Biomedical Engineering Reference
In-Depth Information
The stress field can be written [ 36 , 39 ]:
s rr ¼ X
4
2
l 1
ðl þ 1 ÞlA ðmÞ
k
ð sin y þ M ðmÞ
k
ð cos y þ M ðmÞ
k
r l 1
cos
sin
1
s yy ¼ X
4
1
ðl þ 1 ÞlA ðmÞ
k
ð cos y þ M ðmÞ
k
r l 1
sin
1
s ry ¼ X
4
l
ðl þ 1 ÞlA ðmÞ
k
ð sin y þ M ðmÞ
k
cos yÞð cos y þ M ðmÞ
k
r l 1
sin
1
(3.20)
in which the eight constants A ðmÞ k (four in each material) are found from the
boundary and continuity conditions on stress, and on the displacement field:
u r ¼ X
4
r l ðl þ 1 ÞA ðmÞ
k
ðp ðmÞ
k
cos y þ q ðmÞ
k
sin yÞð cos y þ m ðmÞ
k
sin l
1
(3.21)
u y ¼ X
4
r l ðl þ 1 ÞA ðmÞ
k
ðp ðmÞ
k
sin y þ q ðmÞ
k
cos yÞð cos y þ m ðmÞ
k
sin l
1
where the constants p and q depend on the components of the compliance matrix:
p ðm k ¼ S ðmÞ
ðm ðmÞ
k
2
þ S ðmÞ
12
q k ¼ S ðmÞ
21 m ðm k þ S ðmÞ
=m ðmÞ
k
Þ
;
(3.22)
11
22
As above, the continuity conditions simply require that there be no jump in the
displacement or stress when crossing the boundary between the two materials, and
the free edge boundary conditions require that the exposed surfaces near the
attachment point be traction free. This leads to a set of 32 equations for the eight
constants A ðm k , and the existence of a nontrivial solution requires that the determi-
nant of the coefficient matrix vanishes. As above, this leads to an equation that is
transcendental in l . Despite all of this additional complexity, the problem is as
simple as the others considered in this chapter: Equation ( 3.21 ) shows that the stress
field can be singular if the lowest root l is less than 1, and the tradition is to discard
all solutions for which l is less than 0 as non-physical because ( 3.21 ) shows that
these lead to unbounded displacements at r ¼ 0.
As an example, we repeat here the solution that we described in Liu et al. [ 39 ]of
a transversely isotropic tendon attaching to an isotropic material. The axis of
axisymmetry in the tendon is the y -axis in Fig. 3.7 . The properties chosen for the
tendon are based upon those reported by [ 14 , 40 , 41 ]: E x ¼ 450 MPa, E y ¼ E z
¼ 45 MPa, m xy ¼ m xz ¼ 0.75 E x /1,000. Following Liu et al. [ 39 ], we took for
the remaining two independent constants n yz ¼ 0 and n xz ¼ n xy ¼ 2. The remaining
constants needed to populate the compliance matrix S ij can be found from the
five constants prescribed: symmetry of the compliance matrix requires that n ij ¼ n ji
E i =E j (in this relation i and j range from x through z , and no summation occurs over
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