elasticity

2D Elasticity Examples

Dr. Robert Gracie University of Waterloo

CIVE422 2015

Dr. Gracie, University of Waterloo 2015

T3 Example Point Loads

2

1MN

2MN

1

2

I x y 1 0 0 2 1.5 0 3 0 1 4 0 -1

Nodal Coordinate

e

1 1 2 3 2 1 4 2

Element connectivity

E=200GPa, ! =0.35


T3 Example K1

3

1.31 0.71 -0.73 -0.38 -0.58 -0.33! 0.71 1.90 -0.33 -0.26 -0.38 -1.65!-0.73 -0.33 0.73 0 0 0.33!-0.38 -0.26 0 0.26 0.38 0!-0.58 -0.38 0 0.38 0.58 0!-0.33 -1.65 0.33 0 0 1.65!

K1 = 1011!

K1 = A B1TDB1

B1 =N1,x 0 N2,x 0 N3,x 00 N1,y 0 N2,y 0 N3,yN1,y N1,x N2,y N2,x N3,y N3,x

!

"

####

$

%

&&&&

=12A

y23 0 y31 0 y12 00 x32 0 x13 0 x21x32 y23 x13 y31 x21 y12

!

"

####

$

%

&&&&

-0.67 0 0.67 0 0 0! 0 -1.00 0 0 0 1.00!-1.00 -0.67 0 0.67 1.00 0!

B1 =

[1] [2]

[1]

[2]

[3]

[3]


T3 Example K2

4

1.31 -0.71 -0.58 0.33 -0.73 0.38! -0.71 1.90 0.38 -1.65 0.33 -0.26! -0.58 0.38 0.58 0 0 -0.38! 0.33 -1.65 0 1.65 -0.33 0! -0.73 0.33 0 -0.33 0.73 0! 0.38 -0.26 -0.39 0 0 0.26!

K2 = 1011!

K2 = A B1TDB1

B2 =N1,x 0 N2,x 0 N3,x 00 N1,y 0 N2,y 0 N3,yN1,y N1,x N2,y N2,x N3,y N3,x

!

"

####

$

%

&&&&

=12A

y23 0 y31 0 y12 00 x32 0 x13 0 x21x32 y23 x13 y31 x21 y12

!

"

####

$

%

&&&&

-0.67 0 0 0 0.67 0! 0 1.00 0 -1.00 0 0! 1.00 -0.67 -1.00 0 0 0.67!

B2 =

[1] [4]

[1]

[4]

[2]

[2]


T3 Example Assemble K

5

K =1011!

2.62 0 -1.47 0 -0.58 -0.33 -0.58! 0 3.81 0 -0.51 -0.38 -1.65 0.38! -1.47 0 1.47 0 0 0.33 0! 0 -0.51 0 0.51 0.38 0 -0.38! -0.58 -0.38 0 0.38 0.58 0 0! -0.33 -1.65 0.33 0 0 1.65 0! -0.58 0.38 0 -0.38 0 0 0.58! 0.33 -1.65 -0.33 0 0 0 0!

0.33! -1.65! -0.33! 0! 0! 0! 0! 1.65!

22

4

3

11

K1 = 1011!

[1] [2]

[1]

[2]

[3]

[3]

1.31 0.71 -0.73 -0.38 -0.58 -0.33! 0.71 1.90 -0.33 -0.26 -0.38 -1.65!-0.73 -0.33 0.73 0 0 0.33!-0.38 -0.26 0 0.26 0.38 0!-0.58 -0.38 0 0.38 0.58 0!-0.33 -1.65 0.33 0 0 1.65!


T3 Example Assemble K

6

K =1011!

2.62 0 -1.47 0 -0.58 -0.33 -0.58! 0 3.81 0 -0.51 -0.38 -1.65 0.38! -1.47 0 1.47 0 0 0.33 0! 0 -0.51 0 0.51 0.38 0 -0.38! -0.58 -0.38 0 0.38 0.58 0 0! -0.33 -1.65 0.33 0 0 1.65 0! -0.58 0.38 0 -0.38 0 0 0.58! 0.33 -1.65 -0.33 0 0 0 0!

0.33! -1.65! -0.33! 0! 0! 0! 0! 1.65!

12

4

3

12

1.31 -0.71 -0.58 0.33 -0.73 0.38! -0.71 1.90 0.38 -1.65 0.33 -0.26! -0.58 0.38 0.58 0 0 -0.38! 0.33 -1.65 0 1.65 -0.33 0! -0.73 0.33 0 -0.33 0.73 0! 0.38 -0.26 -0.39 0 0 0.26!

K2 = 1011!

[1] [4]

[1]

[4]

[2]

[2]


Forces

7

f =

f1xf1yf2 xf2 yf3xf3yf4xf4y

!

"

############

$

%

&&&&&&&&&&&&

=

00

1106

2106

0000

!

"

##########

$

%

&&&&&&&&&&

1

22

4

3

1

1MN

2MN

1

2


Apply BCs and Solve

8

1011!

2.62 0 -1.47 0 -0.58 -0.33 -0.58! 0 3.81 0 -0.51 -0.38 -1.65 0.38! -1.47 0 1.47 0 0 0.33 0! 0 -0.51 0 0.51 0.38 0 -0.38! -0.58 -0.38 0 0.38 0.58 0 0! -0.33 -1.65 0.33 0 0 1.65 0! -0.58 0.38 0 -0.38 0 0 0.58! 0.33 -1.65 -0.33 0 0 0 0!

0.33! -1.65! -0.33! 0! 0! 0! 0! 1.65!

u1xu1yu2 xu2 yu3xu3yu4xu4y

!

"

############

$

%

&&&&&&&&&&&&

1MN

2MN

1

21

22

4

3

1

=

0+ R1x0+ R1y1106

2106

0+ R3x0+ R3y0+ R4x0+ R4y

#

$

%%%%%%%%%%%%

&

'

((((((((((((


Apply BCs and Solve

9

1011!

2.62 0 -1.47 0 -0.58 -0.33 -0.58! 0 3.81 0 -0.51 -0.38 -1.65 0.38! -1.47 0 1.47 0 0 0.33 0! 0 -0.51 0 0.51 0.38 0 -0.38! -0.58 -0.38 0 0.38 0.58 0 0! -0.33 -1.65 0.33 0 0 1.65 0! -0.58 0.38 0 -0.38 0 0 0.58! 0.33 -1.65 -0.33 0 0 0 0!

0.33! -1.65! -0.33! 0! 0! 0! 0! 1.65!

= 106

0+ R1x0+ R1y12

0+ R3x0+ R3y0+ R4x0+ R4y

"

#

$$$$$$$$$$$

%

&

'''''''''''

u1xu1yu2 xu2 yu3xu3yu4xu4y

!

"

############

$

%

&&&&&&&&&&&&

u1x = u1y = u3x = u3y = u4x = u4y = 0

KE KEFKFE KF


Plot Displacements

10

the displacements at node 2 are (0.006825,-0.039) mm. the reactions at node 1 are (-1000,2000) kN. the reactions at node 3 are (-1500,225) kN. the reactions at node 4 are (1500,-225) kN. ELEMENT #1 the strain (Exx,Eyy,2Exy) =(4.55e-06,0,-2.6e-05). the stress (Sxx,Syy,Sxy) = (1,0.3,-2) MPa. ELEMENT #2 the strain (Exx,Eyy,2Exy) =(4.55e-06,0,-2.6e-05). the stress (Sxx,Syy,Sxy) = (1,0.3,-2) MPa.


T3 Example Point Loads

11

1

22

4

3

1

%% Define Material Properties E = 200e9; nu = 0.3; %% Define Coordinates of the nodes x1 = 0; x2 = 1.5; x3 = 0; x4 = 0; x = [x1; x2; x3; x4]; y1 =0; y2 = 0; y3 = 1; y4 = -1; y = [y1; y2; y3; y4]; %% Connectivity conn = [1, 2, 3; 1, 4, 2];


K = zeros(8,8);!for e=1:2! enodes= conn(e,:);! sctr = [2*enodes(1)-1,2*enodes(1),2*enodes(2)- ! 1,2*enodes(2),2*enodes(3)-1,2*enodes(3)];! ye = y(enodes); ! xe = x(enodes);! Be = 1/(2*A)*[ye(2)-ye(3),0,ye(3)-ye(1),0,ye(1)-ye(2),0;! 0 ,xe(3)-xe(2), 0 , xe(1)-xe(3), 0, xe(2)-xe(1);! xe(3)-xe(2),ye(2)-ye(3),xe(1)-xe(3),ye(3)-ye(1),xe(2)- ! xe(1),ye(1)-ye(2)];!

! D = E/(1-nu^2)*[1 ,nu,0; nu,1 ,0; 0 ,0 ,(1-nu)/2];! Ke = A*Be'*D*Be;! K(sctr,sctr)=K(sctr,sctr)+Ke;!end!

12


Dam Example

13

1m

2m

5m

3/1000 mkg=

2.040

/2700 3

=

=

=

GPaEmkgc


Dam Example

14

g( )yt

y

x

n

1 25

34

6

7

8 9

1 2

34


Body Loads

n Body load due to gravity

n Nodal forces due to body load

15

b =bxby

!

