4.1 Uniaxial Extension of a Bar
157
4.1.2 Solution
To describe the undeformed and deformed configurations, we introduce the Cartesian coordinates (X, Y, Z) and (x, y, z), respectively (Fig. 4.1). Symmetry arguments indicate that the global Cartesian system coincides with the principal
directions of stress and strain at each point. Thus, we use the specialized equations
in Sect. 3.7.3. Here, we work in terms of first Piola-Kirchhoff stress P i and then
convert to Cauchy stress σ i .
Case 1: Incompressible Bar Equilibrium analysis of isolated sections of the bar
shows that the force acting on each internal cross section is F = P x A 0 , where A 0
is the undeformed cross-sectional area, while the forces on sections normal to the y
and z directions are zero. Thus, the only nonzero stress component is the axial stress
P x = F /A 0 , which is constant because the bar is not tapered. With body forces and
inertia omitted, the differential equations of equilibrium (3.243) 2 reduce to
∂P x
∂X
=
∂P y
∂Y
=
∂P z
∂Z
= 0,
(4.3)
which are identically satisfied since the stresses are uniform.
Symmetry considerations imply λ y = λ z , and so the incompressibility condition
J = λ x λ y λ z = 1 yields
λ y = λ z = λ
−1/2
x
.
(4.4)
Importantly, the entire deformation field is now known in terms of the specified λ x
without needing to consider material properties.
With J = 1, Eq. (3.249) 2 gives
P x =
∂W
∂λ x
−
p
λ x
P y =
∂W
∂λ y
−
p
λ y
P z =
∂W
∂λ z
−
p
λ z
.
(4.5)
To find the Lagrange multiplier, we set P z = 0 to get
p = λ z
∂W
∂λ z
.
(4.6)