374
7 Remodeling
involve only σ r , and the strain-energy density functions of (7.71) show that collagen
and elastin do not contribute directly to the radial stress. Thus, as in the tendon
problem of Sect. 7.4.3, p is lumped into the partial cell stress σ m
i , yielding
σ
m
i = φ
m (λ
m∗
i )
2 ∂W m∗
∂E m∗
i
− p
σ
n
θ (t) =
J n (0)
J (0)
λ
n∗
θ (t, 0)
∂W n∗
∂λ n∗
θ
(t, 0) q
n (t, 0)
+
1
J (t)
t
0
˙
J
n + (τ ) λ
n∗
θ (t, τ )
∂W n∗
∂λ n∗
θ
(t, τ ) q
n (t, τ ) dτ.
(7.79)
The survival function is
q
n (t, τ ) = e
−k n − (t−τ ) ,
(7.80)
and, because the matrix does not grow in this example, we take ˙
J n + = k n + J n (0) =
k n + φ n (0) and set k n + = k n − .
Fluid Shear Stress:
τ f =
4μQ
πa 3
(7.81)
Incompressibility:
J = λ r λ θ λ z =
∂r
∂R
r
R
λ → r
2
= a
2
+
2
λ
R
a 0
J (R) R dR
(7.82)
Boundary Conditions:
r = a :
σ r = −P
r = b :
σ r = 0
(7.83)
Lagrange Multiplier and Pressure: With the stresses ¯
σ i given by Eqs. (7.78)
and (7.79) with p excluded, Eqs. (6.137) and (6.138) apply, i.e.,
p(r) = ¯
σ r (r) +
b
r
( ¯
σ θ − ¯
σ r )
dr
r
.
P =
b
a
( ¯
σ θ − ¯
σ r )
dr
r
.
(7.84)