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 = 690 (100) MPa (kpsi), S yt = 580 (84) MPa (kpsi) Ans. (c) AISI 1141 Q&T at 540 C (1000 F): S ut = 896 (10) MPa (kpsi), S yt = 765 (111) MPa (kpsi) Ans. (d) 04-T4: S ut = 446 (64.8) MPa (kpsi), S yt = 96 (4.0) MPa (kpsi) Ans. (e) Ti-6Al-4V annealed: S ut = 900 (10) MPa (kpsi), S yt = 80 (10) MPa (kpsi) Ans. - (a) Maximize yield strength: Q&T at 45 C (800 F) Ans. (b) Maximize elongation: Q&T at 650 C (100 F) 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 70( 10 ) 47.4 kn m/kg Ans. ρ = 76.5( 10) = 011-T6 aluminum: Tables A- and A-5 S y 169( 10 ) 6. kn m/kg Ans. ρ = 6.6( 10) = Ti-6Al-4V titanium: Tables A-4c and A-5 S y 80( 10 ) 187 kn m/kg Ans. ρ = 4.4( 10) = 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( 6.89)( 10 ) 40.7 kn m/kg Ans ρ = 70.6( 10) = -4 AISI 1018 CD steel: Table A-5 6 E 0.0( 10 ) 6 106( 10 ) γ = 0.8 in Ans. 011-T6 aluminum: Table A-5 6 E 10.4( 10 ) 6 106( 10 ) γ = 0.098 in Ans. Shigley s MED, 10 th edition Chapter Solutions, Page 1/
Ti-6Al-6V titanium: Table A-5 6 E 16.5( 10 ) 6 10( 10 ) γ = 0.160 in Ans. No. 40 cast iron: Table A-5 6 E 14.5( 10 ) 6 55.8( 10 ) γ = 0.60 in Ans. -5 E G G(1 + v) = E v = G Using values for E and G from Table A-5, 0.0 ( 11.5) Steel: v = = 0.04 Ans. ( 11.5) The percent difference from the value in Table A-5 is 0.04 0.9 0.9 = 0.0411 = 4.11 percent Ans. ( ) 10.4.90 Aluminum: v = = 0. Ans. (.90) The percent difference from the value in Table A-5 is 0 percent Ans. Beryllium copper: ( ) ( ) 18.0 7.0 v = = 0.86 Ans. 7.0 The percent difference from the value in Table A-5 is 0.86 0.85 0.85 = 0.0051 = 0.51 percent Ans. ( ) 14.5 6.0 Gray cast iron: v = = 0.08 Ans. ( 6.0) The percent difference from the value in Table A-5 is 0.08 0.11 = 0.014 = 1.4 percent Ans. 0.11-6 (a) A 0 = π (0.50) /4 = 0.1987 in, σ = P i / A 0 Shigley s MED, 10 th edition Chapter Solutions, Page /
For data in elastic range, = l / l 0 = l / l l l0 l A0 For data in plastic range, ò = = = 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 σ = 0.5(10 6 ) 61 000 (1) The equation for the line between data points 8 and 9 is σ = 7.60(10 5 ) + 4 900 () Solving Eqs. (1) and () simultaneously yields σ = 45.6 kpsi which is the 0. percent offset yield strength. Thus, S y = 45.6 kpsi Ans. The ultimate strength from Figure (c) is S u = 85.6 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 σ 1 0 0 0 0 1000 0.0004 0.0000 50 000 0.0006 0.0000 10065 4 000 0.001 0.00050 15097 5 4000 0.001 0.00065 010 6 7000 0.00 0.00115 57 7 8400 0.008 0.00140 47 8 8800 0.006 0.00180 4485 9 900 0.0089 0.00445 4698 10 8800 0.1984 0.00158 4485 11 900 0.1978 0.00461 4698 1 9100 0.196 0.019 45795 1 100 0.194 0.081 6648 14 1500 0.1875 0.05980 7649 15 17000 0.156 0.716 85551 16 16400 0.107 0.507 851 17 14800 0.1077 0.84506 74479 Shigley s MED, 10 th edition Chapter Solutions, Page /
Stress (psi) 50000 y =.05E+07x - 1.06E+01 40000 0000 0000 10000 0 0.000 0.001 0.001 0.00 Strain (a) Linear range Series1 Linear (Series1) Stress (psi) 50000 45000 Y 40000 5000 0000 5000 0000 15000 10000 5000 0 0.000 0.00 0.004 0.006 0.008 0.010 0.01 0.014 (b) Offset yield Strain Stress (psi) 90000 80000 U 70000 60000 50000 40000 0000 0000 10000 0 0.0 0.1 0. 0. 0.4 0.5 0.6 0.7 0.8 0.9 Strain (c) Complete range Shigley s MED, 10 th edition Chapter Solutions, Page 4/
