w u w u dy dx dy wy · 2014. 9. 2. · x y n,n Components of unit normal vector, n &, to body boundary. Also: --cos sin y x n n Q: What about the body forces on the triangle? y dy

CE380P. 4 – BOUNDARY ELEMENT METHODS – Review of Mechanics of Solids © S.A. Kinnas 2014

1 Updated: 9/2/2014

MECHANICS OF SOLIDS:

DISPLACEMENTS – STRAINS

2-D

u: displacement along x; u(x,y)

v: displacement along y; v(x,y)

HTensor of strains:

yyyx

xyxx

'''2

1

2

1

2

1

'''

'''

BADDABx

v

y

u

AD

ADDA

y

v

AD

ADDA

AB

ABBA

x

u

AB

ABBA

yxxy

yy

xx

Distortion of ABCD yxxyBADDAB 22

dxx

udy

y

uu

dyy

uu

dyy

vv

dyy

vdx

x

vv

dyy

vdy

D

D

C

C

dy

A

B

dxx

vv

v

Au

x

y

dxx

udx

dxx

uu

dx

y

u

dyy

vdy

dyy

u

DA

DD

x

v

dxx

udx

dxx

v

BA

BB

'''

'''tan

'''

'''tan

B

''B

''D


2 Updated: 9/2/2014

WHY IS “H” A TENSOR?

Define:

Tensor of changes of displacements:

y

v

y

u

x

v

x

u

is a tensor because: (due to definition of tensor)

T

v

u

y

xvu

y

x

T: symbol for transpose of a vector, matrix

So it can be easily shown that

)1(2

T

H

Thus H, being the sum of two tensors, is also a tensor.

By using (1) it can be easily shown that THH or that:

H: symmetric tensor


3 Updated: 9/2/2014

Change in “volume” (area) of ABCD:

)(

)(111

))(()1)()(1)((

))(())((

))((

))((

2

yyxx

yyxxyyxx

yyxx

V

OVV

ADABADAB

ADABDABAVVV

DABAV

ADABV

So: )(HtrV

Vyyxx

tr: trace of tensor

volumetric strain: )(HtrV

Vyyxx

Note: trace of tensor does not change with coordinate transformation. What is the physical

meaning of this for tr(H)?


4 Updated: 9/2/2014

CHANGE OF COORDINATES SYSTEM:

According to tensor analysis:

)2(TRTRT

where R is the matrix of direction cosines ,...)('xx

cossin

sincos

yyxy

yxxxR

For example axisxwithaxisyofanglexy

cos

For vectors:

v

uR

v

u

Prime ( ′ ) corresponds to the rotated system:

vuv

vuu

cossin

sincos

y

ox

y

x

x

o

y


5 Updated: 9/2/2014

(2) may also be written (for H):

cossin

sincos

cossin

sincos

yyyx

xyxx

yyxy

yxxx

Principal coordinates system is defined as the one at which

0 xyyx (3)

It can be shown that P

(=angle by which the original system has to rotate, positive in the

counter-clockwise direction, so that it becomes a principal coordinates system) is given as:

yyxx

xy

P

22tan (4)

The corresponding strains are called the principal strains


6 Updated: 9/2/2014

EXAMPLE: PURE SHEAR

0,

xy

yyxxyx

xyyx

yyxxyx 0

45

,

Distortion of 1111

DCBA = 2

yy

x

x

D

D

1D

1D

A

A 1A

1A

B

B

1B

1B

C

C

1C1

C

45

45


7 Updated: 9/2/2014

MOHR’S CIRCLE FOR STRAINS:

PPyx , principal axes

P

yy

P

xx , principal strains

y

Py

x

Px

P

yy

P

xx

xy

xx

P

yy

P

xx

yy

2

xy yyxx ,


8 Updated: 9/2/2014

STRAINS IN 3-D

333231

232221

131211

H

yxDin 2,12

i

j

j

i

ijx

u

x

u

2

1

yxxxvuuuDin 2121

,,,2

iiHtr

332211)(

(according to Einstein’s notation)

3

3

33

2

2

22

1

1

11,,

x

u

x

u

x

u

jiforjiij


9 Updated: 9/2/2014

ELEMENTARY FORCES – STRESSES – 2D

ij = stress applying on plane normal to axis i, with direction parallel to axis j

yyxx , normal stresses

yxxy , shear stresses

y

x

dy

dx

dy

dFxx

xx

yy

yx

xy

xx

dx

dFyyyy

dy

dFxy

xy

dx

dFyx

yx

y

x

dy

dx

yydF

yFd

yxdF

xFd

xxdF

xydF


10 Updated: 9/2/2014

TENSOR OF STRESSES (2-D)

2221

1211

yyyx

xyxx

ij

Laws of tensors hold!

