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Two Dimensional Elements
Ara Arabyan
Weeks 9 and 10
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Two-Dimensional Problems
q Two-dimensional problems in structural mechanics occur in a varietyof circumstances
q The most commonly encountered two-dimensional problem is that ofthin plate subjected to in-plane edge loads
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Two-Dimensional Elements
q Two-dimensional problems are typically modeledusing triangular or quadrilateral elements
Triangularelements
Quadrilateralelements
Two-dimensionaldomain
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Constant-strain Triangular Element
q One of the simplest two-dimensional elements toformulate in FEA is the constant-strain triangular(CST) or the linear triangular element
xj
X
i
j
qi
vjvi
qj
xi
pi
uj
ui
k uk
vk
pj
qk
pk
Y
xk
yi
yj
yk
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CST Element (contd)
q The CST element is assumed to have three cornernodes (i, j, k) with two translational DOF at each node(6 DOF for the element)
q
The nodal displacement vector for the element isdefined by
q The locations of the nodes are defined by x and y
coordinates relative to a global reference frameq The triangle can have arbitrary proportions as
defined by the locations of its nodes
T
i i j j k ku v u v u v = u
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CST Assumed Displacement Field
q To use the Rayleigh-Ritz method we need to assumedisplacement fields in the x and y directions such thatwe have exactly six undetermined coefficients
q
Complete displacement fields with six undeterminedcoefficients are
q Note that both functions vary with x and y
( ) 1 2 3
4 5 6
,
( , )
u x y a a x a y
v x y a a x a y
= + +
= + +
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CST Displacement Field (contd)
q These relations can also be written as
q More concisely
( )( )
1
2
3
4
5
6
1 0 0 0,0 0 0 1,
a
a
x y au x yx y av x y
a
a
=
( ) ( )=u x X x a
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CST Shape Functions
q Since the assumed displacements must equal thenodal displacements at the three nodes we have:
q More concisely (as with other element types)
1
2
3
4
5
6
1 0 0 0
0 0 0 11 0 0 0
0 0 0 1
1 0 0 0
0 0 0 1
i i i
i i i
j j j
j j j
k k k
k k k
x y a u
x y a vx y a u
x y a v
x y a u
x y a v
=
=Xa u
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CST Shape Functions (contd)
q This results in
q Substituting in the displacement field expression weobtain
where N(x)=N(x,y) is the shape function matrix for theCST element
1=a X u
( ) ( ) ( )1= =u x X x X u N x u
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CST Shape Functions (contd)
q The shape function matrix that results is given by
where is the area of the triangle and
( )( ) ( ) ( )
( ) ( ) ( )1 2 3
1 2 3
, 0 , 0 , 01
0 , 0 , 0 ,2
N x y N x y N x y
N x y N x y N x yA
=
N x
( ) ( ) ( )1 1 1 1 2 2 2 2 3 3 3 3
1 2 3 3 2 2 3 1 1 3 3 1 2 2 1
1 2 3 2 3 1 3 1 2
1 3 2 2 1 3 3 2 1
, , ,N x y x y N x y x y N x y x y
x y x y x y x y x y x yy y y y y y
x x x x x x
= + + = + + = + +
= = = = = = = = =
( )detA = X
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CST Strain Vector
q The relationship obtained thus far can be written inconcise form as
q Recall that strains in a two-dimensional domain aregiven by
or
( ) ( )=u x N x u
( )
( )
0
,0 ,
x
y
xy
xu x y
v x yy
y x
=
( )u x =
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CST Strain-Displacement Matrix
q Applying this definition to displacement field for theCST element we obtain
where
( ) ( ) ( )= =u x N x u B x u =
31 2
31 2
3 31 1 2 2
0 0 0
0 0 0
NN N
x x xNN N
y y yN NN N N N
y x y x y x
=
B
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CST Strain-Displacement Matrix (contd)
q Performing the required differentiations B isobtained more explicitly as
where and have the same definitions as those in N
q Clearly the strain-displacement matrix B is a constantmatrix (no dependence on x or y)
1 2 3
1 2 3
1 1 2 2 3 3
0 0 0
1 0 0 02A
=
B
i i
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CST Strain Energy
q The strain energy of the CST element can now beformulated
q If the element has thickness t at any point across itsarea dV = tdA
q Since B and E are constant matrices and assuming t is
constant throughout the element
1 1 1
2 2 2
T T T T
V V
dV dV
= = =
Ee u B EB u u ku
T T T T
V A A
dV tdA t dA tA= = = B EB B EB B EB B EB
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Plane Stress and Plane Strain
q Recall that for a state of plane stress
and for a state of plane strain
2
1 0
1 0
1 10 0
2
E
=
E
( )( )
1 01 0
1 1 21 2
0 02
E
= +
E
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Performance of CST Element
q The CST element is found to perform poorly in modeling bendingstress and associated deflections
q It is thus not included in the libraries of any professional FEAprograms
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Linear Strain Triangular Element
q A higher order but still simple two-dimensionalelement is the 6-node linear strain triangular (LST) orthe quadratic triangular element
X
i
jvi
pi ui
kY
l
mn
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LST Displacement Field
q Complete displacement fields for the LST triangle are
