Mechanical Behaviour of Engineering Materials - Metals, Ceramics, Polymers and Composites 2010 Part 13 docx

D The J integral In this appendix, the J integral, introduced in chapter 5, will be derived and discussed in detail. The derivation starts by introducing some concepts of vector calculus. It is based on Gross / Seelig [58] and Rice [118]. D.1 Discontinuities, singularities, and Gauss’ theorem Gauss’ theorem relates the surface integral and the volume integral of a vector- valued function F (x):  V  ∇ · F (x)  dx =  S F (x) · n dS . (D.1) V is the integration volume with surface S, and n is the normal vector on this surface. Gauss’ theorem can be interpreted as follows: A vector field F (x) can be visualised as consisting of flux lines filling space as shown in figure D.1. The divergence ∇·F (x) of a vector field is a measure of its source strength. If the divergence is zero in a region of space, the number of flux lines entering and leaving the volume is the same (region 1 in figure D.1). If this is the case in the whole space, the flux lines do not end or begin anywhere; they are closed. If the divergence is non-zero, flux lines begin at this point (a source) or they end there (a sink), as in region 2 of figure D.1. This is exactly the statement of Gauss’ theorem: The volume integral is a measure of the total source strength within V , the surface integral counts the flux lines entering and leaving the volume. The normal vector in the surface integral ensures that ingoing and outgoing flux lines are counted with opposite signs. Gauss’ theorem is not universally valid, but only if the function F (x) is con- tinuously differentiable with continuous derivative. If, however, the function possesses a singularity (i. e., if its value approaches infinity), Gauss’ theorem cannot be used and the left-hand and the right-hand side of equation (D.1) are not identical anymore. 474 D The J integral vector field region 1 region 2 Fig. D.1. Plot of a vector field F (x). In region 1, ∇ · F (x) = 0 holds, in region 2 ∇ · F (x) = 0 Q x 1 x 2 E Fig. D.2. Electrical field on the surface of a sphere sur- rounding a point charge This fact can be exploited by using Gauss’ theorem to check whether a function with divergence zero contains a discontinuity or singularity in a certain region. If the surface integral  S F (x)·n dS of a function F (x) with ∇· F (x) = 0 yields a n on-z ero value, there must be a discontinuity or singularity in the volume V enclosed by S. This can be illustrated using an example from electrostatics: The diver- gence of the electrical field E(x) vanishes in vacuum, ∇ ·E(x) = 0, implying that the flux lines of the field do not end in free space . If, however, there is an electrical charge, it acts – depending on its sign – as a source or sink of flux lines. The electrical field of a small sp here with charge Q situated in the origin of our coordinate system (figure D.2) fulfils the equation E(x) = Q 4πε 0 x |x| 3 (D.2) away from the charge. A simple calculation shows that the divergence of this function vanishes. To prove this, we write E(x) in Cartesian coordinates for each compo- nent: E i (x) = Q 4πε 0 · x i (x 2 1 + x 2 2 + x 2 3 ) 3 /2 . In Cartesian coordinates, the divergence operator is ∇ = (∂/∂x 1 , ∂/∂x 2 , ∂/∂x 3 ) . Applying this to the field, we find ∇ · E(x) = ∂E i ∂x i = ∂E 1 ∂x 1 + ∂E 2 ∂x 2 + ∂E 3 ∂x 3 = 0 D.2 Energy-momentum tensor 475 for x = 0. The function E(x) contains a singularity at x = 0. If we integrate the electrical field over the surface S with radius R, we find  S E(x) · n dS = Q 4πε 0  S x n |x| 3 dS = Q 4πε 0  S R R 3 dS = Q 4πε 0 1 R 2  S dS = Q 4πε 0 1 R 2 · 4πR 2 = Q ε 0 . As can be seen, the surface integral does not vanish although the divergence of the field vanishes at the integration surface. Thus, the surface integral can be used to probe for charges in a volume. This is especially useful if the charge is a point charge because in this case integrating over the volume would be problematic because of the singularity. A similar problem occurs in media containing cracks. In this case, there is a singularity (for example in the stress field) at the crack tip. Furthermore, the detailed conditions near the crack tip may be unknown, although they may be known at some distance. Thus, if we can find a quantity with normally vanishing divergence that