7.4 Examples: Remodeling in 1D
359
ˆ
σ x (t) = ˆ
σ o (t) + ˆ
σ n (t)
ˆ
σ o (t) = 2
λ
∗2
x (t, 0) −
1
λ ∗
x (t, 0)
e
−k − t
ˆ
σ n (t) =
2k + e −k − t
J (t)
t
o
λ
∗2
x (t, τ ) −
1
λ ∗
x (t, τ )
e
k − τ dτ,
(7.41)
where ˆ
σ o and ˆ
σ n are stresses in the original and new fibers, respectively. Note that
ˆ
σ o (0) is the stress caused by elastic deformation immediately after the weight is
attached. In these expressions, Eq. (7.20) gives the elastic stretch ratio
λ
∗
x (t, τ ) = λ 0
λ x (t)
λ x (τ )
G x (τ )
G x (t)
,
(7.42)
where λ 0 is the deposition stretch ratio. With J (t) and G x (t) known, substituting
these relations into (7.40) produces one equation to be solved at each time step for
λ x (t).
Illustrative Results To focus first on the effects of turnover without the complicating effects of pre-stretch, results are shown for λ 0 = 1, i.e., all new fibers are
initially stress-free. For t ≤ 0, we assume that the bar is unloaded (w = σ x = 0)
and is in a state of homeostasis, i.e., turnover occurs without growth (k
− = k
+ ) so
that λ ∗
x = λ x = 1. It is instructive to first examine what happens if the unloaded
bar begins to grow or atrophy at t = 0 via turnover alone (k
− = k
+ ). Note that the
growth ratios depend only on the values of k
− , k
+ , and γ .
With k
+ = 1 and γ = 1 (isotropic growth), G x and λ x are plotted for
k
− = 0.5 (growth) and k
− = 2 (atrophy). As expected from previous results
(see Fig. 7.4a), the growth and stretch ratios approach new equilibrium values as
t increases (Fig. 7.7a). Note, however, that the magnitude of λ x is considerably
less than that of G x . This result may seem surprising, because λ x = G x for an
unloaded, unconstrained bar undergoing volumetric growth (see Example 6.1, page
260). The difference here is that positive growth of the bar stretches old fibers, which
resist elongation of the bar. Likewise, atrophy compresses old fibers, which resist
shortening. Of course, to keep the total stress at zero, at least some of the new
fibers must be in compression during growth or tension during atrophy. As shown
in Fig. 7.7b, the total stress in the new fibers (σ n ) exactly cancels that in the original
fibers (σ o ) at each time point, until the latter effectively degrade away.
In other words, for the case of net growth (k − = 0.5), new fibers deposited in
the unloaded bar add volume that elongates the bar and stretches old fibers, which,
in turn, compress the new fibers to maintain mechanical equilibrium. When the
original fibers disappear, the bar returns to the homeostatic state, with the stress
being uniformly zero.
359
ˆ
σ x (t) = ˆ
σ o (t) + ˆ
σ n (t)
ˆ
σ o (t) = 2
λ
∗2
x (t, 0) −
1
λ ∗
x (t, 0)
e
−k − t
ˆ
σ n (t) =
2k + e −k − t
J (t)
t
o
λ
∗2
x (t, τ ) −
1
λ ∗
x (t, τ )
e
k − τ dτ,
(7.41)
where ˆ
σ o and ˆ
σ n are stresses in the original and new fibers, respectively. Note that
ˆ
σ o (0) is the stress caused by elastic deformation immediately after the weight is
attached. In these expressions, Eq. (7.20) gives the elastic stretch ratio
λ
∗
x (t, τ ) = λ 0
λ x (t)
λ x (τ )
G x (τ )
G x (t)
,
(7.42)
where λ 0 is the deposition stretch ratio. With J (t) and G x (t) known, substituting
these relations into (7.40) produces one equation to be solved at each time step for
λ x (t).
Illustrative Results To focus first on the effects of turnover without the complicating effects of pre-stretch, results are shown for λ 0 = 1, i.e., all new fibers are
initially stress-free. For t ≤ 0, we assume that the bar is unloaded (w = σ x = 0)
and is in a state of homeostasis, i.e., turnover occurs without growth (k
− = k
+ ) so
that λ ∗
x = λ x = 1. It is instructive to first examine what happens if the unloaded
bar begins to grow or atrophy at t = 0 via turnover alone (k
− = k
+ ). Note that the
growth ratios depend only on the values of k
− , k
+ , and γ .
With k
+ = 1 and γ = 1 (isotropic growth), G x and λ x are plotted for
k
− = 0.5 (growth) and k
− = 2 (atrophy). As expected from previous results
(see Fig. 7.4a), the growth and stretch ratios approach new equilibrium values as
t increases (Fig. 7.7a). Note, however, that the magnitude of λ x is considerably
less than that of G x . This result may seem surprising, because λ x = G x for an
unloaded, unconstrained bar undergoing volumetric growth (see Example 6.1, page
260). The difference here is that positive growth of the bar stretches old fibers, which
resist elongation of the bar. Likewise, atrophy compresses old fibers, which resist
shortening. Of course, to keep the total stress at zero, at least some of the new
fibers must be in compression during growth or tension during atrophy. As shown
in Fig. 7.7b, the total stress in the new fibers (σ n ) exactly cancels that in the original
fibers (σ o ) at each time point, until the latter effectively degrade away.
In other words, for the case of net growth (k − = 0.5), new fibers deposited in
the unloaded bar add volume that elongates the bar and stretches old fibers, which,
in turn, compress the new fibers to maintain mechanical equilibrium. When the
original fibers disappear, the bar returns to the homeostatic state, with the stress
being uniformly zero.