a b
Tensile test Plane stress
s
3
= 0, e
3
= −e
1
2
s
1
, e
1
s
2
= 0, e
2
= −e
1
2 s
3
= 0, e
3
= − 1 + be
1
s
1
, e
1
s
2
= as
1
e
2
= be
1
Figure 2.2 Principal stresses and strains for elements deforming in a uniaxial tension and b a
general plane stress sheet process.
principal directions so that σ
1
σ
2
and the third direction is perpendicular to the surface where σ
3
= 0. The deformation mode is thus: ε
1
; ε
2
= βε
1
; ε
3
= −1 + βε
1
2.6 σ
1
; σ
2
= ασ
1
; σ
3
= 0 The constant volume condition is used to obtain the third principal strain. Integrating the
strain increments in Equation 2.3 shows that this condition can be expressed in terms of the true or natural strains:
ε
1
+ ε
2
+ ε
3
= 0 2.7
i.e. the sum of the natural strains is zero. For uniaxial tension, the strain and stress ratios are β
= −12 and α = 0.
2.4 Yielding in plane stress
The stresses required to yield a material element under plane stress will depend on the current hardness or strength of the sheet and the stress ratio α. The usual way to define the
strength of the sheet is in terms of the current flow stress σ
f
. The flow stress is the stress at which the material would yield in simple tension, i.e. if α
= 0. This is illustrated in the true stress–strain curve in Figure 2.3. Clearly σ
f
depends on the amount of deformation to which the element has been subjected and will change during the process. For the moment,
we shall consider only one instant during deformation and, knowing the current value of σ
f
the objective is to determine, for a given value of α, the values of σ
1
and σ
2
at which the element will yield, or at which plastic flow will continue for a small increment. We
consider here only the instantaneous conditions in which the strain increment is so small that the flow stress can be considered constant. In Chapter 3 we extend this theory for
continuous deformation.
There are a number of theories available for predicting the stresses under which a material element will deform plastically. Each theory is based on a different hypothesis
about material behaviour, but in this work we shall only consider two common models and apply them to the plane stress process described by Equations 2.6. Over the years,
many researchers have conducted experiments to determine how materials yield. While no single theory agrees exactly with experiment, for isotropic materials either of the models
presented here are sufficiently accurate for approximate models.
Sheet deformation processes 17
50 100
150 200
250 300
350 400
0.000 0.050
0.100 0.150
0.200 0.250
True stress MPa
True strain s
f
s
f
e
u
e
1
Figure 2.3 Diagram showing the current flow stress in an element after some strain in a tensile test.
With hindsight, common yielding theories can be anticipated from knowledge of the nature of plastic deformations in metals. These materials are polycrystalline and plastic
flow occurs by slip on crystal lattice planes when the shear stress reaches a critical level. To a first approximation, this slip which is associated with dislocations in the lattice is
insensitive to the normal stress on the slip planes. It may be anticipated then that yielding will be associated with the shear stresses on the element and is not likely to be influenced
by the average stress or pressure. It is appropriate to define these terms more precisely.
2.4.1 Maximum shear stress
On the faces of the principal element on the left-hand side of Figure 2.4, there are no shear stresses. On a face inclined at any other angle, both normal and shear stresses will
act. On faces of different orientation it is found that the shear stresses will locally reach a maximum for three particular directions; these are the maximum shear stress planes and
are illustrated in Figure 2.4. They are inclined at 45
◦
to the principal directions and the maximum shear stresses can be found from the Mohr circle of stress, Figure 2.5. Normal
stresses also act on these maximum shear stress planes, but these have not been shown in the diagram.
The three maximum shear stresses for the element are τ
1
= σ
1
− σ
2
2 ;
τ
2
= σ
2
− σ
3
2 ;
τ
3
= σ
3
− σ
1
2 2.8
From the discussion above, it might be anticipated that yielding would be dependent on the shear stresses in an element and the current value of the flow stress; i.e. that a yielding
condition might be expressed as
f τ
1
, τ
2
, τ
3
= σ
f
We explore this idea below. 18 Mechanics of Sheet Metal Forming
