Fundamentals of Materials Science and Engineering Solutions Ch 9-10

CH A PTE R 9 FA ILU RE 9.7 We are asked for the critical crack tip radius for an A l2 O 3 material. From E quation (9.1b) m Fracture will occur when thus m =2 o a t  1 /2 reaches the fracture strength of the material, which is given as E/10; E 10 =2 o a t  1 /2 O r, solving for  t t = 400a E2 o 2 From Table 7.1, E = 393 G Pa, and thus, t = (400)(2 × 10−3 mm)(275 MPa) 2 (393 × 103 MPa) 2 = 3.9 × 10−7 mm = 0.39 nm 9.8 We may determine the critical stress required for the propagation of a surface crack in soda-lime glass using E quation (9.3); taking the value of 69 G Pa (Table 7.1) as the modulus of elasticity, we get  c =  = 9.12* as 2E −5 × 102 )(0.30 m) N/m) (2)(69 ×()(5 109 N/m = 16.2 × 106 N/m 2 = 16.2 MPa This problem deals with a tensile specimen, a drawing of which is provided. (a) In this portion of the problem it is necessary to compute the stress at point P when the applied stress is 100 MPa (14,500 psi). In order to determine the stress concentration it is necessary to consult Figure 9.8c. From the geometry of the specimen, w/h = (25 mm)/(20 mm) = 1.25; furthermore, the r/h ratio is (3 mm) /(20 mm) = 0.15. U sing the w/h = 1.25 curve in Figure 9.8c, the Kt value at r/h = 0.15 is 1.7. A nd since Kt = ␴m ␴ , then o m 9.15* = Kt o = (1.7)(100 MPa) = 170 MPa (24,650 psi) This problem calls for us to determine the value of B, the minimum component thickness for which the condition of plane strain is valid using E quation (9.12) for the metal alloys listed in Table 9.1. 32 For the 2024-T3 aluminum alloy K lc 2 = (2.5) √ 44 MPa m 2 = 0.041 m = 41 mm (1.60 in .) B = 2 .5 345 MPa y For the 4340 alloy steel tempered at 260◦ C B = (2.5) 9.19 1640 MPa For this problem, we are given values of Klc , ␴, and Y for a large plate and are asked to determine the minimum length of a surface crack that will lead to fracture. A ll we need do is to solve for ac using E quation (9.14); therefore 1  √ 2= 55 MPa m 2 = 0.024 m = 24 mm (0.95 in .) K lc  (1)(200 MPa)  Y This problem first provides a tabulation of Charpy impact data for a ductile cast iron. ac = 1 (a) The plot of impact energy versus temperature is shown below. 140 120 100 Impact Energy, J 9.26 √ 50 MPa m 2 = 0.0023 m = 2.3 mm (0.09 in.) 80 60 40 20 0 -200 -150 -100 -50 0 Temperature, C (b) This portion of the problem asks us to determine the ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. From these data, this average is Average = 124 J + 6 J = 65 J 2 A s indicated on the plot by the one set of dashed lines, the ductile-to-brittle transition temperature according to this criterion is about −105◦ C. (c) A lso as noted on the plot by the other set of dashed lines, the ductile-to-brittle transition temperature for an impact energy of 80 J is about −95◦ C. 33 9.31 We are asked to determine the fatigue life for a cylindrical red brass rod given its diameter (8.0 mm) and the maximum tensile and compressive loads (+7500 N and −7500 N, respectively). The first thing that is necessary is to calculate values of max and min using E quation (7.1). Thus max = F max = Ao 2 do  2 7500 N = min F max 6 2 2 = 150 × 10 N/m = 150 MPa (22,500 psi) 8.0 × 210 ( ) F min = −3 m 2 do 2 −7500 N  = 6 2 2 = −150 × 10 N/m = −150 MPa ( −22,500 psi) ( ) 8.0 × 210−3 m Now it becomes necessary to compute the stress amplitude using E quation (9.23) as a = max − 2 min = 150 MPa − (−150 MPa) 2 = 150 MPa (22,500 psi) From Figure 9.46 for the red brass, the number of cycles to failure at this stress amplitude is about 1 × 105 cycles. This problem first provides a tabulation of fatigue data (i.e., stress amplitude and cycles to failure) for a brass alloy. (a) These fatigue data are plotted below. 