EQUILIBRIUM OF STRESSES

(5)

(6a)

(6b)

y

dyxx

xy

yx yy

dx

x

dxx

xy

xy

dxx

xx

xx

dyy

yy

yy

dyy

yx

yx

Body force vector per

unit “volume”

B

B

Y

GX

Moments about 0G xyyxxy

0x

F 0

X

yx

yxxx

0yF 0

Y

yx

yyxy


11 Updated: 9/2/2014

TENSOR OF STRESSES (3-D):

333231

232221

131211

ij

jijiij

; (Due to equilibrium of moments)

;33

332211 iip

hydrostatic tension (7)

EQUILIBRIUM OF STRESSES (3-D):

0

i

j

jiB

x

(8)

Also written as: 0, ijji B


12 Updated: 9/2/2014

BOUNDARY TRACTIONS (2-D)

Boundary tractions are defined as:

ds

dFp x

x (x force per unit “area” {arc length in 2-D})

ds

dFp

y

y (y force per unit “area”)

Balance of forces on triangle ABC:

Along x:

(9a)

Along y:

(9b)

yxnn , Components of unit normal vector, n

, to body boundary.

Also:

cos

sin

y

x

n

n

Q: What about the body forces on the triangle?

y

dy

xdx

ydF Fd

n

xdF

yx yy

A C

B

ds

n

1 nds

dxn

y

ds

dyn

x

),(yx

dFdFFd

is force acting on the boundary of the body

(over BC)

xy

xx

xyxxxdFdxdy

yxyyydFdydx

yyxxxxxnnp

yyyxxyynnp


13 Updated: 9/2/2014

BOUNDARY TRACTIONS (3-D):

Boundary tractions are defined as:

dS

dFp

dS

dFp

dS

dFp 3

3

2

2

1

1,,

Balance of forces on pyramid ABCD:

3332321313

3232221212

3132121111

nnnp

nnnp

nnnp

or:

3

2

1

333231

232221

131211

3

2

1

n

n

n

p

p

p

(10)

3x

1x

B

A

D

C

21

22

23

),,(321

dFdFdFFd

2x

(Force acting on the

boundary of the body,

over BCD)

1dxAC

3dxAB

2dxAD

dSBCD

dxdxABD

BCD

ABDn

BCD

ABDn

BCD

ABDn

n

)(

2

1)(

)(

)(

)(

)(

)(

)(

1

23

3

2

1

),,( 321 nnnn


14 Updated: 9/2/2014

CONSTITUTIVE RELATIONS:

ijijij 2 (11)

332211

ii

:, Lamé’s constants

ij

Kronecker delta ( 1,;0 ijij

ji for ji )

211

,)1(2

EE

G

G = Shear Modulus

E = Modulus of Elasticity (Young’s Modulus)

= Poisson’s ratio

(11) can also be written as:

ijijij

E

211 (12)

Inverting (12) we get:

ijijijE

pE

13

(13)

3/3/332211

ii

p


15 Updated: 9/2/2014

DEVIATORIC STRESSES AND STRAINS:

They are defined as:

03

iiijijij

ee

0iiijijij

sps

Then (11) becomes:

Kp

Gesijij

2 (14)

K= Bulk Modulus

)21(33

2

EK

ALTERNATIVE EXPRESSIONS FOR EQUATION (13):

33221111

1

E

33112222

1

E (15)

22113333

1

E

1313131313

2323232323

1212121212

/2

/2

/2

G

G

G


16 Updated: 9/2/2014

SPECIAL CASES:

PURE SHEAR:

Then from (15) we have:

G

121323332211,0,0


17 Updated: 9/2/2014

PURE (AXIAL) TENSION

0

0

132312

3322

From (15) we have:

11

11

33

11

11

22

132312

11

110,

E

E

E

Definition of Poisson’s ratio:

11

33

11

22

Change in sectional area: E

11

3322

2

Change in volume of specimen: 1133221121

Eii


18 Updated: 9/2/2014

PLATE STRETCHING (2-D):

0,0231333

122211,, functions of

21, xx

Then from (15) we get:

G

E

E

2

1

1

1212

112222

221111

(16)

0221133

E

01323


19 Updated: 9/2/2014

PLANE STRAIN (2-D):

0,0231333

122211, functions of

21, xx

From (15) we get:

1212

2313

1122

2

22

2211

2

11

2

01

1

1

1

G

E

E

(17)

Note: Equation (17) can be brought to the form of equation (16) by putting:

1

1 2E

E

22113333

10

E

0221133


20 Updated: 9/2/2014

INVERSION OF (16) & (17)

