section 12 volume elements.ppt - cleveland state university

Section 12: VOLUME ELEMENTS

Washkewicz College of Engineering

Tetrahedral Element

We now consider volume elements by first considering the volume element counterpart of a triangular element, i.e., the tetrahedral element. One is shown in the figure below corner nodes numbered 1 through 4. The nodes are numbered in such a way that the first three nodes are numbered in a counterclockwise fashion when viewed from the last node.



The nodal displacements are expressed in a matrix format as

4

4

4

3

3

3

2

2

2

1

1

1

vvuwvuwvuwvu

d

zayaxaazyxw

zayaxaazyxvzayaxaazyxu

1211109

8765

4321

,,,,,,

We formulate linear displacement functions as follows:



The general displacement function can be expressed in matrix notation as

12

11

10

8

78

6

6

5

4

3

2

1

1211109

8765

4321

100000000000010000000000001

aaaaaaaaaaaa

zyxzyx

zyx

zayaxaazayaxaazayaxaa

wvu



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

41241141094

31231131093

21221121092

11211111091

48474654

38373653

28272652

18171651

44434214

34333213

24232212

14131211

zayaxaawzayaxaawzayaxaaw

zayaxaawzayaxaavzayaxaavzayaxaav

zayaxaavzayaxaauzayaxaauzayaxaau

zayaxaau



We can solve for the first four coefficients from the system of equations experssed as

Inverting this expression leads to

4

3

2

1

444

333

222

111

4

3

2

1

1111

aaaa

zyxzyxzyxzyx

uuuu

4

3

2

11

444

333

222

111

4

3

2

1

1111

uuuu

zyxzyxzyxzyx

aaaa



The method of cofactors is used to invert the 4 x 4 matrix, i.e.,

where

and V represents the volume of the tetrahedron, i.e., one third the area of the base times the height.

4321

4321

4321

43211

444

333

222

111

61

1111

Vzyxzyxzyxzyx

444

333

222

111

1111

6

zzzyyxzyxzyx

V



The coefficients are the following determinants

444

333

222

1

zyxzyxzyx

44

33

22

1

111

zyzyzy

44

33

22

1

111

zxzxzx

44

33

22

1

111

yxyxyx

444

333

111

2

zyxzyxzyx

44

33

11

2

111

zyzyzy

44

33

11

2

111

zxzxzx

44

33

11

2

111

yxyxyx



Similarly

444

222

111

3

zyxzyxzyx

44

22

11

3

111

zyzyzy

44

22

11

3

111

zxzxzx

44

22

11

3

111

yxyxyx

333

222

111

4

zyxzyxzyx

33

22

11

4

111

zyzyzy

33

22

11

4

111

zxzxzx

33

22

11

4

111

yxyxyx



This leads to

44444

33333

22222

11111

61

61

61

61,,

uzyxV

uzyxV

uzyxV

uzyxV

zyxu



44444

33333

22222

11111

61

61

61

61,,

vzyxV

vzyxV

vzyxV

vzyxV

zyxv

Simarly



44444

33333

22222

11111

61

61

61

61,,

wzyxV

wzyxV

wzyxV

wzyxV

zyxw

and

where

zyxV

N

zyxV

N

zyxV

N

zyxV

N

44444

33333

22222

11111

61

61

61

61



This leads to

44332211

44332211

44332211

,,

,,

,,

wNwNwNwNzyxw

vNvNvNvNzyxv

uNuNuNuNzyxu



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

zu

xw

yw

zv

xv

yu

zwyvxu

zx

yz

xy

z

y

x



with

44

33

22

11

44332211

ux

Nux

Nux

Nux

N

uNuNuNuNxx

u

and

V

zyxVxx

N

Vzyx

VxxN

Vzyx

VxxN

Vzyx

VxxN

661

661

661

661

44444

4

33333

3

22222

2

11111

1



then

44332211

44

33

22

11

61

6666

uuuuV

uV

uV

uV

uVx

u

Similar derivations lead to

)

(61

)

(61

61

61

44332211

44332211

44332211

44332211

44332211

44332211

wwww

vvvvVy

wzv

vvvv

uuuuVx

vyu

wwwwVz

w

vvvvVy

v



and finally

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

)

