Introduction to the J-integral

Instructor: Ramsharan Rangarajan

February 24, 2016

The purpose of this lecture is to briefly introduce the J-integral, which iswidely used in fracture mechanics. To the extent possible, we shall keep thelecture self-contained. This will mean a repetition of familiar concepts for somestudents, but we hope that the review will be beneficial overall. Ideally, wewould like to see lots of applications/examples of the concept to become fa-miliar with it. We will not be able to do so in class today but this can bedone through homeworks. One of the principal objectives of this lecture is topresent a unification of concepts being learnt in concurrent courses. The ma-terial covered will serve as a useful point of departure when we subsequentlystart learning about numerical methods to simulate fracture. Above all, theJ-integral is a central aspect of (nonlinear) fracture mechanics and it behoovesus to understand its origin instead of accepting it as a given. Such understand-ing is essential to recognize and conceive of its applications when solving newproblems.

1 What is the J-integral

Consider a two-dimensional elastic body as shown in Figure 1 and let Γ be aclosed contour that encloses a simply connected region. The J-integral is defined

(a) (b)

Figure 1: Coordinate system and normal used in defining the J-integral in (1).

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as

J ,∫

Γ

(Wdx2 − t · ∂u

∂x1ds

), (1)

where W is the strain energy density, t is the traction, u represents the displace-ment and we have chosen the usual coordinate system (x1, x2). It is fair to askwhat significance, if any, does the quantity defined in (1) have. The followingdiscussions will help answer this question to some extent.

1.1 Elastostatic equilibrium, no singularities =⇒ J = 0

We claim that in the absence of any sources of singularities (such as cracks) inthe area enclosed by Γ, the value of J equals zero irrespective of the choice ofthe contour Γ. Let us demonstrate this. Observe that we can rewrite (1) as

J =

∫Γ

(Wn1 − (σn) · ∂u

∂x1

)ds, (2)

where n = (n1, n2) is the unit outward normal to Γ and σ is the Cauchy stress.By appealing to the divergence theorem, the first term in (2) simplifies as∫

Γ

Wn1ds =

∫Γ

(We1) · n ds =

∫Ω

div(We1) dΩ =

∫Ω

∂W

∂x1dΩ.

By the chain rule,

∂W

∂x1=∂W

∂εij

∂εij∂x1

= σij∂2ui∂xj∂x1

,

so that we get ∫Ω

Wn1 ds =

∫Ω

σijui,1j dΩ. (3)

The second term in (2) simplifies similarly:∫Γ

(σn) · ∂u∂x1

ds =

∫Γ

(ui,1σijej) · n ds =

∫Ω

div(ui,1σijej) dΩ

=

∫Ω

(ui,1σij),j dΩ

=

∫Ω

(ui,1jσij + ui,1σij,j) dΩ

=

∫Ω

ui,1jσij dΩ, (4)

From (3) and (4), we get conclude that

J =

∫Γ

(Wn1 − (σn) · ∂u

∂x1

)ds =

∫Ω

(σijui,1j − ui,1jσij) dΩ = 0. (5)

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(a) (b)

Figure 2: Setup for defining the J-integral for a cracked body

Questions: In the above arguments,

(i) where did we use the fact that the body was elastic?

(ii) where did we use the assumption that the body was in static equilib-rium?

(iii) where did we use the assumption that there were no singularities suchas cracks?

1.2 J for a cracked body at static equilibrium

When the contour Γ encloses a region where the solution fields have singularities,such as in the presence of cracks, the J-integral may not vanish. However, auseful property still holds. Referring to figure 2, this time we consider any twopaths Γ1 and Γ2 that start and end on opposite faces of the crack. Notice thatΓ1 and Γ2 are not closed since they are disconnected at the crack faces. Let usexamine the values of the J-integral evaluated over these contours.

To this end, first we construct a closed contour Γ = Γ1 ∪ Γ+ ∪ Γ2 ∪ Γ− asindicated in figure 2. Notice that Γ is now a closed curve enclosing a regionwith no singularities. Therefore by our previous arguments, we know JΓ = 0.Splitting the contour integral over Γ into contributions from each of its foursubsets, we get

JΓ = JΓ1 + JΓ+ − JΓ− − JΓ2 = 0 (6)

Let us choose the coordinate system (x1, x2) in alignment with the crack asshown in figure 2. Notice that along the curves Γ±, we have t = 0 and n1 = 0.

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Therefore

JΓ± =

∫Γ±

(Wn1 − t · u,1) ds = 0. (7)

Eq. (6) hence shows that JΓ1 − JΓ2 = 0, i.e., that JΓ1 = JΓ2 . This property ofJ evaluated along contours emanating on one crack face and terminating on theother, is frequently referred to as the path-independence of property.

Questions: Examine the above discussion carefully.