q This field results in a shape function matrix N that isquadratic in x and y
q The quadratic shape function results in a strain-displacement matrix B that varies linearly with x andy
( ) 2 21 2 3 4 5 62 2
7 8 9 10 11 12
,
( , )
u x y a a x a y a x a xy a y
v x y a a x a y a x a xy a y
= + + + + +
= + + + + +
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LST Stiffness Matrix
q Applying the same procedure as before for a constantthickness element we obtain a 12 12 stiffness matrixgiven by
q The integration shown is in general laborious toperform analytically; as a result a numerical method
such as Gaussian quadrature is used to obtain thematrix
maxmax
min min
12 312 12 3 3 3 12yx
T
x yt dxdy
= k B E B
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LST Element in ANSYS
q The LST element in ANSYS is called the PLANE2 2-D6-Node Triangular Structural Solid Element
q The shape functions of the PLANE2 element are
given in terms of natural or area coordinates
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Area Coordinates
q Area or natural coordinates are defined in terms of arearatios
1A
2A
3A
1
1
22
33
1 2 3
AL
AA
LAA
LA
A A A A
=
=
=
= + +i
j
k
l
m
n
Quadratic in L1, L2, L3
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Area Coordinates (contd)
q The area coordinates defined for a general triangle varybetween 0 and 1 as the point P moves from an edge to anarbitrary point in the interior of the triangle (see Figs. 7.8 and7.9 in Moaveni)
q
The shape functions can be expressed in terms of only L1 and L2since the three coordinates are not independent
q This transforms the expression for the stiffness matrix to
1 2 3 1L L L+ + =
1 1
1 2
0 0
Tt dL dL= k B EB
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Plane Quadrilateral Bilinear Element
q The 4-node quadrilateral element is the simplestfour-sided two-dimensional element
xj
X
i
j
qi
vjvi
qj
xi
pi
uj
ui
kuk
vk
pj
qk
pkY
xk
yi
yj
yl
ul
l vl
ql
pl
yk
xl
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Bilinear Displacement Field
q The assumed displacement field for this element is
or
( ) 1 2 3 4
5 6 7 8
,
( , )
u x y a a x a y a xy
v x y a a x a y a xy
= + + +
= + + +
( )
( )
1
2
3
4
5
6
7
8
1 0 0 0 0,
0 0 0 0 1,
a
a
a
x y xy au x y
x y xy av x ya
a
a
=
Note that theassumeddisplacementfield is notcomplete(neither linearnor quadratic)
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Bilinear Displacement Field (contd)
q Writing these expression more concisely andperforming the usual operations we obtain
where N(x)=N(x,y) is the shape function matrix for thethe plane quadrilateral bilinear (PQB) element
=Xa u 1=a X u
( ) ( ) ( )1= =u x X x X u N x u
( )( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
1 2 3 4
1 2 3 4
, 0 , 0 , 0 , 0
0 , 0 , 0 , 0 ,
N x y N x y N x y N x y
N x y N x y N x y N x y=
N x
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PQB Strain-Displacement Matrix
q The strain in this element can now be computed from
where( ) ( ) ( )= =u x N x u B x u =
31 2 4
31 2 4
3 31 1 2 2 4 4
0 0 0 0
0 0 0 0
NN N N
x x x xNN N N
y y y y
N NN N N N N N y x y x y x y x
=
B
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PQB Strains
q Note that because of the assumed displacement field
q This strain distribution may be unsuitable for someapplications
( )
( )
( ),
x x
y y
xy xy
y
x
x y
=
=
=
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PQB Stiffness Matrix
q Using this matrix the stiffness matrix for a constantthickness h PQB element can be computed
q The integration is complicated and is performedusing Gaussian quadrature
maxmax
min min
8 38 8 3 3 3 8yx
T
x yh dxdy
= k B E B
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PQB Element in ANSYS
q The PQB element in ANSYS is called the PLANE422-D Structural Solid
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Natural Coordinates
q ANSYS (and most other FE programs) use naturalcoordinates to define shape functions
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Natural Coordinates (contd)
q Natural coordinates for quadrilaterals are defined as
X
Y s
t
1
1
s
t
= +
=
1
1
s
t
= += +
1
1
s
t
= =
1
1
s
t
= = +
1
2t = +
1
2s =
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Natural Coordinates (contd)
q This transforms the expression for the stiffnessmatrix to
q Additional mathematical operations required totransform strains and stresses from natural to global
coordinates
1 1
1 1
Th dsdt
=
k B EB
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Axisymmetric Stresses
q When circular geometric and loading symmetry exists in athree-dimensional structure it is reasonable to assume thatstresses are independent of the coordinate
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Axisymmetric Stresses (contd)
q In axisymmetric stress cases the meaning of the x andy stresses changes to radial (r) and axial (z) stresses;shear (rz) is still assumed to exist
r
i
j wj
wi ujui
k uk
wk
z
ul
lwl
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Example 1: Beam Model Using 2D Elements
q A cantilevered beam 1 m long, 0.1 m wide, and 0.2 mhigh is loaded by an end load of 1000 N. The Youngsmodulus for the material is 200 GPa. Model a sectionof the beam using PLANE2 and PLANE42 elements
and compare the maximum stresses produced withthe exact solution.
