Because of the way in which these relations have been derived, it can be seen that the work done per unit volume in any process is given by
W vol.
=
ε
σ ε 2.20
It is also evident that because the stress function has been chosen as the von Mises stress which is equal in magnitude to the flow stress when the material is deforming, the effective
strain function will be equal to the strain in uniaxial tension when equal amounts of work are done in the general process and in uniaxial tension. Thus we have identified a general
stress–strain relation for an isotropic material deforming plastically, namely the effective stress–strain curve, σ
= f ε; this is coincident with the tensile test true stress–strain curve for an isotropic material.
The key to this principle of the equivalence of plastic work done is illustrated in Figure 2.11. To reiterate, in a plane stress process, there are two stress strain curves.
These must continuously satisfy the yield criterion and the condition that the work done in the process, i.e. the area under both curves, is equal to the work done in uniaxial tension.
This work done determines the current yield stress in uniaxial tension which is also the flow stress. The effective stress and strain functions ensure that these conditions are met
and enable the current flow stress for a material element deformed in any process to be determined from an experimental stress–strain curve obtained in a tension test. Material
properties can also be obtained from other tests, provided that the test enables an effective stress–strain curve to be obtained.
2.9 Summary
In this chapter, it is shown that for simple monotonic, proportional, plane stress processes, it is possible to determine at any instant the principal membrane stresses required for
deformation provided the current flow stress σ
f
is known and also that either the stress ratio α or strain ratio β are known. The current flow stress can be determined from the
tensile test stress strain curve using the effective stress and strain functions that are based on the equivalence of work. In practice, a process is often defined by the strain ratio β
obtained from measurement of final strains. The assumption is that this point is reached by a proportional process, but if only the initial and final conditions are known, care should
be taken in assuming that the strain ratio is constant.
The theory given in this chapter applies only for an instantaneous state in which the strain increment is small and the flow stress constant. In the next chapter, entire loading
paths are studied using the theory established here. It cannot be emphasized too strongly that while the theory of deformation given here is useful and practical, it is a simple approx-
imation of a very complex process. It is useful in process design and failure diagnosis in industry, but in some studies, more elaborate theories may be necessary.
2.10 Exercises
Ex. 2.1
A square element 8 × 8 mm in an undeformed sheet of 0.8 mm thickness becomes
a rectangle, 6.5 × 9.4 mm after forming. Assume that the stress strain law is:
σ = 6000.008 + ε
0.22
MPa Sheet deformation processes 27
and that the stress normal to the sheet is zero. Determine: a
the final membrane stresses; b
the final thickness; c
the principal strains. Sketch these in the stress or strain space.
Also determine: d
the stress and strain ratios assumed constant and the hydrostatic stress and the deviatoric stresses at the end of the process, and
e the plastic work of deformation in the element.
[Ans: a 151.3 MPa, −336.6 MPa; b 0.838 mm; c 0.161, −0.208; d −2.29, −1.29,
−61.7 MPa, 213 MPa, −274.9 MPa, 61.7 MPa; e 4.01 J]
Ex. 2.2
For a stress state in which the intermediate stress is σ
2
= 0.5 σ
1
+ σ
3
, show that in yielding with a von Mises criterion and flow stress, σ
f
, the diameter of the Mohr circle of stress is 2
√ 3σ
f
.
Ex. 2.3 Show that in uniaxial deformation in which σ
1
= σ
f
, σ
2
= σ
3
= 0, that the effec- tive stress and effective strain increment are:
σ = σ
1
= σ
f
dε = dε
1
Ex. 2.4 Three principal stresses are applied to a solid where σ
1
= 400 MPa, σ
2
= 200 MPa, and σ
3
= 0. a
What is the ratio dε
1
dε
3
? b
If a fluid pressure produces an all-around hydrostatic stress of −250 MPa that is
superimposed upon the original stress state, how does the ratio in a change ? Explain the result.
[Ans: a −1; b no change.]
Ex. 2.5 Compare two plane stress deformation processes for a sample of high-strength
low alloy steel of 1.2 mm thickness. In a, biaxial tension, the strain ratio is 1 and in the other, b, shear or drawing, the strain ratio is
−1. A square element whose sides are aligned with the principal directions is initially 10 by 10 mm. In each case, one side is
extended to 12 mm. The material properties are described by, σ
= 850ε
0.16
MPa Compare the final effective strains and stresses, the principal strains and stresses and the
final sheet thickness. [Ans:
ε σ
ε
1
, ε
2
σ
1
, σ
2
t a
0.365, 723 MPa,
0.182, 0.182, 723, 723 Mpa,
0.833 mm. b
0.211 662 MPa
0.182, −0.182, 382, −382 MPa 1.2 mm]
28 Mechanics of Sheet Metal Forming
