governing equation chapter 5

Chapter 5General Formulation for 2-D elasticity problems

5.1 Fundamental equation elasticity problems

� �,ji j i if u� V U

� �, ,

,

b) Strain-displacement relation1 2

c) Compatibility of strain ------------------------------------6 equations 0

ij j i i j

ij ipr jqs rs pq

e u u

R e e e

½ � °°¾° °¿

� �

*

* *

1 2 1

----------------6 equationsor 2

1 2

ij ij kk ij ij

ij ij ij kk kk ij

ve ev

ve e e ev

V V GP

V P G

½ ½ � �® ¾ °�¯ ¿ °¾

½° � � �® ¾°�¯ ¿¿

a) Equilibrium equation.

----------------------------------------3 equations

d) Constitutive equations

สรปุ Governing Equation

↳%ดุ , + ดุะ tf =ผู่งุ๋ ( 3 สมการ)

Itefuation15 mknown

ea = µµ + วฺหุ) = หิ๋ '

, Gij ⇒ 6 ตำ

eij ⇒ 6 ตัวyli23=1 p E. ข

|ไ

-

c t¥หู๋ P % =

_

แตะกาย""

= o Thhtghnrrrs ✗i.µ gtrain

retetdimn yinelastic us ห< Lstrain

Solution methods: 2 approaches

1. Displacement formulation : Substitute (b) into (d2) then into (a)

2. Stress formulation : assume (statics problem)

substitute (d1) into (c). So, there are 9 eq. for 6

unknowns of stress

� � 0iu U

ijV

� �

� �

� �

,

, ,

,

*

* *

( )

1( ) 2

( ) 0

1( 1) 2 1

( 2) 21 2

ji j i i

ij j i i j

ij ipr jqs rs pq

ij ij kk ij ij

ij ij ij kk kk ij

a f u

b e u u

c R e e e

vd e ev

vd e e e ev

V U

V V GP

V P G

�

�

½ � �® ¾�¯ ¿ ½ � � �® ¾�¯ ¿

แนวคดิในการแกปั้ญหา

}bi = ะแ

[T-

}Gij = fcui )

f(Ui ) = ° คะ =µ {a} }

Tef(d) tef (a)fcGij ) = 0

5.2 Plane strain problems

Plane perpendicular to x3

x1 x2

x3

, 0ji j if� V(a) Equilibrium equation

11,1 21,2 31,3 11, 0i fV V V � � �

12,1 22,2 32,3 22, 0i fV V V � � �

13,1 23,2 33,3 33, 0i fV V V � � �

So,

ปัญหา � มติแิกย้าก

simplify เป็นปัญหาในระนาบStress-strain relationship

สาํหรบัปัญหา plane strain① U3=0 => 033=0

② ดุตา ,น 4,4 = fncn.

✗2) Lปูนi!-ญิ๋ญึ๋

③ ดุ /= 0,533 = fncy , ×2)\ง ×.io ④ % = 0

, f ,£ =1 fท์×3)⑤ ยิ๋3

= ผั๋ = 0g ผิ๋.อหํ๋

, 9ยุ๋ยุ๋ =/ fnc✗3)

43 = {ญู๊ำผู๋งุ้ ⇒ 93=0 ⇒ 5,3=0

/0

023 = { CYำ+%ำ ⇒ emo ⇒ นะ 0

f°

11°

/°

/°

fตาดุ2) 522

Gpa,ptfa = ° 2,13=1,2 2 equatias


Hooke’s law :

So,

*12 1ij ij kk ij ij

ve ev

V V GP ½ � �® ¾�¯ ¿

Strain in the x3 are not present, so � � *33 33 11 22 33 33

1 02 1

ve ev

V V V VP ½ � � � � ® ¾�¯ ¿

That is : � � *33 33 11 22 33 33

1 02 1

ve ev

V V V VP ½ � � � � ® ¾�¯ ¿

33V

From Hooke’s law : � � *11 11 11 22 33

12 1 ij

ve ev

V V V VP ½ � � � �® ¾�¯ ¿

11H

Stress-strain relationship

สาํหรบัปัญหา plane strain

V19HE} -2C 1 +บ)µ ผั๋/ eyp ⇒ ดุ p = 1,20

tu {ดาบ (คแ +ก } + ผํ๋ +ขยุ๋


From Hooke’s law : � � *22 22 11 22 33

12 1 ij

ve ev

V V V VP ½ � � � �® ¾�¯ ¿

22H

Similarly: 12H

Thus: � � * *33

12

e v e veDE DE JJ DE DE DEV V G GP

� � �


สาํหรบัปัญหา plane strain

ย {ษแกดู2) } + eหิ๋ + บผํ๋

lgญิ่ด te

2,13=1,2 30ยู้


Compatibility of strain :

