Towards Mechanochemistry of Fracture and Cohesion:

Mechanics of a Catenary Process Zone

by Michael A. Grinfeld and Steven B. Segletes

ARL-TR-5310 September 2010

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Army Research Laboratory

Aberdeen Proving Ground, MD 21005-5066

ARL-TR-5310 September 2010

Towards Mechanochemistry of Fracture and Cohesion:

Mechanics of a Catenary Process Zone

Michael A. Grinfeld and Steven B. Segletes

Weapons and Materials Research Directorate, ARL

Approved for public release; distribution is unlimited.
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ARL-TR-5310

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October 2009-November 2009

Towards Mechanochemistry of Fracture and Cohesion: Mechanics of a Catenary

Process Zone

Michael A. Grinfeld AH80

Steven B. Segletes

U.S. Army Research Laboratory

ATTN: RDRL-WMP-C

Aberdeen Proving Ground, MD 21005-5066

This document is part of a series of reports.

In this report, a one-dimensional model of cohesion and fracture is examined for flexible substrates (such as tape or hooked

fasteners). The process zone is modeled to contain cohesive springs that apply a constant force per unit cracklength, up to the

point where a critical displacement of decohesion is reached. This particular model resulted in a crack that takes on a catenary

geometry in the process zone. Interestingly, the process zone was found to have a deterministic finite size. More importantly,

however, the force relations that resulted from the model were found to match exactly those of a simple cohesion model

formulated using the variational method with a macroscopic formulation.

process zone, fracture, film, bond, catenary, adhesion

UU 36

Steven B. Segletes

Unclassified Unclassified Unclassified 410-278-6010

ii
Contents

List of Figures v

1. Introduction 1

2. The Idealized Problem 1

3. Formulating the Boundary Conditions and Governing Equation 4

3.1 The Domain x < Sf : The Detached Zone ............................................... 4

3.2 The Domain x > Si: The Bonded Zone ................................................. 4

3.3 The Domain Sf ≤ x ≤ Si: The Process Zone........................................... 5

3.3.1 The Boundary Conditions ........................................................ 5

3.3.2 The Governing Equation.......................................................... 5

4. The Solution 7

4.1 The Catenary Solution Applied to the Domain Sf ≤ x ≤ Si............................ 7

4.2 Process-Zone Cracklength................................................................. 9

4.3 Bonding Force F0 .......................................................................... 11

4.4 The Relation of Applied Tension T0 to Load Angle θ0 .................................. 13

4.5 T∞ and an Interesting Relationship ....................................................... 14

4.6 Total Field Energy of the Bond and the Work of Deformation .......................... 16

4.7 Axial Cable Displacement with Changes in θ0 ........................................... 19

4.8 The Question of Asymptotes .............................................................. 20

iii
5. Conclusions 21

6. References 22

Appendix. The Catenary Solution 23

Distribution List 25

iv
List of Figures

Figure 1. The system under consideration, characterized by two bonded cables being debonded

under an applied load. ............................................................................ 2

Figure 2. The +y configuration of an idealized experiment [the −y configuration (not

shown) is symmetric about the x-axis], in which two inextensible but flexural cables

that have been bonded and attached to the laboratory at point P are “unzipped” by applying an ever-increasing tension T0 through points H and −H (not shown). ................ 3

Figure 3. Force balance on an element of bonded cable in the idealized process zone........... 6

Figure 4. The process-zone shape (aspect ratio 1:1), with the load angle θ0 as a parameter. ... 10

Figure 5. The process-zone shape, where the x-coordinate has been normalized by ∆, with

the load angle θ0 as a parameter................................................................. 10

Figure 6. Normalized x-component (∆/u0) and pathwise (s/u0) size of process zone vs. angle of applied load θ0, expressed on a (a) linear and (b) semi-logarithmic scale. ............ 12

Figure 7. The averaged cable angle within the process zone vs. the load angle θ0. ............. 13

Figure 8. Normalized applied force T0/σ and bonding force F0/σ vs. angle of applied load

θ0, expressed on a (a) linear and (b) semi-logarithmic scale................................... 15

Figure 9. Field energy (E/0) in process-zone bond vs. load angle θ0 (energy σs/2 normalized by 0 given for reference). ................................................................. 17

Figure 10. Schematic showing the origin of the field energy, in terms of external work applied. 18

Figure 11. Schematic showing lengths associated with cable displacement. ..................... 20

v
INTENTIONALLY LEFT BLANK.

vi
1. Introduction

The problem of bonding two surfaces is important in many aspects of Army technology. The

need for bonding may arise when two dissimilar materials require mutual attachment, or when

geometric complexity and/or cost preclude the manufacture of an item as a single piece. Bonding

requirements may call for a permanent adhesive attachment or, alternately, they may require the

ability to debond the items on demand, as in the case of tape or velcro.1In some cases, the

substrate materials being bonded may be stiff, relative to the bonding material, while in other

cases, the substrate materials may be wholly flexible (e.g., tape, velcro). This report focuses on

bonds involving flexible substrates.

