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ME5603 Metal Forming TechnologyWIRE DRAWING

By: A/P Lee Kim Seng

Wire DrawingThe main function of wire drawing is to produce wire of specific size with good surface finish.

In industrial practice, the die has a trumpet-shaped bore, but since the curvature near the working surface is small, this may be simplified into a conical portionwhich serves to deform the wire and a cylindrical portion which is mean to preserve the size of the bore in the face of wear.

Total die angle vary between 5 to 25°°°°.

Wire Drawing

Wire Drawing Dies

Movie by:North American Wire Products Corp

Wire Drawing Process

Examples of Drawn Wires

Applications of Drawn Wires

Stress Evaluation for Wire DrawingAssumptions:-

1. State of stress is constant at planes where x = constant

2. x direction is a principal direction. I.e σσσσx is a principal stress.

3. It follows from 1 and 2 that planes at right angle to x direction are also principal planes and we will assume that pressure between die and wire is a principal stress. (applicable only when αααα is small)

4. Friction obeys Coulomb’s law.

5. Isotropic, homogeneous, rigid non-strain hardening material.

Stress Evaluation for Wire Drawing (Cont..)

Consider the equilibrium of a small element in the working zone.

(σσσσx + dσσσσx) (D + dD)2 - σσσσx D2ππππ4

ππππ4 p (ππππD ) sin αααα

dD2 sin αααα

+

µµµµp (ππππD ) cos ααααdD

2 sin αααα+ = 0

Ddσσσσx + 2σσσσxdD + 2pdD (1 + ) = 0µµµµ

tan αααα

(σσσσx + dσσσσx) (D + 2 dD) - σσσσxππππD4

ππππD2

4

+ + = 0pππππDdD2

dD2 tan αααα

µµµµpππππD


Ddσσσσx + 2 [σσσσx + p(1 + µµµµ cot αααα)] dD = 0 ----- (1)

Assume Tresca Yield Criteria σσσσx ≥≥≥≥ σσσσy ≥≥≥≥ σσσσz

σσσσ1 - σσσσ3 = YIn this case σσσσx + p = Y

Let ββββ = µµµµ cot αααα Equation (1) becomes

Ddσσσσx + 2 [ σσσσx + (Y - σσσσx)(1 + ββββ) ] dD = 0

Ddσσσσx + 2 [ Y (1 + ββββ) - βσβσβσβσx ] dD = 0


2 dDD βσβσβσβσx - Y (1 + ββββ)

dσσσσx=⌠⌠⌠⌠⌡⌡⌡⌡Db

Da ⌠⌠⌠⌠⌡⌡⌡⌡σσσσb

σσσσa

Upon Integration

βσβσβσβσb - Y (1 + ββββ)βσβσβσβσa - Y (1 + ββββ)1

ββββln ln

Da

Db2 =

=2 dD

Dβσβσβσβσx - Y (1 + ββββ)dσσσσx ----- (2)


=βσβσβσβσb - Y (1 + ββββ)βσβσβσβσa - Y (1 + ββββ)2ββββDa

Db

=2ββββDa

Db 1 + ββββββββ

1 + ββββββββ

σσσσb

Y

σσσσa

Y

-

------ (3)

2ββββDa

Db=

2ββββDa

Db


σσσσb

Yσσσσa

Y+1 - ----- (4)


=2ββββDa

Db


σσσσa

Y1 - ----- (5)

For zero back pull I.e σσσσb = 0 then

2Da

DbIf RA = 1 - ; RA = reduction in area

= ββββ1 + ββββββββ

σσσσa

Y1 - (1 - RA)Then


Limiting condition is given by σσσσmax = Y

Reduction in diameter RD =Da

Db= 1 -

Db - Da

Db----- (6)

From Equation (3)

=1 + ββββ

ββββ


σσσσb

Y

σσσσa

Y

-

-2ββββ

(1 - RD)


= 1 + ββββββββ

σσσσa

Y +2ββββ

(1 - RD) 1 + ββββββββ

σσσσb

Y-

RD =1 + ββββ

ββββ


σσσσb

Y

σσσσa

Y

-

-1 -

2ββββ1

----- (7)


An increase in σσσσb (back pull) will decrease reduction, though not proportionally.

Its main advantage can be seen from equation (σσσσx + p = Y) where the die pressure p = Y - σσσσx . Thus at exit, as σσσσa →→→→ Y, p →→→→ 0

But with no back pull, at entry, p = Y with the obvious disadvantage in die wear and in restricting inflow of lubricant.

If the back pull is increased to Y/2, the die pressure will be Y - Y/2 = Y/2 .

The situation will also benefit from a reduction in µµµµ.


Limiting Reduction in Wire Drawing

Let n =σσσσb

σσσσa< 1 and let σσσσa = Y

2ββββDa

Db=

2ββββDa

Db


σσσσb

Yσσσσa

Y+1 -

From Equation (4)

2ββββDa

Db∴∴∴∴ 1 =

2ββββDa

Db


σσσσb

Y+1 - ----- (8)

1 =2ββββDa

Db


+1 + ββββ

ββββn -

ββββ1-

nββββ - 1 - ββββββββ

2ββββDa

Db=

2ββββDa

Db=

1

1 + ββββ (1 - n)

Limiting Reduction in Wire Drawing Cont..

Da

Db∴∴∴∴ 1 + ββββ (1 - n) 2ββββ

1-=

Da

Db= 1 - ½ (1 - n) -

2ββββ1

2ββββ1

- - 1 ββββ2 (1 - n)2

2!

2ββββ1

2ββββ1

- - 12ββββ1

- - 2- ββββ3 (1 - n)3

3!….