300 Stress amplitude, MPa 9.33 200 100 5 6 7 8 Log cycles to failure 34 9 10 (b) A s indicated by one set of dashed lines on the plot, the fatigue strength at 5 × 105 cycles [log (5 × 105 ) = 5.7] is about 250 MPa. (c) A s noted by the other set of dashed lines, the fatigue life for 200 MPa is about 2 × 106 cycles (i.e., the log of the lifetime is about 6.3). 9.34 We are asked to compute the maximum torsional stress amplitude possible at each of several fatigue lifetimes for the brass alloy, the fatigue behavior of which is given in Problem 9.33. For each lifetime, first compute the number of cycles, and then read the corresponding fatigue strength from the above plot. (a) Fatigue lifetime = (1 yr)(365 days/yr)(24 h/day)(60 min/h)(1500 cycles/min) = 7.9 × 108 cycles. The stress amplitude corresponding to this lifetime is about 130 MPa. (c) Fatigue lifetime = (24 h)(60 min/h)(1200 cycles/min) = 2.2 × 106 cycles. The stress amplitude corresponding to this lifetime is about 195 MPa. 9.48 This problem asks that we determine the total elongation of a low carbon-nickel alloy that is exposed to a tensile stress of 40 MPa (5800 psi) at 538◦ C for 5000 h; the instantaneous and primary creep elongations are 1.5 mm (0.06 in.). From the 538◦ C line in Figure 9.43, the steady-state creep rate, ˙s , is about 0.15% /1000 h (or 1.5 × 10−4 % /h) at 40 MPa. The steady-state creep strain, s , therefore, is just the product of ˙s and time as s = ˙s × (time) = (1.5 × 10−4 % /h)(5000 h) = 0.75% = 7.5 × 10−3 Strain and elongation are related as in Equation (7.2); solving for the steady-state elongation, ls , leads to ls = lo s = (750 mm)(7.5 × 10−3 ) = 5.6 mm (0.23 in.) Finally, the total elongation is just the sum of this ls and the total of both instantaneous and primary creep elongations [i.e., 1.5 mm (0.06 in.)]. Therefore, the total elongation is 7.1 mm (0.29 in.). 9.52* The slope of the line from a log ˙ s versus log that is n= plot yields the value of n in E quation (9.33);  log ˙s  log We are asked to determine the values of n for the creep data at the three temperatures in Figure 9.43. This is accomplished by taking ratios of the differences between two log ˙ s and log values. Thus for 427◦ C n=  log ˙s  log = log (10−1 ) − log (10−2 ) log (85 MPa) − log (55 MPa) = 5 .3 and for 538◦ C n=  log ˙s  log = log (1.0) − log (10−2 ) = 4 .9 log (59 MPa) − log (23 MPa) 35 9.55* This problem gives ˙ s values at two different temperatures and 70 MPa (10,000 psi), and the stress exponent n = 7.0, and asks that we determine the steady-state creep rate at a stress of 50 MPa (7250 psi) and 1250 K. Taking the natural logarithm of E quation (9.34) yields ln ˙s = ln K 2 + n ln − Qc RT With the given data there are two unknowns in this equation—namely K2 and Q c . U sing the data provided in the problem we can set up two independent equations as follows: ln[1.0 × 10−5 (h) −1 ] = ln K 2 + (7.0) ln(70 MPa) − Qc (8.31 J/mol-K)(977 K) ln[2.5 × 10−3 (h) −1 ] = ln K 2 + (7.0) ln(70 MPa) − Qc (8.31 J/mol-K)(1089 K) Now, solving simultaneously for K2 and Q c leads to K2 = 2.55 × 105 (h) −1 and Q c = 436,000 J/mol. Thus it is now possible to solve for ˙ s at 50 MPa and 1250 K using E quation (9.34) as ˙s = K 2 n exp − Qc RT  436000 J/mol 5 ˙s = [2.55 × 10−1 (h) −1 ](50 MPa) 7.0 exp − (8.31 J/mol-K)(1250 K) = 0.118 (h)  9.D 1* This problem asks us to calculate the minimum Klc necessary to ensure that failure will not occur for a flat plate given an expression from which Y(a/W) may be determined, the internal crack length, 2a (20 mm), the plate width, W (90 mm), and the value of (375 MPa). First we must compute the value of Y(a/W) using E quation (9.10), as follows: Y(a/W) =  a tan W a  1 /2 W ()(10 mm) 1/2 = 1.021  ()(10 mm) tan 90 mm = 90 mm Now, using E quation (9.11) it is possible to determine Klc ; thus K lc = Y(a/W) √ a  √ √ = (1.021)(375 MPa) ()(10 × 10−3 m) = 67.9 MPa m (62.3 ksi in .) 