能量释放率理论

Chapter Three Energy release rate theory

Energy release rate G is another important concept in fracture mechanics. There is a relationship between G and K. One is known. Another can also be known. Sometime, it is difficult to obtain K factor. However, it is perhaps easy to calculate G .

§3-1 Concept of energy release rate

1. Basic concepts (1) Strain energy

Work done by the external force is changed to the strain energy stored in the elastic deformation. The strain energy can be released and the elastic deformation then disappears.

Strain energy density w : strain energy per unit volume.

ij ij ij ij ij d w εσ=εσ=?ε2

10*

*, (for linear elastic body) Total strain energy U (Internal force potential): total energy stored in the volume V

?=V

wdV U

(2) External force potential U P : the negative value of virtual work done by the external force. Assume that the body force is i B and the surface force is i T on the stress boundary. The external force potential P U is

)(??σ

+-=S i i V

i i P dS u T dV u B U

(3) Total potential ∏:

)(???σ

+-=+=∏S i i V

i i V

P dS u T dV u B wdV U U

2. Energy release rate G

The body force i B , the surface force i T on σS and the displacement i u on u S are given. Assume that the crack size is changed from a to a +?a . Accordingly, the

displacement, strain, stress, stain energy density, internal force potential, external force potential and total potential are also changed. The total potential ∏ is changed to

?∏+∏. ?∏ is the increment caused by the crack growth a ?. Assume that the plate

thickness is t . a t S ??=? denotes the single surface area increment. The energy release rate G is defined as

dS

d S G a ∏

-=??∏-=→?0lim

If the plate thickness t is a constant, the energy release rate G is

da

d t a t G a ∏

-=???∏-=→?1lim

3. Constant force and constant displacement conditions

(1) Constant force condition

A plate with a crack is applied by a constant force F as shown in Fig. 3.1(b). The external force virtual work W and external force potential U P , respectively, are

δF W =, δF W U P -=-=

The total strain energy U (internal force potential) is

δF U 2

1

=

Then, the total potential ∏ is

W U U U P -=+=∏

Now, U W 2=，U -=∏. The energy release rate G under the constant force condition can be written as

F S

U

S G )(??=??-

=∏，for constant F . In this case, 0>?U , the strain energy in the body in fact increases rather than releases with the crack growth. G can not be called as the strain energy release rate. (2) Constant displacement condition

After a displacement δ occurs, the plate is clamped. This is the constant displacement condition as shown in Fig. 3.1(c). In this case, there is

W ＝0，U W U =-=∏

δ∏)(S

U S G ??-=??-

=, for constant displacement δ. It is seen that only for the constant displacement condition, G can be called as the strain energy release rate. Since W =0, the energy needed by the crack growth comes from the release of strain energy stored in the body. That is the strain energy stored in the body decreases with the crack growth. Foe the constant force condition, the increment of external work is δFd dW = in which a part is used to increase strain energy dU while the other part is used for crack growth.

However, the values of G for two cases are equal. Constant force case:

2212C F F U ==δ，S

C

F S U

G F ??=??=2)(2. C is the compliance of the plate.

Constant displacement case:

C F U 2212δδ==，S

C P S C C C S S E G ??=??--=??-=??-=212)12()(2222δδδ

It is seen that the values of G for two cases are equal.

§3-2 Relation between G and K

The relationship between G and K is one of the most important relations in fracture mechanics.

1. Mode I crack

By consideration of stress and energy fields ahead of the crack tip, the relation

between K and G can be established.

