Marder M.P. Condensed Matter Physics

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Cracks

403

where φ is analytic, and ζ = x + iy.

The asymptotic behavior of φ is easy to determine. Far from the hole, the stress

goes to a constant value Σμ, where Σ is dimensionless. So

du μ

σ,

= μ~ = ^

[ίφ'(χ

iy)

- ίφ'{χ +

iy)]

(14.80)

=>0'(C)->-ffifOrC->oc. (14.81)

How does the presence of the hole affect the stress field? Because the edges of

the hole are free, all stresses normal to the edge must vanish. Let

be a variable

that parametrizes the edge of the hole, so that

(x(t),y(t)) (14.82)

travels around the boundary of the hole as t moves along the real axis. Then

dx dy\ -. f dy dx\ _ _

—, — ] and N = I

—

— , — I It is easy to verify that N

■

T =0, and because

dt dt J \ Ot Ot J f

is tangent

must

be „ormai.

(14.83)

are tangent vector and normal vectors along the edge of the hole, so requiring

normal stress to vanish means that

(σ,

,σ

,)·;ν = 0

(du du\ ( dy dx\ dudx du dy

\dx'

V dt' dt J dy dt dx dt

( άφ δφ\ dx ίάφ άφ\ dy

— -J — = —L- L- -L Insert (14.79) for«.

\ dix dix) dt \diy diy

θφ dò

—— = —— This equality holds only on the boundary.

dt dt

(14.84)

(14.85)

(14.86)

(14.87)

=> φ(£) = φ(ζ) Because φ is arbitrary up to a constant any- (14.88)

way, there is no worry about dropping a con-

stant of integration.

when ζ lies on the boundary.

Equation (14.88) appears innocent, but is in fact a powerful relation that can

lead to rapid solution of complicated boundary-value problems. Suppose, to be

definite, that the hole is elliptical and thus can be described by

(" = io -\- — p is a number lying between 0 and 1. (14.89)

Lü'

with ω lying on the unit circle,

ω = ε

ίθ

, (14.90)

and

real. When p = 0, the boundary is circular, and when p= 1, the boundary is

a cut along the real axis. Considering φ as a function of

ω,

one has

_ ] The notation φ(ω) means that if φ(ω) is ex-

φ(ΐθ) = Φ(ί-ϋ) = Φ( ) panded as a power series in ω, one should Π4 91)

CO take the complex conjugate of all the

coef-

ficients in the expansion.

404

Chapter 14. Dislocations and Cracks

because ώ = Ι/ω on the unit circle. Equation (14.91) is a relation between two an-

alytic functions that holds over the whole unit circle. The difference between φ{ω)

and φ{\/ω) is an analytic function; the Taylor series of the difference vanishes on

the unit circle, and by analytic continuation the difference vanishes everywhere, so

the two functions must be equal everywhere in the complex plane.

Now φ(ω) can be determined completely by analyzing its asymptotic behavior.

Outside of

the

hole φ must be completely regular, except for the fact that it diverges

as — ίΣζ for large ζ. The relation between ω and

If the other sign of the square root were cho-

v/C

—

sen

here, the argument below would have to

= .

be modified in several places, but the final

an-

(14.92)

2 swer would be the same.

Therefore, when ζ is large, large, ω = ζ, and in accord with Eq. (14.81) one obtains

φ{ω)

-f

-ίΣω for

-> oo. (14.93)

Consulting Eq. (14.91) one concludes also that

φ{\/ω)^-ίΣω

ΐοτω^οο,

(14.94)

which means that

φ{ω)^

^—

foru;->0 (14.95)

φ(ω)^

— forw-^O. (14.96)

However, φ(ω) can have no other singularities within the unit circle, or else φ(ω)

would have corresponding singularities outside the unit circle, which is forbidden

is to be smooth away from the hole.

Having determined all the possible singularities of φ within and without the

unit circle, it is determined up to a constant. It must be given by

φ(ω)

= -ϊΣω + ΐ—

Add Eqs. (14.96) and (14.93). (14.97)

0(C)

-is|(l

^/l-4

7/C

)

ffi^(l

V/1-4P/C

). (14.98)

Use Eq. (14.92) and 1/ω

ζ(1

- y/]

-4ρ/ζ

)/2ρ.

