132 HOMEWORK PROnMEMs DRAW THE coMEPIETK s N DIAGRAM he text

13.2

HOMEWORK PROnMEMs, DRAW THE coMEPIETK s N DIAGRAM, he textbook problems. Please let Dr Trease know hese problems A13.Ir Estimate the enduran trength ofa -in-diameter m er AISI 40 steel having a machined finish and heat-treated to a ten strength kyst in hwdini. Next estimate the endu \"estrength for the same speci der repeated shoulder fillet, reducing the shaft diameter to 1-3 diameter solid round bar has a 01-n ra of 1700 lbf-in, ublected to a pan ly reversing tom elationship given in section 6.13 a torsion problem.be sure to be a notch in Fig 6-21) timate consider (a) Esti ent with a temperature of 863 F the surface of venus he bar is also plaped in an en stimate the number of cycles to failure, wersed axial load A13.3: The cold drawn AISI 1045 steel bar shown in the ngure ls subjected to a completely on to 25 kN in ten fluctuating between 25 kN in Estimate the fatigue factor ofsafety based on achevinginfinite life andthe yielding factor of safety. nfimite lure is not predicted, estimate the mul ber of cycles to failure. 20 mm Page

Solution

solution:

1)here for given rod subjected to torsion of 1700lbf in or 192.07*10^3 N mm

with smallest diameter of 1.3 in or 33.02 mm

2)here for AISI 1030cold drawn steel has properties as

Sut=525 mpa or 76100psi

Syt=63800 psi or 440 mpa

3)here endurance limit is given by

Se\'=.5*Sut==38050 psi

where factor are

surafce finish factor for CD are

ka=4.51(Sut)^-.265

where sut=525 mpa

so we get that

Ka=.8576

size factor

Kb=1.2587(d)^-.1133 d in mm

Kb=.8469

load factor

Kc=1

temperature factor

Kd=1

stress concentration factor for q=.8

for fillet radius r=.1 in

here r/d=.076

for that value by interpolation we get that

stress concentration factor=Kt=1.84

Kf=fatigue factor=q(Kt-1)+1=1.672

Ke=1/Kf=.5980

hence endurance strength is

Se=Ka*Kb*Kc*Kd*Ke*Se\'=16526.19 psi

4)for finite life ,life of rod is depend stress induced

sf=tmax=tm=16*T/pi*d^3=3940.84 psi

5)hence life of rod is

log(sf)-log(Se)/6-log(N)=Log(.9Sut)-log(Se)/6-3

we get that number of cycles as

n=1059064854 cycle

6)where number of cycles at temperature of 863 F or 461.667C is responsible for variation of endurance strength only,hence temperature factor

Kd=.8227 by interpolation

so value of Se1=Se*Kd=13596.63 psi

and for same stress level life would be

N1=198532190.2 cycles

6)hence at high temperature life of component get reduce