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1 Lecture Note 3 Lecture Note 3 Stress Functions First Semester , Academic Year 2012 Department of Mechanical Engineering Chulalongkorn University

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Page 1: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

1

Lecture Note 3Lecture Note 3

Stress Functions

First Semester, Academic Year 2012,Department of Mechanical Engineering

Chulalongkorn University

Page 2: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Objectives #1

Describe the equation of equilibrium using 2D/3D stress elements

Set well-posed elasticity problems (e.g. for finite element analyses)S l i l l ti bl b th i d i Solve simple elastic problems by the inverse and semi-inverse method

2

Page 3: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Concepts #1

Equation of equilibrium is the governing equation. Well-posed problems can be formulated by substituting Well posed problems can be formulated by substituting

appropriate constitutive relationships, loads, initial conditions and boundary conditions into the governing equationsWith i t diti f tit ti l ti th With appropriate conditions from constitutive relations, the stress distribution can be found by fitting the boundary conditions into the appropriate stress functions

3

Page 4: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Equilibrium ElementEquilibrium X, Y, Z are body

forces per unit lvolume

4

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Equilibrium Complementary Principal of ShearEquilibrium

0zM

x x

( )( ) ( )( )

2 2

( )( ) ( )( ) 0

xyxy xy

yx

x xy z x y zx

y yx z y x z

( )( ) ( )( ) 02 2

02 2 2 2 2 2

yx yx

xy yxxy xy yx yx

x z y x zy

V V V V V Vx y

2 2 2 2 2 2

2 2 0

xy xy yx yx

xy yxxy yx

x y

x yx y

5

As 0 and 0, , etcxy yx

x yx y .

Page 6: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Equilibrium 3D EquationsEquilibrium X, Y, Z are body

forces per unit l

0 ( )( ) ( )x

x x xF x y z y zx

volume

( )( ) ( )yxyx yxy x z x z

y

( )( ) ( ) ( ) 0zxzx zxz x y x y X x y z

z

0xyx xz Xx y z

0

0

yx y yz

zyzx z

Yx y z

Z

6

0yzx z Zx y z

Page 7: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Equilibrium 2D Plane StressEquilibrium 2D Reduction

z 0yz zx

z yz zx

xzxyx

x y

0X

zy

yzyx y

x y

0Y

z

zx

zy

x

z

y

0Z

z

0xyx Xx y

7

0yx y Y

x y

Page 8: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Equilibrium 2D Plane StrainEquilibrium 2D Reduction

z const, 0yz zx

z , yz zx

xzxyx

x y

0X

zy

yzyx y

x y

0Y

z

zx

zy

x

0z Zy z

0xyx X

x y

0yx y Yx y

8

0z Z

z

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Equilibrium Components on an Inclined PlaneEquilibrium

9

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Equilibrium Boundary ConditionsEquilibrium 3D Scale Up

0xF

Unit thickness

1 02

x

x yxX s y y X x y

As 0,

x yx

xdy dxXds ds X l

x yx

ds ds

X l m

x yx zx

xy y zy

X l m n

Y l m n

xy yY l m xz yz zZ l m n

10

coscos

x

y

lm

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Equilibrium Strain Compatibility #1Equilibrium

2 2 2xy

xyv u v ux y x y x y x x y y

2 2 2

2 2xy

yz

x y x y x y x x y y

w v v uy z x y y xx y

zx

y z x y y xx yu wz x

2 2 2 2 2

2 22

xy y x

yz y

x y x y

2 2

2 2 2

yz yz

zx x z

y z y z

11

2 2z x z x

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Equilibrium Strain Compatibility #2Equilibrium

2 2 2 2

( ) ( )xy zx v u u wz x x y z x x y x y z x

2 2 2

2( ) ( )xy zx

z x x y z x x y x y z xu v w u

x z y z x y z y x y zx

y y y y

2

2 ( )yz xyx zx

2

2 ( )

2 ( )

yz xyx zx

y yz xyzx

y z x x y z

2

2 ( )

2 ( )yz xyz zx

z x y x y z

12

( )

x y z x y z

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Equilibrium Strain Compatibility #3Equilibrium

2 2 2

2 2xy y x

x y x y

2 2

2 22

2 2yz yz

x y x y

y z y z

2 2 2

2 2zx x z

y z y z

z x z x

2

2

2 ( )yz xyx zx

y z x x y z

2

2

2 ( )y yz xyzx

z x x x y z

13

2 ( )yz xyz zx

x y x x y z

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Equilibrium 2D Strain CompatibilityEquilibrium 2D Reduction

z z0 or const, 0yz zx yz zx

2 2 2

2 2

2

xy y x

x y x y

2

2 x

y z

(yz

x

zx

x

xy

y z)

2yz

2

2z

y z y

2

2y

z2

2

2 y

z x

(yz

x

zx

x

xy

y z)

