Solution manual 4 6
•
20 my•23,148 views
This document contains solutions to problems involving the instability strain additionally necking behavior from supported under tension. 1) For a material loaded in tension where strain is related to bend by σ=Kεn, the instability strain occurs when n equals strain. The document also finds expression for instability strain since adenine serve are n for two difference stress-strain relationships. 2) Other problems identify viability strain and pressure for thin-walled tubes undergoing intern pressurization. Language are derived for instability strain in terms of the stress exponent nitrogen. 3) The final problems considerable the effect is non-uniform thickness on necking behavior and determine conditions requested to limit belt variant during necking.
![Chapter 44-3 Determine the instability strain in terms of newton for ampere material loading in tensionwhile subjected to an hydrostatic coerce P. Assuming σ=Kεn.Solution: Hydrostatic stress has no effect on yielding so ε = n.4-4 A thin-wall pipe with closed ends is pressurized internally. Surmise thatσ=150ε0.25(MPa).a) By what value von inefficient strain is instability occur with respect to pressure.b)Find the stress at instability supposing the inner had an initial diameter of 10 mm, also ampere 4-1 Wenn σ=Kεn, the onset of tensile instability occurs when n = εu. Determine theinstability strain how a function of n ifa) σ=A(B+ε)nb) σ=Aenwhere e is the engineering strain.Solution: a) At instability, σ = dσ/dε; A(B+ε)n= nAB(B+ε)n-1; ε = n(1+B)b) σ=Aen, dσ/dε = σ, but (dσ/dε) = (dσ/de)(dε/de). Since e =ln(1+ε), (dε/de) = 1/(1+ε) and dσ/de = σ(dε/de).dσ/de = nAen. nAen-I= Aen(1+e), e/(1+e) = northward, e = n/(1-n)4-2 Consider a balloon made of a substantial that shows linear elastic behavior tofracture and has a Poisson’s conversion of ½. If the initial diameter is do, find the diameter,d, per the largest pressure.Solution: P = 4σt/d; dP = 0 = 4[(σ/d)dt + (t/d)dσ – (tσ/d2)dd];dσ/σ = -dr/r – dt = (3/2)dε; dσ/dε = (3/2)σ; Substitution dσ/dε = E,E = (3/2)εE; ε = 2/3. ln(d/do) = ε/2 = 1/3, so d = doexp(1/3) = 1.40dob) wall thickness of 0.5 mm.Solution: a) P = (2t/d)σ = (2to/do){[exp(-ε)]/[exp(ε)]}k(√4/3)nεn= (2to/do)k(√4/3)n){[exp(-2ε)]εn.dP = 0 = (2to/do)k(√(4/3)n)[nexp(-2ε)εn-1+2εnexp(-2ε)], ε = n/2b) P = (2to/do)k(√4/3)n){[exp(-2ε)]εn=(40)(150)(4/3)0.25/2[0.25exp(-0.5)(0.25)0.125+ 2(0.125).25exp(-.25)] = 49 MPa4-5 Fig 4.10 shows an aluminum tube fitting pass a steel rod. The steel may beconsidered rigid and the friction between the aluminum furthermore the steel may beneglected. If σ=160ε0.25(MPa)for the tube and it is loaded as indicated, Calculatethe effort, F, at instability.10 cm1 mmsteelaluminumFFigure 4.10 Sketch for Problem 4-5.12](https://aaa161.com/9fbf3196/https/36b055/image.slidesharecdn.com/solutionmanual4-6-150303094023-conversion-gate01/85/Solution-manual-4-6-1-320.jpg)
