4.4 Extension and Inflation of a Circular Tube
177
With λ and a specified, Eqs. (4.52) and (4.54) provide the stretch ratios as known
functions of R. Therefore, even in this semi-inverse problem, incompressibility
allows us to compute the entire deformation field without considering stress. This
would not be the case, however, if p i is given, rather than a.
Let b be the deformed outer radius. Then, setting r(b 0 ) = b in Eq. (4.54) and
rearranging terms yield
λ
b
2
− a
2
= b
2
0 − a
2
0 .
(4.55)
To better understand the meaning of this relation, we note that
V 0 = π
b
2
0 − a
2
0
L 0
V = π
b
2
− a
2
L
are the wall volume before and after deformation, respectively. For this incompressible tube, setting V = V 0 and L/L 0 = λ reproduces (4.55).
Stress and Equilibrium Since the stretch ratios depend only on the radial coordinate, the stress components also are functions of r. In polar coordinates, the Cauchy
stress tensor can be written in the form
σ = σ r e r e r + σ θ e θ e θ + σ z e z e z .
(4.56)
The equilibrium equation is
∇ · σ = 0,
where Eqs. (2.3) and orthogonality of the base vectors yield
∇ · σ =
e r
∂
∂r
+ e θ
1
r
∂
∂θ
+ e z
∂
∂z
· [σ r (r) e r (θ )e r (θ ) + σ θ (r) e θ (θ )e θ (θ ) + σ z (r) e z e z ]
= e r ·
∂σ r
∂r
e r e r +
1
r
e θ ·
σ r
e r
∂e r
∂θ
+
∂e r
∂θ
e r
+ σ θ
∂e θ
∂θ
e θ + e θ
∂e θ
∂θ
=
∂σ r
∂r
e r +
1
r
e θ · [σ r (e r e θ + e θ e r ) + σ θ (−e r e θ − e θ e r )]
=
∂σ r
∂r
+
1
r
(σ r − σ θ )
e r .
(4.57)
Thus, only the radial equilibrium equation
∂σ r
∂r
+
σ r − σ θ
r
= 0
(4.58)