Question

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 at locations A and B. The shaft is made from AISI 1045 cold-drawn steel. At steady- state speed, the gear transmits a radial load of 300 lbf and a tangential load of 700 lbf at a pitch diameter of 8 in. Assume that the weight and axial load of the fan is negligible this allows for combining the forces of the gear into a single plane. Determine fatigue factor of safety at location D of the keyway using the DE-Goodman failure criterion with a 90% reliability. 12.87 8.30 2.20 2.75 0.20 0.485 1.750 0.75 1.181 1.70 1.40 2.0 1.000 1 11.181 C D E B R0115 R. x keyway 0.1 R. -0.15 BER R R. |-.1.0-26-1.37-4 6.98 3.14 -11-038 Sut=91 kpsi Sy=77kpsi Ka=0.75 What is the maximum bending moment? Be sure to show the moment and torsion diagram.

Answer 1

→ Consider the figure given in question. As, there is only one gear, i am combining both radial + tangential force to a single resultant force, plus a corresponding torque. Now, on calculating the reactions, Rax 209.90 tof The ancial force is neglected which might Bot 464.5 lbf be produced by the fan. 3002 + 7002 - 761.577 lbf Resultant force at D and torque at D 700 X 8 Na 2800 lbf-in

- Reaction at B 761.577 X 6.98 10.12 = 525.27 16f Answer: Man B.M. = 1649.422 Ibftin RA= - 761.577 - 525.27 = 236.307 lbf Maximum B.M will be at D MD 236.307 x 6.98-7649.422 lbf-in Due to rotation of shaft, the bending stress on any element will be completely reversed. and torsional stress will be steady. B.MiD 1649.422 (lbf-in) 1263.27 0 С A D E MF = ME (6.1 +0.45 +2.2-0.38 x 525.27 = 1263.27 lbf-in (2.2 - 0.38) × 525.27= 955.9914 ebf-in Torque diagram r Me=0 MB=0 T -Ibf-in) Como . ) 2800 O A DEF B

Assume k = 0.02 / e = 0.035 in. 1.8 Now, the ancwers are required for slot at D and hence Calculations are done for that only! For typical end-milled cutter for keyway त with d=1.75 in (9th ed) Reference for values below - Mechanical design by 'shigley! use table 7-1 (Page 373); Answer for kts kt = 2.14 and (kts=3.0 fig (6-20) - q = 0.66, Kf = 1+qCkt-1) Eig- 6.21) qs = 0.72 Ifs=1+qs (kts-1 = 2.4. Giren, Sut = 91 kpsi, Sy= 77 kpsi ka=0.75 0.5 X 91 = 45.5 kpsi. 0.828 ke=0.897 (90% Reliability) Hence, se Kaxky x Se = 0.75 X 0.828 X 45.5X0.897 xke (25:345 kpsi = 25.345 kpl Answer using DE- Goodman criterion (Man-Ta=0) for se (eq 77-7) - 0.107 1.75 0.3 1 The te [4Cky ma? 43 (ks ta)] Sut Put Mm = Ta=0; + 1 [4Cky MM) +3lHtsta)) [4 (18 X 1649.422)3 (2.4x2800))))) 16 (1.753 x ñ 25345 91000

0.95028 X { 0-23425 + 0-13790 N In = 2.9054 Booos Required Factor of safety! And Answer n=2.9054

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