20
1 Summary of Elasticity Theory: Basic Concepts
or
d 2 u
dr 2 +
1
r
du
dr
−
u
r 2 = 0.
(1.31)
By direct checking, we can easily make sure that the common integral of the
linear homogeneous differential equation (1.31) is the function
u = Ar +
B
r
, (A,B − const).
(1.32)
The constant values A and B are defined from the conditions
σ r
r=a
= −p a , σ r
r=b
= −p b ,
whereas p a and p b are set (see. p. 18) values. Using formula (1.30), we obtain
A =
1 − ν ∗
E ∗
a 2 p a − b 2 p b
b 2 − a 2 ; B =
1 + ν ∗
E ∗
p a − p b
b 2 − a 2 a
2 b
2 .
(1.33)
By substituting the function (1.32) and (1.33) and its derivative to formula (1.30),
we obtain
σ r =
a 2 p a − b 2 p b
b 2 − a 2
−
p a − p b
b 2 − a 2
a 2 b 2
r 2 .
(1.34)
With the known radial stress σ r , we find the annular stress from formula (1.29)
σ t =
a 2 p a − b 2 p b
b 2 − a 2
+
p a − p b
b 2 − a 2
a 2 b 2
r 2 .
(1.35)
It was suggested above (p. 18) that the cylinder was in the conditions of
plane strain. e. g. there was no strain towards the cylinder axis. By combining the
coordinate axis Oz with the cylinder axis, we have ε z = 0 and find the axial stress
using formula (1.19)
σ z = 2ν
a 2 p a − b 2 p b
b 2 − a 2 .
(1.36)
From the last result, we see that the axial stress in the considered case does not
depend on the coordinate r. Hence, axial stresses can be eliminated by imposing a
respective stress or compression of the cylinder in the axial direction. In this case,
the cylinder will be in the conditions of a plane stressed state.
1 Summary of Elasticity Theory: Basic Concepts
or
d 2 u
dr 2 +
1
r
du
dr
−
u
r 2 = 0.
(1.31)
By direct checking, we can easily make sure that the common integral of the
linear homogeneous differential equation (1.31) is the function
u = Ar +
B
r
, (A,B − const).
(1.32)
The constant values A and B are defined from the conditions
σ r
r=a
= −p a , σ r
r=b
= −p b ,
whereas p a and p b are set (see. p. 18) values. Using formula (1.30), we obtain
A =
1 − ν ∗
E ∗
a 2 p a − b 2 p b
b 2 − a 2 ; B =
1 + ν ∗
E ∗
p a − p b
b 2 − a 2 a
2 b
2 .
(1.33)
By substituting the function (1.32) and (1.33) and its derivative to formula (1.30),
we obtain
σ r =
a 2 p a − b 2 p b
b 2 − a 2
−
p a − p b
b 2 − a 2
a 2 b 2
r 2 .
(1.34)
With the known radial stress σ r , we find the annular stress from formula (1.29)
σ t =
a 2 p a − b 2 p b
b 2 − a 2
+
p a − p b
b 2 − a 2
a 2 b 2
r 2 .
(1.35)
It was suggested above (p. 18) that the cylinder was in the conditions of
plane strain. e. g. there was no strain towards the cylinder axis. By combining the
coordinate axis Oz with the cylinder axis, we have ε z = 0 and find the axial stress
using formula (1.19)
σ z = 2ν
a 2 p a − b 2 p b
b 2 − a 2 .
(1.36)
From the last result, we see that the axial stress in the considered case does not
depend on the coordinate r. Hence, axial stresses can be eliminated by imposing a
respective stress or compression of the cylinder in the axial direction. In this case,
the cylinder will be in the conditions of a plane stressed state.