(i) What is the reason behind the signs appearing in (6)?

(ii) Consider the contour Γ1 shown in figure 2. Will the sign of JΓ1 changeif we traverse Γ1 clockwise instead of counter clockwise?

(iii) In demonstrating the path independence property, did we explicitlyuse the form of the singularity of the crack tip fields?

(iv) Similarly, did we assume anything about the shape of the crack? Doesit have to be straight or can it be curved? Does the crack tip have tobe sharp or can we also permit notches?

Although we have shown that the value of J is independent of the path, wedo not know its value. In the context of LEFM, it can in fact be shown thatJ = G, the strain energy release rate! The computation is quite straightforwardbut occasionally tedious. The simpler case of a mode III crack may help convinceyou of this fact.

Questions: The mode III displacement field around a crack tip is given by

u3(x, y) =1

µIm

(2KIII√

2π

√x+ i y

).

From the G-K relationship, we know G = K2III/2µ. We would like to verify

our claim that J = G using the known displacement field around the crack.To this end, choose Γ to be a circular arc radius r and centered at the cracktip. Working in polar coordinates will help.

(i) What should be the angular coordinates of the end points of Γ?

(ii) Write down the unit outward normal to Γ at a generic point (r, θ).

(iii) Compute the value of the strain energy density W at (r, θ).

(iv) Compute the necessary derivatives of displacements, the strain andthen the stress.

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(v) Conclude that J = K2III/2µ

Solution: Although it is convenient to work in a polar coordinate sys-tem, notice that our definition of J explicitly refers to a Cartesian coordinatesystem at the crack tip. This will be a minor hassle and requires attention.From the given expression for the displacement, we have

u3(r, θ) =1

µIm

(2KIII√

2π

√reiθ

)=

2KIII√2πµ

√r sin

θ

2. (8)

The only non-zero components of stress and strain are σ31, σ32 and ε31, ε32

respectively. It is straightforward to verify that

σ32 + i σ31 =KIII√

2πz. (9)

Our objective then is to compute

J =

∫ π

θ=−π

(Wn1 − σk3

∂u3

∂x1nk

)rdθ. (10)

Let us evaluate the strain energy density:

W =1

2σijεij =

1

2(σ31ε31 + σ32ε32)

=1

2µ

(σ2

31 + σ232

)=|σ32 + i σ31|2

2µ

=K2III

4µπ|z|. (11)

The unit normal to Γ has components (n1, n2) = (cos θ, sin θ). Therefore,∫ π

θ=−πWn1 dΓ =

∫ π

θ=−π

K2III

4µπrcos θ rdθ = 0.

The first term in (10) hence makes no contribution to J. Towards computingthe remaining terms, we have

∂u3

∂x1= 2ε31 =

σ31

µ=

1

µIm (σ32 + i σ31) =

1

µIm

(KIII√

2πz

)= − KIII

µ√

2πrsin

θ

2

Next, we have

σk3nk = σ3r = µ∂u3

∂r=

KIII√2πr

sinθ

2. (12)

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We can now compute J:

J = −∫ π

θ=−πσk3

∂u3

∂x1nk rdθ

=

∫ π

θ=−π

(KIII

µ√

2πrsin

θ

2

)(KIII√

2πrsin

θ

2

)rdθ

=K2III

2πµ

∫ π

θ=−πsin2 θ

2dθ =

K2III

4πµ

∫ π

θ=−π(1− cos θ) dθ

=K2III

2µ,

which coincides with G for a pure mode III crack and is exactly the resultwe wanted to prove. The cases of mode I and II cracks can similarly beverified.

While it is reassuring to see that J is a familiar quantity in LEFM, wealso realize that it does not add any new information. Indeed, some of themain applications of J are in nonlinear fracture, and in particular, for studyingfracture of nonlinearly elastic and monotonically loaded elasto-plastic materials.This is a topic by itself and can be taken up in future lectures depending on theinterest of the class. We will not pursue such applications of J any further atthis point.

2 J-integral as a conservation law

Where did the J-integral come from? The proposed definition (1) appears tobe the work of a wizard, which is quite unsettling. Without providing all thenecessary details, we state here a useful result called Noether’s theorem that re-lates symmetries (coordinate tranformations) with conservation laws for certainvariational systems. We will first state the result in a simplified form, under-stand the statement, and see how it yields the J-integral in one stroke. At thevery least, this exercise should provide some understanding on the origin of theJ-integral as well as motivate the fact that there may be (and are) more suchintegrals in existence.