A successful program in adhesive design requires a knowledge not only of how the intact bond

successfully performs, but also of how and when the bond will fail. This report is one in a series

of reports (1, 2) in which we examine the nature of bonding between two surfaces, and the energy

interchange involved with the creation of new free surfaces (i.e., fracture). In an effort to reduce

the problem to its essentials, we attempt to formulate scenarios that are maximally simplified. In

this way, the problem can be more fully understood without being lost in mathematical

complexity. In this report, a notional experiment is described and analyzed, involving the

progressive fracture of a one-dimensional (1–D) idealized bond under controlled loading

conditions.

2. The Idealized Problem

In figure 1, we display the closed system that we intend to address in this report. It consists of

two cables that have been bonded together, which are being separated in a controlled manner.

The word “cable” is used here to convey one-dimensionality, though other arrangements that

physically approximate one-dimensionality would apply, such as velcro straps or adhesive tape.

The goal of such a notional experiment is to study the manner and degree to which a given load

mg will bring about the fracture of the bond, altering the load angle θ0.

1Velcro is a hook-and-loop fastener system, which is a registered trademark of Velcro Industries B.V. In this report,

we use the lowercase term “velcro” to refer to generic hook-and-loop fasteners in the style of Velcro.

1
H

P

 H

Process Zone

Symmetry Plane

mg mg

! 0

Figure 1. The system under consideration, characterized by two

bonded cables being debonded under an applied load.

In figure 1, point P is a rigid attachment point for the bonded cable system. Points H and −H

are rigid, frictionless pins, over which the debonded cables respectively pass. The load applied to

the cables, brought about by mass m under the influence of gravity g, can and will be, hereafter,

characterized by a magnitude of force equal to T0.

For the analysis, we focus on a subset of figure 1, as shown in figure 2. In addition to the zoom,

figure 2 has been rotated, for convenience, 90◦clockwise relative to figure 1. Here, one half of a

symmetric system is depicted (i.e., one of the cables), because of planar symmetry across the

x-axis. The idealized cables are assumed to be massless, linearly inextensible, yet fully flexible

(i.e., incapable of supporting a bending moment). It is important to remember when considering

figure 2 that the cable is not attached to point H, but merely directs the applied load T0 through

point H.

The artifice of planar symmetry was introduced to this problem merely to ensure that we were

describing a physical system in which the deforming bonds applied force purely in the

y-direction. Mathematically, and with no loss of generality, we may solve the half-problem

actually shown in figure 2 by considering the cable to lie upon a rigid substrate, under the

influence of a perpendicular force-field of specified characteristics. Note, therefore, the

descriptive similarity to the thermomechanical problem described in reference 1.

2
x

y

T0

 0

Sf

Si

To

u0

f = f0

H

 (x)

o P

0

!

Figure 2. The +y configuration of an idealized experiment [the −y configuration

(not shown) is symmetric about the x-axis], in which two inextensible but

flexural cables that have been bonded and attached to the laboratory at

point P are “unzipped” by applying an ever-increasing tension T0 through

points H and −H (not shown).

In this analysis, we assume the bond provides a fixed resistive force (f0) per unit length of cable

up to the point at which the bond breaks, characterized as when the y-displacement exceeds the

critical value for fracture, u0. Such process-zone mechanics could be representative of several

types of bonding models, for example, a rigid-plastic bond, an electric field, or even a

gravitational field. Nonetheless, for a bond resulting from a continuum adhesive, the model is

highly simplified, in that transverse (e.g., Poisson) influences on the stress state are ignored.

For points x < Sf , the cable is either initially unbonded or has become debonded and remains

under load T0, applied at an angle θ0 with respect to the x-axis. The angle θ0 will change

(i.e., decrease) with the progression of the bond fracture, as the load level T0 is increased.

For cable in the domain Sf ≤ x ≤ Si, the deforming bond remains intact, but under resistive

stress. The restoring force per unit cable-length, given by f0, is independent of the

y-displacement, u. The tension in the cable, T(x), will not remain fixed in this region, since the

stressed bond is applying forces to the cable which are not strictly perpendicular to it. This part

3
of the problem domain is of x-length ∆, though the pathlength of cable in this domain will always

exceed ∆ (in fact, it will always be greater than or equal to q∆2 + u

2

0

).

For the bonded cable, we admit the possibility that beyond a certain domain, x > Si, the y-forces

establish a precise balance, such that the cable displacement, u, remains exactly zero. This

domain of the cable is still, however, subject to axial tension in the x-direction, given by T∞.

The idealized problem, therefore, consists of these three domains: (1) x < Sf ; (2) Sf ≤ x ≤ Si;

and (3) x > Si. While the problem envisions that the stress states over these domains and indeed

the domain limits themselves will change with increasing load, T0, the problem we solve here is

quasi-static, in which the load T0 is increased slowly and incrementally, allowing progressive

equilibrium states to establish themselves throughout the system. In this manner, we intend to

establish a relation that links the magnitude of the applied tension, T0, to the resultant level of

debonding as characterized by domain boundaries Sf and Si (alternately expressed as a function

of θ0), in terms of the relevant material parameters (e.g., f0 and u0).

3. Formulating the Boundary Conditions and Governing Equation

3.1 The Domain x < Sf : The Detached Zone

The domain x < Sf represents that portion of cable that was not bonded to begin with, as well as

the additional portion that debonds from the application of force T0 upon it. The only force acting

upon the cable in this domain is the tension applied through point H. Equilibrium considerations

dictate that the cable in this domain of the problem remain straight and under constant load, T0.

There is no requirement that the cable remain forever fixed at point H. Indeed, quite the contrary

is required. . . namely, that the cable be pulled through point H as the load T0 is increased.

3.2 The Domain x > Si: The Bonded Zone

At the far end of the cable, for x > Si out to point P, the situation is likewise simple to analyze.

The cable, in this part of the problem domain, lies upon the x-axis. The force applied where the

cable is rigidly attached to point P is given as T∞. Force equilibrium dictates, that at all points in

this part of the problem domain, the cable tension remain fixed at T∞.

Furthermore, by considering the complete length of cable for x > 0 as a free body, and knowing

that all external forces upon it in the intermediate region (Sf ≤ x ≤ Si) are perpendicular to the x

4
axis, a force equilibrium on the cable, applied in the x-direction, leads one to conclude that

T∞ = T0 cos θ0 . (1)

3.3 The Domain Sf ≤ x ≤ Si: The Process Zone

We now proceed to analyze the most important part of the problem—the process zone.

3.3.1 The Boundary Conditions

This part of the problem domain represents the key part of the problem: the deforming bond on

its way toward eventual fracture. We may conclude several salient boundary conditions on this

domain. Because there are no concentrated forces applied at Si and/or Sf (only a distributed

force along the entire domain), equilibrium applied across the domain boundaries will require a

continuity of cable tension, displacement, and slope.

The cable tension (T) changes along the length of cable in this part of the problem domain.

However, the tension must be T = T0 at x = Sf and T = T∞ at x = Si, in order to match the

adjacent domains. Since the coordinates and slopes of the cable must also match the adjacent

domains, this requires that, at x = Sf , the y-displacement u = u0 and u

0 = − tan θ0. At the other

end of the domain, x = Si, we must satisfy the requirement that u = 0, and u

0 = 0.

3.3.2 The Governing Equation

To develop the governing relations for the process zone, we must satisfy the equilibrium

requirements for an arbitrary section of cable in that zone, such as that depicted in figure 3.

In the x-direction, the only forces arise from the tension in the cable, and therefore

d

dx(T cos θ) = 0 . (2)

From this, we conclude that T cos θ must remain constant, and that constant, from matching the

conditions at the end of the domain, is given by

T cos θ = T∞ = T0 cos θ0 . (3)

In the y-direction, the situation is slightly more complex, in that the independent variable will still

be taken as x, even as the forces being summed are y-forces. If F represents the integral along

5
T(x)

f

0

 (x)

ds

dx

Figure 3. Force balance on an element

of bonded cable in the

idealized process zone.

the x-direction of the distributed bond force f0, then the condition of equilibrium in the

y-direction becomes

d

dx(T sin θ + F) = 0 . (4)

One must note, however, that the bonds which produce the distributed force f0 are distributed

along the cable length ds. Therefore, as the angle θ(x) changes, the density of bonds along dx

changes as well. Thus, the quantity dF/dx must be considered as dF/ds · ds/dx. The quantity

dF/ds is merely the distributed force of the bond, namely f0, while ds/dx is obtained from the

geometry as sec θ, alternately expressible as √

1 + u

02

. Using equation 3, we can reexpress

T sin θ as T∞ tan θ, alternately expressible as −T∞u

0

.

Combining all these substitutions into equation 4 yields the following governing relation:

u

00 −

f0

T∞

√

1 + u

02 = 0 . (5)

Equation 5 should be immediately recognized as being associated with the so-called “catenary”

solution, which is the curve that governs the geometry of a free-hanging chain or cable under the

force of gravity (indeed, the word catenary comes from the word catena, Latin for chain). Such a

connection should not be surprising when one considers that the distributed force of the

deforming bonds, f0, constant per unit length of cable, is directly analogous to the distributed

force of a cable’s weight under the force of gravity.

6
It may, therefore, be instructive to visualize the solution to this mechanics problem in terms of the

analogous gravitational problem, in which a long, heavy chain is laid in a straight line upon a

table (the rigid substrate). As a free end of the chain is lifted to a given height above the table

(analogous to u0), the shape in which the hanging chain establishes itself will vary with the level

of tension applied (itself being a function of the vertical (lifting) and lateral (stretching)

components of that force). The derivation of the catenary solution is provided in the appendix,

for the reader’s convenience.

