# NUMERICAL INVESTIGATION OF THE ELASTIC-PLASTIC LINEAR HARDENING STRESS-STRAIN STATE OF THE FRAME ELEMENT CROSS-SECTION.

IntroductionSeveral material models can be used for evaluation of plastic strains in the structural elements: perfectly rigid-plastic, perfectly elastic-plastic, linearly hardened (Cyras et al. 2004; Kalanta 2007; Jaras, Kacianauskas 2001, 2002; Davies 2002). As it is investigated by experiments (Sawko 1964; Byfield et al. 2005), the last model has the best correspondence to the real steel stress-strain dependency in the structure.

The effect of linear hardening and general nonlinear material models of stress-strain state of cross-sections are analysed in many papers (Atkociunas, Cizas 2009; Dulinskas et al. 2010; Van Long, Dang Hung 2008a, 2008b). Some of them are related to the pure bending only, in other ones an effect of axial forces is evaluated, though only for very simple--perfectly elastic-plastic material model. Straightening of nonlinear stress distribution in cross-sections using equivalent or averaged rectangular stress blocks is suggested in paper (Dulinskas et al. 2010).

In most of the literature related to physically non-linear beam structures, simplified approaches for evaluation of cross-section plasticity are investigated. Most common are: plastic hinge approach (Landesmann 2010; Kalanta 2007) and semi-rigid connection approach (Hu, Zhou 2012; Sekulovic, Nefovska-Danilovic 2008). The techniques of such approaches have significant computation advantage, but analysis in most cases is suitable for preliminary approximate study as it is unable to deal with the spread of material plasticity.

The main aim of this paper is to create and apply a methodology for an exact determination of elastic and plastic zones in a cross-section, when linear hardening material model and combined effect of axial force and bending moment are evaluated. To achieve this goal, several systems of nonlinear equations must be created and solved, however solutions of these equations are complicated and still not analysed well enough (Juriev 1977; Webster, Ellison 1967). Mathematical software MATLAB (Moore 2014) was used to overcome some of these difficulties in the present paper. Programming code was created for solving such problems, i.e. performing numerical and graphical analysis of stress-strain state of a rectangular cross-section.

1. Main equations of stress-strain state

Equilibrium of forces and bending moments must be valid in any cross-section subjected to bending moment M and axial force N:

[mathematical expression not reproducible], (1)

where A is area of the cross-section; y-distance to neutral axis of the cross-section; [sigma]-normal stress.

Linearly hardening material model is described by stress-strain diagram (Fig. 1) composed of two deformation stages--elastic and elastic-plastic. Normal stresses [sigma] and longitudinal strains [epsilon] in this diagram are related by such equations:

E = [[sigma].sub.0]/[[epsilon].sub.0]; [E.sub.h] = ([sigma] - [[sigma].sub.0]) /([epsilon] - [[epsilon].sub.0]). (2)

where E and [E.sub.h] are elastic and hardening modules; [[sigma].sub.0] and [[epsilon].sub.0]--limit elastic stress and strain of the material.

Elastic-plastic stage (when [sigma] > [[sigma].sub.0]) can be expressed by hardening ratio [alpha] = [E.sub.h]/E:

[sigma] = [[sigma].sub.0] + [alpha]E ([epsilon] - [[epsilon].sub.0]). (3)

Since hypothesis of plane sections must be satisfied, following relation have to be valid:

[kappa] = -[[epsilon].sub.max] - [[epsilon].sub.min]/h, (4)

where [kappa] is curvature; [[epsilon].sub.max], [[epsilon].sub.min]--longitudinal strains at the top and bottom of the cross-section; h-cross-section height.

1.1. Pure bending

In this case axial force is zero-N = 0, therefore formation of the force equilibrium is unnecessary. Equilibrium of bending moments can be written as equality between moment M and internal moment generated by normal stresses in respect of neutral line of a rectangular cross-section (Fig. 2):

[mathematical expression not reproducible], (5)

where b is width of the section.

