assignment 7.pdf - Assignment Print View Score 41.65\/100 Points 41.65 A shaft is loaded in bending and torsion such that Ma = 74 N\u00b7m Ta = 47.647 N\u00b7m

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Unformatted text preview: 4/11/2020 Assignment Print View Score: 41.65/100 Points 41.65 % A shaft is loaded in bending and torsion such that Ma = 74 N·m, Ta = 47.647 N·m, Mm = 58.144 N·m, and Tm = 37 N·m. For the shaft, Su = 700 MPa and Sy = 560 MPa, and a fully corrected endurance limit of Se = 210 MPa is assumed. Let Kf = 2.2 and Kfs = 1.8. With a design factor of 2.0, determine the minimum acceptable diameter of the shaft using the methods mentioned in the subsection. 1/24 4/11/2020 1. Assignment Print View Award: 8.33 out of 8.33 points Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Gerber criterion. The diameter of the shaft is 26.338 mm. Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Gerber criterion. The diameter of the shaft is 26.339 ± 2% mm. Explanation: A = B = √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Ma ) + 3(Kf s Ta ) = √ ‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 74 N·m) + 3(1.8 × 47.647 N·m)‾ = 357.8857 N·m √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Mm ) + 3(Kf s Tm ) = √‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 58.144 N·m) + 3(1.8 × 37 N·m)‾ = 280.6378 N·m Using the DE-Gerber, Eq. (7-10) can be shown to be 8(2)(357.8857 N·m) d = ( 6 π(210)(10 ) Pa 6 1 + ( 1 + ( ( 2(280.6378 N·m)(210)(10 ) Pa 6 357.8857 N·m(700)(10 ) Pa 2 1/3 1/2 ) ) )) d = 0.026 m = 26.339 mm 2/24 4/11/2020 2. Assignment Print View Award: 8.33 out of 8.33 points Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Ellipitic criterion. The diameter of the shaft is 26.252 mm. Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Ellipitic criterion. The diameter of the shaft is 26.253 ± 2% mm. Explanation: A = B = √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Ma ) + 3(Kf s Ta ) = √ ‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 74 N·m) + 3(1.8 × 47.647 N·m)‾ = 357.8857 N·m √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Mm ) + 3(Kf s Tm ) = √‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 58.144 N·m) + 3(1.8 × 37 N·m)‾ = 280.6378 N·m Using DE-Elliptic, Eq. (7-12) can be shown to be 1/3 d = 16n ( π √ 2 2 ‾ ‾‾‾ ‾‾‾ A B‾ + 2 2 S e S y ) 1/3 = 16(2) ( π 2 2 (280.6378 N·m) ‾ ‾(357.8857 N·m) ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + 2 2 6 6 ) √ (210) 10 Pa (560) 10 Pa ( ( ) ) ( ( ) ) d = 0.026 m = 26.253 mm 3/24 4/11/2020 3. Assignment Print View Award: 8.33 out of 8.33 points Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Soderberg criterion. The diameter of the shaft is 28.215 mm. Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Soderberg criterion. The diameter of the shaft is 28.216 ± 2% mm. Explanation: A = B = √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Ma ) + 3(Kf s Ta ) = √ ‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 74 N·m) + 3(1.8 × 47.647 N·m)‾ = 357.8857 N·m √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Mm ) + 3(Kf s Tm ) = √‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 58.144 N·m) + 3(1.8 × 37 N·m)‾ = 280.6378 N·m Using DE-Soderberg, Eq. (7-14) can