4.5 Extension and Torsion of a Cylindrical Bar
181
P θ =
σ θ
λ θ
=
1
λ θ
( ¯
σ θ − p).
(4.67)
Clearly, when these stresses are inserted into (4.66), p remains in the second
term rather than canceling out as it does in Eq. (4.62), thus making the integration
considerably more complicated.
For this reason, working in terms of Cauchy stress has a particular advantage for
this problem. In fact, unlike numerical solutions where the second Piola-Kirchhoff
stress offers certain benefits, the use of Cauchy stress often facilitates finding
analytical solutions to nonlinear problems involving incompressible materials.
4.5 Extension and Torsion of a Cylindrical Bar
To a first approximation, the main muscles in the arms and legs can be treated
as cylindrical bars containing longitudinally aligned fibers. The same is true of
papillary muscles, which connect the wall of the heart to the valves between the atria
and ventricles. These muscles are loaded primarily in uniaxial tension, but they also
can twist, e.g., by twisting an arm or leg. Here, we consider combined extension and
torsion of a bar containing longitudinal fibers. 2
4.5.1 Problem Statement
A solid cylinder of undeformed radius b 0 and length L 0 is composed of transversely
isotropic, incompressible, hyperelastic material consisting of longitudinal fibers
embedded in an isotropic matrix (Fig. 4.11a). The strain-energy density function
is given by Eq. (4.1) with c 2 = 0 and
I 1 = 3 + 2(E rr + E θθ + E zz )
I 4 = λ
2
z = 1 + 2E zz ,
(4.68)
as given by Eqs. (3.69) and (3.228) 1 with λ z being the fiber stretch ratio. With these
relations, Eq. (4.1) becomes
W = 2c 1 (E rr + E θθ + E zz ) +
c 3
2c 4
e
4c 4 E 2
zz − 1
.
(4.69)
End loads cause the bar to stretch to the length L (stretch ratio λ = L/L 0 ) and
twist by an amount ψ, which is the (constant) angle of twist per unit undeformed
2 Only passive deformation is considered here; active contraction is explored in the next chapter.
181
P θ =
σ θ
λ θ
=
1
λ θ
( ¯
σ θ − p).
(4.67)
Clearly, when these stresses are inserted into (4.66), p remains in the second
term rather than canceling out as it does in Eq. (4.62), thus making the integration
considerably more complicated.
For this reason, working in terms of Cauchy stress has a particular advantage for
this problem. In fact, unlike numerical solutions where the second Piola-Kirchhoff
stress offers certain benefits, the use of Cauchy stress often facilitates finding
analytical solutions to nonlinear problems involving incompressible materials.
4.5 Extension and Torsion of a Cylindrical Bar
To a first approximation, the main muscles in the arms and legs can be treated
as cylindrical bars containing longitudinally aligned fibers. The same is true of
papillary muscles, which connect the wall of the heart to the valves between the atria
and ventricles. These muscles are loaded primarily in uniaxial tension, but they also
can twist, e.g., by twisting an arm or leg. Here, we consider combined extension and
torsion of a bar containing longitudinal fibers. 2
4.5.1 Problem Statement
A solid cylinder of undeformed radius b 0 and length L 0 is composed of transversely
isotropic, incompressible, hyperelastic material consisting of longitudinal fibers
embedded in an isotropic matrix (Fig. 4.11a). The strain-energy density function
is given by Eq. (4.1) with c 2 = 0 and
I 1 = 3 + 2(E rr + E θθ + E zz )
I 4 = λ
2
z = 1 + 2E zz ,
(4.68)
as given by Eqs. (3.69) and (3.228) 1 with λ z being the fiber stretch ratio. With these
relations, Eq. (4.1) becomes
W = 2c 1 (E rr + E θθ + E zz ) +
c 3
2c 4
e
4c 4 E 2
zz − 1
.
(4.69)
End loads cause the bar to stretch to the length L (stretch ratio λ = L/L 0 ) and
twist by an amount ψ, which is the (constant) angle of twist per unit undeformed
2 Only passive deformation is considered here; active contraction is explored in the next chapter.