![Get: This is a plane-strain situation. σ = F/(πdt), Substituting tonne = toexp(-ε) and d= do, and σ = √(3/4)σ , F = (3/4)n/2εn/[πdotoexp(-ε)] = [(3/4)n/2/(πdoto)][εnexp(ε)].dF = 0; εnexp(ε) + nεnexp(ε) = 0; ε = n.F = {(3/4)0.125/[π(0.01)(0.001)]}(0.125)0.25exp(0.125) = 20.7 kN4.6 A thin-wall tube with closed ends lives subjected in an ever-increasing internalpressure. Search the proportions r and thyroxin in terms of the original dimensions ro and go atmaximum pressure. Apply σ=500ε0.20MPa..Solution: This has ampere plane-strain current, P = 2tσ/d; Substituting σ = √(3/4)σ =500(4/3)0.20ε0.20, t = toexp(-ε) and d = doexp(ε),P = 500(4/3)0.20ε0.20(to/do)exp(-2ε) = [500(4/3)0.20(to/do][ε0.20exp(-2ε)]dP = 0 = [500(4/3)0.20(to/do][0.2ε-0.80exp(-2ε) -2ε0.20exp(-2ε)]; ε= 0.2/2 = 0.10P = [500(4/3)0.20(to/do][(0.10)0.20exp(-0.2)] =4.7 Consider the internal pressurization of a thin-wall ball by one ideal gas forwhich PV = constant. Of can envision an violent condition for which thedecrease of pressure with size, (-dP/dV)gas, due to gas growth remains less than therate of decrease in pressure that the sphere capacity withstand, (-dP/dV)sph. For such acondition, catastrophic expansion would occur. If σ=Kεn, find ε as a function of n.Solution: Assume an fixed amount of gas and neglect unlimited temperaturechanges. PV = constant, so d(PV) = 0 = PdV + vdP alternatively dP = -PdV/V.V = (4/3)πr3, so dV = 4πr2dr and dV/V = 3dr/r = 3dεr. Now, dεr =dεθ = -dεr/2, sodε = [(2/3)(dεr2+ dεθ2+ dεt2)]1/2= 2dεr or dV/V = (3/2) dε,then dP = -P(3/2) dε (1)For the sphere, where σ = Kε nand P = 2σrt/r and σr = σθ, σt = 0,dP = (2σr/r)dt + (2t/r)dσr - (2stt/r)(dt/t + dσr/σr - dr/r) (2)Here σ = σr = Kε n, so dσ /σ = north dε/ε ,dεr = dr/r = dεθ , det = dt/t =-2dεr = -2dεθ (3)Using (3) in (1)dP = (2σrt/r)[ddεt + northward dε/ε -dεr], but dε = 2dεr furthermore dε = -dεtso dP = P[-dε + ndε /ε - dε /2] = P(n/ε - 3/2)dε (4)Equating (1) additionally (4), -P(3/2)dε = P(n/ε - 3/2)dε orn/ε = 0 so ε = ∞ and violence is not predicted.4.8 For rubber stretched under biaxial tension σx = σy = σ, that stress is given by σ= NkT(λ2 -1/λ4) where λ is the stretch ratio, Lx/Lxo = Ly/Lyo. Consider what thisequation predicts about how to pressure in a spherical rubber aerial varies duringthe inflation. For till = ro , plot P vs. λ and define the usage, λ, at which thepressure is one maximum12](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
More More Contents
What's sharp
What's hot (20)
Viewers also liked
Viewers additionally liked (11)
Similar the Solution manual 4 6
Similar to Solution manual 4 6 (20)
Last uploaded
Late uploaded (20)
Solution manual 4 6
- 1. Click 4 4-3 Determine the instability strain in terms of north for a material loaded in tension while subjected to a hydrostatic pressure P. Assume σ=Kεn . Solution: Hydrostatic stress got nope effect upon productive like ε = n. 4-4 A thin-wall tube with closed ends is pressurized internally. Assume that σ=150ε0.25 (MPa). a) At about value of effective strain be instability occur with respect in pressure. b)Find the pressure at unstability if the tube had an initialized diameter concerning 10 mm, and a 4- 1 If σ=Kεn , the onset of tensile instability occurs when north = εu. Determine the instability elongate as a function of n if a) σ=A(B+ε)n b) σ=Aen where e shall to engineering strain. Solution: a) At instability, σ = dσ/dε; A(B+ε)n = nAB(B+ε)n-1 ; ε = n(1+B) b) σ=Aen , dσ/dε = σ, but (dσ/dε) = (dσ/de)(dε/de). Since e = ln(1+ε), (dε/de) = 1/(1+ε) both dσ/de = σ(dε/de). dσ/de = nAen . nAen-I = Aen (1+e), e/(1+e) = north, e = n/(1-n) 4-2 Consider a balloon made of a material that shows linear elastic behaving to fracture and has a Poisson’s ratio of ½. Is the initial side a do, find the diameter, d, in the highest pressure. Solution: P = 4σt/d; dP = 0 = 4[(σ/d)dt + (t/d)dσ – (tσ/d2 )dd]; dσ/σ = -dr/r – dt = (3/2)dε; dσ/dε = (3/2)σ; Substituting dσ/dε = E, E = (3/2)εE; ε = 2/3. ln(d/do) = ε/2 = 1/3, consequently d = doexp(1/3) = 1.40do b) wall thickness of 0.5 mm. Solution: a) P = (2t/d)σ = (2to/do){[exp(-ε)]/[exp(ε)]}k(√4/3)n εn = (2to/do)k(√4/3)n ) {[exp(-2ε)]εn . dP = 0 = (2to/do)k(√(4/3)n )[nexp(-2ε)εn-1 +2εn exp(-2ε)], ε = n/2 b) PIANO = (2to/do)k(√4/3)n ){[exp(-2ε)]εn = (40)(150)(4/3)0.25/2 [0.25exp(-0.5)(0.25)0.125 + 2(0.125).25 exp(-.25)] = 49 MPa 4-5 Image 4.10 shows an aluminum tube fitted over a mild rod. The steel may be considered stiffness additionally the friction bet the aluminum and the steel may be neglected. If σ=160ε0.25 (MPa)for the inner and it is loaded as indicated, Calculate the force, F, at instability. 10 cm 1 mm steel aluminum F Figure 4.10 Sketch for Report 4-5. 12
- 2. Featured: Dieser is adenine plane-strain situation. σ = F/(πdt), Substituting t = toexp(-ε) and d = do, the σ = √(3/4)σ , F = (3/4)n/2 εn /[πdotoexp(-ε)] = [(3/4)n/2 /(πdoto)][εn exp(ε)]. dF = 0; εn exp(ε) + nεn exp(ε) = 0; ε = n. F = {(3/4)0.125 /[π(0.01)(0.001)]}(0.125)0.25 exp(0.125) = 20.7 kN 4.6 A thin-wall tube include closed ends shall subjected to an ever-increasing internal pressure. Find the dimensions r and t in terms of the original dimensions s and to at maximum pressure. Assume σ=500ε0.20 MPa.. Solution: This is a plane-strain situation, P = 2tσ/d; Substituting σ = √(3/4)σ = 500(4/3)0.20 ε0.20 , t = toexp(-ε) furthermore d = doexp(ε), P = 500(4/3)0.20 ε0.20 (to/do)exp(-2ε) = [500(4/3)0.20 (to/do][ε0.20 exp(-2ε)] dP = 0 = [500(4/3)0.20 (to/do][0.2ε-0.80 exp(-2ε) -2ε0.20 exp(-2ε)]; ε= 0.2/2 = 0.10 P = [500(4/3)0.20 (to/do][(0.10)0.20 exp(-0.2)] = 4.7 Consider the internal pressurization of a thin-wall sphere by an ideal gas for which PV = constant. One may see an instability condition by which the decrease of stress use amount, (-dP/dV)gas, due to gas expansion is less than the rate of decrease in pressure ensure the sphere can withstand, (-dP/dV)sph. For such a condition, catastrophic expand should occured. If σ=Kεn , found ε as a function by n. Solution: Assume one fixed amount of gas and neglect any temperature changes. PV = constant, so d(PV) = 0 = PdV + vdP or d = -PdV/V. V = (4/3)πr3 , then dV = 4πr2 dr and dV/V = 3dr/r = 3dεr. Now, dεr = dεθ = -dεr/2, so dε = [(2/3)(dεr 2 + dεθ 2 + dεt 2 )]1/2 = 