section 7 constant strain triangular elements.pptacademic.csuohio.edu/duffy_s/cve_512_7.pdf · ·...

Section 7: CONSTANT STRAIN TRIANGULAR ELEMENTS

Washkewicz College of Engineering

IntroductionWe have considered “line” elements up to this point, i.e., springs, rods and beams. Grid elements could be added to that category. Line elements have geometric properties associated with the position along the line, for example cross sectional area and moments of inertia. Hence, a local coordinate axis along the line of the element is a necessity.

Now two dimensional planar elements are considered. These elements require two coordinate axes to define the deformation at a position within the element. At first it is quite easy to utilize the global coordinate system for element formulations – at least for constant strain elements. For linear strain elements the use of the global coordinate system is extremely difficult, and we will resort to utilizing a local coordinate system.

With two dimensional elements we will be able to analyze components that can be classified as plane stress problems, or plane strain problems.

The stiffness matrix for the constant strain element is derived and a simple stress problem is considered to illustrate assembling the global stiffness matrix for a component as well as how externally applied tractions are modeled for a component.

1



Discretizing a Linear RegionThe initial objective is the division of a one dimensional region into linear elements and then deriving an equation for each element. The element equation is then generalized so that a continuous piecewise equation can be written along the length of the region. The linear element is used to obtain the approximate solution to the governing differential equation for the problem

02

2

Q

dxydD

The division into elements is straight forward. The elements do not have to be equal in length. For instance consider the cantilever beam subjected to the following temperature profile along the top of the beam



There are rules that guide the placement of nodes. They are

1. Place the nodes closer together in regions where the unknown parameter (in this case temperature throughout the beam) changes rapidly and further apart where the unknown is relatively constant.

2. Place a node wherever there is a stepped change in the differential equation coefficients D and Q. Consider the beam deflection problem where

0MQEID

When there is an abrupt change in the cross sectional area a node is required. In addition nodes are required at the point of application of loads. Consider the uniaxially loaded bar subjected to a uniform temperature all around the surface.

3. Finally place the nodes where you need the value of something.



Discretization

The concepts associated with finite element analysis are strongly tied to the use of meshes (information networks really). The Civil Engineering student has been exposed to information and mesh networks early on in their studies when surveying principles are introduced. The Civil Engineer understands how contour (terrain) plots are generated from grid (raster) field data or from a triangulated irregular network (TIN) of elevation data. Examples of both are depicted on the next overhead.



TINS are comprised of irregularly distributed nodes and lines with three-dimensional coordinates (x, y, and z) that are arranged in a network of non-overlapping triangles. An advantage of using a TIN model over a raster model is that the points (nodes of the triangles) in a TIN are distributed in an irregular fashion in order to efficiently capture an accurate representation of the terrain.



Contrast the terrain model TIN in the previous figure with the finite element mesh for an automobile bumper.

Both models pass information to the rest of the network through nodes at the vertex of each triangle. Therefore information is exact at the nodes and the nodal information is interpolated through each triangle. Interpolated values will have an error associated with it. For a TIN the nodal information is elevation. In finite element analysis we will require that the strong formulation for an engineering mechanics be satisfied at each node.



NotationTo introduce the basic equations necessary for the analysis of constant strain elements consider the following thin plate with an applied tensile traction. This plate is subsequently discretized using triangular elements:

The coordinate axes in the figures above are global coordinate axes.

The discretized plate has been divided into triangular elements with nodes labeled generically as i, j and m. The nodes for each triangular element are ordered in a counterclockwise fashion. Each node has two degrees of freedom – an x and y displacement identified as uiand vi at i’th node, respectively.

