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Finite element method for structural dynamic
and stability analyses
1
Prof C S Manohar Department of Civil Engineering
IISc, Bangalore 560 012 India
Module-7
Analysis of 2 and 3 dimensional continua
Lecture-21 3D solid element
2
, , ,x u x y t
, , ,y v x y t
,x y
1 11 ,x y
4 44 ,x y
3 33 ,x y
2 22 ,x y
1, 1 1, 1
1,1 1,1
,
Linear quadrilateral element
4 3
21
A
B
4
1
4
1
, ,
, ,
t
e
A
t
e
A
j j
j
j j
j
K hB DBdA
M h N NdA
x N x
y N y
3
, , ,x u x y t
, , ,y v x y t
1
43
2
1, 1 1, 1
1,1 1,1
4 3
211
1
1
1
4
1
4
1
, ,
, ,
j j
j
j j
j
x N x
y N y
References
• M Petyt, 1998, Introduction to finite element vibration
analysis, Cambridge University Press, Cambridge
• S S Rao, 2011, The finite element method in
engineering, 5th Edition, Elsevier, Amsterdam.
4
5
4 4
1 1
4 4
1 1
1 1
1 1
1 1
1 1
Isoparametric formulation
, , & , ,
, , , & , , ,
Use Gauss quadrature to evaluate
e
j j j j
j j
j j j j
j j
t t
e
A
t t
e
A
x N x y N y
u t N u t v t N v t
M h N N dA h N N J d d
K hB DBdA hB DB J d d
these integrals.
A polynomial of order is integrated exactly by employing
=smallest integer greater than 0.5 1 .
Choice of order of integration needs to be made carefully.
p
n p
6
Convergence and choice of order of interpolation polynomial.
What happens if we reduce the element size successively? Does
the FE solution converge?
Discussion & miscellaneous rema
Requirements t
rks
o be
A The displacement field must be continuous within the element domains.
This is automatically satisfied if since we are using polynomials as
interpolation funct
satisfied by the interpolation functions
ions.
B Consider the Lagrangian - . This would be a function of the field
variables and its spatial derivatives. Let =highest order of partial derivative
of field variable that appears in . All the
L T V
n
L
uniform states of the field variable
and its derivatives up to order must be correctly represented in the limit of
element size going to zero.
C The displacement field and its derivatives up to order
n
-1,
must be continuous at the element boundaries.
n
7
B
If all nodal displacements are identical, the field variable must be constant
within the element, that is the element must permit rigid body state.
Requirement on derivative requirement that the el
ement must permit
constant strain state.
2P t
1u t 2u t
1P t ,u x t1 2
2
0
1 2
0 1 2 0 0 0 0 0
1Axially deforming element ; Field variable: ,
2
Highest order of derivative: 1; Interpolation used: , 1
, constant= 1 OK
Lu
V AE dx u x tx
x xu x t u t u t
l l
x xu x t u t u t u t u u u u u
l l
2 1
, constant the requirement on 1st derivative is satisfied.u t u t
u x tl
8
1s
2s
1
32
2
2
1
1 2
,
,
xu x t u t
l
u l t u t
2 3
2 2
2
, 1
0,
x xu x t u t u t
l l
u t u t
1 1 2 2
Element 1 Element 2
, 0, @ 0x l u l t u t u t u t x
9
Elements which satisfy conditions A and C are called compatible or
conforming elements.
Elements which satisfy B are called complete elements.
The field variable is said to possess C continuir
Remarks
th
1
ty if its derivative is
continuous.
Completeness requirement field variable has C continuity within the
element.
Compatibility requirement field variable has C continuity across
element interfac
n
n
r
e.
10
If the requirments A,B and C are satisfied, the FE approximation converges
to the correct solution if the FE mesh is refined (that is if we use
inreasing number of elements with smaller dimensions).
Note:
While the mesh is refined, the form of the interpolation function
must remain unchanged and the mesh refinement must be such that the mesh
with larger number of elements contains the mesh with smaller
number of elements.