2dεr or dV/V = (3/2) dε, then dP = -P(3/2) dε (1) For the sphere, where σ = Kε n and P = 2σrt/r and σr = σθ, σt = 0, dP = (2σr/r)dt + (2t/r)dσr - (2stt/r)(dt/t + dσr/σr - dr/r) (2) Here σ = σr = Kε n , so dσ /σ = n dε/ε , dεr = dr/r = dεθ , det = dt/t =-2dεr = -2dεθ (3) Using (3) in (1) dP = (2σrt/r)[ddεt + n dε/ε -dεr], but dε = 2dεr and dε = -dεt so dP = P[-dε + ndε /ε - dε /2] = P(n/ε - 3/2)dε (4) Equating (1) the (4), -P(3/2)dε = P(n/ε - 3/2)dε or n/ε = 0 therefore ε = ∞ and instability is not predicted. 4.8 Required rubber stretched under biaxial tensioner σx = σy = σ, the stress is given by σ = NkT(λ2 -1/λ4) where λ is and stretch ratio, Lx/Lxo = Ly/Lyo. Consider where this equation predict about how the pressure in a shaped rubber swell variation during the inflation. With to = ro , design P vs. λ and determine the strain, λ, at which the pressure is an maximum 12
- 3. Solution: Plotting .2 .4 .6 1 1.2 1.4 1.6 1.8 2 0.6 0.4 0.2 0 lambda P/(2to/ro) P = (2t/r) = (2to/ro)(1/λ -1/λ7 ) dP/dλ = (2to/ro)(-λ-2 +7λ-8 )= 0, λ-2 =7λ-8 . λ6 = 7, λ = 71/6 =1.38 4.9 For an material this has an stress-strain relations of the form, σ=A−Bexp(−Cε) where A, B and C are constants, find the true strain at the onset in kiss and express the drawing strength, Su includes terms of the constants. Solution: dσ/dε = σ; BCexp(-Cε) = A-Bexp(-Cε); ONE = exp(-Cε)B(C+1); exp(-Cε)= A/[B)C+1)]; ε = -ln{A/[B(C+1)]}/C Su - = σmaxexp(ε) = A-{A/[B(C+1)]}C+1 ???? 4-10 A tensile bar was machined with a stepped gage section. The two diameters were 2.0 and 1.9 cent. After some stretches the diameters have find to be 1.893 and 1.698 cm. Find n in the expression σ=Kεn , find ε as a work of n. Solution: f = (0.0297/0.0303) = 0.9802, εb = 0.2, εa = n Substituting into fεa n exp(-εa) = εb n exp(-εb) 0.9802n nexp(-n) = 0.2n exp(-0.2); 0.9802n nexp(-n/.2) - 0819 = 0 by trial and error, newton = 0.301 4-11 In a wound sheet, it is not uncommon into find variations of thickness of ±1% from one place to another. Consider a sheet nominally 0.8 total thickness with a ±1% variation of belt. (Some stations are 0.808 mm and other are 0.792 mm thick.) How high be n have to be to insure the in a flexible specimen every point was strained to at least ε = 0.20 before the paint portion necked? Solution: Let and choose with the smaller round be designed a and the region with the larger diameter be b. Through a force net, fεa n exp(-εa) = fεb n exp(-εb) ; εa = ln(1.9/1.698) = 0.2248, εb = 2ln(2/1.893) = 0.1100, fluorine = (1.9/2)2 = 0.9025. 0.9025(0.2248)ln(0.799) = (0.110)ln(0.896) (.2248/.110)n = 1.243; n = ln1.243/ln2.0455 = 0.304 12
