7.8 Case Study: Changing Fiber Orientation During Cyclic Stretch
393
when the rate of fibers added matches the rate of fibers lost. For k 0 = 1 s −1 , the SFs
reach steady state within a few seconds, but slower turnover takes more time and
leads to higher peaks in J n (Fig. 7.18b).
The curves in Fig. 7.18c show that J n peaks twice during each loading cycle. One
peak occurs near the middle (maximum λ x ) and the other near the end (minimum
λ x ) of each cycle. This pattern reflects the different deformations experienced by
SFs deposited at different times. For example, longitudinal fibers created near the
beginning of a cycle quickly stretch, while those deposited near the middle of a
cycle are compressed. Both cases lead to increased disassembly rate according to
Eq. (7.101).
Finally, elastic stretch ratios for the classic uniaxial case are plotted as functions
of deposition time at t = 4 s for α n
0 = 0 ◦ and ±55 ◦ (Fig. 7.18d). Comparison with
Fig. 7.18a shows that fibers deposited near the middle (maximum λ x ) of each cycle
shorten as the membrane returns toward its reference configuration. Fibers oriented
initially at ±55 ◦ undergo relatively little deformation (λ n∗ ≈ 1), corresponding to
minimal disassembly and leading to a net accumulation of fibers at this orientation.
This is the reason fibers become aligned near ±90 ◦ during strip biaxial stretch, for
which the deformation in the y-direction is zero.
Problems
7.1 Equations (7.7) and (7.13) provide two expressions for the net fiber volumeproduction rate J n (t). Assume the fiber volume decays at a rate proportional
to the current volume, i.e., J n − = k n − J n (t). For this case, integrate (7.7) and
show that it is equivalent to (7.13) with the survival function given by (7.12).
7.2 The true volumetric production rate for fiber family n is
˙
J
n + = k
n + J
n
0 (1 + at),
where J n
0 = J n (0), and a and k n + are constants. With q n (t, τ ) given by
Eq. (7.12), integrate (7.13) to determine the volume ratio J n (t) in closed form.
7.3 A bar is composed of axially aligned fibers (x-direction) embedded in an
isotropic, incompressible matrix. At t = 0, the volume fractions are φ n
0 , with
n = (m, f ) denoting matrix and fibers, respectively. The constitutive relations
are
¯
σ
m
x = c m [(λ
m∗
x )
2
− (λ
m∗
x )
−1
]
¯
σ
f
x = c f (λ
f ∗
x − 1),
where λ n∗
x are elastic stretch ratios and ¯
σ n
x Cauchy stresses relative to the current
area of constituent n. (The Lagrange multiplier is already included in ¯
σ m
x .) The
393
when the rate of fibers added matches the rate of fibers lost. For k 0 = 1 s −1 , the SFs
reach steady state within a few seconds, but slower turnover takes more time and
leads to higher peaks in J n (Fig. 7.18b).
The curves in Fig. 7.18c show that J n peaks twice during each loading cycle. One
peak occurs near the middle (maximum λ x ) and the other near the end (minimum
λ x ) of each cycle. This pattern reflects the different deformations experienced by
SFs deposited at different times. For example, longitudinal fibers created near the
beginning of a cycle quickly stretch, while those deposited near the middle of a
cycle are compressed. Both cases lead to increased disassembly rate according to
Eq. (7.101).
Finally, elastic stretch ratios for the classic uniaxial case are plotted as functions
of deposition time at t = 4 s for α n
0 = 0 ◦ and ±55 ◦ (Fig. 7.18d). Comparison with
Fig. 7.18a shows that fibers deposited near the middle (maximum λ x ) of each cycle
shorten as the membrane returns toward its reference configuration. Fibers oriented
initially at ±55 ◦ undergo relatively little deformation (λ n∗ ≈ 1), corresponding to
minimal disassembly and leading to a net accumulation of fibers at this orientation.
This is the reason fibers become aligned near ±90 ◦ during strip biaxial stretch, for
which the deformation in the y-direction is zero.
Problems
7.1 Equations (7.7) and (7.13) provide two expressions for the net fiber volumeproduction rate J n (t). Assume the fiber volume decays at a rate proportional
to the current volume, i.e., J n − = k n − J n (t). For this case, integrate (7.7) and
show that it is equivalent to (7.13) with the survival function given by (7.12).
7.2 The true volumetric production rate for fiber family n is
˙
J
n + = k
n + J
n
0 (1 + at),
where J n
0 = J n (0), and a and k n + are constants. With q n (t, τ ) given by
Eq. (7.12), integrate (7.13) to determine the volume ratio J n (t) in closed form.
7.3 A bar is composed of axially aligned fibers (x-direction) embedded in an
isotropic, incompressible matrix. At t = 0, the volume fractions are φ n
0 , with
n = (m, f ) denoting matrix and fibers, respectively. The constitutive relations
are
¯
σ
m
x = c m [(λ
m∗
x )
2
− (λ
m∗
x )
−1
]
¯
σ
f
x = c f (λ
f ∗
x − 1),
where λ n∗
x are elastic stretch ratios and ¯
σ n
x Cauchy stresses relative to the current
area of constituent n. (The Lagrange multiplier is already included in ¯
σ m
x .) The