8



The nodal displacements are expressed in a matrix format as

Displacements within the component are functionally dependent upon position, and this is represented by the vector

For a constant strain element u and v must be linear as a function of position.

m

m

j

j

i

i

vu

v

uvu

d

yxvyxu

,,

9



We postulate that displacements vary linearly through the elements, i.e.,

These displacement functions are expressed in matrix notation as

yaxaayxv

yaxaayxu

654

321

,,

6

5

4

3

2

1

654

321

10000001

,,

aaaaaa

yxyx

yaxaayaxaa

yxvyxu

10



To obtain the coefficients we substitute the coordinates of the nodes into the previous equations and this yields the following six equations in six unknown coefficients

We can solve for the first three coefficients from the system of equations expressed as

mmm

jjj

iii

mmm

jjj

iii

yaxaav

yaxaavyaxaavyaxaau

yaxaauyaxaau

654

654

654

321

321

321

3

2

1

1

11

aaa

yx

yxyx

u

uu

mm

jj

ii

m

j

i

11



Inverting the last expression leads to

The method of cofactors is used to invert the 3 x 3 matrix, i.e.,

where 2A is a scalar quantity defined by the following determinant

m

j

i

mm

jj

ii

u

uu

yx

yxyx

aaa

1

3

2

1

1

11

mji

mji

mji

mm

jj

ii

Ayx

yxyx

21

1

11

1

mm

jj

ii

yx

yxyx

A

1

11

2

12



This determinant is

Note that A is the area of the triangular element (civil engineering students should recall area by coordinates from Surveying). With the following elements of the adjoint matrix

then

jimimjmji yyxyyxyyxA 2

ijmmijjmi

jimimjmji

jijimmimijmjmji

xxxxxx

yyyyyy

xyyxyxxyxyyx

m

j

i

mji

mji

mji

u

uu

Aaaa

21

3

2

1

m

j

i

mji

mji

mji

v

vv

Aaaa

21

5

4

3

13



Utilizing the last two matrix expressions we formulate a linear displacement function for the displacement u(x,y) as follows

mmjjii

mmjjii

mmjjii

m

j

i

mji

mji

mji

uuu

uuu

uuu

Ayx

u

uu

Ayx

aaa

yx

yaxaayxu

211

211

1

,

3

2

1

321

14



This leads to

If we identify the following as shape functions for the constant strain element

then

mmmmjjjj

iiii

uyxA

uyxA

uyxA

yxu

21

21

21,

yxA

N

yxA

N

yxA

N

mmmm

jjjj

iiii

21

21

21

mmjjii uNuNuNyxu ,15



Similarly

mmjjii

mmjjii

mmjjii

m

j

i

mji

mji

mji

vvv

vvv

vvv

Ayx

v

vv

Ayx

aaa

yx

yaxaayxv

211

211

1

,

6

5

4

654

16



and

Once again

These are the same shape functions developed for u(x,y). Using the shape function notation leads to

mmmmjjjj

iiii

vyxA

vyxA

vyxA

yxv

21

21

21,

yxA

N

yxA

N

yxA

N

mmmm

jjjj

iiii

21

21

21

mmjjii vNvNvNyxv , 17



m

m

j

j

i

i

mji

mji

mmjjii

mmjjii

vu

v

uvu

NNN

NNN

vNvNvN

uNuNuN

yxvyxu

000

000

,,

If we identify i=1, j=2 and m=3, then the following graphical representation of the shape functions can be depicted for a constraint strain (CST) element:

Now the displacements within the element can be expressed in matrix notation as

18



or

The displacements within an element have now been expressed as functions of the nodal displacements relative to the global coordinate system.

Finally, take note that the coefficients a1 through a6 are implicitly functions of the nodal displacements since

dN

yxvyxu

,,

m

m

j

j

i

i

mji

mji

vu

v

uvu

NNN

NNN

aaaaaa

yxyx

vu

000

000

10000001

6

5

4

3

2

1

19



The inverse of a non-square matrix does not exist and the components of the matrix

are not functionally dependent on the nodal displacements. So the expression on the previous page indicates that the coefficients a1 through a6 are dependent on the nodal displacements. That functional dependence was delineated earlier through the following two matrix expressions:

Thus the unknown coefficients are dependent upon the nodal displacements. However, we don’t really need to find the unknown coefficients to define displacements within a constant strain triangular element. We only have to define the shape functions Ni through Nm.

yxyx

10000001

m

j

i

mji

mji

mji

u

uu

Aaaa

21

3

2

1

m

j

i

mji

mji

mji

v

vv

Aaaa

21

5

4

3

20



Strain Displacement Relationships The element stiffness matrix will be formulated via the potential energy function. We need to define strain in order to formulate the potential energy function for this element. The strains associated with a two dimensional element are

