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 8.50 0.20 2.20 0.75 2.75 2.0 1.000 0.485 1.750 1.181 1.70 1.40 + LIR A DE B A fr. 0.IR 1 x keyway 0.15 AR R 1-10715-7 6.98 -3.14 - 0.38 Sut=91 kpsi Sy=77kpsi Ka=0.75 What is the maximum bending moment? Be sure to show the moment and torsion diagram. What is Se? What is Kts? What is the factor of safety?

Answer 1

→ Consider the figure given in question As, there is only one gear, i am combining both sadial + tangential force to a single resultant force, plus a corresponding torque. Now, on calculating the reactions,

Resultant force at D = 13002 + 7002 a 7610577 lbf and torque at D 700 X 8 2 2800 lbf-in

2230 - Reaction at B RB 761.577 X 6.98 10.12 = 525.27 16F RA= 761.577 - 525:27 = 236.307 lbf Ans. O Maximum B.M will be at D MD = 236.307 x 6.98-1649.422) lbf-in Due to rotation of shaft, the bending stress on any element completely revessed, and torsional stress will be steady. B.MiD will be 1649.422 M lbf-in) 1263.27 E F B 0 D С ME (6.170.485 + 2.2 -0.38)x525.27 = 1263.27 lbf-in (2.2 - 0.38) x 525.27= 955.9914 ebf-in Tosque diagray T Cbf-in) 2800 V DEF B A

Assume } = 0-02 2 with d = 1.25 in k = 0.035 in. Ans kt = 2.14 and Now the answers are required for slot at D and hence Calculations are done for that only! For typical end-milled cutter for keyway (9th ed) Reference for values below - Mechanical design by 'shigley! use table 7-1 (Page 373); (kts = 3.0 fig (6-20) q = 0.66, lkf= 1+qCkt-1)=1.08 Eig-6-21) 9 = 0.72 Ifs=1+95 (kts-D = 2.4. Giren, Sut = 91 kpsi, Sy= 77 kps i ka=0.75 0.5 x 91 45.5 kpsi. 1.75 0.828 ke=0.897 (90% Reliability) Hence, se kaxky a se = 0.75 X 0.828 X 45.5 X0-897 xke (25:345 kpsil = 25.345 kpl using DE-Goodman criterion (Man-Ta=0) (eq 77-7) 9 :-se - 0.107 0.3 12 | n 16 Td3 { [4 Ck4 Mas +3 (kfs ta)] 4 / [ACK & M M ² Sut +3(Efs Tom] Put Mm = Ta = 0; | - 16 (੧) [4 (13X 1849.412) | 3 (3.4X2800)}]}} N (1.753 xk 25345 91000

0.95028 X } h 0.23478 + 0.12790 12770 Required Factor of safety! 2.9054 n = Bodon Ang n=2.9054

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