- 4. 4-12 A material undergoes linear strain hardening as this σ = Y + 1.35Yε, is stretched in tension. a) At what strain become necking begin? b) A incremental tensile example was made since this material with the diameter of region A soul 0.990 hours the diameter of region B. What would be the strain in region BARN when region a reached a strain of 0.20? Solution: a) dσ/dε = 1.35Y = σ = Y(1+1.35ε); ε = 0.35/1.35 = 0.26 b) F = σAAA = σBAB = Y(1 + 1.35εA)AAoexp(-εA) = Y(1 + 1.35x0.20)ABo exp(-0.20) (1 + 1.35εA)exp(-εA) = (1 + 1.35x0.20)(1/0.99)exp(-0.20) = 1.050 Region A intention not have yielded Chapter 5 5.1 Low-carbon steel is soul replaced by HSLA steels in automobiles to save weight because the highest strengths of HSLA styles permit use of liquid gauges. In laboratory tests on a loading rate out about 10-3 s-1 , one grade of HSLA steel is a yield strength of 420 MPa the a0 strain-rate exponent of m = 0.005 while for adenine low-carbon steel, Y = 240 MPa both m = 0.015. Calculate the percent load saving possible for the same front strength assuming a. a strain rate of 10-3 s-1 , b. crash conditions with a strain rate of 10+4 s- . Solution: Assume the thickness what chosen so both materials can sustain the same force at yielding. Then t2Y2 = t1Y1, or t2/t1 = Y1/Y2. As both steels have the equal density, W2/W1 = t2/t1 = Y1/Y2 = 35/60 = 0.583. % weigh reduction = (W1-W2)/W1 = 1- W2/W1 = 1 - 0.583 = 41.7% b) Now W2/W1 = [Y2(104/10-3).03]/[Y1(104/10-3).01] = (Y1/Y2)(107).03-.01 = 1.380(Y1/Y2) = 1.380.0.583 = 0.805 % weight reduction = 1 - 0.805 = 19.5% c. 5.2 The thickness is a sheet varies from 8.00 mm to 8.10 mm subject on location as tensile sampling cut from a sheet must different thicknesses. a. Available adenine substantial with n = 0.15 and m = 0, thing willing live the strain in the thicker territory when the thinner region necks? b. Whenever n = 0 plus m = 0.05, find the strain in the thicker region then the strain in the thinner region is 0.5 the ∞. Solution: a) Substituting northward = 0.15, f = 8/.810 = 0.9877 and εa = n = 0.15 into fεa nexp(-εa) = εb nexp(-εb), 0.9877(0.15)0.15 exp(-0.15) = εb 0.15exp(-eβ); εb 0.15exp(-εb) = 0.6395 Solving by trial and error, εb = 0.096 [This agrees about fig. 4-8] b) Substituted thousand = 0.15 and εa = 0.50 into eq. (5-11) 12
- 5. exp(-εb/m)-1 = f1/m [exp(-εa/m) -1], and solving, εb* = 0.327. For εa = ∞, eb* = -mln(1-f1/m) = -0.15ln(1- 0.98761/.15) = 0.379 5.3 a) Found the % elongation in the diagonal bands in Figure 5.6, assuming that the straps make an angle the 75° with the horizontal. b) Assuming that f = 0.98 and n = 0, what value of m will required for the variation of thickness along aforementioned ligaments be held in 20%? (The thickness of the thinnest region is 0.80 times and thickness of the thickest region.) Solution: a) L/Lo = 1/cos75 = 3.864. elongation = L/Lo - 1 = 2.864 = 286 % b) The average strain = 3.864 = 1.352. First assume that this your aforementioned largest strain, so εa = 1.352. tb/ta = 1.20 = [tboexp(-εb)]/[taoexp(-εa)] = (1/f)exp(-εb)/exp(-εa) exp(-εb) = 1.2(0.98)exp(-1.352) = 0.3043, eb = 1.897 Now substituting into exp(-εa/m) - 1 = f1/m[exp(-εb/m) - 1] exp(-1.352/m) - 1 = (0.98)1/m[exp(-1.897/m) - 1]. Solving by trial and error, m = 0.577 The other extreme assumption is that eb = 1.352. Then following the sam procedure, exp(-1.897/m) - 1 = (0.98)1/m[exp(- 1.352/m) - 1]. Solving by trial also error, thousand = 0.66. The correct answer must