First term of Equation (5) corresponds to the internal moment generated by stresses of an elastic core [h.sub.el]; second and third terms correspond to the moments generated by stresses in elastic-plastic zones 0.5[h.sub.pl] (Fig. 2).

For a rectangular cross-section, relations satisfying plane sections hypothesis are:

[[epsilon].sub.max] = -[kappa] [h/2]; [kappa] = - 2[[sigma].sub.0]/[Eh.sub.el]. (6)

Nonlinear system of four equations (3), (5) and (6) can be expressed by any of the four unknowns--[[epsilon].sub.max], [gamma], [h.sub.el] or [[sigma].sub.max]. For instance, expression of longitudinal strain [[epsilon].sub.max] is obtained as follows:

[bar.a][[epsilon].sup.3.sub.max] - [bar.b][[epsilon].sup.2.sub.max] - [bar.c] = 0, (7)

where [bar.a] = 2[alpha][E.sup.3]W; [bar.b] = [E.sup.2] (2M - 3W[[sigma].sub.0] (l - [alpha])); [bar.c] = W[[sigma].sup.3.sub.0] (1 - [alpha]) and W is a resistance moment of rectangular cross-section.

If bending moment sign is taken to be positive, the only positive root of the cubic Equation (7) defines true maximum strain of the cross-section [[epsilon].sub.max].

1.2. Combined effect of bending moment and axial force

Five cases of bending moment and axial force equilibrium must be analysed for definition of five different stress-strain states in the cross-section. Four limit values of axial force restrict these states. In this paper, all five cases are analysed assuming that bending moment and axial force are positive (positive directions are shown in Fig. 3). For these positive directions, the curvature [kappa] and the coordinate of neutral line [y.sub.0] will always have negative values:

1. First case. If M < [M.sub.0] and N [less than or equal to] [N.sub.lim1] (only elastic strains in the cross-section),

where [M.sub.0] is limiting bending moment of the cross-section (it corresponds to the elastic stress limit); [N.sub.lim1] is the first axial force limit. This limit defines maximum axial force which, together with acting bending moment, develops only the elastic strains in the cross-section. For a rectangular cross-section these values are:

[M.sub.0] = [[sigma].sub.0] [[bh.sup.2]/6]; [N.sub.lim1] = 6([M.sub.0] - M)/h. (8)

Plastic strains are not developing in this case, therefore normal stresses and longitudinal strains are calculated by well-known equations of elastic state:

[mathematical expression not reproducible]. (9)

2. Second case. If N [less than or equal to] [N.sub.lim2] and [[sigma].sub.max] [greater than or equal to] [[sigma].sub.0] (Fig. 3a), where [N.sub.lim2] is second axial force limit. This limit defines axial force which, together with acting bending moment, develops normal stress [[sigma].sub.min] = -[[sigma].sub.0] (Fig. 4b). Second axial force limit can be expressed using equilibrium Eq. (1) and the equations of plane section hypothesis:

[mathematical expression not reproducible], (10)

and the elastic core [h.sub.el] is calculated from the cubic equation:

[mathematical expression not reproducible], (11)

where [mathematical expression not reproducible].

Values of stress-strain state when [[sigma].sub.min] [less than or equal to] - [[sigma].sub.0] (Fig. 3a) are calculated as follows. At first, the coordinate of neutral line [y.sub.0] is calculated from fifth order non-linear equation:

[mathematical expression not reproducible], (12)

where [mathematical expression not reproducible].

Hereafter, elastic core is calculated from the cubic equation:

[bar.a][h.sup.3.sub.el] + [bar.b][h.sub.el] + [bar.c] = 0, (13)

where [mathematical expression not reproducible].

