Page 205 - Biaxial Multiaxial Fatigue and Fracture

Estimation of the Fatigue Life of High Strength Steel Under ...

At +log 2
z'f,=lO mt

respectively.
Similarly to Eq.(9) we can write the equation for pure shearing

2, = Tf (2Nf )"

where zIf and b are the coefficient and exponent of strength under shearing, respectively.

Under uniaxial tension-compression the relationship

0, = df (2Nf )b (13)
can be written with use of shear stresses in the plane of the maximum shearing as

0'
T, = --f-(2Nf )" = Yf (2Nf)b.
2
Thus
'SIf = 5
,
2
From the ratio of shear stress amplitudes under shearing Eq. (12) and torsion Eq. (9) for a
given number of cycles Nt., we obtain

and using Eq. (15) we have

For a chosen number of cycles Nt, from Eq. (17) we can calculate the shear stress amplitudes
under shearing (in the section of uniform stress distribution), 2, equivalent to the maximum
shear stress amplitude under torsion (non-uniform stress distribution in the section), Tat.
Under random loading, application of Eq. (17) is time - consuming because it must be used
many times, for each amplitude of cycle counted from the stress history (for example, with use
of the rain flow algorithm). In order to reduce and simplify calculations we can rescale all the
random history of shear stress with use of one constant coefficient.
Applying the equation