be shown to be 1/3 1/3 d = 16n ( π A ( Se + B Sy )) = 16(2) ( π 357.8857 N·m ( 6 210(10 ) Pa + 280.6378 N·m 6 560(10 ) Pa )) d = 0.0282 m = 28.216 mm 4/24 4/11/2020 4. Assignment Print View Award: 8.33 out of 8.33 points Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Goodman criterion. The diameter of the shaft is 27.78 mm. Determine the diameter of the shaft for the given loadings, material properties, and fatigue stress concentration factors using the DE-Goodman criterion. The diameter of the shaft is 27.782 ± 2% mm. Explanation: A = B = √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Ma ) + 3(Kf s Ta ) = √ ‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 74 N·m) + 3(1.8 × 47.647 N·m)‾ = 357.8857 N·m √ 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 4(Kf Mm ) + 3(Kf s Tm ) = √‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.2 × 58.144 N·m) + 3(1.8 × 37 N·m)‾ = 280.6378 N·m Using DE-Goodman: Eq. (7-8) can be shown to be d = ( 16n π A ( Se + B Sut 1/3 1/3 )) = 16(2) ( π 357.8857 N·m ( 6 210(10 ) Pa + 280.6378 N·m 6 700(10 ) Pa )) = 0.0278 m = 27.782 mm 5/24 4/11/2020 Assignment Print View A rotating step shaft is loaded as shown, where the forces FA and FB are constant at 650 lbf and 325 lbf, respectively, and the torque T alternates from 0 to 1800 lbf·in. The shaft is to be considered simply supported at points O and C and is made of AISI 1045 CD steel with a fully corrected endurance limit of Se = 40 kpsi. Let Kf = 2.1 and Kfs = 1.7. Take the value of design factor to be 2.5. 6/24 4/11/2020 5. Assignment Print View Award: 0 out of 8.33 points Determine the minimal acceptable diameter of section BC using the DE-Gerber criterion. The minimal acceptable diameter of section BC is 7.38 in. Determine the minimal acceptable diameter of section BC using the DE-Gerber criterion. The minimal acceptable diameter of section BC is 1.3754 ± 2% in. Explanation: Given: AISI 1045 CD steel, Se = 40 kpsi, Kf = 2.1, Kfs = 1.7 and nf = 2.5. From Table A-20, Sy = 77 kpsi and Sut = 91 kpsi. (∑ MO )y = 0 = −18 in × RCz + 12 in × 300 lbf RCz = 216.6667 lbf (∑ MO )z = 0 = −18 in × RCy + 6 in × 600 lbf RCy = 216.6667 lbf (MB )y = −6 in × (216.6667 lbf) = -1300 lbf·in (MB )z = 6 in × (216.6667 lbf) = 1300 lbf·in 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ MB = √ ‾ (-1300 lbf·in) + 1300 lbf·in‾ = 1838.477631 lbf⋅in Mm = 0, Ma = 1838.477631 lbf٠in, 7/24 4/11/2020 Assignment Print View Tm = Ta = 1800 lbf·in 2 = 900 lbf·in Using the DE-Gerber criterion, 2 2 A = √‾ 4‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ (2.1 (1838.477631 lbf·in)) + 3(1.7 (900 lbf·in))‾ = 8163.694017 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ B = √‾ 3(1.7 (900 lbf·in))‾ = 2650 8(2.5)8163.694017 lbf·in d = ( π(40 kpsi)10 3 lbf⋅in 2 3 1 + ( lbf⋅in 1 + ( ( 2(2650 lbf·in)40 kpsi(10 ) 8163.694017 lbf·in(91 kpsi)10 3 1/3 1/2 ) ) = 1.3754 in )) 8/24 4/11/2020 6. Assignment Print View Award: 0 out of 8.33 points Determine the minimal acceptable diameter of section BC using the DE-Goodman criterion. The minimal acceptable diameter of section BC is n/r in. Determine the minimal acceptable diameter of section BC using the DE-Goodman criterion. The minimal acceptable diameter of section