(61

44332211

44332211

uuuu

wwwwVz

uxw

4

4

4

3

3

3

2

2

2

1

1

1

44332211

44332211

44332211

4321

4321

4321

00000000

000000000000

0000000000000000

61

wvuwvuwvuwvu

V

zx

yz

xy

z

y

x

dB



Stress Strain Relationships

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

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D

where no distinction is made between plane strain and plane stress:

22100000

02210000

00221000

0001

0001

0001

211

ED



Element Stiffness Matrix – Tetrahedral Element

The stiffness matrix is determined with the following expression:

However, the [B] matrix is not functionally dependent on spatial coordinates. The [B]matrix will have the form

The element stiffness matrix is now a 12x12 matrix.

VBDBk TT

44332211

44332211

44332211

4321

4321

4321

00000000

000000000000

0000000000000000

61

VB



Accounting for Body Forces and Surface Tractions

Body forces at the nodes are defined through the expression

where

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

dVXNfV

Tb

b

b

b

ZYX

X



then for constant body forces the nodal components of the body forces are distributed to the nodes as follows:

b

b

b

b

b

b

b

b

b

b

b

b

bz

by

by

bz

by

bx

bz

by

bx

bz

by

bx

b

ZYXZYXZYXZYX

f

f

ff

fff

fff

ff

f41

3

3

3

3

3

3

2

2

2

1

1

1

3

3

3

2

2

2

1

1

1

000000

000000

000000

NN

NN

NN

NN

N

N T

With



For surface tractions recall that

where {Ns} is the shape function matrix evaluated at the surface where the tractions are physically applied. Consider the case where a uniform pressure p is applied to face 123 of a tetrahedral element. The expression above takes the form

S

Tss dSTNf

S

z

y

x

surfaceonevaluatedT

ss dS

p

pp

Nf 123|



then

where S123 is the area of the tetrahedral where the uniform surface p traction is applied.

000

3123

4

4

4

3

3

3

2

2

2

1

1

1

z

y

x

z

y

x

z

y

x

sz

sy

sx

sz

sy

sx

sz

sy

sx

sz

sy

sx

s

p

ppp

ppp

pp

S

f

fff

fff

fff

ff

f



Quadratic Tetrahedral Element

The quadratic tetrahedral element has 10 nodes with three degrees of freedom at each node. This yields a total of 30 degrees of freeedom (dof). The element is depicted below:

The displacement field equations (u, v, w) will contain quadratic terms in the expressions and thus strains (which are a derivative of displacements) will vary linearly through the element.



A complete quadratic polynomial in terms of spatial coordinates x, y, and z with no additional cubic or higher-order terms has exactly 10 terms. Thus, the ten-node tetrahedral element employs complete quadratic function to interpolate the displacement field across the element from the nodal values. This means that the strain tensor for the element is linear (i.e., each of the various strain components can vary as a linear polynomial over the element volume), and the element performs much better under bending types of loads. It also turns out that the element can now take on a quadratic shape. Thus, for example, the edges of the element may be curved to fit a portion of a circular arc.

When a finite element model with 10 node tetrahedral elements are compared to a model with 4 node tetrahedral elements solutions change significantly. Increasing the order of the displacement field functions across the element by adding additional nodes without changing the number of elements is called p- refinement. In the literature the variable p is often used to denote the order of the polynomial function describing the element displacement field.

Increasing the number of elements without changing the element order by decreasing the element size and re-meshing is called h-refinement. The variable h is often used in the literature to designate the element size. Research has shown that a combination of h- and p-refinement will result in the fastest convergence rate, especially if the refinement is done locally where most needed, such as in a region of high strains and stresses.



Linear Hexahedral Element – 8 Node Brick

While the 10 node tetrahedral element performs pretty well, the downside is the additional degrees of freedom required. Further, it takes a significant number of tetrahedral elements to mesh a volume. Consider a brick-like object. A minimum of 5 tetrahedral elements are required to mesh it.



The formulation for this element is a direct extension of the 4 node quadrilateral element. The 8 node brick can be formulated in the x-y-z Cartesian coordinate system, or the s-t-z isoparametric coordinate system. The notation from Logan’s text book is followed relative to the naming of the isoparametric coordinate axes. Both global and local (isoparametric) configurations are shown below:

If we use the isoparametric natural coordinate system (s, t, z then the procedure for developing shape functions proceeds in a manner similar to the 4 node quadrilateral element.