1 m
1000 N
0.1 m
0.2 m
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Example 1 (contd)
q The maximum stress in the beam using exactmethods is
q The maximum deflection (at the free end) usingexact methods is
( )( )( )
( )( ),max 3
1000 1 0.11.500 MPa
1 0.1 0.212
x
Mc
I = = =
( )( )
( ) ( )( )
335
max39
1000 12.5 10 m
13 3 200 10 0.1 0.212
FLv
EI= = =
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Example 1 (contd)
q When the beam is modeled as a set of plane stress elements theload at its end must be divided by its width to produce a loadper unit width; thus the load applied at the end must be(1000/0.1 = 10000 N/m)
q
Using PLANE2 elements and mapped meshing (with anelement side of 0.05 m) we obtain in ANSYS
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Example 1 (contd)
q ANSYS produces the following solution for thismodel (contour plot of SX on deformed shape)
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Example 1 (contd)
q The maximum axial stress predicted by this model is1.68 MPa, which is 12% higher than the exact solution
q
Similarly the maximum deflection predicted by thismodel is 2.59 10-5 m, which is 3.6% higher than theexact solution
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Example 1 (contd)
q Using PLANE42 elements and mapped meshing (with anelement side of 0.05 m) we obtain in ANSYS
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Example 1(contd)
q ANSYS produces the following solution for thismodel (contour plot of SX on deformed shape)
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Example 1 (contd)
q The maximum axial stress predicted by this model is1.51 Mpa, which is 0.7% higher than the exactsolution
q Similarly the maximum deflection predicted by thismodel is 2.58 10-5 m, which is 3.2% higher than theexact solution
q Note also that the x-direction stresses vary only in they direction in each element
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Example 1 (contd)
q The PLANE42 element provides a betterapproximation with fewer elements and DOF thanthe PLANE2 element in this case because the stressvariation in the y direction is linear in the exact
solutionq The PLANE2 element may provide a better
approximation in instances when the bendingmoment varies quadratically or at a higher order in
the x direction
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Example 2: Axisymmetric Model of PressureVessel
q Using axisymmetric modeling determine the stressdistribution in a pressure vessel made of aluminumtubing of outside diameter 8 in and a wall thicknessof in subjected to an internal pressure ofpi = 5000
psi. Compare FE results to the exact solution.
pi
Aluminum
pipe 8
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Example 2 (contd)
q The dimensions of the pipe are such that it has beconsidered a thick-walled vessel
q The exact solution for the radial stress is given by
resulting in maximum and minimum values of
2 2
2 2 2 2
161 16,333 1 psii i or
o i
p r r
r r r r
= =
max
min
0 psi (at outer fiber)
5000 psi (at inner fiber)
r
r
==
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Example 2 (contd)
q Similarly the exact solution for the tangential orhoop stress is given by
resulting in maximum and minimum values of
2 2
2 2 2 2
16
1 16,333 1 psi
i i o
o i
p r r
r r r r
= + = +
max
min
37,667 psi (at inner fiber)
32,667 psi (at outer fiber)
=
=
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Example 2 (contd)
q A model of the pipe wall using axisymmetric PLANE42elements is shown below
The pressure is applied as apressure of 5000 on nodes
The top andbottom nodesareconstrained inthe y directiononly to ensurethat all
displacementsare only in thex direction
The left surface of the cross
section must be 3.5 from theorigin of reference from to obtaincorrect results
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Example 2 (contd)
q A contour plot of SX (corresponding to r) is shown below
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Example 2 (contd)
q The FE results predict the maximum and minimumradial stresses as 8.363 psi and 4987 psi, respectively,which are not too far from the exact results
q The contour plot shows clearly that the maximumstress occurs at the outer fiber and the minimum atthe inner fiber
q The FE results predict a linear increase of the radialstress from the inner to the outer fiber; the exact
solution predicts a quadratic variation
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Example 2 (contd)
q A contour plot of SZ (corresponding to ) is shown below
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Example 2 (contd)
q The FE results predict the maximum and minimumhoop stresses as 37,653 psi and 32,659 psi,respectively, which are very close to the exact results
q The contour plot shows clearly that the maximumstress occurs at the inner fiber and the minimum atthe outer fiber
q The FE results predict a linear decrease of the hoopstress from the inner to the outer fiber; the exact
solution predicts a quadratic variation