1. Wire and rod drawing

2. Variables in Wire Drawing

3. Various methods of tube drawing

4. 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.

5. Wire Drawing/ Rod drawing
s = Ken
(Equation 6.10)
The drawing force then is
F = YAfIn (equation 6.62)

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

7. Die Pressure
Die pressure p at any diameter along the die can be conveniently obtained from
p = Yf – s
Yf = flow stress at given point
s = stress applied at that point
(variation in drawing stress and die con…)

8. ( forces acting on an element…..)

9. 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

10. Maximum reduction per pass
s = Y ln = Y, for perfectly plastic materials
thus ln = 1 or = e
= 0.63
(effect of reduction of cs) (effect of friction factor)

11. Derive the maximum reduction perpass
For 1- efficiency =0.9
2- Power law material
3- drawing stress with friction

12. Optimum die angle
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)
Effect of CS area on the optimum die angle in drawing of copper wire
Increases with reduction

13. Defects
Seams
Stresses
residual stresses in cold drawing

14. Residual stresses in cold drawn AISI 1045 carbon steel round
T : transverse
L : longitudinal
R : radial direction

15. Initial rod is shaped to a point Speed 30feet/min to 10000 feet/min
Die angle : º
Sizing pass
Reduction > 45%
Bad surface
Bad lubrication
Initial rod is shaped to a point
Speed 30feet/min to feet/min
Lubrication
Copper coating
Surface clean
Lubrication oil

16. Terminology Typical die for drawing round rod or wire
Typical wear pattern in wire drawing die

18. Problem Data: D0 = 10mm Df = 8mm Speed is 0.5 m/s
Friction/redundant force 40%
Find power and exit die pressure
Answer:
The true strain in this operation is
e1 = ln(102/82) = 0.446

19. From table 2.3 K = 1300 MPa and n = 0.3
= 1300(0.466)0.30/1.30 = 785 MPa
From eq ,
F = Afln
Af = p(0.0082)/4 = 5x10-5 m2
F = 785* 5x10-5 (0.446) = MN

20. Power = force x velocity
= (0.5) = MW
=8.75KW
Exit Die pressure
p = Yf –s
Yf being the flow stress of the material at exit
Yf = Ken1 = 1300(0.446)0.30 = 1020 MPa
s is the drawing stress sd = F/Af = 1.4(0.0175)/ = 490 MPa
Therefore the die pressure at the exit is
p= 1020 – 490
= 530MPa

21. Topics In Rolling

22. 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.

23. Where quantity H is defined as
At entry f = a; hence H =H0 with f replaced by a.
At exit f = 0 hence H = Hf =0

24. Determination of neutral point
We determine neutral point by simply by equating Entry and Exit
Substituting eq into eq we have

25. Simpler method to find roll force
L : arc of contact
we can approximate L as the projected area
Dh= hf - ho

26. Pressure distribution
Function of front and back tension
As tension increases the neutral point shifts and there is reduction in area under the curves

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

28. Reduce Pressure Roll force can be reduced by Smaller radii
Smaller reduction
Higher workpiece temperature
Lower friction
Front and back tension

29. 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

30. F is given by eq 6.37
L is given by eq. 6.36
Thus L = ((12)( ))1/2 = 1.55 in = 0.13 ft.
w = 9in
For 6061-O Al. K=30000 psi and n=0.2 (table 2.4)
e1 = ln(1.0/0.8) = 0.223

31. Thus from equation 6.10 ,
Y= (30000)(0.223)0.2/1.2 = psi
And =1.15(18500) = psi