Two Dimensional Elements

7/30/2019 Two Dimensional Elements

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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

Two Dimensional Elements

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