Finite Element Method
Chapter 9
Axisymmetric Elements
Definition of an axisymmetric solid
An axisymmetric solid (or a thick-walled body) of revolution is defined as a 3-D body that is generated by rotating a plane and is most easily described in cylindrical coordinates. Where z is called the axis of symmetry.
If the geometry, support conditions, loads, and material properties are all axially symmetric (all are independent of ), then the problem can be idealized as a two-dimensional one.
Problems such as soil masses subjected to circular footing loads, thick-walled pressure vessels, and a rocket nozzle subjected to thermal and pressure loading can often be analyzed using axisymmetric elements.
Examples of an axisymmetric solid
Examples of an axisymmetric solid
Enclosed pressure vesselEngine valve stem
axisymmetric problems can be analyzed by a finite element of revolution, called axisymmetric elements. Each element consists of a solid ring, the cross-section of which is the shape of the particular element chosen (triangular, rectangular, or quadrilateral elements). An axisymmetric element has nodal circles rather than nodal points
FE axisymmetric elements
Equations of Equilibrium:
The three-dimensional elasticity equations in cylindrical coordinates can be summarized as follows
011
021
01
bzrzzzr
rzr
rrzrrr
Zrzrr
Yrzrr
Xrzrr
Equations of Equilibrium:
The three-dimensional strain-displacement relationships of elasticity in cylindrical coordinates were u, v, w are the displacements in the r, ,dz , respectively, are:
w
rz
v
z
w
z
u
r
w
r
uv
r
r
v
r
vu
rr
u
zz
zr
rr
1,
,1
1,
Equations of Equilibrium:
The three-dimensional stress-strain relationships for isotropic elasticity are:
z
zr
r
z
r
z
zr
r
z
r
E
2
210000
02
21000
002
21000
0001
0001
0001
)21()1(
Axisymmetric Stresses and Strains
In axisymmetric problems, because of the symmetry about the z -axis, the stresses are independent of the coordinate. Therefore, all derivatives with respect to vanish and the circumferential (tangent to direction) displacement component is zero; therefore,
0 0
, , ,
r z r z
r z r z
and
u u w u w
r r z z r
Axisymmetric Stresses and Strains
,
,
r
z r z
u u
r r
w u w
z z r
The stress-strain relationship for isotropic axisymmetric problems
zr
z
r
zr
z
r
E
2
21000
01
01
01
)21()1(
}{][}{ D
Derivation of the Stiffness Matrix and Equations
Typical slice through an axisymmetric solid Discretized into triangular elements
Step 1: Discretize and Select Element Type
Step 1: Discretize and Select Element Type
m
m
j
j
i
i
m
j
i
w
u
w
u
w
u
d
d
d
d}{
(ui , wi ) displacement components of node i in the r and z directions,respectively.
Step 2: Select Displacement Functions
zaraazrw
zaraazru
654
321
),(
),(
6
5
4
3
2
1
654
321
1000
0001
),(
),(}{
a
a
a
a
a
a
zr
zr
zaraa
zaraa
zrw
zru
Step 2: Select Displacement Functions
mmm
jjj
iii
mmm
jjj
iii
zaraaw
zaraaw
zaraaw
zaraau
zaraau
zaraau
654
654
654
321
321
321
In Matrix Form
m
j
i
mm
jj
ii
u
u
u
zr
zr
zr
a
a
a1
3
2
1
1
1
1
m
j
i
mm
jj
ii
w
w
w
zr
zr
zr
a
a
a1
6
5
4
1
1
1
mm
jj
ii
yx
yx
yx
A
1
1
1
2
mji
jimimjmji yyxyyxyyxA
)()()(2
Solving for the a’s
A is the area of the triangle
m
j
i
mji
mji
mji
u
u
u
A
a
a
a
2
1
3
2
1
m
j
i
mji
mji
mji
w
w
w
A
a
a
a
2
1
6
5
4
ijmmijjmi
jimimjmji
jijimimimjmjmji
rrrrrr
zzzzzz
rzzrrzzrrzzr
3
2
1
1}{
a
a
a
zru
m
j
i
mji
mji
mji
u
u
u
zrA
u
12
1}{
1{ } 1
2
1( , ) ( ) ( ) ( )
2
1( , ) ( ) ( ) ( )
2
( , ){ }
( , )
i i j j m m
i i j j m m
i i j j m m
i i i i j j j j m m m m
i i i i j j j j m m m m
i i j
u u u
u r z u u uA
u u u
u r z r z u r z u r z uA
w r z r z w r z w r z wA
N u Nu r z
w r z
j m m
i i j j m m
u N u
N v N v N v
( , ){ }
( , )
1
2
1
2
1
2
i i j j m m
i i j j m m
i i i i
j j j j
m m m m
N u N u N uu r z
N v N v N vw r z
N r zA
N r zA
N r zA
[ ]
0 0 0{ }
0 0 0
{ } { }
0 0 0
0 0 0
i
i
i j m j
i j m j
m
m
i j m
i j m
u
w
N N N u
N N N w
u
w
d
N N N
N
N
N NN
Step 3: Define the Strain/Displacement and Stress/Strain Relationships