A version of Noether’s theorem: Let an elastic body occupy region Ω.Let u denote an admissible displacement field, ∇u its gradient, and let thestrain energy density be of the form W (x) ,W (x,u,∇u) so that the totalstrain energy of the body is given as

Φ(u) =

∫Ω

W (x,u,∇u) dΩ. (13)

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Consider a smooth family of coordinate transformations

ξ(x, η) = x + η f(x) (14)

and a family of vector transformations

h(u, η) = u + η g(u) (15)

each parameterized by the scalar η. If the one-parameter family of func-tionals

Φη(w) =

∫Ωη

W (ξ,h,∇ξh) dΩη. (16)

is such that Φη(v) = Φ(v) for all η and admissible v, and if u is a stationarypoint of the functional Φ, then∫

Γ

(Wfj + (gk − f`uk,`)

∂W

∂uk,j

)nj dΓ = 0 (17)

for every closed surface Γ bounding a regular subregion of Ω and n beingthe unit outward normal to Γ.

The statement of the theorem makes two important assumptions. First, thatu is a stationary point of the energy functional Φ and second, that there arecertain transformations of x and u that leave Φ invariant in some sense. Noticethat coordinate change x → ξ and the transformation v → h are possiblydifferent mappings for each value of η. With these assumptions, we get that acertain integral depending on the solution u vanishes over any arbitrary closedsurface Γ.

In the next section, we will examine what it means for u to be a stationarypoint, at least within the context of linear elasticity. For the moment, let us seehow we can recover the J-integral for a linearly elastic material from the aboveresult. For such a material, we know

W (x,v,∇v) =1

2∇v : C : ∇v.

Consider the coordinate transformation

ξ = x + η ei, for a fixed i,

which is just a translation along the direction ei, and the identity vector trans-formation h(v) = v. We see that

∇ξh(u) = ∇u, (18)

and hence

W (ξ,h(v),∇ξh(v)) = W (x,v,∇v).

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Furthermore, the Jacobian of the transformation x → ξ equals 1. Hence wetrivially get that

Φη(v) = Φ(v) =1

2

∫Ω

∇v : C : ∇v dΩ.

We have thus found one-parameter families of coordinate and vector transfor-mations that leave Φ invariant. Noether’s theorem then tells us that a certainintegral quantity equals zero. Let us see what it is. In the notation of thestatement, we have

f(x) = ei ⇒ fj(x) = δij ,

g(v) = 0⇒ gk − f`uk,` = −δi`uk,` = −uk,i, (19)

so that∫Γ

(Wfj + (gk − f`uk,`)

∂W

∂uk,j

)nj dΓ =

∫Γ

(Wδij − uk,iσk,j)nj dΓ

=

∫Γ

(Wni − t · ∂u

∂xi

)dΓ = 0, (20)

which coincides with out definition of the J-integral for the case i = 1. Hencewe have demonstrated to ourselves that:

(i) the conservation property of the J-integral is a natural consequence of acertain kind of symmetry inherent in elasticity,

(ii) there is not just one J-integral, but in fact one for each component i =1, 2, 3. Hence we can consider a vector valued J-integral, of which the onewe have discussed so far corresponds to one component.

(iii) the conservation property holds not just over contour integrals in two-dimensions, but also over surfaces in three-dimensions. This provides ahelpful way of computing stress intensity factors along the propagatingedge of a three dimensional crack surface.

Observe that we used a very simple coordinate transformation, namely a rigidbody translation along ei to arrive at what appears to be a nontrivial conserva-tion law. It definitely seems plausible that if we can find more such symmetrytransformation, we may be able to discover new conservation laws with deeperconsequences. You are strongly encouraged to find out about L and M integrals,which are conservation laws with important applications in micro-mechanicswith defects, and can be derived in this way.

Question:

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(i) The conservation property of the J-integral holds also for nonlinearlyelastic solids, with appropriate stress and strain measures. Indeed,notice that there is no mention of linear elasticity in the statement ofthe theorem. In our demonstration of J as a consequence of Noether’stheorem however, we made essential use of the assumption that thematerial was linearly elastic. Where?

(ii) Did we assume anything about the symmetry of the material? Forexample, did we assume that the material was isotropic?

(iii) Did we assume that the material distribution in the body was ho-mogeneous? That is, can C = C(x) or did we implicitly assumeC = a constant?

3 Linearized elastostatics from a variational prin-ciple

Noether’s theorem applies to systems that follow a variational principle. In thecurrent context, this means that the equations of elastostatic equilibrium bederivable as the minimization of an energy functional. We demonstrate herethat the equilibrium equations for a linearly elastic solid can indeed by derivedthis way. Equilibrium equations for bodies composed of hyper-elastic materialscan also be derived this way, but we omit this case to avoid introducing newstress/strain measures and concepts from continuum mechanics that may notbe familiar to all the students in the class.

In the following, we consider a linearly elastic body Ω composed of a materialwith constitutive relation σ = C : ε. Under imposed displacements u0 alongthe boundary ∂Ω, we would like to compute the response of the body, i.e., itsdisplacement field u.

