Page 238 - Intro to Tensor Calculus

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Figure 2.3-16. Displacement due to strain

i
Compressible and Incompressible Material With reference to figure 2.3-16, let x , i =1, 2, 3denote
the position vector of an arbitrary point P in a continuum before there is a state of strain. Let Q be
i
i
a neighboring point of P with position vector x + dx , i =1, 2, 3. Also in the figure 2.3-16 there is the
1
2
3
displacement vector ~u. Here it is assumed that ~u = ~u(x ,x ,x ) denotes the displacement field when the
continuum is in a state of strain. The figure 2.3-16 illustrates that in a state of strain P moves to P and Q
0
moves to Q . Let us find a relationship between the distance PQ before the strain and the distance P Q when
0
0
0
~
~
~
the continuum is in a state of strain. For E 1 , E 2 , E 3 basis functions constructed at P we have previously
shown that if
1 2 3 i ~ i j ~
~u(x ,x ,x )= u E i then d~u = u dx E i .
,j
Now for ~u + d~u the displacement of the point Q we may use vector addition and write
PQ + ~u + d~u = ~u + P Q . (2.3.63)
0
0
i ~
i ~
Let PQ = dx E i = a E i denote an arbitrary small change in the continuum. This arbitrary displacement
i ~
0
gets deformed to P Q = A E i due to the state of strain in the continuum. Employing the equation (2.3.63)
0
we write
j
i
i
i
i
j
dx + u dx = a + u a = A i
,j ,j
which can be written in the form
i
i
i
i
i
i
δa = A − a = u a j where dx = a ,i =1, 2, 3 (2.3.64)
,j
denotes an arbitrary small change. The tensor u i ,j and the associated tensor u i,j = g it u t ,j are in general
not symmetric tensors. However, we know we can express u i,j as the sum of a symmetric (e ij )and skew-
symmetric(ω ij ) tensor. We therefore write
i
i
or u i = e + ω ,
u i,j = e ij + ω ij
,j j j
where
1 1 m m 1 1 m m
e ij = (u i,j + u j,i )= (g im u ,j + g jm u ,i ) and ω ij = (u i,j − u j,i )= (g im u ,j − g jm u ,i ) .
2 2 2 2
i
i s
i
i
The deformation of a small quantity a can therefore be represented by a pure strain A − a = e a followed
s
i
i s
i
by a rotation A − a = ω a .
s