Displacements and stresses can now be determined from Eqs. (14.79) and (14.80).

Figure 14.18 shows

plot of a

along the line

= 0 for

narrow ellipse with

p = 0.9. Problem 4 uses Eq. (14.98) to verify Eq. (14.77).

14.4.3 Stress Intensity Factor

The limit

—>

1 is particularly interesting. The elliptical hole closes down and

becomes a thin crack reaching from x = — 2 to x

—

2. The function φ acquires

Cracks

405

branch cut over exactly the same region. The displacement u is finite everywhere,

but the stress o

becomes singular, diverging on the x axis as

du μΣ,χ μΣ,

'dy ~ yfx-2Λ/Χ + 2 ~* y/x-2

The stress intensity factor K is a coefficient of the diverging stress, defined by

K = lim v2irra

r = x

~ 2 is the distance from the crack tip. (14.100)

r—»0

■

= μ— = , \ , _

■ -> ]

as*-^2.

(14.99)

In the case of Eq. (14.99), K =

^2-ημΥ,.

The stress intensity factor serves for fracture the same purpose as the Peach-

Kohler force served for dislocations. A crack is presumed to become mobile when

the stress intensity factor reaches a particular critical value that can be tabulated

for each material. This criterion is equivalent to assuming, as in the argument

accompanying Figure 14.16, that the crack moves when the energy transmitted to

its tip reaches a critical value. The relationship between energy flux to crack tips

and stress intensity factor is determined in Problem 3.

14.4.4 Atomic Aspects of Fracture

Problem 5 shows that a microscopic theory of crack motion has some features that

are difficult to understand from continuum analysis. Cracks can be trapped by a

lattice even when continuum calculations predict they should have enough energy

to move, as discussed by Thomson (1986).

When a crack begins running in a brittle material, it excites phonons that radiate

off behind it, shown in Figure 14.20. The amplitude of these phonons is difficult to

calculate, but their frequencies are easily determined by the following argument:

Suppose any object moves at velocity v in a medium that supports phonons

of frequency ω(κ). Far from the moving object, radiation must take the form of

traveling waves. Take the motion of the object to be along a line of atoms, so that

at time intervals a/v the object moves from r to r + R, where R is a lattice vector

of magnitude a. Because the motion of the object is steady, the radiation it emits

must be invariant under the replacements

r^r + R (14.101a)

t->t + a/v. (14.101b)

At the same time, because the radiation far away is of the form exp[ik

■

—

ίωί],

the

symmetries in Eq. (14.101) require

ik-(r+R)-iui(t+a/v) _

ά·7-ίωί (14.102)

^/k-R-iwa/v

= χ

(14.103)

=> (k + K) -V = Uj(k) Because Rv/a = v, with K a reciprocal lattice (14.104)

vector.

=> k · V = U)(k). Working now in the extended zone scheme, (14.105)

so k is not restricted to the first Brillouin zone.

406 Chapter 14. Dislocations and Cracks

4 6

Wave number k

Figure 14.19. Graphical solution of Eq. (14.105), showing the resonances excited by a

steadily moving object as it travels through a lattice.

The radiation frequencies emitted by a moving crack are therefore determined by

drawing the phonon dispersion relation along the direction of crack motion and

then finding where the straight line k

■

v crosses the dispersion relation. This graph-

ical construction is illustrated in Figure 14.19.

The condition (14.105) for radiation emitted by moving objects, often ascribed

to Cherenkov and Landau, applies to many different situations, including damping

in liquid helium (Section 15.5.4) and plasmas (Problem 4 in Chapter 23), and to

electrons participating in the anomalous skin effect (Section 23.2.1).

Problems

Tensile stress: Carry out the computations needed to reproduce Figure 14.2,

and estimate the maximum tensile stress of silicon.

Interacting dislocations:

(a) Show that for the displacement field in Eq. (14.55) it is legitimate to write

υ = -ψ

ruV

(14.106)

Why would this equation be incorrect for the displacement field (14.51)?

