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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. 74,
Mxy = —209.3 lbfin, Mch = —293.01bf‘in, :r = 192 lbfin 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. A158: Km = 1.8
Fig. A—159: K, = 2.3
Fig. 620: q = 0.65
Fig. 621: (15 = 0.70
Eq. (632): Kf =1+0.65(2.3—1)=1.85
Kfs =1+0.70(1.8—1) = 1.56 Using DEASME Elliptic, Eq. (711) 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 crownedteeth in the gears will eliminate this problem. 77 through 716 _
These are design problems, which can have many acceptable designs. See the solution for
Prob. 717 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} lbfé (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, =21781bfin Tm =25001bf8in Mm :7; :0 From Table 71, estimate stress concentrations for the endmilled 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 621, 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. (632): Kf =1+0.75(2.14—1) = 1.9
K}, =1+0.8(3.0—1)= 2.6
Eq. (619): k, = 2.70(68)‘°‘265 = 0.883
For estimating kb, guess d = 2 in.
Eq. (620) k, = (2/0.3)““°7 = 0.816
Eq. (618) 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(1.92178)2]U2 [3(2.62500)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 noncrowned 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.
 Spring '10
 Dr.Yang

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