Chapter 9 9-1 Using the slip-line field in Figure 9.8 for frictionless indentation it was found that P ⊥ /2k = 2.57. Figure 9.35 shows an alternate field for the same problem proposed by Hill. (a) Find P ⊥ /2k for this field. (b) Construct the hodograph. (c) What percent of the energy is expended along lines of intense shear? Figure 9.35 Slip-line field for plane-strain indentation. Solution : a)Along ED, σ 1 = 0, σ 2 = -k, Along EC σ 2 = -k, DCFBA is an α-line so in triangle ABE σ 2 = -k +2k∆φ = -k +2k(-π/2) = -k(1+π). σ 3 = σ 2 -k = P ⊥ . P ⊥ /2k = 1+π/2 = 2.57. b) V p = 1 V AE V* AB V F V* CD =1, Intense shear occurs along ABFCD. Note that along BE and EC, the deformation is gradual.. Letting A E =1, A B =1/√2= B E = C D , B C = ( π /2) B E , V* AB = Vp√2 = V* BC = V* CD so the rate of energy dissipation on these lines is k( A B V* AB + B C V* BC + C D V* CD ) or k[(1/√2)(√2) + π((1/√2)(√2) +(1/√2) (√2)]V p . = (2 + π/2)k = 3.57k.
Transcript
Chapter 9
9-1 Using the slip-line field in Figure 9.8 for frictionless indentation it was found that
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P⊥/2k = 2.57. Figure 9.35 shows an alternate field for the same problem proposed by Hill. (a) Find
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P⊥/2k for this field.(b) Construct the hodograph.(c) What percent of the energy is expended along lines of intense shear?
Figure 9.35 Slip-line field for plane-strain indentation.
Solution: a)Along ED, σ1 = 0, σ2 = -k, Along EC σ2 = -k, DCFBA is an α-line so intriangle ABE σ2 = -k +2k∆φ = -k +2k(-π/2) = -k(1+π). σ3 = σ2 -k = P
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⊥ . P
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⊥ /2k = 1+π/2 = 2.57.b)
Vp = 1
VAE
V*AB
VF
V*CD
=1, Intense shear occurs along ABFCD. Note that along BE and EC, the deformation is gradual.. Letting
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A E =1,
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A B =1/√2=
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B E =C D ,
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B C =(π/2)B E , V*AB = Vp√2 = V*BC = V*CD so the rate of energy dissipation on these lines is k(
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A B V*AB +
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B C V*BC +
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C D V*CD) or k[(1/√2)(√2) + π((1/√2)(√2) +(1/√2)(√2)]Vp. = (2 + π/2)k = 3.57k.
The total rate of energy dissipation from (a) is 5.14k so the energy dissipation from theses lines is 3.57/5.14 = 69.5%9-2. Figure 9.36 shows a slip-line field with a frictionless punch. Construct the hodograph and find
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P⊥/2k for this field.
Solution:
Vp
V* GO
VG
VFV* DE =VE
9-3. Plane-strain compression of a hexagonal rod is shown in Figure 9.37 together with a possible slip line field.
(a) Determine whether this field or penetrating deformation will occur.(b) Find
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P⊥/2k.
Figure 9.36 Slip-line fieldfor Problem 9-2.
Figure 9.37. Possible slip-line field for Problem 9-3.
On the other hand, for penetrating deformation, P⊥/2k = 1.15 according to Fig. 9-20, which
is lower and therefore appropriate
b) For von Mises, 2k = 2Y/√3 so P^= 2.25,000/√3 = 33,200 psi
9-4 Figure 9.38 is the slip-line field for plane-strain wedge indentation. For thevolume of the side mounds to equal the volume displaced, the angle ψ must be related to θ by cos(2θ-ψ) = cosψ/(1 + sinψ). Determine F/x in terms of 2k for θ = 120°.
Figure 9.38 Slip-line field for plane-strain wedge indentation.Solution: a)
Substituting 2θ = 120o into cos(2θ - ψ) = cos ψ/(1+siny), and solving by trial
and error, ψ = 50.84o = 0.887 radians. Starting on the left-hand side, at the surface σ1 = 0,
so P = k. Now rotating on a β-line through ∆φβ = +0.887, the pressure in the D next to
indenter is P∆ = k + 2k(0.887); P⊥ = P∆ + k = 2k(1.887). Let the length of contact between indenter
and material be 1, then F⊥ = P⊥, Fx = F⊥ sin60 = P⊥sin60, σx = Fx/sin60 = P⊥ =
2k(1.887), σx/2k = 1.887
9-5. Figure 9.39 shows the slip-line field for a 2:1 reduction by indirect or backward frictionless extrusion. (a) Determine
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P⊥/2k..(b) Construct a hodograph for the lower half of the field.(c) Find V*AD/V0. (d) What percent of the energy is expended by the gradual deformation in the centered fans?