4. The Solution

4.1 The Catenary Solution Applied to the Domain Sf ≤ x ≤ Si

To solve equation 5, let us first make the substitution of convenience that α = f0/T∞. Next, we

introduce a shifted coordinate system, ξ = x − Sf , which allows equation 5 to remain unaltered

with respect to ξ, while at the same time redefining the bounds of the domain as 0 ≤ ξ ≤ ∆,

where ∆ = Si − Sf .

The general solution to equation 5 is

u = Ae−αξ + Beαξ + C , (6)

where the constants A and B are subject to the constraint

2

√

AB =

1

α

. (7)

Note that equation 7 requires that A and B be of the same sign, so that their product is positive.

The boundary conditions developed in section 3.3.1 require that

u(0) = A + B + C = u0 (8)

u

0

(0) = (−A + B)α = − tan θ0 (9)

u(∆) = Ae−α∆ + Beα∆ + C = 0 (10)

u

0

(∆) = −Aαe−α∆ + Bαeα∆ = 0 . (11)

7
Equation 11, upon multiplication by e

α∆ and upon elimination of α by way of equation 7, may be

directly solved for ∆ as

∆ = √AB ln(A/B) . (12)

From this and equation 7, the following two results follow:

e

α∆ =

q

A/B

e

−α∆ =

q

B/A

. (13)

Substituting these results into equation 10 reveals that

C = −2

√

AB , (14)

which also happens to equal the negative reciprocal of α.

Starting with equation 9 and eliminating α with a substitution of equation 7 produces the equation

q

A/B −

q

B/A = 2 tan θ0 . (15)

This may be solved quadratically, in light of the requirement that A and B be of the same sign, as

s

A

B

=

1 + sin θ0

cos θ0

. (16)

We are now prepared to substitute equation 14 into equation 8, while isolating B, to obtain

B



(A/B) + 1 − 2

q

A/B= u0 . (17)

This conveniently reduces to

B

q

A/B − 1

2

= u0 . (18)

The term B may now be directly solved by substituting equation 16 into equation 18 in order to

obtain

B =

u0(1 − sin θ0)

2(1 − cos θ0)

. (19)

The quantity A directly follows as

A =

u0(1 + sin θ0)

2(1 − cos θ0)

. (20)

8
Now that A and B are explicitly known in terms of the boundary conditions, C may also be

explicitly calculated by substitution into equation 14:

C =

−u0

sec θ0 − 1

. (21)

The quantity ∆ follows from equations 12 as:

∆ = u0

2(sec θ0 − 1) ln 

1 + sin θ0

1 − sin θ0

!

. (22)

Thus, our four unknowns, A, B, C, and ∆, have been fully characterized.

These parameters may be employed through equation 6 to trace the shape of the cable within the

process zone, where the distributed force f0 is applied. Figure 4 shows the actual shape of the

cable for different load angles θ0. For smaller load angles, the curves exit the right side of the

graph, because the x-measure of the process-zone size, ∆, exceeds u0. Therefore, we provide

figure 5, which shows the typical cable geometry for various load angles, except that the x-axis

has been normalized by ∆.

Note from figure 5 that the normalized shapes of the curves are all very similar for load angles

below 60◦. As the load becomes more perpendicularly oriented to the substrate, however, the

shape of the cable acquires much more of a knee-like shape to it and the tail becomes longer.

4.2 Process-Zone Cracklength

Now that the shape of the deforming surface of the process zone has been fully defined through

the calculation of the parameters A, B, C, and ∆, the process-zone cracklength can be calculated.

The process zone is defined by the rectangle Sf ≤ x ≤ Si and 0 ≤ y ≤ u0 (refer to figure 2). It

spans an x distance of Si − Sf , which we have called ∆ and a y distance of u0. Equation 22

gives us an explicit measure of the x-component of the zone size. However, ∆ is not, strictly

speaking, the process-zone cracklength. From figure 3, we note that the length of the crack

formed is the integral of ds, not that of dx. Therefore, the full process-zone cracklength s, is

obtained as follows:

s =

Z

ds =

Z ∆

0

√

1 + u

02 dξ . (23)

From directly differentiating and squaring equation 6, it may be verified, in light of equation 7,

that

√

1 + u

02 = α(Ae−αξ + Beαξ) . (24)

9
 !"u0

0.0 0.2 0.4 0.6 0.8 1.0

u/u0

0.0

0.2

0.4

0.6

0.8

1.0

70o

80o

89o

60o

45o

30o

#0=1o

Figure 4. The process-zone shape (aspect ratio 1:1), with

the load angle θ0 as a parameter.

 !"#

0.0 0.2 0.4 0.6 0.8 1.0

u/u0

0.0

0.2

0.4

0.6

0.8

1.0

89o

80o

89.99o

60o

0.01o

Figure 5. The process-zone shape, where the x-coordinate

has been normalized by ∆, with the load angle

θ0 as a parameter.

