metal forming-1.pdf

80 WORK BALANCE

6.4. The drawing limit corresponds to theintersection of the plots of drawing stress

with the stressstrain curve.

For all conditions (strain hardening rules and efficiencies), the maximum drawing

strain corresponds to d = / as illustrated by Figure 6.4.The conditions are different in the initial start-up. Undrawn material must be fed

through the die. If it is annealed, the maximum drawing force will correspond to the

tensile strength, or

d(max) = K nn (6.14)and the maximum strain is

= [(n + 1)nn]1/(n+1). (6.15)

More likely, however, the end of the wire to be fed through the die will be reduced

by swaging (see Section 6.6). In this case it will have been strain hardened at least as

much as if it had been drawn.

6.5 EFFECTS OF DIE ANGLE AND REDUCTION

Figure 6.5 shows schematically the dependence of the friction and redundant work

terms on die angle. For a given reduction, the contact area between the die and material

decreases with increasing die angle. The pressure the die exerts on the material in the

die gap is almost independent of the die angle, so the force between the die and work-

piece increases with greater area of contact at die angles. With a constant coefficient of

friction, wf increases as decreases. The redundant work term, wr, increases with die

angle. The ideal work term, wi, doesnt depend on the die angle. For each reduction,

there is an optimum die angle, , for which the work is a minimum.The efficiency and optimum die angle are functions of the reduction. In general

the efficiency increases with reduction. Figures 6.6 and 6.7 are adapted from the

measurements of Wistreich. The optimum die angle increases with reduction.

J. Wistreich, Metall. Rev., 3 (1958), pp. 97142.

6.6. SWAGING 81

wr

wi wfwa= + + wr

wi wf

Semi-die angle,

wa

, w

i ,w

r, w

f

6.5. Dependence of the various work terms on the die angle.

0.8

0.6

0.4

0.2

0 0.1 0.2 0.3 0.4 0.5

4

8

1216

Reduction, r

Def

orm

atio

n ef

ficie

ncy,

4

6.6. Wistreichs data show an increase of efficiency with increasing reduction.

In extrusion the same general effects of die angle and reduction on efficiency apply.

However, there is no theoretical limit to reduction, though excessive extrusion forces

may be beyond the capacity of the extruder. Therefore the reductions and die angles in

extrusion tend to be much larger than in drawing. Area reductions of 8:1 (r = 0.875)are not uncommon.

6.6 SWAGING

The diameters of wires and tubes may be reduced by rotary swaging. Shaped dies are

rotated around the wire or tube and hammer the work piece as they rotate. This causes

a swirled microstructure. In a wire drawing plant, swaging is used to reduce the ends

of wires so they can be fed through drawing dies in the initial setup. Figure 6.8 is a

schematic of the tooling used in swaging.

82 WORK BALANCE

0 2 4 6 8 10 12 14 16

0.2

0.4

0.6

0.8

0

Def

orm

atio

n ef

ficie

ncy,

Semi-die angle,, derees

0.05

0.10

0.20

0.25

0.30

0.35

locus of *

reduction

0.15

6.7. Replot of data in Figure 6.6 showing the increase of optimum die angle with reduction.

spindle spindle

diedie

wire wire

6.8. Tooling for swaging. As the spindle rotates it causes the die to hammer the work at different

angles (in this case 30).

REFERENCES

ASM Metals Handbook: Forging and Forming, vol. 14. 9th ed., ASM 1988.

W. A. Backofen, Deformation Processing, Addison-Wesley, 1972.

PROBLEMS

6.1. The diameter, D0, of a round rod can be reduced to D1 either by a tensile force

of F1 or by drawing through a die with a force, Fd, as sketched in Figure 6.9.

Assuming ideal work in drawing, compare F1 and Fd (or 1 and d) to achieve

the same reduction.

PROBLEMS 83

Fd F1

Do

D1

Do

D1

F1

1

1

= Kn

6.9. Sketch for Problem 6.1.

6.2. Calculate the maximum possible reduction, r, in wire drawing for a material

whose stressstrain curve is approximated by = 2000.18 MPa. Assume anefficiency of 65%.

6.3. An aluminum alloy billet is being hot extruded from 20-cm diameter to 5-cm

diameter as sketched in Figure 6.10. The flow stress at the extrusion temperature

is 40 MPa. Assume = 0.5.a) What extrusion pressure is required?

b) Calculate the lateral pressure on the die walls.

die

die

billetPext 5 cm20 cm

6.10. Aluminum billet being extruded.

6.4. An unsupported extrusion process (Figure 6.11) has been proposed to reduce

the diameter of a bar from D0 to D1. The material does not strain harden. What

is the largest reduction, 1D/D0, that can be made without the material yielding

before it enters the die? Neglect the possibility of buckling and assume =60%.

