Solution manual 7 8

Chapter 7

7-1 A coil of steel, 252 mm wide and 3 mm thick is drawn though a pair of dies of semi-angle 8° to a final thickness of 2.4 mm in a single pass. The outlet speed is 3.5 m/s. The average yield strength is 700 MPa and the friction coefficient is 0.06. Calculate the power in kw consumed. Solution: σd = 2k[(1+B)/B][1-exp(-Bε)]. Substituting B = µcotα = 0.06cot8° = 0.427, 2k = 1,154(700) and ε = ln(3/2.4) = 0.223, σd = 245 MPa. Power =σd(velocity)(crossectional area) = (245 MPa.)(3.5m/s)(0.252 m)(0.0024) = 52 kw.

7-2 An efficiency of 65% was found in a rod-drawing experiment with a reduction of 0.2 and a semi-die angle of 6°.a) η = σi/σd = Yave/σd Substituting eq. 7-14 for σd, h = Yave/{Yave[1+B)/B][1-exp(-Bε)]} = εB/(1+B)]/[1-exp(-Bε)]. Now substituting e = _n[1/(1-r) = 0.223 and h = .65, 0.65/0.223 = [B/{(1+B))[1-exp(-0.223B)]}Solving by trial and error, B = 0.654 but B = m/tanα, so with a = 6o, m= Btan6o = 0.0687b) Substituting B = 0.654 and ε = _n[1/(1-.4) = 0.5108 into η= εB/(1+B)]/[1-exp(-Bε)] = 0.711c) Sachs analysis ignores the redundant work, so the friction coefficient calculated to explain the experimental data is too high. This is particularly true for low reductions where the redundant work is a large fraction of total work. Use of this friction coefficient for higher reductions then tends to overestimate the actual work (Hence low predicted efficiency) 7-3 Estimate the force required to coin a U. S. 25¢ piece. Assume that the mean flow stress is 30,000 psi, the diameter is 0.95 in. and the thickness after forming is 0.060 in. Solution: Assume sticking friction because embossing won’t allow sliding. P = Y[1+1.154(0.475)/2x0.60)] = 132,800 psiF = 132.8x103(π(0.475)2 = 94,200 lbs

7-4 Figure 7.26 shows a billet before and after hot forging from an initial size of 2.5 mm x2.5 mm x25 mm to 5 mm x1.25 mm x 10 mm. This is accomplished by using a flat-face drop hammer. Sticking friction can be assumed. For the rate of deformation and the temperature, a flow stress of 180 MPa can be assumed.a) Find the force necessary.b) Find the work required. (Remember that work = ∫FdL and that F changes with L.)c) From what height would the hammer of 3 kg have to be dropped?d) Compute the efficiency, η.

Chapt. 7 Slab analysis and Friction 87

Figure 7.26 Compression in Problem 7-4.

Solution: a) Using the final dimensions, F = Pav(0.025)(0.005) = 1.154(18)[1 + 5/(4x1.25)] = 41.5 MPab) b = (bo)(ho)/hWork = ∫Fdh where F = 2k(1+b/4h)bh = 2k(1+(bo)(ho)/h2)L(bo)(ho)/h = 2k(1+(bo)(ho)/h2)L(bo)(ho)/h = 2kL(bo)(ho)/h + 2k(bo)

2(ho)2/h3)L

Work = 2kL(bo)(ho)∫dh/h + 2kL(bo)2(ho)

2)∫dh/h3) = 2kL(bo)(ho)∫ln(h/ho) + 2kL(bo)2(ho)

2)[-ho

-2+h-2) = (1.154)(180x106)(0.025)(0.0025)(0.0025){ln(2.5/1.25) + [0.0025)-2- 0.00125)-

2]/(-2)}= 0.00325{0.693 + 1.5) = 71.3 Jc) 71.3 = ghm; h = 71.3J/[(10 m/s2)(3 kg)] = 2.4 md) Wi = σε(vol) = 180MPa(0.0025)(0.0025)(0.025) = 28 J; η = 2.8/7.13 = 39%

7-5 Two steel plates are brazed as shown in Figure 7.27. The steel has a tensile yield strength of 70 MPa and the filler material a with tensile yield strength of 7 MPa. Assuming that the bonds between the filler and the steel do not break. Determine the force necessary to cause yielding of the joint.

Solution: P = 2k(1+b/4h) = 2k(1+3/8) = 70 MPa(0.003m)(0.050m)(1.154)(1.375) = 16.6 kN


Figure 7.27 Brazed joint for Problem 7-5.

7-6 Figure 7.28 shows a thin lead ring being used as a gasket. To insure an acceptable seal the gasket must be compressed to a thickness of 0.25mm. Assume that the flow stress of lead is 15 MPa and strain hardening is negligible. Find the required force.