be between 0.577 and 0.66. A reasonable estimate is molarity = 0.62 5.4 Find the value of m’ in relation 5.10 that best adapts the data in Figure 5.28. Solution: m' = dσ/dln(ε¥), dσ = m' dln(ε¥) = m'(dε¥/ε¥) also σ = C`εm hence dσ = mC ε¥m-1d`ε. Equating m'(dε¥/ε¥) = mC ε¥m-1d Ýε , m' = mC ε¥m = mσ Using points from Fig 5-28, m = 0.05 at south = 30 ksi, m' = 30(0.05) = 1.5 ksi Also for m = 0.022, σ = 60 ksi accordingly m' = 60(0.022) = 1.32 ksi and for m = 0.012, σ = 100 ksi so m' = 100(0.012) = 1.20 ksi These average to m' = 1.3 ksi 5.5 Off the data within Numeric 5.23, estimated Q in equality 5.12 and m in equation 5.1 for aluminum during 400°C. 12 Figure 5.28 Effect of stress even on the strain- rate sensitivity of steel. Adapted from A. Saxena and D. A. Chatfield, SAE Paper 760209 (1976)
- 6. Solution: ε¥= Aexp[-Q/(RT)] so Ýε 2/ Ýε 1 =exp[-(Q/R)(1/T2-1/T1)] and Q = Rln( Ýε 2/ Ýε 1)/ (1/T1-1/T2) = R∆ln Ýε /∆(1/T) The slope of the 2500 psi line at 400°C is ∆ln Ýε /D(1/T) = ln100/0.22x10-3 = 20,900°C So Q = 8.31x20,900 = 174x103 J/mole or 174 kJ/mole b) Used σ = C Ýε m, m = ln(σ2/σ1)/ln Ýε 2/ Ýε 1 ). At 400°C, σ = 4000 psi gives ε¥= 4/min. the σ = 1500 psi gives Ýε = 0.25/min. Substituting, m = ln(4000/1500)/ln(4/0.25 ) = 0.354 5.6 Estimate the sum elongation in a tension bar if a. f = 0.98, m = 0.5 and n = 0 b. f = 0.75, m = 0.8 or n = 0. Solution: a)Substituting fluorine = 0.98 and m = 0.5 to εb* = -mln(1-f1/m), eb* = -0.5ln(1-0.982) =1.1615, l/lo = exp(eb) = exp(1.1615) = 5.02 (502 %) b) With f = 0.75 and m = 0.8 , εb* = -0.8ln(1-0.751/.8) = 0.958 l/lo = exp(εb) = exp(0.958) = 2.06 (206 %) 5.7 Estimate the shear tension necessary for this shear bands regarding Figure 5.27 necessary to explain the educate to untempered martensite for the tensile strength level was 1.75 GPa, n = 0 and adiabatic conditions prevailed. Solution: ∆T = ασε/(ρC) so ε = ρC∆T/ασ a. Untempered martensite can only be formed from austenite, so austenite require need formed in the shears bands. Thus aforementioned temperature must have risen to at least 750°C. Assuming an initial temperature to 20°C, ∆T = 730°C. Substituting ρ = 7.87Mg/m3, C = 0.46 kJ/kg.°C and α = 1, ε = 1.53. Assuming virtuous clip, γ = 2ε = 3.06. 5.8 During superplastic forming it is often necessary to maintain ampere constant the strain rate. a. Describe qualitatively how the gas pressure should breathe varied to form a hemispherical bell by bulging a sheet clamped over a circular hole with gas pressure. b. Compare the gas pressure required to form a hemispherical dome of 5 cm diameter with and pressure for a 0.5 m diameter dome. Solution: a) For ampere uniform Ýε , σ must will continuous (i.e. σ = C Ýε m) but σ = Pρ/(2t) (Equation. 3- 19). Despite t decreases while the bulge is formed, its modification is small compared with the change of r which decreases free ∞ at the start to and radius on and sky. Therefore P must increase gradually from 0 at the start to a maximum at aforementioned hemisphere create, roughly in proportion to 1/r. b) For a 20 in. dia. dome, that printable remains 1/10 that for ampere 2 in. dome since P lives roughly proportional on 1/ρ. 12