84 WORK BALANCE

6.11. Unsupported extrusion.

6.5. A sheet, 1 m wide and 8 mm thick, is to be rolled to a thickness of 6 mm in

a single pass. The strain-hardening expression for the material is = 2000.18MPa. A deformation efficiency of 80% can be assumed. The von Mises yield

criterion is applicable. The exit speed from the rolls is 5 m/s. Calculate the

power required.

6.6. The strains in a material for which = 3500.20 MPa are 1 = 0.200 and2 = 0.125. Calculate the work per volume.

6.7. You are asked to plan a wire-drawing schedule to reduce copper wire from 1

mm to 0.4 mm in diameter. How many wire drawing passes would be required

to be sure of no failures, if the drawing stress never exceeds 80% of the flow

stress, and the efficiency is assumed to be 60%?

6.8. Derive an expression for at the initiation of drawing when the outlet diameteris produced by machining.

6.9. For a material with a stressstrain relation = A + B, find the maximumstrain per wire drawing pass if = 0.75.

7 Slab Analysis and Friction

Slab analyses are based on making a force balance on a differentially thick slab of

material. It is useful in estimating the role of friction on forces required in drawing,

extrusion, and rolling. The important assumptions are these:

1. The principal axes are in the directions of the applied loads.

2. The effects of friction do not alter the directions of the principal axes or cause

internal distortion. The deformation is homogeneous so plane sections remain

plane.

7.1 SHEET DRAWING

Figure 7.1 illustrates sheet or strip drawing. A force F pulls a strip through a pair of

wedges. The strip width w is much greater than the thickness t so the width doesnt

change and plane strain prevails. The drawing stress is d = F/wte.Figure 7.2 shows the differential slab element. Pressure normal to the die, P, acts

on two areas of w dx/cos , and the corresponding component of force in the x-direction

is 2Pw dx(sin /cos ). The normal pressure also causes a frictional force on the work

metal, 2wP dx/cos , with a horizontal component 2wP dx(cos /cos ), where

is the coefficient of friction. Both of these forces act in the positive x-direction. The

drawing force, x wt , acts in the negative x-direction, while the force in the positive

x-direction is (x + dx )(t + dt)w. An x-direction force balance gives(x + dx )(t + dt)w + 2w(tan )P dx + 2Pw dx = x wt. (7.1)

Simplifying and neglecting second order differentials,

x t + x dt + tdx + 2(tan )Pdx + 2Pdx = x t. (7.2)Substituting 2dx = dt/tan and B = cot , P dt + x dt + t dx+(cot )P dt = 0,or

tdx + [x + P(1 + B)]dt = 0, or (7.3)dx

x + P(1 + B)= dt

t. (7.4)

85

86 SLAB ANALYSIS AND FRICTION

!

"

!

#

$

%

&

'

()*+

,%

7.1. Plane-strain drawing of a sheet.

Pds

ds

dx

PcosdsPds

x x+dx

P

PP

Pdt/2

dx

t+dtt

7.2. Slab used for force balance in sheet drawing.

To integrate equation 7.4, the relation between P and x must be found and

substituted. The fact that P does not act in a direction normal to x will be ignored

and P will be taken as a principal stress, P = x . For yielding in plane strain,x y = 2k, where k is the yield strength in shear. For Tresca 2k = Y and for vonMises 2k = (2/

3)Y . Substituting P = 2k x into equation 7.4,

dx

Bx 2k(1 + B)= dt

t. (7.5)

Integrating between x = 0 and x = d and between t = t0 and te, and solvingfor d,

d

2k= (1 + B)

B

[1

(te

t0

)B]. (7.6)

7.2. WIRE AND ROD DRAWING 87

Finally substituting h = ln(t0/te) as the homogeneous strains,

d

2k= (1 + B)

B[1 exp(Bh)]. (7.7)

Several assumptions are involved in the derivation of equation 7.7. First, it is

assumed that 2k is constant, which is not true if the material work hardens. The effect

of work hardening can be approximated by using an average value of 2k. Another is

that the friction coefficient is constant. The assumption that P is a principal stress is

reasonable for low die angles and low coefficients of friction, but the assumption gets

progressively worse at high die angles and high friction coefficients.

EXAMPLE 7.1: A 2.5 mm thick metal sheet 25 cm wide is drawn to a thickness of 2.25

mm through a die of included angle 30. The flow stress is 200 MPa and the frictioncoefficient is 0.08. Calculate the drawing force using the von Mises criterion.

