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Wire and rod drawing

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.

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Wire and rod drawing

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  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…..) • s =

  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. 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

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

  17. 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

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

  19. Power = force x velocity = 0.0175(0.5) = 0.00875 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)/0.00005 = 490 MPa • Therefore the die pressure at the exit is p= 1020 – 490 = 530MPa

  20. Topics In Rolling

  21. 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.

  22. 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

  23. Determination of neutral point • We determine neutral point by simply by equating Entry and Exit Substituting eq. 6.30 into eq. 6.29 we have

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

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

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

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

  28. 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

  29. 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) • e1 = ln(1.0/0.8) = 0.223

  30. Thus from equation 6.10 , Y= (30000)(0.223)0.2/1.2 = 18500 psi And =1.15(18500) = 21275 psi

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