"

##

$

%

&&=

0g

!

"##

$

%&&

fe = NeTbd

e =

N1e ,( ) 00 N1

e ,( )N2

e ,( ) 00 N2

e ,( )N3

e ,( ) 00 N3

e ,( )N4

e ,( ) 00 N4

e ,( )

!

"

##############

$

%

&&&&&&&&&&&&&&

1

1(

1

1

( 0 g!

"##

$

%&&

Je ,( ) dd=

N1e 0

0 N1e

N2e 0

0 N2e

N3e 0

0 N3e

N4e 0

0 N4e

!

"

#############

$

%

&&&&&&&&&&&&&

0 ! g

!

"##

$

%&&dxdy

( e)

1 2 5

3 4

6 8 9

1! 2!

3!4!


Body Loads

n Compute Integral using numerical quadrature. Note that in general the Jacobian is not constant but a

function of the parent coordinates

16

fe =

N1e i, j( ) 00 N1

e i, j( )N2

e i, j( ) 00 N2

e i, j( )N3

e i, j( ) 00 N3

e i, j( )N4

e i, j( ) 00 N4

e i, j( )

"

#

$$$$$$$$$$$$$$$

%

&

'''''''''''''''

0g

"

#$$

%

&''J e i, j( )

j=1

nQ

i1

nQ

WiW j = NeT i, j( )b J e i, j( )j=1

nQ

i1

nQ

WiW j

J e ,( )


Body Loads

n Since in general the Jacobian is not constant but a function of the parent coordinates then

17

fe =

f1xe

f1ye

f2 xe

f2 ye

f3xe

f3ye

f4xe

f4ye

"

#

$$$$$$$$$$$$$

%

&

'''''''''''''

J e ,( )

Ag4

01010101

"

#

$$$$$$$$$

%

&

'''''''''

This occurs when opposite edges of the element are not parallel (As in this example)


Tractions (surface loads)

n If the surface tractions can be reasonably approximated by a linear function along the edges of the domain (almost always the case) then

18

f2 =

f1xe

f1ye

f2 xe

f2 ye

f3xe

f3ye

f4xe

f4ye

"

#

$$$$$$$$$$$$$

%

&

'''''''''''''

=l6

00

2t e2 x + te3x

2t e2 y + te3y

t e2 x + 2te3x

t e2 y + 2te3y

00

"

#

$$$$$$$$$$$

%

&

'''''''''''

Where the summation is only over the nodes which are on the boundary of the domain t = NI

e

i=1

2

tIe

21

32

4[5] [2]

[6]

[9]

t 22 =t 22 xt 22 y

!

"

##

$

%

&&= 5 y2

e( )wgnxe

nye

!

"

##

$

%

&&

t 23 =t 23xt 23y

!

"

##

$

%

&&= 5 y3

e( )wgnxe

nye

!

"

##

$

%

&&


Tractions (surface loads)

n If the surface tractions can be reasonably approximated by a linear function along the edges of the domain (almost always the case) then

19

f4 =

f1x4

f1y4

f2 x4

f2 y4

f3x4

f3y4

f4x4

f4y4

"

#

$$$$$$$$$$$$$

%

&

'''''''''''''

=l6

00

2t 42 x + t43x

2t 42 y + t43y

t 42 x + 2t43x

t 42 y + 2t43y

00

"

#

$$$$$$$$$$$

%

&

'''''''''''

Where the summation is only over the nodes which are on the boundary of the domain t = NI

e

i=1

2

tIe

21

34

4[9] [6]

[3]

[7]

t 42 =t 42 xt 42 y

!

"

##

$

%

&&= 5 y2

e( ) ! wgnxe

nye

!

"

##

$

%

&&

t 43 =t 43xt 43y

!

"

##

$

%

&&= 0

0

!

"#

$

%&


Max Normal and Shear Stress

20


Dam Example Q9

21

g( )yt

y

x

n

1 25

34

6

7

8 9


Q9

22

-1 -0.5 0 0.5 10

1

2

3

4

5Max Principal Shear Stress - Deformation x10000

0.5

1

1.5

2

2.5

3x 105

-1 -0.5 0 0.5 10

1

2

3

4

5Max Principal Stress - Deformation x10000

0

2

4

6

8

10

12

14x 104


1 x Q9 vs 4 x Q4

23

-1 -0.5 0 0.5 10

1

2

3

4

5Max Principal Stress - Deformation x10000

0

2

4

6

8

10

12

14x 104

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