Page 61 - Analysis and Design of Machine Elements

Strength of Machine Elements
Again, the actual endurance limit diagram of an element can be expressed by line 39
equations. For line EF ,the equation is thesameasthatofline EF,thatis
1
′
+ ′ me = s (2.22)
ae
While in section A F , the line equation can be obtained by the coordinates of points
1 1
A (0, /K )and C ( /2, /2K ), as
0

−1
0
1
1

′
= K + ′ (2.23)
−1 ae me
or
−1 ′ ′
= = + (2.24)
−1e ae e me
K
where ′ ae and ′ me are the stress amplitude and mean stress of an element, respectively,
′
and = 1 2 −1 − 0 = tan is the mean stress influence factor of an element. These
•
e
K 0
equations are applicable to shear stress by substituting with .
2.3.4 Fatigue Strength for Uniaxial Stresses with Constant Amplitude
Like static stress state, the strength criterion for constant fluctuating stresses is also
S ≥ [S]. The calculated safety factor S is defined as the material strength divided by
ca
ca
maximum working stress, that is,
lim
S = (2.25)
ca
max
3
For any number of cycles of N (10 < N < N ), the endurance strength is determined
0
√
by Eq. (2.16) as = m N 0 ; while for any stress ratios, the endurance limit is deter-
rN N r
mined by a − diagram.
m
a
Generally, when analysing fatigue strength of a machine element, the maximum and
minimum stress at critical cross sections are first determined by load variations. These
stresses are then converted to mean stress and stress amplitude ,and thelocation
m a
of working point M ( , ) can be indicated in the − diagram. The maximum
m a m a
working stress is the sum of mean stress and stress amplitude . the endurance
m
a
limit locates on the line A F E in Figure 2.7. If the location of endurance limit on the
1 1
line A F E can be determined, the safety factor can be easily obtained.
1 1
The determination of endurance limit on the line A F E is based on structural con-
1 1
straints and stress variation as loads would fluctuate to cause element failure in service.
The typical cases of stress variation include constant stress ratio r = Const., constant
mean stress = Const. and constant minimum stress = Const. [14], as illustrated
m min
in Figure 2.7.
A practical example of a constant stress ratio involves the stresses in a rotating shaft. In
the endurance limit diagram, the working point M (or N) indicates the combination of
mean stress and stress amplitude representing the critical stress in an element. Connect
the origin O and M (or N) and extend the line to intersect with the endurance limit line
A F or F E at M (or N ), we have
1 1 1 1 1
a ( max − min )∕2 1 − r
tan = = = = Const. (2.26)
m ( max + min )∕2 1 + r
Therefore, points on line MM (or NN )havethe same stress ratioasthatofworking
1
1
point of M (or N), that is, the stresses at point M (or N ) represent the endurance
1 1