# ebc - transformation

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• INDIVIDUAL STUDY

EBC Transformation

Submitted by:

PRUTHVI R VENKAYALA

UFID: 9831-1764

• IBFEM formulation for linear elasticity problems:

A plate as shown above is subjected to horizontal force. Boundary conditions are shown accordingly in the

above figure. Ux is the displacement of the points in X direction. UY is the displacement of the points in Y

direction. So along the X axis, the points are constrained to move only in X direction. Along the Y axis, the

points are constrained to move only in Y direction. The solution structure for displacement, in IBFEM, is

constructed as:

{}XYZ = []{}XYZ + {}XYZ ----------(1)

Where []is essential boundary function and {} is grid variable defined by piecewise interpolation. {} is

boundary value function

{}XYZ = []{}XYZ + {}XYZ

s a g aU U U HU U ;

gU H U ;

{} is the grid variable that is defined by interpolating nodal values with in the elements of grid and {} is

boundary value function which is a vector field and whose value at boundary is equal to prescribed boundary

conditions. H is a diagonal matrix such that its diagonal components Hii are essential boundary functions

satisfying all the essential properties.

1122

0

0

HH

H

The stress and strain tensor can be decomposed in the following form:

{ } [ ]{ } [ ]{ } { } { }g a g aC C

For the 2D case, if displacement vectors are { , }T

s s sU U V and { , }T

a a aU U V then strain vectors can be

defined as follows:

Y

X UY = 0

UX =0

• Using above solution structure, finite element formulation is developed for 2D elasticity problems.

s

ss

s s

U

x

V

x

U V

y x

&

a

aa

a a

U

x

V

x

U V

y x

s

ss

s s

U

x

V

x

U V

y x

&

a

aa

a a

U

x

V

x

U V

y x

1

1

0 ... 0{ } [ ]{ }

0 ... 0

Nu us

g g

s v v N

H N H NUX N X

V H N H N

1

1

0 ... 0{ } [ ]{ }

0 ... 0

Na

a a

a N

N NUX N X

V N N

Where { }gX and { }aX are nodal values of the grid variable boundary value functions respectively.

• 1 1

1 1

. .

. .

. .

1

1

1 2

1

11

{ } &{ }

.([ ] [ ])

.

0 ... 0

0 ... 0

g a

g a

g a

g a

N N

g a

N N

g

g

s g

g

N

g

N

N

U U

V V

X X

U U

V V

UU

Vdx

VB B B X

y

U V U

dy dx V

NNH H

x x

NB H H

y

1 1

1

2 1

1 1

...

0 ... 0

0 ... 0

...

N

N N

N

N

N N

N

y

N NN NH H H H

y x y x

H HN N

x x

H HB N N

y y

H H H HN N N N

y x y x

1 1 1

1

e

ei e

e

NE NBE NETT T T

g g g g

e e e

NET

g a

e

X B C B X d X N t d X N b d

X B d

• Now if the plate, as shown below, is rotated by an angle with respect to X and Y axes, and a force F is

applied parallel to X axis in the new transformed coordinate system, we wish to apply boundary conditions in

the local coordinate system. In the other words we would like to define the transformed boundary conditions

with respect to new coordinate system. The rotated block is shown below.

XU & YU were the boundary conditions in the initial illustration where the block was held parallel to the X

axis. After the block is rotated by an angle , let the new co-ordinate axes be X and Y. So consequently, the

boundary conditions in this new coordinate system will be 'XU & 'YU .

To define ' '&

X YU U in terms of &X YU U or in other words, to obtain boundary conditions in local coordinates

from global coordinates, we need to derive appropriate transformation matrix.

Let (X,Y) be coordinates of the point with respect to global co-ordinate system and let (X,Y) be coordinates of

the same point in the local coordinate system. (This at an angle to the global coordinate system.)

Now X = Xcos + Ysin ;

Y = -Xsin + Ycos ;

Writing above set of equations in the matrix form, we obtain;

' cos sin

' sin cos

X X

Y Y

So the transformation matrix is cos sin

sin cosT

.