With

xv

yu

yvxu

xy

y

x

mm

jj

ii

mmjjii

ux

Nux

Nu

xN

uNuNuNxx

u

21



this leads to

and

A

yxAxx

N

Ayx

AxxN

Ayx

AxxN

mmmm

m

jjjj

j

iiii

i

221

221

221

mmjjii

mm

jj

ii

uuuA

uA

uA

uAx

u

21

222

22



this leads to

and

A

yxAxx

N

Ayx

AxxN

Ayx

AxxN

mmmm

m

jjjj

j

iiii

i

221

221

221

mmjjii

mm

jj

ii

uuuA

uA

uA

uAx

u

21

222

23



Similar derivations lead to

In a matrix format the strain in a two dimensional element takes the following form

mmjjiimmjjii

mmjjii

vvvuuuAx

vyu

vvvAy

v

21

21

m

m

j

j

i

i

mmjjii

mji

mji

xy

y

x

vu

v

uvu

A

000

000

21

24



In more compact notation

where

dB

d

dd

BBB

m

j

i

mji

xy

y

x

mm

m

m

m

jj

j

j

j

ii

i

i

i AB

AB

AB

0

0

210

0

210

0

21

m

mm

j

jj

i

ii v

ud

v

ud

vu

d25



Stress Strain Relationships

The constitutive relationship for plane stress/plane strain elements is given by

where

for plane stress and plane strain, respectively.

xy

y

x

xy

x

x

D

2100

01

01

1 2

ED

26

22100

01

01

211

ED



Element Stiffness Matrix via Potential Energy RelationshipAt this point a force displacement matrix equation has not been specified for this two dimensional element. This will be accomplished through the use by minimizing potential energy. With

where the strain energy is given by

the potential energy of body forces is given by

where {X} is a 2 x 1 matrix that represents body forces per unit volume.

spb

mmjjii

U

vuvuvu

,,,,,

V

T

V

T

dVD

dVU

2121

V

T

b dVXyxvyxu

,,

27



The potential energy of concentrated loads applied at the nodes is

The potential energy due to distributed load, or surface tractions is given by

where {T} represents surface tractions (with units of force per unit length). We can now express the potential energy of the element as

Pd Tp

S

s dSTyxvyxu

,,

PddSTNd

dVXNddVdBDBd

T

S

TT

V

TT

V

TTT

21

28



Since the nodal displacements are independent of the x and y coordinates, then

From the last three terms we can express the loads on the element in the following manner

such that

PddSTNd

dVXNdddVBDBd

T

S

TT

V

TT

V

TTT

21

S

T

V

T dSTNdVXNPf

fdddVBDBd T

V

TTT

2

1

29



Taking the partial derivative of the potential energy with respect to the nodal displacements leads to

Setting this expression equal to zero leads to

This is the relevant force displacement relationship where the element stiffness matrix is

fddVBDB

fdddVBDBddd

V

TT

T

V

TTT

21

dkddVBDBf

fddVBDB

V

TT

V

TT

0

V

TT dVBDBk30



For an element with constant thickness the element stiffness matrix can be recast as

It is now readily apparent that the element stiffness matrix is a function of the nodal coordinates and material properties. We can expand the matrix above such that

ABDBt

dydxBDBt

dABDBtk

TT

TT

A

TT

mmmjmi

jmjjji

imijii

TT

kkk

kkk

kkk

AtBDBk

31



where

AtBDBk

kAtBDBk

AtBDBk

kAtBDBk

kAtBDBk

AtBDBk

mTT

mmm

mjmTT

jjm

jTT

jjj

mimTT

iim

jijTT

iij

iTT

iii

To obtain the global component stiffness matrix use the direct stiffness method, i.e.,

In the formulation of the element stiffness matrix keep in mind we used the global, or component coordinate axes. Therefore no transformation from local to global coordinates is necessary.

