Chapter -1 From Tables A-0, A-1, A-, and A-4c, (a) UNS G1000 HR: S ut = 80 (55) MPa (kpsi), S yt = 10 (0) Mpa (kpsi) Ans. (b) SAE 1050 CD: S ut = 90 (100) MPa (kpsi), S yt = 580 (84) Mpa (kpsi) Ans. (c) AISI 1141 Q&T at 540C (1000F): S ut = 89 (10) MPa (kpsi), S yt = 75 (111) Mpa (kpsi) Ans. (d) 04-T4: S ut = 44 (4.8) MPa (kpsi), S yt = 9 (4.0) Mpa (kpsi) Ans. (e) Ti-Al-4V annealed: S ut = 900 (10) MPa (kpsi), S yt = 80 (10) Mpa (kpsi) Ans. - (a) Maximize yield strength: Q&T at 45C (800F) Ans. (b)maximize elongation: Q&T at 50C (100F) Ans. - Conversion of kn/m to kg/ m multiply by 1(10 ) / 9.81 = 10 AISI 1018 CD steel: Tables A-0 and A-5 S y 7010 47.4 kn m/kg Ans. 7.510 011-T aluminum: Tables A- and A-5 S y 1910. kn m/kg Ans..10 Ti-Al-4V titanium: Tables A-4c and A-5 S y 8010 187 kn m/kg Ans. 4.410 ASTM No. 40 cast iron: Tables A-4a and A-5.Does not have a yield strength. Using the ultimate strength in tension S ut 4.5.8910 40.7 kn m/kg Ans 70.10-4 AISI 1018 CD steel: Table A-5 E 0.010 1010 0.8 in Ans. 011-T aluminum: Table A-5 E 10.410 1010 0.098 in Ans. Ti-Al-V titanium: Table A-5 Chapter - Rev. D, Page 1/19
E 1.5 10 0.10 1010 in Ans. No. 40 cast iron: Table A-5 E 14.510 55.810 0.0 in Ans. -5 From Table A-5 E G G(1 v) E v G 0.0 11.5 Steel: v 0.04 Ans. 11.5 Aluminum: Beryllium copper: 10.4.90 v 0. Ans..90 18.0 7.0 v 0.8 Ans. 7.0 14.5.0 Gray cast iron: v 0.08 Ans..0 - (a) A 0 = (0.50) /4, = P i / A 0 For data in elastic range, = l / l 0 = l / l l l0 l A0 For data in plastic range, 1 1 l0 l0 l0 A On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1-7. Figure (b) shows data points 1-1. Figure (c) shows the complete range. Note: The exact value of A 0 is used without rounding off. (b) From Fig. (a) the slope of the line from a linear regression is E = 0.5 Mpsi Ans. From Fig. (b) the equation for the dotted offset line is found to be = 0.5(10 ) 1 000 (1) The equation for the line between data points 8 and 9 is = 7.0(10 5 ) + 4 900 () Chapter - Rev. D, Page /19
Solving Eqs. (1) and () simultaneously yields = 45. kpsi which is the 0. percent offset yield strength. Thus, S y = 45. kpsi Ans. The ultimate strength from Figure (c) is S u = 85. kpsi Ans. The reduction in area is given by Eq. (-1) is A0 A f 0.1987 0.1077 R 100 100 45.8 % Ans. A 0.1987 0 Data Point P i l, A i 1 0 0 0 0 1000 0.0004 0.0000 50 000 0.000 0.0000 1005 4 000 0.001 0.00050 15097 5 4000 0.001 0.0005 010 7000 0.00 0.00115 57 7 8400 0.008 0.00140 47 8 8800 0.00 0.00180 4485 9 900 0.0089 0.00445 498 10 8800 0.1984 0.00158 4485 11 900 0.1978 0.0041 498 1 9100 0.19 0.019 45795 1 100 0.194 0.081 48 14 1500 0.1875 0.05980 749 15 17000 0.15 0.71 85551 1 1400 0.107 0.507 851 17 14800 0.1077 0.8450 74479 (a) Linear range Chapter - Rev. D, Page /19
(b) Offset yield (c) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of S y, S ut, and R is SAE 1045 HR with S y = 45 kpsi, S ut = 8 kpsi, and R = 40 %. Ans. -7 To plot true vs., the following equations are applied to the data. P true A Eq. (-4) Chapter - Rev. D, Page 4/19
l ln for 0 l 0.008 in l0 A0 ln for l 0.008 in A (0.50) where A0 0.1987 in 4 The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log vs log The curve fit gives m = 0.0 log 0 = 5.185 0 = 15. kpsi Ans. For 0% cold work, Eq. (-14) and Eq. (-17) give, A = A 0 (1 W) = 0.1987 (1 0.) = 0.1590 in A 0.1987 A 0.1590 0 ln ln 0.1 m 0.0 Eq. (-18): Sy 0 15.(0.1) 108.4 kpsi Ans. Eq. (-19), with S 85. from Prob. -, u Su 85. S u 107 kpsi Ans. 1W 10. P L A true log log true 0 0 0.198 71 0 0 1000 0.0004 0.198 71 0.000 50.88 -.99 01.701 774 000 0.000 0.198 71 0.000 10 04.78 -.5 94 4.00 804 000 0.001 0.198 71 0.000 5 15 097.17 -.01 14 4.178 895 4000 0.001 0.198 71 0.000 5 0 19.55 -.187 4.0 84 7000 0.00 0.198 71 0.001 149 5.7 -.99 55 4.54 87 8400 0.008 0.198 71 0.001 99 4 7.0 -.854 18 4. 