For ij =11:

Similarly: R22 =R12 =R13 =R23 =0 automatically

, 0ij ipr jqs rs pqR e e e

11 123 123 33,22 132 132 22,33 123 132 32,23 132 123 23,32

33,22 22,33 23,32 2 0

R e e e e e e e e e e e ee e e

� � �

� �

are not zero, automatically.

33 22,11 11,22 12,122R e e e � �Only

22,11 11,22 12,122 0e e e� � So, the compatibility of strain reduced to

� �, ,1 , , 1,22

e u uDE D E E D D E � Strain-displacement relation :

*****There are 5 basic equations (in the red boxes).******

Compatibility of strainและ Strain-disp.

relationshipสาํหรบัปัญหา plane strain

Ij = 1,43

ณื๋"

0 Yo Yo Rij = o ⇒ 6 eq"

lrrไกง Rฑั้°☐

/emfnfn


Compatibility of strain 22,11 11,22 12,122 0e e e� �

� �, ,1 , , 1,22

e u uDE D E E D D E � Strain-displacement relation

There are 5 basic equations (in the red boxes).

Governing equations สาํหรบัปัญหา plane strain

� � * *33

12


� � �

� � � � *33 11 22 332 1v v eV V V P � � �

, 0, , 1, 2fED E DV D E� Equilibrium equation

Constitutive equations

ะ×ะ ¥หุงญุ๋วุ๋ญ๊ยิ๋

๔11,1 t

๔21,2tf

,= 0

en งาน =ญฺว×2ฑุ +ณื้ +f. = o

e☒= inelasticstrainเ

↳€:

็

ปาโน เคย

GDTfBB.EELAT

E"

=p#} คะ 0

5.3 Plane stress problems

x1 x2

x3

P P

, 0fED D DV � Equilibrium equation : same as that of the plane strain problem

Plane stress = ม ีstress ในระนาบเทา่นนัหรอื 3 0iV

11,1 21,2 31,3 1

12,1 22,2 32,3 2

13,1 23,2 33,3 3

000

fff

V V V

V V V

V V V

� � �

� � �

� � �

Equilibrium equation for 3D problem

① ด33= 0g ๔13 = 0,52} =0

② หุ , Uz = fncx, , หู)ะ . ey ande 23 = 0 ande33 /= 0 C33 =%4,42)

ตุ , ,ดุ2 ,ม = fn CY , หะ)

p"""

☐ ญื๊.

☒อ. . ฎื่.

ญื๊ตั๊ตณื๊


x1 x2

x3

P P

Compatibility of strain :

22 231 231 11,33 213 213 33,11 231 213 13,31 213 231 31,13

11,33 33,11 13,31 2R e e e e e e e e e e e e

e e e � � �

� �

are not zero, automatically. R11, R12 and R33 are not zero automatically either.22RSo,