N

e

ekK1

32



In class example

33



Accounting for Body Forces and Surface TractionsThe concept of using shape functions to compute equivalent nodal forces from distributed surface tractions was introduced in the notes for axial force elements. The concept is reinforce here for two dimensional elements. If shape functions are used to interpolate/distribute information about nodal displacements through a two dimensional element it seems eminently reasonable to use the concept in reverse. If we have forces distributed along or over an element we should be able to utilize shape functions to obtain equivalent forces at the element nodes. For body forces distributed through an element the following integral expression

is employed where

Note that typically Xb and Yb are the weight densities ( force/length3) in the x and ydirections, respectively. These forces may arise from gravitational forces, angular velocity, or electromagnetic forces.

dVXNfV

Tb

b

b

YX

X

34



In the first expression on the previous page the vector of shape functions, {N}, are not constant. These functions vary with respect to x and y. Without lack of generality, the integration of those shape functions in order to obtain nodal forces from the distributed bodyforces is simplified if the origin of the coordinates is taken to coincide with the centroid of the element. This foreshadows the future discussion on isoparametric element formulation. Consider the following constant strain element

where the origin of the coordinate system is located at the centroid of the element. With this coordinate system

thus

00 dAydAx

00 dAydAx ii 35



With the geometry of the element from the previous page

In a similar fashion

32

231

31

031

32

21

A

hbAhb

AAhb

hhb

xyyx mjmji

32

32 AxyyxAyxxy jijimmimij

36



Through integration, the body force at node i is (show for homework)

Similar considerations at nodes j and m lead to

Body forces are distributed to the nodes in three equal parts. For the case of gravity loads we have only Yb (Xb = 0).

3tA

YX

ff

fb

b

biy

bixb

3tA

YXYXYX

ff

f

f

ff

f

b

b

b

b

b

b

bmy

bmx

bjy

bjx

biy

bix

b

37



For surface tractions recall that

Use of this equation will be demonstrated by considering a component under plane stress conditions with a uniform stress acting between nodes #1 and #3 on element #1 in the figure below:

Just like the distributed body forces where local element coordinates were a local set of element coordinates are used to assign surface tractions to nodes. The surface traction vector for this applied load is

0p

TT

y

x

S

Ts dSTNf

38



With the transpose of the vector of shape functions given as

then

3

3

2

2

1

1

00

00

00

NN

NN

NN

N T

t L

smy

smx

sjy

sjx

siy

six

s dzdyp

NN

NN

NN

ffffff

f0 0

3

3

2

2

1

1

0

00

00

00

(Evaluated at x coordinate of nodes #1 and #3)39



For plane stress

Noting the location of the coordinate axes in the previous figure and the figure below then the shape function for node #1 is

L

smy

smx

sjy

sjx

siy

six

s dyp

NN

NN

NN

t

ffffff

f0

3

3

2

2

1

1

0

00

00

00

yxA

N 1111 21

(Evaluated at x coordinate of nodes #1 and #3)

40



With

and using the coordinate geometry of the element nodes depicted in the previous figure, then

Similarly

However

mjmji xyyx

0

00032321

axyyx

01

aa

xx

0231

41



Thus

similarly

Aya

yaxA

yxA

N

2

0021

21

1111

AayLxN

AxaLN

2

2

3

2

42



Substituting for the shape functions yields (again, show for homework)

The surface traction from a uniformly distributed load on the edge of a constant strain triangular element can be treated as two concentrated loads acting at the nodes associated with the loaded edge. The concentrated loads are statically equivalent to the uniform surface traction between the nodes.

02

0002

02

000

2

220

00

00

00

22

2

0

3

3

2

2

1

1

pLt

pLt

pLL

pL

aLtady

p

NN

NN

NN

t

ffffff

fL

smy

smx

sjy

sjx

siy

six

s

(Evaluated at x coordinate of nodes #1 and #3)

43



In class example

44



Static Condensation

The concept of static condensation is now introduced in order to produce a quadrilateral element with one internal node. Consider the quadrilateral element shown below comprised of four triangular elements. The element has four external nodes (#1 through #4) and an internal node (#5):

The stiffnesses from the four triangular elements are superimposed to create the an element stiffness matrix for the quadrilateral element. The degrees of freedom associated with node #5 are condensed out in the following manner.