BC is 1.4373 ± 2% in. Explanation: Given: AISI 1045 CD steel, Se = 40 kpsi, Kf = 2.1, Kfs = 1.7, and nf = 2.5. From Table A-20, Sy = 77 kpsi and Sut = 91 kpsi. (∑ MO )y = 0 = −18 in × RCz + 12 in × 300 lbf RCz = 216.6667 lbf (∑ MO )z = 0 = −18 in × RCy + 6 in × 600 lbf RCy = 216.6667 lbf (MB )y = −6 in × (216.6667 lbf) = -1300 lbf·in (MB )z = 6 in × (216.6667 lbf) = 1300 lbf·in 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ MB = √ ‾ (-1300 lbf·in) + 1300 lbf·in‾ = 1838.477631 lbf⋅in Mm = 0, Ma = 1838.477631 lbf٠in, 9/24 4/11/2020 Assignment Print View Tm = Ta = 1800 lbf·in 2 = 900 lbf·in Using the DE-Goodman criterion, 1/3 d = 16(2.5) ( π 8163.694017 lbf·in ( 3 40 kpsi(10 ) + 2650 lbf·in 3 91 kpsi(10 ) )) = 1.4373 in In the double-reduction gear train shown, shaft a is driven by a motor attached by a flexible coupling attached to the overhang. The motor provides a torque of 2500 lbf·in at a speed of 1200 rpm. The gears have 20° pressure angles, with diameters shown in the figure. Use AISI 1040 Q and T cold-drawn steel with Sut = 85 kpsi and Sy = 71 kpsi. Design shaft a using a design factor of 1.5. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 10/24 4/11/2020 7. Assignment Print View Award: 8.33 out of 8.33 points If the shaft’s loading is fully reversed and the torque between points B and C is constant, what are the values of the mean and alternating torques and the mean and alternating moments in lbf·in? (You must provide an answer before moving to the next part.) Tmean = 2550 lbf·in Talternating = 0 lbf·in Mmean = 0 lbf·in Malternating = 2178 lbf·in If the shaft’s loading is fully reversed and the torque between points B and C is constant, what are the values of the mean and alternating torques and the mean and alternating moments in lbf·in? (You must provide an answer before moving to the next part.) Tmean = 2500 ± 2% lbf·in Talternating = Mmean = 0 ± 2% lbf·in 0 ± 2% lbf·in Malternating = 2178 ± 2% lbf·in Explanation: At the gear keyway, approximately 9 in from the left end of the shaft, the bending is completely reversed and the torque is steady. Mmax = 2178 lbf·in The torque is steady with a value of 2500 lbf·in. Using these results, Mmin = −Mmax Mmean = Mmax +(−Mmax ) 2 = 0 11/24 4/11/2020 Assignment Print View Malternating = Mmax −(−Mmax ) 2 = Mmax 2178 lbf·in−(−2178 lbf·in) Maltternating = 2 = 2178 lbf·in Tmax = Tmin Tmean = Tmean = Talt = T max +(T max ) 2 = Tmax 2500 lbf·in+(2500 lbf·in) 2 T max −(T max ) 2 = 2500 lbf·in = 0 12/24 4/11/2020 8. Assignment Print View Award: 0 out of 8.33 points Using the DE-Goodman criteria and a design factor of 1.5, calculate the diameter based on the shaft’s loadings and your guess for the shaft’s diameter at the critical location, what is the shaft diameter and what is the factor of safety against yielding? Assume the notch radius to be 0.02 in. (You must provide an answer before moving to the next part.) The shaft diameter is n/r in. The factor of safety against yielding is n/r . Using the DE-Goodman criteria and a design factor of 1.5, calculate the diameter based on the shaft’s loadings and your guess for the shaft’s diameter at the critical location, what is the shaft diameter and what is the factor of safety against yielding? Assume the notch radius to be 0.02 in. (You must provide an answer before moving to the next part.) The shaft diameter is 1.4631 ± 2% in. The factor of safety against yielding is 3.38 ± 2% . Explanation: Kt = 2.14 Kts = 3.0 The notch sensitivity, q, for reversed bending, assuming a notch radius of 0.02 in, is determined from the following equation: q = 1 1+ √a √r For bending and axial loads, 13/24 4/11/2020 Assignment Print View ‾ = 0. 