Regarding the Ratio of Tensile Strength to Shear Strength in Fundamental Strength Sciences
by
© Ph. D. & Dr. Sc. Lev Gelimson
Academic Institute for Creating Fundamental Sciences (Munich, Germany)
RUAG Aerospace Services GmbH, Germany
Strength and Engineering Journal
of the "Collegium" All World Academy of Sciences
Munich (Germany)
6 (2006), 3
To predict fatigue and fracture of metals [1], fundamental strength sciences [2]generalize strength criteria and fits the substantial influence of the σu/τu ratio of the ultimate normal stress σu to the ultimate shear stress τu unlike common criteria with pre-defined values of this ratio. In combined bending and torsion with a normal stress σ and a shear stress τat a point of a solid [biaxial stress state], the principal stresses σ1 , σ2 , σ3 commonly ordered (σ1 ≥ σ2 ≥ σ3) are
σ1 = (σ + (σ2 + 4 τ2)1/2)/2, σ2 = 0, σ3 = (σ - (σ2 + 4 τ2)1/2)/2.
That pre-defined value is 2 and 31/2 by the Tresca and Mises criteria [1], respectively,
σe = σ1 - σ3 ≤ σu , σe = (σ12 + σ22 + σ32 + σ1σ2 + σ1σ3 + σ2σ3)1/2 ≤ σu
[σe the equivalent uniaxial stress] giving in our case two particular and one unified criteria
σe = (σ2 + 4 τ2)1/2 ≤ σu , σe = (σ2 + 3 τ2)1/2 ≤ σu ; σe = (σ2 + (σu/τu)2 τ2)1/2 ≤ σu .
To understand the naturalness of the last one, divide all its parts by σu :
σe° = ) σe/σu = ((σ/σu)2 + (τ/τu)2)1/2 ≤ σu/σu ( = 1)
with using relative (dimensionless) stresses (with the ° sign) introduced by the author [2].
For the general case of a triaxial stress state by using its three principal directions 1, 2, and 3 with vanishing all the shear stresses, first consider the Mises criterion. To take into account the unique additional constant σu/τu of a material, include one unknown factor k:
σe = (σ12 + σ22 + σ32 + k(σ1σ2 + σ1σ3 + σ2σ3))1/2 ≤ σu
Note that uniaxial tension σ1 > 0, σ2 = 0, σ3= 0 gives that the factor at the sum σ12 + σ22 + σ32 should be 1. To determine k , use combined bending and tension once more:
σe = (σ2 + (2 - k) τ2))1/2 ≤ σu , k = 2 - (σu/τu)2,
σe = (σ12 + σ22 + σ32 + (2 - (σu/τu)2) (σ1σ2 + σ1σ3 + σ2σ3))1/2 ≤ σu .
For the Tresca criterion, it is impossible to keep the criterion’s linear form by correction because its factors 1 and -1 are unique to provide its applicability to a uniaxial tension and compression σ1 = 0, σ2 = 0, σ3 < 0, and introducing σ2brings nothing at least in the particular case of combined bending and torsion [σ2 vanishes] and thus cannot solve, all the more, the general problem. Now it is natural to choose a similar quadratic form of a formula
σe = ((σ1 - σ3)2 + kσ1σ3)1/2 ≤ σu .
The same combined bending and torsion leads to
σe = (σ2 + (4 - k) τ2))1/2 ≤ σu , k = 4 - (σu/τu)2,
σe = ((σ1 - σ3)2 + (4 - (σu/τu)2) σ1σ3)1/2 ≤ σu , σe = (σ12 + σ32 + (2 - (σu/τu)2) σ1σ3)1/2 ≤ σu .
Experimental data on strength of ductile materials is often placed between the curves given by the Tresca and Mises criteria. Expecting more criterion precision by smaller deviation of the σu/τuratio from its pre-defined value gives the unified criterion as the linear combination
σe = ((σu/τu)2 - 3) (σ1 - σ3) +
(4 - (σu/τu)2) (σ12 + σ22 + σ32 + σ1σ2 + σ1σ3 + σ2σ3)1/2 ≤ σu .
The unified criterion using the criteria forms already corrected by the present methods is
σe = ((σu/τu)2 - 3) (σ12 + σ32 + (2 - (σu/τu)2) σ1σ3))1/2 +
(4 - (σu/τu)2) (σ12 + σ22 + σ32 + (2 - (σu/τu)2) (σ1σ2 + σ1σ3 + σ2σ3))1/2 ≤ σu .
Note that such a linear unification approach is not the only. To naturally extend it, a power unification approach can be also used. And the Hosford exponential unification [3]
σe = (((σ1 - σ2)k+ (σ2 - σ3)k+ (σ1 - σ3)k)/2)1/k ≤ σu
of the Tresca and Mises criteria gives them by k = 1 and k = 2. To provide these values of k using the same pre-defined values of the σu/τu ratio for the both criteria, naturally choose k = 5 - (σu/τu)2 . For further generalizations, introduce the notation (with apparently extending for any sets of variables and function values)
f(x, y | f(a, b) = v, f(c, d) = w)
for any function taking, e.g., values v and w by values a, b and c, d of its variables x, y, respectively. Then choose instead of (σu/τu)2 - 3, 4 - (σu/τu)2 and k = 5 - (σu/τu)2, respectively:
f(σu/τu | f(2) = 1, f(31/2) = 0),
g(σu/τu | g(2) = 0, g(31/2) = 1),
h(σu/τu | h(2) = 1, h(31/2) = 2).
[1] Handbuch Struktur-Berechnung. Prof. Dr.-Ing. L. Schwarmann. Industrie-Ausschuss-Struktur-Berechnungsunterlagen, Bremen, 1998
[2] Lev Gelimson. General Strength Theory. Abhandlungen der Wissenschaftlichen Gesellschaft zu Berlin, Publisher Prof. Dr. habil. V. Mairanowski, 3 (2003), Berlin, 56-62
[3] Hosford W. F. A generalized isotropic yield criterion. Trans. A. S. M. E., Ser. E 2 (1972), 290-292