7.8 Case Study: Changing Fiber Orientation During Cyclic Stretch
395
For G m
c (t) given above, compute the solution using Humphrey-Rajagopal
theory for α c = 0, with all other parameters being the same. By iteration,
choose values for k that yield results in qualitative agreement with those
obtained in (a). Note that the results given by the two approaches may not
agree quantitatively. Hint: One way to approach this problem is to step in
time and solve a single equation for λ x at each time step.
7.5 Write a computer program to solve the problem considered in Sect. 7.4.1 for a
bar undergoing a given stretch. Plot σ x /c versus t and check the results shown
in Fig. 7.6. For α = 0, the strain-energy density function of Eqs. (7.32) reduces
to that of (7.31). For this case, verify that the solution computed using the two
integrals of Eq. (7.38) matches that computed using the single integral of (7.39).
7.6 Strategies to construct replacement heart muscle in vitro (tissue engineering)
have encountered numerous obstacles. One issue is how to grow muscle with
realistic mechanical properties, including the ability to develop contractile
forces in the physiological range. Research has shown that contractile properties of differentiating stem cells can be improved by culturing cells on
deformable substrates of appropriate stiffness (Ribeiro et al. 2015). To study
these effects, consider an idealized model for this system consisting of a
rectangular bar of stem cells embedded in a passive incompressible matrix.
The cells are assumed to grow and remodel by synthesizing contractile fibers
(sarcomeres) aligned with the axis of the bar (x-direction) under load-free
conditions without constraint.
All passive constituents, including those in the cells, are lumped into a single
passive component with volume fraction φ p , and the contractile fibers make up
an active component with volume fraction φ a . For incompressible constituents,
the passive and active stresses are given by
¯
σ
p
x = c p (λ
2
x − λ
−1
x )
¯
σ
a
x = c a (λ
a∗
x − 1),
where ¯
σ n
x is stress per unit deformed area of constituent n. (The Lagrange
multiplier is already included in ¯
σ
p
x .) Assume the following:
• In the initial configuration, the bar is stress-free, and no sarcomeres are
present. Therefore, φ p = 1 and φ a = 0 at t = 0.
• The passive component does not grow or remodel. During culture, however,
contractile fibers are produced at the volumetric rate
˙
J
a + (t) = ˙
J
a +
0 [1 + β ¯
σ
a
x (t)],
where ˙
J a +
0 and β are constants. After formation, these fibers do not decay,
but they immediately undergo a constant contraction K, which can be
simulated by taking the deposition stretch as λ a
0 = K −1 . All contractile
fibers are added in parallel, causing the bar to thicken but not grow longer.
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