Upper-Bound Analysis
Calculation of exact forces to cause plastic deformation in metal forming processes isoften difficult. Exact solutions must be both statically and kinematically admissible.That means theymust be geometrically self-consistent as well as satisfying the requiredstress equilibrium everywhere in the deforming body. Frequently it is simpler to uselimit theorems that allow one to make analyses that result in calculated forces that areknown to be either correct or higher or lower than the exact solution.
Lower bounds are based on satisfying stress equilibrium, while ignoring geometricself-consistency. They give forces that are known to be either too lowor correct. As such, they can assure that a structure is “safe.” Conditions in which η = 0 are lower bounds.Upper-bound analyses, on the other hand, predict stress or forces that are known tobe too large. These are usually more important in metal forming. Upper bounds arebased on satisfying yield criteria and geometric self-consistency. No attention is paidto satisfy equilibrium.
1. UPPER BOUNDS
The upper-bound theorem states that any estimate of the forces to deform a body madeby equating the rate of internal energy dissipation to the external forces will equal orbe greater than the correct force. The method of analysis is to
1. Assume an internal flow field that will produce the shape change;
2. Calculate the rate at which energy is consumed by this flow field; and
3. Calculate the external force by equating the rate of external work with the rate ofinternal energy consumption.
The flow field can be checked for consistency with a velocity vector diagram orhodograph. In the analysis, the following simplifying assumptions are usually made:
1. The material is homogeneous and isotropic.
2. There is no strain hardening.
3. Interfaces are either frictionless or sticking friction prevails.
4. Usually only two-dimensional (plane-strain) cases are considered with deformationoccurring by shear on a few discrete planes. Everywhere else the material is rigid.
2. ENERGY DISSIPATION ON PLANE OF SHEAR
Figure 1 (a) shows an element of rigid material, ABCD, moving at a velocity V1 atan angle θ1 to the horizontal. AD is parallel to yy′. When it passes through yy′it isforced to change direction and adopt a new velocity V2 at an angle θ2 to the horizontal.It is sheared into a new shape A′B′C′D′. The corresponding hodograph is shown inFigure 1 (b). The absolute velocities V1 and V2 are drawn from the origin, O. Becausethis is a steady-state process, they both have the same horizontal component, Vx. Thevector is the difference between V1 and V2 and must be parallel to the line ofshear, yy′.
Fig.1: (a) Drawing for calculating energy dissipation on a velocity discontinuity and (b) the correspondinghodograph.
The rate of energy dissipation along the discontinuity equals the volume of materialcrossing the discontinuity per time, SVxtimes the work per volume. The workper volume is w = k dy/dx = k/Vx, so the rate of energy dissipation along thediscontinuity is
dW/dt = (kV/Vx)SVx= kS. ………………………..…………….(1)
For deformation fields with more than one shear discontinuity,
dW/dt =I. …………………………………………………….(2)
3. PLANE-STRAIN FRICTIONLESS EXTRUSION
Consider the plane-strain extrusion through frictionless dies as illustrated in Figure2(a). Only half of the field is shown. There are two planes of discontinuity, AB andBC. The corresponding hodograph, Figure 2(b), is constructed by drawing horizontalvectors representing the initial velocity Vo and the exit velocity Ve.
Fig. 2: (a) Top half of an upper-bound field for plane-strain extrusion and (b) the corresponding hodograph.
Both start at theorigin. The velocity in triangle ABC, V1, is drawn parallel to AC; it also starts at theorigin. The velocity discontinuity, AB, is the difference between Vo and V1, and BCis the difference between V1 and Ve.
The rate of internal work is
dw/dt = k(+ ). …………………………………..……….(3)
The rate of external work is dw/dt = Peh0V0. Equating and solving for Pe/2k,
Pe/2k = (+ ). ………………………………………(4)
Equation 4 may be evaluated physically by measuring and relative to Vo onthe hodograph and measuring and relative to Vo on the physical field. Howeverit is easier to evaluate Pe/2k analytically.
ho/= sin θ, h0/= sin ψ.
For a 50% reduction with a half die angle of 30◦, with the law of sines,
/ sin 30o = Vo/ sin(θ − 30o) or /Vo= sin 30o/ sin(θ − 30o) and
/ sin θ = / sinψ so /V0 = (sin θ/ sinψ)/Vo.
The magnitude of Pext/2k depends on θ. If θ = 90o, /Vo= sin 30o/sin(90o– 30o)= 0.577, and ψ = 30o. /Vo= (sin 90o/sin 30o)0.577 = 1.154, = = ho.Therefore Pext/2k = (0.577 + 1.154)/2 = 0.866.
Figure 3shows the calculated variation of Pe/2k with θ. The lowest value ofPe/2k ≈ 0.78 occurs when θ ≈ 72o. A lower bound can be found as Pe==2k ln(2) so Pe/2k = 0.693. The true solution of Pe/2k = 0.762 liesbetween these.
Fig. 3: Variation of calculated extrusion pressurewith the angle θ in the upper-boundfield of Figure 2.
Ex.
Figure belowshows an upper-bound field for a plane-strain extrusion.
Change its thickness from 25 mm to 8 mm, sime angle 45o, friction factor 0.2 and yield stress 150 MPa. Estimate the required extrusion force.
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