section 8 linear strain triangular elements.ppt · 2014. 11. 4. · section 8: linear strain...

Section 8: LINEAR STRAIN TRIANGULAR ELEMENTS

Washkewicz College of Engineering

Introduction

We now consider the development of higher order elements, specifically the linear strain triangular element (LST). As its name implies strain varies linearly through the element, hence displacements must vary quadratically.

The LST element has 6 nodes and twelve displacements, i.e., 12 degrees of freedom.

The procedure for developing the equations associated with the LST elements follows the same path as the CST element.

1



The nodal displacements are expressed in a matrix format as

This element has 3 corner nodes and 3 mid-side nodes. Compatibility at the nodes and along the sides of adjoining elements are insured. Along the sides 3 points define the line of a parabola, and displacements will vary in a parabolic manner with this element.

6

6

5

5

4

4

3

3

2

2

1

1

vuvuvuvuvuvu

d

Notation

2



We formulate quadratic displacement functions as follows:

The general displacement function can be expressed in matrix notation as

2

12112

10987

265

24321

,

,

yaxyaxayaxaayxv

yaxyaxayaxaayxu

aMaaaaaaaaaaaa

yxyxyxyxyxyx

yxvyxu

*

10000000000001

,,

12

11

10

9

8

7

6

5

4

3

2

1

22

22

3



When considering triangular elements, we typically use complete polynomials in Cartesian coordinates to describe the displacement field within an element (see table to the right). Polynomials are used because they are easily differentiated.

The types of polynomials can be grouped as follows:

a) Lagrange polynomials - complete polynomial expansions

b) Serendipity polynomials - incomplete polynomial expansions

c) Hermitian polynomials - polynomials including derivatives4



To obtain the coefficients we substitute the coordinates of the nodes into the previous equations which yields

We set up the same system of equations for the CST elements to find the coefficients.

12

11

7

6

2

1

2666

2666

2555

2555

2111

2111

2666

2666

2222

2222

2111

2111

6

5

1

6

2

1

10000001000000

10000000000001

00000010000001

aa

aa

aa

yyxxyxyyxxyx

yyxxyxyyxxyx

yyxxyxyyxxyx

vv

vu

uu

5



Inverting the last expression leads to

or

Note that only the 6x6 portion of the [] matrix must be inverted for the LST element. The coefficients a1 through a6 and the coefficients a7 through a12 are different then the coefficients for the CST element.

6

5

1

6

2

11

2666

2666

2555

2555

2111

2111

2666

2666

2222

2222

2111

2111

12

11

7

6

2

1

10000001000000

10000000000001

00000010000001

vv

vu

uu

yyxxyxyyxxyx

yyxxyxyyxxyx

yyxxyxyyxxyx

aa

aa

aa

da 1

6



The matrix [] has the following block format

This matrix can be inverted block wise using the following analytic formula from matrix algebra

with

GCFA

000000000000000000000000000000000000

CF

11111

111111111

FACGACFACGFACGFAACFACGFAA

7



then

with

1

1

11111

111111111

00

000000000000

GA

AGAAGAGAAAGAA

2666

2666

2555

2555

2444

2444

2333

2333

2222

2222

2111

2111

111111

yyxxyxyyxxyxyyxxyxyyxxyxyyxxyxyyxxyx

A

8



Using the notation

and utilizing MatLab’s symbolic algebra tool box, the elements of the matrix above appear in a pdf that can be found with the class notes on the web site. Element A11 is seven pages long. The entire file containing all elements is 247 pages long.

A similar derivation could be obtained for the inverse of block matrix [G]. There is no need to go through the details. The point is that coding the inverse in general terms is not worthwhile. This provides the motivation for developing the solution matrices in a coordinate system other than the global coordinate system. This is accomplished through the use of a coordinate system local to the element with the additional wrinkle that computations are conducted in a transformed coordinate space (isoparametric elements).