Ignore R11, R12 and R22

Keep only, 33 0R

Compatibility of strain ในปัญหา Plane stress

๖0 b#

น R22 = 033,11 = 0

✓ ✓ ✓

Rแญื๋t" plmestress # exactRn = 0 Solทฺ

R2นะ 0

งอไT R33 = 0


Substitute in e33, we get

Alternatively: � �* *21 2ij ij ij kk kk ij

ve e e ev

V P G ½ � � �® ¾�¯ ¿

33 0V and

� �* * * *33 33 33 11 22 33 11 22 332 0

1 2ve e e e e e e e

vV P ½ � � � � � � � ® ¾

�¯ ¿That is

Rewrite � � � �* * *33 11 22 33 11 221 1

v ve e e e e ev v

� � � � �

� �

Hooke’s law for plane stress : *12 1

ve evDE DE JJ DE DEV V G

P§ · � �¨ ¸�© ¹

Since : for plane stress problem 33 31 32, , and V V V

From Hooke’s law in term of stress shown above : * 12 1

ve evDD DD DD JJ DDV V G

P§ ·� �¨ ¸�© ¹

That is � �*11 11 11 11 22

12 1

ve ev

V V VP ½� � �® ¾

�¯ ¿� �*

22 22 22 11 221

2 1ve e

vV V V

P ½� � �® ¾

�¯ ¿and

� � � � *33 11 22 332 1

ve ev

V VP�

� ��


สาํหรบัปัญหา plane stress=° ได้30 b

→ 011,022,92@ 13

ะ C23

= 0 C33 =/ 0-

วาง↳ = tspl

= า¥ Ceii ฑํ๋ + ยeษิ๋ ) +ผุ๋

.LI

Summary of 2-D fundamental equations

� �, ,12

e u uDE D E E D �

33 22,11 11,22 12,122 0R e e e � �

, 0fED E DV �

� � * *33

31 , 3 4 , for plane strain2 4

e e e v vDE DE JJ DE DE DE

NV V G K G N K

P�§ ·

� � � � ¨ ¸© ¹

� � � �� * * *33

1 32 3 4 , , 0 for plane stress2 1 1

ve e e e evDE DE DE JJ JJ DEV P N K G N K

N ½ �° °ª º � � � � � ® ¾¬ ¼� �° °¯ ¿

Compatibility of strain

Strain-displacement relation

Equilibrium equation

Constitutive equations

� � * *33

12


� � �

� � � � *33 11 22 332 1v v eV V V P � � �

*12 1

ve evDE DE JJ DE DEV V G

P§ · � �¨ ¸�© ¹

� � � � *33 11 22 332 1

ve ev

V VP�

� ��

Unified governing equations สาํหรบั

ปัญหา plane stress & plane strain

/

/

✓

v0ง .

/ nwnifiedcmstitiutiveeqh

Compatibility in term of stress� � � � � �* *

22 22 11 22 22 33

32 2

4e e e

NP V V V P K

� � � � �

� � � � � �* *11 11 11 22 11 33

32 2

4e e e

NP V V V P K

� � � � �

*12 12 122 2e eP V P �

33 22,11 11,22 12,122 0R e e e � �

� � � � � ��

* *33 22,11 11,11 22,11 22,11 33,11

* * *11,22 11,22 22,22 11,22 33,22 12,12 12,12

32

43

2 2 2 04

R e e

e e e

NV V V P K

NV V V P K V P

� � � � �

��

� � � � � �2 * * * 2 *22,11 11,22 12,12 11 22 22,11 11,22 12,12 33

32 2 2 2 0

4e e e e

NV V V V V P PK

��

พยายามรวมสมการทงัหมดเขา้ดว้ยกนั

mified Hooke} lw diff บห ,rtimes

=7 022,11

| หื๋ " """""

}

straincompzbtcgmdgetsimplify .

⇒

� � � � � �2 * * * 2 *22,11 11,22 12,12 11 22 22,11 11,22 12,12 33

32 2 2 2 0

4e e e e

NV V V V V P PK

��

21,12 11,11 1,1fV V � �

12,12 22,22 2,2fV V � �

(1)

(a)

(b)

(a)+(b) : 12,12 22,22 11,11 2,2 1,12 f fV V V� � � � (2)

Subs (2) into (1) : � � � �2 2 2 * 2 *11 22 1,1 2,2 11 22 33 33

32 2 0

4f f R e

NV V V V P PK

��

Simplify:� � � �^ `2 * 2 *

, 33 334 21

f R eDD D DV P KN�

� � � ��

If no body force and no inelastic strain: 2 0DDV�

Compatibility in term of stressพยายามรวมสมการทงัหมดเขา้ดว้ยกนั

☐

ดุµ + ญµ + f =0 ⇒ ดุแ +ดูµ2 +§ ,= o (a)

๔12,1 t 522,2 t { = 0 ⇒ ดุ2,124ดุ22 +§2 = o (b)

⇒

อึ๊ Oinekอื๋ strain"

⇒ Laplaa operatorขั้ = ฑํ๋ +ฑื่ฑั่

ขั้น = วง⇒ + ฒู +ญฺ + ม้ญื๋ = 0 •0.ruคะ×แ×2)

\ม(×, ×2)

Airy stress function

211,11 11,22 22,11 22,22DDV V V V V� � � �

2211 11 2222 22 1111 11 1122 22 , , , , , , , ,V V V V ) � �) � �) � �) �

� �4 2

1111 1122 2222 11 22

2

, 2 , , 2 , ,V

V V� ) �

) � ) �) � �

4 2 2 V � )� �

Introduction a function such that stresses are defined as � �1 2,x x) )

11 22, VV ) �

22 11, VV ) �

12 12,V �)

Do they satisfied the equilibrium equation? …………………..

