16.21 Techniques of Structural Analysis and
Design
Spring 2005
Unit #3 ­ Kinematics of deformation
Raúl Radovitzky
February 9, 2005
Figure 1: Kinematics of deformable bodies
Deformation described by deformation mapping :
x� = ϕ(x) = x + u
1
(1)
We seek to characterize the local state of deformation of the material in a
neighborhood of a point P . Consider two points P and Q in the undeformed:
P : x = x 1 e1 + x 2 e 2 + x 3 e3 = x i e i
Q : x + dx = (xi + dxi )ei
(2)
(3)
P � : x� = ϕ1 (x)e1 + ϕ2 (x)e2 + ϕ3 (x)e3 = ϕi (x)ei
(4)
�
�
Q� : x� + dx� = ϕi (x) + dϕi ei
(5)
and deformed
configurations. In this expression,
dx� = dϕi ei
(6)
Expressing the differentials dϕi in terms of the partial derivatives of the
functions ϕi (xj ej ):
dϕ1 =
∂ϕ1
∂ϕ1
∂ϕ1
dx2 +
dx3 ,
dx1 +
∂x1
∂x2
∂x3
(7)
and similarly for dϕ2 , dϕ3 , in index notation:
dϕi =
∂ϕi
dxj
∂xj
(8)
Replacing in equation (5):
�
�
∂ϕi
Q� : x� + dx� = ϕi +
dxj ei
∂xj
∂ϕi
dx� i =
dxj ei
∂xj
(9)
(10)
−→
We now try to compute the change in length of the segment P Q which
−−
→
deformed into segment P � Q� . Undeformed length (to the square):
ds2 = �dx�2 = dx · dx = dxi dxi
2
(11)
Deformed length (to the square):
(ds� )2 = �dx� �2 = dx� · dx� =
∂ϕi
∂ϕi
dxj
dxk
∂xk
∂xj
(12)
−→
The change in length of segment P Q is then given by the difference between
equations (12) and (11):
(ds� )2 − ds2 =
∂ϕi
∂ϕi
dxj
dxk − dxi dxi
∂xj
∂xk
(13)
We want to extract as common factor the differentials. To this end we observe
that:
dxi dxi = dxj dxk δjk
(14)
Then:
∂ϕi
∂ϕi
dxj
dxk − dxj dxk δjk
∂xj
∂xk
� ∂ϕ ∂ϕ
�
i
i
=
− δjk dxj dxk
∂x ∂x
� j ��k
�
2�jk : Green­Lagrange strain tensor
(ds� )2 − ds2 =
(15)
Assume that the deformation mapping ϕ(x) has the form:
ϕ(x) = x + u
(16)
where u is the displacement field. Then,
∂ϕi
∂xi
∂ui
∂ui
=
+
= δij +
∂xj
∂xj ∂xj
∂xj
(17)
and the Green­Lagrange strain tensor becomes:
�
∂um ��
∂um �
2�ij = δmi +
δmj +
− δij
∂xi
∂xj
∂ui ∂uj ∂um ∂um
=� δij +
+
+
− � δij
∂xi
∂xi ∂xj
∂xj
3
(18)
Green­Lagrange strain tensor :
�ij =
1 � ∂ui ∂uj ∂um ∂um �
+
+
2 ∂xj
∂xi
∂xi ∂xj
(19)
When the absolute values of the derivatives of the displacement field are
much smaller than 1, their products (nonlinear part of the strain) are even
smaller and we’ll neglect them. We will make this assumption throughout
this course (See accompanying Mathematica notebook evaluating the limits
of this assumption). Mathematically:
� ∂u �
∂um ∂um
� i�
∼0
�
�
� 1 ⇒
∂xj
∂xi ∂xj
(20)
We will define the linear part of the Green­Lagrange strain tensor as the
small strain tensor :
1 � ∂ui ∂uj �
(21)
�ij =
+
2 ∂xj
∂xi
Transformation of strain components
Given: �ij , ei and a new basis ẽk , determine the components of strain in the
new basis �˜kl
1 �
∂ u˜i ∂ u˜j �
(22)
�˜ij =
+
2 ∂ x˜j
∂ x˜i
We want to express the expressions with tilde on the right­hand side with
their non­tilde counterparts. Start by applying the chain rule of differentia­
tion:
∂ u˜i
∂ u˜i ∂xk
=
∂ x˜j
∂xk ∂ x˜j
(23)
Transform the displacement components:
u = u˜m˜
em = ul el
u˜m (˜
ei ) = ul (el · ˜
ei )
em · ˜
ei )
u˜m δmi = ul (el · ˜
u˜i = ul (el · ˜
ei )
4
(24)
(25)
(26)
(27)
take the derivative of ũi with respect to xk , as required by equation (23):
∂ul
∂ ũi
=
(el · ˜
ei )
∂xk
∂xk
(28)
and take the derivative of the reverse transformation of the components of
the position vector x:
x = xj ej = x˜k ˜
ek
ek · ei )
xj (ej · ei ) = x˜k (˜
xj δji = x˜k (˜
ek · ei )
xi = x˜k (˜
ek · ei )
(29)
(30)
(31)
(32)
∂ x̃k
∂xi
=
(˜
ek · ei ) = δkj (˜
ek · ei ) = (ẽj · ei )
∂ x˜j
∂ x˜j
(33)
Replacing equations (28) and (33) in (23):
∂ u˜i
∂ u˜i ∂xk
∂ul
=
=
(el · ˜
ei )(˜
ej · ek )
∂ x˜j
∂xk ∂ x˜j
∂xk
Replacing in equation (22):
�
1 � ∂ul
∂ul
�˜ij =
(el · ˜
ei )(˜
ej · ek ) +
(el · ˜
ej )(˜
ei · ek )
2 ∂xk
∂xk
Exchange indices l and k in second term:
�
1 � ∂ul
∂uk
�˜ij =
(el · ˜
ei )(˜
ej · ek ) +
(ek · ˜
ej )(˜
e i · el )
2 ∂xk
∂xl
�
�
∂uk
1 ∂ul
ei )(˜
ej · ek )
=
+
(el · ˜
2 ∂xk
∂xl
(34)
(35)
(36)
Or, finally:
�˜ij = �lk (el · ˜
ei )(˜
ej · ek )
(37)
Compatibility of strains
Given displacement field u, expression (21) allows to compute the strains
components �ij . How does one answer the reverse question? Note analogy
with potential­gradient field. Restrict the analysis to two dimensions:
�11 =
∂u1
∂u2
∂u1 ∂u2
, �22 =
, 2�12 =
+
∂x1
∂x2
∂x2 ∂x1
5
(38)
Differentiate the strain components as follows:
�
∂ 3 u1
∂ 3 u2
∂2 �
2�12 =
+
∂x1 ∂x2
∂x1 ∂x22 ∂x21 ∂x2
(39)
∂ 2 �11
∂ 3 u1
=
∂x22
∂x1 ∂x22
(40)
∂ 2 �22
∂ 3 u2
=
∂x21
∂x2 ∂x21
(41)
∂ 2 �11 ∂ 2 �22
∂ 2 �12
2
=
+
∂x1 ∂x2
∂x22
∂x21
(42)
and conclude that:
6