290
6 Growth
Ideally, all surfaces should be stress-free, but the zero-stress condition on the ends
of the tube can only be satisfied in an average sense if we neglect end effects. Thus,
we set the resultant axial force to zero, i.e.,
N = 2π
b
a
σ z r dr = 0,
which, for computational purposes, can be transformed into the more convenient
form [see Eq. (5.66)]
b
a
(2 ¯
σ z − ¯
σ r − ¯
σ θ ) r dr = 0.
(6.76)
After substituting for the ¯
σ i (r; a, λ), Eqs. (6.75) and (6.76) provide two integral
equations to solve for a and λ.
Illustrative Results Like a metal experiencing isotropic thermal expansion, an
unloaded and unconstrained body undergoing uniform, isotropic growth remains
stress-free, because the element zero-stress configurations remain geometrically
compatible as they grow. Growth in the present problem is inherently anisotropic,
however, since there is no axial growth. Nevertheless, because uniform axial growth
of a tube, like a bar, does not produce stress, we expect that uniform, isotropic
growth in the cross section (G r = G θ ) also would not generate stress. The solution
confirms this behavior (not shown).
In contrast, if growth is uniform but anisotropic in the plane of the cross section
(G r = G θ ), the situation is quite different. Figure 6.11a,b shows residual stress
distributions caused by uniform circumferential growth only (G θ = 1, G r = 1)
and uniform radial growth only (G r = 1, G θ = 1). Both cases yield transmural
gradients in σ θ . The gradient is positive for G θ < 1 or G r > 1 and negative for
G θ > 1 or G r < 1. Moreover, the gradients in both cases are steeper for 50%
atrophy than for 50% positive growth.
To understand these trends, consider the schematics shown in Fig. 6.12. In
the upper part of the figure, the tube is cut radially, and then it grows in the
circumferential direction. For G θ > 1, the tube lengthens circumferentially, creating
a region of overlap. Restoring the intact section requires applying equal and opposite
moments that bend the wall outward until the cut surfaces are again coincident.
This deformation puts the inner and outer parts of the wall into circumferential
tension and compression, respectively, in agreement with the curve for G θ = 1.5
in Fig. 6.11a. Similarly, negative circumferential growth requires inward bending to
restore compatibility, reversing the locations of tension and compression (G θ = 0.5
in Fig. 6.11a). Figure 6.12 also indicates that the radius of the tube increases for
G θ > 1 and decreases for G θ < 1, consistent with the results in Fig. 6.11a.
Positive radial growth (G r > 1) causes the artery wall to thicken, pushing the
outer surface outward and the inner surface inward (Fig. 6.12, bottom). Increasing
the outer perimeter causes tension in the outer part of the wall, while shortening
6 Growth
Ideally, all surfaces should be stress-free, but the zero-stress condition on the ends
of the tube can only be satisfied in an average sense if we neglect end effects. Thus,
we set the resultant axial force to zero, i.e.,
N = 2π
b
a
σ z r dr = 0,
which, for computational purposes, can be transformed into the more convenient
form [see Eq. (5.66)]
b
a
(2 ¯
σ z − ¯
σ r − ¯
σ θ ) r dr = 0.
(6.76)
After substituting for the ¯
σ i (r; a, λ), Eqs. (6.75) and (6.76) provide two integral
equations to solve for a and λ.
Illustrative Results Like a metal experiencing isotropic thermal expansion, an
unloaded and unconstrained body undergoing uniform, isotropic growth remains
stress-free, because the element zero-stress configurations remain geometrically
compatible as they grow. Growth in the present problem is inherently anisotropic,
however, since there is no axial growth. Nevertheless, because uniform axial growth
of a tube, like a bar, does not produce stress, we expect that uniform, isotropic
growth in the cross section (G r = G θ ) also would not generate stress. The solution
confirms this behavior (not shown).
In contrast, if growth is uniform but anisotropic in the plane of the cross section
(G r = G θ ), the situation is quite different. Figure 6.11a,b shows residual stress
distributions caused by uniform circumferential growth only (G θ = 1, G r = 1)
and uniform radial growth only (G r = 1, G θ = 1). Both cases yield transmural
gradients in σ θ . The gradient is positive for G θ < 1 or G r > 1 and negative for
G θ > 1 or G r < 1. Moreover, the gradients in both cases are steeper for 50%
atrophy than for 50% positive growth.
To understand these trends, consider the schematics shown in Fig. 6.12. In
the upper part of the figure, the tube is cut radially, and then it grows in the
circumferential direction. For G θ > 1, the tube lengthens circumferentially, creating
a region of overlap. Restoring the intact section requires applying equal and opposite
moments that bend the wall outward until the cut surfaces are again coincident.
This deformation puts the inner and outer parts of the wall into circumferential
tension and compression, respectively, in agreement with the curve for G θ = 1.5
in Fig. 6.11a. Similarly, negative circumferential growth requires inward bending to
restore compatibility, reversing the locations of tension and compression (G θ = 0.5
in Fig. 6.11a). Figure 6.12 also indicates that the radius of the tube increases for
G θ > 1 and decreases for G θ < 1, consistent with the results in Fig. 6.11a.
Positive radial growth (G r > 1) causes the artery wall to thicken, pushing the
outer surface outward and the inner surface inward (Fig. 6.12, bottom). Increasing
the outer perimeter causes tension in the outer part of the wall, while shortening