300
6 Growth
according to the growth law
˙
G z = α( ˆ
σ z − ˆ
σ 0 )G z ,
(6.91)
where the target stress ˆ
σ 0 is a constant. Determine ˆ
σ z = σ z /c as a function of time.
Solution
In Example 6.2 (page 261), we considered specified growth in a rectangular neoHookean bar with fixed ends. It is easy to show that Eq. (6.21) in terms of the elastic
stretch ratios applies to a circular bar as well. Setting λ ∗
z = λ z /G z yields
ˆ
σ z = 2
λ
∗2
z −
1
λ ∗
z
= 2
λ 2
z
G 2
z
−
G z
λ z
.
(6.92)
Substituting this relation and λ z = λ into (6.91) provides the nonlinear differential
equation
˙
G z = α
2
λ 2
G 2
z
−
G z
λ
− σ 0
G z .
(6.93)
Given λ(t) and the initial condition G z (0) = 1, solving this equation gives G z (t).
This differential equation can be solved rather easily using the routine ode45 in
MATLAB. Then, (6.92) yields ˆ
σ z (t).
Results
Illustrative results are shown for the following cases:
Case I:
λ(t) = 1 + (( − 1)(1 − e
−βt )
Case II:
λ(t) = 1 + (( − 1) sin ωt.
Case I specifies a monotonic increase or decrease in λ (β is a decay constant), with
λ → as t → ∞. Case II is an oscillation of amplitude − 1 about λ = 1 and
circular frequency ω = 2πf , with f being the frequency in Hz. Since this problem
contains two characteristic rate constants, α and either β or ω, the response depends
on the ratio α/β for Case I and α/ω (or α/f ) for Case II.
According to the growth law (6.91) with α > 0, G z increases when ˆ
σ z > ˆ
σ 0
and decreases when ˆ
σ z < ˆ
σ 0 . First, we examine Case I with the bar stretched
( > 1) and ˆ
σ 0 set to zero; thus, the bar is initially in tension with ˙
G z > 0. As
shown in Fig. 6.16a, G z increases with λ but at a slower rate. Thus, growth lags
the specified deformation by an amount that decreases as α/β increases, i.e., as the
growth-rate constant increases. As t becomes large, λ → = 1.2 and changes
little, giving growth time to catch up (G z → ). Thus, λ ∗
z = λ/G z → 1, and the
stress approaches the target stress of zero (Fig. 6.16a ). Moreover, as the growth rate
increases, allowing G z (t) to stay closer to λ(t), the peak stress drops.
6 Growth
according to the growth law
˙
G z = α( ˆ
σ z − ˆ
σ 0 )G z ,
(6.91)
where the target stress ˆ
σ 0 is a constant. Determine ˆ
σ z = σ z /c as a function of time.
Solution
In Example 6.2 (page 261), we considered specified growth in a rectangular neoHookean bar with fixed ends. It is easy to show that Eq. (6.21) in terms of the elastic
stretch ratios applies to a circular bar as well. Setting λ ∗
z = λ z /G z yields
ˆ
σ z = 2
λ
∗2
z −
1
λ ∗
z
= 2
λ 2
z
G 2
z
−
G z
λ z
.
(6.92)
Substituting this relation and λ z = λ into (6.91) provides the nonlinear differential
equation
˙
G z = α
2
λ 2
G 2
z
−
G z
λ
− σ 0
G z .
(6.93)
Given λ(t) and the initial condition G z (0) = 1, solving this equation gives G z (t).
This differential equation can be solved rather easily using the routine ode45 in
MATLAB. Then, (6.92) yields ˆ
σ z (t).
Results
Illustrative results are shown for the following cases:
Case I:
λ(t) = 1 + (( − 1)(1 − e
−βt )
Case II:
λ(t) = 1 + (( − 1) sin ωt.
Case I specifies a monotonic increase or decrease in λ (β is a decay constant), with
λ → as t → ∞. Case II is an oscillation of amplitude − 1 about λ = 1 and
circular frequency ω = 2πf , with f being the frequency in Hz. Since this problem
contains two characteristic rate constants, α and either β or ω, the response depends
on the ratio α/β for Case I and α/ω (or α/f ) for Case II.
According to the growth law (6.91) with α > 0, G z increases when ˆ
σ z > ˆ
σ 0
and decreases when ˆ
σ z < ˆ
σ 0 . First, we examine Case I with the bar stretched
( > 1) and ˆ
σ 0 set to zero; thus, the bar is initially in tension with ˙
G z > 0. As
shown in Fig. 6.16a, G z increases with λ but at a slower rate. Thus, growth lags
the specified deformation by an amount that decreases as α/β increases, i.e., as the
growth-rate constant increases. As t becomes large, λ → = 1.2 and changes
little, giving growth time to catch up (G z → ). Thus, λ ∗
z = λ/G z → 1, and the
stress approaches the target stress of zero (Fig. 6.16a ). Moreover, as the growth rate
increases, allowing G z (t) to stay closer to λ(t), the peak stress drops.