With

After solving for the ai’s we will then have in a matrix format, the isoparametric transformation, that is:

stzaszaztastazatasaazstzaszaztastazatasaay

stzaszaztastazatasaax

2423222120191817

161514131211109

87654321

Tzyxzyxzyxzyxzyxzyxzyxzyx

NNNNNNNNNNNNNNNN

NNNNNNNN

zyx

888777666555444333222111

87654321

87654321

87654321

000000000000000000000000000000000000000000000000



The individual shape functions can be characterized by a single expression:

where

and the exact value is determined by the position of the node relative to the natural coordinate system. For example, referring back to the original figure for the 8 node brick element, for node #1:

8,,2,18

)1)(1)(1(),,(

izzttssztsN iiii

111

i

i

i

zts

8

)1)(1)(1(8

)11)(11)(11(8

)1)(1)(1(),,( 1111

ztszts

zzttssztsN



The displacement functions in terms of the degrees of freedom use the same isoparametric formulation. Thus

To construct the element stiffness matrix we must have expressions for strains which are theoretically derived in terms of derivatives of the displacements with respect to the x-y-zcoordinate system. If you use the s-t-z coordinate system to find displacements, the displacements are functions of s, t as well as z and not x, y, z. Therefore we need to apply the chain rule for differentiation.

Twvuwvuwvuwvuwvuwvuwvuwvu

NNNNNNNNNNNNNNNN

NNNNNNNN

wvu

888777666555444333222111

87654321

87654321

87654321

000000000000000000000000000000000000000000000000



This means the derivatives of the displacements are

zz

zv

zy

yv

zx

xv

zv

tz

zv

ty

yv

tx

xv

tv

sz

zv

sy

yv

sx

xv

sv

zz

zu

zy

yu

zx

xu

zu

tz

zu

ty

yu

tx

xu

tu

sz

zu

sy

yu

sx

xu

su



as well as

zz

zw

zy

yw

zx

xw

zw

tz

zw

ty

yw

tx

xw

tw

sz

zw

sy

yw

sx

xw

sw

Focusing on

zz

zu

zy

yu

zx

xu

zu

tz

zu

ty

yu

tx

xu

tu

sz

zu

sy

yu

sx

xu

su



we solve this system of equations for

Similarly we solve

for

xu

x

zz

zv

zy

yv

zx

xv

zv

tz

zv

ty

yv

tx

xv

tv

sz

zv

sy

yv

sx

xv

sv

yv

y



Finally to complete the derivations of normal strains we solve the following system of equations:

for

zw

z

zz

zw

zy

yw

zx

xw

zw

tz

zw

ty

yw

tx

xw

tw

sz

zw

sy

yw

sx

xw

sw



Using Cramer’s rule

zz

zy

zz

tz

ty

tx

sz

sy

sx

zz

zy

zu

tz

ty

tu

sz

sy

su

xu

x

zz

zy

zz

tz

ty

tx

sz

sy

sx

zz

zv

zx

tz

tv

tx

sz

sv

sx

yv

y



and

zz

zy

zz

tz

ty

tx

sz

sy

sx

J

where we can identify the determinant of the Jacobian as

zz

zy

zz

tz

ty

tx

sz

sy

sx

zw

zy

zx

tw

ty

tx

sw

sy

sx

zw

z



For the shear strains

zz

zy

zz

tz

ty

tx

sz

sy

sx

zz

zy

zv

tz

ty

tv

sz

sy

sv

zz

zy

zz

tz

ty

tx

sz

sy

sx

zz

zu

zx

tz

tu

tx

sz

su

sx

xv

yu

xy



in addition

zz

zy

zz

tz

ty

tx

sz

sy

sx

zz

zw

zx

tz

tw

tx

sz

sw

sx

zz

zy

zz

tz

ty

tx

sz

sy

sx

zv

zy

zx

tv

ty

tx

sv

sy

sx

yw

zv

yz



and finally

zz

zy

zz

tz

ty

tx

sz

sy

sx

zu

zy

zx

tu

ty

tx

su

sy

sx

zz

zy

zz

tz

ty

tx

sz

sy

sx

zz

zy

zw

tz

ty

tw

sz

sy

sw

zu

xw

zx



These six equations lead to the following three expressions for normal strains:

sy

tu

zz

su

tz

zy

zu

sz

ty

zy

tu

sz

zu

tz

sy

zz

su

ty

Jxu

x1

zz

sv

tx

zv

tz

sx

zx

tv

sz

zv

tx

sz

zx

tz

sv

zz

tv

sx

Jyv

y1

zw

tx

sy

zy

tw

sx

zx

ty

sw

zy

tx

sw

zx

tw

sy

zw

ty

sx

Jzw

z1



The student is required to develop the three expressions for shear strains. In a matrix format

wvu

J

zu

xw

yw

zv

xv

yu

yvyvxu

zx

yz

xy

z

y

x

6361

5352

4241

33

22

11

00

000

0000

1



where

and the student is required to derive the other components for homework (see previous page).

sy

tzz

stz

zy

zsz

ty

zy

tsz

ztz

sy

zz

sty

11

zz

stx

ztz

sx

zx

tsz

ztx

sz

zx

tsv

zz

tsx

22

ztx

sy

zy

tsx

zx

ty

s

zy

tx

szx

tsy

zty

sx

33



Utilizing

then

where

dND

dB

Twvuwvuwvuwvuwvuwvuwvuwvu

NNNNNNNNNNNNNNNN

NNNNNNNN

wvu

888777666555444333222111

87654321

87654321

87654321

000000000000000000000000000000000000000000000000

6361

5352

4241

33

22

11

00

000

0000

1J

D



87654321

87654321

87654321

000000000000000000000000000000000000000000000000

NNNNNNNNNNNNNNNN

NNNNNNNNN

and

again

where

Twvuwvuwvuwvuwvuwvuwvuwvud 888777666555444333222111

dND

NDB



Element Stiffness Matrix – Isoparametric Hexahedral Element

For an 8 node isoparametric hexahedral element, the 24 x 24 element stiffness matrix is given by the volume integration:

Again, the stiffness matrix is evaluated numerically using Gauss quadrature. Here a 2 x 2 x 2 quadrature rule is utilized, i.e.,

where the Gauss points and the weights are given in a table on the next page.

1

1

1

1

1

1

zddtdsJBDBk TT

8

1

,,,,,,i

kjiiiiiiiTT

iii WWWztsJztsBDztsBk



As is true with bilinear quadrilateral element the 8 node linear hexahedral element does not model beam bending well because element sides remain straight during element deformation.

In beam bending these type of elements are stretched and are subject to shear locking. The concepts of shear locking are described by Cook et al (2002, Concepts and Applications of Finite Element Analysis, 4th Edition, Wiley) and ways of remedying this problem.

The quadratic, 20 node, hexahedral element described briefly in the next section does not have shear locking problems.



Quadratic Hexahedra Element – 20 Node Brick

The formulation for this element is a direct extension of the 8 node quadrilateral element. The 20 node is formulated in the s-t-zsoparametric coordinate system. The configuration for this element is shown below:

For this element we have 12 mid-side nodes and 8 corner nodes.



260

259

25857

256

255

254

253

252

251

250

249

248

47464544434241

240

239

23837

236

235

234

233

232

231

230

229

228

27262524232221

220

219

21817

216

215

214

213

212

211

210

29

28

7654321

zstastza

tszazstaszazsazta

ztastatsazatasa

szaztastazatasaazzstastza

tszazstaszazsazta

ztastatsazatasa

szaztastazatasaayzstastza

tszazstaszazsazta

ztastatsazatasa

szaztastazatasaax

Now



Without going through any of the details of deriving the ai’s we simply state that the shape functions at the corner nodes are given by the expression

For the mid-side nodes for si = 0

For the mid-side nodes for ti = 0

For the mid-side nodes for zi = 0

8,,2,128

)1)(1)(1(),,(

izzttsszzttssztsN iiiiii

i

20,19,18,174

)1)(1)(1(),,(2

izzttsztsN iii

16,14,12,104

)1)(1)(1(),,(2

izztssztsN iii

15,13,11,94

)1)(1)(1(),,(2

izttssztsN iii

section 12 volume elements.ppt - cleveland state university

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