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

1

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

A

9



We will proceed on as If the vector of coefficients {a} were easily obtained knowing that it is not. Back substitution into the matrix expression below

yields the displacement field throughout the LST element in terms of the global coordinate system.

aMaa

aaa

yxyxyxyxyxyx

yxvyxu

*

10000000000001

,,

12

11

3

2

1

22

22

10



Finally, the general displacement expression can be obtained in terms of shape functions as follows:

thus

Here two of the shape functions (N2 and N5) are plotted. We will develop specific formulations in an example problem.

dNdM

dNaMvu

1*

*

1* MN

11



Strain Displacement Relationships The strains associated with a two dimensional element are

Rewriting in a more compact form leads to:

With

then

12

11

3

2

1

02010201002010000000000000002010

aa

aaa

yxyxyx

yx

xv

yu

yvxu

xy

y

x

aM *

da 1

dB

dM 1*'

12



or in full notation

1

2666

2666

2555

2555

2111

2111

2666

2666

2222

2222

2111

2111

1

1000000

1000000

1000000

0000001

0000001

0000001

02010201002010000000000000002010

*'

yyxxyx

yyxxyx

yyxxyx

yyxxyx

yyxxyx

yyxxyx

yxyxyx

yxMB

13



In the next example problem we will derive formulations for the various shape functions and we will see that

Given the coordinate system designated in the problem.

665544332211

654321

654321

000000000000

B

14



In class example

15



Stress Strain Relationships

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

where

for plane stress and plane strain, respectively.

x x

y y

xy xy

D

2100

01

01

1 2

ED

22100

01

01

211

ED

16



In class example

17



Element Stiffness Matrix

The stiffness matrix is determined once again with the following expression:

However, the [B] matrix is functionally dependent on x and y for the LST element. Thus the expression above must be integrated. The [B] matrix will have the form

where the ’s and ’s are now functions of x and y. The stiffness matrix is very awkward to obtain in explicit form, however, if the origin of the reference axes is located at the centroid of the element the integration becomes more amenable.

V

TT dVBDBk

665544332211

654321

654321

000000000000

B

18



Accounting for Body Forces and Surface Tractions

Body forces at the nodes are defined through the expression

where

and Xb and Yb are the weight densities in the x and y directions, respectively. These forces may arise from gravitational forces, angular velocity, or electromagnetic forces.

For surface tractions recall that

V

Tb dVXNf

b

b

YX

X

S

Ts dSTNf

19



In class example

20



Comparison of CST and LST Elements

For a given number of nodes a better representation of true stress and displacement are given by linear strain elements relative to constant strain elements. Consider the figure below that depicts one LST element and four CST elements for the same number of nodes and the same element area:

Using one LST element (the element on the right) yields better stress results than using four CST elements with the same number of nodes (thus the same number of degrees of freedom). The elements yield the same results for the constant stress problem. Assume a linear stress distribution and draw plot lines through the CST elements. One will see bar charts for a stress plot along a line. 21



Consider a cantilever beam with E = 30,000 ksi, = 0.25, and a thickness of one inch.

The following tables lists relevant information regarding a mesh density study for this beam:

The mesh depicted above is 4 x 16.

22



The following table lists results from the mesh densities defined on the previous overhead:

The table compares the free end deflections and the normal stress at the centroid of the element occupying the upper left hand corner of the mesh. Several observations:

• The larger the number of degrees of freedom for a given element type the closer the mesh converges on the true value, either deflection or stress.

• For a given number of nodes the LST mesh provides better values then a CST mesh.

• In most commercial codes LST and CST elements are available, but they are used as transition elements (see the axisymmetric mesh for the Abrams tank barrel in the next section of notes). The isoparametric quadrilateral (four sided) elements is used most frequently in commercial codes to model plane strain and plane stress models. 23

section 8 linear strain triangular elements.ppt · 2014. 11. 4. · section 8: linear strain...

Documents