2zx

2

2x

z x z

2

2z

x

2

2 z

x y

(yz

x

zx

x

xy

y z)

2 2 2xy y

14

2 2xy y x

x y x y

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2D Elastic Problems 3D Set Up #1 Problem

Governing equations

0xyx xz Xx y z

0yx y yz Yx y z

0zyzx z Z

x y z

15

Page 16: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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2D Elastic Problems 3D Set Up #2 Problem

Constitutive equations

1

xuxv

1 ( )

1 ( )

x x y zE

y

z

yw

( )

1 ( )

y y z x

z z x y

E

E

z

xy

zu vy x

y

xyxy

E

G

yz

yv wz y

yzyz G

16

zxw ux z

zxzx G

Page 17: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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2D Elastic Problems 3D Set Up #3 Problem

2 2 2

2 2xy y x

Compatibility condition

2 2

2 22

2 2yz yz

x y x y

y z y z

2 2 2

2 2zx x z

y z y z

z x z x

2

2

2 ( )yz xyx zx

y z x x y z

2

2

2 ( )y yz xyzx

z x x x y z

17

2 ( )yz xyz zx

x y x x y z

Page 18: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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2D Elastic Problems Plane Stress Set Up #1 Problem

1

Governing equations

Constitutive equations

1 ( )

1 ( )

x x yE 0xyx X

x y

( )

1 2(1 )

y y x

xy xy xy

E

G E

0yx y Yx y

xy xy xyG E

2 2 X

2

2

2 2

0xy xy xyx x x XX Xx y y x x y xx

Y

18

20yx y xy y xy y YY Yx y x y x y yy

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2D Elastic Problems Plane Stress Set Up #2Problem

Compatibility condition2 2

2 2xy x X

2 2 2

2 2

2 2 2

xy y x

x y x y

2

2 2

2xy y

x y xx

Y

2 2 2

2 2

2 2 22 2

2(1 ) ( ) ( )

2

xyy x x yx y x y

2x y yy

2 2 2 2

22

2(1 )

(1 )( )

xy y yx x

yx

x y x x y y

X Y

2 2

2 22 2

(1 )( )yx

y yx x

x yx y

19

2 2 2 2x x y y

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2D Elastic Problems Plane Stress Set Up #3Problem

2 2 22 2 2

2 2 2 2 2 2(1 )( )y y yx x xX Yx yx y x x y y

2 2 2 2 2 2

2 22 2

2 2 2 2 (1 )( )y yx x

x yx y x x y y

X Yx yx y x y

2 22 2

2 2 2 2y yx x

x yx y x y

x x y y

2 22 2

2 2 2(1 )( )y yx x

y y

X Yx yx y x y 2

2 22 2

2 2 2 2

2 22 2

(1 )( ) y yx xX Yx y x y x y

20

2 22 2

2 2 2 2(1 )( ) y yx xX Yx y x y x y

Page 21: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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2D Elastic Problems Plane Stress Set Up #4Problem

Compatibility condition

2 22 2

2 2 2 2(1 )( ) y yx xX Yx y x y x y

x y x y x y

2 2

2 2( )( ) (1 )( )x yX Y

2 2( )( ) ( )( )x y x yx y

21

Page 22: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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2D Elastic Problems Plane Stress Set Up #5Problem

Neglect the body force

2 2

2 2( )( ) 0x yx y

0xyx

x y

x y 2 2

( )( ) 0

xy x

y x

2 2

2 2 2 2

( )( ) 0

( )( ) 0

x yx y

0yx y

x y

2 2 2 2

4 4 4 4

2 2 4 4 2 2

( )( ) 0

0

x y y x

Assume a function

xy y

x y

2 2 4 4 2 2

4 4 4

4 2 2 42 0

x y x y y x

x x y y

22

2 2 2

2 2, , x y xy x yy x

x x y y

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Stress Function Airy Stress Function Airy

Neglect the body force2 2

( )( ) 0

To solve the equations use stress functions that satisfy

2 2( )( ) 0x yx y

To solve the equations, use stress functions that satisfy relationships

2 2 2

Substitute into the compatibility condition

2 2, , x y xy x yy x

Substitute into the compatibility condition

4 4 4

4 2 2 42 0x x y y

23

x x y y

Page 24: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Inverse MethodPlate

Procedure Select the stress function Select the stress function Check compatibility Fit in the boundary condition

AdvantageSimplified assumptions Simplified assumptions

Disadvantage Find problems that fit solution

24

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Example Timoshenko 1 #1 Plate

For an elastic plate subjected to static loads shown, determine the stress function. The A, B and C are constantsconstants.