Let ββββ →→→→ 0

= 1 - ½ (1 - n) +41 (1 - n)2

2! 81- (1 - n)3

3!+ ….Da

Db limit

= exp[ - ½ (1 - n)]

Da

Db limit= exp[ - ½ (1 - n)] ----- (9)


2.086.563.2n = -1

1.063.239.3n = 0

0.2522.111.75n = ¾

0.7552.731.3n = ¼

0.539.322.1n = ½

EntryRA %RD %n D1

D2 limit= exp[ ½ (n - 1) ]

pY

14

exp - = 0.779

exp - 1 = 0.367

38

exp - = 0.687

18

exp - = 0.882

12

exp - = 0.607

NO Back Pull


Distribution of Die Pressureσσσσx + p = YFrom above

σσσσx - Y = p

dσσσσx = - dp

Substitute dσσσσx = - dp into Equation (1)

Ddσσσσx + 2 [σσσσx + p(1 + ββββ)] dD = 0 ββββ = µµµµ cot αααα

- Ddp + 2 [(Y - p) + p(1 + ββββ)] dD = 0

dp =2dD

D[ Y + ββββp ] ----- (10)

Distribution of Die Pressure (Cont..)Upon Integration

=Y + ββββp

dp⌠⌠⌠⌠⌡⌡⌡⌡pb

pa dDD

⌠⌠⌠⌠⌡⌡⌡⌡Db

Da

2

lnDa

Db2=

Y + ββββpb

Y + ββββpa1ββββ

ln

2ββββDa

Db=

Y + ββββpb

Y + ββββpa∴∴∴∴ ----- (11)

Distribution of Die Pressure (Cont..)

2ββββDa

Db=

1 +Yββββ pa

Yββββ pb1 +

2ββββDb

Da=

1 +Yββββ pb

Yββββ pa1 +

2ββββDb

Da=1 +

Yββββ pb

Yββββ pa1 +

2ββββDb

Da= - 1

Yββββ pb

Yββββ pa1 +

2ββββDb

Da= - 1

ββββYpb

Yββββ pa1 +

2ββββDb

Da= - 1

ββββY

Yββββ pa+

2ββββDb

Da


If σσσσa = Y , then pa = 0

Hence pb = - 1ββββY 2ββββDb

Da

----- (13)

pb

2ββββDb

Da= - 1

ββββY pa+

2ββββDb

Da

----- (12)∴∴∴∴


43.5%24.8%0.751.332

49%28.5%0.711.41

55.6%33.3%0.671.5½

63.2%39.3%0.6061.650

RARDββββ Da

Db

Db

Da

Da

DbRD = 1 -

Da

Db

2

RA = 1 -


Drawing of Wide, Non-hardening, Strip through Wedge-shaped Dies

Drawing of Wide, Non-hardening, Strip through Wedge-shaped Dies (Cont..)

Consider the force equilibrium of an element dx

+ 2µµµµp(wdh

2 sin αααα) cos αααα = 0

(σσσσx + dσσσσx)(h + dh)w - σσσσxhw + 2p(w dh

2 sin αααα) sin αααα

σσσσx dh + h dσσσσx + p dh + µµµµp dh cot αααα = 0

h dσσσσx + [σσσσx + p(1 + µµµµ cot αααα)] dh = 0 ----- (1)

Let ββββ = µµµµ cot αααα

h dσσσσx + [σσσσx + p(1 + ββββ)] dh = 0 ----- (2)∴∴∴∴

Consider the equilibrium of the forces perpendicular to the direction of drawing.

σσσσy dx = - pdx

cos ααααdx

cos ααααcos αααα + µµµµp sin αααα

= - p dx + µµµµp tan αααα dx

= - p (1 - µµµµ tan αααα) dxσσσσy dx ----- (3)


Assume µµµµ tan αααα << 1 , then σσσσy ≈≈≈≈ - p

For µµµµ = 0.05, αααα = 10°°°°, then µµµµ tan αααα = 0.009 << 1

∴∴∴∴ σσσσy = σσσσ3 = - p

∴∴∴∴ σσσσ1 = σσσσx ; σσσσ3 = - p

Under plane strain conditions.

σσσσ1 - σσσσ3 = 2k = s 2√√√√ 3

Y=∴∴∴∴ -- Von Mises

σσσσx + p = s p = s - σσσσx&∴∴∴∴


Substitute in Equation (2)

h dσσσσx + [σσσσx + (s - σσσσx)(1 + ββββ)] dh = 0

=σσσσx + (s - σσσσx)(1 + ββββ)

dσσσσx dhh

-

=βσβσβσβσx - s (1 + ββββ)

dσσσσx dhh


Upon Integration

=⌠⌠⌠⌠⌡⌡⌡⌡σσσσb=0

σσσσa

βσβσβσβσx - s (1 + ββββ)dσσσσx ⌠⌠⌠⌠

⌡⌡⌡⌡hb

ha dhh

=- s (1 + ββββ)

βσβσβσβσa - s (1 + ββββ) ββββha

hb

- s (1 + ββββ)βσβσβσβσa =ββββha

hbs (1 + ββββ)

=ββββha

hb


σσσσa

s1 - ----- (4)∴∴∴∴


Wire Drawing -- SummaryFor Rectangular slab

----- (5)=ββββha

hb


σσσσa

s1 -

ββββha

hb

σσσσb

s+

=ββββha

hb


σσσσa

s1 - For Zero Back Pull

σσσσb = 0

2ββββDa

Db=

2ββββDa

Db


σσσσb

Yσσσσa

Y+1 -

For Round Bar

=2ββββDa

Db


σσσσa

Y1 - For Zero Back Pull

σσσσb = 0

me5603 metal forming technology wire drawingcourses.nus.edu.sg/course/mpeleeks/personal/ppt/me...

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