EM 424: Compatibility Equations
1
STRAIN COMPATIBILITY EQUATIONS
Consider a body with displacements at points P1 and P2 given by u1 and u2, respectively,
as shown in Figure 1:
P
C
2
u2
P
1
u1
Figure 1
Let ∆ = u2 − u1 = the relative displacement of P2 with respect to P1
Breaking that relative displacement into its components, we have
3
∆ = ∑ ∆i ei
i =1
Then those components can be written in terms of integrals of the local strains and
rotations as:
P2
3 P2
P
3 2
∂ui
dx j = ∑ ∫ ( ε ij + ωij ) dx j
j =1 P1 ∂x j
j =1 P1
∆ i = ∫ dui = ∑ ∫
P1
But,
ω ijdx j = d (ω ij x j ) − ∑ x j
3
k =1
∂ω ij
dx
∂x k k
so that
3
P2
j =1
P1
∆ i = ∑ ω ij x j
3 P2 
3
∂ω ij 
+ ∑ ∫  ε ik − ∑ x j
 dx
k =1 P1 
j =1
∂x k  k
By differentiating the strain-displacement relationship it can be verified that
EM 424: Compatibility Equations
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these derivatives of the local rotation can also be written in terms of derivatives of the
strains:
∂ε ik ∂ε jk 1 ∂  ∂ui ∂u j  ∂ω ij
−
=
−

=
2 ∂x k  ∂x j ∂xi  ∂x k
∂x j
∂xi
giving
3
P2
j =1
P1
∆ i = ∑ ω ij x j
where
3 P2
+ ∑ ∫ ξik dx k
k =1 P1
3
 ∂ε ik
j =1
 ∂x j
ξik = ε ik − ∑ x j 
−
∂ε jk 

∂x i 
The first term in the above ∆ i expression only depends on P1 and P2 . The second term in
∆ i is independent of the path C between those two points (and hence only depends on P1
and P2) if, for a simply connected region (a region with no holes) we have
∂ξik ∂ξil
=
∂xl
∂xk
(1)
(see Wylie, C.R., Advanced Engineering Mathematics, 4th 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
∂2 ε jk 
∂ξik ∂ε ik  ∂ε ik ∂ε lk  3  ∂2 ε ik
=
−
−
−

 − ∑x 
∂xl
∂xl  ∂xl
∂xi  j =1 j  ∂x j ∂xl ∂x i∂x l 
∂2 ε jl 
∂ξil ∂ε il  ∂ε il ∂ε kl  3  ∂2 ε il
=
−
−
−

 − ∑x 
∂xk ∂x k  ∂xk ∂xi  j =1 j  ∂x j∂x k ∂xi ∂xk 
(2)
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:
EM 424: Compatibility Equations
3
2
2
 ∂2 ε
∂ ε jl
∂ ε jk
∂2 ε il 
ik
∑ x j
+
−
−
 =0
j =1  ∂x j ∂xl
∂x i∂x k ∂xi ∂xl ∂x j ∂xk 
3
But, since the xj are independent, the quantity in the brackets must vanish, and
∂ ε jl
∂ ε jk
∂2 ε ik
∂ 2ε il
+
−
−
=0
∂x j∂x l ∂x i∂x k ∂xi ∂xl ∂x j∂x k
2
2
(3)
We will write Eq.(3) symbolically as
Rijkl = 0
(4)
Since i, j, 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:
Rijkl = Rklij
Rijkl = − R jikl = −Rijlk
it turns out that there are really only six distinct terms in Eq. (4) given by the
compatibility conditions:
S11 ≡ R2323 = 0
S22 ≡ R3131 = 0
S33 ≡ R1212 = 0
S21 = S12 ≡ R2331 = 0
(5)
S23 = S32 ≡ R3112 = 0
S13 = S31 ≡ R1223 = 0
which we have written in terms of the components of a symmetric S matrix:
 S11

S21
S31
S12
S22
S32
S13 

S23 
S33 
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:
EM 424: Compatibility Equations
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∂Sij
= 0 (i = 1,2,3)
j =1 ∂x j
3
∑
so that these six compatibility equations really represent only three independent
conditions that the strains must satisfy. Explicitly, these six equations are
2
∂ ε 23
∂ ε 22 ∂ ε 33
−
−
=0
∂x 2 ∂x3
∂x32
∂x22
2
∂2 ε 31 ∂ 2ε 33 ∂2 ε11
−
−
=0
2
2
∂x 3 ∂x1
∂x1
∂x3
2
2
2
∂2 ε12
∂2 ε11 ∂ 2 ε 22
2
−
2 −
2 =0
∂x1 ∂x2
∂x2
∂x1
∂ 2 ε 33 ∂ 2ε 12
∂2 ε 23
∂2 ε 31
+
−
−
=0
2
∂x1 ∂x 2
∂x3
∂x3 ∂x1 ∂x 3∂x 2
∂ 2ε 11 ∂2 ε 23
∂2 ε 31
∂2 ε12
+
−
−
=0
∂x 2 ∂x3
∂x12
∂x1∂x2 ∂x1∂x3
∂2 ε 22
∂2 ε 31
∂2 ε12
∂ 2ε 23
+
−
−
=0
∂x 3∂x1 ∂x 22
∂x 2 ∂x3 ∂x2 ∂x1
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
u1 = u1 (x1 , x2 )
u2 = u2 (x1 , x2 )
u3 = 0
which implies that
ε 33 = ε 13 = ε 23 = 0
ε11 = ε 11(x1 , x2 )
ε 22 = ε 22 (x1, x 2 )
ε12 = ε12 (x1 , x 2 )
so there is only one compatibility equation that is not identically zero given by
2
∂2 ε11 ∂ 2 ε 22
∂2 ε12
−
−
=0
∂x1 ∂x2
∂x22
∂x12
EM 424: Compatibility Equations
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For plane stress problems, on the other hand, we have the conditions
σ 13 = σ 23 = σ 33 = 0
σ 11 = σ 11 (x1 , x 2 )
σ 22 = σ 22 (x1 ,x 2 )
σ 12 = σ 12 (x1 ,x 2 )
which implies that
ε13 = ε 23 = 0 but ε 33 ≠ 0
ε11 = ε11 ( x1 , x2 )
ε 22 = ε 22 ( x1 , x2 )
ε12 = ε12 ( x1 , x2 )
so that the compatibility equations reduce to
∂ 2ε 33
=0
∂x22
∂ 2ε 33
=0
∂x12
∂ 2 ε 33
=0
∂x1 ∂x 2
2
∂2 ε11 ∂ 2 ε 22
∂2 ε12
−
−
=0
∂x1 ∂x2
∂x22
∂x12
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 justified. 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.2:
EM 424: Compatibility Equations
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C1
P
D
P
C2
Fig.2
where the point P on one side of the split is fixed (displacement = 0) and the xdisplacement of the same point P on the other side of the split = D. In this case, we see
that
∫ du
j
=0
C1
∫ du
C2
j
D
=
0
j =1
j = 2,3
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,
∫ du
j
=0
( i = 1, 2, ..., m )
Ci
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
EM 424: Compatibility Equations
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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.
C1
C2
Fig. 3
Cm