Page 69 - Analysis and Design of Machine Elements

Steps Computation Strength of Machine Elements 47
Results
Units
4. Locate working point From Eq. (2.3)
and fatigue limit in the
− diagram = r =0.25 × 480 = 120MPa
m a min max
Therefore
+ 480 + 120
= max 2 min = 2 =
m
300MPa
− 480 − 120
= max min = = 180MPa
a
2 2
Locating working point M by ( , ).
m a
Connect points O and M and extend
the line to intersect with − diagram
m
a
′′
′
to obtain M and M ,which are
the endurance limit and yield limit,
respectively.
5. Determine the safety Find the coordinates of points M ′ S = 1.46 < [S]
′′
factor (437,262) and M (500,300).
Safety factor for infinite life
′
S = + ′ a = 437 + 262 = 1.46
m
+ a 300 + 180
m
Safety factor for static strength S = 1.67 > [S]
′′
+ ′′ 500 + 300
S = m a = = 1.67
+ a 300 + 180
m
Discussion
The results obtained from the − diagram are similar to those from calculation.
a
m
Using the - diagram, one can easily find out the reason for failure. From the − a
m
m
a
diagram, the element is prone to yield rather than fatigue; therefore, it is not necessary
to analyse fatigue strength.
Example Problem 2.2
An initial design of a shaft is shown in Figure E2.2. The shaft is made of tempered
medium carbon steel, with = 640 MPa, = 275 MPa, = 155 MPa, = 500 MPa,
b −1 −1 0
= 295 MPa and rotates at a speed of n = 20 rpm. Assume the critical number of cycles
0
7
is N = 10 and the material constant of the S−N curve is m = 9. The shaft is designed
0
to work 8 hours daily, 300 days yearly for 2.5 years. Operating stresses on the cross
section H are: bending stress = 5.48 MPa, axial tensile stress = 0.26 MPa
bending tension
and torsional shear stress = 11.0 MPa. An allowable safety factor is [S] = 1.6.
Calculate the endurance strength and check the strength of the shaft at the cross
section H.