Solution:a) By inspection, A'B'CD is an α-line and CD and A'D' are β-lines.Along CD', σx = 0 so PCD' = k, rotating on a-lines -π/2 from CD' to A'D',
PA'D' = PCD' -2k(-π/2) = k +2k(π/2)
This is the pressure in triangle A'E'D', so Pext = P⊥ = PA'D' + k = 2k(1+π/2)
pext/2k = 2.57
c) V*AD = V*CD = (√2/2)Vo = √2/2 = .707
Figure 9.39 Slip-line field forthe indirect extrusion of Problem 9-5.
d) rate of energy expenditure along AD and CD = k(
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A D V*AD+
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C D V*CD) = 2k
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A D V*AD = 2k(√√2)(√2/2)] = 2krate of energy expenditure along arc ABC = k
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A B C V*ABC
arc length
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A B C = √2π/2), the velocity discontinuity along this arc V*ABC = V*AD = (√2/2)Vo = √2/2, so rate of energy = kπ/2 =2k(π/4)total rate of work = PextVox1 = 2k(1+π/2)
Thefraction expended on lines AD, CD & arc ABC = (1+π/4)/(1+π/2)= 69.4%The rest is expended in fan = (π/4)/(1+π/2) = 30.6%
9-6 Fig. 9.40 shows the slip-line field for the 3:1 frictionless extrusion. Find
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P⊥
/2k and η.
Solution:This is one of the special fields covered in sect 9-6 for which sinα = r/[2(1-r)]. (Here α =
90o=π/2 so sinα = 1 and r = 2/3. For these fields Pext/(2k) = r(1+α) = (2/3)(1+ π/2) = 1.714
η = wi/wa = 2kε/Pext = 2kε/[2kr(1+α)] = ln3/[(2/3)(1+ π/2)] = η = 64.1%
Note this problem can also be solved by realizing that along PCD = k, and rotating on an α-line
through ∆φα = -π/2, PABD = k + 2k(π/2)
P⊥ = PABD + k = 2k(1+π/2), 3Pext = 2P^, so Pext/2k = (2/3)(1+π/2) etc.
9-7 Indentation of a step on a semi-infinite hill is shown in Figure 9.41. Show that thisfield is not possible and draw a correct field
Figure 9.40 The slip-line field for the 3:1 frictionless extrusion in Problem 9-6.
Solution: σ1FG = 0, σ2FGE = -k, FEDA is an α-line.
a) starting from the right side of the field, σ2D = σ2E + 2k∆φDE = -k + 2k(-φR) (clockwise) so
σ2D = -k -2kφR. Starting from the left side of the field, σ1AB = 0,σ2ABC = -k, BCDG is a β-line.
σ2D = σ2C - 2k∆φDC = -k - 2k(φL) (counterclockwise) so σ2D = -k -2kφLSince φR > φL, the value of σ2D calculated from the two fields is different. The full field cannot
be correct.b) Simply use the portion GDCBA which gives the lower value of P/(2k) 9-8 What is the highest level of hydrostatic tension, expressed as σ2/2k, that can be induced by two opposing flat indenters as shown in Figure 9.42?
Solution: The highest midplane tensile stress occurs in the field with the highest H/L for which deformation penetrates. From fig 9-20, H/L = 8.75. For this field P⊥/2k = (1 + π/2). The
value of P in the triangle below the indentor is P⊥ - k. The field corresponds to a centered fan with
9-9 Consider the back extrusion in Figure 9.43. Assume frictionless conditions.(a) Find
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P⊥/2k.
Figure 9.41 An incorrect field for indentation on a semi-infinite hill
Figure 9.42 Indentation by two opposing flat indenters.
(b) Construct the hodograph.
Solution:
9.10 Consider punching a long, thin slot as shown in Figure 9.44. For punching, shear must occur along AB and CD. (a)Find
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P⊥/2k. as a function of t.
Figure 9.43 Back extrusion in Problem 9-9.
(b) If the ratio of t/w is too great, an attempt to punch will result in a plane-strain hardness indentation. What is the largest ratio of t/w for which a slot can be punched?(c) Consider punching a circular hole of diameter, d. Assume that for a can be punched
Solution: a) The area of shear = 2lt, The internal work rate is
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Ý W int = k ≥ 2lt ≥ Vp. The external work
rate is
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Ý W ext= P ≥ wl ≥Vp. Equating, P ≥ wl ≥ Vp = k≥2lt≥Vp, P/(2k) = t/w.
b) Hardness indentation will occur if P/(2k) = 1 +p/2 = 2.57. Therefore since P/(2k) = t/w, if t/w > 2.57, a hardness indentation will occur instead of punching of the slot.
c) Now the upper-bound gives Wint = kpdtVp = Wext= Pp(d2/4)Vp. Therefore, P/
(2k) = 2t/d. Equating with P/(2k) = 3, 2t/d = 3; d/t = 2/3. NOTE : This is a minimum value of d/t for hole punching, not a maximum as implied
9-11 Deeply notched tensile specimen much longer than its width and very deep in thedirection normal to the drawing is shown in Figure 9.45. Calculate σx/2k for the field
where σx = Fx/tn.