45



The force – displacement equation for the four triangular elements is partitioned as follows:

Here “i” represents “imaginary” and “a” represents “actual.” Thus di and Fi represent the displacements and the forces at “imaginary node #5. Expanding the matrix format above leads to:

Solving the latter expression for di leads to:

i

a

i

a

dd

kk

kk

FF

dKF

|

|

iai

iaa

dkdkFdkdkF

2221

1211

iai Fkdkkd 12221

122

46



Substitution of this back into the first expression leads to:

or

where

and are referred to as the “condensed load vector” and the “condensed stiffness matrix” respectively.

iaa

iaa

Fkkdkkkdk

dkdkF1

2212211

221211

1211

acc

aaia

dkFdkkkdkFkkF

211

2212111

2212

21

1221211

12212

kkkkk

FkkFF

c

iac

47



If we identify the stiffness matrix of triangular element #1 in the previous figure as:

then the stiffness matrix for the quadrilateral element before condensation is as follows:

)1(

55)1(

52)1(

51

)1(25

)1(22

)1(21

)1(15

)1(12

)1(11

)1(

kkk

kkk

kkk

k

)4(55

)3(55

)2(55

)1(55

)4(54

)3(54

)3(53

)2(53

)2(52

)1(52

)4(51

)1(51

)4(45

)3(45

)4(44

)3(44

)3(43

)4(41

)3(35

)2(35

)3(34

)3(33

)2(33

)2(32

)2(25

)1(25

)2(23

)2(22

)1(22

)1(21

)4(15

)1(15

)4(14

)1(12

)4(11

)1(11

0

0

0

0

kkkkkkkkkkkk

kkkkkk

kkkkkk

kkkkkk

kkkkkk

k

48



)4(44

)3(44

)3(43

)4(41

)3(34

)3(33

)2(33

)2(32

)2(23

)2(22

)1(22

)1(21

)4(14

)1(12

)4(11

)1(11

0

0

0

0

kkkk

kkkk

kkkk

kkkk

kc

In the process of condensing the stiffness matrix, we eliminate the fifth row and fifth column of subelement matrices. These subelement matrices are 2 x 2 matrices. Essentially we eliminate the rows and columns associated with the u and v displacements of node #5. After condensation we have:

Before condensation the size of the stiffness matrix is (10x10) – recall that each node has two degrees of freedom. After condensation the stiffness matrix is (8x8).

49



Equilibrium &Compatibility of Strains and Displacements

An approximate solution for the stress analysis of a component using finite element analysis does not produce a “perfect” distribution of stresses that an exact solution from elasticity theory would produce. One should also be cognizant that there are a limited number of exact elasticity solutions available.

The finite element method will yield approximate, but reasonable solutions. The following list describes issues associated with the fidelity of a solution obtained from a finite element analysis.

1. Equilibrium of nodal forces is satisfied. The reason for this is that the solution for displacements from the global force displacement equation

imposes equilibrium at each node. As we saw earlier structural reactions, i.e., free body diagram reaction forces, are explicitly included in the global force displacement equation above. So boundary conditions are incorporated

dKF

50



2. Equilibrium within an element is not guaranteed except for the constant strain element (CST) triangular element. The linear strain element (studied next) and the axisymmetric elements usually approximate element equilibrium.

3. Equilibrium is not satisfied between or across elements. Consider the two element problem in the last example. If we look at a differential element situated along the common boundary of the two elements to the right we see that interelement equilibrium is not satisfied. The coarseness of the mesh causes this as well. The stresses are not zero at the boundary of the element, i.e., along the top and bottom of the component.

114.2

3011005

psipsipsi

xy

y

x

224.22.1

995

psipsi

psi

xy

y

x

51



3. Compatibility is satisfied within an element. Constant strain elements never “tear apart.” The compatibility equations are conditions placed on strain fields that guarantee displacement solutions obtained are single valued displacement fields. For a two dimensional plane strain/stress problem the equation of compatibility is

For three dimensional problems there are six compatibility equations.

3. In the formulation of element equations compatibility is invoked at the nodes. Multiple elements framing into a node are required to stay connected at the shared nodes.

4. In general, compatibility may or may not be satisfied between elements along boundaries. Most times this is a desirable trait and for constant strain elements neighboring elements will have boundaries that are compatible. Incompatible elements produce gaps or overlaps. Contact stress elements are examples where two elements will overlap one another in an acceptable manner. Special fracture mechanics elements are another example of incompatible elements where gaps are allowed to form between and across nodes. 52

02 21

222

21

122

22

112

xxxx



Compatibility in MeshingIn a very general sense compatibility means that things fit together when deformed.