3.1 Admissible displacements

Clearly, we need to write down the equations of static equilibrium and solve foru. This can be achieved with free body diagrams and invoking Newton’s secondlaw. Here, we follow a different idea, namely that of strain energy minimization.Recall that the strain energy corresponding to a displacement field v is givenby

Φ(v) =

∫Ω

W (ε[v]) dΩ, (21)

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where ε[v] = v(i,j) is the strain associated with v, and the strain energy densityW is defined as

W (ε) =1

2σ : ε =

1

2ε : C : ε. (22)

Observe that (21) defines Φ as functional, i.e., Φ assigns a scalar value Φ(v) to adisplacement field v. We follow the principle that the actual displacement field uof the body is one that in fact minimizes the strain energy Φ. How can be searchfor such a displacement? Well, to claim that u is a solution, we need to compareΦ(u) against the values of Φ realized over all other possible displacements v.In performing this comparison, we shall restrict our attention to displacementfields v that satisfy the prescribed boundary conditions. Additionally, we shallrequire that all candidate displacements v be sufficiently smooth so that allthe derivatives appearing in our computations will make sense. We call suchcandidate displacement fields as the set of admissible solutions. Hence ourobjective is to find u such that Φ(u) ≤ Φ(v) for all admissible solutions v.

3.2 Static equilibrium: Euler-Lagrange equations

We approach the minimization problem with tools from calculus. We shallrequest that u be a stationary point of the functional Φ by insisting that

〈δΦ(u),w〉 , d

dηΦ(u + ηw)

∣∣∣∣η=0

= 0, (23)

for all smooth vector fields w satisfying w = 0 along the boundary of Ω. Letus examine (23) closely. First, notice that any admissible displacement v canbe expressed in the form v = u + ηw with w satisfying homogeneous Dirichletbcs. Hence we are sampling the functional Φ along all possible perturbationsof the solution u. The derivative condition should be very reminiscent of thefirst derivative test in calculus. The intuition behind it is indeed the same—Φ has a stationary point at u if its directional derivative along all admissibleperturbations is zero. Make sure you understand (23) or consult your peers whohave seen this material before.

We understand (23) to be a necessary condition for u to be a solution of the

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linear elasticity problem. Let us simplify (23). We have

0 = 〈δΦ(u),w〉 =d

dη

(1

2

∫Ω

ε[u + ηw] : C : ε[u + ηw] dΩ

) ∣∣∣∣η=0

(24a)

=d

dη

(1

2

∫Ω

(ui,j + ηwi,j)Cijk`(uk,` + ηwk,`) dΩ

) ∣∣∣∣η=0

(24b)

=1

2

∫Ω

(wi,jCijkluk,` + ui,jCwk,`) dΩ (24c)

=

∫Ω

ui,jCijk`wk,` dΩ (24d)

=

∫Ω

σk`wk,` dΩ (24e)

=

∫Ω

((σk`wk),` − σk`,`wk) dΩ (24f)

=

∫Γ

w · (σn) dΓ−∫

Ω

w · divσ dΩ (24g)

= −∫

Ω

w · divσ dΩ. (24h)

In this way, we have arrived at a condition that for u to be a stationary pointof Φ, we should have

〈δΦ(u),w〉 = 0 =

∫Ω

w · divσ dΩ ∀w s.t. w|Γ = 0. (25)

Notice that (25) imposes a tremendous constraint on u, because we are com-pletely free to choose any smooth w that satisfies homogeneous Dirichlet bcs.A well known result called the fundamental lemma of the calculus of variationsin fact helps us conclude that the only way to satisfy (25) is if the integrand isitself zero, i.e., divσ = 0.

Hence we have found an alternate way of deriving the equilibrium equationsfor linear elasticity without ever drawing a free body diagram. The form ofthe equilibrium equation remains unchanged for nonlinear materials as well;the essential arguments remain the same as what we have discussed above.But to avoid introducing new concepts from continuum mechanics, we haveintentionally omitted the general case.

Questions:

(i) Justify the computation from (24a) to (24b).

(ii) Justify the computation from (24b) to (24c). In particular, evaluatethe the derivatives carefully.

(iii) What properties of C were used in going from (24c) to (24d).

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(iv) Justify the manipulation from (24e) to (24f).

(v) How did we switch from a domain to boundary integral in(24f)→(24g)?

(vi) Why does the boundary integral in (24g) vanish?

(vii) Justify the fundamental lemma of calculus of variations in one di-mension. That is, suppose that f : [0, 1] → R is a smooth function

and∫ 1

0f(x)g(x)dx = 0 for all smooth functions g : [0, 1] → R with

g(0) = g(1) = 0. Show that f(x) = 0 for all x ∈ (0, 1). Looking thisup on Wikipedia is OK.

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