(b) Show that V

« is proportional to a delta function when applied to

Eq.

(14.51

and find the coefficient of proportionality.

in Eq. (14.54), and cannot be derived.

Relation between stress singularity and energy flux: This problem demon-

strates that once the strength of the stress singularity around a crack tip is

known, the energy flowing to the crack can be determined as well, and the

two are simply related.

Problems 407

(a) Consider a two-dimensional elastic medium subjected to out-of-plane distor-

tions u

, so that its total energy is

£ =

—

/ dxdy

By considering dtjdt, show that

L(ù

)

(vu

J = ßü

gives the flux of energy at every point in the elastic medium.

(b) By working with Eqs. (14.98) and (14.100) in the limit p

near a crack tip

=-'Js^-^

ζ = χ + ίγ.

(14.107)

(14.108)

show that

(14.109)

Thus ù = —vdu/dx. Suppose that the crack is straight and lies along the

negative x axis; also suppose that its tip has just reached the origin. Using

Eq. (14.109) in (14.108), find the integrated flux of energy into the crack tip.

The most convenient contour for the calculation is one running from x = — oc

to x = oo while y

—

a, and then from x = oo back to x

—

— oo for y

—

—a;

the calculation is carried out in the limit where a is very small. Because the

contour is horizontal, only the y component of the energy flux is needed.

Stresses at end of an elliptical hole:

(a) The curvature κ—

/R of a general plane curve is defined by

K =

ds'

(14.110)

where

is the angle of a tangent vector, and s is arc length along the curve.

Show that for a general parametrization of the curve one obtains

■

dx d

y dy d

'dt~dt

~~dt~dt

[d)

(i)l

3/2

(14.111)

(b) Show that the radius of curvature R at the tip of an ellipse described by

Eq. (14.89) is

R={

-\y/{p+\).

(14.112)

1 —

p).

(d) Verify Eq. (14.77) for ellipses with small radii of curvature R.

408

Chapter 14. Dislocations and Cracks

Figure 14.20. This one-dimensional model mimics the motion of a crack in a strip, incor-

porating effects of discreteness. One can view it as a model for the atoms lying just along

the surface of a crack. The mass points are only allowed to move vertically, and they are

tied to their neighbors with springs that break when they exceed a certain extension. The

lower portion of the figure shows a dynamical solution of the model, where the crack is

emitting phonons.

5. Lattice trapping of

cracks:

This problem concerns a one-dimensional model

of crack propagation, analogous to the Frenkel-Kontorova model of Section

14.2.3.

Consider the model shown in Figure 14.20. One can view it as a

model for the atoms lying just along a crack surface. They are tied to nearest

neighbors by elastic springs, with spring constant % = 1, and tied to a line

of atoms on the other side of the crack line by similar springs, which snap

when extended past some breaking point. The lines of atoms are being pulled

apart by weak springs of spring constant X =

/N. These weak springs are

References 409

meant schematically to represent N vertical rows of atoms pulling in series.

The equation that describes the upper row of mass points is

—

+ u

~\ Elastic coupling to neighbors

+jj(S

—

) Driven by dispk

+(«

—

m>+

)ö(2

—

). Bonds that snap

+jj

(δ —

) Driven by displacing edges of strip

(14.113)

(a) Consider equilibrium solutions, so that

= 0. What is the potential energy

per bond far to the right of the crack, as a function of the imposed opening

extension

<5?

(b) What is the potential energy per bond far to the left of the crack, including

the energy needed to snap the bonds?

14.4.1,

show that the crack should be able to

move when δ

—

^2N + 1.

(d) Following the steps that solve the Frenkel-Kontorova model, find an explicit

expression for function of δ.

(e) Show that the bond at 0 reaches height

and snaps when

-N/3

+ 1

S = V2N+l r- . (14.114)

6. Conformai mapping: Repeat the calculations leading to Eq. (14.98) but for

an elliptical hole of arbitrary length /.

Positional fluctuations in three dimensions: Return to Eq. (14.26), but con-

sider a three-dimensional crystal rather than a two-dimensional one. Show

that the mean square displacement of atoms vanishes at sufficiently low tem-

perature.

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