278
6 Growth
In B 0 :
∇ 0 = e ρ
∂
∂ρ
+
e ϑ
ρ
∂
∂ϑ
+ e ζ
∂
∂ζ
.
(6.43)
With these relations, we get
F 1 = (∇ρ)
T
=
∂ρ
∂R
e ρ e R +
πρ
φ 0 R
e ϑ e + e ζ e Z
F 2 = (∇ 0 r)
T
=
∂r
∂ρ
e r e ρ +
r
ρ
e θ e ϑ + λ e z e ζ
F = F 2 · F 1
= λ r e r e R + λ θ e θ e + λ z e z e Z ,
(6.44)
where the stretch ratios in F are those given by Eqs. (6.40). Thus, both approaches
yield the same result.
Note that the manipulations leading to Eqs. (6.44) involve differentiating base
vectors in cylindrical polar coordinates. Since most of these differentiations have
been addressed in earlier problems, the details are omitted here. However, it is
instructive to examine differentiation of e ρ , defined in the unloaded state B 0 , with
respect to the coordinates (R, ,, Z), defined in the ZSS B. In terms of Cartesian
base vectors, we have
e ρ = e x cos ϑ + e y sin ϑ
e ϑ = −e x sin ϑ + e y cos ϑ,
where ϑ = ππ/φ 0 . Clearly, derivatives of e ρ with respect to R and Z are zero, but
∂e ρ
∂∂
=
∂e ρ
∂ϑ
∂ϑ
∂∂
= e ϑ
π
φ 0
,
which was used above in deriving F 1 .
Substituting (6.40) into the incompressibility condition J = det F = λ r λ θ λ z = 1
for the total deformation gives
r dr =
φ 0
πλλ
R dR.
Integrating both sides and using the boundary condition r(a 0 ) = a yield
r(R) =
a
2
+
φ 0
πλλ
R
2
− a
2
0
1
2
,
(6.45)
which reduces to Eq. (4.54) for no residual stress (φ 0 = π , = 1).