Biomedical Engineering Reference
In-Depth Information
2
F
T
+
F
−
2
I
,
1
ε
=
ε
(
e
0
)
=
e
0
·
ε
·
e
0
with
ε
=
(10.43)
where the symmetrical tensor
ε
is called the
linear strain tensor
. In displace-
ments this tensor can also be expressed as
∇
0
u
T
.
u
∇
0
1
2
ε
=
+
(10.44)
The strain tensor
is linear in the displacements and can be considered (with
respect to the displacements) as a linearized form of the Green Lagrange strain
tensor
E
. In component form this results in the well-known and often-used
formulation:
ε
2
∂
u
x
x
0
2
∂
u
x
x
0
⎡
⎣
⎤
⎦
y
0
+
∂
u
y
∂
u
x
∂
1
1
+
∂
u
z
∂
x
0
∂
∂
∂
z
0
2
∂
u
y
∂
y
0
2
∂
u
y
∂
y
0
∂
u
y
∂
y
0
+
∂
u
z
1
∂
x
0
+
∂
u
x
1
ε
=
.
(10.45)
∂
z
0
2
∂
u
z
∂
z
0
2
∂
u
z
∂
z
0
∂
y
0
+
∂
u
y
∂
u
z
∂
z
0
1
∂
x
0
+
∂
u
x
1
The components of the (symmetrical, 3
×
3) linear strain matrix
ε
can be
interpreted as follows:
•
The terms on the diagonal are the linear strains of material line segments of the ref-
erence configuration in the
x
-,
y
-and
z
-directions respectively (the component on the
first row in the first column is the linear strain of a line segment that is oriented in the
x
-direction in the reference configuration).
•
The off-diagonal terms determine the shear of the material (the component on the
first row of the second column is a measure for the change of the angle enclosed by
material line segments that are oriented in the
x
-and
y
-direction in the undeformed
configuration. See Fig.
10.3
.
y
∂
u
x
∂
dy
y
∂
u
y
∂
x
dx
dy
u
x
u
y
x
dx
Figure 10.3
Interpretation of linear strain components.
