6.4 Fundamental Growth Mechanics
269
σ
(1)
x = 2c
λ 2
G 2
1 −
G 3
λ 3
σ
(2)
x = 2c λ
2
1 −
1
λ 3
,
(6.31)
where λ = λ
(1)
x = λ
(2)
x is to be determined.
With G(t) given, we find λ(t) by considering axial equilibrium. In the qualitative
analysis of this problem, we showed that the net axial force on the bar must be zero,
and this requires 2f 2 − f 1 = 0, where f 1 is the compressive force resultant in the
middle layer and f 2 is the tensile force resultant in both outer layers. In terms of first
Piola-Kirchhoff stresses, these forces are f 1 = −P
(1)
x A 0 and f 2 = P
(2)
x A 0 , with the
minus sign needed because normal stress is defined to be positive in tension. Thus,
the equilibrium equation takes the form
P
(1)
x + 2P
(2)
x = 0,
and substituting Eq. (6.12) 1 gives
J
(1) σ
(1)
x + 2J
(2) σ
(2)
x = 0,
(6.32)
since λ
(1)
x = λ
(2)
x . For J ∗(L) = 1, Eq. (6.8) yields the volume ratios J (L) = J
(L)
G =
G
(L)
x G
(L)
y G
(L)
z or
J
(1)
= G
(1)
x = G,
J
(2)
= 1,
and (6.32) becomes
Gσ
(1)
x + 2σ
(2)
x = 0.
Finally, substituting Eqs. (6.31) and solving for λ give
λ =
G(2 + G 2 )
1 + 2G
1
3
,
in which G(t) is given by (6.24).
Results
If all layers grow equally, we would expect the bar to lengthen stress-free with
λ = G. In this example, however, λ < G (Fig. 6.4b) because the nongrowing
outer layers resist extension of the growing middle layer. Growth of the middle
layer generates tension in the outer layers and compression in the middle layer,
with P
(1)
x = −2P
(2)
x (Fig. 6.4c). Compression increases the cross-sectional area of
the middle layer, while tension decreases the area of the outer layers. This causes
the magnitudes of the Cauchy stresses to be lower and higher, respectively, than the
corresponding values of P x in these layers (Fig. 6.4c). All these results are consistent
with our qualitative analysis.
269
σ
(1)
x = 2c
λ 2
G 2
1 −
G 3
λ 3
σ
(2)
x = 2c λ
2
1 −
1
λ 3
,
(6.31)
where λ = λ
(1)
x = λ
(2)
x is to be determined.
With G(t) given, we find λ(t) by considering axial equilibrium. In the qualitative
analysis of this problem, we showed that the net axial force on the bar must be zero,
and this requires 2f 2 − f 1 = 0, where f 1 is the compressive force resultant in the
middle layer and f 2 is the tensile force resultant in both outer layers. In terms of first
Piola-Kirchhoff stresses, these forces are f 1 = −P
(1)
x A 0 and f 2 = P
(2)
x A 0 , with the
minus sign needed because normal stress is defined to be positive in tension. Thus,
the equilibrium equation takes the form
P
(1)
x + 2P
(2)
x = 0,
and substituting Eq. (6.12) 1 gives
J
(1) σ
(1)
x + 2J
(2) σ
(2)
x = 0,
(6.32)
since λ
(1)
x = λ
(2)
x . For J ∗(L) = 1, Eq. (6.8) yields the volume ratios J (L) = J
(L)
G =
G
(L)
x G
(L)
y G
(L)
z or
J
(1)
= G
(1)
x = G,
J
(2)
= 1,
and (6.32) becomes
Gσ
(1)
x + 2σ
(2)
x = 0.
Finally, substituting Eqs. (6.31) and solving for λ give
λ =
G(2 + G 2 )
1 + 2G
1
3
,
in which G(t) is given by (6.24).
Results
If all layers grow equally, we would expect the bar to lengthen stress-free with
λ = G. In this example, however, λ < G (Fig. 6.4b) because the nongrowing
outer layers resist extension of the growing middle layer. Growth of the middle
layer generates tension in the outer layers and compression in the middle layer,
with P
(1)
x = −2P
(2)
x (Fig. 6.4c). Compression increases the cross-sectional area of
the middle layer, while tension decreases the area of the outer layers. This causes
the magnitudes of the Cauchy stresses to be lower and higher, respectively, than the
corresponding values of P x in these layers (Fig. 6.4c). All these results are consistent
with our qualitative analysis.