Page 242 - Biaxial Multiaxial Fatigue and Fracture

226 C. GAIER AND H. DANNBAUER

For the second step the stress vector is normal-projected onto a given line, whose direction
is specified by its unit direction vector d (Fig. 3). The resulting scalar quantity denotes the
stress component &,At) in direction d and can be obtained by simply applying the inner product
on d and S,(t):

There are two special cases:
0 With d-71 one obtains the normal component N.(t) of the stress vector S,(t).
0 With d-n=O one obtains the resolved shear stress T,(t). For 3D-stress states, a unique
direction d cannot be derived from the relation d.n=O. This is true due the fact, that the shear
stress can point in any direction inside the plane and for nonproportional loads it can change
its direction or even rotate.

THE “CRITICAL PLANE - CRITICAL COMPONENT” - APPROACH
For given unit vectors n and d, one obtains a scalar stress quantity &,At) by Eqs (1) and (2), for
which rainflow counting and damage analysis can be performed following Miner’s rule of
linear damage accumulation. This procedure can be applied for each combination of n and d.
This combination, where the damage reaches its maximum value, is denoted as “critical plane -
critical component” n, and d,. Analogous to the simple critical plane criterion it is assumed,
that the stress component in direction d,, acting on the plane ne, is responsible for local crack
initiation.

DEPENDENCY OF FATIGUE PARAMETERS ON THE POLAR ANGLE

For a given line specified by its direction d, the projected stress vector Sn,djt) = Sn&d =
(d.S,(t))d can be resolved into its normal and shear component (Fig. 3):

The type of loading can be characterized by the ratio of shear stress to normal stress, which
depends only on the polar angle 0, but not on the azimuth one:

Three basic load types can be simply specified by its polar angle:
8 = 0” ... Tensile Stress
0
8 = 90” ... Shear Stress
8 = 180” . . . Compressive Stress
0
Other load types are combinations of these basic ones. Because all directions d with same polar
angle B represent the same load type, the stress &At) can be compared with the same fatigue
limit for all possible directions d lying on a cone with opening angle 26. Therefore the fatigue
limit is a fimction only of the polar angle 8, if the material is assumed to be isotropic. This