Assignment Print View 6.1.pdf - Assignment Print View Score 72.15\/100 Points 72.15

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Unformatted text preview: 4/10/2020 Assignment Print View Score: 72.15/100 Points 72.15 % 1/24 4/10/2020 1. Assignment Print View Award: 5.55 out of 5.55 points A steel rotating-beam test specimen has an ultimate strength, Sut, of 120 kpsi. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude, σrev, of 73 kpsi. Take the value of the fatigue strength fraction, f, from Fig. 6-23. The life of the specimen is 64815 cycles. A steel rotating-beam test specimen has an ultimate strength, Sut, of 120 kpsi. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude, σrev, of 73 kpsi. Take the value of the fatigue strength fraction, f, from Fig. 6-23. The life of the specimen is 64815 ± 2% cycles. Explanation: Sut = 120 kpsi, σrev = 73 kpsi The endurance limit can be calculated as follows: ⎧ ⎪ ′ Se = 0.5Sut Sut ≤ 200 kpsi (1400 MPa) ⎨ 100 kpsi S > 200 kpsi ut ⎪ ⎩ 700 MPa Sut > 1400 MPa The value of the tensile strength is 120 kpsi, which is less than 200 kpsi, hence ′ Se = Se = 0.5(120 kpsi) = 60 kpsi From Fig. 6-23, at 120 kpsi, we have f = 0.82. The values of a and b can be calculated as follows: 2 a = (f Sut ) b = − Se 1 3 = log [(0.82)(120 kpsi)] 60 kpsi fS ( 2 ut Se ) = − 1 3 = 161.376 kpsi log (0.82)(120 kpsi) ( 60 kpsi ) = −0.0716 When a completely reversed stress, σrev, is given, the number of cycles-to-failure can be expressed as 1 N = ( σrev a 1/b ) = ( 73 kpsi 161.376 kpsi −0.0716 ) = 64815 cycles 2/24 4/10/2020 Assignment Print View 2. Award: 5.55 out of 5.55 points Estimate the endurance strength (Se) of a 1.47-in-diameter (d) rod of AISI 1040 steel having a machined finish and heat-treated to a tensile strength ( Sut) of 104 kpsi, loaded in rotating bending. Obtain the parameters for Marin surface modification factor from Table 6–2. The endurance strength is 32 kpsi. Estimate the endurance strength (Se) of a 1.47-in-diameter (d) rod of AISI 1040 steel having a machined finish and heat-treated to a tensile strength ( Sut) of 104 kpsi, loaded in rotating bending. Obtain the parameters for Marin surface modification factor from Table 6–2. The endurance strength is 32 ± 2% kpsi. Explanation: d = 1.47 in, Sut = 104 kpsi From Eq. (6 − 10), S ′ e = 0. 5(104 kpsi) = 52 kpsi. From Table 6 − 2, a = 2.00 and b = -0.217. From Eq. (6 − 18), ka = aSut b = 2(104 kpsi) From Eq. (6 − 19), kb = 0. 879d From Eq. (6 − 17), Se = ka kb S ′ e -0.107 −0.217 = 0.73 = 0. 879(1.47 in) -0.107 = 0.844 = 0.73(0.844)(52 kpsi) = 32 kpsi 3/24 4/10/2020 Assignment Print View Two steels are being considered for the manufacture of as-forged connecting rods subjected to bending loads. One is AISI 4340 Cr-Mo-Ni steel capable of being heat-treated to a tensile strength of 260 kpsi. The other is a plain carbon steel AISI 1040 with an attainable Sut of 113 kpsi. Each rod is to have a size giving an equivalent diameter de of 0.55 in. Determine the endurance limit for each material. Is there any advantage of using the alloy steel for this fatigue application? 