Assume that a segment of crack length is closure. To this end, a distributed stress σyy is applied. This distributed stress σyy is the stress field in the vicinity in the crack tip for

θ=0, r=x , i.e.

x

K I yy πσ2=

In addition, for the plane strain case, the displacement produced in the crack closure process can be known from the asymptotic solution of displacement field in the vicinity of the crack tip. The point o’ is taken as the original point. For πθ=, x a r -=, the displacement along y-axis is

π

ννπν2)1(4)1(221)(2x

a K E r K E x v I

I --=-+= Assume that the plate thickness is 1. The applied force F in the area dx ?1 is

dx F y ??=1σ

As shown in Fig. 3.2, the total crack closure displacement is v 2=δ. Since the actual work done by the external force σy in the length dx is equal to the strain energy, there is

dx v v dx F U yy yy ?=???==

σσδ212

1

21 The crack closure length is a so that the change of system strain energy is

?a

y

vdx 0

σ

The change of system strain energy by closing the unit crack area is

??=??a

y vdx a

S U 011σ On the other hand, the crack growth case is similar to the crack closure case. The energy for crack growth is equal to the energy for crack closure. Therefore, in the case of constant displacement, the dissipative energy for the unit area increment of crack area, i.e. G I , can also be written as

?=??-

=a

y I vdx a a U G 0

1σ The expression of y σ and v can be inserted into the above eq. to give

22

2

2

0201212)1(4211I a I a I I a y I K E

dx x x a a K E dx

x a K E x K a vdx a G ν

π

νπνπσ-=--=--==?

??

i.e.

2

21I I K E

G ν-=, for plane strain case

For the plane stress case, there is

2

1I I K E

G =

, for plane stress case. 2. Mode II crack

For the mode II crack, the relative slide displacement of crack surfaces along the x -axis by the shear stress xy τ is 2u . The energy release rate G II is

?=a

xy II udx a G 0

1τ

The shear stress τxy and displacement u are respectively known as

x

K II xy πτ2=

, 0=θ, x r =

π

ν2)1(42x

a K E u II

--=, 0=θ, x a r -= We have

2

20

22121II a II

II K E

dx x x a a K E G νπ

ν-=--=?

, for plane strain case 2

1II II K E

G =

, for plane stress case. 3. Mode III crack

For the mode III crack, the relative displacement by the anti-plane shear stress zy τ along the z -axis is 2w . The energy release rate G III can also be written as

?=

a

zy III wdx a

G 01τ The anti-plane shear stress zy τ and anti-plane displacement w are known as

x

K III

zy πτ2=

, 0=θ, x r = π

ν2)1(4x

a K E w III

-+=

, 0=θ, x a r -= We have

2

012)1(421III a III III III

K E

dx x a K E x K a G νπνπ+=-+=? For the complex crack, K I ≠0, K II ≠0, K III ≠0, the total energy release rate is

22221)(1III II I I K E

K K E G νν+++-=

§3-3 Bi-cantilever beam problem

It is a mode I crack problem.

1. Long crack case: h l >>

When h l >>, the crack segment is equivalent to a cantilever beam.

EI

Fl d 3223

==δ, 3121th I = (rectangular section)

3

3

8Eth

Fl d = Fd U 21=, Fd U P -=, 3

3

2421Eth l F Fd U U P -=-=+=∏

3

22

2121h

Et l F dl d t G =-=∏ For the plane stress problem, there is

21I K E

G =

The stress intensity factor can be known as

2

/33

2th Fl

K I =

2. Short crack case:

In this case, the crack segment is equivalent to a dumpy beam (短粗梁). The shear deformation of the crack segment and the deformation in the uncracked segment must be considered. The displacement d can be divided into three parts.

321d d d d ++=

d1 is produced by bend and d 2 by shear. It is known that

31l d ∝, 22l d ∝

d 3 is th

e body displacement induced by the deformation o

f the uncracked segment.

l d ∝3

l l l d 32231

ααα'+'+'= A normalized compliance coefficient λ is introduced.

)

()()()()()()(322313222331

h

l h l h l h

l

h F tE h l h F tE h l h F tE tE F d ααααααλ++='+'+'==

The coefficients 1α, 2α and 3α can be determined by theory or experiment. If the normalized compliance λ is known, the displacement d can be determined.