(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 σ true vs.ε, the following equations are applied to the data. P σ true = A Eq. (-4) l ε = ln for 0 l 0.008 in (0 P 8400 lbf ) l 0 A0 ε = ln for l > 0.008 in ( P > 8 400 lbf ) A π (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 σ vs log ε The curve fit gives m = 0.06 log σ 0 = 5.185 σ 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 ε = = = A 0.1590 0 ln ln 0.1 m 0.06 Eq. (-18): S y = σ 0ε = 15.(0.1) = 108.4 kpsi Ans. Eq. (-19), with S = 85.6 from Prob. -6, u Su 85.6 S u = = = 107 kpsi Ans. 1 W 1 0. Shigley s MED, 10 th edition Chapter Solutions, Page 5/
P l A ε σ true log ε log σ true 0 0 0.198 7 0 0 1 000 0.000 4 0.198 7 0.000 5 0.71 -.699.70 000 0.000 6 0.198 7 0.000 10 065.4 -.5 4.00 000 0.001 0 0.198 7 0.000 5 15 098.1 -.01 4.179 4 000 0.001 0.198 7 0.000 65 0 10.9 -.187 4.04 7 000 0.00 0.198 7 0.001 15 5 9 -.940 4.547 8 400 0.00 8 0.198 7 0.001 4 4 74.8 -.854 4.66 8 800 0.198 4 0.001 51 44 54.8 -.81 4.647 9 00 0.197 8 0.004 54 46 511.6 -.4 4.668 9 100 0.196 0.01 15 46 57.6-1.915 4.666 1 00 0.19 4 0.0 68 607.1-1.49 4.86 15 00 0.187 5 0.058 0 81 066.7-1.6 4.909 17 000 0.156 0.40 0 108 765-0.60 5.06 16 400 0.10 7 0.418 89 15 478-0.78 5.099 14 800 0.107 7 0.61 45 17 419-0.1 5.18 Shigley s MED, 10 th edition Chapter Solutions, Page 6/
-8 Tangent modulus at σ = 0 is σ 5000 0 6 E = = 5( 10 ) psi ò 0. 10 0 ( ) Ans. At σ = 0 kpsi Shigley s MED, 10 th edition Chapter Solutions, Page 7/
E 0 )( 10 ) ( 1.5 1)( ) ( 6 19) (10 - ) σ (kpsi) 0 0 0.0 5 0.44 10 0.80 16 1.0 19 1.5 6.0.8 40.4 46 4.0 49 5.0 54 6 ( ) = 14.0 10 psi Ans. 10-9 W = 0.0, (a) Before cold working: Annealed AISI 1018 steel. Table A-, S y = kpsi, S u = 49.5 kpsi, σ 0 = 90.0 kpsi, m = 0.5, f = 1.05 After cold working: Eq. (-16), u = m = 0.5 A0 Eq. (-14), A = 1 1 1.5 i 1 W = 1 0.0 = A0 Eq. (-17), ε i = ln = ln1.5 = 0. < ε u A Eq. (-18), Eq. (-19), (b) Before: i ( ) 0.5 S m y = σ 0ε i = 90 0. = 61.8 kpsi Ans. 9% increase Ans. S S 49.5 u = u 61.9 kpsi Ans. 1 W = 1 0.0 = 5% increase Ans. Su 49 S = 9.5 1.55 y = After: S u 61.9 = = 1.00 S 61.8 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 = 61.5 kpsi, σ 0 = 110 kpsi, m = 0.4, f = 0.85 After cold working: Eq. (-16), u = m = 0.4 y Ans. Shigley s MED, 10 th edition Chapter Solutions, Page 8/
Eq. (-14), A0 1 1 1.5 A i 1 W 1 0.0 Eq. (-17), A0 εi = ln = ln1.5 = 0. < εu A m Eq. (-18), ( ) 0.4 Eq. (-19), (b) Before: i S y = σ 0ε i = 110 0. = 76.7 kpsi Ans. 174% increase Ans. Su 61.5 S u = = = 76.9 kpsi Ans. 5% increase Ans. 1 W 1 0.0 Su 61.5.0 S = y 8 = After: S u 76.9 = = 1.00 S 76.7 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 = 64.8 kpsi, σ 0 = 100 kpsi, m = 0.15, f = 0.18 After cold working: Eq. (-16), u = m = 0.15 A0 1 1 Eq. (-14), 1.5 A = i 1 W = 1 0.0 = A0 Eq. (-17), εi = ln = ln1.5 = 0. > ε 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. y Ans. Shigley s MED, 10 th edition Chapter Solutions, Page 9/
-15 For the data given, converting H B to S u using Eq. (-1) H B S u (kpsi) S u (kpsi) 0 115 15 116 1456 116 1456 4 117 1689 5 117.5 1806.5 5 117.5 1806.5 5 117.5 1806.5 6 118 194 6 118 194 9 119.5 1480.5 ΣS u = 117 ΣS u = 177 Eq. (1-6) S u Su 117 = = = 117. 117 kpsi Ans. N 10 Eq. (1-7), 10 Su NSu 177 10 1 ( 117. i= ) ss = = = 1.7 kpsi Ans. u N 1 9-16 For the data given, converting H B to S u using Eq. (-) H B S u (kpsi) S u (kpsi) 0 40.4 16.16 40.86 1669.54 40.86 1669.54 4 41. 1707.4 5 41.55 176.40 5 41.55 176.40 5 41.55 176.40 6 41.78 1745.568 6 41.78 1745.568 9 4.47 180.701 ΣS u = 414.1 ΣS u = 1715.6 Shigley s MED, 10 th edition Chapter Solutions, Page 10/