10
Integrating equation 24 across the process zone (i.e., from ξ = 0 to ∆), in light of equation 13,

gives the following value for the cracklength s within the process zone:

s =

Z ∆

0

√

1 + u

02 dξ = A − B . (25)

If B and A are explicitly substituted from equations 19 and 20, one obtains the cracklength in the

process zone as

s = u0 cot(θ0/2) . (26)

A comparison of s to ∆ is presented in figure 6. Naturally, s ≥ ∆ for all loading angles θ0. A

specific point of note on figure 6a is the behavior as θ0 approaches 90◦(i.e., applied loading

perpendicular to the plane of symmetry). We note that the x-width of the process zone (∆)

approaches zero length. However, the pathlength of cable in the process zone (i.e., the

process-zone cracklength s) can never fall below u0, because even the limiting case of a

process-zone rectangle approaching dimensions 0 × u0, the cable, oriented perpendicular to the

plane of symmetry, will retain a cracklength of u0.

Knowledge of ∆ allows the contents of figure 6 to be presented in a slightly different manner.

Namely, tan−1(u0/∆) represents the average angle of the cable within the process zone. This is

shown in figure 7.

4.3 Bonding Force F0

The bonding force F0 is also of interest. We define it as the total resistive force in the y-direction

applied to the cable in the process zone Sf ≤ x ≤ Si. We should note that, because of the

perpendicularity of F0 and T∞, equilibrium upon the cable dictates that T

2

0 = F

2

0 + T

2

∞. The

resistive force comprising F0 is uniformly distributed along the cable length with a constant

magnitude of f0 per unit cable length. Therefore, calculating the total bonding force is

straightforward:

F0 = f0s . (27)

Substituting from equation 26, and introducing the term grouping

σ = f0u0 , (28)

the total bonding force is obtained as

F0 = σ cot(θ0/2) . (29)

11
 0

(deg.)

0 10 20 30 40 50 60 70 80 90

Normalized Distance

0

1

2

3

4

5

!/u0

s/u0

(a)

 0

(deg.)

0 10 20 30 40 50 60 70 80 90

Normalized Distance

0.1

1

10

100

!/u0

s/u0

(b)

Figure 6. Normalized x-component (∆/u0) and pathwise

(s/u0) size of process zone vs. angle of applied load

θ0, expressed on a (a) linear and (b) semi-logarithmic

scale.

12
 0

(deg.)

0 10 20 30 40 50 60 70 80 90

tan!"

(u0

/#)

0

10

20

30

40

50

60

70

80

90

Figure 7. The averaged cable angle within the process

zone vs. the load angle θ0.

In equation 28, the product of the distributed bonding force per unit cracklength and the critical

displacement to bring about full fracture represents the work per unit cracklength required to

completely fracture the bond. Thus, σ is a material property that may be termed the specific bond

energy (e.g., the bond energy per unit cracklength).

Therefore, equation 29 shows the total bonding force to be directly proportional to the specific

bond energy σ and the cotangent of the half-angle of the applied force. In the limiting case of a

θ0 = 90◦applied load, the total bonding force (F0) exactly equals the specific bond energy σ.

This result should not be surprising in light of section 4.2, which showed the minimum size of the

process-zone cracklength to be u0 as the load angle θ0 approaches 90◦. Since the bonding force

F0 is proportional to the process-zone cracklength s and not merely the x-component of

cracklength ∆, the bonding force can never fall below the specific bond energy σ, even as ∆

approaches zero length.

4.4 The Relation of Applied Tension T0 to Load Angle θ0

The bond force F0, which is known explicitly, may be used to calculate the ultimately desired

quantity, the tension in the cable required to bring about separation angle θ0. From geometry, we

know that, in order to satisfy y-equilibrium, the y-component of tension must exactly equal the

13
bonding force F0. Thus,

T0 =

F0

sin θ0

=

σ

2 sin2(θ0/2) . (30)

Therefore,

sin 

θ0

2

!

=

s

σ

2T0

. (31)

This is precisely the relation obtained in reference 1 where the analogous problem was solved

using variational principles of energy exchange between a hanging mass m and the specific

chemical energy σ of the bond, which supports the mass through a cable bonded to a rigid

substrate. In that solution, the tension applied to the bond at an angle θ0 arises from a weight,

given by mg. However, in the variational approach, the mechanics of the problem are not even

addressed and, thus, it was a bit unexpected (though fully welcomed) to achieve exactly the same

result.

4.5 T∞ and an Interesting Relationship

The quantity T∞ is the only force quantity left to calculate. It is easily obtained as

T∞ = T0 cos θ0, which reduces to the following:

T∞ =

σ

sec θ0 − 1

(32)

Figure 8 presents the relation of the applied cable tension T0, the total bonding force F0, and T∞

as a function of the applied-load angle θ0. Above load angles of 45◦, F0 will exceed T∞ in

magnitude. Below 45◦, the reverse is true. In all cases, T0 exceeds the other forces because it is

the hypotenuse associated with the components F0 and T∞.