Domain &

25

Boundary conditions

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Example Timoshenko 1 #2 Plate

Describe stress distributions within the platewithin the plate

2 2

4 4 4 4 4 4

4 2 2 4 4 2 2 40, 0, 0 2 0

Ax Bxy Cy

x x y y x x y y

4 2 2 4 4 2 2 4

2 2 2

2 22 , 2 , x y xy

x x y y x x y y

C A Bx yy x x yy x

26

Page 27: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Timoshenko 1 #3 Plate

2 2, 2 , 2 , x y xyAx Bxy Cy C A B

Let 1, 2, 3

6 MPa, 2 MPa, 2 MPa x y xy

A B C

27

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Example Timoshenko 1 #4 Plate

6 MPa, 2 MPa, 2 MPa x y xy

Plane stress, 0 MPa

6 83 MP

z

1

2

6.83 MPa 1.17 MPa 0 MPa

3

1 3

0 MPa

6.83 MPaT

1 3

2 2 21 2 2 1( ) ( )

2

T

v

28

6.32 MPav

Page 29: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Timoshenko 2 #1 Plate

For an elastic plate subjected to static loads shown, determine the stress function. The D is a constant.

Boundary conditionsFor left & right edges: 0Dy

29

For left & right edges: , 0

For top & bottom edges: 0, 0x xy

y xy

Dy

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Example Timoshenko 2 #2 Plate

Describe stress distributions

3 2 2 3

6 2 2 6A B C Dx x y xy y

4 4 4

4 2 2 4

4 4 4

0, 0, 0x x y y

4 4 4

4 2 2 42 0x x y y Satisfy the condition if

F l ft & i ht d

2

2x Cx Dyy

For left & right edges

x C

For top & bottom edgesx Dy Dy

2

2

2

y Ax Byx

For top & bottom edges

y A x B 0

For all edges

y

30

2

xy Bx Cyx y

For all edges

xy B x C 0y

Page 31: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Timoshenko 2 #3 Plate

0 0Dy

, 0, 0Let 1 MPa/m, 2 m, 1 m

1 , 0, 0

x y xy

x y xy

Dy

D a byx y xyy

31

Page 32: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Timoshenko 3 #1 Plate

For an elastic plate subjected to static loads shown, determine the stress function. The B is a constant.

State the domain & boundary conditionsconditions

32

Page 33: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Timoshenko 3 #2 Plate

4 3 2 2 3 4

12 6 2 6 12A B C D Ex x y x y xy y

4 4 4

4 2 2 42 , 2 , 2A C Ex x y y

4 4 4

4 2 2 42 0

2 4 2 0x x y yA C E 2

2 2

2 4 2 0Thus, (2 )A C E

E C A

2 22

2 2(2 )

x Cx Dxy EyyCx Dxy C A y

22 2

2

2

y Ax Bxy Cyx

B D

33

22 22

2 2xyB Dx Cxy y

x y

Page 34: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Timoshenko 3 #3 Plate

Describe stress distributions

2 2(2 )x Cx Dxy C A y

22 2

2

2

y Ax Bxy Cyx

B D

22 22

2 2xyB Dx Cxy y

x y

Satisfy the condition if

x C 2x D (2xy C A )y

y A 2x Bxy C 2

2 2

y

B x C Dxy 2y

34

22xy x C xy

2y

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35

Example Timoshenko 3 #4 Plate

2

Let 0, 2 MPa0 MPa 2 MPa MPaA C D B

xy x 20 MPa, 2 MPa, MPax y xyxy x

2 MPay xy 2 MPaxy x

35

y

Page 36: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Timoshenko 3 #5 Plate

2 2 2 2

Let 1 MPa, 2 MPa, 3 MPa, 4 MPa3 4 7 MPa 2 3 MPaA B C Dx xy y x xy y

2 2

3 4 7 MPa, 2 3 MPa,

6 2 MPax y

xy

x xy y x xy y

x xy y

36

x y xy

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Semi-Inverse MethodBeam

Procedure Select the stress function by assuming reasonable Select the stress function by assuming reasonable

assumptions Check compatibility Fit in the boundary condition

Advantage Advantage More versatile stress functions

Disadvantage Inaccuracies in regions which assume the St. Venant’s

principle

37

Page 38: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Beam 1#1 Beam

For an elastic plate subjected to static loads shown, determine the stress distribution and displacements. Assume unit thicknessAssume unit thickness.