Eq. (1-6) S u Su 414.1 = = = 41.4 kpsi Ans. N 10 Eq. (1-7), 10 Su NSu 1715.6 10 1 ( 41.4 i= ) ss = = = 1.0 Ans. u N 1 9-17 (a) Eq. (-9) ur 45.6 = (0) 4.7 in lbf / in Ans. (b) A 0 = π(0.50 )/4 = 0.19871 in P L A (A 0 / A) 1 σ = P/A 0 0 0 0 0 1 000 0.000 4 0.000 5 0. 000 0.000 6 0.000 10 070 000 0.001 0 0.000 5 15 100 4 000 0.001 0.000 65 0 10 7 000 0.00 0.001 15 5 0 8 400 0.00 8 0.001 4 4 70 8 800 0.00 6 0.001 8 44 90 9 00 0.008 9 0.004 45 46 00 9 100 0.196 0.01 8 0.01 8 45 800 1 00 0.19 4 0.0 80 0.0 80 66 40 15 00 0.187 5 0.059 79 0.059 79 76 500 17 000 0.156 0.71 4 0.71 4 85 550 16 400 0.10 7 0.50 5 0.50 5 8 50 14 800 0.107 7 0.845 0 0.845 0 74 480 From the figures on the next page, 5 1 ut Ai = (4 000)(0.001 5) + 45 000(0.004 45 0.001 5) i= 1 1 + ( 45 000 + 76 500 ) (0.059 8 0.004 45) + 81 000 0.4 0.059 8 + 80 000 0.845 0.4 ( ) ( ) ( ) 66.7 10 in lbf/in. Ans Shigley s MED, 10 th edition Chapter Solutions, Page 11/
Shigley s MED, 10 th edition Chapter Solutions, Page 1/
-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: For diameter, F F 4F σ = = = S y d = A d π S ( π / 4) Weight/length = ρa, Cost/length = $/in = ($/lbf) Weight/length, Deflection/length = δ /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 0.7.060 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 1.6 0.55 0.1.667E-0 4140 0 0.8 165 0.80 0.878 0.1709 0.14 5.500E-0 Al 10.4 0.098 50 1.10 1.596 0.1960 0. 4.808E-0 Ti 16.5 0.16 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 Al [ π (1 in) / 4](6 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 A-0, perform a Brinell hardness test, then use Eq. (-1) to estimate an ultimate strength Shigley s MED, 10 th edition Chapter Solutions, Page 1/
of Su = 0.5H B = 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, 1060, 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 Al = [ π (1 in) / 4](6 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 Al [ π (1 in) / 4](6 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 100 4 E = 17.7 Mpsi Iy = (1) 64 (17 / ) = 4 ( π ) 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. -6 For strength, σ = F/A = S A = F/S Shigley s MED, 10 th edition Chapter Solutions, Page 14/
For mass, m = Alρ = (F/S) lρ Thus, f (M ) = ρ /S, and maximize S/ρ (β = 1) In Fig. (-19), draw lines parallel to S/ρ The higher strength aluminum alloys have the greatest potential, as determined by comparing each material s bubble to the S/ρ guidelines. Ans. -7 For stiffness, k = AE/l A = kl/e For mass, m = Alρ = (kl/e) lρ =kl ρ /E Thus, f (M) = ρ /E, and maximize E/ρ (β = 1) In Fig. (-16), draw lines parallel to E/ρ Shigley s MED, 10 th edition Chapter Solutions, Page 15/
From the list of materials given, tungsten carbide (WC) is best, closely followed by aluminum alloys. 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, σ = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (-6b), p. 104 ]. 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 π. Then, for strength, Eq. (1) is Fl CA / Fl = S A = CS / () Shigley s MED, 10 th edition Chapter Solutions, Page 16/
For mass, / / Fl F 5/ ρ ρ / m = Alρ = l = l CS C S Thus, f (M) = ρ /S /, and maximize S / /ρ (β = /) In Fig. (-19), draw lines parallel to S / /ρ 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. (-6), p. 77, 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 (a constant). Thus, Eq. (-7) becomes Shigley s MED, 10 th edition Chapter Solutions, Page 17/