Other unknowns in this state are calculated as follows:

[kappa] = -2[[sigma].sub.0]/[h.sub.el]E; (14)

[[epsilon].sub.max] = -[kappa](0.5h - [y.sub.0]); [[epsilon].sub.min] = [lambda](0.5h + [y.sub.0]); (15)

[[sigma].sub.max] = [[sigma].sub.0] (1 - [alpha]) + [E.sub.h][[epsilon].sub.max]; [[sigma].sub.min] = -[[sigma].sub.0](1 - [alpha]) + [E.sub.h][[epsilon].sub.min]. (16)

3. Third case. If N [less than or equal to] [N.sub.lim3] and -[[sigma].sub.0] [less than or equal to] [[sigma].sub.min] [less than or equal to] 0 (Fig. 3b), where [N.sub.lim3] is third axial force limit. This limit defines axial force which, together with acting bending moment, develops zero normal stress [[sigma].sub.min] = 0 (Fig. 4b):

[mathematical expression not reproducible]. (17)

The elastic core [h.sub.el] is calculated from the cubic equation:

[bar.a][h.sup.3.sub.el] = [bar.b][h.sup.2.sub.el] + [bar.c][h.sub.el] - [bar.d] = 0, (18)

where [bar.a] = [[sigma].sub.0]b(1 - [alpha]); [bar.b] = 1.5[[sigma].sub.0]bh([alpha] - 1); [bar.c] = 6M; [bar.d] = 0.5[alpha][[sigma].sub.0]b[h.sup.3]. The values of stress-strain state when -[[sigma].sub.0] [less than or equal to] [[sigma][greater than or equal to]sub.min] [less than or equal to] 0 and [[sigma].sub.max] [greater than or equal to] [[sigma].sub.0] (Fig. 3b) are calculated by solving the nonlinear system of seven equations:

[mathematical expression not reproducible]; (19)

[mathematical expression not reproducible]; (20)

[mathematical expression not reproducible]; (21)

[y.sub.0] = 0.5h = [h.sub.el1]; 0.5h - [y.sub.0] - [h.sub.el2] = [h.sub.pl]. (22)

Equation (19) describes equilibrium of forces in the direction perpendicular to the cross-section plane; equation (20) is the equilibrium of bending moments in respect to neutral axis; equations (21) relates stresses and curvature in various zones of the cross-section height (according to the hypothesis of plane sections); equations (22) relates geometrical parameters of the cross-section height (Fig. 3b).

4. Fourth case. If [N.sub.lim3] < N < [N.sub.lim4] and 0 [less than or equal to] [[sigma].sub.min] [less than or equal to] [[sigma].sub.0] (Fig. 4d),

where [N.sub.lim4] is fourth axial force limit. This limit defines axial force which, together with acting bending moment, develops normal stress [[sigma].sub.min] = [[sigma].sub.0] (Fig. 4c):

[N.sub.lim4] = [[sigma].sub.0]bh + 6M/h. (23)

Characteristics of stress-strain state, when 0 [less than or equal to] [[sigma].sub.min] [less than or equal to] [[sigma].sub.0] and [[sigma].sub.max] [greater than or equal to] [[sigma].sub.0] (Fig. 4d) are calculated as follows. At first, elastic core [h.sub.el] is calculated from cubic equation:

[bar.a][h.sup.3.sub.el] + [bar.b][h.sup.2.sub.el] - [bar.c][h.sub.el] + [bar.d] = 0, (24)

where [bar.a] = 1/6 b (E - [E.sub.h])(N - [[sigma].sub.0]bh);

[bar.b] = 0.25b (E - [E.sub.h])(6[[sigma].sub.0]W - 2M - Nh);

[bar.d] = [E.sub.h]W (3[[sigma].sub.0]W - 0.5Nh + 3M);

[bar.c] = M[E.sub.h]bh.

Hereafter, curvature is calculated from quadratic equation:

[bar.a][[kappa].sup.2] + [bar.b][kappa] + [bar.c] = 0, (25)

where

[mathematical expression not reproducible].