Also, the mesh refinement must ensure that all points in the stucture are within
an element.
11
Mesh with larger number of elements contains the mesh with with smaller
number of elements
12
All points in the stucture are not
within an element.
13
Accuracy with which the structure geometry is represented
Choice of polynomials used for interpolation
Distribution of elements and node
Factors contributing to the development of an accurate FE model
s
Details of integration used in time marching
Reduce the element size ( -refinement)
Increase the order of polynomial ( -refinement)
Locate node points
h
p
How to refine the FE model to improve accuracy?
differently in a fixed element topology ( -refinement)
Alter the mesh having differing element distributions
Improvments to the time integration schemes
Alternatives invovling a combination of the abo
r
ve strategies.
14
The polynomials should satisfy to the extent possible conditions A,B and C.
The representation of the field variable must be invariant with respect to
change
Selection of the interpolation polynomial
in the local coordinate system of the element.
Geometric invariance
Spatial isotropy
Geometric isotropy
The number of generalized coordinates must match the number of nodal dofs
of the element
Geometric invariance can be achieved if the poynomial contains terms
which do not violate symmetry in the Pascal trianlge (two dimensions)
or Pascal pyramid (three dimensions).
15
2 2
3 2 2 3
4 3 2 2 3 4
5 4 3 2 2 3 4 5
1
x y
x xy y
x x y xy y
x x y x y xy y
x x y x y x y xy y
0
1
2 2 2
3 3 2 2 3
1 2 2 1
0 1 2 1
1
2
3 3
... n n n n n n
n n
x y
x y x y
x y x xy y
x y x x y xy y
x y a x a x y a x y a xy a y
16
1 2 3
1 2 3 4
2
1 2 3 4
2
1 2 3 4
Triangle element: , ,
Rectangular element: , ,
Not appropriate., ,
Interchanging of and , ,
would change the representation
u x y t x y
u x y t x y xy
u x y t x y xx y
u x y t x y y
1
2 3
2 2
4 5 6
3 2 2 3
7 8 9 10
3 3
11 12
Four noded element with 3 dofs/node
, ,
u x y t
x y
x xy y
x x y xy y
x y xy
17
1
x y
2x 2y
z
2z
xy
xz yz
2x y
3z
3y3x
2x z 2xz2z y
2zy
xyz
2xy
0
1
2 2 2 2
1
2 2 2
x y z
x y z x y z
x y z x xy yz xz y z
Pascal’s
pyramid
1
2 3 4
5 6 7
8
Eight noded 3d element with 2 dofs per node
, , ,
u x y z t t
t x t y t z
t xy t yz t xz
t xyz
18
How to decide on dof-s?
Inspect the functional in the variational formulation.
Identify the field variables and the order of their highest derivatives ( )
DOFS: field variables and their derivatives u
n
p to order -1. n
2
0
Axially deforming element
1
2
Field variable: ,
Highest order of derivative: 1
DOF: ,
Lu
V AE dxx
u x t
u x t
22
2
0
Euler Bernoulli beam
1
2
Field variable: ,
Highest order of derivative: 2
DOF: , &
Lv
V t EI x dxx
v x t
vv x t
x
19
0
0
, , , , ; , , , ,
, ,
0
,, 0
,
1 2 2
2 2 2
Field variables: , , & , ,
t t
xx yy xy xx yy xy
t t t
V A A
x y σ x y σ x y σ x y x y x y x y x y
x y D x y
x
u x yx y
v x yy
y x
h hV dV dA D dA
u x y t v x y t
Plane stress element
Highest order of derivative: 1
