrangian
formulation:
Element Matrix
Matrices Required
tC
,
~Bl
t'T , ~BNl
tAt
'T , tBl
We define
tC
so that
tSl1
tel1
tS22
tS12
te22
te12
= te
2
tS33
Transparency
tSt
te33
For example, we may choose
(axisymmetric analysis),
1
Ct_ -
(1
E(1 - v)
+ v)(1 - 2v)
7-36
analogous to
= tCtrs tars
v
1- v
o
v
1- v
1
0
0
0
1 - 2v
2(1 - v)
0
v
1- v
v
1- v
0
1
v
1-
V
7-22 1\\'0- and Three-Dimensional Solid Elements
Transparency
7-37
We note that the incremental strain
components are, in two-dimensional
analysis,
te11 =
aU1
-at
X1
= t U1,1
te22 = t U2,2
2 te12
te33
= tU1,2
=
U1/X1
and
Transparency
7-38
~ ((tU1,1)2 + (t U2,1)2)
t'TI11
=
tT)33
= 21 (U1)2
t
X1
+ t U2,1
Topic Seven 7-23
Transparency
7-39
We construct ~BL so that
- te11
t e 22
2 te12
t
=
e33
te
=
ttB L
A
U
only included for
axisymmetric analysis
Entries in ~BL:
Transparency
node k
-
th k ,1
...
.....
0
I
I
I
I
I
th k ,2
hk /x1 I
7-40
.
.
-
0
th k ,2
th k ,1
u~
...
u~
0
-
only included for axisymmetric analysis
This is similar in form to the
B matrix used in linear analysis.
t
node k
1
7-24 Two- and Three-Dimensional Solid Elements
Transparency
7-41
We construct ~SNL and tT so that
~
A
u~
T tsT t,.,. ts
t,.,. ~
t_NL...!.- t_NL ~ = • y.Ut'Tly.
A
Entries in t'T:
r--
t
'T11
12
t
t
'T21 'T22
trr
0
0
0
0
0
0
0
0
0
0
0
0
-
t
t
'T11 'T12
t
t
'T21 'T22
0
0
0
0
t
'T33
included only
for axisymmetric
analysis
Entries .In ts
t NL:
Transparency
7-42
node k
...
l-.
I'
u~
!,
u~
0
0
I
I
I
I
I
I
th k,1
th k,2
hk/X1
I
0
thk,1
th k,2
0
0
.I ...
u~
u~
-
~
t
node k
l
included only for
axisymmetric analysis
Topic Seven 7-25
Transparency
t"
'T is constructed so that
Entries in
7-43
tf:
included only for
axisymmetric analysis
Three-dimensional elements
X2
~ node k
(ox~, 0xt °x~)
Transparency
7-44
7-26 Two- and Tbree-Dimeusional Solid Elements
Transparency
7-45
Here we now use
N
oX1 =
0X3
=
~
£J
k=1
hk
°
X1k
N
°
o
~ hkX2
k
,X2=£J
k=1
N
~ hk O~,
k=1
where the hk's are the isoparametric
interpolation functions of the three­
dimensional element.
Transparency
Also
7-46
N
1<1 = k=1
~
1cs =
hk
1<~
N
, 1<2 = k=1
~
hk 1<~
N
~ hk tx~
k=1
and then all the concepts and
derivations already discussed are
directly applicable to the derivation of
the three-dimensional element matrices.
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Finite Element Procedures for Solids and Structures
Klaus-Jürgen Bathe
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