kl A = CE and Eq. (-9) becomes Thus, minimize ( ) 1/ 1/ k 5/ ρ ρ 1/ m = Al = l C E ρ f M =, or maximize M 1/ E 1/ E =. From Fig. (-16) ρ 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 = Alρ = (kl/e) lρ =kl ρ /E So, f (M) = ρ /E, and maximize E/ρ. Thus, β = 1. Ans. -1 For strength, σ = F/A = S A = F/S Shigley s MED, 10 th edition Chapter Solutions, Page 18/
For mass, m = Alρ = (F/S) lρ So, f (M ) = ρ /S, and maximize S/ρ. Thus, β = 1. Ans. - Eq. (-6), p. 77, 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 the circular cross section (see p.77), C = (4π) 1. Another 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, a constant. Thus, Eq. (-7) becomes kl A = CE and Eq. (-9) becomes 1/ 1/ k 5/ ρ ρ 1/ m = Al = l C E 1/ ρ E So, minimize f ( M ) =, or maximize M =. Thus, β = 1/. Ans. 1/ E ρ - For strength, σ = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (-6b), p. 104 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in (ips) or m (SI). The area of the cross section has the units in or m. Thus, for a given cross section, Z =C (A) /, where C is a number. For example, for a circular cross section, Z = πd /()and the area is A = πd /4. This leads to C = ( 4 π ) 1. So, with Z =C (A) /, for strength, Eq. (1) is Fl CA / Fl = S A = CS / () For mass, / / Fl F 5/ ρ ρ / m = Alρ = l = l CS C S So, f (M) = ρ /S /, and maximize S / /ρ. Thus, β = /. Ans. Shigley s MED, 10 th edition Chapter Solutions, Page 19/
-4 For stiffness, k=ae/l, or, A = kl/e. Thus, m = ρal =ρ (kl/e )l = kl ρ /E. Then, M = E /ρ and β = 1. From Fig. -16, lines parallel to E /ρ for ductile materials include steel, titanium, molybdenum, aluminum alloys, and composites. For strength, S = F/A, or, A = F/S. Thus, m = ρal =ρ F/Sl = Fl ρ /S. Then, M = S/ρ and β = 1. From Fig. -19, lines parallel to S/ρ 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. -5 See Prob. 1-1 solution for x = 1.9 kcycles and s x = 0. kcycles. Also, in that solution it is observed that the number of instances less than 115 kcycles predicted by the normal distribution is 7; whereas, the data indicates the number to be 1. From Eq. (1-4), the probability density function (PDF), with µ = x and ˆ σ = sx, is 1 1 x x 1 1 x 1.9 f ( x) = exp = exp s sx 0. 0. x π π (1) The discrete PDF is given by f /(Nw), where N = 69 and w = 10 kcycles. From the Eq. (1) and the data of Prob. 1-1, the following plots are obtained. Shigley s MED, 10 th edition Chapter Solutions, Page 0/
Range midpoint (kcycles) Observed PDF Normal PDF Frequency x f f /(Nw) f (x) 60 0.00898551 0.0015649 70 1 0.00144975 0.0086804 80 0.0044786 0.0048507 90 5 0.0074677 0.00704 100 8 0.0115940 0.009895407 110 1 0.0179104 0.010566 10 6 0.00869565 0.0110645 10 10 0.01449754 0.01809861 140 8 0.0115940 0.0118104 150 5 0.0074677 0.00886008 160 0.00898551 0.006189 170 0.0044786 0.00996 180 0.00898551 0.00004 190 1 0.00144975 0.0011847 00 0 0 0.000517001 10 1 0.00144975 0.0001141 Plots of the PDF s are shown below. 0.0 0.018 0.016 0.014 Probability Density 0.01 0.01 0.008 0.006 0.004 0.00 Normal Distribution Histogram 0 0 0 40 60 80 100 10 140 160 180 00 0 L (kcycles) Shigley s MED, 10 th edition Chapter Solutions, Page 1/
It can be seen that the data is not perfectly normal and is skewed to the left indicating that the number of instances below 115 kcycles for the data (1) would be higher than the hypothetical normal distribution (7). Shigley s MED, 10 th edition Chapter Solutions, Page /