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
Stress Evaluation for Wire Drawing (Cont..)
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
Stress Evaluation for Wire Drawing (Cont..)
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)
Stress Evaluation for Wire Drawing (Cont..)
=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
1 + ββββββββ
σσσσb
Yσσσσa
Y+1 - ----- (4)
Stress Evaluation for Wire Drawing (Cont..)
=2ββββDa
Db
1 + ββββββββ
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
Stress Evaluation for Wire Drawing (Cont..)
Limiting condition is given by σσσσmax = Y
Reduction in diameter RD =Da
Db= 1 -
Db - Da
Db----- (6)
From Equation (3)
=1 + ββββ
ββββ
1 + ββββββββ
σσσσb
Y
σσσσa
Y
-
-2ββββ
(1 - RD)
Stress Evaluation for Wire Drawing (Cont..)
= 1 + ββββββββ
σσσσa
Y +2ββββ
(1 - RD) 1 + ββββββββ
σσσσb
Y-
RD =1 + ββββ
ββββ
1 + ββββββββ
σσσσb
Y
σσσσa
Y
-
-1 -
2ββββ1
----- (7)
Stress Evaluation for Wire Drawing (Cont..)
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.
Stress Evaluation for Wire Drawing (Cont..)
Limiting Reduction in Wire Drawing
Let n =σσσσb
σσσσa< 1 and let σσσσa = Y
2ββββDa
Db=
2ββββDa
Db
1 + ββββββββ
σσσσb
Yσσσσa
Y+1 -
From Equation (4)
2ββββDa
Db∴∴∴∴ 1 =
2ββββDa
Db
1 + ββββββββ
σσσσb
Y+1 - ----- (8)
1 =2ββββDa
Db
1 + ββββββββ
+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!….
Limiting Reduction in Wire Drawing Cont..
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)
Limiting Reduction in Wire Drawing Cont..
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
Limiting Reduction in Wire Drawing Cont..
Limiting Reduction in Wire Drawing Cont..
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
Distribution of Die Pressure (Cont..)
If σσσσa = Y , then pa = 0
Hence pb = - 1ββββY 2ββββDb
Da
----- (13)
pb
2ββββDb
Da= - 1
ββββY pa+
2ββββDb
Da
----- (12)∴∴∴∴
Distribution of Die Pressure (Cont..)
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 -
Distribution of Die Pressure (Cont..)
Distribution of Die Pressure (Cont..)
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)
Drawing of Wide, Non-hardening, Strip through Wedge-shaped Dies (Cont..)
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&∴∴∴∴
Drawing of Wide, Non-hardening, Strip through Wedge-shaped Dies (Cont..)
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
Drawing of Wide, Non-hardening, Strip through Wedge-shaped Dies (Cont..)
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
1 + ββββββββ
σσσσa
s1 - ----- (4)∴∴∴∴
Drawing of Wide, Non-hardening, Strip through Wedge-shaped Dies (Cont..)
Wire Drawing -- SummaryFor Rectangular slab
----- (5)=ββββha
hb
1 + ββββββββ
σσσσa
s1 -
ββββha
hb
σσσσb
s+
=ββββha
hb
1 + ββββββββ
σσσσa
s1 - For Zero Back Pull
σσσσb = 0
2ββββDa
Db=
2ββββDa
Db
1 + ββββββββ
σσσσb
Yσσσσa
Y+1 -
For Round Bar
=2ββββDa
Db
1 + ββββββββ
σσσσa
Y1 - For Zero Back Pull
σσσσb = 0