Biomedical Engineering Reference
In-Depth Information
where
h
is the thickness of the cube. The stress field
σ
xx
(
x
0
,
y
) may be approxi-
mated by
x
=
x
0
,
y
=
y
0
σ
xx
(
x
0
,
y
)
≈
σ
xx
(
x
0
,
y
0
)
+
∂σ
xx
∂
(
y
−
y
0
) ,
(8.13)
y
which is allowed if
y
is sufficiently small. Notice that the partial derivative of
σ
xx
in the
y
-direction is taken at a
fixed
value of
x
and
y
,i.e.at
x
x
0
and
y
=
y
0
(for readability reasons, in the following the addition
|
x
=
x
0
,
y
=
y
0
to the
partial derivatives is omitted from the equations). Consequently, the force
F
lx
can be written as
F
lx
=
h
y
0
+
y
y
0
=
(
y
−
y
0
)
dy
.
+
∂σ
xx
∂
y
σ
xx
(
x
0
,
y
0
)
(8.14)
To compute this integral it must be realized that
σ
xx
(
x
0
,
y
0
),
∂σ
xx
/∂
y
and
y
0
denote
quantities that are not a function of the integration parameter
y
. Bearing this in
mind, it is straightforward to show that
y
2
2
F
lx
=
σ
xx
(
x
0
,
y
0
)
h
y
+
∂σ
xx
∂
h
.
(8.15)
y
The stress on the right face of the cube,
σ
xx
(
x
0
+
x
,
y
) may be integrated to give
the force in the (positive)
x
-direction on this face:
h
y
0
+
y
y
0
F
rx
=
σ
xx
(
x
0
+
x
,
y
)
dy
.
(8.16)
Clearly, the stress component
σ
xx
on the right face, hence at constant
x
0
+
x
,may
be approximated by
σ
xx
(
x
0
+
x
,
y
)
≈
σ
xx
(
x
0
,
y
0
)
+
∂σ
xx
∂
x
x
+
∂σ
xx
(
y
−
y
0
) .
(8.17)
∂
y
Therefore the force
F
rx
can be computed as
y
2
2
F
rx
=
σ
xx
(
x
0
,
y
0
)
h
y
+
∂σ
xx
∂
h
x
y
+
∂σ
xx
∂
h
.
(8.18)
x
y
A similar exercise can be performed with respect to the forces in the
x
-direction
on the top and bottom faces, giving
h
x
2
2
+
∂σ
xy
∂
y
+
∂σ
xy
∂
x
F
tx
=
σ
xy
(
x
0
,
y
0
)
h
x
h
x
y
,
(8.19)
while
h
x
2
2
F
bx
=
σ
xy
(
x
0
,
y
0
)
h
x
+
∂σ
xy
∂
.
(8.20)
x