contains a singularity or discontinuity at the crack tip, this quantity can be used to gain information on the crack by integrating over a surface far away. This idea is pursued in the following. D.2 Energy-momentum tensor Our task is to find a physical quantity that can be defined for any elastic- plastic material, that has a vanishing divergence, and that becomes discontin- uous or singular at a crack tip. The most obvious choice would be the stress tensor σ. If there are no volume forces, the stress tensor fulfils the equation ∇·σ = 0 because stresses are generated only where forces act. Using the stress tensor, however, has a severe disadvantage because the stress is frequently prescribed by the external load. This is illustrated in the following example: Consider a tensile test specimen made of two materials with different Young’s modu li E 1 and E 2 (figure D.3). The specimen is loaded with con- stant force at its end. In this case, the stress is constant anywhere within the s pecime n so that any surface integral over the stress vanishes within the material. 1 The discontinuity in the material properties thus cannot be found using such an integral. Another quantity whose values differ in both halves of the specimen is the energy density w =  σ ij dε ij . At a given stress, the strain is larger in 1 For simplicity, transversal contraction is neglected here. 476 D The J integral F F S E 1 E 2 Fig. D.3. Elastic medium with a discontinuous change in Young’s modulus. Also shown is an integration surface S the region with smaller Young’s modulus, corresponding to a larger energy density. The energy density itself is not a suitable quantity, however, because Gauss’ theorem requires a vector-valued function. Nevertheless, it is a good idea to use a quantity containing the energy density. One such quantity is the energy-momentum tensor T, defined by T ij = w ·δ ij − σ jk ∂u k ∂x i . (D.3) Here, w is the energy den sity, σ the stress tensor, u the displacement vector, and δ ij is the Kronecker delta introduced in appendix A.6. The derivation of the energy-momentum tensor is way beyond the scope of this book. The name stems from classical field theory which deals with arbitrary physical fields like the electromagnetic field, the velocity field in a fluid, or the strain field in an elastic medium. In field theory, a generalised tensor is used, with s ome components describing the en- ergy and momentum density of the system. In the context of elasticity theory, the name energy-momentum tensor is misleading because none of its components are the energy or momentum density. A detailed, but mathematically involved, introduction to the subject can be found in Landau / Lifschitz [86]. The energy-momentum tensor has the desired property of a vanishing diver- gence:  ∇ · T  i = ∂ ∂x j T ij = 0 (D.4) for j = 1 . . . 3. This is easy to show if we take the equation ∂σ ij /∂x j = 0 into account that is valid if there are no volume forces. We thus get not only one, but three quantities to be used in Gauss’ theorem to test for discontinuities or singularities. D.3 J integral To further study the energy-momentum tensor, we take another look at the example of the medium with two different Young’s moduli (see figure D.3). The D.3 J integral 477 stress in the specimen is everywhere equal to the external stress σ; the strain is constant in each half of the specimen, with 11-component ε (n) = σ/E (n) , where the superscript ‘n’ denotes the two halves. Using the energy density of a linear-elastic material, w = σε/2, and the Poisson’s ratio ν (assumed to be the same in both parts), the energy-momentum tensor is T (n) ij = w (n) δ ij − σ (n) jk ∂u (n) k ∂x i i. e., T (n) = σ 2 2E (n)   1 0 0 0 1 0 0 0 1   −   σ 0 0 0 0 0 0 0 0   ·   ε (n) 0 0 0 −νε (n) 0 0 0 −νε (n)   = σ 2 2E (n)   1 0 0 0 1 0 0 0 1   −   σ 0 0 0 0 0 0 0 0   · σ E (n) ·   1 0 0 0 −ν 0 0 0 −ν   = σ 2 2E (n)   1 0 0 0 1 0 0 0 1   − σ 2 E (n)   1 0 0 0 0 0 0 0 0   = σ 2 2E (n) ·   −1 0 0 0 1 0 0 0 1   . The energy-momentum tensor is thus rather simple and is constant in each half of the material. What happens if we apply Gauss’ theorem? As already stated, we have three possibilities, one for each column of the matrix representation of the energy-momentum tensor. We consider the surface S in figure D.3 and define the three quantities J 1 , J 2 , and J 3 , called the J integrals, J i =  S T ij · n j dA , (D.5) where n is the normal vector on the surface [118]. Since the tensor is constant in each half of the material, the four side faces of the integration volume cancel (for each face, there is an opposite face with opposite normal vector). We have to look at the two end faces with surface area A only. This results in J 1 =  A 1 −σ 2 2E (1) · n 1 dA +  A 2 −σ 2 2E (2) · n 2 dA = σ 2 A 2  − 1 E (1) + 1 E (2)  , J 2 = 0 , J 3 = 0 . (D.6) 478 D The J integral As can be seen, two of the three J values vanish, and only J 1 is non-zero. Because the discontinuity surface has a normal vector in the 1-direction, this is a plausible result. The J integral can also be interpreted directly: Looking at its unit, we see that it has the unit of a force, but it is not too obvious what kind of force it is and what it is applied to. A better interpretation can be found if we multiply the J integral with a distance dx 1 . The resulting quantity dΠ = J 1 dx 1 is an energy. If we remember that the energy density in the elastic media is w = σ 2 /2E (n) , we see that dΠ is the energy that would be released if we were to shift the interface in 1 direction by a distance dx 1 . The same interpretation is valid for the other two J integrals: If we shift the interface in the 2 or 3 direction, nothing changes, and the energy release is zero. The J integral can thus be interpreted as the differential energy release, the change in energy when the discontinuity in the system is shifted by an infinitesimal distance. So far, we considered the J integral only for the case of a medium with an elastic discontinuity. The results, however, are universally applicable. Thus, if S is an arbitrary surface with normal vector n enclosing a discontinuity or singularity of the system, each component of the J integral J is defined as the surface integral J i =  S T ij · n j dA . (D.7) The differential energy release during an infinitesimal displacement dx of the position of the ‘disturbance’ is dΠ = J ·dx. This simple interpretation of this quantity as energy release is only valid in elastic (linear) media. In elastic- plastic media, dΠ is the difference of the energy of two systems in which the position of the disturbance differs by dx, but it is not always ensured that this energy would in fact be released if the disturbance would actually move by this distance (for example, when a crack propagates). This will be explained in more detail below. One important property of the J integral, following directly from what we saw so far, is that it is independent of the integration surface. As long as it encloses the disturbance, the exact choice of the surface is irrelevant. This was already obvious in the example of the surface integral containing a point charge: As long as the charge is enclosed, the value of the integral is indepen- dent of the radius of the sphere and always equals Q/ε 0 . The same holds for the medium with elastic discontinuity; again, the exact shape of the surface does not matter, and only the enclosed surface of the disruption is important. This property is important because it allows to perform numerical calculations (for example, using finite elements) without calculating the detailed conditions near the crack tip – it is sufficient if the stress and displacement fields in some distance from the crack tip are correct. D.4 J integral at a crack tip 479 h x 2 x 1 C Fig. D.4. Simple configuration to evaluate the J integral D.4 J integral at a crack tip In the following, we will use the J integral to understand material beh aviour at a crack tip. The considerations are limited to the two-dimensional case i. e., a system in a state of plane strain or stress. In this case, the integration surface is chosen to have the same cross section at any x 3 position. If we fix an arbitrary x 3 value, the integral is not a surface integral anymore, but only a path integral 2 . In two dimensions, the J integral in i direction is thus defined as J i =  C T ij · n j ds . (D.8) Here, C denotes the integration path (or contour), n is the normal vector on this path, and ds is an infinitesimal path element along the curve. The path of integration has to enclose the crack tip. Apart from that, it is arbitrary as explained above. This path independence explains why the J integral can be used to