9.D 7* We are asked in this problem to estimate the maximum tensile stress that will yield a fatigue life of 2.5 × 107 cycles, given values of ao , ac , m, A , and Y. Since Y is independent of crack length we may utilize E quation (9.31) which, upon integration, takes the form Nf = 1 −m/2 da  a ca ao A  m/2 ( ) m Y m 36 A nd for m = 3.5 1 Nf = −1.75 da  a ca ao 3.5 3.5 A  1.75( 1.33) Y 1 =− a 0.75 − a 0.75  c1  1.75( yields ) 3.5 Y 3.5 from this A expression Now, solving for    = 1 a 0.75 − a 0.75 1/3.5 1.33 o1 N f A  1.75Y 3.5 c  1.33 = = 178 .5 × 107 )(2 × 10−14 )( ) 1.75(1.4) 3.5 (2MPa o 1  (1.5 × 10−4 ) 0.75 − 1 (4.5 × 1/3.5 10−3 ) 0.75 This 178 MPa will be the maximum tensile stress since we can show that the minimum stress is a compressive one—when min is negative,  is taken to be max . If we take max = 178 MPa, and since m is stipulated in the problem to have a value of 25 MPa, then from Equation (9.21) min Therefore min =2 m − max = 2(25 MPa) − 178 MPa = −128 MPa is negative and we are justified in taking max to be 178 MPa. 9.D 16* We are asked in this problem to calculate the stress levels at which the rupture lifetime will be 5 years and 20 years when an 18-8 Mo stainless steel component is subjected to a temperature of 500◦ C (773 K). It first becomes necessary, using the specified temperature and times, to calculate the values of the Larson-Miller parameter at each temperature. The values of tr corresponding to 5 and 20 years are 4.38 × 104 h and 1.75 × 105 h, respectively. H ence, for a lifetime of 5 years T(20 + log t r ) = 773[20 + log (4.38 × 104 )] = 19.05 × 103 A nd for tr = 20 years T(20 + log t r ) = 773[20 + log (1.75 × 105 )] = 19.51 × 103 U sing the curve shown in Figure 9.47, the stress values corresponding to the five- and twenty-year lifetimes are approximately 260 MPa (37,500 psi) and 225 MPa (32,600 psi), respectively. 37 CH A PTE R 10 PHA SE D IAGRA MS 10.5 This problem asks that we cite the phase or phases present for several alloys at specified temperatures. (a) For an alloy composed of 90 wt% Z n-10 wt% Cu and at 400◦ C, from Figure 10.17, phases are present, and and ␩ C = 87 wt% Z n-13 wt% Cu C ␩ = 97 wt% Z n-3 wt% Cu (c) For an alloy composed of 55 wt% A g-45 wt% Cu and at 900◦ C, from Figure 10.6, only the liquid phase is present; its composition is 55 wt% A g-45 wt% Cu. 10.7 This problem asks that we determine the phase mass fractions for the alloys and temperatures in Problem 10.5. (a) For an alloy composed of 90 wt% Z n-10 wt% Cu and at 400◦ C, and and ␩ phases are present, C o = 90 wt% Z n C = 87 wt% Z n C ␩ = 97 wt% Z n Therefore, using modified forms of E quation (10.2b) we get W = 97 − 90 C␩ − Co = 0.70 = 97 − 87 C␩ − C W␩ = 90 − 87 Co − C = = 0.30 C␩ − C 97 − 87 (c) For an alloy composed of 55 wt% A g-45 wt% Cu and at 900◦ C, since only the liquid phase is present, then WL = 1.0. 