Question

The rotating shaft shown below is supported by self aligning bearings at A and D. The diameter of the shaft changes from D-1.5" to d-1.0" at the midpoint of the shaft. The radius of the fillet at the diameter change is r-0.2" and all surfaces are machined. The shaft is made of normalized 1020 steel having . 64.0 ksi and S, = 50.3 ksi . The shaft is loaded with 250-lb forces at B and C as shown to create pure bending between B and C, with no torsion along the shaft. a) Estimate the fatigue factor of safety at the fillet for 10° or more cycles. b) Estimate the fatigue factor of safety at the fillet for 5x10 cycles. 250 lb 250 lb cs r.o" 1.5" G1 1" トー"4" 10" 10"

Answer 1

Determine:

1. Fatigue factor of safety if component is subjected to infinite life (consideration point will endurance limit).

2. Fatigue factor of safety if component is subjected to finite life (consideration point will be amplitude stress from S-N curve).

Solutíon :- betermmine i) ao of safeby (f) at fllet for o cyclet Fillet fo ycles cFor 아 caFe V 5xlo cycles Given Data filletrad i us (r)= 0.2 inch Surface condition - Machinec Normali zed 20 steel proper ti es i- Su (ulti mate tencile strengh yield strength (Sy) 50.3 ksi -50.3 xử psi Load at point ' and c - 64 ksi - 64xlo psi =250 lb

2501 b 250 b Now calculate the bearing reactions : RA/RD- 2 Rat RD = 50016 Due to symmehrycity of the loadi n Now we have to calculate the moment at fillet ơ)-02 Mg - looo L -inch Bendliuq stress ( fatijue strecs) at Hhat point- > eSs Fatigue stress (F) = 3 (σε) 32x1000 3 CE 10.186 ksi

Calculation for fati ue safer factorf。r infinite afe Žtoccycle Endurauce limit (EnJurance strenath 2- Corrected Endurance Umit -S<e CI) where ka - Surface Anich factor kc = Modification factor due to stress concentrati at Fillet point- KaMachined surface ka - a (Sut ー> Sut= 64-ksi = 441. 27 MPa For machined surface au. s1 and b-o.265 -o.24S . ka = 4.51 441.27 ka - o-29g ko -» size facfor d. 1.0 tinch-25.4 mm kc → modification factor dl stress concentrato factor (usivy aph belo w)

3.0 2.4 2.2 D/d- 2.0 D/d- 1.5 2.0 Dld-1.2 1:.8 1.6 1.4 Dld- 1.1 1.0 0.0 0.1 0.2 0.3 (r/d) Fig. 5.5 Stress concentration factor Round shaft with shoulder fillet in bending)

匕 l'니 kc = 0.72 corrected enJuronce shregth Sé - ka kb ke Se ..Su= 64 ksi 2. 2 Next part:- calculation of safety fact ar Infinite ife If- any、 component can sus ta in 10 in the compouent should not exceed the corrected enduranie limit. Henca loe the stress imit ex ce fatizue safeby fackor 17-586 10 186 si 1-7 3 uswer of part (i) Part 'B Calculation fsafety factor fer Sxlot cycles ing(동) 辽 ug (Se= 121.25) = 2.084 log (0.9%) = 2.6

usìuq the ciwila tiaugle or üu equation (3-4-?) 3 -6 2.6 - Sao = 2.6-0.2924 2.3076 cl = 203.0486 MPa ~ 203 . 1 MPa よ

here thi component is runniug Sxio'cyclas , so covrespondli ta ús ît can sustain more stressthat is Sa= 2o 3.1 Mpa , The stress develo ped dz to bendiug Voad (calcuted befs te) ena Iu ơf- 10-186 psi-70.23 MPa Hence fatigue facts. of safety(f)ウ Sa 203. 70.23 2.89 Solu tion of part B