220
16 Solution of the Simplest Problems for the Strain Theory of Plasticity
a
c
b
Fig. 16.1 Pure elastic–plastic bending of a straight beam
the component of stress σ z differs from zero, and it does not depend on the axial
coordinate z.
In this manner, the problem is solved by the single ratio between the bending
moment and curvature that is obtained from the equilibrium of moments of inner
and outer forces:
M =
h
−h
σ (ε) · y · b
y
h
dy.
(16.4)
The substitution of the dependency σ = ϕ(ε) to formula (16.4) gives
M =
h
−h
ϕ(ε)b
y
h
ydy =
h
−h
ϕ(χy)b
b
h
ydy.
(16.5)
Let us make the following transformations in the integral (16.5):
ε = χy; y =
ε
χ
; b
y
h
= b
ε
ε h
,
where ε h = χh is the strain of the marginal (lower) fiber of the beam. By
substituting these values into formula (16.5), we can write as follows:
M =
2
χ 2
ε h
0
ϕ(ε)b
ε
ε h
εdε =
2h 2
ε 2
h
ε h
0
ϕ(ε)b
ε
ε h
εdε.
(16.6)
Let us designate
1
ε 2
h
ε h
0
ϕ(ε)b
ε
ε h
εdε = h ).
(16.7)
16 Solution of the Simplest Problems for the Strain Theory of Plasticity
a
c
b
Fig. 16.1 Pure elastic–plastic bending of a straight beam
the component of stress σ z differs from zero, and it does not depend on the axial
coordinate z.
In this manner, the problem is solved by the single ratio between the bending
moment and curvature that is obtained from the equilibrium of moments of inner
and outer forces:
M =
h
−h
σ (ε) · y · b
y
h
dy.
(16.4)
The substitution of the dependency σ = ϕ(ε) to formula (16.4) gives
M =
h
−h
ϕ(ε)b
y
h
ydy =
h
−h
ϕ(χy)b
b
h
ydy.
(16.5)
Let us make the following transformations in the integral (16.5):
ε = χy; y =
ε
χ
; b
y
h
= b
ε
ε h
,
where ε h = χh is the strain of the marginal (lower) fiber of the beam. By
substituting these values into formula (16.5), we can write as follows:
M =
2
χ 2
ε h
0
ϕ(ε)b
ε
ε h
εdε =
2h 2
ε 2
h
ε h
0
ϕ(ε)b
ε
ε h
εdε.
(16.6)
Let us designate
1
ε 2
h
ε h
0
ϕ(ε)b
ε
ε h
εdε = h ).
(16.7)