)]()()([32231h

l h l h l tE F d ααα++=

Total potential of the system is

)]()()([2121122312h

l h l h l tE F Fd U ααα∏++-=-=-=

The energy release rate is

)]()(2)(3[2113

223122h

l h l h l l Et F dl d t G ααα∏++=-= For the plane stress problem, the stress intensity factor can be determined from

21I K E

G =

that

21

32231)]()(2)(3[21h l h l h l t F l K I ααα++=

3. Determination of α1, α2 and α3 by experiment

There are three specimens. The crack lengths are 1l , 2l and 3l respectively. Three F versus d curves can be obtained by test. F and d are the generalized force and displacement. In light of

)()()()(32231h

l

h l h l tE F d αααλ++==

we can have three equations

)()()(1321231111h l h l h l F d Et

ααα++= )()()(2322232122h l h l h l F d Et

ααα++= )()()(3323233133h

l h l h l F d Et

ααα++=

The coefficients 1α, 2α and 3α now can be known. Then, we calculate the energy release rate G and the stress intensity factor K .

§3-4 Determination of SIF by experiment

The method in the last section can be applied to the general case.

Consider a Mode I crack problem. The crack can be a central or edge crack. F and d denote the generalized force and displacement respectively. For the different crack sizes

1a , 2a , …, n a , we can obtain a group of F versus d straight lines by experiment.

For a constant applied force F , we can have the displacement values 1d , 2d , …, n d . The total potential can be calculated for different crack sizes.

i i Fd 2

1-=∏, for i a , n i ,,2,1 =

Finally, a group of data (∏i , a i ) is obtained. The ∏i versus a i can be plotted.

For the Mode I and plane stress problem, in view of

2

11I K E

da d t G =∏-

= the stress intensity factor I K is

)1(da

d t E K I ∏

-= where da

d ∏

-

is the slope at the point a . Another treatment:

Assume that the compliance λ(a ) of the system depends on the crack size a . The generalized displacement q and force F are related by λ as

F a d )(λ=

)(2

121

2a F Fd λ∏-=-=

When F =constant, the derivative is

da

a d F da d )

(212λ∏-= da

a d t E F

da d t E K I )

(2)1(λ∏=-

= For a constant applied force F , we can have the displacement values 1d , 2d , …, n d from Fig. 3-8. By using F d i i /=λ, a group of data ),(i i a λ can be obtained. The λ versus a curve can be depicted. By using the slope at the point a , we can calculate the stress intensity factor for the crack size a and the applied load F .

§3-5 An infinite plate with a central crack under uniaxial tension

This problem has been solved in chapter two. Now we resolve it by the energy approach.

Case (a) can be divided into case (b) plus case (c).

)()()(c I b I a I K K K +=

Case (b) is the same as no crack case, 0)(=b I K , such that

)()(c I a I K K =

Case (a) is equal to case (c).

1. Solution for case (c)

For the case (c), assume that the displacement distribution on the upper crack surface is an ellipse.

1)()(

220=+a

v The relation r a x -= is substituted to give

2

2

22222021)(1)(1)(a r ar a a r a a x v v +--

=--=-= When the distance r is very small, r 2 can be ignored. On the crack surface near the crack tip, the displacement distribution is a parabola.

a

r

v v 20

= It has been known in chapter two that the displacement asymptotic solution is

)2

cos 21(2sin 222θ

-+κθπμ=

r K v I

On the upper crack surface, π=θ, we have

)1(22+κπ

μ=r

K v I a

r

v r K I 2)1(220

=+κπμ

For plane stress, ν+ν-=κ13, )

1(2ν+=μE

, the stress intensity factor can be obtained that

a

v E K I π

=0

2 and

20241v a

E K E G I π==

The displacement 0v remains determined.

2. The expression of energy release rate G

In what follows, the expression of energy release rate G is derived from its definition. Then, the displacement 0v can be determined and the solution for the problem can be

The system total potential is

]2

1[40?σ-

=-=∏a

tdx v U 20)(1a

x

v v -=

002

20

2

12tav dx x a a tv a σπ-=-σ-=∏? Note that 0v is related to a .

