Solutions_for_Problem_7-17

Solutions_for_Problem_7-17 - The simplified model yielded...

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Unformatted text preview: The simplified model yielded reasonable results. Strength S“, = 72 kpsi, Sy = 39.5 kpsi At the shoulder atA, x = 10.75 in. From Prob. 7-4, Mxy = —209.3 lbf-in, Mch = —293.01bf‘in, :r = 192 lbf-in M = ./(—209.3)2 + (—293)2 = 360.0 lbf ~in S; = 0.5(72) = 36 kpsi ka = 2.70(72)‘°'265 — 0.869 1 —0.107 kb = (m) = 0.879 0.3 a=a=a=a=1 Se = 0.869(0.879)(36) = 27.5 kpsi D/d=1.25,r/d=0.03 Fig. A-15-8: Km = 1.8 Fig. A—15-9: K, = 2.3 Fig. 6-20: q = 0.65 Fig. 6-21: (15 = 0.70 Eq. (6-32): Kf =1+0.65(2.3—1)=1.85 Kfs =1+0.70(1.8—1) = 1.56 Using DE-ASME Elliptic, Eq. (7-11) with M m = I; = 0, 1/2 1_ 16 4 1.85(360) 2+3 1.56092) 2 72 7:03) 27 500 39 500 n=3.9l Perform a similar analysis at the profile keyway under the gear. The main problem with the design is the undersized shaft overhang with excessive slope at the gear. The use of crowned-teeth in the gears will eliminate this problem. 7-7 through 7-16 _ These are design problems, which can have many acceptable designs. See the solution for Prob. 7-17 for an example of the design process. 7 ~17 (a) One possible shaft layout is shown in part (e). Both bearings and the gear will be located against shoulders. The gear and the motor will transmit the torque through the Chapter 7 - Rev. A, Page 10/45 keys. The bearings can be lightly pressed onto the shaft. The left bearing will locate the shaft in the housing, While the right bearing will float in the housing. (b) From summing moments around the shaft axis, the tangential transmitted load through the gear will be W, =T/(d/2) 22500/(4/2) 21250 lbf The radial component of gear force is related by the pressure angle. Wr : W, tan¢ =1250tan20” = 4551bf 1/2 1/2 W = (W,2 +Wf) = (4552 +12502) =13301bf Reactions R A and RB, and the load W are all in the same plane. From force and moment balance, RA = 1330(2 /11) = 2421bf RB =1330(9/11)=10881bf Mmax = RA (9) = 242(9) = 2178 lbf -in Shear force, bending moment, and torque diagrams can now be obtained. Ans. {(133 BTW 3.175% Ebt'in M 25th} lbf-é (c) Potential critical locations occur at each stress concentration (shoulders and keyways). To be thorough, the stress at each potentially critical location should be evaluated. For Chapter 7 — Rev. A, Page 11/45 now, we will choose the most likely critical location, by observation of the loading situation, to be in the keyway for the gear. At this point there is a large stress concentration, a large bending moment, and the torque is present. The other locations either have small bending moments, or no torque. The stress concentration for the keyway is highest at the ends. For simplicity, and to be conservative, we will use the maximum bending moment, even though it will have dropped off a little at the end of the keyway. ((1) At the gear keyway, approximately 9 in from the left end of the shaft, the bending is completely reversed and the torque is steady. M, =21781bf-in Tm =25001bf8in Mm :7; :0 From Table 7-1, estimate stress concentrations for the end-milled keyseat to be K; = 2.14 and Kt, = 3.0. For the relatively low strength steel specified (A181 1020 CD), roughly estimate notch sensitivities of q = 0.75 and qs = 0.80, obtained by observation of Figs. 6- 20 and 6-21, assuming a typical radius at the bottom of the keyseat of r / d = 0.02 (p. 373), and a shaft diameter of up to 3 inches. Eq. (6-32): Kf =1+0.75(2.14—1) = 1.9 K}, =1+0.8(3.0—1)= 2.6 Eq. (6-19): k, = 2.70(68)‘°‘265 = 0.883 For estimating kb, guess d = 2 in. Eq. (6-20) k, = (2/0.3)““°7 = 0.816 Eq. (6-18) S, = O.883(O.816)(0.5)(68) = 24.5 kpsi Selecting the DE—Goodman criteria for a conservative first design, 72 S S e 111‘ 1/3 1/2 1/3 1605) [4(1.9-2178)2]U2 [3(2.6-2500)2] d = —————+ 7: 24 500 68 000 d=1.57 in Ans. With this diameter, the estimates for notch sensitivity and size factor were conservative, but close enough for a first iteration until deflections are checked. Check yielding with this diameter. ‘ Chapter 7 - Rev. A, Page 12/45 1/2 3219M, 2 1619,er 2 Eq.(7—15): cam: — +3 m" 75(l.57)3 7z(1.57)3 ny = s, mg,“ =57/18.4 = 3.1 Ans. 2 21/2 6. 218389,”. =18,4kpsi (e) Now estimate other diameters to provide typical shoulder supports for the gear and bearings (p. 372). Also, estimate the gear and bearing widths. (f) Entering this shaft geometry into beam analysis software (or Finite Element software), the following deflections are determined: Left bearing slope: 0.000 532 rad Right bearing slope: —— 0.000 850 rad Gear slope: — 0.000 545 rad Right end of shaft slope: — 0.000 850 rad Gear deflection: — 0.001 45 in Right end of shaft deflection: 0.005 10 in Comparing these deflections to the recommendations in Table 7 -2, everything is within typical range except the gear slope is a little high for an uncrowned gear. (g) To use a non-crowned gear, the gear slope is recommended to be less than 0.0005 rad. Since all other deflections are acceptable, we will target an increase in diameter only for the long section between the left bearing and the gear. Increasing this diameter from the proposed 1.56 in to 1.75 in, produces a gear slope of —— 0.000 401 rad. All other deflections are improved as well. Chapter 7 — Rev. A, Page 13/45 ...
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This note was uploaded on 03/07/2012 for the course MAE 3090 taught by Professor Dr.yang during the Spring '10 term at FIT.

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Solutions_for_Problem_7-17 - The simplified model yielded...

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