charac- terise a crack and is independent of other aspects of the material’s state. If we choose the integration path close to the crack tip, it is plausible that the value of the J integral d epe nds only on the state there. Due to the path indepen- dence, the value of the integral does not change if we continuously move the path away from the crack tip, allowing us to use a distant integration path. We now want to study the J integral using the simple example from fig- ure D.4. The figure shows a crack in x 1 direction in a material that is infinite in the x 1 and x 3 direction and has a height h. The material is clamped at the upper and lower end, with constantly prescribed displacement u on both boundaries. The integration path C is chosen as shown in the figure. It starts at one crack surface and ends at the other. In principle, the contour has to be closed but because there is no material within the crack, the energy- momentum tensor vanishes there. Since the crack is in the x 1 direction, we choose the J 1 integral J = J 1 =  C  w dx 2 −  σ · ∂u ∂x 1  · n ds  . (D.9) 2 Care has to be taken in evaluating this path integral: In most path integrals o ccurring in mathematics or physics, the integration variable is a vector tangential to the curve describing the integration path. This is different here: because the path integral is in fact a dimensionally reduced surface integral, the quantity ds is a scalar and the vector n j ds is perpendicular to the curve. 480 D The J integral If we move the x 1 coordinate of the vertical parts of C to ±∞, it is easy to calculate the value of the J integral. The two vertical parts above and below the crack at x 1 = −∞ do not contribute to the integral because the energy density and the derivative of the displacement are zero. The integration along the clamped ends does not contribute as well because the integration over the energy density vanishes (dx 2 is zero along a path in x 1 direction) and because the second term also vanishes due to ∂u/∂x 1 = 0. The only remaining part is the path at +∞. In this region, ∂u/∂x 1 = 0 holds, so it is only the energy density w ∞ that contributes to the integral: J 1 = w ∞ h . (D.10) It is somewhat problematic to use the energy interpretation of the J integral in an infinitely extended system because the total energy of the system is infinite. We can try to argue as follows: If we shift the crack tip in x 1 direction by dx 1 , some energy is released, and since the configuration after the shift is the same as before (only disp laced by dx 1 ), this energy release can be written as w ∞ h dx 1 . This argument is a bit obscure, however, because if the configuration is the same, its energy must be the same as well. The problem is that we c onside r the differen ce between two infinite quantities. Nevertheless, this simple example shows that the J integral may serve to characterise a crack in an elastic material. According to the statements made above, the J integral can only be non- zero if there is a singularity in the energy-momentum tensor. This is indeed the case if there is a crack tip in an e lastic material As shown in section 5.2.1, stresses and strains become infinite if we approach the crack tip, see equa- tion (5.1): 3 σ ∝ 1 √ r , ε ∝ 1 √ r . (D.11) r denotes the distance from the crack tip. Such a singularity may seem unphysical. However, this is not the case be- cause stresses, strains, and energy densities are only relative quantities that cannot be measured directly. The strain, for example, is the normalised differ- ence of the displacements at two points. The displacement itself cannot take infinite values, but its change may, if normalised to an infinitesimally small distance. The stress, defi ned as f orce per unit area, may become singular as long as the forces in the medium stay finite. 4 The energy density may also 3 A simplified argument for this is given below. 