6.9 Mechanical Feedback
303
where α z and α t are positive constants. Neglecting the weight of the bar, determine
G z , G t , and ˆ
σ z as functions of time.
Solution
In terms of the elastic stretch ratios λ ∗
i , the total axial and transverse stretch ratios,
respectively, are
λ z = λ
∗
z G z ,
λ r = λ θ = λ t = λ
∗
t G t .
(6.97)
Both elastic deformation and growth affect the cross-sectional area, which changes
from A 0 at t = 0– (before the weight is attached) to
A = λ r λ θ A 0 = λ
2
t A 0 = λ
∗2
t G
2
t A 0
for t ≥ 0 [see Eq. (3.79)]. Enforcing incompressibility gives
J
∗
= λ
∗
r λ
∗
θ λ
∗
z = λ
∗2
t λ
∗
z = 1,
or λ ∗2
t = (λ ∗
z ) −1 = G z /λ z . Combining these relations yields
A = (G
2
t G z /λ z )A 0 .
Axial equilibrium gives σ z = w/A or
ˆ
σ z =
σ z
c
=
λ z
G 2
t G z
ˆ
w,
(6.98)
where ˆ
w = w/A 0 c is the dimensionless load, with c being the modulus. The
constitutive relation (6.92) gives
ˆ
σ z = 2
λ 2
z
G 2
z
−
G z
λ z
.
(6.99)
Given ˆ
w and ˆ
σ 0 , substituting (6.98) into Eqs. (6.96) and (6.99) provides two
differential equations and one algebraic equation to solve for G z (t), G t (t), and
λ z (t) with G z (0) = G t (0) = 1. This differential-algebraic system of equations
can be solved numerically using the MATLAB function ode15s.
Results
Results are shown for ˆ
w = 1, α z = 1 h −1 , α t = 2 h −1 , and two values of the
target stress ˆ
σ 0 , one smaller and one larger than the initial stress (Fig. 6.17). For
ˆ
σ 0 = 0.5 (blue curves), the bar grows longer and thicker, as the stress decreases
and approaches the target stress. For ˆ
σ 0 = 2 (red curves), the bar grows shorter and
thinner, as the stress increases toward the target stress. After the initial elastic stretch