One very interesting relation that may be shown is that

T0 − T∞ = σ , (33)

regardless of the angle of applied force, θ0. That the difference of T0 and T∞ should remain

invariant may be understood in the following way. Every segment of cable in the process zone is

subjected to a distributed y-force of magnitude f0. For each segment, there is a component of

that distributed force which is directed along the cable length, thereby changing the magnitude of

the cable tension. This distributed component directed along the cable is of magnitude f0 sin θ,

where θ varies from 0 to θ0 through the process zone. The overall change in the cable tension,

therefore, is R s

0

f0 sin θ ds. The quantity f0, being constant, may be taken out of the integral,

14
 0

(deg.)

0 10 20 30 40 50 60 70 80 90

Normalized Force

0

1

2

3

4

5

6

7

8

9

10

T0 /!

F0 /!

T 

o 

/! o

(a)

 0

(deg.)

0 10 20 30 40 50 60 70 80 90

Normalized Force

0.1

1

10

100

T0 /!

F0 /!

T 

o 

/! o

(b)

Figure 8. Normalized applied force T0/σ and bonding force

F0/σ vs. angle of applied load θ0, expressed on a

(a) linear and (b) semi-logarithmic scale.

15
while the remaining integrand (sin θ ds) is simply du. The change in cable tension is, therefore,

f0

R u0

0

du = f0u0 = σ, which is independent of the particular geometry of the process zone.

Note that the relationships for F0, T0, and T∞ are functions of the term grouping σ, rather than

functions of f0 alone or u0 alone. What this indicates is that, while f0 and u0 were introduced as

fundamental model parameters, the calculation of equilibrium loads is really not dependent on

their individual respective values, but rather only upon the load angle and the specific bond

energy, σ. Where u0 will play a direct role is in the calculation of the field energy within the

process zone itself, as will be shown in section 4.6 of this report.

4.6 Total Field Energy of the Bond and the Work of Deformation

In section 4.5, the specific bond energy, σ, was introduced to describe the energy required to

create a unit of cracklength (i.e., to grow the crack). We also have the ability to integrate what

would traditionally be called the elastic energy of the bond. In the current model, however, the

term “elastic” might not be appropriate, because the force-displacement relationship here is a

constant, and not linear with displacement in the traditional sense of elasticity. Additionally, the

bonding force may be dissipative, as in the case of a plastic or velcro bond. Therefore, rather

than call it “elastic” energy, we will refer to it as the total field energy of the bond. It refers to the

integration of energy throughout the process zone, arising from the displacement of the cable

from the u = 0 configuration:

E =

Z

f0uds = f0

Z ∆

0

u

√

1 + u

02 dξ . (34)

In the simplest case, if the cable geometry were linear within the process zone (rather than

catenoid), such that the cable extended linearly from the (x, y) point (Sf , u0) to (Si, 0), the total

field energy of the bond would be σs/2.

Except when the load is applied at 90◦, the present solution does not have a linearly varying

displacement in the process zone. For the special case of 90◦loading, the cable approaches in the

limit a y-oriented line in the process zone, of cracklength s = u0. Let us define a baseline field

energy as 0 = σu0/2 = f0u

2

0

/2. It represents the energy stored (or dissipated) in the process

zone of a bond that is subjected to a y-oriented load (i.e., perpendicular to the plane of symmetry)

of limiting magnitude T0 = F0 = σ.

For load angles other than θ0 = 90◦, we would expect (because of the displacement field’s

concavity) the total field energy to be less than σs/2. Let us proceed to obtain it. Equation 34

may be directly integrated through substitution of equations 6 and 24, in light of equation 13.

16
The result is given by

E = f0

A − B

2

(A + B + C) + A − B

2

C + ∆√

AB, (35)

which may be reduced to

E =

1

2

[σs − T∞(s − ∆)] , (36)

consistent with our expectations. This energy relationship is displayed in figure 9, normalized by

0.

To understand the physical origin of equation 36, and how the field energy arises from externally

applied work, figure 10 is provided. The point O represents the location x = Si, where ξ = ∆.

The actual path representing the cable’s deformed state is given by the arc OB (of length s). If

the cable lying along arc OB were not subject to the tension T0 at point B, the cable would

instead lie (unstressed) along the line OA.

 0

(deg.)

0 10 20 30 40 50 60 70 80 90

Normalized Energy

0

1

2

3

4

5

6

7

8

9

10

E/!

0

"s/(2!

0

)

Figure 9. Field energy (E/0) in process-zone bond

vs. load angle θ0 (energy σs/2 normalized by

0 given for reference).

We contemplate the application of tension −→T to point A of the unstressed cable, applied at a

constant angle of θ0 (in this case, 60◦). The magnitude of the tension would start as 0 when the

17
 !"u0

-0.5 0.0 0.5 1.0 1.5

u/u0

0.0

0.5

1.0

1.5

2.0

#0 =60o

$ O

%

C

&

'!(!')