DOFs: , , & , ,u x y t v x y t
20
0 0
2 2
0 0
2 2 2
1 0 0 0
1 0 0 0
1 0 0 0
1 20 0 0 0 0
1 1 21 2
0 0 0 0 0
1 20 0 0 0 0
1 1 1 = &
2 2 2
t
xx yy zz xy xz yz
t
xx yy zz xy xz yz
t t
V V
D
ED
V dV D dV T u v
3D solid elements
0
2
0
V
w dV
21
0 0 0
2 2 2
0 0 0
1 1 1 = &
2 2 2
0 0
0 0
0 0
0
0
0
t t
V V V
V dV D dV T u v w dV
x
y
uz
v
wy x
z x
z y
22
0 0
0 0
0 0
0 0
0 0
0
0
0
1 1&
2 2
e
e e e e
e
t t t t
e e e e
V V
x
y
u uz
v N v Nu Nu Bu
w wy x
z x
z y
V u B DBu dV T u N Nu dV
23
8
54 3
21
7
6
Rectangular hexahedron
Pentahedron
Isoparametric hexahedron
Tetrahedron
24
Tetrahedron element
4
3
2
1
x
z
y
1 2 3 4
5 6 7 8
9 10 11 12
4 noded element with 3 dof-s per node
Dofs=12
, , ,
, , ,
, , ,
u x y z t t t x t y t z
v x y z t t t x t y t z
w x y z t t t x t y t z
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4
1
4
1
At , , , , , , , , , , , , ,
At , , , , , , , , , , , , ,
At , , , , , , , , , , , , ,
, , , , , ;
, , , , ,
i i
i
i i
i
x y z u x y z t u t v x y z t v t w x y z t w t
x y z u x y z t u t v x y z t v t w x y z t w t
x y z u x y z t u t v x y z t v t w x y z t w t
u x y z t u t N x y z
v x y z t v t N x y z
4
1
, , , , ,i i
i
w x y z t w t N x y z
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2
0
3 3 3 3 3 3 3 30
4 4 4 4 4 4 4 4
2 2 2 2 2 2 2 2 2
1 3 3 3 1 3 3 1 3 3 1 3 3
4 4 4 4 4 4 4 4
11
11 1;
16 6
1
1 1 1
; 1 ; 1 ; 1
1 1 1
N a b c d x y z
N a b c d x y zxV
N a b c d x y zyV
N a b c d x y zz
x y z y z x z x y
a x y z b y z c x z d x y
x y z y z x z x y
4
3 1 2 4
1
Obtain other constants by cyclic interchange of subscripts in the order
1,2,3,4. Sign of will be same as and both and will have sign
opposite to that of . Similar rules apply to the coef
a a a a
a ficients , , and . b c d
27
1 1 1 4 4 4
1 4
1 4
1 4
, , ,
, , ,
, , ,
0 0 0 0
0 0 0 0
0 0 0 0
This integral can be evaluated exactly.
e
e
t
e
t
e e
V
u x y z t
v x y z t N u
w x y z t
u u v w u v w
N N
N N N
N N
M N N dV
28
0
2 0 0 1 0 0 1 0 0 1 0 0
0 2 0 0 1 0 0 1 0 0 1 0
0 0 2 0 0 1 0 0 1 0 0 1
1 0 0 2 0 0 1 0 0 1 0 0
0 1 0 0 2 0 0 1 0 0 1 0
0 0 1 0 0 2 0 0 1 0 0 1
1 0 0 1 0 0 2 0 0 1 0 020
0 1 0 0 1 0 0 2 0 0 1 0
0 0 1 0 0 1 0 0 2 0 0 1
1 0 0 1 0 0 1 0 0 2 0 0
0 1 0 0 1 0 0 1 0 0 2 0
0 0 1 0 0 1 0 0 1 0 0 2
e
VM
29
0 0 0 0
0 0 0 0
0 0 0 0
0 0
0 0
0 0
e e
x x
y y
uz z
v N u B u
wy x y x
z x z x
z y z y
30
1 2 3 4
1 2 3 4
1 2 3 4
1 1 2 2 3 3 4 40
1 1 2 2 3 3 4 4
1 1 2 2 3 3 4 4
0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 01
0 0 0 06
0 0 0 0
0 0 0 0
Note: is independent of , , and .
e
t t
e e
V
e
b b b b
c c c c
d d d dB
c b c b c b c bV
d c d c d c d c
d b d b d b d b
B x y z
K B DBdV V B DB
Bu
DB
constant over the elementeu
31
Rectangular hexahedron element
8
54 3
21
7
6
,x
,z
,y
2c 2b
2a
, ,x y z
a b c
Field variables
, , ,
, , ,
, , ,
u x y z t
v x y z t
w x y z t
8 noded element with 24 dofs
Dof-s at each node: , , u v w
32
1 2 3 4
Each of the variables , , , , , , , ,& , , ,
needs to be represented by a polynomial with 8 terms.
