Page 251 - Biaxial Multiaxial Fatigue and Fracture

Fatigue Analysis of Multiaxially Loaded Components with the FE-Postprocessor FEMFAT-MAX 235

The polar angle 6, where SFe becomes a minimum, defines the critical stress component. q.0 is
the fatigue limit taken from Eq. (12). By a suitable modification of Eq. (12), the result of Eq.
(13) can be adapted to test results. Sn,d denotes the amplitude of the stress component Sn,a(t),
with d.n=CosQ. Nevertheless, Sn,d is not unique in Eq. (13), because it depends not only on the
polar angle 6, but also on the plane specified by its unit normal vector n. Our main interest has
to be focused on SFo for a given 6, which is a minimum over all planes. Therefore we have to
look firstly for a minimum of SFe over all planes and secondly for an additional minimum
("minimum of minimum") over all inclinations 8 of direction d. As a result we will obtain both
the critical plane 8, and the critical component in direction d,.
For this purpose, Eq. (13) has to be redefined:

sn,d,- denotes the maximum stress over all planes for a given 0, with d.n = cos6 = constant.
This stress can be found by constructing the tangent to Mohr's circle, which is perpendicular to
a line through the origin of the coordinate system, with inclination 0 to the 0-axis (Fig. 16).
Using the following geometric relations, a formula can be derived for Sn,d,-:

s, =Jm (17)
a = arctan( z)

a denotes the angle between the stress vector Sn and the plane's normal n, as it can be seen in
Fig. 3. As an interesting result it can be pointed out, that the plane, where Sn,d,max is acting on, is
inclined by an angle of 8/2 to the first principal axis.
Figure 17 shows the normalized fatigue limit uj,e /up, the normalized stress amplitude
S,,~,,,,Jn~ and the resulting safety factor SFe for the ratio a3 /u, = -0.085. The minimum of SFe
can be found at 6 = 50". This value was also delivered by FEMFAT. It means, that the angle
between the critical stress component and the plane's normal is 50". The result for in phase
loading in Fig. 15 can be improved by increasing the fatigue limit at 6 = 50". For this case we
have extended Eq. (12) by an additional term: