16.4 Symmetric Strain of a Cylindrical Tube
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Taking into account formulas (16.49) and (16.50), the last dependency can be
written as follows:
σ r = −p a + 2
r
a
f (γ )
dr
r
.
Let us substitute the integration variable r with γ in this formula. To do it, let us
differentiate formula (16.54); we will obtain
dr
r
= −
dγ
2γ
.
By substituting from the last formula, we obtain
σ r = −p a −
γ
γ a
f (γ )
dγ
γ
.
(16.55)
Having in mind that
σ ϕ − σ r
2
= τ = f (γ ),
and taking into account formula (16.55), we will obtain
σ ϕ = −p a −
γ
γ a
f (γ )
dγ
γ
+ 2f (γ ).
(16.56)
Due to formula (16.54), the integration limits in formulas (16.55) and (16.56) are
related by the dependency
γ = γ a
a 2
r 2 .
Let us now determine the axial normal stress σ z . To do it, let us use an intact
boundary condition at the outer outline of the tube section. From formula (16.55),
we obtain
p b − p a =
γ b
γ a
f (γ )
dγ
γ
.
(16.57)
Let us write the ratio (16.49) as
2σ i
3ε i
=
σ ϕ − σ r
ε ϕ − ε r
.