F

E

D

G

#)

Figure 10. Schematic showing the origin of the field

energy, in terms of external work applied.

cable was in the OA configuration, increasing to a magnitude of T0 as the cable reoriented itself

to the deformed configuration OB. Because the magnitude of tension −→T can be shown directly

proportional to the u-displacement of the endpoint, one may assert that the averaged value of

tension −→T applied when the point A is deformed to point B is exactly (0 + T0)/2 = T0/2.

The work W done by this action, as point A moves to B under increasing force −→T , is the integral

of the force times the component of displacement in the direction of action, which is R −→T ·d

−→AB,

given more simply as W = (T0/2)|CB|. Because OE is parallel with −→T , while AE and GF are

perpendicular to it, we can assert that |CB| = |DB| − |OE| + |OF|. We may further assert, on

geometrical grounds, that 6 F OG = 6 BGD = θ0. Therefore, |DB| = |GB|sin θ0 = u0 sin θ0.

Likewise, |OE| = |OA| cos θ0 = s cos θ0. Finally, we have that |OF| = |OG| cos θ0 = ∆ cos θ0.

Substituting all these metrics into the equation representing the work done to deform the cable

from the unstressed configuration OA to the fully-stressed configuration OB gives the following

result:

W = (T0/2)[u0 sin θ0 − (s − ∆) cos θ0] . (37)

18
Knowing that T0u0 sin θ0 = F0u0 = σs and that T0 cos θ0 = T∞ allows equation 37 to be reduced

to

W =

1

2

[σs − T∞(s − ∆)] . (38)

The identical forms of equations 36 and 38 tell us that the amount of field energy stored in (or

dissipated by) the bonds along the cable in the region OB exactly equals the external work that

would have been required to deform the cable from its stress-free configuration OA to its fully

loaded configuration OB.

4.7 Axial Cable Displacement with Changes in θ0

Examining the underlying premise of figure 1, it is apparent that, as the cable bond becomes

progressively “unzipped,” the weights (each of adjustable mass m) will give up potential energy

as they lower in height. In this section, we seek to determine the amount of axial cable

displacement (i.e., decrease in height of mass m) that occurs as a function of the load angle θ0.

As a reference state, we chose the situation where the initial value of θ0 is 90◦and the initial cable

tension exactly equals the associated equilibrium value of T0 = σ. As the masses lower and the

cables begin to debond, the decrease in cable length between points H and P corresponds exactly

to the change in the elevation, ∆z, of the masses.

We employ figure 11 to assist in determining cable lengths. Initially, we assume that u0 (and thus

the process zone) is vanishingly small, so that the fractured geometry associated with load angle

θ0 is composed of two segments: the detached cable segment (of length c) and the bonded cable

segment (of length p − a). Under this assumption, the elevation change of the masses, ∆z, when

the load angle, θ0, deviates from 90◦, is given by

∆z = h + a − c (when u0  h) , (39)

where a = h/ tan θ0 and c = h/ sin θ0.

However, when the size of the process zone is not considered negligible, a further refinement to

equation 39 is required to obtain the elevation change of the masses, ∆z, when the load angle, θ0,

deviates from 90◦. Employing the relation for s from equation 26 and seeing that b = u0/ sin θ0

and d = ∆ − u0/ tan θ0, we obtain

∆z = (h + a − c) + (b + d − s) = h

sin θ0

(sin θ0 + cos θ0 − 1) + (∆ − 2u0 cot θ0) . (40)

19
x

y

 a 0

u0

H

P

0

h c

b

d

s

Figure 11. Schematic showing lengths associated with cable displacement.

To repeat, the value of ∆z represents the finite change in elevation level of the masses of figure 1

as a function of the load angle θ0. An initial value of θ0 = 90◦is assumed. The first term of

equation 40, proportional to h, is the dominant term, arising from the macroscopic geometry of

the system. The second term, on the order of u0, provides a secondary correction, to account for

the catenoid shape of the process zone. Further insight could be pursued, for example, by

comparing the displacement, ∆z, of the adjustable masses with the debonded cable length, given

by Sf = (h − u0)/ tan θ0.

4.8 The Question of Asymptotes

One of the questions that arose (in the minds of the authors) during the course of this investigation

was whether or not the eventual solution would be asymptotic. That is to say, whether the

location of Si (see figure 2) would approach infinity, even for finite values of T0. The present

solution shows that, for the current set of governing assumptions, it does not. One mathematical

fact regarding asymptotes is that their second derivative approaches zero, since an asymptote’s

graph approaches a straight line.

Considering equation 5, however, one can see that, as long as the specific bonding force f0 is

nontrivial, u

00 must be greater than zero. Thus, the governing equation 5 is incapable of

producing an asymptotic result, wherein u

00 = 0.