3. Award: 5.55 out of 5.55 points What are the estimated values of the endurance limits for the 4340 and 1040 steels? The endurance limit for the 4340 steel is 100 The endurance limit for the 1040 steel is 56.5 kpsi. kpsi. What are the estimated values of the endurance limits for the 4340 and 1040 steels? The endurance limit for the 4340 steel is 100 ± 2% kpsi. The endurance limit for the 1040 steel is 56.5 ± 2% kpsi. Explanation: Since the 4340 steel has an ultimate tensile strength greater than 200 kpsi, the estimated endurance limit for this steel is 100 kpsi. The 1040 steel, having an ultimate tensile strength below 200, has an estimated endurance limit of Se = 0.5Sut = 0.5 (113 kpsi) = 56.5 kpsi 4/24 4/10/2020 4. Assignment Print View Award: 5.55 out of 5.55 points What is the surface modification factor value for each of the materials? The surface modification factor value for the 4340 steel is 0.188 . The surface modification factor value for the 1040 steel is 0.353 . What is the surface modification factor value for each of the materials? The surface modification factor value for the 4340 steel is 0.188 ± 2% . The surface modification factor value for the 1040 steel is 0.353 ± 2% . Explanation: The surface condition for both components is as-forged. For the as-forged conditions, −0.758 ka = 12.7(Sut ) For the 4340 steel, ka = 12.7(260 kpsi) −0.758 = 0.188 For the 1040 steel, ka = 12.7(113 kpsi) −0.758 = 0.353 5/24 4/10/2020 5. Assignment Print View Award: 5.55 out of 5.55 points What is the size modification factor value for each of the components? The size modification factor value for the 4340 steel is 0.937 . The size modification factor value for the 1040 steel is 0.937 . What is the size modification factor value for each of the components? The size modification factor value for the 4340 steel is 0.937 ± 2% . The size modification factor value for the 1040 steel is 0.937 ± 2% . Explanation: The equivalent diameter for each component is the same, 0.55 inches, so the size modification factor will be the same. kb = ( 0.55 in 0.30 −0.107 ) kb = 0.937 6/24 4/10/2020 6. Assignment Print View Award: 5.55 out of 5.55 points Given that all other modifying factors are equal to 1, except surface modification factor and size modification factor, what are the endurance limits for each of the components? The endurance limit for the 4340 steel is 17.615 kpsi. The endurance limit for the 1040 steel is 18.687 kpsi. Given that all other modifying factors are equal to 1, except surface modification factor and size modification factor, what are the endurance limits for each of the components? The endurance limit for the 4340 steel is 17.619 ± 2% kpsi. The endurance limit for the 1040 steel is 18.692 ± 2% kpsi. Explanation: For the 4340 steel, ′ Se = ka kb Se Se = (0.188) (0.937) (100 kpsi) Se = 17.619 kpsi For the 1040 steel, ′ Se = ka kb Se Se = (0.353) (0.937) 113 kpsi ( 2 ) Se = 18.692 kpsi 7/24 4/10/2020 Assignment Print View A 1.1-in-diameter solid round bar has a groove 0.1-in deep with a 0.1-in radius machined into it. The bar is made of AISI 1020 CD steel and is subjected to a purely reversing torque of 1500 lbf·in. 7. Award: 5.55 out of 5.55 points Determine the theoretical stress concentration for this geometry and loading. The theoretical stress concentration is 1.4 . Determine the theoretical stress concentration for this geometry and loading. The theoretical stress concentration is 1.4 ± 2% . Explanation: The ratios D/d and r/d are required to read the stress-concentration value from the following chart (refer to Figure A-15-15): d = D – 2r r = 0.1 in d = 1.1 in.