Terms: 1, , , , , , ,
Ensures geometric invariance
, , ,
u x y z t v x y z t w x y z t
x y z xy xz yz xyz
u x y z t t t x t y t z
5 6 7 8
Select , 1, 2, ,8 so that the value of , , ,
matches with its respective nodal values at the 8 nodes.
i
t xy t yz t xz t xyz
t i u x y z t
33
8
1
8
1
8
1
, , , , ,
, , , , ,
, , , , ,
1, , 1 1 1 , 1,2, ,8
8
Remark
Continuity of field variables across element boundaries
is ensured. Why?
j j
j
j j
i
j j
i
j j j j
u t N u t
v t N v t
w t N w t
N j
u
v N
w
e
u
34
0 0
1
1
1 8 1
1 8
1 8 8
8
8
0 0
1 1 1
1 1 1
0 0 0 0
0 0 0 0
0 0 0 0
1 1 1
2 2 2
, , , ,
e
t t t t t
e e e e e ee
V V
i jeij
u
v
u N N w
v N u N N
w N N u
v
w
T u N Nu dV u N NdV u u m u
m abcN N d d d
35
1 1 1
1 1 1
1 1
1 1
1 1
1 1
1
1
, , , ,
1 1 1 1 1 164
1 1 1 164
1 1
1 11 1 1
8 3 3
i jeij
i i i j j j
i j i j
i j
i j i j
m abcN N d d d
abcd d d
abcd d
d
abc
1
3
2
2
i j
e
m mm
m m
36
4 0 0 2 0 0 1 0 0 2 0 0
0 4 0 0 2 0 0 1 0 0 2 0
0 0 4 0 0 2 0 0 1 0 0 2
2 0 0 4 0 0 2 0 0 1 0 0
0 2 0 0 4 0 0 2 0 0 1 0
0 0 2 0 0 4 0 0 2 0 0 1
1 0 0 2 0 0 4 0 0 2 0 027
0 1 0 0 2 0 0 4 0 0 2 0
0 0 1 0 0 2 0 0 4 0 0 2
2 0 0 1 0 0 2 0 0 4 0 2
0 2 0 0 1 0 0 2 0 0 4 0
0 0 2 0 0 1 0 0 2 0 0 4
abcm
37
0 0 0
0
0 0 0
0
0 0
0 0
0 0
&
0
0
0
1 1 1 =
2 2 2
1 1
2 2
e e e
tt t t
e e
V V V
t tt
ee e e e
V
x
y
uz
v N u N u B u
wy x
z x
z y
V dV D dV u B DB u dV
u B DBdV u u K u
38
0
1 1 1
0
1 1 1
1 8
1
2
0 0
0 0
0 0
with
0
0
0
t
ee e
t t
e
V
i
i
i
i
i i
i i
i i
V u K u
K B DBdV abcB DBd d d
N
x
N
y
N
zB B B B
N N
y x
N N
z x
N N
y z
39
0
0
1 1 1
1 1 1
11 1
8
11 1
8
11 1
8
can be evaluated.