246 − 3. 08 (10 √a ‾ = 0. 246 − 3. 08 (10 √a −3 −3 )Sut + 1. 51 (10 −5 2 )Sut − 2. 67 (10 ) (85 kpsi) + 1. 51 (10 −5 −8 2 )(85 kpsi) 3 )Sut − 2. 67 (10 −8 3 )(85 kpsi) For torsion, ‾ = 0.190 − 2.51 (10 √a ‾ = 0.190 − 2.51 (10 √a −3 −3 )Sut + 1.35 (10 −5 2 )Sut − 2.67 (10 ) (85 kpsi) + 1.35 (10 −5 −8 2 )(85 kpsi) 3 )Sut − 2.67 (10 −8 3 )(85 kpsi) q~0.65 The notch sensitivity, qs, for reversed torsion, assuming a notch radius of 0.02 in, is determined from the previous equation. Using Neuber’s constant for torsion. qs ~0.71 The values for the fatigue stress-concentration factors for bending and shear stresses in the keyseat are Kf = q (Kt − 1) + 1 Kf = 0.65 (2.14 − 1) + 1 Kf = 1.74 Kf s = qs (Kts − 1) + 1 Kf s = 0.71 (3 − 1) + 1 Kf s = 2.42 For a machined surface, the endurance strength correction factor is -0.217 Ka = 2(85 kpsi) Ka = 0.762687 14/24 4/11/2020 Assignment Print View The value of the size modification factor, kb, based on your guess of a diameter between 1.00 inch and 2.00 inches is kb = 0.879(dguess ) kb = 0.879(2 in) -0.107 -0.107 = 0.8162 The infinite life value of the endurance strength for the steel shaft is Sf = ka kb (0.5) (85 kpsi) Sf = 26.4554 kpsi 1 ⎛ d = 1 ⎛⎛ ⎜ 16×1.5 ⎜⎜ ⎜ π ⎜⎜ ⎝ ⎞ 2 4(Kf Ma ) ( ) 2 ⎟ + ⎟ Sf ⎝⎝ 1 ⎛ ⎠ ⎜ ⎞ ⎞ ⎞3 2 3(Kf s T m ) ( ) ⎜ 2 ⎟⎟⎟ ⎟⎟⎟ Sut ⎝ ⎠⎠⎠ 1 1 2 (4(1.74(2178 lbf·in)) ) 16×1.5 d = ( π (3(2.42(2500 lbf·in)) ) 2 + 3 26.4554×10 psi (( 1 2 ) 3 85×10 psi ( 3 2 ))) d = 1.4631 in The factor of safety against yielding is determined as shown below. 1 σ' = 32Kf Ma (( πd 3 2 ) + 3 ( 16Kf s T m πd 3 2 2 ) ) 1 σ' = 32(1.74)(2178 lbf·in ) (( π(1.46 in) 3 2 ) + 3 ( 16(2.42)(2500 lbf·in ) π(1.46 in) 3 2 ) ) 2 × 10 −3 15/24 4/11/2020 Assignment Print View σ ' = 21.04 kpsi ny = Sy σ' = 71 kpsi 21.04 kpsi ny = 3.38 The shaft shown in the figure is driven by a gear at the right keyway, drives a fan at the left keyway, and is supported by two deep-groove ball bearings. The shaft is made from AISI 1020 cold-drawn steel. At steadystate speed, the gear transmits a radial load of 230 lbf and a tangential load of 633 lbf at a pitch diameter of 10 in. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 16/24 4/11/2020 9. Assignment Print View Award: 0 out of 8.33 points Determine the magnitudes of the bearing reactions at the bearings (points A and B). (You must provide an answer before moving to the next part.) |RA| = n/r lbf |RB| = n/r lbf Determine the magnitudes of the bearing reactions at the bearings (points A and B). (You must provide an answer before moving to the next part.) |RA| = 208.97 ± 2% lbf |RB| = 464.52 ± 2% lbf Explanation: Since there is only one gear, we can combine the radial and tangential gear forces into a single resultant force with an accompanying torque and handle the statics problem in a