If the defined stresses will be a solution of the elastic problem, they must satisfy the compatibility condition

นิยาม stress functionGn =ฐู่ + Vf = - V, 1

AAIrystressfmctim £ =-บุ 2

→

ดุµ +ดุ 2 tf = o ⇒ 0 +Y -ยู้µ-Y = 0

ำ 1 +ำะ +£ = o ⇒ -_# + gzyg_y= ๐} satstied

Cgantmatiaay -

ง่ด =% {fay +µ 1ณํ๋ +ทุย้ผํ๋ )}

ยัง =0↳ ¢ , 2211 +" "

2eปไไกื๋mmicopemtor224ขั๋= ญํ๋µ + ฑุ๋ว×ข้ + ศั 4๋o

Airy stress function

� � � �^ `4 2 2 * 2 *33 33

42 21

V V R eP KN�

� ) � � ��

� �� 4 2 2 * 2 *

33 33

2 14 21 4

V V R eN

P KN

� ½�� ) �� ® ¾

� ¯ ¿

� �� 4 2 * 2 *

33 33

14 21 2

V R eN

P KN

� ½�� ) � � � �® ¾

� ¯ ¿

� � � � � �^ `4 2 * 2 *33 33

2 1 41

V R eN P KN�

� ) � � � � ��

So, the problem is reduced to finding ) that satisfies the boundary conditions

If no body force and no inelastic strain

นิยาม stress function,

ปัญหาลดเหลอืการหา

stress function ท ีsatisfy

boundary conditions

ไม่ใช่ sdwtionofelasticityproblemนุ๋ = หุ้หุ

⇐ pbiharmonicfmctimข๋¢ = o

Bomdary Condition

µµµ _

④ Fh ⇒ าyะ0Zxy

= 0

ไtyjmnggwc Nlm

) @ ห=L ⇒๔×× = 0

Zxy =0

tnrynnn→ ×

④ ห= 0 U,= 0

U2=0ไv1 เเนาว์วะ

" เท้า |¢

{ .

☐<¥]-% ✗

II.→%×

@ y = ihandgn < L ⇒ Gyy = Number

⇒ zxy = 0

⑨ y =-1h andอะแว ⇒ Gyy = o

Zxg = 0

Uniqueness of elastic solution

11,1 21,2 31,2 1 0fV V Vc c c� � �

� � � � � �211 22 1,1 2,2

41

f fV VN�c c� � ��

� �1 11 1 21 2 31 3

nT n n nV V Vc c c � �

11,1 21,2 31,2 1 0fV V Vcc cc cc� � �

� � � � � �211 22 1,1 2,2

41

f fV VN�cc cc� � ��

� �1 11 1 21 2 31 3

nT n n nV V Vcc cc cc � �

� Airy stress function: If stresses from the stress function match stresses of the problem on the boundaries, that stress function will be a solution of the problem.� However, we must assure that, for a given stress condition on the boundary, there is only a set of stresses that satisfies the governing equation.

� �1T , surface traction

� �3T

� �4T

� �2T

f , body force

คาํตอบของสมการ

ทงัหลายมคีาํตอบเดยีว

"Gij

|§

set ± ⇒ Gij§

Set # ⇒ ดุj"

0 0

✓

set I - Set I

±ตุ๊ด ,

"

) + ฒิ๊ - คะ"

า + ฑึ๋ว่า"

ง = o ⇐ ดุj =ดู"

¢

Thercanbeonlyonesetfsolevtim

Example: Uniform stress field in a rectangular plate

Consider

c

a

-b x1

x2

2 2

2 2a cx bxy y) � �

4¢ = 0 ⇒ stressfmction

ศื๋4 + มญื๋หิ๋ญื๋ = o ⇒§ isastressfmctias

Nobodyfora .