Simpler alternative: evaluate
using Gaussian quadrature.
i i ii i
i i ii i
i i ii i
t
e
V
t
e
N N
x a a
N N
y b b
N N
z c c
K B DBdV
K abcB DBd d d
1 1 1
1 1 11 1 1
, , , ,
Use 2 2 2 quadrature.
n n n
i j k i j k
i j k
I f d d d w w w f
40
Isoparametric hexahedron element
1
2
1
3
23
4
4
5
5
6
7
68
78
x
z
y
8 noded element with 3dofs/node
41
8 8 8
1 1 1
18
1
1
1 8 18
1 8
1
1 8 88
81
8
Coordinates
, , ; , , ; , ,
, ,
0 0 0 0
, , 0 0 0 0
0 0 0 0
, ,
i i i i i i
i i i
i i
i
i i
i
i i
i
x N x y N y z N z
x
yN x
x N N z
y N y N N
z N N x
N z y
z
1
, , 1 1 1 , 1,2, ,88
j j j jN j
42
1 2 8
0 0
0 0
0 0
&
0
0
0
e e e
x
y
uz
v N u N u B u
wy x
z x
z y
B B B B
6 3
0 0
0 0
0 0
0
0
0
i
i
i
i
i i
i i
i i
N
x
N
y
N
zB
N N
y x
N N
z x
N N
z y
43
i i ii i
i i i i
i i i i
N N Nx y zN x y z N
x y z x
N N N Nx y z x y z
x y z
N x y zN N Nx y z
x y z
i
i i
i i
N
x
N NJ
y y
N N
z z
8 8 8
1 1 1
8 8 8
1 1 13 3
8 8 8
1 1 1
i i ii i i
i i i
i i ii i i
i i i
i i ii i i
i i i
N N Nx y zx y z
N N Nx y zJ x y z
x y z N N Nx y z
44
8 8 8
1 1 1
8 8 8
1 1 13 3
8 8 8
1 1 1
1, , 1
8
i i ii i i
i i i
i i ii i i
i i i
i i ii i i
i i i
i i
N N Nx y zx y z
N N Nx y zJ x y z
x y z N N Nx y z
N
1 1 , 1,2, ,8
11 1 , 1,2, ,8
8
11 1 , 1,2, ,8
8
11 1 , 1,2, ,8
8
Elements of J are tri-quadratic functions
i i
ii i i
ii i i
ii i i
i
Ni
Ni
Ni
1
ii
i i
ii
NN
x
N NJ
y
NN
z
45
0
1 1 1
0
1 1 1
2
1
Elements of this matrix ratios of triquadratic functions of , ,& .
These elements cannot be evaluated exactly.
Use 2 2 2 Gauss quadrature to get
t t
e
V
t
ers i j k
k
K B DBdV B DB J d d d
K w w w B DB
2 2
1 1
1 1 1
1 1 1
, ,
:
The integrand is a triquadratic function and hence the integral can
be evaluated exactly using 2 2 2 Gauss quadrature.
i j krsi j
e
J
Note
V J d d d
46
Cantilever block
E=210 GPa
Nu=0.3
Rho=7800 kg/m3
108 dofs
16 elements
8-noded hexahedron elements
Mode 1, 84.80 Hz Mode 2, 122.81 Hz Mode 3, 379.27 Hz
48
DOFs=600
128 elements
8 noded hexahedron elements
Mode 1, 83.33 Hz Mode 2, 120.43 Hz Mode 3, 357.38 Hz
50
2D approximations
Object is so thin that stresses across the thickness are
neglected (plane stress)
Object so thick that displacements across the thickness
are neglected (plane strain)
Objects possessing rotational symmetry about an axis and
loaded and supported in an axisymmetric manner.
51
Axisymmetric problems
3D axisymmetric solid
Not necessarily prismatic
Not necessarily thin or thick
Geometry
Surface tractions , , = ,
Body force: , , 0,
, , = , , , , = ,r r z z
f r z f r z
F r z
F r z F r z F r z F r z
Loads
, , 0
, , ,
, , ,
v r z
u r z u r z
w r z w r z
Displacements
Linear, homogeneous
elastic, isotropic
Material
Rotational
Symmetry
about an axis
,r u
,z w,v
52
x
y
z
middle plane thicknesst
transverse loadlongitudinal load
face edge
least lateral dimension
Plate bending element