05 8800 0.00 0.198 4 0.001 575 44 54.84 -.80 1 4.4 941 900 0.0089 0.197 8 0.004 04 4 511. -. 85 4.7 5 9100 0.19 0.01 1 4 57. -1.91 05 4. 11 100 0.19 4 0.0 84 8 07.07-1.491 01 4.8 9 1500 0.187 5 0.058 08 81 0.7-1.5 9 4.908 84 17000 0.15 0.40 08 108 75.0-0.19 4 5.0 49 1400 0.10 7 0.418 95 15 478.0-0.77 8 5.098 58 14800 0.107 7 0.1 511 17 418.80-0.1 89 5.18 04 Chapter - Rev. D, Page 5/19
-8 Tangent modulus at = 0 is E 5000 0 0. 10 0 5 10 psi Ans. At = 0 kpsi Chapter - Rev. D, Page /19
E (10 - ) 1910 1.5 110 0 14.0 10 psi Ans. (kpsi) 0 0 0.0 5 0.44 10 0.80 1 1.0 19 1.5.0.8 40.4 4 4.0 49 5.0 54-9 W = 0.0, (a) Before cold working: Annealed AISI 1018 steel. Table A-, S y = kpsi, S u = 49.5 kpsi, 0 = 90.0 kpsi, m = 0.5, f = 1.05 After cold working: Eq. (-1), u = m = 0.5 Eq. (-14), A0 1 1 1.5 A i 1W 10.0 Eq. (-17), A0 i ln ln1.5 0. u Ai Eq. (-18), S m 0.5 y 0 i 90 0. 1.8 kpsi Ans. 9% increase Ans. Eq. (-19), Su 49.5 S u 1.9 kpsi 1W 10.0 Ans. 5% increase Ans. Su 49.5 (b) Before: 1.55 S y After: S u 1.9 1.00 Ans. S 1.8 y Lost most of its ductility -10 W = 0.0, (a) Before cold working: AISI 11 HR steel. Table A-, S y = 8 kpsi, S u = 1.5 kpsi, 0 = 110 kpsi, m = 0.4, f = 0.85 After cold working: Eq. (-1), u = m = 0.4 Chapter - Rev. D, Page 7/19
Eq. (-14), A0 1 1 1.5 A i 1W 10.0 Eq. (-17), A0 i ln ln1.5 0. u A i 0.4 m Eq. (-18), S y 0 i 110 0. 7.7 kpsi Ans. 174% increase Ans. Su 1.5 Eq. (-19), S u 7.9 kpsi Ans. 5% increase Ans. 1W 10.0 Su 1.5 (b) Before:.0 S y 8 After: S u 7.9 1.00 Ans. S 7.7 y Lost most of its ductility -11 W = 0.0, (a) Before cold working: 04-T4 aluminum alloy. Table A-, S y = 4.0 kpsi, S u = 4.8 kpsi, 0 = 100 kpsi, m = 0.15, f = 0.18 After cold working: Eq. (-1), u = m = 0.15 A0 1 1 Eq. (-14), 1.5 A i 1W 10.0 A0 Eq. (-17), i ln ln1.5 0. f Material fractures. Ans. Ai -1 For H B = 75, Eq. (-1), S u =.4(75) = 95 MPa Ans. -1 Gray cast iron, H B = 00. Eq. (-), S u = 0.(00) 1.5 =.5 kpsi Ans. From Table A-4, this is probably ASTM No. 0 Gray cast iron Ans. -14 Eq. (-1), 0.5H B = 100 H B = 00 Ans. Chapter - Rev. D, Page 8/19
-15 For the data given, converting H B to S u using Eq. (-1) H B S u (kpsi) S u (kpsi) 0 115 15 11 145 11 145 4 117 189 5 117.5 180.5 5 117.5 180.5 5 117.5 180.5 118 194 118 194 9 119.5 1480.5 S u = 117 S u = 177 S u Su 117 117. 117 kpsi Ans. N 10 Eq. (0-8), 10 Su NSu 177 10 1 117. i ss 1.7 kpsi Ans. u N 1 9-1 For the data given, converting H B to S u using Eq. (-) H B S u (kpsi) S u (kpsi) 0 40.4 1.1 40.8 19.54 40.8 19.54 4 41. 1707.4 5 41.55 17.40 5 41.55 17.40 5 41.55 17.40 41.78 1745.58 41.78 1745.58 9 4.47 180.701 S u = 414.1 S u =1715. Chapter - Rev. D, Page 9/19
S u Eq. (0-8), Su 414.1 41.4 kpsi Ans. N 10 10 Su NSu 1715. 10 1 41.4 i ss 1.0 Ans. u N 1 9-17 (a) (b) ur 45.5 (0) 4.5 in lbf / in. Ans P L A A 0 / A 1 = P/A 0 0 0 0 0 1000 0.0004 0.000 5 0.9 000 0.000 0.000 10 04.78 000 0.0010 0.0005 15 097.17 4000 0.001 0.000 5 0 19.55 7000 0.00 0.001 15 5.7 8400 0.008 0.0014 4 7.0 8800 0.00 0.0018 44 85.0 900 0.0089 0.004 45 4 97.97 9100 0.19 0.01 91 0.01 91 45 794.7 100 0.194 0.0 811 0.0 811 47.5 1500 0.1875 0.059 80 0.059 80 7 49.0 17000 0.15 0.71 55 0.71 55 85 550.0 1400 0.107 0.50 7 0.50 7 8 51.17 14800 0.1077 0.845 059 0.845 059 74 479.5 From the figures on the next page, 5 1 ut Ai (4 000)(0.001 5) 45 000(0.004 45 0.001 5) i1 1 45 000 7 500 (0.059 8 0.004 45) 81 000 0.4 0.059 8 80 000 0.845 0.4.7 10 in lbf/in. Ans Chapter - Rev. D, Page 10/19
