Biomedical Engineering Reference
In-Depth Information
Force equilibrium in the
x
-direction now yields
−
F
lx
−
F
bx
+
F
rx
+
F
tx
=
0.
(8.21)
Use of the above results for the force components gives
−
σ
xx
(
x
0
,
y
0
)
h
y
−
∂σ
xx
∂
h
2
y
2
−
σ
xy
(
x
0
,
y
0
)
h
x
y
−
∂σ
xy
∂
h
2
x
2
+
σ
xx
(
x
0
,
y
0
)
h
y
+
∂σ
xx
∂
h
x
y
+
∂σ
xx
∂
h
2
y
2
x
x
y
+
σ
xy
(
x
0
,
y
0
)
h
x
+
∂σ
xy
∂
h
x
y
+
∂σ
xy
∂
h
2
x
2
=
0,
(8.22)
y
x
which implies that
∂σ
xx
∂
x
+
∂σ
xy
y
=
0.
(8.23)
∂
This should hold for any position in space of the prism, hence for all values of
x
and
y
. Performing a similar exercise in the
y
-direction yields the full set of partial
differential equations, known as the local equilibrium equations:
∂σ
xx
∂
x
+
∂σ
xy
∂
y
=
0
(8.24)
∂σ
yx
∂
x
+
∂σ
yy
y
=
0.
(8.25)
∂
σ
xy
and
σ
yx
. However, by using
equilib-
Two different shear stresses are present:
rium of moment
it can be proven that
σ
xy
=
σ
yx
.
(8.26)
This result is revealed by considering the sum of moments with respect to the
midpoint of the infinitesimal cube of Fig.
8.7
. With respect to the midpoint, the
moments due to the normal forces,
F
lx
,
F
rx
,
F
ty
and
F
by
are equal to zero,
while the shear forces generate a moment, hence enforcing the sum of moments
with respect to the midpoint to be equal to zero gives
−
x
2
(
F
ly
+
F
ry
)
+
y
2
(
F
tx
+
F
bx
)
=
0.
(8.27)
