Wire and rod drawing

Variables in Wire Drawing

Various methods of tube drawing

Variation in strain and flow stress

Variation in strain and flow stress in the deformation zone in drawing.

•Note that the strain increases rapidly toward the exit.

•When the exit diameter reaches zero, the strain reaches infinity.

•Ywire represents the yield stress of the wire.

Wire Drawing/ Rod drawing

• = Kn

• (Equation 6.10)

• The drawing force then is

F = YAfIn (equation 6.62)

11

n

kY

n

Af

Ao

Stresses acting on an element in drawing of a cylindrical rod or wire

Die Pressure

• Die pressure p at any diameter along the die can be conveniently obtained from

p = Yf –

• Yf = flow stress at given point

• = stress applied at that point

• (variation in drawing stress and die con…)

• ( forces acting on an element…..)

• s =

cot

1tan

1Ao

AfY

Variation in the drawing stress and die contact pressure

•Along the deformation zone

•As the drawing stress increases the die pressure decreases

•Refer to yield criteria

Maximum reduction per pass

• =Y ln = Y, for perfectly plastic materials

• thus ln = 1 or = e

• = 0.63• (effect of reduction of cs) (effect of friction

factor)

Af

Ao

Af

Ao

Af

Ao

eAo

AfAo 11

• Derive the maximum reduction perpass

• For 1- efficiency =0.9

• 2- Power law material

• 3- drawing stress with friction

Optimum die angle

Effect of CS area on the optimum die angle in drawing of copper wire

•Increases with reduction

Effect of friction factor m and die angle on maximum possible reduction

•m = 1 indicates complete sticking

•Maximum possible reduction is 63% (Eq. 6.72)

Defects

• Seams

• Stresses

• residual stresses in cold drawing

Residual stresses in cold drawn AISI 1045 carbon steel round

•T : transverse

•L : longitudinal

•R : radial direction

• Die angle : 6 - 15º• Sizing pass• Reduction > 45%

– Bad surface– Bad lubrication

• Initial rod is shaped to a point

• Speed 30feet/min to 10000 feet/min

• Lubrication – Copper coating– Surface clean– Lubrication oil

Terminology

Typical die for drawing round rod or wire

Typical wear pattern in wire drawing die

Problem

Data:

D0 = 10mm

Df = 8mmSpeed is 0.5 m/sFriction/redundant force 40%Find power and exit die pressureAnswer:The true strain in this operation is

1 = ln(102/82) = 0.446

• From table 2.3 K = 1300 MPa and n = 0.3

= 1300(0.466)0.30/1.30 = 785 MPa

From eq. 6.62 ,

F = Afln

Af = (0.0082)/4 = 5x10-5 m2 F = 785* 5x10-5 (0.446) = 0.0175 MN

11

n

kY

n

Af

Ao

'_

Y

• Power = force x velocity = 0.0175(0.5) = 0.00875 MW

=8.75KWExit Die pressure p = Yf –Yf being the flow stress of the material at exitYf = Kn

1 = 1300(0.446)0.30 = 1020 MPa is the drawing stress d = F/Af = 1.4(0.0175)/0.00005 =

490 MPa• Therefore the die pressure at the exit is

p= 1020 – 490 = 530MPa

Topics In Rolling

Relative velocity distribution between roll and strip surfaces

•Note the difference in the direction of frictional forces.

•The arrows represent the frictional forces acting on the strip.

Vr

VrVfSlipForward

_

)cos1(2 Rhh f

)(

0

0': HHf eh

hYpEntry

Hf eh

hYpExit

0':

Where quantity H is defined as

ff h

R

h

RH 1tan2

At entry = ; hence H =H0 with f replaced by .

At exit = 0 hence H = Hf =0

Determination of neutral point

• We determine neutral point by simply by equating Entry and Exit

)2(2

0 0

0

n

n

HHH

H

fe

e

e

h

h

fn

h

hHH

00 ln1

2

1

2tan

nff H

R

h

R

hn

Substituting eq. 6.30 into eq. 6.29 we have

Simpler method to find roll force

L : arc of contact

• we can approximate L as the projected area

h= hf - ho

avLwpF

hRL

Pressure distribution

•Function of front and back tension

•As tension increases the neutral point shifts and there is reduction in area under the curves

Roll forces: the area under pressure contact length curves multiplied by the strip width

H

ff e

h

hfYp )( '

)0(' HH

obf eh

hYp

n

n

wpRdwpRdF0

Reduce Pressure

Roll force can be reduced by

• Smaller radii

• Smaller reduction

• Higher workpiece temperature

• Lower friction

• Front and back tension

Calculate power required in rolling

Q: A 9 in. wide 6061-O aluminum strip is rolled from a thickness of 1.0 in. to 0.80 in. if the roll radius is 12 in. and the roll rpm is 100, estimate the horse power required for this operation

A:

The power needed for a set of two rolls is given by eq. 6.42 hp

FLNPower

33000

2

• F is given by eq 6.37

• L is given by eq. 6.36

• Thus L = ((12)(1.0-0.8))1/2 = 1.55 in = 0.13 ft.

• w = 9in

• For 6061-O Al. K=30000 psi and n=0.2 (table 2.4)

• 1 = ln(1.0/0.8) = 0.223

'_

YLwF

hRL

• Thus from equation 6.10 ,

Y= (30000)(0.223)0.2/1.2 = 18500 psi

And =1.15(18500) = 21275 psi'_

Y

11

n

kY

n