Chapter - Rev. D, Page 11/19
-18, -19 These problems are for student research. No standard solutions are provided. -0 Appropriate tables: Young s modulus and Density (Table A-5)100 HR and CD (Table A- 0), 1040 and 4140 (Table A-1), Aluminum (Table A-4), Titanium (Table A-4c) Appropriate equations: F F 4F For diameter, Sy d A /4 d S Weight/length = A, Cost/length = $/in = ($/lbf) Weight/length, Deflection/length = /L = F/(AE) With F = 100 kips = 100(10 ) lbf, y Material Young's Modulus Density Yield Strength Cost/lbf Diameter Weight/ length Cost/ length Deflection/ length units Mpsi lbf/in^ kpsi $/lbf in lbf/in $/in in/in 100 HR 0 0.8 0 $0.7.00 0.9400 $0.5 1.000E 0 100 CD 0 0.8 57 $0.0 1.495 0.4947 $0.15 1.900E 0 1040 0 0.8 80 $0.5 1. 0.55 $0.1.7E 0 4140 0 0.8 15 $0.80 0.878 0.1709 $0.14 5.500E 0 Al 10.4 0.098 50 $1.10 1.59 0.190 $0. 4.808E 0 Ti 1.5 0.1 10 $7.00 1.00 0.1 $0.9 7.7E 0 The selected materials with minimum values are shaded in the table above. Ans. -1 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three would favor steel, cast iron, or maybe a less common ferrous material. The expectation would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 7.95 lbf, the unit weight is determined to be W 7.95 lbf w 0.81 lbf/in 0.8 lbf/in Al [ (1 in) / 4]( in) which agrees well with the unit weight of 0.8 lbf/in reported in Table A-5 for carbon steel. Nickel steel and stainless steel have similar unit weights, but surface finish and darker coloring do not favor their selection. To select a likely specification from Table Chapter - Rev. D, Page 1/19
A-0, perform a Brinell hardness test, then use Eq. (-1) to estimate an ultimate strength of Su 0.5HB 0.5(00) 100 kpsi. Assuming the material is hot-rolled due to the rough surface finish, appropriate choices from Table A-0 would be one of the higher carbon steels, such as hot-rolled AISI 1050, 100, or 1080. Ans. - First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of.90 lbf, the unit weight is determined to be W.9 lbf w 0.10 lbf/in 0.10 lbf/in Al [ (1 in) / 4]( in) which agrees reasonably well with the unit weight of 0.098 lbf/in reported in Table A-5 for aluminum. No other materials come close to this unit weight, so the material is likely aluminum. Ans. - First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To further distinguish the material, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 9 lbf, the unit weight is determined to be W 9.0 lbf w 0.18 lbf/in 0. lbf/in Al [ (1 in) / 4]( in) which agrees reasonably well with the unit weight of 0. lbf/in reported in Table A-5 for copper. Brass is not far off (0.09 lbf/in ), so the deflection test could be used to gain additional insight. From the measured deflection and utilizing the deflection equation for an end-loaded cantilever beam from Table A-9, Young s modulus is determined to be Fl 1004 E 17.7 Mpsi 4 Iy (1) 4 (17 / ) which agrees better with the modulus for copper (17. Mpsi) than with brass (15.4 Mpsi). The conclusion is that the material is likely copper. Ans. -4 and -5 These problems are for student research. No standard solutions are provided. Chapter - Rev. D, Page 1/19
- For strength, = F/A = S A = F/S For mass, m = Al = (F/S) l Thus, f (M ) = /S, and maximize S/ ( = 1) In Fig. (-19), draw lines parallel to S/ From the list of materials given, both aluminum alloy and high carbon heat treated steel are good candidates, having greater potential than tungsten carbide or polycarbonate. The higher strength aluminum alloys have a slightly greater potential. Other factors, such as cost or availability, may dictate which to choose. Ans. -7 For stiffness, k = AE/l A = kl/e For mass, m = Al = (kl/e) l =kl /E Thus, f (M) = /E, and maximize E/ ( = 1) In Fig. (-1), draw lines parallel to E/ Chapter - Rev. D, Page 14/19
