Question

6 In the figure shown, shaft A, made of AISI 1010 hot-rolled steel, is welded to a fixed support and

is subjected to loading by equal and opposite forces F via shaft B. A theoretical stress concentration
Kts of 1.6 is induced by the 3 mm fillet. The length of shaft A from the fixed support to the connection
at shaft B is 1 m. The load F cycles from 0.5 to 2 kN.
a) For shaft A, find the factor of safety for infinite life using the modified Goodman fatigue failure
criterion.
b) Repeat part(a) using Gerber fatigue failure criterion.

Answer 1

Step-by-step explanation:

Calculate the factor of safety for infinite life using modified Goodman fatigue failure.

Tmax =16KfsTmax /πd^3

HB =200, r=3 mm, qs = 1

Kfs=1+qs(Kts-1)= 1+1(1.6-1) = 1.6

Calculate the maximum and minimum torque.

Tmax=2000(0.05) = 100 N. m

Tmin=500 /2000 (100)

Tmin= 25 N.m

Calculate the maximum and minimum shear stress on the shaft.

Tmax= 16(1.6)(100) (10-6) /π(0.02)3 = 101.9 MPa Tmax

Tmin=500 /2000 (101.9)

Tmin=7.4 N. m

Kb=(7.4/7.62) ^-0.107

Kb=0.59

Se=0.917(1.003)(0.59)(161.3) = 87.5 MPa

nf = 1/ (Ta/Se)+(Tm/Ssu)

=1 /(38.22/87.5)+(63.68/214.4)

= 1.36

b)

Calculate the factor of safety for infinite life using modified Gerber fatigue failure.

nf=1/2 (Ssu/Tm) ^2 Ta/Se [ -1+square root of 1+

(2Tm Se/Ssu Ta) ^2]

= 1/2 (214.4/63.68) ^2 (38.22/87.6) {-1+square root of 1+ [2(63.68) (87.5) /214.4 (38.22) ]^2

=1.70

= 25.46 MPa

Calculate the mean stress.

Tm = 1/2 (101.9+25.46)

= 63.68 MPa

Calculate the average stress.

Ta= 1/2

Ta= 1/2 (101.9-25.46)

=38.22 MPa

Ssu=0.67Sut

= 0.67(320)

= 214.4 MPa

Ssy = 0.577Sy

= 0.577(180)

= 103.9 MPa

Se = 0.504(320)

= 161.3 MPa

ka =57.7(320)-0.718

= 0.917

de = 0.370(20)