– 2 (0.1 in) = 0.9 in D d r d = = 1.1 in 0.9 in 0.1 in 0.9 in =1.2222 = 0.1111 Kts = 1.4 8/24 4/10/2020 8. Assignment Print View Award: 5.55 out of 5.55 points What are the values of the notch sensitivity qs and the fatigue stress-concentration factor Kfs? The notch sensitivity is 0.812 . The fatigue stress concentration is 1.325 . What are the values of the notch sensitivity qs and the fatigue stress-concentration factor Kfs? The notch sensitivity is 0.812 ± 2% . The fatigue stress concentration is 1.325 ± 2% . Explanation: Using the equation for the Neuber's constant for torsion with Sut = 68 kpsi, ‾ = 0.190 - 2.51 (10 √a qs = 1 1 + a √ √r −3 1 = 1 + ) (68 kpsi) + 1.35 (10 −5 2 )(68 kpsi) - 2.67 (10 −8 3 ‾‾ )(68 kpsi) = 0.0733 √in = 0.812 0.0733 √in √ 0.1 in Kf s = qs (Kts - 1) + 1 Kf s = 0.812 (1.4 - 1) + 1 = 1.325 9/24 4/10/2020 9. Assignment Print View Award: 5.55 out of 5.55 points Determine the endurance modification factors for the surface finish ka, the size kb, and the loading kc. The value of ka is 0.801 . The value of kb is 0.889 . The value of kc is 0.59 . Determine the endurance modification factors for the surface finish ka, the size kb, and the loading kc. The value of ka is 0.801 ± 2% . The value of kb is 0.889 ± 2% . The value of kc is 0.59 ± 2% . Explanation: For a machined surface condition, the surface finish modification factor ka is −0.217 ka = 2(68 kpsi) = 0.801 The size modification factor kb is kb = 0.879(d) −0.107 −0.107 = 0.879(1.1 in - 2 (0.1 in)) = 0.889 The loading modification factor kc is kc = 0.59 10/24 4/10/2020 10. Assignment Print View Award: 5.55 out of 5.55 points Determine the estimated endurance limit of the rotating beam specimen Se' and fatigue strength for the conditions and materials in this application. Also, determine the ultimate strength in shear for the material, Ssu. Assume N = 1 million cycles. The endurance limit is 34 kpsi. The fatigue strength is 14.284 kpsi. The ultimate strength in shear is 45.56 kpsi. Determine the estimated endurance limit of the rotating beam specimen Se' and fatigue strength for the conditions and materials in this application. Also, determine the ultimate strength in shear for the material, Ssu. Assume N = 1 million cycles. The endurance limit is 34 ± 2% kpsi. The fatigue strength is 14.284 ± 2% kpsi. The ultimate strength in shear is 45.56 ± 2% kpsi. Explanation: The estimated endurance limit for this material is Se = 0.5 (Sut ) = 0.5 (68 kpsi) = 34 kpsi Applying each of the modifying factors to the estimate for the endurance limit results in the fatigue strength: Sse = (ka ) (kb ) (kc ) (0.5Sut ) Sse = Sf = (0.801) (0.889) (0.59) (34 kpsi) Sse = Sf = 14.284 kpsi The ultimate strength in shear is given by Ssu = 0.67 (68 kpsi) 11/24 4/10/2020 Assignment Print View Ssu = 45.56 kpsi 12/24 4/10/2020 11. Assignment Print View Award: 5.55 out of 5.55 points Estimate the number of cycles to failure. The number of cycles to failure is 1.205 × 106. Estimate the number of cycles to failure. 1.205 ± 2% × 106. The number of cycles to failure is Explanation: The estimated ultimate shear strength at 1000 cycles is Ssu = (0.67) (Sut ) Ssu = (0.67) (68 kpsi) = 45.6 kpsi The estimated endurance limit Sse is Sse = Sf = (0.801) (0.889) (0.59) (0.5 × 68 kpsi) Sse = 14.284 kpsi a = (f Ssu ) 2 = Sse b = − 1 3 log (0.9 × 45.6 kpsi) 14.284 kpsi f Ssu ( Sse = − ) 2 = 117.914 kpsi 1 3 log 0.9 × 45.6 kpsi ( 14.284 kpsi = -0.153 ) Now that a and b are known, the number of cycles to failure can be determined. τa = aN N = ( b a 1 1 τa ) b = ( 13.882 kpsi 117.914 kpsi -0.153 ) 6 N = 1.205 × 10 cycles 13/24 4/10/2020 Assignment Print View 14/24 4/10/2020 12. Assignment Print View Award: 5.55 out of 5.55 points If the bar is also placed in an environment with a temperature of 750°F, estimate the number of cycles to failure. The number of cycles to failure is 0.554 × 106. If the bar is also placed in an environment with a temperature of 750°F, estimate the number of cycles to failure. 