258
17 Additions and Generalizations to the Strain Theory of Plasticity
closed path and direction of going around it are shown in Fig. 17.7 with arrows. The
gain of the stress dσ will cause a respective gain of strain de.
The Drucker postulate states that the work of additional stresses on the closed
loading path is not negative:
(σ ij − σ ∗
ij )de ij 0.
(17.18)
Let us show that formula (17.18) results in a number of important consequences.
Assume that loading is done over a closed path shown in Fig. 17.7. The loading
point M ∗ corresponding to the stress vector σ ∗ lies within ∗ or maybe on the
surface as shown in the figure. The path goes from the point M ∗ to the point M 1
as directed by the arrow for any trajectory fully lying inside ∗ . Then we impart
gain dσ to stress σ in the direction of the point M located on the external side of
the enclosed area ∗ . Now a new yield surface goes through the point M, so
returning from the point M to the point M ∗ along any path fully lying inside
means unloading. In this manner, strain is elastic in the sections M ∗ M 1 and MM ∗
and elastic–plastic in the section M 1 M and consists of elastic (de y ) and plastic
(de p ) parts:
de = de
y
+ de
p .
(17.19)
The work of additional loading on elastic displacements with closed loading path
equals zero. Indeed,
(σ − σ ∗ )de y =
σ de y − σ ∗
de y .
The first integral in the right part of this equation represents the complete work of
stress in elastic displacement with a closed path and, according to the definition of
the elasticity property, equals zero. The second integral also equals zero due to the
unambiguity of elastic strains. It follows that an irreversible work is done only in
the case of gaining plastic strain. Therefore, from formulas (17.18) and (17.19) we
can write as follows:
(σ
ij − σ
∗
ij )de
p
ij 0.
(17.20)
In the case of continuous active strain from the point M ∗ to an infinitely close point
M past the intermediate loading, we have a gain of plastic strain during the entire
period of additional loading. If we return from the point M to the point M ∗ via an
arbitrary path fully lying within , the irreversible work is done again on the plastic
component of strain only. Therefore,
dσ ij de
p
ij 0.
(17.21)
By considering the left part of the condition (17.20) as a scalar product of
the vectors σ − σ ∗ and de p , we conclude that these vectors form a sharp angle
17 Additions and Generalizations to the Strain Theory of Plasticity
closed path and direction of going around it are shown in Fig. 17.7 with arrows. The
gain of the stress dσ will cause a respective gain of strain de.
The Drucker postulate states that the work of additional stresses on the closed
loading path is not negative:
(σ ij − σ ∗
ij )de ij 0.
(17.18)
Let us show that formula (17.18) results in a number of important consequences.
Assume that loading is done over a closed path shown in Fig. 17.7. The loading
point M ∗ corresponding to the stress vector σ ∗ lies within ∗ or maybe on the
surface as shown in the figure. The path goes from the point M ∗ to the point M 1
as directed by the arrow for any trajectory fully lying inside ∗ . Then we impart
gain dσ to stress σ in the direction of the point M located on the external side of
the enclosed area ∗ . Now a new yield surface goes through the point M, so
returning from the point M to the point M ∗ along any path fully lying inside
means unloading. In this manner, strain is elastic in the sections M ∗ M 1 and MM ∗
and elastic–plastic in the section M 1 M and consists of elastic (de y ) and plastic
(de p ) parts:
de = de
y
+ de
p .
(17.19)
The work of additional loading on elastic displacements with closed loading path
equals zero. Indeed,
(σ − σ ∗ )de y =
σ de y − σ ∗
de y .
The first integral in the right part of this equation represents the complete work of
stress in elastic displacement with a closed path and, according to the definition of
the elasticity property, equals zero. The second integral also equals zero due to the
unambiguity of elastic strains. It follows that an irreversible work is done only in
the case of gaining plastic strain. Therefore, from formulas (17.18) and (17.19) we
can write as follows:
(σ
ij − σ
∗
ij )de
p
ij 0.
(17.20)
In the case of continuous active strain from the point M ∗ to an infinitely close point
M past the intermediate loading, we have a gain of plastic strain during the entire
period of additional loading. If we return from the point M to the point M ∗ via an
arbitrary path fully lying within , the irreversible work is done again on the plastic
component of strain only. Therefore,
dσ ij de
p
ij 0.
(17.21)
By considering the left part of the condition (17.20) as a scalar product of
the vectors σ − σ ∗ and de p , we conclude that these vectors form a sharp angle