20
This also raises the possibility, however, that were the bonding-force relation something other

than a constant, the conclusion regarding the possibility of an asymptotic solution would change

as well. For example, instead of a constant specific bond force f0, the bond might be

characterized by an spring-type f = ku specific bonding force. With such a physical model, the

governing relation would become u

00 − (k/T∞)u

√

1 + u

02 = 0. In this circumstance, it would be

mathematically possible to achieve an asymptotic solution, in which u approached zero only as x

approached infinity. In such a case, u

00 would likewise approach zero, which is the requirement

for an asymptote.

5. Conclusions

In this report, a notional experiment was presented involving the progressive debonding of two

flexible surfaces (e.g., tape or velcro). An idealized model was developed which, when solved,

predicts the shape of the flexible substrate in the process zone where debonding occurs. More

importantly, it provides the force and energy relationships that govern the problem as a function of

the loading angle, which will provide the means to test the model against future experimental data.

The assumptions contained in the formulation are that (1) the (tape or velcro) substrate is fully

flexible; (2) the bonds are one-dimensional in nature, such that there is no lateral (i.e., Poisson)

interaction in the process zone; (3) the bonding force in the process zone is independent of

separation up to the point of fracture (in the manner of a perfectly-plastic bond); and (4) the bonds

are uniformly distributed along the substrate length.

Based on these assumptions to the problem, the process zone was seen to take on the shape of a

catenary (i.e., a free-hanging cable under the influence of gravity), for which an analytical

solution presents itself. Importantly, the force relations predicted for this model identically

matched those developed for the problem using an alternate energy-based approach (1).

21
6. References

1. Grinfeld, M. A.; Segletes, S. B. Towards Mechanochemistry of Fracture and Cohesion:

General Introduction and the Simplest Model of Velcro; ARL-TR-XXXX; U.S. Army

Research Laboratory: Aberdeen Proving Ground, MD, XXX 2010.

2. Grinfeld, M. A.; Segletes, S. B. Towards Mechanochemistry of Fracture and Cohesion:

Two-Component Spring Models of Velcro Detachment; In progress; U.S. Army Research

Laboratory: Aberdeen Proving Ground, MD, April 2010.

22
Appendix. The Catenary Solution

The catenary is a textbook problem in the teaching of differential equations. Historically, it refers

to the shape a chain (or cable) assumes under the influence of gravity, being attached only at its

endpoints. We rederive the result here for the convenience of the reader.

The governing equation to solve is functionally identical to that given in equation 5, namely:

u

00 − a

√

1 + u

02 = 0 . (A-1)

We reduce the equation order by casting u

00 as du0/du · du/dx, knowing that du/dx is the same as

u

0

. In so doing, we obtain

u

0du0

√

1 + u

02

= adu . (A-2)

This is readily integrated as

√

1 + u

02 = au + C1 . (A-3)

Now we note that, in equation A-1, a vertical shift of magnitude ushift, such that u = u − ushift

produces the identical equation in u. This indicates that any solution, u(x) is likewise a solution

u(x). The vertical shift is related to constant C1. Therefore, we may choose any particular value

of C1, for mathematical convenience, knowing that any solution subsequently obtained for u(x)

can be subjected to an arbitrary vertical shift. We choose C1 so that u

0

is identically zero, when u

is zero. Such a choice forces C1 to a value of 1. With this substitution, equation A-3 becomes

du

√

u

q

u + 2/a

= adx . (A-4)

Introducing the substitution v

2 = u reduces the subsequent integral of equation A-4 to a

convenient form:

Z

dv

q

v

2 + 2/a

=

Z

a

2

dx . (A-5)

23
The left-side integral may be solved in the following way:

Z

dx

√

x

2 + b

=

Z

(x +

√

x

2 + b2

)dx

(x +

√

x

2 + b2

)

√

x

2 + b

=

Z

 

1 +

x

√

x

2 + b2

!

dx

x +

√

x

2 + b2

= ln(x +

√

x

2 + b2

) + C

(A-6)

With this integration form, equation A-5 may be integrated as

ln(v +

q

v

2 + 2/a) = (a/2)x + ln C2 . (A-7)

Equation A-7 may be explicitly solved for v to yield

v =

C2

2

e

ax/2 −

1

aC2

e

−ax/2

. (A-8)

Squaring this equation and substituting u for v

2 gives the following:

u =

C

2

2

4

e

ax +

1

a

2C2

2

e

−ax − 1/a . (A-9)

Equation A-9 is equivalent to the form

u = Ae−ax + Beax + C , (A-10)

where A and B are subject to the constraint 2

√

AB = 1/a. We see that the solution, given by

equation A-10, matches the form presented in the report as equation 6, subject to the identical

constraint on A and B. Often, in the literature, the result is given in terms of the equivalent

hyperbolic cosine function.

For the general catenary solution (which may be shifted vertically), C is a free constant. If the

constraint is enforced to make u

0 = 0 when u = 0 (as was enforced when we chose C1), then C

will take on a value of −1/a, as dictated by equation A-9. This constrained value of C was

indeed obtained in the report as equation 14, in which C was revealed as −1/α.

24
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