232
16 Solution of the Simplest Problems for the Strain Theory of Plasticity
2τ =
2σ i
3ε i
(ε ϕ − ε r ).
(16.50)
According to the strain theory of plasticity (see p. 197)
τ =
σ i
3ε i
γ.
Then the previous formula looks as follows:
γ = ε ϕ − ε r .
(16.51)
Let us write the non-compressibility condition:
ε ϕ + ε r + ε z = 0.
(16.52)
Using the Cauchy formulas (16.42), we obtain as follows from the condition (16.52):
du
dr
+
u
r
= −ε z .
(16.53)
A common solution of equation (16.53) is the function
u = −
1
2
ε z r +
B
r
, (B − const).
Using formulas (16.42), we will find
ε r =
du
dr
= −
1
2
ε z −
B
r 2 ; ε ϕ =
u
r
= −
1
2
ε z +
B
r 2 .
By substituting the obtained strain expressions into formula (16.51), we obtain
γ =
2B
r 2 .
(16.54)
Let us integrate the equilibrium equation (16.41)
r
a
dσ r =
r
a
(σ r − σ ϕ )
dr
r
,
or, using the boundary condition at r = a
σ r = −p a +
r
a
(σ ϕ − σ r )
dr
r
.
16 Solution of the Simplest Problems for the Strain Theory of Plasticity
2τ =
2σ i
3ε i
(ε ϕ − ε r ).
(16.50)
According to the strain theory of plasticity (see p. 197)
τ =
σ i
3ε i
γ.
Then the previous formula looks as follows:
γ = ε ϕ − ε r .
(16.51)
Let us write the non-compressibility condition:
ε ϕ + ε r + ε z = 0.
(16.52)
Using the Cauchy formulas (16.42), we obtain as follows from the condition (16.52):
du
dr
+
u
r
= −ε z .
(16.53)
A common solution of equation (16.53) is the function
u = −
1
2
ε z r +
B
r
, (B − const).
Using formulas (16.42), we will find
ε r =
du
dr
= −
1
2
ε z −
B
r 2 ; ε ϕ =
u
r
= −
1
2
ε z +
B
r 2 .
By substituting the obtained strain expressions into formula (16.51), we obtain
γ =
2B
r 2 .
(16.54)
Let us integrate the equilibrium equation (16.41)
r
a
dσ r =
r
a
(σ r − σ ϕ )
dr
r
,
or, using the boundary condition at r = a
σ r = −p a +
r
a
(σ ϕ − σ r )
dr
r
.