Question

• Design a well-roundedshoulder at Hand the keyway near G • Use AISI 1045 CD steel • Reliability 95% • Gerber n>3 • Langer line n>3 Problem Statement Shown is the intermediate shaft in a double reduction gearbox. The tangential and radial forces through the gears to the shaft are as follows: Ws = 540 lbf Was = 197 lbf W = 2431 lbf W. = 885 lbf where superscripts and r represent tangential and radial components, respectively. G 167 СА РЕГон

Answer 1

PAGE No DATE 4 get W 23 wist Solution Perform free body diagram analysis to reacho Forces the bearing "We23 at RAZ = 115 abr Ray 21356.7 1bf RBZ 21776 16F RBY = 753 725.3 lbf Roy A Luisa G IJ K 16 RAZ RB2 z 3240 from EM find the oil 2 1 torque in shaft bet" gears T = W / 23 (d3/2) -540 (122) - 3240 Sbf in V X2 plane -1776 13996 3341 М. Generate Shear- moment diagrama for too planes 230 2420 - 1725 plane -- M +4721 1632 713 907 1

PAGE No. | DATE Combine orthogonal planes 4316 1 2 365.1 - as vector to get total moment. Trot egat j = √3998², 183 2² 4376 01bf 10 749 4 Start with point I, where the bending moment is high, there is stress concentrarion! at the shoulder and forque is present At I, Ma: 3651bf. in Tm: 3240 lbfin Mm : Ta = 0 Assume generous Hillet iadius for gear I from table we found stittnen value kt 17 kes 1.5 for quick, conservative first pass assume kf=kt kh :kt Choose inexpensive steel 1020 (D with sut: 68kpsi Forse, ka a sus 21668) 0265 = 0.893 Guess tb = 0.9 (check later when d is known) ke': kd : ke: ses (0.983) (0.9160.5) (68) - 27 kps) Foo Arst estimate of the small diameter at the shoulder at pt I use the DE- Goodman citessa. 327/21 173 de 159 (2009 Ma) [5(kts Troy? ] Se 1776 3996 + ut Vz ws dos 1605) 2 x 1:7x3651 27010 + 240] 68000 1901 whild a 1.65 int All estimates have probaby been conservative, so seletet next std size . Id=1.625 in

PAGE NO DATE 3 A typical old rasio for support at a showdes is old = 12 thus = 1.2 (1.621) 1.92 in. Increase to: 2.0 in 이 이 211025 1.23 Assume fillet radius y=d/10 & 0-16 in rid: 0.1 Kt=1.6 9=0.82 ke & 1 +0.82 1.6-1) 2 1.49 to its i 1.35 KES = 1 + 0.85 (-35-1) 1.30 kad 0.885 2825 0.135 8.3 Se 0.883 x 6-835 x 0.5888 - 25.1 Epsi -0./07 Kb fillones ? 32 KEMA - 32(1.49) 3651 12910gsi H (1.625) 279 V3 (16) (1.30) (9240) 865915 (1.6253 34 16kGTm rds -) usind Goodma Giteria in 8' + ne ut 1 12910 25 100 + 8659 0.642 63000 netist Now check yielding Sy Sy 57000 : 2.64 I Lobsera s'man bat i'm 12900 + 86.59 Also check this diameter at end of keyway just the right of point I and a the groove of point K. from moment diagram estimate M at end of keyway MA 3750 tbr in.. 40 be

I DATE 4 Assume the radius at the bottom of keyway will be the standard rod behat Ild 0.02,7 = 0.02d = 0.02(1.625 ) - 0.0325 in kt: 2.14 9 = 0.65 Kf = 1 +0.65 (2.14-1) = 174 kts kis : If 0.2 03-D) : 2.42 - 3.0 + 9 = 0.71 z'a: 32 KIM a 32 -32 (1.749 3750 M (1.825) 154908 psi 8'm = √3 (16) Kism V3 (16) (2-42) 3240 16 120 psi red Hu M (1.6253 I 25 15490 + + 16120 - 0.854 nf sut 25100 68000 use = 100 lg si -2.625 The key way turn to be more critical than the shoulder We can either increase the diameter o bigben strength material. Toy 10s0 CD with Recalculate factor Sut, Se ; qk4 8'a la: 2.7 x 100 20.197 0.797 *0.336x0:5 x 100 = 333 kps, kf : 182 Ba. 32 ):82 3750 tx1.825 16200 + 16120 of 33300 100 doa Se 9 = 0.72 18200 kapsi 3 0.648. nf. 1.54 accepting close value 22:15

PAGE No. DATE 5 Check at the groore at k, since kt for that bottomed ottes vary high from torque diagrom a Note that no torque is groove. from moment Ma: 2398 Ibtin Mm Tam - grooves are present at diag. 20 : 32 kp Ma - 32 (5) 2398 28460psi d3 x 1.625 33300 ne: Se ba 3.28 460 This is loo We will look up data for a specte retaining obtain kf more accurately. kt: 4.3 kf 21+ 0.65 14 3 1) = 3.15 'ring to 92 0.651 conuntiation is high za 32kp Mo 32 (315) 2398 17930 psi Hd 3 1 (1.125) Se 33 3001 ba 17930 Quickly check it poich M might be Critical. Only bending is present, and the moment is small, but the diam. is small & stress for a sharp fillet required for a bearing. from moment dagrom: Ma: 959 1bfin f Mm : Tm tao KE: 2.7 Id: 0.02 r: 0.02*1:0.02 q = 0.7 E 1 +0.7 (2.1 -0 : 2.19 32 kf Ma 82C2-10 6959) 21396 Psi 7CD 1 10"

PAGE NO DATE sa 83 300 oa 21390 show di be ok close enough to recheck afte bearing is selected. With the diameters specified for the critical locations, fill in trial values for rest of the diameter, tellung into account typical shoulder beights for bearing & gear support al & the DI 17 a 1.0 in D2 z brola 11.4 in D3Ds a 1.625 in D4 220 in