SOLUTION: Use equation 7.7 and take B = cot = 0.08 cot(15) = 0.299 andh = ln(2.5/2.25) = 0.1054. From equation 7.7, d = 1.15(200)(1.299/0.299)[1 exp(0.299 0.1054)] = 31 MPa. Fd = 31 MPa (0.25m)(0.0225) = 175 kN.

Note that if = 0 (i.e., B = 0) were substituted into equation 7.7, with LHospitalsrule it can be shown that

d

2k= h, (7.8)

which is exactly what would be predicted by the work balance.

7.2 WIRE AND ROD DRAWING

Sachs analyzed wire or rod drawing by an analogous procedure. The basic differentialequation is

d

B (1 + B) = 2dD

D, (7.9)

where D is the diameter of the rod or wire. Realizing that 2dD/D = d, d0

d

B (1 + B) = h

0

d. (7.10)

Integrating,

d = a(

1 + BB

)[1 exp(Bh)], (7.11)

where a is the average flow stress of the material in the die. This analysis neglects

redundant strain, has the same limitations as the plane-strain drawing in Section 7.1,

and becomes unrealistic at high die angles and low reductions.

G. Sachs, Z. Angew. Math. Mech., 7 (1927), p. 235.


7.3. Essentials for a slab analysis.

7.3 FRICTION IN PLANE-STRAIN COMPRESSION

The slab analysis can also be used for plane-strain compression. Figure 7.3 shows

a specimen with h < b being compressed with a constant coefficient of friction, .

Making a force balance on the differential slab, x h + 2P dx (x + dx )h = 0.This simplifies to

2P dx = hdx . (7.12)Again take y = P and x as principal stresses and realize that for plane strain,x y = 2k, so dx = dP . Equation 7.12 becomes 2P dx = h dP or

dP

P= 2 dx

h. (7.13)

Integrating from P = 2k at x = 0 to P at x, ln(P/2k) = 2x/h orP

2k= exp

(2x

h

). (7.14)

This predicts a friction hill, which is illustrated in Figure 7.4.

This is valid from the edge (x = 0) to the centerline (x = b/2). The average pressure,Pa, can be found by integrating P over half of the block to find the force and dividing

that by the area of the half-block.

Fy = b/2

0

P dx = b/2

0

2k exp

(2x

h

)dx = 2k

(h

2

)[exp

(b

h

) 1

], (7.15)

Pav

2k=

(h

b

)[exp

(b

h

) 1

]. (7.16)

A simple approximate solution can be found by expanding exp(b/h) 1 = (b/h) +(b/h)2/2 + . For small values of (b/h),

Pav

2k 1 + b

2h. (7.17)

7.3. FRICTION IN PLANE-STRAIN COMPRESSION 89

7.4. Friction hill in plane-strain compression with a constant coefficient of friction.

EXAMPLE 7.2: Plane-strain compression is conducted on a slab of metal 20 cm wide

and 2.5 cm high, with a yield strength in shear of k = 100 MPa. Assuming a coefficientof friction of = 0.10,

(a) Estimate the maximum pressure at the onset of plastic flow;

(b) Estimate the average pressure at the onset of plastic flow.

SOLUTION:

(a) From equation 7.14, at x = b/2, Pmax = 2k exp(b/h) = 200 exp(0.1 20/2.5) = 45 MPa

(b) Using the exact solution (equation 7.16), Pav = 200[2.5/(0.1)(20)]{exp[(0.10)(0.20)/2.5 1]} = 253 MPa. The approximate solution (equation 7.17)gives

Pav = 200[

1 + (0.1)(20)2(2.5)

]= 280 MPa.


1.0

2.0

3.0

P/2k Pav/2k

P/2k

b/h = 4

x/h

b h

7.5. Friction hill in plain-strain compression

with sticking friction.

7.4 STICKING FRICTION

There is an upper limit to the shear stress on the interface. It cannot exceed 2k. This

limits equations 7.16 and 7.17 to

x h2

ln(2) andb

h ln(2)/. (7.18)

Otherwise the tool and work piece will stick at the interface, and the work piece will be

sheared. If sticking occurs, the shear stress will be k instead of P and equation 7.14

will become

P/2k = 1 + x/h. (7.19)The pressure distribution is shown in Figure 7.5.

The average pressure is given by

Pav

2k= 1 + b

4h. (7.20)

EXAMPLE 7.3: Repeat Example 7.2 for sticking friction.

SOLUTION:

(a) Using equation 7.19, Pmax = 200(1 + 20/5) = 1,000 MPa.(b) Using equation 7.20, Pav = 200(1 + 20/10) = 600 MPa.

7.5 MIXED STICKINGSLIDING CONDITIONS

If b/h ln(2)/, sticking is predicted at the center and sliding at the edges.Equation 7.14 predicts P for x x where x = h ln(2)/(2). From x to the

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