(a) Plate Stretching: 033

1212

2211222

2211211

21

1

G

E

E

(16a)

(b) Plane Strain: 033

1212

221122

221111

2

1211

1211

G

E

E

(17a)


21 Updated: 9/2/2014

COMPATIBILITY OF STRAINS (2-D):

yxxy

xyyyxx

2

2

2

2

2

2 (18)

(18) valid

COMPATIBILITY OF STRESSES (2-D)

PLATE STRETCHING 033

From (16)

12;/2

EGGxyxy

So (18) yxxy

xy

xxyyyyxx

2

2

2

2

2

12 (19)

From equations (6a) and (6b)

062

2

2

x

X

yxxa

x

yxxx

062

22

y

Y

yyxb

y

yyxy

2

3

2

2

yx

u

yx

u xxxx

2

3

2

2

xy

v

xy

v yyyy

yx

v

xy

u

yxx

v

y

u xyxy

2

3

2

32

22

1

xxyyyy

yyxxxx

E

E

1

1


22 Updated: 9/2/2014

So (19) becomes:

y

Y

x

X

yxxy

yyxx

xxyyyyxx 2

2

2

2

2

2

2

2

1

(20)

For 0YX (negligible body forces):

02

2

2

2

yyxx

yx or 02 p (21)

and the equilibrium equations become:

0

yx

xyxx

(21a); 0

yx

yyxy

(21b)

Defining: (22)

(21a), (21b) are satisfied automatically

y

Y

yyx

x

X

xyx

yyxy

xxxy

2

22

2

22

012

2

2

2

y

Y

x

X

yxyyxx

yx

F

x

F

y

Fxyyyxx

2

2

2

2

2

,,


23 Updated: 9/2/2014

In order for (21) to be also satisfied we must have:

02

2

2

2

2

2

2

2

y

F

x

F

yx

or 04 F or 024

4

22

4

4

4

y

F

yx

F

x

F (23)

F = stress function of Airy

F = biharmonic function (satisfies 23)

vu

yxFbaequationscb

xyyyxxxyyyxx ,,,,,

),()9&9.(.23


24 Updated: 9/2/2014

EQUILIBRIUM EQUATIONS IN TERMS OF DISPLACEMENTS:

(2-D, PLATE STRETCHING)

0

X

yx

xyxx

(6a)

0

Y

yx

yyyx

(6b)

Equ. (16a):

yx

v

y

uE

y

x

v

yx

uE

x

x

v

y

uE

x

v

y

uG

y

v

yx

uE

yy

v

x

uE

yx

v

x

uE

xy

v

x

uE

x

v

y

u

y

v

x

u

E

GE

xy

xy

xy

yy

yy

xx

xx

xyyyxx

yyxxyy

xyxyyyxxxx

2

2

2

2

22

2

22

22

2

2

2

22

2

2

12

12

12

11

11

2

1,,

1

21


25 Updated: 9/2/2014

0

12

1

12 2

2

22

2

2

2

EX

yx

v

y

u

yx

v

x

u

01

22

22

2

2

2

22

2

2

G

X

x

u

yx

v

x

u

y

u

yx

v

x

u

01

1 2

2

2

2

G

X

yx

v

x

uu

Thus (6a) becomes:

0

121

2

2

22

2

2

2

X

yx

v

y

uE

yx

v

x

uE

(24a)

Similarly (6b) becomes:

01

1 2

2

2

2

G

Y

yx

u

y

vv

(24b)

Equations (24a) and (24b) are the equilibrium equations for u, v in the case of plate stretching.


26 Updated: 9/2/2014

GOING FROM PLATE-STRETCHING TO PLANE-STRAIN AND VICE-VERSA

EQUILIBRIUM EQUATIONS IN TERMS OF DISPLACEMENTS:

(2-D, PLANE STRAIN)

Put 1/ in (24a) & (24b) to get:

021

1 2

2

2

2

G

X

yx

v

x

uu

(25a)

021

1 2

2

2

2

G

Y

yx

u

y

vv

(25b)

Equations (25a) and (25b) are the equilibrium equations for u, v in the case of plane strain


27 Updated: 9/2/2014

BOUNDARY CONDITIONS (B.C.s)

According to (9a), (9b)

yyxxxxxnnp

(9a)

yyyxxyynnp

(9b)

Natural b.c.s involve yx

, of u, v

Essential b.c.s involve u, v

y

A

B

n yn

xn

px, py known

(natural b.c.s)

u, v known

(essential b.c.s)

x

Documents

w u w u dy dx dy wy · 2014. 9. 2. · x y n,n Components of unit normal vector, n &, to body boundary. Also: --cos sin y x n n Q: What about the body forces on the triangle? y dy