From the list of materials given, tungsten carbide (WC) is best, closely followed by aluminum alloys, and then followed by high carbon heat-treated steel. They are close enough that other factors, like cost or availability, would likely dictate the best choice. Polycarbonate polymer is clearly not a good choice compared to the other candidate materials. Ans. -8 For strength, = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (-b), p. 90 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in (ips) or m (SI). Thus, for a given cross section, Z =C (A) /, where C is a number. For example, for a circular cross section, C = 1 4. Then, for strength, Eq. (1) is Fl CA / Fl S A CS / () Chapter - Rev. D, Page 15/19
For mass, / / Fl F 5/ m Al l l / CS C S Thus, f (M) = /S /, and maximize S / / ( = /) In Fig. (-19), draw lines parallel to S / / From the list of materials given, a higher strength aluminum alloy has the greatest potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly not a good choice compared to the other candidate materials..ans. -9 Eq. (-), p. 5, applies to a circular cross section. However, for any cross section shape it can be shown that I = CA, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a Chapter - Rev. D, Page 1/19
constant. The moment of inertia is I = bh /1, and the area is A = bh. Then I = h(bh )/1 = cb (bh )/1 = (c/1)(bh) = CA, where C = c/1 (a constant). Thus, Eq. (-7) becomes 1/ kl A CE and Eq. (-9) becomes 1/ k 5/ m Al l 1/ C E 1/ E Thus, minimize f M, or maximize M. From Fig. (-1) 1/ E From the list of materials given, aluminum alloys are clearly the best followed by steels and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other candidate materials. Ans. -0 For stiffness, k = AE/l A = kl/e For mass, m = Al = (kl/e) l =kl /E Chapter - Rev. D, Page 17/19
So, f (M) = /E, and maximize E/. Thus, = 1. Ans. -1 For strength, = F/A = S A = F/S For mass, m = Al = (F/S) l So, f (M ) = /S, and maximize S/. Thus, = 1. Ans. - Eq. (-), p. 5, applies to a circular cross section. However, for any cross section shape it can be shown that I = CA, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a constant. The moment of inertia is I = bh /1, and the area is A = bh. Then I = h(bh )/1 = cb (bh )/1 = (c/1)(bh) = CA, where C = c/1. Thus, Eq. (-7) becomes 1/ kl A CE and Eq. (-9) becomes 1/ k 5/ m Al l 1/ C E 1/ E So, minimize f M, or maximize M. Thus, = 1/. Ans. 1/ E - For strength, = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (-b), p. 90 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in (ips) or m (SI). Thus, for a given cross section, Z =C (A) /, where C is a number. For example, for a circular cross section, C = 1 4. Then, for strength, Eq. (1) is Fl CA / Fl S A CS / () For mass, / / Fl F 5/ m Al l l / CS C S So, f (M) = /S /, and maximize S / /. Thus, = /. Ans. -4 For stiffness, k=ae/l, or, A = kl/e. Chapter - Rev. D, Page 18/19
Thus, m = Al = (kl/e )l = kl /E. Then, M = E / and = 1. From Fig. -1, lines parallel to E / for ductile materials include steel, titanium, molybdenum, aluminum alloys, and composites. For strength, S = F/A, or, A = F/S. Thus, m = Al = F/Sl = Fl /S. Then, M = S/ and = 1. From Fig. -19, lines parallel to S/ give for ductile materials, steel, aluminum alloys, nickel alloys, titanium, and composites. Common to both stiffness and strength are steel, titanium, aluminum alloys, and composites. Ans. Chapter - Rev. D, Page 19/19