CONTINUUM MECHANICS for ENGINEERS

Consider once more the two neighboring particles which were at positions
P and Q in the undeformed configuration, and are now at positions p and
q, respectively, in the deformed configuration (see Figure 4.2). In general, an
arbitrary displacement will include both deformation (strain) and rigid body
displacements. To separate these we consider the differential displacement
vector du. Assuming conditions on the displacement field that guarantee the
existence of a derivative, the displacement differential du i is written
⎛ ∂u
⎞
i dui = ⎜ dXj
(4.7-26)
⎝ ∂X
⎟
⎠
where the derivative is evaluated at P as indicated by the notation. From
this we may define the unit relative displacement of the particle at Q with
respect to the one at P by the equation
(4.7-27)
where N j is the unit vector in the direction from P toward Q. By decomposing
the displacement gradient in Eq 4.7-26 into its symmetric and skew-symmetric
parts we obtain
= (ε ij + ω ij)dX j
(4.7-28)
in which ε ij is recognized as the infinitesimal strain tensor, and ω ij is called
the infinitesimal rotation tensor.
If ε ij happens to be identically zero, there is no strain, and the displacement
is a rigid body displacement. For this case we define the rotation vector
which may be readily inverted since ω =−ω
to yield
Therefore, Eq 4.7-28 with ε ij ≡ 0 becomes
j P
du u dX u
dX X dX X N
i ∂ i j ∂ i
= = j
∂ ∂
j
⎡ 1 ⎛ ∂u
∂u
⎞ 1 ⎛ ∂u
∂u
⎞⎤
i j
i j
dui = ⎢
dX
2
⎜ +
∂Xj
∂X
⎟ +
⎝
i ⎠ 2
⎜ −
⎝ ∂Xj
∂X
⎟⎥
⎣⎢
i ⎠⎦⎥
ω = ε ω
1 2
i ijk kj
ωij =
εkjiωk j
kj jk
(4.7-29)
(4.7-30)
du i = ε kjiω kdX j = ε ikjω kdX j or du = × dX (4.7-31)
j