4 Strictly speaking, continuum mechanics cannot be used at the crack tip because we must not neglect the fact that matter consists of atoms. This is discussed in exercise 13. D.5 Plasticity at the crack tip 481 become infinite provided the energy of the system (the integral over the en- ergy density) stays finite. According to equation (D.11), the energy density w =  σ dε is proportional to 1/r. In cylindrical coordinates, we can write w(r, φ) = w φ (φ)/r. If we integrate the energy density within a circle C with radius R around the crack tip, we find  C w(x) dA =  π 0  R 0 w(r, φ)r dr dφ =  π 0  R 0 w φ (φ) r r dr dφ =  π 0 w φ (φ) dφ ·  R 0 dr = R  π 0 w φ (φ) dφ . The energy stored in the region near the crack tip is thus finite. D.5 Plasticity at the crack tip The considerations made so far were valid in an elastic material. The J integral is also to be used if the material is yielding close to the crack tip. Again, stresses, strains, and energy densities become singular in this case. The kind of singularity can be analysed – in a slightly simplified argument – as follows: If we integrate the J integral along a circular path with radius R, we find J 1 =  π −π  wn 1 −  σ ∂u ∂x 1  n  R dφ . (D.12) The J integral must be path-independent as we saw above and thus have the same value for all values of R. If we consider a small vicinity of the crack tip, it is plausible to assume that the influence of the stress state far away from the crack tip (i. e., the influence of the exact geometry of the component) becomes less and less important and everything is determined by the geometry of the crack (and the mode of loading). 5 Furthermore, we assume that the energy-momentum tensor can be written as T (r, φ) = T r (r)T φ (φ). The independence of the integral on the radius of the integration path thus means that the value of the energy-momentum tensor must be proportional to 1/R. Thus, we find T r (r) = 1/r. The energy den- sity and the product σ ij ε ij are each proportional to 1/r. In the linear-elastic case, where σ ∝ ε, this implies that σ and ε are proportional to 1/ √ r (see equation (D.11)). For the case of a plastic material, we can assume the following relation between stress and strain, using equation (3.16), ε = K −1/n σ 1/n , (D.13) 5 This assumption is justified with hindsight by the fact that we find a singularity in the energy-momentum tensor, showing that external influences become negligible if we approach the crack tip. 482 D The J integral if elastic parts of the strain are small compared to the plastic parts. Using equation (D.13) and σε ∝ 1/r yields 6 σ(r, φ) ∝ 1 r n/(n+1) σ φ (φ) , (D.14) ε(r, φ) ∝ 1 r 1/(n+1) ε φ (φ) (D.15) for the singularity near the crack tip. The smaller the exponent n becomes, the stronger is the singularity in the strain and the weaker is the singularity in the stress. What is the role of the J integral in this context? Because of the indepen- dence of the stress- and strain fields from external influe nce s far away from the crack tip, all stress fields that we can find for a given crack geometry and mode of loading (mode I, mode II, or mode III) are similar. They just differ by a factor specifying the amount of loading. This factor is nothing but the J integral. Thus, we can write T (r, φ) = (J/r)T φ (φ). If we raise the external stress, the value of the energy-momentum tensor changes accordingly. The equations for the stress and strain are thus σ(r, φ) = c σ  J 1 r  n/(n+1) σ φ (φ) , (D.16) ε(r, φ) = c ε  J 1 r  n/(n+1) ε φ (φ) . (D.17) The constants c σ and c ε depend on the geometry, the exponent n, and the mode of loading, bu t not on the load strength. The J integral is thus a measure quantifying the stress and strain field. The same is true for the linear-elastic case from the previous section. D.6 Energy interpretation of the J integral As already detailed in section D.3, the J integral can be interpreted as specific energy difference between two systems that differ by an infinitesimal displace- ment of the singularity or discontinuity. 7 This will be proven here. To do so, we consider two three-dimensional bodies with an axially sym- metric cavity. The task is to calculate the energy difference between the bodies (see figure D.5). Both bodies are assumed to be absolutely identical except for the cavity being larger in body B than in body A by an infinitesimal volume ∆V . We assume both bodies to be elastic, but not necessarily linear- elastic. This assumption is important because the work done by a deformation on an elastic body is independent of the deformation history. Furthermore, we 6 Frequently, an exponent m = 1/n is used ins tead of n in these equations. 7 This may be, for instance, due to crack propagation. [...]... 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