242
16 Solution of the Simplest Problems for the Strain Theory of Plasticity
σ =
N(x)
F (x)
=
P 0
F (x)
−
1
F (x)
x
0
Q(ζ )dζ.
(16.82)
By using expressions (16.80) and (16.81), formula (16.82) can be re-written as
follows:
σ = E
du
dx
1 − ω
du
dx
=
P 0
F (x)
−
1
F (x)
x
0
Q(ζ )dζ.
(16.83)
Let us express as follows:
du
dx
=
P 0
EF
−
1
EF
x
0
Q(ζ )dζ + ω
du
dx
·
du
dx
.
(16.84)
Let us integrate two parts of Eq. (16.84) upon the variable x from zero to x; now,
we have
u − u 0 =
P 0
E
ψ(x) −
1
E
x
0
[ψ(x) − ψ(η)] Q(η)dη +
x
0
ω
du
dx
du
dx
dx,
(16.85)
where u 0 = u(0) and ψ(x) designates the integral
ψ(x) =
x
0
dx
F (x)
.
The dependencies (16.77), (16.84), and (16.85) form a system of ratios sufficient
to solve the problems under consideration.
16.6.2 Specification of Problem Setting
By having Eqs. (16.77), (16.84), and (16.85), we can set the following problems.
Problem 1 P 0 , P 1 , and Q(x) are set. It is required to find stresses, strains, and
displacements.
As for stresses, this task is statically definable.
Problem 2 P 0 and P 1 are not defined in a clear form. Some restrictions that impede
displacement are placed on the left and right end of the rod, e.g., the following
function is set:
u 1 − u 0 = f (P 0 , P 1 ), [u 1 = u(l)].
(16.86)