Page 267 - Intro to Tensor Calculus

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A similar situation occurs in two dimensions for a material in a state of strain where e zz = e zx = e zy =0,
called plane strain. In this case, are we allowed to arbitrarily assign values to the strains e xx ,e yy and e xy and
from these strains determine the displacement field u = u(x, y)and v = v(x, y)in the x− and y−directions?
Let us try to answer this question. Assume a state of plane strain where e zz = e zx = e zy =0. Further, let
us assign 3 arbitrary functional values f, g, h such that

∂u 1 ∂u ∂v ∂v
e xx = = f(x, y), e xy = + = g(x, y), e yy = = h(x, y).
∂x 2 ∂y ∂x ∂y

We must now decide whether these equations are consistent. That is, will we be able to solve for the
displacement field u = u(x, y)and v = v(x, y)? To answer this question, let us derive a compatibility equation
(integrability condition). From the given equations we can calculate the following partial derivatives
2 3 2
∂ e xx ∂ u ∂ f
= =
∂y 2 ∂x∂y 2 ∂y 2
2 3 2
∂ e yy ∂ v ∂ h
= =
∂x 2 ∂y∂x 2 ∂x 2
2 3 3 2
∂ e xy ∂ u ∂ v ∂ g
2 = + =2 .
∂x∂y ∂x∂y 2 ∂y∂x 2 ∂x∂y
This last equation gives us the compatibility equation

2 2 2
∂ e xy ∂ e xx ∂ e yy
2 = +
∂x∂y ∂y 2 ∂x 2
or the functions g, f, h must satisfy the relation

2
2
2
∂ g ∂ f ∂ h
2 = + .
∂x∂y ∂y 2 ∂x 2
Cartesian Derivation of Compatibility Equations

If the displacement field u i ,i =1, 2, 3 is known we can derive the strain and rotation tensors

1 1
e ij = (u i,j + u j,i ) and ω ij = (u i,j − u j,i ). (2.4.50)
2 2
Now work backwards. Assume the strain and rotation tensors are given and ask the question, “Is it possible
to solve for the displacement field u i ,i =1, 2, 3?” If we view the equation (2.4.50) as a system of equations
with unknowns e ij ,ω ij and u i and if by some means we can eliminate the unknowns ω ij and u i then we
will be left with equations which must be satisfied by the strains e ij . These equations are known as the
compatibility equations and they represent conditions which the strain components must satisfy in order
that a displacement function exist and the equations (2.4.37) are satisfied. Let us see if we can operate upon
the equations (2.4.50) to eliminate the quantities u i and ω ij and hence derive the compatibility equations.
Addition of the equations (2.4.50) produces

∂u i
u i,j = = e ij + ω ij . (2.4.51)
∂x j