Resumen y recomendaciones
CÓDIGOS DE PRÁCTICAS Y LINEAMIENTOS RECIENTES Y OTROS INSTRUMENTOS
In Section 2.2.4 I mentioned that not all strain fields are compatible and
Compatibility equations insure integrability of strains
therefore there is no guarantee that a given strain field can be integrated to a displacement field. The so-called Strain Compatibility Equations must be satisfied in order for this to be the case. In principle, the derivation goes as follows, see Fig. 4.1:
x2 x3 x1 Γ1 Q P Γ2
Fig. 4.1: Two integration paths.
1. Assume that the displacements uj(xPk) at point P are known.
2. Find the displacements uj(xQk) at another point Q by integrating the
strains εij and the rotations ωij or, equivalently, the displacement
gradients ui,j.
3. Demand that the integral is path independent. This is an obvious requirement because the displacements of a point must not depend on the integration path one has traveled to get to that point. In Fig. 4.1 two such paths, Γ1and Γ2are shown.
4.1It may be confusing that the same quantity is denoted ε
mnas well as emn. The reason that both notations are kept here is that they both are very common in the inter- national literature. To make the situation even more complex the permutation symbol is denoted eijkand eijin the two- and three-dimensional case, respectively, see (31.19) and (31.12).
Kinematics and Deformation
There is another way of getting the compatibility equations where different Other ways of obtaining the compatibility equations
expressions for the strains are differentiated and combined to get the re- quired result. Personally, I have never liked that procedure because it does not address the fundamental issue of integrability, i.e. path independence, directly, but see e.g. (Malvern 1969).
Among the procedures that are based on requiring path independence Sokolnikoff (1956) and Pearson (1959) are worth mentioning. Here, however, the derivation below lies closer to the ones found in the books by Fung (1965), by Fung & Tong (2001) and by Reismann & Pawlik (1980).4.2
As you will see, not all of the manipulations below are straightforward— actually a number of them seem strange, but I intend to explain them as far as is possible for me.
In order to find the displacements uQj at point Q simply write uQj = uP j + Z Q P duj with uQj ≡ uj(xQi) and uPj ≡ uj(xPi) (4.6) or uQj = uPj + Z Q P uj,kdxk= uPj + Z Q P εjkdxk+ Z Q P ωjkdxk (4.7)
where we have exploited the fact that (4.5) and (4.3) combined provide εjk+ ωjk=12(uj,k+ uk,j) +12(uj,k− uk,j) = uj,k (4.8)
The question of path independence is now equivalent to establishing the conditions for path independence of the sum of the two integrals in (4.7).
If we can get rid of the rotation term in (4.7)—and hopefully write it in terms of the strain—it seems likely that we can get the relation between the strains or strain derivatives that is necessary for integrability. Therefore, we try to rewrite in terms of the displacement gradients in the hope that we may end up with an expression involving only the strains, possibly their derivatives. First, however, write
ωjkdxk= d(ωjkxk) − xkωjk,ldxl (4.9)
With this in hand, get uQj = uPj +ωjkxkQP+
Z Q P
εjl− xkωjk,ldxl (4.10)
We wish to write the term involving the derivative of the rotation in terms of the derivatives of the strain.
4.2Originally, my inspiration was notes found on the Internet and written by L. Schmerr of Iowa State University for the course EM 424. Dr. Schmerr was so kind as pointing me in the direction of Fung (1965) and Reismann & Pawlik (1980).
Therefore, note that
ωjk,l=12(uj,kl− uk,jl) =12(uj,kl− uk,jl) +12(ul,jk− ul,kj) (4.11)
where the last term obviously vanishes because of symmetry in indices k and j. Rearranging of terms provides
ωjk,l=12(uj,lk− ul,jk) −12(uk,lj− ul,kj)
=1
2(uj,l− ul,j)k−12(uk,l− ul,k)j
= εjl,k− εkl,j
(4.12)
Thus, (4.6) or (4.7) may be written uQj = uP j + ωjkxk Q P + Z Q P εjl− xk εjk,l− εkl,jdxl (4.13)
where the bracketed term [ωjkxk]QP contains two terms that are associated
with the end-points, not the integration path. Therefore, we may concen- trate on the integral, and inspection of (4.13) may show us that it is of the form
Z Q P
Φjldxl, where Φjl≡ εjl− xk εjl,k− εkl,j (4.14)
The necessary and sufficient condition for path independence of the inte- gral in (4.14) is that the integrand is an exact differential, see e.g. (Kreyszig 1993) Φjl,m= Φjm,l or 0 = Φjl,m− Φjm,l (4.15) Therefore, compute Φjl,m= εjl,m− δkm εjl,k− εkl,j− xk εjl,km− εkl,jm = εjl,m− εjl,m− εml,j− xk εjl,km− εkl,jm Φjm,l= εjm,l− δklεjm,k− εkm,j− xk εjm,kl− εkm,jl = εjm,l− εjm,l− εlm,j− xk εjm,kl− εkm,jl (4.16)
When we exploit the symmetry in the indices of εijthe first two terms
on the right-hand sides vanish and, utilizing the same symmetry, the third terms in Φjl,mand Φjm,lcancel each other when inserted into (4.15b). Thus,
the requirement for integrability becomes
0 = xk εjl,km− εkl,jm− εjm,kl+ εkm,jl (4.17)
Kinematics and Deformation equations ensue
Compatibility equations
0 = εjl,km− εkl,jm− εjm,kl+ εkm,jl (4.18)
but some of these are just identities and therefore do not provide useful information, and others are mere repetitions since many of the expressions exhibit symmetry, and in the end there are only six seemingly independent equations The 6 (independent?) compatibility equations ε11,23= (−ε23,1+ ε31,2+ ε12,3),1 ε22,31= (−ε31,2+ ε12,3+ ε23,1),2 ε33,12= (−ε12,3+ ε23,1+ ε31,2),3 2ε12,12= ε11,22+ ε22,11 2ε23,23= ε22,33+ ε33,22 2ε31,31= ε33,11+ ε11,33 (4.19)
Most authors claim that in reality there are only three independent rela-
tions, while others insist that there are four. I do not intend to investigate Number of independent compatibility equations between 3 and 6
this problem, and conclude with the comment that if a given strain field sat- isfies all six equations, then it is compatible and may therefore be integrated and thus provide a displacement field. In that case, it is less important if the six relations are not all independent. Furthermore, in many cases we determine the displacement field without integrating the strains and, in that case, the strains are obviously compatible.
Although it does not appear explicitly in the above derivations the con- ditions for strain compatibility presuppose that the body—the domain—is simply connected, i.e. that it does not contain holes. In the case of a mul- tiply connected region it may be converted into a simply connected one by application of the trick used in Section 13.6.3.2, see especially Fig. 13.6. However, since most of this book is concerned with methods that determine the displacements without first postulating a strain field we shall leave the question of compatibility here, but refer to the book by Fung (1965).
4.2.3
Kinematic Boundary Conditions
Especially for the present case with infinitesimal displacements the kine- Kinematic boundary conditions
matic boundary conditions for a three-dimensional body are usually quite self-evident. Therefore, we do not go into any detail at this point but refer to Part II, where the problem is much more prominent.