0.554 ± 2% × 106. The number of cycles to failure is Explanation: A modification factor accounting for the impact of temperature can be included to determine a new value of Sse. For an operating temperature of 750°F, the temperature modification factor, from Eqs. 6-26 and 6-27, is kd = 0.8881. Sse = (ka ) (kb ) (kc ) (kd ) (0. 5Sut ) Sse = Sf = (0.801) (0.889) (0.59) (0.8881) (34 kpsi) Sse = 12.686 kpsi Sut = 0.8881 × 68 kpsi = 60.39 kpsi Ssu = 0.67 × 60.39 kpsi = 40.5 kpsi a = (f Ssu ) 2 = Sse b = − 1 3 τa = aN log (0.9 × 40.5 kpsi) 12.686 kpsi f Ssu ( Sse = − ) 2 = 104.733 kpsi 1 3 log 0.9 × 40.5 kpsi ( 12.686 kpsi = -0.153 ) b 15/24 4/10/2020 Assignment Print View 1 13.882 kpsi ( 104.733 kpsi -0.153 ) 6 =0.554 × 10 cycles A solid square rod is cantilevered at one end. The rod is 0.5 m long and supports a completely reversing transverse load at the other end of ±1.6 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 4.5 × 105 cycles with a design factor of 1.5, what dimension should the square cross section have? Neglect any stress concentrations at the support end. 16/24 4/10/2020 13. Assignment Print View Award: 0 out of 5.55 points Determine the endurance limit for this application. Assume that the size modification factor kb is 0.85 until a size for the section is known. The endurance limit is n/r MPa. Determine the endurance limit for this application. Assume that the size modification factor kb is 0.85 until a size for the section is known. The endurance limit is 167.97 ± 2% MPa. Explanation: The fatigue strength fraction f is found to be 0.83, noting that the value of Sut, 770 MPa, needs to be converted to kpsi. For a hot-rolled surface finish, the surface modification factor is ka = 38.6(770 MPa) −0.65 Se = (0.5133)(0.85) ( = 0.5133 770 MPa 2 ) = 167.97 MPa 17/24 4/10/2020 14. Assignment Print View Award: 0 out of 5.55 points What will the strength of the metal be at 4.5 × 105 cycles? The strength of the metal at 4.5 × 105 will be n/r MPa. What will the strength of the metal be at 4.5 × 105 cycles? The strength of the metal at 4.5 × 105 will be 196.144 ± 2% MPa. Explanation: To determine the strength of the metal at 4.5 × 105 cycles, the values of a and b in the equation Se = aNb need to be determined. 2 a = (f Sut ) Se b = − 1 3 2 = log (0.83 × 770 MPa) 167.97 MPa fS ( ut Se = − ) 1 3 = 2431.677 MPa log ( 5 Sf = 2431.677 MPa(4.5 × 10 ) 0.83 × 770 MPa 167.97 MPa −0.1934 ) = − 0.1934 = 196.144 MPa 18/24 4/10/2020 15. Assignment Print View Award: 0 out of 5.55 points Determine the size of the cross section. The size of the cross section is n/r mm. Determine the size of the cross section. The size of the cross section is 25.3625 ± 2% mm. Explanation: The loading on the bar is fully reversed with ±1.6 kN acting at the end of the beam. The mean moment, as a result, is 0.0 kN·m. The alternating moment is Ma = (0.5 m × 1.6 kN) - (-0.5 m × 1.6 kN) 2 = 800 N·m b Ma σa = 2 1 = 3 6M a b b b 3 12 σa Sf = 1.5 6Ma ( ) b3 