= the relative displacement of P 2 with respect to P 1 Breaking that relative displacement into its components, we have

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1 EM 44: ompatibility Equations 1 STRAIN OMATIBILITY EQUATIONS onsider a body with displacements at points 1 and given by u 1 and u, respectively, as shown in Figure 1: 1 u 1 u Figure 1 Let = u u 1 = the relative displacement of with respect to 1 Breaking that relative displacement into its components, we have Then those components can be written in terms of integrals of the local strains and rotations as: But, = 3 e i i =1 u = du = dx = + dx 3 3 i ( ε ω ) i i i i = 1 x = 1 3 ω ω i dx = d( ω i x ) x i k=1 dx k so that 3 3 = ω i x + εik 3 x ω i dx =1 1 =1 x k k=1 1 k By differentiating the strain-displacement relationship it can be verified that

2 EM 44: ompatibility Equations these derivatives of the local rotation can also be written in terms of derivatives of the strains: ε ik x ε k = 1 u i x u = ω i giving 3 3 = ω i x + ξik dx k =1 1 k=1 1 3 ε where ξ ik = ε ik x ik =1 x ε k The first term in the above expression only depends on 1 and. The second term in is independent of the path between those two points (and hence only depends on 1 and ) if, for a simply connected region (a region with no holes) we have ξ ik = ξ il (1) (see Wylie,.R., Advanced Engineering Mathematics, 4 th Ed., McGraw-Hill, p. 684). Thus, if Eq. (1) is satisfied everywhere in a simply connected region the displacement will have a single value everywhere in that region (the displacement is single valued but not unique since we can always add a rigid body displacement that does not change the strains). The derivatives contained in Eq. (1) can be written as ξ ik ξ il = ε ik ε ik = ε il ε il ε lk ε kl 3 ε x ik ε k =1 x 3 ε x il ε l =1 x () However, the first two terms in the above equations cancel and the third terms in the above equations also cancel when placed back into Eq. (1), reducing Eq. (1) to:

3 EM 44: ompatibility Equations 3 ε ik + ε l ε k ε il =0 x x 3 x =1 But, since the x are independent, the quantity in the brackets must vanish, and We will write Eq.(3) symbolically as ε l ε ik + ε il x x x (3) k Since i,, k, and l can all have values ranging from 1 to 3, Eq. (4) looks like a total of 81 equations. However, because of the following symmetries and anti-symmetries: ε k R ikl (4) R ikl = R kli R ikl = R ikl = R ilk it turns out that there are really only six distinct terms in Eq. (4) given by the compatibility conditions: S 11 R 33 S R 3131 S 33 R 11 S 1 = S 1 R 331 S 3 = S 3 R 311 S 13 = S 31 R 13 (5) which we have written in terms of the components of a symmetric S matrix: S 11 S 1 S 13 S 1 S S 3 S 31 S 3 S 33 All six of the compatibility equations listed in Eq. (5), however, are not independent since the components of the S matrix can be shown to satisfy the three equations:

4 EM 44: ompatibility Equations 4 3 S i i = 1,,3 =1 x so that these six compatibility equations really represent only three independent conditions that the strains must satisfy. Explicitly, these six equations are ε 3 x ε ε 33 x ε 31 ε 33 ε 11 ε 1 ε 11 ε x x ε 33 + ε 1 ε 3 ε 31 x x ε 11 + ε 3 ε 31 ε 1 x x ε + ε 31 ε 1 ε 3 x x x Because these six equations are not independent, it is difficult in general 3-D problems to use these compatibility equations directly. However, for plane strain problems, we have only two non-zero displacements since plane strain conditions require which implies that u 1 = u 1, x u = u, x u 3 ε 33 = ε 13 = ε 3 ε 11 = ε 11, x ε = ε, x ε 1 = ε 1, x so there is only one compatibility equation that is not identically zero given by ε 1 x ε 11 x ε

5 EM 44: ompatibility Equations 5 For plane stress problems, on the other hand, we have the conditions σ 13 = σ 3 = σ 33 σ 11 = σ 11, x σ = σ,x σ 1 = σ 1,x which implies that ε = ε but ε ( x, x ) ( x, x ) ( x, x ) so that the compatibility equations reduce to ε ε ε = ε = ε = ε ε 33 x ε 33 ε 33 x ε 1 x ε 11 x ε In general, the first three of the above conditions cannot be satisfied exactly, so that plane stress conditions can only be approximately true in a body. For thin bodies, however, this approximation can be usually ustified. Note that the compatibility equation for the inplane strains in the case of plane stress is identical to the compatibility equation for plane strain. For bodies with holes these compatibility equations are not sufficient to guarantee that the strains can be obtained from a single-valued displacement field. In fact, in bodies with holes there may be some cases where we want to have displacements that are not single valued such as the split ring shown in Fig.:

6 EM 44: ompatibility Equations 6 1 D Fig. where the point on one side of the split is fixed (displacement ) and the x- displacement of the same point on the other side of the split = D. In this case, we see that 1 du du D = 1 = 0 =,3 To ensure that a multiply connected body like the ring shown above cannot split in this fashion, we must supplement the compatibility equations by additional conditions. For a body with m holes as shown in Fig. 3 if, in addition to the compatibility equations, we require the m subsidiary conditions, i du i = 1,,..., m where the integrals are taken around each hole, then the displacements will be singlevalued. Note that in solving any problem, if we end up obtaining directly a displacement field that represents the solution to the desired problem, then compatibility is not an issue since we can always generate a set of strains that are compatible with those displacements simply by taking the appropriate displacement derivatives. When we directly solve a

7 EM 44: ompatibility Equations 7 problem only for the strains (or corresponding stresses), however, it is not automatically guaranteed that a well-behaved, single-valued displacement field can be obtained through integrating those strains. The compatibility equations (and any subsidiary conditions needed for multiply connected bodies) provide the guarantee that a well-behaved, singlevalued displacement field can be obtained from the given strain field. 1 m Fig. 3

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