6.7 Growth-Induced Residual Stress in Tubes
289
For the special case where J G is uniform in the tube, this relation becomes
r(R) =
a
2
+
J G
λ
R
2
− a
2
0
1
2
,
which reduces to Eq. (4.54) when there is no growth (J G = 1). The deformed outer
radius is given by b = r(b 0 ). In terms of r(R), Eqs. (6.59) provide the elastic stretch
ratios
λ
∗
r =
1
G r
∂r
∂R
λ
∗
θ =
1
G θ
r
R
λ
∗
z =
1
G z
λ.
(6.73)
In the present problem, the growth ratios, G r (R) and G θ (R), are specified, so
J G (R) is known a priori. Thus, Eqs. (6.72) and (6.73) yield r and the λ ∗
i at each
point in terms of the unknown geometric variables a and λ. To determine these
quantities, we need to consider stress and equilibrium.
With the strain-energy density function relative to the ZSS given by
W
∗
= W (E
∗
i ) = c(e
Q ∗ − 1)
Q
∗
= α 1 E
∗2
r + α 2 E
∗2
θ + α 3 E
∗2
z + 2α 4 E
∗
r E
∗
θ + 2α 5 E
∗
θ E
∗
z + 2α 6 E
∗
z E
∗
r ,
(6.74)
in which E ∗
i =
1
2 (λ ∗2
i − 1), the ¯
σ i are computed at each point using (6.63) 2 (with
J ∗ = 1) and (6.73). Next, enforcing radial equilibrium yields Eq. (6.64), which
gives the Lagrange multiplier in the form
p(r) = ¯
σ r (r) +
r
a
( ¯
σ θ − ¯
σ r )
dr
r
,
where the limits on the integral have been chosen so that the radial stress
σ r = ¯
σ r − p = −
r
a
( ¯
σ θ − ¯
σ r )
dr
r
satisfies the boundary condition σ r = 0 at r = a in the unloaded tube. The boundary
condition at the outer surface gives
σ r (b) = −
b
a
( ¯
σ θ − ¯
σ r )
dr
r
= 0.
(6.75)