Figure 7-28. Lead gasket for Problem 7-6. Solution: Substituting b = 0.004m and h = 0.00025m, into P = 2k(1+b/4h) = (1.154)(15)(1+4/0.25) = 294 MPa. F = 294x106π(0.025)(0.004) = 92.4 kN

7-7 The derivation of equation 7.11 assumed a constant coefficient of friction. Derive an equivalent expression assuming sticking friction. Solution: Making a force balance, π[σx(r+dr)2 + (r+dr)2]dσx +mk(2πr)dx + P(2πr)tanαdx = πr2σx

Expanding, neglecting second order differentials and simplifying2σxrdr + r2dσx + 2rσxdr +2mkrdx + 2rPtanα = 0Substituting P = Y - σx and dx = dr/tanα,

7-8 Magnetic permalloy tape is produced by roll flattening of drawn wire. The final cross section is 0.2 mm by 0.025 mm. It is physically possible to achieve this cross section with different rolling schedules. However it has been found that the best magneticproperties result with a maximum amount of lateral spreading. For production the rolling direction must be parallel to the wire axis. Describe how you would vary each of the parameters below to achieve the maximum spreading.

a) roll diameterb) reduction per passc) the frictiond) back and front tension


rolling direction

0.2 mm

0.025 mm

Figure 7.29 Permalloy tape (Problem 7-8)Solution: To maximize spreading, the contact length should b a long as possible. a) use as large roll diameter as possibleb) maximize the reduction/pass. (use only one pass)c) If the contact length is greater than the width, high friction will promote spreadingd) Use no back or front tension

*7-9A metal with a flow stress of 35 MPa is to be drawn from a diameter of 25 mm to 20mm through a die of 15 semi-angle. Calculate the necessary drawing stress if:a) The conditions are frictionlessb) µ = 0.20.Solution: ε = ln(do/d) = 0.223a) σd = σε = (35)(0.223) = 7.8 MPa

b) Using Sachs’ analysis with B = (0.20)(cot15°) = 0.746, σd = (1.154)(35)[(1+0.746)/0.746][1-exp(-0.46x0.223)] = 14.5 MPa

7-10 Consider the rolling of a sheet 15 cm wide from a thickness of 1.8 mm to 1.2 mmin a single pass by steel rolls 20 cm in diameter. Assume a friction coefficient of 0.10 anda flow stress of 125 MPa. a) Calculate the roll pressure if roll-flattening is neglected. b) Calculate the roll pressure taking into account roll flattening.c) Estimate the minimum thickness that could be achieved.Solution: a) Substituting µ = 0.1, h = 0.015 and L = √(R∆h) = √(0.010x0.0006 =0.025 so µh/L = 0.1667 and σ = 125 MPa into P = [h/(µL)][exp(µh/L) – 1]σ , P = 136 MPa

Chapter 8

8-1 Find Pe/2k for Figure 8.2 if θ is 80° and compare with Figure 8.3.Solution:

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A B = 1/sin80° = 1.015. V*AB/sin30 = V0 /sin(50), V*AB/V0 = sin30/sin50 = 0.653. The horizontal component of BC is √3/2-1/2 = 0.732. ψ = (1/2)/arctan (√3-1)/2] =36.2°

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B C = √[(1/2)2+ (0.7322] = 0.886, V*BC/sin80 = V0 /sin63.8; V*BC/V0 = 1.098dW/dt = (1.015)(0.653) + (0.886)(1.098)k = 1.635kPe/2k = 0.818. This agrees with Fig 8-3

8-2 Calculate Pe/2k for the plane-strain frictionless extrusion illustrated in Figure 8.14. Triangles ABC and CDE are equilateral.


Solution: The hodograph is V0

V*AB V*BC

V*CD

V*ED

30 30 30

60

60 60

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A B =

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B C =6,

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C D =

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E D =3, V*AB =V0(1/√3), V*BC =V0(2/√3, V*CD = (2/√3)V0, V*ED

=V0(4/√3). Pe/2k = (1/2)[6(1/√3) + 6(2/√3) + 3(2/√3) + 3(4/√3)]0 = (18√3)10.4 =

8.3 On which discontinuity in Figure 8.14 is the largest amount of energy expended?Solution:

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A B and

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E D

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A B .

8.4 *Draw the hodograph corresponding to the frictionless indentation illustrated in Figure 8.15.

Solution:

Vp V*AC

V*AB

V*BC

V*DC

V*DE

8-5 For the plane-strain compression illustrated in Figure 8.16, calculate Pe/2k for L/H values of 1, 2, 3, and 4. Assume sticking friction.


Figure 8.14 Upper-bound field for plane strain drawing for Problem 8-2.

Figure 8.15 Upper-bound field for indentation for Problem 8-4.