Gxx = อุ๋ 22+ฟู๋ =L

Gyy = อู๋µ = d

Gxy = -อุ๋ 12 =- b

Example: consider 3 2 2 3

6 2 2 6a b c dx x y xy y) � � �

Case 1: only d z 0 Î pure bending

x

y

x

y

Case 2: only a z 0 Î pure bending by normal stresses on y =

/ /4¢ = o ⇒ 0k

ต = ¥ = Cแ + dyด2 ะหื่ = au + by ,

ดุ2 = -ฅู =- bn - cy

¢ =ฎู่☒⇐ ฐํ๋ด = dy

¢ = หูห้nit

%#

Example: consider 3 2 2 3

6 2 2 6a b c dx x y xy y) � � �

Case 3: only b z 0 Î pure bending

x

y

22,xx cx dyV ) �

11,yy ax byV ) �

12,xy bx cyW �) � �

Gxx = 0

Gyy = byGxy = -bn

mnnnmmrnormalandshearstress

ฎู๋ktlttgry = - bc

เ

II"

ht-j.tt#tt_!

Example: Using to determine stress distribution in a square

shown below

2 31 2C x C y) �

VR

VR

VR

VR

x

y

a

a

BE Gyy (กอ) = -ดู , Gyycy a) = -Go

Zxy C 1,0) = 0 Zxycn, a) ะ 0

Gxx C 0,4 ) = ¥ ด×Ca, g)

= ด9g Ey -- -

←ณํ๊

= เอ

¢ = 4ห้+ dzy3⇒ ดู ×

= §, yy ะ 6 dsy= 10

Gyy = ¢, ×× ะ 24Gij C } 3)

Txy =

y= 0

Apply B- C. :(yyc "เอา = -8 ⇒ 24 =-ดู ⇒ 4 = ¥

Gxx เอง 4) = ¥ ⇒ Gdzy = # ⇒ dนะยะ. ¢ = -ญฺหิ้ญุ๊y 3 ⇒ ด×=ๆ

"" เ า

, Gyy =- Go

, Zxy = 0 AE

Example: A cantilever beam is load with shear force P at the free end, determine the stress

distributions.

G = ญูง Z = ¥อู

¢ hastsatisfy v4¢ = o13¥ (6×g) y = ±h= ° ฏื๋tญื่ญื๋4 "ฑิ๋yะ + ญื๋ = °(G) y = ±h = 0

P = -fcxytdy ฎํ๋หื ,๋+ ศื .๋" = o

-h

ึ """" "mt """ """ "" " µสั๋หํ๋"" " หื๋หิ๊ "

- ๔×× atany sectimdependony ะ tix ) = Cหู้+↳ห้+ cuietds

andfglx) = Cittd >ห้+ Cg ห tdgi. we can assumc ดู = ว้อุ่

Iy Heyi. ¢ = ยู่ uy

}+ y (d.ห้+↳ห้+14ktG)

Int.

twice 24Iy= 4หู๋ + t.cn)

+ แห่+ C>ห้+ cgktdg¢ = 4หฺ

3

+ yf.cm +£a)

Example: A cantilever beam is load with shear force P at the free end, determine the stress

distributions. 45,48,4 ⇒ tririalsd

ทั้ ¥ cyEh}h

+ µµµผู๋iy

→%"" ¥น ⇒ ยนำ"ฏื๋

¢ = # uy}+yyiy ห้+µ +ญื๋ APII-B.CH%×gtdy =ฏุ่ษื๋ cyihdy

- h

I = §thyahny ⇒ . . =ฏิ ÷๊Hpply_B.ci Gy =¥ = 6diey +2dg y + แ6k +2dg ะ -๔×=-ญื๋

:< Cz = dg = d เ = Cy = ° Gy = 0

Zxycy = b) = 0 : Zxy =-

#2- 3นู๊ห้ -2ญํ๋-Cy Zxy = ญั๋ Chtyy-¥ท้- dy = o ⇒ C4 =¥ hh

Example: Investigate what problem of plane stress is satisfied by the stress function

applied to the region included in y = 0, y = d, x = 0 on the side x positive.3

22

34 3 2F xy pxy yd dª º

) � �« »¬ ¼

y =-d, y

-

- d,w -- o andn =L

Example: Stress function in form of a Fourier series

Additional paper : Elasticity Based Stress Analysis under Arbitrary Load using Fourier Series

'Et¥

governing equation chapter 5

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