Sf = 1.5 1 b = 3 6 × 800 N·m ( 1.5 × 196.1438 × (10 6 ) Pa ) ×1000 b = 25.3625 mm 19/24 4/10/2020 Assignment Print View A solid round bar with a diameter of 2.46 in has a groove cut to a diameter of 2.216 in, with a radius of 0.124 in. The bar is not rotating. The bar is loaded with a repeated bending load that causes the bending moment at the groove to fluctuate between 0 and 25,000 lbf·in. The bar is hot-rolled AISI 1095, but the groove has been machined. Determine the factor of safety for fatigue based on infinite life using the modified Goodman criterion and the factor of safety for yielding. 20/24 4/10/2020 16. Assignment Print View Award: 5.55 out of 5.55 points Determine the endurance limit for this material as well as the value after correction for surface finish, sizing, and loading. The endurance limit for the material is 60 kpsi. The endurance limit for the material after correction is 38.126 kpsi. Determine the endurance limit for this material as well as the value after correction for surface finish, sizing, and loading. The endurance limit for the material is 60 ± 2% kpsi. The endurance limit for the material after correction is 38.125 ± 2% kpsi. Explanation: The endurance limit for hot-rolled AISI 1095 is ′ Se = 0.5Sut = 0.5 (120 kpsi) = 60 kpsi For a cold-drawn surface finish, the surface modification is ka = 2(120 kpsi) −0.217 = 0.7077 The size modification factor for bending requires finding an equivalent diameter because it is not rotating: de = 0.370 × 2.216 in = 0.82 in kb = 0.879(0.82 in) -0.107 = 0.898 The loading modification factor for bending is kc = 1.0 Se = 0.7077 × 0.898 × 1 × 120 kpsi 2 Se = 38.125 kpsi 21/24 4/10/2020 Assignment Print View 22/24 4/10/2020 17. Assignment Print View Award: 0 out of 5.55 points What are the values of the theoretical stress-concentration factor, the notch sensitivity, and the fatigue stress-concentration factor? The value of the theoretical stress-concentration factor is n/r The value of the notch sensitivity is n/r . . The value of the fatigue stress-concentration factor is n/r . What are the values of the theoretical stress-concentration factor, the notch sensitivity, and the fatigue stress-concentration factor? The value of the theoretical stress-concentration factor is The value of the notch sensitivity is 2.1 ± 2% . 0.881 ± 2% . The value of the fatigue stress-concentration factor is 1.969 ± 2% . Explanation: For the effect of bending on the geometry, kt is 2.1. (Refer to Figure A−15–14.) r d = 0.056 D d = 1.11 Neuber's constant, which is used to calculate the notch sensitivity, is ‾ = 0.246 - 3.08 (10 √a ‾ = 0.246 - 3.08 (10 √a −3 −3 )Sut + 1.51 (10 −5 2 )Sut - 2.67 (10 ) (120 kpsi) + 1.51 (10 −5 −8 3 )Sut 2 )(120 kpsi) - 2.67 (10 −8 )(120 kpsi) 3 ‾‾ ‾ = 0.0477 √in √a 23/24 4/10/2020 Assignment Print View 1 q = = 0.881 0.0477 √in 1 + √ 0.124 in Kf = q (Kt - 1) + 1 Kf = 0.881 (2.1 - 1) + 1 Kf = 1.969 18. Award: 0 out of 5.65 points What is the factor of safety against yielding? The factor of safety against yielding is 0.361 . What is the factor of safety against yielding? The factor of safety against yielding is 2.821 ± 2% . Explanation: The maximum stress for this application is σmax = 32Mmax πd 3 = 32(25000 lbf·in) π(2.216 in) 3 × 1 1000 σmax = 23.394 kpsi The factor of safety against yielding is ny = Sy σmax = 66 kpsi 23.394 kpsi = 2.821 24/24 ...
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