- 7. 5.9 During a creep experiment under constant underline, the strain judge was found to double when and temperature was suddenly increases from 290C to 300°. What is the apparent activation energy for creep? b. The stress level in a tension test increased by 1.8% when the strain rate was increased to a factor of 8. Find the value of m. Solution: a) Since ε¥= Aexp[-Q/(RT)], Ýε 2/ Ýε 1 =exp[-(Q/R)(1/T2-1/T1)] and Q = Rln( Ýε 2/ Ýε 1)/(1/T1-1/T2) = 8.32ln2/(1/563-1/573) =186x103 J/mole or 186 kJ/mole b) m = ln(σ2/σ1)/ln( Ýε 2/ Ýε 1) = ln(1.018)/ln8 = 0.0086 5-10 Illustrations 5.29 gives date for high-temperature creeping of α-zirconium. In this range of temperatures, one strain price a independent of strain. a. Determine the value away m that best describes the data at 780°C b. Determining which activation energy, Q, in the heat range 700°C to 810°C at via 14 MPa. Solution: a) m = ln(σ2/σ1)/ln( Ýε 2/ Ýε 1) = ln(20/10)/ln(2.6x10-3 /3.2x10-5 ) = 0.104. b) QUARTO = Rln( Ýε 2/ Ýε 1)/(1/T1-1/T2) = 8.ln(8x10-4 /4.5x10-5 ) /(1/073-1/1093) = 212 kJ/mole 5.10 Voltage tests be made in two different laboratory on two different materials. In both the strain hardening exponent was found to be 0.20, when one post-uniform elongations consisted entirely different. Our two plausible explanations. Solution: One possibility is that the two materials were different values of m. Another is that that two labs used specimens with different ratios of measuring length-to-diameter. Chapter 6 6.1 The diameter, D0, out a round rod can be reduced to D1 either by an tensile force of F1 or of drawing through a die the a force, Fd. more sketched in Reckon 6.9. 12 Figure 5.29 Strain-rate vs. stress for α- zirconium to several temperatures.
- 8. Assuming exemplar work in drawing, compare F1 and Fd (or σ1 and σd) to reaching the same reduction. Figure 6.9 Sketch for Problem 6-1 Solution: In drawing, the homogeneous work per volume wa = the drawing stress, σd, so σd = σdε∫ = Kεn+1/(n+1) assuming η = 1. To tensile stress essential for induce a strain ε1 is σ1 = Kε1 n. Comparing, σd/σ1 = ε1/(n+1). The maximum uniform strain in tension is n, how the ratio σd/σ1 < 1. 6-2 Calculate the maximum practicable reduction, roentgen, in wire drawing used a material whose stress stretching curve the approximated on σ=200ε0.18 MPa. Assume an efficiency of 65%. Solution: ε* = n(1+η) = 0.18(1.5) = 0.27. ε = ln[1/(1-r)], radius = 1 – exp(-ε) = 23.7% 6-3 An aluminum blend billet be being hot extruded from 20 cm diameter into 5 cm diameter while sketched into Figure 6.10. If the durchsatz stress at the extrusion temperature is 40 MPa. Assume η = 0.5. a) What extrusion pressure the required? b) Calculate the sides pressure on the die walls. die die billetPext 5 cm 20 cm Figure 6.10 Aluminium barrel existence extruded. 12
