Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Two Dimensional Analysis – Plane Stress and Plane StrainIn a large class of every day engineering problems certain approximations are made to simplify the structural analysis of three dimensional components. Recall that with the strong formulation there are 15 equations to solve in terms of 15 unknowns. One approach to reduce the computational effort in solving this system of equations is realizing that certain problems are really two dimensional. This reduces the number of equations to solve.
Three types of two dimensional idealizations are employed:1. Plane stress2. Plane strain3. Axisymmetry
For example, simplifying approximations can be made in analyzing deformations in a thin plate subjected to in-plane forces. Consider a prismatic structural member with a very short length or thickness (h). The mid plane contains the x and y coordinate axes and the thickness extends along the z axis to +h/2 and –h/2. If the member is not loaded on surfaces perpendicular to the z axis then
on these surfaces perpendicular the z axis and through the thickness. Assuming the remaining stress components are not dependent on z, then this is known as plane stress. 1
0 yzxzz
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Here the stress matrix and the strain matrix take the following forms
For plane strain one dimension (say along the z axis) is exceedingly large relative to the other two dimensions. Applied forces act in the x – y plane and do not vary along the zdirection. Some practical examples include dams and tunnels as well as bars that are compressed along their length. Here
on these surfaces perpendicular the z axis. Assuming the remaining strain components are not dependent on z, then this is known as plane strain.
2
0 yzxzz
yxz
yxy
xyx
yxy
xyx
E
yxyxyxyx
yxyxyxyx
00
0,,0,,
0000,,0,,
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Note that for plane strain the functional dependence of the displacements are
Thus
The axisymmetric formulation has the same geometric implication, i.e., a two dimensional problem with the additional wrinkle of posing the problem in a different coordinate system.
0
,
,
w
yxvv
yxuu
yxz
yxy
xyx
yxy
xyx
yxyxyxyx
yxyxyxyx
000,,0,,
0000,,0,,
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
AxisymmetryWe have considered “line” elements and two dimensional elements. We now turn our attention to axisymmetric elements. This is a special two dimensional element utilized when there is a specific type of geometric symmetry and load symmetry present in the problem. We use cylindrical coordinates (r, , z) to describe all aspects of the problem, i.e., displacements, strains and stresses. The z-axis is the axis of symmetry.
Examples of axisymmetric problems include, but are not limited to
• Pressure vessels
• Cylindrical shafts
• Hertzian contact between spheres
Use of axisymmetric elements provides computational efficiency relative to a full three dimensional analysis.
4
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Quasi-Axisymmetric NotationSome fundamental concepts are presented first. An axisymmetric element is essentially a triangular (or quadrilateral) torus. Each node traces a circular line which are depicted as dashed lines in the figure below on the left:
The z-axis is the axis of symmetry. The component being modeled must have geometric as well as load symmetry with respect to this axis. These problems are best modeled in a polar coordinate system (r, , z). Corresponding displacements will be designated (u, v, w) and the two dimensional stress state is indicated in the figure on the right.
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
The following two load applications are acceptable for an axisymmetric analysis:
The next two are not because the load is functionally dependent on . The figure on the left represents the wind loads (windward and leeward) on a cooling tower.
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
As a previous figure indicates, we wish to use the notation, and to some extent the derivations, developed for the two dimensional constant strain element. That element will be “spun around” an axis of symmetry (the z-axis) and thereby generate an axisymmetric element.
In that context the “two dimensional” axisymmetric element presents itself in an r-z plane much the same way the constant strain element presents itself in the x-y plane. For the two dimensional constant strain and linear strain element we ignored displacements in the zdirection. Here we do not. We could wave our hands over the constant strain and linear strain elements and mumble something regarding plane stress or plane strain and somewhat ignore displacements in the z direction. We will rectify that when solid (tetrahedron) elements are developed. For axisymmetric elements the “out of plane” displacements are a bit more troubling.