)(2

0a tav σπ

-=∏

)(200da

dv a v t da d +σπ

-=∏ )(421)2(00da

dv a v da d t ta d d dS d G +σπ

=∏-=∏-=∏-

= The early result is

2

04v a

E G π=

A differential equation can be obtained that

2

0200v a

E a v da dv σ=+, Bernoulli equation The general form of the Bernoulli equation is

n y x q y x p y )()(=+'

The solution can be found in the mathematic handbook. The result is

C

a E a

v 2022σ+σ=

C is an integral constant. When ∞→a , the displacement ∞→0v . This requires that C =0.

E

a

v σ=

20 a a

v E K I πσ=π=

02 Finally, the solution is

a K I πσ=, 20)(1a

x

v v -=, E a v σ=20 It is identical to the early result in chapter two.

§3-6 Mode I crack in the general case solved by energy approach

1. Symmetric case

For this case, it is difficult for the direct analytical method. The case (b) is the same as no crack case.

)()(c I a I K K =

Case (c) can also be divided into case (1) and case (2). Case (1) is the uniform applied loading case.

21)()(I I c I a I K K K K +==

1I K has been known. The problem is to solve the case (2). The energy release rate G is written as

2112122212112211212

)2(1)(11I I I I I I I I I K K E G G K K K K E K K E K E G ++=++=+==

2

1111I K E G =

, 221

21I K E G = In addition, the energy release rate G can be derived from the definition.

???????-∏+∏=+---=++-=-=-=∏a

a

a

a

a

a a

tdx

v p tdx v p tdx v p tdx v p tdx v p tdx v v p p pvtdx U 0

1221 theorem

Betti equal,0120210220110212104)22(22))((2)2

1(

4

)

(2)

4(21210

122101221??++=--+=∏-=∏-=a a dx v p da d

G G tdx v p da

d

t G G da d t dS d G

Comparison with

211

212

I I K K E G G G +

+= results

)(22012211?=a

I I dx v p da d K K E ])([)(0121211012112??+==

a I a I I dx da

dv

p a v p K E dx v p da d K E K ,

0)(1=a v , the displacement at the crack tip vanishes.

?=

a I I dx da

dv p K E K 01

2112, Conclusion: the stress intensity factor 2I K can be determined from the stress intensity factor 1I K for another case that has been known.

Example:

The solution for case (1) has been known.

a p K I π=11, 221

1

12x a E p v -=

?=a I I dx da dv p K E K 01

3113

, ??

?≠==b x b x p x p ,0 ,)(3 2

2

2211111

1132)2(b

a a a

p b a E p da d p

a

p E da

dv p K E K b

x I I -π=-π=

=

=

The stress intensity factor 2I K can be derived from case (3) by integral operation.

2

2

222)(b

a a a

dx x p dK I -π=

??-π=-π=a a I x a dx

x p a a b a db b p a a K 02

2202222

)(2)(2 If the distribution )(2x p is known, the SIF 2I K can be determined. Therefore, we can know the stress intensity factor )(a I K for case (a).

21)()(I I c I a I K K K K +==

2. General case: asymmetric tension

21)()(I I c I a I K K K K +==

In the same way, we can derive that

???

=

12),,()(2111

2

112l l I I dx l l x v l x p K E K

Know that

a p K I π=11, 221

112

),(x a E p x a v -=, )(21

21l l a -=, 12l x l ≤≤

???

=

12),,()(21113113l l I I dx l l x v l x p K E K , ???++≠++==b l a x b l a x p x p 223,0,)( b

a b

a a p

l l x v l p K E K b l a x I I -+π=??=

++=2),,(2111113 b a b

a a

db b p dK I -+π=

)(22 By integration, we have

?

+--+π=

a a

I db b

a b

a b p a

K )

(122 第三章完