Page 65 - Analysis and Design of Machine Elements

Strength of Machine Elements
Therefore, the safety factor and fatigue strength of the element under variable uniaxial 43
stresses is
√
lim −1Nv √ N 0
√
S = = = √ ≥ [S] (2.41)
ca
max 1 −1 m √ z
√ ∑
n m
i i
i=1
2.3.6 Fatigue Strength for Combined Stresses with Constant Amplitude
During machine operation, an element is most likely subjected to a combination of ten-
sion, bending and torsion, so both tensile and shear stresses will generate. Under static
stress states, yield occurs when the equivalent stress obtained from the maximum dis-
tortion energy theory equals the uniaxial yield strength expressed by
1 2 2 2 1∕2
√ [( − ) +( − ) +( − ) ] = s (2.42)
3
2
2
1
3
1
2
where the three principle normal stresses are
1 √
= + + 4 2 (2.43)
2
1
2 2
= 0 (2.44)
2
1 √ 2 2
= − + 4 (2.45)
3
2 2
Substituting the three principle normal stresses back into Eq. (2.42), we have
( ) 2 ( ) 2
3
+ = 1 (2.46)
s s
And from Eqs. (2.12) and (2.46), can be further expressed as
( ) 2 ( ) 2

+ = 1 (2.47)

s s
This relation was validated by experiment when all the stress components are com-
pletely reversed and always in the same phase [16]. By introducing a safety factor, the
fatigue strength under completely reversed stress can be evaluated by [17]
( ) 2 ( ) 2
S S
ca a ca a
+ = 1 (2.48)
−1e −1e
Here we define −1e = S and −1e = S , which are associated with completely reversed

a a
tensile and shear stress, respectively. Therefore, we have
( ) 2 ( ) 2
S ca S ca
+ = 1 (2.49)
S S

Thus
S S

ca
S = √ (2.50)
2
S + S 2