Determination critical stresses of buckling on basis of stresses and geometrical parameters analyses

Solving the problem of unstability for constructional elements in case of stress conditions, especially in case of triaxial loading appears to be insufficient. Loads, not stresses, are concerned most of all in unstability solutions. Therefore, physico-mechanical characteristics of certain materials are partly omissed. Critical equivalent stresses appearing in cases of biaxial or triaxial stress conditions are not valued correctly. Classical solutions obtained by Euler and other scientists in cases of simple loading are applied for loads not exceeding the limit of elasticity. Hutchiston [1] made an important hypothesis to the nonlinear branching theory of structures loaded in the plastic range. The important theory of shear-flexural buckling of columns presented Timoschenko [2]. Galambos [3] introduced the stability criteria for steel structures. Nowadays, the studies [4] pay most attention to avoiding the stability loss in thin-wall constructions using new achievements in engineering. Standards are also intended for this [5, 6]. However, theoretical solutions for stability criteria are insufficient. Thus, this study aims to obtain the criterion of stability loss on the analogy of strength criteria first of all using the probability prognosis.


Introduction
Solving the problem of unstability for constructional elements in case of stress conditions, especially in case of triaxial loading appears to be insufficient.Loads, not stresses, are concerned most of all in unstability solutions.Therefore, physico-mechanical characteristics of certain materials are partly omissed.Critical equivalent stresses appearing in cases of biaxial or triaxial stress conditions are not valued correctly.
Classical solutions obtained by Euler and other scientists in cases of simple loading are applied for loads not exceeding the limit of elasticity.Hutchiston [1] made an important hypothesis to the nonlinear branching theory of structures loaded in the plastic range.The important theory of shear-flexural buckling of columns presented Timoschenko [2].Galambos [3] introduced the stability criteria for steel structures.Nowadays, the studies [4] pay most attention to avoiding the stability loss in thin-wall constructions using new achievements in engineering.Standards are also intended for this [5,6].However, theoretical solutions for stability criteria are insufficient.Thus, this study aims to obtain the criterion of stability loss on the analogy of strength criteria first of all using the probability prognosis.

Beams loaded by flexural-torsional buckling
In case of flexural-torsional buckling or lateral buckling (Fig. 1) the unstability equation is recorded as [7]: where  is angle of torsion; Mx, Mz are moments of bending and torsion about axes x and z; Iy are minimum moment of inertia with respect to axis y; Is are inertial moment of torsion; G is modulus of shear.According to Eq. ( 1) the critical force can be expressed [7]: where Fc is critical unstability force in case of flexural-torsional buckling.According to Eq. ( 2) instability appears on influence of two kinds of stresses, i.e. normal (bending) and tangential (torsion).Therefore, the case of lateral buckling is the case of complicated loading.Then, stress intensity i is calculated as follows [7]: where x, y, z, are normal stresses in directions x,y,z; xy, yz, zx are tangential stresses in plane surfaces.
In case of torsional bending that makes where 1, 2, are principal stresses.
Thus, the stability and strength as well is evaluated using criteria that depend on normal and tangential influences.These criteria are determined by the fact if the material is brittle or plastic.In order to apply universal criteria, characteristics of determination instability referring to probability analysis must be described.

Accidental events of stability loss
Let us mark the event of unstability A, and the event of stability A .These two events will be opposite and impossible to happen at the same time.This can be shown in a diagram (Fig. 2).If certain element is under influence of three critical stresses 1  2  3, multitude of events } { i A and } { i A in all three directions of stresses (i = 1, 2, 3) using the principle of independence between the effects of certain forces is to be analyzed (Fig. 2).Area covered in Fig. 2, e show the identification of multitudes {A1}, {A2}, {A3} as A1A2A3 = C2 and the fact that all these events are happening at the same time influenced by all three stresses 1, 2, 3.

Relation between probable instability conditions and critical stress
Let us mark critical stresses for instability ci, then main influencing stresses i(i = 1, 2, 3) are distributed according to functions p(ci) and p (i) (Fig. 3).

Fig. 3 Connection between the chance of instability and certain stresses
Let us suggest the fact when P(Ai) reaches   i ekv P the condition of statistical equivalence is met and the probability for instability in case of linear stress condition P equals to equivalent probability Pekv in case of complicated stress condition.
123 ekv e where Pe is experimentally obtained probability for instability in case of complicated stress condition.
Referring to the theory of probability, the probability for instability can be expressed as: Each main stress 1, 2, 3 after the analogy of the theory of probability [7] equals to critical instability stress multiplied by probability P(Ai) and can be expressed as: Then the Eq. ( 7) considering the events as independent is can be expressed as:

e P P A P A P A P A A P A A P A A P
Considering Eq. ( 7) and putting the members of Eq. ( 8) as functions of first stress invariant I1, second stress invariant I2 and third stress invariant I3 we have the following equation: ; .

P A P A P A F σ σ σ F I P A P A P A P A P A P A F σ σ σ σ σ σ F I P A P
Thus: where Pall is allowed probability.Instability condition can be expressed in stresses: where ekv is equivalent stress corresponding to the instability chance Pekv in case of triaxial stress condition; 123 is stress obtained in case of linear stress condition corresponding to the instability chance P123; c,all is possible instability stress corresponding to normative instability probability Pall.
Equivalent ekv stresses in case of complicated condition depend on stress invariants I1, I2, I3.As shown in (9, 10) equational relationship between stresses and instability probability is unlinear.