10.9 This problem asks that we determine the phase volume fractions for the alloys and temperatures in Problem 10.5a, b, and c. This is accomplished by using the technique illustrated in E xample Problem 10.3, and the results of Problem 10.7. (a) This is a Cu-Z n alloy at 400◦ C, wherein C = 87 wt% Z n-13 wt% Cu C ␩ = 97 wt% Z n-3 wt% Cu W = 0.70 W ␩ = 0.30 ␳ Cu = 8.77 g/cm 3 ␳ Zn = 6.83 g/cm 3 38 U sing these data it is first necessary to compute the densities of the E quation (5.10a). Thus ␳ = = ␳␩ = = and ␩ phases using 100 C Zn( ) C Cu( ) + ␳ Zn ␳ Cu 100 = 7.03 g/cm 3 87 13 + 3 3 8.77 g/cm 6.83 g/cm 100 C Zn(␩) C Cu(␩) + ␳ Zn ␳ Cu 100 = 6.88 g/cm 3 97 3 + 6.83 g/cm 3 8.77 g/cm 3 Now we may determine the V and V ␩ values using E quation 10.6. Thus, W ␳ V = W W␩ + ␳ ␳␩ 0.70 7.03 g/cm 3 = 0.70 = 0.30 0.70 + 7.03 g/cm 3 6.88 g/cm 3 W␩ ␳␩ V␩ = W W␩ + ␳ ␳␩ 0.30 6.88 g/cm 3 = 0.30 = 0.70 0.30 + 6.88 g/cm 3 7.03 g/cm 3 10.12 (a) We are asked to determine how much sugar will dissolve in 1500 g of water at 90◦ C. From the solubility limit curve in Figure 10.1, at 90◦ C the maximum concentration of sugar in the syrup is about 77 wt% . It is now possible to calculate the mass of sugar using E quation (5.3) as m sugar × 100 m sugar + m water m sugar 77 wt% = × 100 m sugar + 1500 g C sugar (wt% ) = Solving for msugar yields msugar = 5022 g. (b) A gain using this same plot, at 20◦ C the solubility limit (or the concentration of the saturated solution) is about 64 wt% sugar. 39 (c) The mass of sugar in this saturated solution at 20◦ C (m′ sugar) may also be calculated using E quation (5.3) as follows: 64 wt% = m ′sugar m ′sugar + 1500 g × 100 which yields a value for m′sugar of 2667 g. Subtracting the latter from the former of these sugar concentrations yields the amount of sugar that precipitated out of the solution upon cooling m′′sugar; that is m ′′sugar = m sugar − m ′ sugar = 5022 g − 2667 g = 2355 g 10.21 U pon cooling a 50 wt% Pb-50 wt% Mg alloy from 700◦ C and utilizing Figure 10.18: (a) The first solid phase forms at the temperature at which a vertical line at this composition intersects the L-(␣ + L) phase boundary—i.e., about 550◦ C; (b) The composition of this solid phase corresponds to the intersection with the ␣-( ␣ + L) phase boundary, of a tie line constructed across the ␣ + L phase region at 550◦ C—i.e., 22 wt% Pb-78 wt% Mg; (c) Complete solidification of the alloy occurs at the intersection of this same vertical line at 50 wt% Pb with the eutectic isotherm—i.e., about 465◦ C; (d) The composition of the last liquid phase remaining prior to complete solidification corresponds to the eutectic composition—i.e., about 66 wt% Pb-34 wt% Mg. 10.24 (a) We are given that the mass fractions of ␣ and liquid phases are both 0.5 for a 30 wt% Sn-70 wt% Pb alloy and asked to estimate the temperature of the alloy. U sing the appropriate phase diagram, Figure 10.7, by trial and error with a ruler, a tie line within the ␣ + L phase region that is divided in half for an alloy of this composition exists at about 230◦ C. (b) We are now asked to determine the compositions of the two phases. This is accomplished by noting the intersections of this tie line with both the solidus and liquidus lines. From these intersections, C␣ = 15 wt% Sn, and CL = 42 wt% Sn. 10.28 This problem asks if it is possible to have a Cu-A g alloy of composition 50 wt% A g-50 wt% Cu that consists of mass fractions W␣ = 0.60 and W␤ = 0.40. Such an alloy is not possible, based on the following argument. U sing the appropriate phase diagram, Figure 10.6, and, using E quations (10.1) and (10.2) let us determine W␣ and W␤ at just below the eutectic temperature and also at room temperature. A t just below the eutectic, C␣ = 8.0 wt% A g and C␤ = 91.2 wt% A g; thus, W␣ = 91.2 − 50 C␤ − Co = = 0.50 C␤ − C␣ 91.2 − 8 W ␤ = 1.0 − W ␣ = 1.0 − 0.5 = 0.50 Furthermore, at room temperature, C␣ = 0 wt% A g and C␤ = 100 wt% A g; employment of E quations (10.1) and (10.2) yields W␣ = 100 − 50 C␤ − Co = = 0.50 C␤ − C␣ 100 − 0 A nd, W␤ = 0.50. Thus, the mass fractions of the ␣ and ␤ phases, upon cooling through the ␣ + ␤ phase region will remain approximately constant at about 0.5, and will never have values of W␣ = 0.60 and W␤ = 0.40 as called for in the problem. 