single plane. The resultant force is 2 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ P = √ ‾ (230 lbf) + (633 lbf)‾ = 673.49 lbf Σ F = 0 = RA - 673.49 lbf + RB Σ M = 0 = -673.49 lbf (6.98 in.) + RB (10.12 in.) RB = 673.49 lbf(6.98 in.) 10.12 in. = 464.52 lbf RA = 673.49 lbf - 464.52 lbf = 208.97 lbf The maximum bending moment at D is MD = 208.97 lbf (6.98 in.) = 1459 lbf·in. and the torque transmitted from D to C is T = 633 lbf × ( 10 in. 2 ) = 3165 lbf·in. Due to the shaft rotation, the bending stress on any stress element will be completely reversed, while the torsional stress will be steady. Since we do not have any information about the fan, we will ignore any axial load that it may introduce. It may not contribute as much as the bending anyway. Potentially critical locations are identified as follows: 17/24 4/11/2020 Assignment Print View The keyway at C, where the torque is high, the diameter is small, and the keyway creates a stress concentration The keyway at D, where the bending moment is maximum, the torque is high, and the keyway creates a stress concentration The groove at E, where the diameter is smaller than at D, the bending moment is still high, and the groove creates a stress concentration. There is no torque here. The shoulder at F, where the diameter is smaller than at D or E, the bending moment is still moderate, and the shoulder creates a stress concentration. There is no torque here. The shoulder to the left of D, which can be eliminated since the change in diameter is very slight. The stress concentration will undoubtedly be much less than at D. 18/24 4/11/2020 10. Assignment Print View Award: 0 out of 8.33 points What is the factor of safety at the keyway at C using the maximum shear stress theory? (You must provide an answer before moving to the next part.) The factor of safety at the keyway at C is 2.21 . What is the factor of safety at the keyway at C using the maximum shear stress theory? (You must provide an answer before moving to the next part.) The factor of safety at the keyway at C is 1.768 ± 2% . Explanation: 10 in × 633 lbf τ = Tr J = ( 2 π(1.00 in) 4 1.00 in )( 2 ) = 16.12 kpsi ×1000 32 Using Equation 5–3, ny = Sy /2 τ = 57 kspi/2 16.12 kpsi = 1.768 19/24 4/11/2020 11. Assignment Print View Award: 0 out of 8.33 points What is the factor of safety at the keyway at D using the DE-Gerber failure criterion? (You must provide an answer before moving to the next part.) The factor of safety at the keyway at D is n/r . What is the factor of safety at the keyway at D using the DE-Gerber failure criterion? (You must provide an answer before moving to the next part.) The factor of safety at the keyway at D is 3.071 ± 2% . Explanation: Consider the keyway at D. From Table 7–1, Kt = 2.14, Kts = 3.0 From Figure 6–26, q = 0.66 From Figure 6–27, qs = 0.72 Using Equation 6–32, Kf = 1 + q (Kt - 1) = 1 + 0.66 (2.14 - 1) = 1.8 Kf s = 1 + qs (Kts - 1) = 1 + 0.72 (3.0 - 1) = 2.4 Using Equation 6–19, kb = ( 1.75 0.30 -.107 ) = 0.828 Using Equation 6–17, Se = 0.8005 (0.828) (34.0 kpsi) = 22.5 kpsi Using Equation 7–6 with Mm = Ta = 0, 20/24 4/11/2020 Assignment Print View 2 ‾‾‾‾‾‾‾‾‾‾ 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ A = 4 = √ ‾ 4((1.8) (1459 lbf·in.))