``Figure 9.45 Deeply-notched tensile specimen for Problem 9-11.
Solution: First label the a and b-lines as shown . This is easily done at point C where obviously the most tensile direction (and therefore the 1 axis) is vertical.
Figure 9.44 Slot punching.
In triangle ABF, σy = σ3 = 0. σ2(FB) = k. Moving along a β line to FC, σ2(FC) = σ2(FB) -
2k(∆φβ) = k - 2k(-π/2)Every where within region CFIG, σ2 is the same and σ1 = σ2 + k = 2k +
2k(π/2) = 2k(1+ π/2). The stress σ = F/x = σ1= 2k(1+ pπ/2), so σ/2k = (1+ π/2) = 2.57Note that half of this field is exactly the same as the field for plane- strain indentation except that all of the stresses are tensile here instead of compressive. Our answer for the stress here is the same as for the indentation problem (1+ π/2 = 2.57) except that the stress is tensile.
9-12 Consider a plane-strain tension test on the notched specimen in Figure 9-46.
Figure 9.46 Notched tensile specimen for Problem 9-12. Solution: At the surface, the stress normal to the surface, σ3 =0, so σ2 = k.
Rotating 45o = π/4 to the center, σ2 = k + 2k(π/4). and here σ1 = σ2 + k = 2k(1+π/4).
The stress is the same everywhere along a cut through the center so σx = σ1 = 2k(1+π/4).
σx/2k = (1+π/4) = 1.785
9-13 If the notches in Problem 9-12 are too shallow (i.e. to/tn is too small) the specimenmay deform by shear between the base of one notch and the opposite side as shown in Figure 9.47. What ratio of to/tn is needed to prevent this?
Solution: From problem 9-12, the value of σx necessary to operate the field is 1.785, or σx = =
1.785(2k) = 3.571k. Alternatively, shear could occur on a 45o plane from the base of the notch to the opposite surface. The shear stress on such a plane,
(b) If this were a drawing operation the drawing stress would be σd/2k = 2k(0.178) and an x-
direction force balance gives σd(1-r) = P⊥r, so P⊥= σd(1-r)/r = 2k(0.178)(1- .070)/.076 = 2.164(2k).
Now P in ∆ABO,O = 1.646(2k) and moving to the centerline,
P = 1.646(2k) - (2k)(9π/12) = -0.710(2k) or σ2 = 0.710(2k).
(c) The higher hydrostatic tension at centerline in drawing would cause more centerline porosity and internal cracks, lowering the toughness.
9.15 Figure 9.49 shows two slip-line fields for the compression of a long bar with an octagonal cross section. Which field is appropriate? Justify your answer.
Figure 9.48 Slip-line field for Problem 9-14.
Solution: For an octagon the value of h/L = 1 + 2√2/2 = 1 + √2 = 2.414. For penetrating flow with h/L = 2.4, Figure 9-19 indicates that P⊥/(2k) = about 1.4.The non-penetrating flow field is very
similar to the hardness indentation except that the centered fans are now π/4 instead of π/2. Consequently the pressure required is P⊥/(2k) = 1 + π/4 = 1.785. Since this is higher than the
pressure for penetrating flow, we should expect penetrating flow (right-hand figure).
9-16 Consider an extrusion with in a frictionless die with a = 30° and such that point (2,4) in Figure 9.18 is on the centerline.
(a) What is the reduction?(b) Calculate Pext/2k. (c) Calculate η. (d) Find the hydrostatic stress, σ2, at the centerline.
Solution: Refer to Figs. 9-18 and 9-20.
Take a cut from A through 0,2, and 1,3 to 2,4 and make an x-direction force balance. This cut is easy to analyse because everywhere along it the a and b lines are at 45° to x, so σx = σ1 and the net force is Fx = ∫σxdy = 0.
Using figure 9-19 and letting x and y in that figure be x' and y',
Point x' y' y = y'cos30 + (x'-1)sin30
Figure 9.49 Two possible slip-line fields for plane-strain compression of an octagonal rod. For Problem 9-15.
