16.6 Rod of a Variable Section: Method of Elastic Solutions
243
For example, if both ends of the rod are terminated, f = 0. In the case of rigid
termination on the left end and elastic attachment of the right end u 0 = 0; u 1 =
−kP 1 , where k is the elastic compliance coefficient.
In the case of Problem 2, let us re-write the expression (16.85) as
f (P 0 , P 1 ) =
P 0
E
ψ(l) −
1
E
l
0
[ψ(x) − ψ(η)] Q(η)dη +
l
0
ω
du
dx
du
dx
dx.
(16.87)
16.6.3 Algorithm of the Elastic Solutions Method
Let us consider the primary initial system of equations (16.77), (16.87), and (16.84)
for Problem 2. In the case of the elastic work of the rod ω = 0. Then Eqs. (16.87)
and (16.77) give P 0 and P 1 , formula (16.84) is used to define strain, and then we
find stresses knowing strains under Hooke’s law.
In the case of elastic–plastic work, to solve the system of equations (16.77),
(16.87), and (16.84) we use the method of successive approximations based on the
assumption that ω is a small parameter. As the first approximation, let us assume the
elastic solution obtained at ω = 0
P
(1)
0 , P
(1)
1 , ε
(1) , σ
(1) .
If |ε (1) | < ε s , the rod shows no plastic strains and the first approximation is
the accurate solution to the elastic problem. Otherwise, there is an elastic–plastic
state. Then the condition |ε (1) | ε s is used to find the area (1) (x) where the
plastic state has occurred. With known ε (1) (x) from the diagram σ ∼ ε, we find
ω (1) (ε (1) ) = ω (1) [x].
The found value ω (1) [x] is substituted into Eqs. (16.87) and (16.84) so as to
obtain the second approximation.
P
(2)
0 , P
(2)
1 , ε
(2) , σ
(2) , ω
(2)
[x].
Integration in the interval [0; l] is done for x ∈ (1) (x) only. The iteration
process is continued until the difference of values obtained from two successive
approximations is within the permitted limits. The calculations show that usually
two or three approximations are enough.
To conclude, we shall note that the convergence of the method of elastic solutions
is not rigorously proven yet.
243
For example, if both ends of the rod are terminated, f = 0. In the case of rigid
termination on the left end and elastic attachment of the right end u 0 = 0; u 1 =
−kP 1 , where k is the elastic compliance coefficient.
In the case of Problem 2, let us re-write the expression (16.85) as
f (P 0 , P 1 ) =
P 0
E
ψ(l) −
1
E
l
0
[ψ(x) − ψ(η)] Q(η)dη +
l
0
ω
du
dx
du
dx
dx.
(16.87)
16.6.3 Algorithm of the Elastic Solutions Method
Let us consider the primary initial system of equations (16.77), (16.87), and (16.84)
for Problem 2. In the case of the elastic work of the rod ω = 0. Then Eqs. (16.87)
and (16.77) give P 0 and P 1 , formula (16.84) is used to define strain, and then we
find stresses knowing strains under Hooke’s law.
In the case of elastic–plastic work, to solve the system of equations (16.77),
(16.87), and (16.84) we use the method of successive approximations based on the
assumption that ω is a small parameter. As the first approximation, let us assume the
elastic solution obtained at ω = 0
P
(1)
0 , P
(1)
1 , ε
(1) , σ
(1) .
If |ε (1) | < ε s , the rod shows no plastic strains and the first approximation is
the accurate solution to the elastic problem. Otherwise, there is an elastic–plastic
state. Then the condition |ε (1) | ε s is used to find the area (1) (x) where the
plastic state has occurred. With known ε (1) (x) from the diagram σ ∼ ε, we find
ω (1) (ε (1) ) = ω (1) [x].
The found value ω (1) [x] is substituted into Eqs. (16.87) and (16.84) so as to
obtain the second approximation.
P
(2)
0 , P
(2)
1 , ε
(2) , σ
(2) , ω
(2)
[x].
Integration in the interval [0; l] is done for x ∈ (1) (x) only. The iteration
process is continued until the difference of values obtained from two successive
approximations is within the permitted limits. The calculations show that usually
two or three approximations are enough.
To conclude, we shall note that the convergence of the method of elastic solutions
is not rigorously proven yet.