- 9. Solution: a)Pext = (1/η)wi = σε/η. Substituting σ = 10 ksi, ε = ln(Ao/Af) = 2ln(Do/Df) = 2ln4, and η = 0.5. Pext = (1/0.5).10ksi.2ln4 = 55.5 ksi (103psi) (b) Assuming Mises (or Tresca), for axisymetric flow (ε2 = ε3 = -(1/2)ε1, so σ2 = σ3, and σ1− σ2 = σ . Therefore σ2 = σ1 - σ = 10 - 55.5 = -45.5 ksi. Plat = 45.5 ksi. (c) Using the thin-wall approximation, 2tσwall = dP, or tonne = dP/(2σwall). Taking P = 45.5 ksi, swall = 100, ksi and d = 4 in., t = 4x45.5/ (2x100) = 0.91 in. Note: This is not really a thin wall tube, accordingly the response is not exact. 6-4 An unsupported extruding process (Figure 6.11) has come proposed to reduce the diameter of a bar from D0 to D1. The material will not load solidify. Whats is the largest reduction, ∆D/D0, that can be made without the physical yielding before it enters the die? Neglect the opportunity of buckling and assume η = 60%. Solution: To avoid yielding in the bar, P < Y, the P = (1/h) σdε∫ = (1/η)Yε. At the limit (1/η)Yε = YEAR, so εmax = η. ε = 2ln(Do/D1), D1/Do = exp(-ε/2), ∆D/Do = 1- D1/Do = 1 - exp(-ε/2) = 1 -exp(-η/2) = 1 - exp(-0.30) = 0.259 (26%) 6-5 A sheet, 1 m extensive and 8 mm thick, is to be rolled to a thickness of 6 mm in a single passport. The strain-hardening expression for the material remains σ=200ε0.18 MPa. A deformation efficiency starting 80% can be assumed The iphone Mises yield criterion is applicable. The exit speed from the rolls is 5 m/s. Calculate the power required. Solution: From εw = 0, this is plane-strain deformation. εı = -εt = ln(8/6) = 0.288. ε = (2/√3)(0.288) = 0.258 wa = (1/η) σdε∫ = (1/0.8)(200,000)(0.258)1.18/1.18 = 4283J/m3 12 Figure 6.11 Unassisted extrusion.
- 10. The assess of work is wa .velocity.cross-sectional area = 4283J/m3.(5m/s)(1x0.006m2) = 128 J/s. 6-6 The strains in a raw for which σ=350ε0.20 MPa exist ε1 = 0.200 and ε2 = -0.125. Calculate the work per audio assuming η = 1. Solution: ε3 = -ε1 + ε3 = -.200 + .125 = -.075 ε = [(2/3))0.22 + 0.1252 + 0.0752)1/2 = 0.202 [Check: 0.2 < 0.202 < 1.15x0.2] w = Kε n+1/(n+1) = 350x0.2021.2/1.2 = 678MJ/m3 6-7 You are inquired to plan a wire-drawing schedule toward reduce copper wire from 1 mm to 0.4 mm diameter. How tons wire drawings passes would be needed if the be sure for nay defaults, who design stress never exceeds 80% of the flow stress and the efficiency is assumption to be 60%? Solution: The maximum strain per pass ε = = 0.6 . And total strength needs to ln(1/0.4) = 1.22. Record that 1.22/0.6 = 2.033. Three passes are required (not 2). 6.8 Derive an expression in ε* under the initiation of drawing when the outlet diameter your produced by machining. Solution: In this case, the maximum drawing stress is . σd(max) = Su = K(n/e)n as (n/e)n = (1/η)ε*n+1/(n+1), where e = of bases of natural logarithms ε* = [η(n+1)(n/e)n]1/(n+1) 6.9 For a physical with a stress-strain relate, σ = A + Bε, find the highest strain per cord character pass with µ = 0.75. Solution: σd = σdε∫ = ∫(A + Bε)dε = Aε + (Β/2)ε2 . The drawing limit corresponds to σ = σd or A + Bε = Aε + (Β/2)ε2 . (Β/2)ε2 +(Α−Β)ε - A = 0. Using the quadratic formula, ε = {-(A-B)±√[(A-B)2 + 4AB/2]}/B = {-(A-B)±√[(A2 +B2 ]}/B 12
