Upper-Bound Analysis
Calculation of exact forces to cause plastic deformation in metal forming
processes is often difficult. Exact solutions must be both statically and
kinematically admissible. That means they must be geometrically selfconsistent as well as satisfying the required stress equilibrium everywhere
in the deforming body. Frequently it is simpler to use limit theorems that
allow one to make analyses that result in calculated forces that are known
to be either correct or higher or lower than the exact solution.
Lower bounds are based on satisfying stress equilibrium, while
ignoring geometric self-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 to be
too large. These are usually more important in metal forming. Upper
bounds are based on satisfying yield criteria and geometric selfconsistency. No attention is paid to satisfy equilibrium.
1. UPPER BOUNDS
The upper-bound theorem states that any estimate of the forces to deform
a body made by equating the rate of internal energy dissipation to the
external forces will equal or be 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 of internal energy consumption.
1
The flow field can be checked for consistency with a velocity vector
diagram or hodograph. 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
deformation occurring 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 at an angle θ1 to the horizontal. AD is parallel to yy′. When it
passes through yy′ it is forced 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 in Figure 1 (b). The
absolute velocities V1 and V2 are drawn from the origin, O. Because this is
a steady-state process, they both have the same horizontal component, Vx.
∗
The vector 𝑉12
is the difference between V1 and V2 and must be parallel to
the line of shear, yy′.
Fig.1: (a) Drawing for calculating energy dissipation on a velocity
discontinuity and (b) the corresponding hodograph.
2
The rate of energy dissipation along the discontinuity equals the
volume of material crossing the discontinuity per time, SVx times the
∗
work per volume. The work per volume is w = k dy/dx = k𝑉12
/Vx, so the
rate of energy dissipation along the discontinuity is
∗
∗
dW/dt = (kV𝑉12
/Vx )SVx = kS 𝑉12
. ………………………..…………….(1)
For deformation fields with more than one shear discontinuity,
dW/dt = ∑i 𝑘𝑆𝑖𝑉𝑖∗ I . …………………………………………………….(2)
3. PLANE-STRAIN FRICTIONLESS EXTRUSION
Consider the plane-strain extrusion through frictionless dies as illustrated
in Figure 2(a). Only half of the field is shown. There are two planes of
discontinuity, AB and BC. The corresponding hodograph, Figure 2(b), is
constructed by drawing horizontal vectors 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.
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Both start at the origin. The velocity in triangle ABC, V1, is drawn
∗
parallel to AC; it also starts at the origin. The velocity discontinuity, 𝑉𝐴𝐵
∗
AB, is the difference between Vo and V1, and 𝑉𝐵𝐶
BC is the difference
between V1 and Ve.
The rate of internal work is
∗ ̅̅̅̅
∗ ̅̅̅̅
dw/dt = k(𝑉𝐴𝐵
𝐴𝐵 + 𝑉𝐵𝐶
𝐵𝐶 ). …………………………………..……….(3)
The rate of external work is dw/dt = Peh0V0. Equating and solving
for Pe/2k,
Pe/2k =
1
2ℎ° 𝑉°
∗ ̅̅̅̅
∗ ̅̅̅̅
(𝑉𝐴𝐵
𝐴𝐵 + 𝑉𝐵𝐶
𝐵𝐶 ). ………………………………………(4)
∗
∗
Equation 4 may be evaluated physically by measuring 𝑉𝐴𝐵
and 𝑉𝐵𝐶
relative to Vo on the hodograph and measuring ̅̅̅̅
𝐴𝐵 and ̅̅̅̅
𝐵𝐶 relative to Vo
on the physical field. However it is easier to evaluate Pe/2k analytically.
̅̅̅̅ = sin θ, h0/𝐵𝐶
̅̅̅̅ = sin ψ.
ho/𝐴𝐵
For a 50% reduction with a half die angle of 30◦, with the law of sines,
∗
∗
𝑉𝐴𝐵
/ sin 30o = Vo/ sin(θ − 30o) or 𝑉𝐴𝐵
/Vo = sin 30o / sin(θ − 30o) and
∗
∗
∗
∗
𝑉𝐵𝐶
/ sin θ = 𝑉𝐴𝐵
/ sinψ so 𝑉𝐵𝐶
/V0 = (sin θ/ sinψ) 𝑉𝐴𝐵
/Vo.
∗
The magnitude of Pext /2k depends on θ. If θ = 90o, 𝑉𝐴𝐵
/Vo= sin
∗
30o/sin(90o – 30o) = 0.577, and ψ = 30o. 𝑉𝐵𝐶
/Vo = (sin 90o/sin 30o)0.577 =
1.154, ̅̅̅̅
𝐴𝐵 = ̅̅̅̅
𝐵𝐶 = ho. Therefore Pext/2k = (0.577 + 1.154)/2 = 0.866.
Figure 3 shows the calculated variation of Pe/2k with θ. The lowest
value of Pe/2k ≈ 0.78 occurs when θ ≈ 72o. A lower bound can be found
as Pe =∫ 𝜎d𝜀 =2k ln(2) so Pe/2k = 0.693. The true solution of Pe/2k =
0.762 lies between these.
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Fig. 3: Variation of calculated extrusion pressure with the angle θ in the
upper-bound field of Figure 2.
Ex.
Figure below shows an upper-bound field for a plane-strain extrusion.
Change its thickness from 25 mm to 8 mm, sime angle 45 o, friction factor
0.2 and yield stress 150 MPa. Estimate the required extrusion force.
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