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Wire and rod drawing
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Variables in Wire Drawing
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Various methods of tube drawing
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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.
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Wire Drawing/ Rod drawing
s = Ken (Equation 6.10) The drawing force then is F = YAfIn (equation 6.62)
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Stresses acting on an element in drawing of a cylindrical rod or wire
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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…)
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( forces acting on an element…..)
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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
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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)
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Derive the maximum reduction perpass
For 1- efficiency =0.9 2- Power law material 3- drawing stress with friction
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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
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Defects Seams Stresses residual stresses in cold drawing
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Residual stresses in cold drawn AISI 1045 carbon steel round
T : transverse L : longitudinal R : radial direction
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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
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Terminology Typical die for drawing round rod or wire
Typical wear pattern in wire drawing die
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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
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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
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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
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Topics In Rolling
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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.
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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
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Determination of neutral point
We determine neutral point by simply by equating Entry and Exit Substituting eq into eq we have
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Simpler method to find roll force
L : arc of contact we can approximate L as the projected area Dh= hf - ho
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Pressure distribution
Function of front and back tension As tension increases the neutral point shifts and there is reduction in area under the curves
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Roll forces: the area under pressure contact length curves multiplied by the strip width
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Reduce Pressure Roll force can be reduced by Smaller radii
Smaller reduction Higher workpiece temperature Lower friction Front and back tension
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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
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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
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Thus from equation 6.10 , Y= (30000)(0.223)0.2/1.2 = psi And =1.15(18500) = psi
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