With finite elements this means displacements agree between neighboring elements, not only at the nodes, but all along common edges in 2D or all over common surfaces in 3D.

This node is not connected to the top element, since the top element does not have a mid-side node.

Hence the node common to the lower elements can move up and down independently of the element above. Typically this is not a compatible mesh.

But it is a possible model of a crack in a component. 53



Transition MeshesTo change mesh refinement between two grid densities, consider the following simplistic mesh refinements.

54



55

4,400 elementsVery thin elements along the inner

and outer surfaces

Contact Surface



56



Element Distortion and the Patch Test

Distorted elements are less accurate, especially as the distance between two sides that parallel diminishes. In automatic meshing algorithms a tolerance is set on element distortion.

A test of how well elements cope with distortion is the patch test. A group of distorted elements (a patch of elements) are loaded in a way that should produce uniform stress, and the actual stress in the elements is examined to see if it is uniform, e.g., consider

ddIf one applies a uniform displacement to the nodes on the top boundary in either the x or y direction, then the internal nodes should have the displacements proportional to the geometry of the element.

57



Spatial Interpretation of Element Stresses

In displacement based finite element analysis nodal displacements are the primary quantities determined in an analysis. Stress and strain are derived quantities that are based on the displacement results. For a bar element or a constant strain element stresses are constant over the entire element. For graphical purposes it is usually convenient to assign the stress value of the element to its centroid.

As has been seen in several example problems the stress state of a component is not predicted as accurately as the displacements. Consider the following cantilever beam:

58



If constant strain elements are used, then constant stress is produced in each element. The figure below depicts the solution from a finite element analysis utilizing four elements from top to bottom. These elements are located at the fixed end of the beam, and the entire beam is modeled with 85 elements.

For higher order elements, such as the linear strain element, stress varies as a function of position through the element. The common practice is to evaluate the stress at the centroid an use this value to represent stress in the element. We will see for isoparametric elements, stress values are exact at the Gauss points and those values will be used.

x stresses are constant throughout the element and are plotted at the mid height of the vertical side of the element.

column of CST elements

59



Sources of Error in the FEM• The three main sources of error in a typical FEM solution are discretization errors,

formulation errors and numerical errors.• Discretization error results from transforming the physical system (continuum) into a

finite element model, and can be related to modeling the boundary shape, the boundary conditions, etc.

Discretization error due to poor geometryrepresentation.

Discretization error effectively eliminated.60



• Formulation error results from the use of elements that don't precisely describe the behavior of the physical problem.

• Elements which are used to model physical problems for which they are not suited are sometimes referred to as ill-conditioned or mathematically unsuitable elements.

• For example a particular finite element might be formulated on the assumption that displacements vary in a linear manner over the domain. Such an element will produce no formulation error when it is used to model a linearly varying physical problem (linear varying displacement field in this example), but would create a significant formulation error if it used to represent a quadratic or cubic varying displacement field.

61



• Numerical error occurs as a result of numerical calculation procedures, and includes truncation errors and round off errors.

• Numerical error is therefore a problem mainly concerning the FEM vendors and developers.

• The user can also contribute to the numerical accuracy, for example, by specifying a physical quantity, say Young’s modulus, E, to an inadequate number of decimal places.

62



Basic Scorecard – CST Element

yxA

N

yxA

N

yxA

N

mmmm

jjjj

iiii

21

21

21

ijmmijjmi

jimimjmji

jijimmimijmjmji

xxxxxx

yyyyyy

xyyxyxxyxyyx


63



Basic Scorecard – CST Element

yxA

N

yxA

N

yxA

N

mmmm

jjjj

iiii

21

21

21

ijmmijjmi

jimimjmji

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64



A

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mmmm

m

jjjj

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221

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AB

000

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yxAyy

N

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m

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221

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65



FKddKF 1

N

e

ekK1

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TT

kkk

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m

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j

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66



xy

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2100

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1 2

ED

22100

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211

ED

67

section 7 constant strain triangular elements.pptacademic.csuohio.edu/duffy_s/cve_512_7.pdf · ·...

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