Other unknowns of this state are calculated as follows:

[mathematical expression not reproducible]; (26)

[mathematical expression not reproducible]. (27)

5. Fifth case. If N [greater than or equal to] [N.sub.lim4] (Fig 4e). In this case, plastic strains are developing in the entire area of the cross-section and therefore [h.sub.el] = 0, [h.sub.pl] = h. Other unknowns are calculated as follows:

[mathematical expression not reproducible]; (28)

[[epsilon].sub.max] = [kappa]([y.sub.0] - 0-5h; [[epsilon].sub.min] = [kappa]([y.sub.0] + 0-5h, (29)

[[sigma].sub.max] = [[sigma].sub.0] - [alpha][[sigma].sub.0] + [E.sub.h][[epsilon].sub.max]; [[sigma].sub.min] = [[sigma].sub.0] - [alpha][[sigma].sub.0] + [E.sub.h][[epsilon].sub.min]. (30)

2. Problem implementation in MATLAB

Suggested methodology was applied for two numerical experiments in MATLAB environment. In the first experiment we analyse a 0.3x0.2 m cross-section subjected to bending moment M = 1000 kNm and axial force N = 10000 kN. Material modulus of elasticity is E = 2100 MPa, hardening modulus--[E.sub.h] = 210 MPa, yield stress--[[sigma].sub.0] = 235 MPa. Numerical and graphical MATLAB results are shown in Figure 5, where red colour indicates plastic stresses and deformations. It is evident that the cross-section is in the second case of stress-strain state, since the acting axial force is smaller than second axial force limit: N < [N.sub.lim2a] = 27249.127 kN. Acting bending moment is M [greater than or equal to] [M.sub.0] = 705 kNm, therefore the first axial force limit does not exist.

Since equations (11) and (18) are of the third degree, in general the second and the third axial force limits can have two or three real values (roots). One of such cases is investigated in the second numerical experiment (Figs 6 and 7; Table 1). In the second numerical experiment we consider a cross-section of the same parameters as in the first experiment except for the acting forces and modulus of elasticity: now E = 2100 MPa, the section is subjected to bending moment M = 250 kNmand axial force, which varies from 0 to 25000 kN. We will analyse the influence of the axial force to the different parameters of the strees-strain state of the cross section. While the axial force gradually increases, MATLAB program finds the corresponding stress-strain state and collects the data in memory. Numerical and graphical results are summarised in Table 1 and Figures 6 and 7.

As seen from the results (Fig. 6a; Table 1), at first, while the axial force increases from zero, Equation (9) of elastic stress-strain state is valid (point A). When axial load reaches and exceeds N [greater than or equal to] [N.sub.lim1] = 9100 kN value, the plastic deformations begins to develop in the top of the cross-section (point B), meanwhile the bottom layers of the cross-section remains in the elastic state and therefore Equations (24)-(27) are used to define the stress-strain state of the section (fourth case of equilibrium). If axial force increases further, the cases of equilibrium vary in such order (Table 1): 3rd case (point C), 2nd case (point D), 3rd case (point E), 4th case and finally 5th case (point F). As seen from the curve of [[sigma].sub.min] variation (Fig. 6a), the stress between points B and D decreases, despite that axial force still increases. At the same time the area of plastic deformations at the top of cross-section increases very fast, meanwhile plastic deformations are zero or very small at the bottom. Such a big difference between development of plastic deformations at the top and the bottom of cross-section causes temporary decrease of [[sigma].sub.min]. Normal stress at the top of the cross-section [[sigma].sub.max] and the absolute value of curvature k non-linearly but constantly increases with the increase of axial force and area of total plastic deformations (Figs 6b, 7).

Conclusions

--The stress-strain state of a rectangular cross-section made of linearly hardening material under combined effect of bending moment and axial force was investigated in detail. Numerical experiments demonstrate non-linear variations of stresses and strains as functions of axial force. An important observation is that a temporary decrease of normal stress at the bottom of the cross-section is possible even if axial force is constantly increasing.

--Stress-strain relation functions obtained in this paper are useful in the applications of analysis and optimization problems of linearly hardening material (e.g. steel) frame structures. These applications are under current investigation by the authors and show promising results.

--Presented derivation techniques and methods of analysis can easily be adapted and applied for other types of cross-sections.

