The number of cyclces to failure for the shaft material is f

The number of cyclces to failure for the shaft material is found by determining the life of the axle (how many miles will it last)

3. A motor vehicle has the given stub axle made of steel having an ultimate strength of 1050 MPa and fatigue limit oft560 MPa. The surface in the region of high stress is ground. The axle does not rotate as the wheel turns; consequently, the stress is substantially constant as long as the vehicle is traveling on a smooth road. Under these conditions the force, F, is 11.0 kN Whenever the wheel strikes an obstruction, the force, F, can be considered to increase by a certain amount and then decrease by the same amount before returning to its constant level. For design purpose the wheel may be considered to encounter two types of bumps, called minor and major. A minor bump, which can be expected once every 2 kilometers, imposes a transient force of 5 kN. A major bump occurs with one-tenth ofthe frequency of a minor bump and results in a transient force of 8 kN. Using Miner\'s cumulative damage theory, estimate the safe life of the axle, in kilometers. 20 mM 100 mm

Solution

solution:

1)here assume that vehicle tyre is rotating at Nt=720 rpm and velocity of V= 50 Km per hr

2)here load of F1= 11 KN is acting on axle of 30 mm,hence bending moment is

M=F1*l=11000*100=11*10^5 N mm

hence bending stress is

Sb=32*M/pi*d^3

S1=414.98 MPa

2)where minor bump stresses are

F2=F1+5=16000 N

M2=16*10^5 N mm

hence bending stress is

Sb=32*M/pi*d^3

S2=603.6 MPa

3)for major bump

here load of F3=F1+8=19 KN is acting on axle of 30 mm,hence bending moment is

M=F3*l=19000*100=19*10^5 N mm

hence bending stress is

Sb=32*M/pi*d^3

S3=716.786 MPa

5)here endurance limit of shaft is given by

Se=Kb*Ke*Se\'

where for radius of fillet/shaft diameter=3/30=.1

Kt=1.68

Ke=1/Kt=.5952

Kb=.85

hence

Se=.5952*.85*560=285.37 MPa

5)where for axle subjected to finite life of completely reverse stresses then life is given for each stress alone is as follows

log(S1)-logSe/6-logN=Log(.9Sut)-logSe/6-3

here Sf1=S1=414.98 MPa

N1=115307.27 cycle

for Sf2=S2=603.6 MPa

N2=13277.16 cycle

for Sf3=S3=716.786 MPa

N3=4926.11 cycle

7)here let assume that this bump takes 5% kilometer length to gradually increased load and vanished for both major and minor bump,hence

a2=percentage of Km take by minor bump=.05*(1/2)=.025 %

a3=pertcentage of Km take by major bump=.05*(1/20)=.0025 %

a1=percentage of Km take by smooth road is=1-a2-a3=.9725%

8)hence by miner equation we get that

a1/N1+a2/N2+a3/N3=1/Nr

Nr=92383.707 cycle

here resultant life of axle for given condition with assumption is Nr=92383.707 cycle

9)here for tyre speed of 720 rpm and speed of 50 KPh we get axle life in Km as follows

life in Km=(Nr/Nt)*V/60=92383.707*50/720*60=106.91 Km=66.2842 mile