4.5 Extension and Torsion of a Cylindrical Bar
183
r(R, ,, Z) = r(R)e r ((, Z) + z(Z)e z .
(4.71)
Computing the deformation gradient tensor using F T = ∇r gives Eq. (4.50), but
with the extra term
re Z
∂e r
∂Z
= re Z
∂e r
∂θ
∂θ
∂Z
= ψre Z e θ ,
since ∂e r /∂θ = e θ . This calculation yields 3
F = F rr e r e R + F θθ e θ e + F zz e z e Z + F θz e θ e Z ,
(4.72)
where
F rr =
∂r
∂R
,
F θθ =
r
R
,
F zz = λ,
F θz = ψr.
(4.73)
Since F includes a shear term, two subscripts are needed to define the tensor
components. The Lagrangian strain tensor is
E =
1
2
F
T
· F − I
=
1
2
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
∂r
∂R
2
− 1
0
0
0
r 2
R 2 − 1
ψr 2
R
0
ψr 2
R
λ 2 + ψ 2 r 2 − 1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(e I e J )
,
(4.74)
where I and J represent the coordinates (R, ,, Z).
Writing F in matrix form reveals that the incompressibility condition det F = 1
again produces Eq. (4.53) with solution
1
2 λr 2 =
1
2 R 2 + C. Because the bar has no
hole, we must set r = 0 at R = 0, giving C = 0 and
r =
R
√
λ
.
(4.75)
The deformed radius of the bar is b = b 0 /
√
λ, and Eq. (4.73) gives
F rr = F θθ =
1
√
λ
,
F zz = λ,
F θz =
ψR
√
λ
.
(4.76)
3 As mentioned previously, we use lowercase indices for tensor components even when their
associated base vectors have indices of upper case. This is done for simplicity, it is hoped without
confusion.