'

'

cos sin

sin cos

X X

Y Y

U U

U U

The solution structure for displacement in terms of global coordinate system (X,Y) is;

g aXYZ XYZXYZU H U U -----------(1)

X

X

Y

Y

• And the solution structure in terms of local coordinate system (X,Y) is;

' ' ' ' ' '' ' 'g aX Y Z X Y ZX Y Z

U H U U -------------(2)

Now using transformation matrix, derived above, (2) can be rewritten as;

g aXYZ XYZXYZT U H T U T U -----------(3)

Now multiplying the above equation with T

T on both sides, we get;

T T T

g aXYZ XYZXYZT T U T H T U T T U , which is equal to;

T

g aXYZ XYZXYZU T H T U U ------------(4)

Equation (4) gives final form of solution structure that gives the displacement when the plate is at an arbitrary

angle with the X axis.

1122

0cos sin cos sin'

0sin cos sin cos

HH

H

2 2

11 22 11 22

2 2

11 22 22 11

cos sin cos sin sin cos'

cos sin sin cos cos sin

H H H HH

H H H H

Now incorporating this 'H instead of H in the original equations, we get the following set of new equations. Finally the new B matrix is derived in the following fashion.

• ' '1 211 12

' '1 221 22

' ' ' ' ' '

11 1 12 1 11 2 12 2 11 12

' ' ' ' ' '

21 1 22 1 21 2 22 2 21 22

1 2

1 2

0 0 ... 0

0 0 ... 0

...

...

0 0 ... 0

0 0 ... 0

S

s N

g

N

N N

g

N N

a N

a

Na

U N N NH HX

V N N NH H

H N H N H N H N H N H NX

H N H N H N H N H N H N

U N N NX

N N NV

1 1

1 1

' ' '

1 2

. .( ) ( )

. .

g g

sg g

ss

g g

s s N N

g g

N N

U UU

V Vdx

VB B B

dy

U V U U

dy dx V V

' ' ' '1 111 12 11 12

' ' ' ' '1 11 21 22 21 22

' ' ' ' ' ' ' '1 1 1 111 21 12 22 11 21 12 22

'

111

'

2

...

...

...

N N

N N

N N N N

N NN NH H H H

x x x x

N NN NB H H H H

y y y y

N N N NN N N NH H H H H H H H

y x y x y x y x

HN

x

B

' ' '

12 11 121

' ' ' '

21 22 21 221 1

' ' ' ' ' ' ' '

11 21 12 22 11 21 12 221 1

...

...

...

N N

N N

N N

H H HN N N

x x x

H H H HN N N N

y y y y

H H H H H H H HN N N N

y x y x y x y x

Modified H matrix can be represented as column matrix, as shown below. This notation is helpful during the

implementation phase. This notation is handy while storing the matrices as sparse matrix form. Also derivatives

of elements in modified H matrix with respect to X and Y are shown below.

• ' 2 2

11 11 22

' 2 2

22 11 22

'

12 11 22

' '

11 11 2 2 2 211 22 11 22

' '

22 22

' '

12 12

cos sin

mod sin cos

cos sin cos sin

cos sin cos sin

H H H

H H H H

H H H

H H H H H H

X Y X X Y

H H

X Y

H H

X Y

2 2 2 211 22 11 22

11 22 11 22

sin cos sin cos

cos sin cos sin cos sin cos sin

Y

H H H H

X X Y Y

H H H H

X X Y Y

Now these changes are subsequently programmed and are implemented in IBFEM. Several test examples are

created and the results are validated using Solidworks.

Example 1:

An AISI steel plate of dimension 0.09mX0.04mX0.001m. It is at an angle of 45 degrees to the x axis. Boundary

conditions and Loads are shown below. A force of 2N acts on the one end of the plate as shown below.

Material properties used are E = 2.0E11 Pa and Poisson ratio is 0.29. Mesh density used is 10*10*1 in

Simugrid. Elements of the mesh are 4 node quadrilateral element.

Model:

On the line marked with red, boundary condition is Uy = 0 and on the line marked with green, boundary

condition Ux = 0. A load of 2N acts on the plate as shown. Following are the stress plots and displacement

plots.

• Displacement Plot (Max. Displacement = 9.326E-13)

Stress plot (Von Mises, Max Stress = 2.925E0 Pa):

If the mesh element is changed from 4 node quadrilateral to 9 node quadrilateral and by keeping the mesh

density same (10 X 10 X 1), following results are obtained.

Stress plot (Von Mises, Max stress = 2.932E0 Pa):

• Displacement plot ( Max. Displacement = (9.361E-13)

Similar analysis is performed using solidworks. Here roller boundary conditions are used. Following results are

obtained:

Stress Plot ( sigma max = 4.896E4 Pa)

• Displacement Plot: (Max. Displacement = 2.269E-005 mm)

Example 2:

An AISI steel pl

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