A particular point on a 34 diameter stepped shaft is loaded

A particular point on a 3/4\" diameter stepped shaft is loaded in torsion of 100in#(1+sin(t)) and a bending moment of 300in#(sin(t)). The shaft is rotating and is reduced by a groove that has a radius of 1/8\" and a depth of 1/8\" cut for a retaining ring. The shaft is machined from AISI1080 HR steel. Determine the fully corrected endurance limit. Determine the Von Misses equivalent representation of the stresses. Determine the factor of safety using the Gerber fatigue failure theory. If the factor of safety is less than one determine the number of cycles the part will last. Determine the factor of safety using the Goodman fatigue failure theory. If the factor of safety is less than one determine the number of cycles the part will last. Determine the diameter, d, of the shaft that is required for infinite life using the gerber failure theory.

Solution

Maximum Torsion, MT= 100*(1+1) = 200 [units] (in is not the full unit of torque and hence units are assumed)

Minimum Torsion, MT’ = 100*(1+(-1))=0 [units]

Maximum Bending Moment, MB = 300*(1) = 300 [units] occurs when torsion is maximum

Minimum Bending Moment, MB’ = 300*(-1) = -300 [units] occurs when torsion is minimum

Diameter of the shaft: D=0.75 [in], Radius of groove, r = 0.125 [in], Depth of grove, h = 0.125 in

Thus, depth of the narrowest section, d = D -2*h = 0.75-2*0.125=0.5 [in]

The groove is of semi-circular type. r/d = 0.125/0.5 = 0.25

Stress concentration factor at r/d=0.25 for semi-circular groove in bending = KB = 1.56

Stress concentration factor at r/d=0.25 for semi-circular groove in torsion = KT = 1.29

Where KB=1 + 1/[0.4*r/(D-d)+5.5(r/d)*(1+2*r/d)^2]^0.5

And        KT=1 + 1/[1.4*r/(D-d)+20.6(r/d)*(1+2*r/d)^2]^0.5

UTS for AISI1080, SUT = 615[MPa], YS for AISI1080, SY = 380 [MPa]

High cycle endurance limit without any stress concentration in bending SBE » 0.50*615 = 307.5 [MPa]

High cycle endurance limit without any stress concentration in torsion STE » 0.29*615 = 178.4 [MPa]

(a)Corrected endurance limit in bending = SBE’ = SBE/KB = 307.5/1.56= 197.1 [MPa]

Corrected endurance limit in torsion = STE’ = STE/KT = 178.4/1.29=138.3 [MPa]

(b)SB = KB*(32*MB*/pi*D³), ST=KT*(16*MT/pi/D³), SEQV = (SB/2)±[ST^2+(SB/2)^2]^0.5

(c)n = design factor, SA = stress amplitude = (SEQVMAX-SEQVMIN)/2, SM = Mean stress = (SEQVMAX+SEQVMIN)/2, SE = Endurance stress. Hence, n(SA/SE)+(nSM/SUT)^2 = 1 as per Gerber Line

(d)(SA/SE) + (SM/SUT) = 1/n

(e)The unit of torsion and bending moment is required to correctly estimate the factor of safety as well as diameter for infinite life.