The term “quasi-axisymmetric” refers to fact that in truly axisymmetric problems the circumferential displacements (v or u depending on the notation used) are identically zero. Where stresses and strains are not functionally dependent on and
this class of problems are labeled quasi-axisymmetric.
0v
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
The most familiar components that can be modeled in an axisymmetric fashion are thin walled pressure vessel (hot dog stresses). In fact any cylindrical pressure vessel, thin walled or otherwise, has the potential of being modeled as an axisymmetric problem. Consider the gun barrel for an Abrams M-1 tank:
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
A portion of the axisymmetric mesh used to model the tank barrel is depicted below:
Note that the mesh is predominantly quadrilateral axisymmetric elements, with a triangular element furnishing a transition as the throat of the barrel thins out. The point here is that there are a lot of elements in this mesh and they all trace out a torus when spun around the z-axis. If three dimensional elements had been used to model the full barrel the problem would have been too big to solve.
triangular transition element
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Recall that in Cartesian coordinates the strain displacement relationships were
In cylindrical coordinates the strain displacement relationships are
Strain Displacement Relationships – Cylindrical Coordinates
yw
zv
zw
xw
zu
yv
xv
yu
yu
yzz
xzy
xyx
wrz
vzw
rw
zuv
rru
rv
rvu
rru
zz
rz
rrr
1
1
1
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
For plane strain in the z-direction
The stress and strain matrices take the following form
Any dependence upon z is suppressed for plane strain, and due to symmetry about the z-axis the strains in an axisymmetric component are independent of . Thus all derivatives with respect z and vanish
keeping in mind that w = 0 for plane strain.
0 zrzyzxzz
00
0
zz
rz
rrr
ru
rv
rv
ru
z
r
rr
0000
00000
r
rr
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
For plane stress in the z-direction
The stress and strain matrices take the following form
Due to symmetry about the z-axis the strains in an axisymmetric component are independent of . Thus all derivatives with respect vanish.
In addition, shear strains associated with the zero shear stresses are similarly zero
0 yzxzzrzz
rrzrrr Eru
rv
rv
ru
00000
r
rr
z
r
rr
0000
00 zrz
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
In axisymmetric problems radial displacements produce circumferential strains, which in turn generate circumferential stresses on face EBDF of the differential element shown below (this is not a depiction of a finite element).
Line AB becomes longer as it moves from its original position, marked in red, to its new position, marked with a solid black line. This lengthening produces strains in the direction.
y – axis into the pagez – axis out of the page
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
The strain in the radial direction is defined by the change in length in the radial direction of line segment BD in the previous figure divided by the original length (dr).
Looking at the change in length of line segment AB divided by its original length we would obtain from the previous figure
ru
udrruu
drr
1
ru
rdrddur
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
If we now look directly at face BDFE in the previous figure we would see the following (the original position of the face is marked with a dashed line)
Focusing on line segment BE, the change in its length divided by its original length would yield the strain in the z-direction, i.e.,
zw
wdzzww
dzz
1
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
m
m
j
j
i
i
wu
w
uwu
d
Notation – Three Node Triangular Axisymmetric ElementWe formulate the linear displacement function as follows
The nodal displacements are
zaraazrw
zaraazru
654
321
,,
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
The general displacement function can be expressed in matrix notation as
6
5
4
3
2
1
654
321
10000001
aaaaaa
zrzr
zaraazaraa