State the domain & boundary

38

boundary conditions

Page 39: Lecture NoteLecture Note 3 - pioneer.netserv.chula.ac.th

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Example Beam 1 #2Beam

3BAxy xy

4 4 4

4 2 2 4

6

0, 0, 0

y y

x x y y

4 4 4

4 2 2 42 0

x x y y

x x y y

2 2 22

2 2, 0, 2x y xyBBxy A y

x yy x

39

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Example Beam 1 #3Beam

2 2 22

2 2, 0, 2x y xyBBxy A y

x yy x

2

Top and bottom edges BC: 0 at 2xy

x yy xhy

2 2

2

02 4 8B h BhA A

Bh B

2, 0, 8 2

Right edge BC: at 0

x y xy

xy

Bh BBxy y

P dA x

2 3/ 2 2

/ 2( )

8 2 12

y

h

h

Bh B BwhP y wdy BI /B P I

40

2 2 2 22 2

12 3, 0, ( 4 ) ( 4 )8 2x y xy

P Pxy B Pxy h y h yI wh wh

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41

Example Beam 1 #4Beam

The shear force over the free end as the integration of shear gstress

No resultant normal force across the sectionsAll sections including the built in end are free to distort All sections, including the built-in end, are free to distort.

Then, find displacement for plane stress problems

41

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42

Example Beam 1 #5Beam

1 ( ) xx x y

PxyE E EI u

1 ( )

y

xy y x

E E EIPxy

E E EI

x

y

uxv

( )

1

x yz E

y

z

ywz

1

xy xyG

xy

zu vy x

yz

v wz y w u

42

zx

w ux z

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43

Example Beam 1 #6Beam

2

( )Pxy u Pxy Px yu dx f y

2

( )2

( )2

x

y

u dx f yEI x EI EIPxy v Pxy Pxyv dy g xEI y EI EI

2 2

2

( 4 )8xy

EI y EI EIP u vh yGI y x

2 22 2( 4 ) ( ) ( )

8 2 2P Px y Pxyh y f y g xGI y EI x EI

2 2 2 ( )8 2 2Ph Py Px f y PyGI GI EI y

2

2 2 2 2

( )2

( ) ( )

g xEI x

Ph P P f P

43

2 2 2 2( ) ( ) ( ) ( )8 2 2 2Ph Px g x Py f y Py F x G yGI EI x EI y GI

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Example Beam 1 #7Beam

2 2 2 2( ) ( )Ph Px g x Py f y Py

2

8 2 2 2

( ) ( )8

GI EI x EI y GI

G x F y G FPhGI

2

( ) ( )

As const, ( ) const and ( ) const.8

8y

Ph F y F G x GGI

GI

2 2

2

( ) ( )2 2Px g x g x PxG GEI x x EI

2

3

(

(

) ( )2

)

Pxg x G dxEI

Px G C

44

( )6Pg x xE

1G x CI

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Example Beam 1 #8Beam

2 2 2 2( ) ( )Ph Px g x Py f y Py

2

( ) ( )8 2 2 2

( ) ( )

g y y yGI EI x EI y GI

G x F y G FPh

2

( ) ( )

As const, ( ) const and ( ) const.8

8G x F y G F

Ph F y F G x GGI

GI

2 22 2

8( ) ( )

2 2 2 2

GIPy f y Py f y Py PyF FEI y GI y GI EI

2

( ) (2Pyf yGI

2

3 3

)2Py F dyEI

P P

45

3 3

26(

6) Py Py F y C

GI Iy

Ef

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46

Example Beam 1 #9Beam

2 2 3 3

2( )Px y Px y Py Pyu f y F y C

2

2 2 3

1

( )2 2 6 6

( )2 2 6

u f y F y CEI EI GI EI

Pxy Pxy Pxv g x G x CEI EI EI

2

2 2 6Boundary condition at the built-in end:

, 0, 0 0

EI EI EI

x L y u v C

3

1

2 3

6PLC G LEI

d PL PL 2 3

1

2

, 0, 0 , 2 3

dv PL PLx L y G Cdx EI EI

Ph 2 2 2Ph PL PhG F F G

46

8G

8 2 8

G F F GI GI EI GI

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47

Example Beam 1 #10Beam

2 3 3 2 2

( )2 6 6 2 8Px y Py Py PL Phu yEI GI EI EI GI

2 3 2 3

2 6 6 2 8

Ans2 6 2 3

EI GI EI EI GIPxy Px PL x PLvEI EI EI EI

The relationship at 0 can be predicted by simple beam.These are displacements without shear of neutral plane

v y These are displacements without shear of neutral plane.

47

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48

Example Beam 1 #11Beam

2P Ph2 2

2

( 4 ) at NA8 8

Deflections of NA due to shear strain ( )

xy xyP Phh yGI GI

Ph L x

3 2 3 2

0

Deflections of NA due to shear strain ( )8

( ) Ans6 2 3 8y

L xGI

Px PL x PL Phv L xEI EI EI GI

48

6 2 3 8y EI EI EI GI

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49

Example Beam 1 #12Beam

Let 5 m, 1 m, 0.3 m, 0.2 MN,

200 GPa 0 3

L h w PEE G

2 2

200 GPa, 0.3, 2(1 )

, 0, ( 4 )81x y xy

E G

Px Py h yI 81x y xyI

x

xy

49