16.5 Torsion of a Beam of Ideally Plastic Material
239
τ
2
= τ
2
x + τ
2
y = τ
2
s ,
(16.73)
where τ s is the shear yield stress.
In the case of elastic–plastic torsion, the stresses τ x and τ y must satisfy the
equilibrium equation (16.63), and this requires fulfillment of Eqs. (16.68). The latter
means that the newly implemented function y) can be interpreted as a surface
supported by the cross-section outline, and it can be called the stress function in
plastic torsion of the beam of this section. Let us use the stress function y)
for the elastic and elastic–plastic strain of the beam. Based on formulas (16.68) and
(16.73), the plasticity condition looks as follows:
∂∂
∂x
2 +
∂∂
∂y
2 = τ
2
s .
(16.74)
It means that the problem of elastic–plastic torsion of the beam comes to finding
the function y) that satisfies Eq. (16.69) in the plastic area, Eq. (16.74) in the
plastic area, and to the boundary condition (16.71).
Furthermore, the stress τ must change continuously at the boundary between
elastic and plastic areas. If we designate belonging of the values of the elastic area
using the indexes “y” and those in the plastic area as “p,” the continuity condition
of tangential stresses at this boundary can be written as follows:
∂∂
∂x
p =
∂∂
∂x
y
,
∂∂
∂y
p =
∂∂
∂y
y
.
(16.75)
The left part of Eq. (16.74) represents a square of the modulus of vector grad
that expresses the highest incline of the surface . In this manner, the following
equation must be fulfilled in all points of the section with plastic strain:
|grad | = τ s = const.
Finally, according to formula (16.70), the tangential stress τ at the boundary of
the plastic zone of cross-section is directed along the tangential line to the outline
y = f (x), which equals the condition = const at this boundary.
Form the above properties of the function , it follows that the stress function
in plastic torsion represents a surface of the highest incline that can be built on the
outline of the beam cross-section. As per formula (16.74), in the plastic area, the
tangent of the highest incline angle equals ±τ s , whereas this tangent in the elastic
area is less than in τ s . This requires using the method of experimental determination
of tangential stresses in elastic and plastic areas of the section and determination of
these areas.
The method consists in the following. A surface (roof) of equal slope must be
built above the cross-section. The basis of this surface (outline L, Fig. 16.8) is pulled
over by a membrane to which even pressure p is applied. If the membrane does not
touch the equal slope surface, there is elastic torsion of the beam.
239
τ
2
= τ
2
x + τ
2
y = τ
2
s ,
(16.73)
where τ s is the shear yield stress.
In the case of elastic–plastic torsion, the stresses τ x and τ y must satisfy the
equilibrium equation (16.63), and this requires fulfillment of Eqs. (16.68). The latter
means that the newly implemented function y) can be interpreted as a surface
supported by the cross-section outline, and it can be called the stress function in
plastic torsion of the beam of this section. Let us use the stress function y)
for the elastic and elastic–plastic strain of the beam. Based on formulas (16.68) and
(16.73), the plasticity condition looks as follows:
∂∂
∂x
2 +
∂∂
∂y
2 = τ
2
s .
(16.74)
It means that the problem of elastic–plastic torsion of the beam comes to finding
the function y) that satisfies Eq. (16.69) in the plastic area, Eq. (16.74) in the
plastic area, and to the boundary condition (16.71).
Furthermore, the stress τ must change continuously at the boundary between
elastic and plastic areas. If we designate belonging of the values of the elastic area
using the indexes “y” and those in the plastic area as “p,” the continuity condition
of tangential stresses at this boundary can be written as follows:
∂∂
∂x
p =
∂∂
∂x
y
,
∂∂
∂y
p =
∂∂
∂y
y
.
(16.75)
The left part of Eq. (16.74) represents a square of the modulus of vector grad
that expresses the highest incline of the surface . In this manner, the following
equation must be fulfilled in all points of the section with plastic strain:
|grad | = τ s = const.
Finally, according to formula (16.70), the tangential stress τ at the boundary of
the plastic zone of cross-section is directed along the tangential line to the outline
y = f (x), which equals the condition = const at this boundary.
Form the above properties of the function , it follows that the stress function
in plastic torsion represents a surface of the highest incline that can be built on the
outline of the beam cross-section. As per formula (16.74), in the plastic area, the
tangent of the highest incline angle equals ±τ s , whereas this tangent in the elastic
area is less than in τ s . This requires using the method of experimental determination
of tangential stresses in elastic and plastic areas of the section and determination of
these areas.
The method consists in the following. A surface (roof) of equal slope must be
built above the cross-section. The basis of this surface (outline L, Fig. 16.8) is pulled
over by a membrane to which even pressure p is applied. If the membrane does not
touch the equal slope surface, there is elastic torsion of the beam.