2
6
312
3 5
{ }
{ }
r
z
rz
r
z
rz
u
r
w
z
u
r
u w
z r
a
a
a zaa
r r
a a
}{][}{ dB
0 0 0
0 0 0
0 0 0
0
0
0
i
i j m
i
i j m
j
j ji i m mji j m
mi i j j m m
m
i
ii
ij
i j m i i ij i
mi i
m
u
w
uzz z
wr r r r r r
u
w
u
w
uB B B B z
wr r
u
w
B is a function of r and z
Stress Strain Relationship
}{][][}{ dBD
zr
z
r
zr
z
r
E
2
21000
01
01
01
)21()1(
Step 4 :Derive the Element Stiffness Matrix and Equations
[ ] [ ] [ ][ ]
[ ] 2 [ ] [ ][ ]
T
V
T
A
k B D B dV
k B D B r dr dz
1) Numerical integration (Gaussian quadrature)2) Explicit multiplication and term-by-term integration.3) Evaluate [B] at a centroidal point of the element
3 3
[ ( , )] [ ]
[ ] 2 [ ] [ ][ ]
i j m i j m
T
r r r z z zr r z z
B r z B
k r A B D B
}{}{][}{][}{ PdSTNdVXNf
S
T
V
T
Concentrated nodal forcesBody forces
Surface Tractions
Step 4 :Derive the Element Stiffness Matrix and Equations
{ } [ ] { } 2 [ ]bT T
b
bV A
Rf N X dV N r dr dz
Z
for a machine part rotating with:a constant angular velocity about the z axis (zero angular acceleration); : material mass density (kg/m3) and r is the radial coordinate.Zb= body force per unit volume due to the force of gravity
(weight density, kN/m3).
Distributed Body ForcesBody forces can either : Gravity force in the direction of z axis, or Centrifugal forces in rotating machine parts in the direction of the r axisThe nodal equivalent body forces can found using
2
bR r
0{ } 2
0
2{ }
3
i b
bi
i bA
bir b
bi
biz b
N Rf r dr dz
N Z
f RArf
f Z
b
b
b
b
b
b
bmz
bmr
bjz
bjr
biz
bir
b
Z
R
Z
R
Z
R
Ar
f
f
f
f
f
f
f3
2}{
rRb 2
Distributed Body Forces
{ } [ ] { } [ ]rT T
s
zS S
pf N T dS N dS
p
dzrp
p
zr
zr
Af j
z
rz
zjjj
jjj
jsm
j
2
0
0
2
1}{
For example: along the vertical face jm of an element, let uniform loads pr and pz be applied, as shown along r=rj.
Surface Forces
dzrp
p
zr
zr
Af j
z
rz
zjjj
jjj
jsm
j
2
0
0
2
1}{
z
r
z
rjmj
smz
smr
sjz
sjr
siz
sir
s
p
p
p
pzzr
f
f
f
f
f
f
f
0
0
2
)(2}{
Surface Forces
evaluated at r = rj ; z = z
Step 5: Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions
N
e
ekK1
)( ][][
}{][}{ dKF
N
e
efF1
)( ][][
Step 6: Solve for the Nodal Displacements
Step 7: Solve for the Element Stresses
Steps 5 through 7
Steps 5 through 7, are analogous to those of Chapter 6 for the CST element, except the stresses are not constant in each element.
They are usually determined by: Determine the centroidal element stresses, Determine the nodal stresses for the element
and then average them. The latter method has been shown to be more
accurate in some cases
Example 1
For the element of an axisymmetric body rotating with a constant angular velocity = 100 rev/min.Evaluate the approximate body force matrix, include the weight of the material, where the weight density w is 0.283 lb/in
3.The coordinates of the element (in inches) are shown in the figure.
The body forces per unit volume evaluated at the centroid of the element are
Zb =0.283 lb/in3
All r-directed and z-directed nodal
body forces are equal
Example 1 -Continue
Example 2
For the long, thick-walled cylinder under internal pressure p equal to 1 psi shown, determine the displacements and stresses.
Example 2
To illustrate the finite element solution for the cylinder, we first discretize the cylinderinto four triangular elements, as shown. A horizontal slice of the cylinderrepresents the total cylinder behavior.
Example 2
Assemblage of the Stiffness Matrix
0.5, 0, 1.0, 0, 0.75, 0.25
( 1; 2,
For element
5 )
1
i i j j m mr z r z r and z
i j and m
Example 2 – Stiffness Matrix for ELEMENT 1
Example 2 – Stiffness Matrix for ELEMENT 1
Example 2 – Stiffness Matrix for ELEMENT 1
1.0, 0, 1.0,
For element 2
0.5, 0.75, 0.25
( 2; 3, 5 )
i i j j m mr z r z r and z
i j and m
Example 2 – Stiffness Matrix for ELEMENT 1
Example 2 – Stiffness Matrix for ELEMENT 2 & 3
Example 2 – Stiffness Matrix for ELEMENT 1
Example 2 – Total Stiffness Matrix
HW:
xxxx
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