‾ = 5252 lbf·in. = 5.252 kip·in. √ (Kf Ma ) 2 2 ‾‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 10×633 B = 3 = √ 3((2.4) ( )) kip·in. = 13.157 kip·in. √ (Kf s Tm ) 2 1 1 n = 8A πd 3 1 + Sf 1 + ( ( ( 2BSf ASut 2 ) ) 2 ) 1 n = ⎛ 8A πd 3 S f ⎜ 1 + 1 + ⎜ ( ( 2BS f ASut 2 1 ⎞ 2 ⎟ ) ) ⎝ ⎟ ⎠ 1 n = ⎛ 8 × (5.252 kip·in.) π(1.75 in.) 3 (22.5 kpsi) ⎜ 1 + 1 + ⎜ ( ( ⎝ 2(13.157 kip·in.)(22.5 kpsi) (5.252 kip·in.)(68 kpsi) 2 ) ) 1 ⎞ 2 ⎟ = 3.071 ⎟ ⎠ 21/24 4/11/2020 12. Assignment Print View Award: 0 out of 8.37 points What is the factor of safety using the DE-Gerber failure criterion at the groove at E and at the shoulder at F? The factor of safety at the groove at E is n/r . The factor of safety at the shoulder at F is n/r . What is the factor of safety using the DE-Gerber failure criterion at the groove at E and at the shoulder at F? The factor of safety at the groove at E is The factor of safety at the shoulder at F is 4.08 ± 2% . 4.76 ± 2% . Explanation: Consider the groove at E. r/d = 0.1 in./1.55 in. = 0.065 and D/d = 1.75 in./1.55 in. = 1.13 Using Figures A–14 and A–15, Kt = 2.1 From Figure 6–26, q = 0.76 Using Equation 6–32, Kf = 1 + q (Kt - 1) = 1 + 0.76 (2.1 - 1) = 1.836 Using Equation 6–19, kb = ( 1.55 0.30 -.107 ) = 0.839 Using Equation 6–17, Se = 0.8005 (0.839) (34.0 kpsi) = 22.84 kpsi Using Equation 7–6 with Mm = Ta = Tm = 0, 22/24 4/11/2020 Assignment Print View 2 ‾‾‾‾‾‾‾‾‾‾ 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ A = 4 = √ ‾ 4((1.836) (1115 lbf·in.))‾ = 4094 lbf·in. = 4.094 kip·in. √ (Kf Ma ) B=0 1 1 n = 8A πd 3 1 + Se 1 + ( ( ( 2 2BSe 2 ) ) ASut ) 1 n = ⎛ 8A πd 3 Se ⎜ 1 + 1 + ⎜ ( ( ⎝ 2BSe ASut 2 1 ⎞ 2 ⎟ ⎟ ) ) ⎠ 1 n = π(1.55 in.) 3 = 4.08 1 8 × (4.094 kip·in.) 2 (22.84 kpsi) 1 + (1 + (0) ) ( 2 ) Consider the shoulder at F. r/d = 0.125 in./1.40 in. = 0.089 and D/d = 2.0 in./1.40 in. = 1.43 Using Figure A–15–9, Kt = 1.7 From Figure 6–26, q = 0.78 Using Equation 6–32, Kf = 1 + q (Kt - 1) = 1 + 0.78 (1.7 - 1) = 1.546 Using Equation 6–19, kb = ( 1.40 0.30 -.107 ) = 0.848 Using Equation 6–17, Se = 0.8005 (0.848) (34.0 kpsi) = 23.08 kpsi Using Equation 7–6 with Mm = Ta = Tm = 0, 2 ‾‾‾‾‾‾‾‾‾‾ 2 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ A = 4 K Ma ) = √ ‾ 4((1.546) (845 lbf·in.))‾ = 2613 lbf·in. = 2.613 kip·in. √ ( f B=0 1 1 n = 8A πd 3 1 + Se 1 + ( ( ( 2 2BSe 2 ) ) ASut ) 1 n = ⎛ 8A πd 3 Se ⎜ 1 + 1 + ⎜ ( ( ⎝ 2BSe ASut 2 ) ) 1 ⎞ 2 ⎟ ⎟ ⎠ 1 n = 8 × (2.613 kip·in.) π(1.40 in.)3 (23.08 kpsi) = 4.76 1 2 1 + (1 + (0) ) ( 2 ) 23/24 4/11/2020 Assignment Print View We will model the shaft with the following three sections: Section 1 2 3 Diameter (in.) 1.00 1.70 1.40 Length (in.) 2.90 7.77 2.20 The deflection problem can readily (though tediously) be solved with singularity functions. Alternatively, shaft analysis software or finite element software may be used. Using any of the methods, the results should be as follows: Location Left bearing A Right bearing B Fan C Gear D Slope (radians) 0.000290 0.000400 0.000290 0.000146 Deflection (in.) 0.000000 0.000000 0.000404 0.000928 Comparing these values to the recommended limits in Table 7–2, we find that they are all within the recommended range. 24/24 ...
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