Determination instability stresses according to invariants of stress condition
Certainly [8], in case of complicated stress condition equivalent stresses ekv are calculated under criterion: where i is stress intensity; 0 is average stress; ma, mb are constant of materials.However, measuring peculiarities of instability it is clear that critical stresses are distributed nonlinearly.Criterion (13) can be expressed as: where c are critical instability stresses.
In case of instability (bending and torsion) that makes: Critical cases of deformation are as follows: 1.The force functions in vertical plane.Critical stress makes yield limit Y  , thus


. Then, Eq. ( 15) can be changed as follows: 2. In case the cross-section is buckled irreversibly and moves horizontally and vertically critical instability stresses depend on the element length and geometrical characteristics of the cross-section.Critical stress Fc in case: where FY is yield calling force calculated as: Referring to Eq. ( 2) we can change this as follows: Constant D can also be obtained from experimental data i.e. applying Eq. ( 19).Constant D can be calculated as: Therefore, Eq. ( 15) can be changed as follows: Evaluating dependence ( 16) that makes: Thus, after obtaining results of stability loss in single research and calculating constant m1 is possible to analyze other cases of complicated loading by modelling under Eq.( 15).

Experimental data
For experiments, two types of steel profiles are chosen, i.e. girded and double-T.Geometric and mechanical characteristics for tested beam profiles is: modulus of elasticity -210 5 MPa ; shear modulus -810 4 MPa; yield limit in bending -325 MPa; strength limit in bending -520 MPa .
Chosen profiles were fixed rigidly as shown in Fig. 1 and loaded with force F concentrated on free end.Complicated loading is obtained without bending force adding torsion moment in the cross-section.Beam stability loss is determined measuring beam displacement and twisting angle.
During the experiment, critical bending forces in various torsion moments were measured While testing the gird 50 × 5 (mm) with length 1.5 m loaded by torsion moment Mtor = 10 Nm, stability loss is obtained with bending force F = 205 N, the beam lost its stability with 1 = 147.17MPa and  = 25.61MPa.The beam of 2 m length, 100 × 6 (mm) loaded with same torsion moment lost its stability with bending force F = 396 N, the beam lost its stability with 1 = 79.2MPa and  = 8.66 MPa.Bending force corresponding to yield limit was FY = 451 N.
Dependencies of critical stability stresses calculated under formula (23) on length of the beam given in Fig. 4. Results obtained during the experiment presented in Table 1.
Dependencies presented in Fig. 4 show that adding extra torsion moment without bending force decreases critical stability force according to the value of the moment.Fig. 4 Dependencies between critical stability stresses 1  (using Eq. ( 23)) and length for steel beam, with tor M as constant: 1 -10 Nm; 2 -5 Nm; 3 -2.5 Nm; a -steel beam 50 × 5 (mm) ; b -steel beam100 × 6 (mm) Then, Eq. ( 23) is presented as:   Dependencies of critical stability stresses calculated under Eq.( 24) and length for double-T profiled IPE EN 10034 presented in Fig. 5. Results obtained during the experiment given in Table 2.
After the experiments, calculating constant 1 m , that is invariable with constant beam cross-section and const   , using Eqs.( 23) or (24) various cases of stability loss are to be foreseen and critical instability stresses found in case of rotated bending.

Conclusions
1. Complicated stress condition obtained in cases of lateral buckling is evaluated on strength criterion.
2. Probability forecast shows the relation between the stress invariants and the probability of instability.
3. Stability loss can be described by nonlinear strength criterion and geometrical characteristics of the constructional element.
4. Constant values for the strength criterion are included into general equation of stability loss are obtained experimentally.
5. After loading the gird subjected to bending with the moment of torsion, stability loss occurs in case of less bending load.

Fig. 1
Fig. 1 Contilever beam of rectangular cross-section in bending

Fig. 2 1 A and   1 A 2 A and   2 A or   3 A and   3 A
Fig. 2 Diagrams of stability events in case of various events multitude If only the first critical stress 1 influences the events multitudes of   1 A and   1 A can be shown in Fig. 2, a.If the events are influenced by 2 or 3 multitudes   2 A and   2 A or   3 A and   3 A as well can be shown in Fig. 2, b, c.Areas covered with horizontal lines correspond to multitude {A1}, vertical -{A2}, diagonal -{A3} shown in Fig. 2, d, represent common area A1A2A3 = C1, where instability appears because of one kind of stress 1, 2, 3 or because of all three stresses.Area covered in Fig.2, e show the identification of multitudes {A1}, {A2}, {A3} as A1A2A3 = C2 and the fact that all these events are happening at the same time influ- D is shelf stiffness in bending) s GI C  is stiffness of the whole profile in torsion.

Table 1
Experimentally obtained data of steel beams

Table 2
Experimentally obtained data of steel beams Mtor = 50 Nm, stability loss was found with bending force F = 1670 N, the beam lost its stability with 1 = 168.26MPa and  = 22.11MPa.