wu
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
To obtain the coefficients we substitute the coordinates of the nodes into the previous equations. This yields
We can solve for the first three coefficients and the last three coefficients from the following two systems of equations;
mmm
jjj
iii
mmm
jjj
iii
zaraaw
zaraawzaraawzaraau
zaraauzaraau
654
654
654
321
321
321
3
2
1
1
11
aaa
zr
zrzr
u
uu
mm
jj
ii
m
j
i
6
5
4
1
11
aaa
zr
zrzr
w
ww
mm
jj
ii
m
j
i
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Inverting the first expression leads to
The method of cofactors is used to invert the 3 x 3 matrix, i.e.,
where
m
j
i
mm
jj
ii
u
uu
zr
zrzr
aaa
1
3
2
1
1
11
mji
mji
mji
mm
jj
ii
Azr
zrzr
21
1
11
1
mm
jj
ii
zr
zrzr
A
1
11
2
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
This determinant is
Note that A is the area of the triangular element. Thus (see appendix in a Calculus book – Cramer’s rule)
Now
jimimjmji zzrzzrzzrA 2
ijmmijjmi
jimimjmji
jijimmimijmjmji
rrrrrr
zzzzzz
rzzrzrrzrzzr
m
j
i
mji
mji
mji
u
uu
Aaaa
21
3
2
1
m
j
i
mji
mji
mji
w
ww
Aaaa
21
6
5
4
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Utilizing the matrix expressions from the previous we once again formulate linear displacement function as follows
mmjjii
mmjjii
mmjjii
m
j
i
mji
mji
mji
uuu
uuu
uuu
Azr
u
uu
Azr
aaa
zr
zaraazru
211
211
1
,
3
2
1
321
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
This leads to
If we identify
then
mmmmjjjj
iiii
uzrA
uzrA
uzrA
zru
21
21
21,
zrA
N
zrA
N
zrA
N
mmmm
jjjj
iiii
21
21
21
mmjjii uNuNuNzru ,
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Similarly
mmjjii
mmjjii
mmjjii
m
j
i
mji
mji
mji
vww
vww
www
Azr
w
ww
Azr
aaa
zr
zaraazrw
211
211
1
,
6
5
4
654
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
and
Once again
thus
mmmmjjjj
iiii
wzrA
wzrA
wzrA
zrw
21
21
21,
zrA
N
zrA
N
zrA
N
mmmm
jjjj
iiii
21
21
21
mmjjii wNwNwNzrw ,
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Now
m
m
j
j
i
i
mji
mji
mmjjii
mmjjii
wu
w
uwu
NNN
NNN
wNwNwN
uNuNuN
zrwzru
000
000
,,
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Strain Displacement Relationships The strains associated with a two dimensional element are
with
rw
zu
ruzwru
rz
z
r
mm
jj
ii
mmjjii
ur
Nur
Nu
rN
uNuNuNrr
u
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
this leads to
and
A
zrArr
N
Azr
ArrN
Azr
ArrN
mmmm
m
jjjj
j
iiii
i
221
221
221
mmjjii
mm
jj
ii
uuuA
uA
uA
uAr
u
21
222
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
For circumferential strain
mm
mm
jj
jj
ii
ii
mmmmjjjj
iiii
ur
zr
urz
r
urz
rA
uzrA
uzrA
uzrA
rru
21
21
21
21
1
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Similar derivations lead to
In a matrix format the strain in a two dimensional element takes the following form
mmjjiimmjjii
mmjjii
uuuuuuAr
wzu
uuuAz
w
21
21
m
m
j
j
i
i
mmjjii
mm
mjj
jii
i
mji
mji
rz
z
r
wu
w
uwu
rz
rrz
rrz
rA
000
000
000
21
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
dB
rz
z
r
Note that the {B} matrix is a function of r and z coordinates. It is not dependent only on the nodal coordinate information. In general, the circumferential strain will not be constant within the element in the radial or z directions. However, the circumferential strain will be constant throughout the element in the circumferential direction.
mmjjii
mm
mjj
jii
i
mji
mji
rz
rrz
rrz
rA
B
000
000
000
21
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Clayton Rencis (2000) Paper
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
One final note. The strain displacement relationships have singularities built into the expressions. As the edge of the element approaches the axis of revolution the (1/r)approaches infinity. How close the edge of the element is to the axis of revolution dictates how much influence this effect has on the numerical calculations.
This can have a pronounced effect on numerical computations for the stiffness matrix. For a three node triangular element the computations for the stiffness matrix have been made at the centroid of the element, which can be very close to the axis of revolution depending on the size of the axisymmetric element.