40 10.35* This problem asks that we determine the composition of a Pb-Sn alloy at 180◦ C given that W␤′ = 0.57 and We = 0.43. Since there is a primary ␤ microconstituent present, then we know that the alloy composition, Co , is between 61.9 and 97.8 wt% Sn (Figure 10.7). Furthermore, this figure also indicates that C␤ = 97.8 wt% Sn and Ceutectic = 61.9 wt% Sn. A pplying the appropriate lever rule expression for W␤′ W ␤′ = C o − C eutectic C o − 61.9 = 0.57 = C ␤ − C eutectic 97.8 − 61.9 and solving for Co yields Co = 82.4 wt% Sn. 10.47* We are asked to specify the value of F for G ibbs phase rule at point B on the pressure-temperature diagram for H 2 O. G ibbs phase rule in general form is P+F =C+N For this system, the number of components, C, is 1, whereas N, the number of noncompositional variables, is 2—viz. temperature and pressure. Thus, the phase rule now becomes P+F =1+2=3 Or F =3−P where P is the number of phases present at equilibrium. A t point B on the figure, only a single (vapor) phase is present (i.e., P = 1), or F =3−P =3−1=2 which means that both temperature and pressure are necessary to define the system. 10.54 This problem asks that we compute the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.94. A pplication of the lever rule [of the form of E quation (10.12)] yields W ␣ = 0.94 = C Fe3 C − C ′ C Fe3 C − C ␣ o = 6.70 − C ′ o 6.70 − 0.022 and solving for C′o C ′o = 0.42 wt% C 10.59 This problem asks that we determine the carbon concentration in an iron-carbon alloy, given the mass fractions of proeutectoid ferrite and pearlite. From E quation (10.20) W p = 0.714 = C ′ o − 0.022 0.74 which yields C′o = 0.55 wt% C. 41 10.64 This problem asks if it is possible to have an iron-carbon alloy for which W␣ = 0.846 and WFe 3 C′ = 0.049. In order to make this determination, it is necessary to set up lever rule expressions for these two mass fractions in terms of the alloy composition, then to solve for the alloy composition of each; if both alloy composition values are equal, then such an alloy is possible. The expression for the mass fraction of total ferrite is W␣ = 6.70 − C o C Fe3 C − C o = = 0.846 C Fe3 C − C ␣ 6.70 − 0.022 Solving for this Co yields Co = 1.05 wt% C. Now for WFe 3 C′ we utilize E quation (10.23) as W Fe3 C ′ = C ′ 1 − 0.76 5.94 = 0.049 This expression leads to C′ 1 = 1.05 wt% C. A nd, since Co = C′ 1 , this alloy is possible. 10.70 This problem asks that we determine the approximate Brinell hardness of a 99.8 wt% Fe-0.2 wt% C alloy. First, we compute the mass fractions of pearlite and proeutectoid ferrite using E quations (10.20) and (10.21), as C ′ o − 0.022 = 0.20 − 0.022 = 0.24 0.74 0.74 ′ o 0.76 − 0. 20 0.76 − C = W ␣′ = = 0.76 0.74 0.74 Wp = Now, we compute the Brinell hardness of the alloy as H B alloy = H B ␣′ W ␣′ + H B p W p = (80)(0.76) + (280)(0.24) = 128 10.73* We are asked to consider a steel alloy of composition 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C. (a) From Figure 10.36, the eutectoid temperature for 6 wt% Ni is approximately 650◦ C (1200◦ F). (b) From Figure 10.37, the eutectoid composition is approximately 0.62 wt% C. Since the carbon concentration in the alloy (0.2 wt% ) is less than the eutectoid, the proeutectoid phase is ferrite. (c) A ssume that the ␣-(␣ + Fe 3 C) phase boundary is at a negligible carbon concentration. Modifying E quation (10.21) leads to W ␣′ = 0.62 − C ′ o 0.62 − 0.20 = 0.62 − 0 0.62 = 0.68 Likewise, using a modified E quation (10.20) Wp = C′ o − 0 0.20 = 0.32 = 0.62 0.62 − 0 42