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O A =A B =B C =C D =w(h/2)(1/√2)

8-6 Reanalyze Problem 8-4 if frictionless conditions prevailed.Solution: No sliding occurs on the platen surface so the solution is the same.

8-7 For the indentation shown in Figure 8.7, Pe/2k = 2.89 if all the angles were 60°. Find Pe/2k if the angles OAB, ABC and BCD are 90° and the other angles are 45°. Solution:

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O A =A B =B C =C D =w/(2√2),

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A C =w/2, V*OA = V*AB = V*BC = V*CD = √2Vp, V*AC = 2Vp

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Ý W =kΣ(V*S)= k[(4)w/(2√2)Vp +wVp) = P(w/2)Vp; P/2k = 2(1+1/√2) = 3.414

8.8 Figure 8.17 shows an upper-bound field for a plane-strain extrusion. There are two dead metal zones ADF and FEG. a. Calculate Pe/2k for the field.b. Determine the velocity inside triangle ABC.c. Determine V*AC.d. Compute the deformation efficiency.

Solution: First draw a hodograph.0.502 or 50.2%AB = √2/2, BC = 1/cos15 = 1.035, AC = (1/2)/sin60 = 0.577V*AB = sin45/sin105 = 0.732, V*BC = sin45/sin30 = 1.414, V*AC = sin75/sin45 = 1.366,


Figure 8.16 Upper-bound field for plane-strain compression for Problem 8-5.

Figure 8.17 Upper-bound field for Problem 8.

a) Pe(1)(1) = k[(√2/2)(0.732) + (1.035)(1.414) +(0.577)(1.366)] = 2.762Pe/2k= 1.38

b) V∆ABC = V*AB = 0.732,

c) V*AB = 0.732,η = ln2/1.38 =

8-9 a. Use equation 8.27 to find the drawing stress, σd, for an axisymmetric rod drawing with reduction of 30%, a semi-die angle of 10° and a constant interfacial shear stress of 0.1k. Assume the Tresca criterion.b. Predict σd using the von Mises criterion.

Solution:

8-10 Consider the upper-bound field in Figure 8.19 for an asymmetric extrusion.a. Draw the corresponding hodograph.

b. Determine the angle, θ.


Figure 8.18 Illustration ofaxisymmetric drawing for Problem 8-9.

Figure 8.19 Illustration of an asymmetric plane-strain drawing for Problem 8-10.

Solution: a)

Vo

V1

V*BD

V*BC

V*AB

V*EFV3

V*DE

V*DC

Ve

b) tanθ = 0.5Vo/2.5Vo = 0.20, θ = 11.3°.

8-11 For the plane-strain compression illustrated in Figure 8.20, calculate Pe/2k for L/h values of 1, 2, 3, and 4. Assume sticking friction.

Figure 8.20 Upper-bound fields for plane-strain compression in Problem 8-11. Solution: For L/h = 1, Pe/2k = 1, Drawing a hodograph for L/h = 2,

Vup

V*AC

V*AB V*BC

V*AC= 2Vp, V*AB = V*BC = √2Vp, AB = BC = AC/√2; PeVpAC = k[V*ABAB + V*BCBC +V*ACAC] = k(√2/√2 + √2/√2 + 2(1)]VpACPe/2k = 2Drawing a hodograph for L/h = 3,

Vup

V*ABV*OA

V*O’C

V*AB V*BC


V*OA = V*AB = V*BC = √2Vp, V*O’C = 3Vp, OA = AB = BC = √2O’C/3PeVpO’C = k[V*OAOA + V*ABAB + V*BCBC + V*O’CO’C] == k[(3√2)(√2/3)VpO’C +

8-12 For the indentation shown in Figure 8.7, Pe/2k = 2.89 if all the angles were 60°. Find Pe/2k if the angles OAB, ABC and BCD are 90° and the other angles are 45°. Solution: From the geometry of the field, OA = BC = CD = (√2/2)OB and AC = OB.From the geometry of the hodograph, V*OA = V*AB = V*BC = √2VP, V*AC = = 2VP, The rate of external work on the right side of the field will be P(Vp)(OD) P(Vp)(OB) = k[(V*OAOA + V*ABAB_+ V*BC BC + V*CD CD + V*AC C )P(Vp)(OB) = k[3(√2VP)((√2/2)OB) + (2VP)(OB)] = k(3 + 2)(Vp)(OB) Pe/2k = 2.5

8-13 A proposed upper-bound field for extrusion is shown in Figure 8.21. Draw a hodograph to scale and determine the absolute velocity of particles in the triangle bounded BCD.

