High Order 1

7/27/2019 High Order 1

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Higher Order Elements

In this section, we will develop a higher-

order triangular element, called the Linear-Strain Triangular (LST).

This element has many advantages over

the CST because it has 6 nodes and 12

displacements DOF.

The displacement function of the triangleis quadratic.

Steps in the formulation of LST element

stiffness equations

The procedure to derive LST element stiffness and element

equations is identical to that used for the CST element.

x

y

2

3

1

L1=

0

L1= 1L2

= 0

L2= 1

L3= 0 L3= 1

L1= 1/2L2= 1/2

L3= 1/2

4

5

6


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Step 1 : Discretize and Select element types

Consider the triangular element shown below: Each node is defined bynodes numbers. Each node has 2 DOFs (displacements in x-,y- directions)

x

y

1

2

3

u1

v1

v2

u3

v3

u2

u4

v4

6

5

v6

4

u6

u

5

v5

Step 2 : Select Displacement Functions

The variation of the displacements over the element may be expressed as

The displacement compatibility among the adjoining elements is satisfiedbecause the 3 nodes defining adjacent sides define a unique parabola.

54322345

432234

3223

22

1

yxyyxyxyxx

yxyyxyxx

yxyyxx

yxyx

yx

PASCAL TRIANGLE

The CST and LST are variations of the Pascal Triangle.


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Step 3 : Define the Strain-Displacement and Stress-Strain Relationships

Elemental Strains: The strains over a 2D element are:

Observe that the strains are linear!! over the triangular element and that is

why it s called a linear-strain tri angle (LST).

Write

The above equation may be written in matrix form as:


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Step 4: Derive the Element Stiffness Matrix and Equations using the Total

Potential Energy Approach

Step 5,6, and 7: Assembling the global stiffness matrix, determining the

global displacements and calculating the stresses, are identical to the

procedures used for CST elements.

1(0,0)

6(0.5b,0)

2(b,0)

4(0.5b,0.5h)

3(0,h)

5

(0,0.5h)

The triangle has a base dimension ofb and a height h, with mid-side nodes.

We can calculate the coefficients a1 through a6 by evaluating the displacement

u at each node.


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Solving for t he as gives

The u displacement equation is


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Comparison of Elements

Observations

For a given number of nodes, a better representation of true stress and

displacement is obtained using LST elements than is obtained using the same no. ofnodes a finer subdivision of CST elements.

The larger the no. of degrees of freedom for a given type of triangular element, the

closer the solution converges to the exact one.

Although the CST is poor in modeling bending, we observe from Table 2 that the

element can be used to model a beam in bending if sufficient no. of elements is

used through the depth of the beam.

In general, both LST and CST analyses yield sufficient results for most plane

stress/strain problems provided a sufficient number of elements are used (LST

model might be preferred over the CST for plane stress when a relatively small no.

of nodes is used).

Most commercial programs incorporate the use of CST and/or LST elements for

plane stress/strain problems although these elements are used primarily as

transition elements (usually during mesh generation).


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some other elements

4-noded rectangle

x

y

a a12

3 4

b

b

In local coordinate system

abybxaN

ab

ybxaN

ab

ybxaN

ab

ybxaN

4))((

4

))((

4

))((

4

))((

4

3

2

1

+=

=

+=

++=

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3223

22

1

yxyyxyxyxx

yxyyxyxx

yxyyxx

yxyx

yx4 node; p=2

Corner nodes

224

))((

224

))((

224

))((

224

))((

784

763

652

851

NN

ab

ybxaN

NN

ab

ybxaN

NN

ab

ybxaN

NN

ab

ybxaN

+

=

=

+

=

++

=

x

ya a 12

3 4

b

b

5

6

7

8

Midside nodes

+=

=

=

+

=

2

22

82

22

7

2

22

62

22

5

2

)(

2

)(

2

)(

2

)(

b

yb

a

xaN

b

yb

a

xaN

b

yb

a

xaN

b

yb

a

xaN

8-noded rectangle

4 node; p=2

8 node; p=3

54322345

432234

3223

22

1

yxyyxyxyxx

yxyyxyxx

yxyyxx

yxyx

yx

STATIC CONDENSATION

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High Order 1