Compatibility Eqs Rev

8/13/2019 Compatibility Eqs Rev

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EM 424: Compatibility Equations 1

STRAIN COMPATIBILITY EQUATIONS

Consider a body with displacements at points P 1 and P 2 given by u 1 and u 2, respectively,as shown in Figure 1:

Figure 1

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

Then those components can be written in terms of integrals of the local strains androtations as:

But,

so that

By differentiating the strain-displacement relationship it can be verified that

P

P

u

u

1

2

1 2

C

= ii =1

3 e

i

( )2 2 2

1 1 1

3 3

1 1

P P P i

i i j ij ij j j j j P P P

udu dx dx

x

= =

= = = +

ijdx j = d ij x j( ) x j ij xk dxk k =13

i = ij x j j =1

3

P 1

P 2+ ik x j

ij xk j =1

3

dxk

P 1

P 2

k =1

3


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these derivatives of the local rotation can also be written in terms of derivatives of thestrains:

giving

The first term in the above i expression only depends on P 1 and P 2 . The second term ini is independent of the path C between those two points (and hence only depends on P 1and P 2) if, for a simply connected region (a region with no holes) we have

(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 displacementwill have a single value everywhere in that region (the displacement is single valued butnot unique since we can always add a rigid body displacement that does not change thestrains). The derivatives contained in Eq. (1) can be written as

However, the first two terms in the above equations cancel and the third terms in theabove equations also cancel when placed back into Eq. (1), reducing Eq. (1) to:

ik x j

jk xi

= 12

xk

ui x j

u j xi

= ij xk

i = ij x j j =1

3

P 1

P 2+ ik dxk

P 1

P 2

k =1

3

where ik = ik x j ik x j

jk x i

j =13

ik xl

= il xk

ik xl

= ik xl

ik xl

lk xi

x j2 ik

x j xl

2 jk x i x l

j =13

il xk

= il xk

il xk

kl xi

x j 2 il

x j xk 2 jl

xi xk j =1

3

(1)

(2)


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But, since the x j are independent, the quantity in the brackets must vanish, and

We will write Eq.(3) symbolically as

Since i, j, k, and l can all have values ranging from 1 to 3, Eq. (4) looks like a total of 81equations. However, because of the following symmetries and anti-symmetries:

it turns out that there are really only six distinct terms in Eq. (4) given by thecompatibility conditions:

which we have written in terms of the components of a symmetric S matrix:

All six of the compatibility equations listed in Eq. (5), however, are not independentsince the components of the S matrix can be shown to satisfy the three equations:

x j2 ik

x j xl +

2 jl x i xk

2 jk xi xl

2 il

x j xk

j =13 =0

2 ik x j x l

+2 jl x i xk

2 jk xi xl

2 il

x j xk = 0 (3)

Rijkl =0 (4)

Rijkl = Rklij Rijkl = R jikl = Rijlk

S 11 S 12 S 13S 21 S 22 S 23S 31 S 32 S 33

(5)

S 11

R2323

=0S 22 R3131 =0S 33 R1212 = 0S 21 = S 12 R2331 = 0S 23 = S 32 R3112 = 0S 13 = S 31 R1223 = 0


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so that these six compatibility equations really represent only three independentconditions that the strains must satisfy. Explicitly, these six equations are

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 haveonly two non-zero displacements since plane strain conditions require

which implies that

so there is only one compatibility equation that is not identically zero given by

S ij x j j =1

3 =0 i =1,2,3( )

22 23 x2 x3

2 22 x32

2 33 x22

= 0

22 31 x3 x1

2 33 x1

2

2 11 x3

2= 0

22 12

x1 x2

2 11

x22

2 22

x12 =0

2 33 x1 x2

+ 2 12 x3

2 2 23 x3 x1

2 31

x3 x2= 0

2 11 x2 x3

+ 2 23 x12

2 31

x1 x2

2 12 x1 x3

=0

2 22 x3 x1

+ 2 31 x22

2 12

x2 x3

2 23 x2 x1

= 0

u1 = u1 x1 , x2( )u2 = u 2 x1, x2( )u3 = 0

33 = 13 = 23 =0

11 = 11 x1, x2( ) 22 = 22 x1, x2( ) 12 = 12 x1 , x2( )

22 12 x1 x2

2 11 x2

2 2 22 x1

2 =0


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For plane stress problems, on the other hand, we have the conditions

which implies that

so that the compatibility equations reduce to

In general, the first three of the above conditions cannot be satisfied exactly, so that planestress conditions can only be approximately true in a body. For thin bodies, however, thisapproximation can be usually justified. Note that the compatibility equation for the in-

plane strains in the case of plane stress is identical to the compatibility equation for planestrain.

For bodies with holes these compatibility equations are not sufficient to guaranteethat the strains can be obtained from a single-valued displacement field. In fact, in bodieswith holes there may be some cases where we want to have displacements that are notsingle valued such as the split ring shown in Fig.2:

13 = 23 = 33 = 0 11 = 11 x1 , x2( ) 22 = 22 x1 , x2( ) 12 = 12 x1, x2( )

( )

( )( )

13 23 33

11 11 1 2

22 22 1 2

12 12 1 2

0 0

,

,,

but

x x

x x x x

= = =

==

2 33 x2

2 = 0

2 33 x1

2 = 0

2 33 x1 x2

=0

22 12 x1 x2

2 11 x22

2 22 x12

=0


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Fig.2

where the point P on one side of the split is fixed (displacement = 0) and the x-displacement of the same point P on the other side of the split = D. In this case, we seethat

1

2

0

1

0 2,3

jC

jC

du

D jdu

j

=

===

To ensure that a multiply connected body like the ring shown above cannot split in thisfashion, 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, werequire the m subsidiary conditions,

where the integrals are taken around each hole, then the displacements will be single-valued.

Note that in solving any problem, if we end up obtaining directly a displacement fieldthat represents the solution to the desired problem, then compatibility is not an issue sincewe can always generate a set of strains that are compatible with those displacementssimply by taking the appropriate displacement derivatives. When we directly solve a

PP

C1

C 2

D

( )0 1, 2, ...,i

jC

du i m= =


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problem only for the strains (or corresponding stresses), however, it is not automaticallyguaranteed that a well-behaved, single-valued displacement field can be obtained throughintegrating those strains. The compatibility equations (and any subsidiary conditionsneeded for multiply connected bodies) provide the guarantee that a well-behaved, single-valued displacement field can be obtained from the given strain field.

Fig. 3

C CC

12

m

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Compatibility Eqs Rev