Solution:Vo

Ve

V*AEV*AD

V*DB V*BC

V*EA = 1/cos(67.8°) = 2.647, V∆AED = tan(67.8°) = 2.45, V∆DAB/sin22.2 = V*EA/sin45,

V∆DAB = 2.647sin22.2/sin45 = 1.414, tan(67.8°) = V∆BCD/V∆DAB, V∆BCD =

V∆DABtan(67.8°) = 1.414tan(67.8) = 3.465 downward

8-14 Figure 8.22 shows an upper-bound field for a plane-strain extrusion. There are two dead metal zones ADF and FEG. a. Calculate Pe/2k for the field.


Figure 8.21 Upper-bound field for the plane strain extrusion of Problem 8-13.

b. Determine the velocity inside triangle ABC.c.Determine V*AC.d. Compute the deformation efficiency.

Solution: First draw a hodographB

A

C

45 105

75

45

30

60

105

V*AB

Ve Vo

V*BC

V*AC

AB = √2/2, BC = 1/cos15 = 1.035, AC = (1/2)/sin60 = 0.577V*AB= sin45/sin105 = 0.732, V*BC= sin45/sin30 = 1.414, V*AC= sin75/sin5 = 1.366a) Pe(1)(1) = k[(√2/2)(0.732) + (1.035)(1.414) + (0.577)(1.366] = 2.77Pe/2 k = 1.385

b) V∆ABC = V*AB= 0.732

c) V*AC= sin75/sin5 = 1.366d) η = ln2/1.385 = 50.2%

8-15 Use equation 8.27 to find the drawing stress, σd, for an axisymmetric rod drawingwith reduction of 30%, a semi-die angle of 10° and a constant interfacial shear stress of 0.1k. a. Assume the Tresca criterion.b. Assume the von Mises criterion.Solution: σd/2k = (σo/2k + m/sinα)ε + (2/3)tanα. Substituting ε = .357 for a reduction of 30%, α =10° and m = 0.1; σd/2k = (σo/2k + 0.576)(.357) + 0.1176.a For Tresca σo/2k = 1, σd/2k = 1.576(.357) + 0.1176 = 0.68


Figure 8.22 A 2:1 extrusion field for Problem 8-14.

b For von Mises, σo/2k = 1.15, σd/2k = 0.74

8-16 Consider the plane-strain indentation illustrated in Figure 8.23. Assume that the deformation in region AA’B’B is homogeneous. There are discontinuities along AA’ and BB’. a. Write an expression for V*AA’ and V*BB’ in terms of Vo, z and t. b. What is the ratio of the energy expended on these discontinuities to the homogeneous work?

Solution: a) V* = (z/t/2)VP = (2z/t)VP

b) rate of work on 1/4 of shear discontinuity = ∫kV*dz = k(2z/t)VPdz = (2kVP/t)(t/2)2/2 = (kVP)(t/2)/2; on all 4 discontinuities dW/dt = kVPt Rate of homogeneous work is dW/dt = (2k)[VP/(2t)](wt) = kVPw shear work/homogeneous work = kVPt/(kVPw) = t/w

8-17 Figure 8.24 shows two different upper-bound fields for a 2:1 reduction by extrusion.Regions ABC and EFG are dead-metal zones.

a. Calculate Pe/2k for both fields.b. Determine the deformation efficiency, η, for both cases.c. What is the absolute velocity of a particle in triangle JGH?

Solution:First draw the hodographs for both fields


Figure 8.23. Figure for Problem 8-16.

Figure 8.24 Two proposed upper-bound fields for a plane-strain extrusion with a 50% reduction for Problem 8-17.

Vo = 1

V*AD = 1V*AC =¦2 V*DC =¦2

A

D

C

Let AD = 1. Then AC = CD = 1/√2; Pext(1)(1) = k[2(1/√2)(√2) + 1(1)] = 3k, Pext/2k = 1.5η = ln(2)/1.5 = 46.2%

Vo = 1

E

J

H

GV*EG

V*JG

V*HG V*EJ V*JH

Let Vo = 1, Then V*EG = sin60/sin75 = 01 + .897 = V*HG. V*EJ = sin45/sin75 = 0.732 =V*JG = V*JH . Let ho = 1. Then EG = HG = √2/2, EJ = JH = 0.5/sin60 = 0.577 and JG/sin75 = EC/sin 60, so JG = (√2/2)sin75/sin60 = 0.7887. Pext/2k = (1/2)[2(.577)(0.732) + 2(0.707)(0.897) +(0.732)(0.7887)] = 1.345η = ln(2)/1.345 = 51.5%

c) The horizontal component of V∆GJH = 1 + V*JG/cos60° = 1.464.

The vertical component of V∆GJH = V*JG/cos60° = 0.845

V∆GJH = [1.4642 + 0.8452]1/2 = 1.690


Solution manual 7 8

Engineering

Transcript of Solution manual 7 8