Noting that for all these points, σx = -3.935Pe + 2k + f(y), Fx = 0 = ∫σxdy may be expressed as Fx =
0 = (-3.935Pe + 2k)(2.935) +∫f(y)dy. Now plotting f(y) vs. y:
It can be seen that ∫f(y)dy = 1.πk/3 + (1.5πk/3)(1.788 -1.00) + (2.5πk/3)(2.935 -1.788) = 5.29kFx = 0 = -3.935Pe(2.935) + 2k(2.935) + 5.29kPe/(2k) = 0.483
c) η = wi/wa where wi = 2kεh = 2k(ln[1/(1-r)] = 0.293 and wa = Peη = 0.293/0.483 = 60.7%
d) at point 2,4, σ2 = -3.935Pe + (1+π)k = -3.935(.483.2k) + (1+π)k
σ2/(2k) = +0.170 Note that this is tensile!
9-17 Consider the slip-line field for an extrusion with a constant shear stress along the die wall as shown in Figure 9.50. (a) Label the α- and β-lines. (b) Draw the Mohr’s circle diagram for the state of stress in triangle ACD showing
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P⊥, τf = mk, α and β(c) Calculate the value of m.
Solution:
c) m = τ/k = sin 30° = 0.500
9-18 At the end of an extrusion in a 90° die, a non-steady state condition develops. Thefield in Figure 9.51(a) is no longer appropriate. Figure 9.51(b) is an upper-bound field fort < 1. Calculate and plot Pext/2k as a function of t for 0.25 ≤ t ≤ 5 for the upper-bound field. Include on your plot the value of Pext/2k for the slip-line field.
Figure 9.50 Slip-line field for extrusion with a constant shear stress die interface. For Problem 9-17.
Solution:For the slip-line field (a): Along CD σ1 = 0, σ2 = k, and CBA is an α-line
π)σ3AD = σ2ACD - k = -2k(1 + π/2) = - P⊥. P⊥ = 2k(1 + π/2); Making a force balance, P⊥.1 =
2Pe so Pe = k(1 + π/2). Pe/(2k) = (1 + π/2)/2 = 1.285
For the upper-bound field, r = 50%, Ve = 2Vo and γ = θ
2PeVo = k(
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A O V*AO +
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O C V*OC) = k(2(t/cosθ)(Vo/cosθ),
Substituting cosθ= t/
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A O and
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A O 2 = t2 +1
2PeVo = 2kVo[t(t2+1)/t2] = 2kVo(t + 1/t), Pe/(2k) = (t + 1/t)/2t Pe/(2k) t Pe/(2k) t Pe/(
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A O 2k)
.25 2.125 .75 1.04 2 1.25
.375 1.52 1.0 1.0 3 1.67
.5 1.25 1.25 1.025 4 2.125
.625 1.11 1.5 1.08 5 2.6
Figure 9.51 (a) Slip-line field for a 2:1 extrusion. (b) Upper-bound field for the end of a 2:1 extrusion. For Problem9-18.
Note the general similarity with Fig. 9-30. Because of the assumption of frictionless conditions, the result is similar to an indirect extrusion. Also prob. 9-19 addresses the question of pipe formation
9-19 In Problem 9-18, either the slip-line field or the upper bound gives a lowersolution. However as discussed in Section 9-12, a pipe may form at the end of an extrusion. Figure 9.52 gives an upper -bound field that leads to pipe formation. Calculate Pext/2k as a function of t for 0.25 ≤ t ≤ 5 and compare with the solution to Problem 9-18.
Solution:The following analysis is based on simplifying assumptions that are not correct. Yet theresults do indicate the basic reason for pipe formation in extrusion. These assumptions are: 1. 2Pe is the force applied to the piston. (It should be 2Pe(1-2x) but
this leads to messy math. 2. What is called Pe is the average value of the extrusion pressure,
which is not uniform.
Figure 9.52 An upper-bound field for pipe formation. (Problem 9-18).
As point C moves upward along the piston face, x increases, and the metal adjacent to the piston face moves faster than Vo so a pipe forms. A general hodograph is shown:
Now Ve depends on x, Ve(1-x) = Vo(2-x) or Ve = (2-x)/(1-x)Now to find the value of x for a
minimum Pe.
2VoPe = k[
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A O V*AO +
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O C V*CO]; V*AO = Vo/cosθ and V*CO = (Ve -Vo)/cosγ,
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A O = t/cosθ,
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O C = t/cosγ Now, V*CO = [(2-x)/(1-x) -1]/cosγ = 1/[(1-x) cosγ]
2VoPe = k{(t2+1)/t + [t2 + (1-x)2]/[t(1-x)]} (a)
Setting dPe/dx = 0 to find the minimum, t/(1-x2) -1/t = 0 or t = 1-x, so when γ = 45°, Pe is a
minimum. Substituting t = 1-x into (a),Pe/(2k) = (1/4)[1-x + 1/(1-x) + 1 +1] = (1/4)[3 - x + 1/(1-x)]. Pe/(2k) = (1/4)[2+t+1/t]
(note this solution is valid only for t ≤ 1 (γ ≤ 45 an x ≥ 0. For t ≤ 1 the solution of Prob. 9-18 gives a lower value than the slip-line field solution)