Clayton and Rencis (2000 and 1998) advocate numerical techniques to circumvent this issue. We will revisit the issue and their approach to the problem when the topic of integration points and quadrature rules are introduced.
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Stress Strain Relationships
The constitutive relationship for axisymmtric elements is given by
where
rz
z
r
rz
z
r
D
221000
0101
01
211
ED
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Element Stiffness Matrix
The stiffness matrix is determined once again with the following expression:
after integrating along the circumferential boundary. However, the [B] matrix is functionally dependent on r and z. Thus the stiffness matrix [k] is a function of r and z and is a 6x6 matrix. We can evaluate the integration above by one of three methods:
1. Numerical integration (Gaussian quadrature presented in a later section)2. Explicit multiplication and term-by-term multiplication3. Evaluate {B} at the centroid of the element. The centroid of an element is defined
by its coordinates, i.e.,
A
TT
V
TT
dzdrrBDB
dVBDBk
2
33mjimji zzz
zrrr
r
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Now define
and as a first approximation take
If the triangular mesh is fine enough then acceptable results can be obtained using this third method. However, most commercial codes will utilize numerical integration, i.e., a quadrature rule.
zrBB ,
BDBArk TT2
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
In class example
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Accounting for Body Forces and Surface Tractions
Gravity loads in the direction of the z-axis or centrifugal loads in the direction of the r-axis dominate axisymmetric analyses. Body forces at the nodes are defined through the expression
where
and
V
Tb dVXNf
b
b
ZR
X
rRb 2
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Consider the body forces at node i
where
Thus
where Ni =1 at node i, and the origin of the coordinate axes has been located at the centroid of the element. Note that the body forces, especially the centrifugal body force, is evaluated at the centroid as well.
dzdrrZR
NfV b
bTibi
2
i
iTi N
NN
00
rAZR
fb
bbi
3
2
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Similar considerations at nodes j and m lead to
where
The expression for { fb } is a first approximation relative to the radially directed body forces.
3
2 Ar
ZRZRZR
ff
f
fff
f
b
b
b
b
b
b
bmz
bmr
bjz
bjr
biz
bir
b
rRb 2
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
For surface tractions recall that
For radial and axial pressure loads, which are typical to the internal surface for an axisymmetric problem
The expression at the top can be evaluated at each node. For node j
S
Ts dSTNf
z
r
pp
T
dzr
pp
zrzr
A
dzrpp
NN
dzrpp
Nf
jz
rz
z iii
iii
jz
rz
z i
i
jz
rz
z
Tisj
m
j
m
j
m
j
20
021
20
0
2
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Performing the integration at the other two nodes lead to
z
r
z
rjmj
smz
smr
sjz
sjr
siz
sir
s
ppppzzr
ff
f
fff
f
00
22
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
In class example
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
Basic Scorecard – Axisymmetric Element
zrA
N
zrA
N
zrA
N
mmmm
jjjj
iiii
21
21
21
jimimjmji zzrzzrzzrA 2
ijmmijjmi
jimimjmji
jijimmimijmjmji
rrrrrr
zzzzzz
rzzrzrrzrzzr
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
zrAxx
NA
zrAxx
NA
zrAxx
NA
mmmmm
jjjjj
iiiii
21
2
21
2
21
2
mmjjii
mm
mjj
jii
i
mji
mji
rz
rrz
rrz
rA
B
000
000
000
21
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
FKddKF 1
N
e
ekK1
AtBDBk TT
221000
0101
01
211
ED
Section 9: AXISYMMETRIC ELEMENTS
Washkewicz College of Engineering
m
m
j
j
i
i
mmjjii
mm
mjj
jii
i
mji
mji
rz
z
r
wu
w
uwu
rz
rrz
rrz
rA
000
000
000
21
rz
z
r
rz
z
r
D
m
m
j
j
i
i
wu
w
uwu
d