--Described methodology can be applied for different and more complicated models of physically nonlinear materials (i.e. linear-softening or stress-strain diagrams composed of more than two linear segments).

doi: 10.3846/2029882X.2016.1217796

References

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Andrius GRIGUSEVICIUS is Assistant lecturer at the Department of Structural Mechanics, Vilnius Gediminas Technical University. His research interests include analysis and optimization of elastic-plastic structural systems, plate on deformable foundation contact problems.

Gediminas BLAZEVICIUS, Dr is Assistant lecturer at the Department of Structural Mechanics, Vilnius Gediminas Technical University. His research interests include nonlinear analysis, optimization and optimal shakedown design of structural systems.

Andrius GRIGUSEVICIUS, Gediminas BLAZEVICIUS

Department of Structural Mechanics, Faculty of Civil Engineering, Vilnius Gediminas Technical University, Sauletekio al. 11, LT-10223 Vilnius, Lithuania

Received 13 June 2016; accepted 25 July 2016

Corresponding author:

A. Grigusevicius E-mail: andrius.grigusevicius@vgtu.lt

Caption: Fig. 1. Stress-strain diagram of linearly hardening material model

Caption: Fig. 2. Distributions of stresses and strains in the cross-section in case of pure bending

Caption: Fig. 3. Distributions of stress and strains in the cross-section resulted by combined effect of axial force and bending moment: (a) second case of equilibrium; (b) third case of equilibrium

Caption: Fig. 4. Distributions of stresses in the cross-section resulted by common effect of bending moment and axial force: (a) when N = [N.sub.lim2]; (b) when N = [N.sub.lim3]; (c) when N = [N.sub.lim4]; (d) when [N.sub.lim3] < N < [N.sub.lim4]; (e) when N > [N.sub.lim4]

Caption: Fig. 5. MATLAB plot of the stress-strain state of the first numerical experiment

Caption: Fig. 6. Curves of stress variation at the bottom (a) and at the top (b) of the cross-section and (c) stress-strain diagrams at the points of interest

Caption: Fig. 7. Variation of the curvature of the cross-section for the second numerical example

Table 1. Results of the second numerical experiment Point in Fig. 6 N, kN [[sigma].sub.min], [[sigma].sub.max], kPa kPa A 3100 -31666.7 135000 [N.sub.lim1] 9100 68333.3 235000 B 10600 86682.7 235401.0 [N.sub.lim3b] 12269.5 0 241092.6 C 12600 -57908.0 246558.1 [N.sub.lim2b] 14322.5 -235000 286223.4 D 15100 -235226.1 305178.6 [N.sub.lim2c] 16426.2 -235000 337618.0 E 18100 -82663.9 378328.9 [N.sub.lim3c] 18542.3 0 388926.5 [N.sub.lim4] 19100 235000 401666.7 F 22100 268333.3 435000 Point in Fig. 6 [kappa] [[epsilon].sub.min] A -0.026455 -0.001508 [N.sub.lim1] -0.026455 0.003254 B -0.029907 0.004128 [N.sub.lim3b] -0.134009 0 C -0.229956 -0.002758 [N.sub.lim2b] -0.887672 -0.011190 D -1.192137 -0.012267 [N.sub.lim2c] -1.703461 -0.011190 E -2.325485 -0.003936 [N.sub.lim3c] -2.480580 0 [N.sub.lim4] -2.645503 0.011190 F -2.645503 0.169921 Point in Fig. 6 [[epsilon].sub.max] A 0.006429 [N.sub.lim1] 0.011190 B 0.013100 [N.sub.lim3b] 0.040203 C 0.066230 [N.sub.lim2b] 0.255111 D 0.345374 [N.sub.lim2c] 0.499848 E 0.693709 [N.sub.lim3c] 0.744174 [N.sub.lim4] 0.804841 F 0.963571 Point in Fig. 6 Comments A 1st stress-strain state; Eq. (9) [N.sub.lim1] 1st limit axial force; Eq. (8) B 4thstress-strain state; Eqs. (24)-(27) [N.sub.lim3b] 3rd limit axial force; Eqs. (17), (18) C 3rd stress-strain state; Eqs. (19)-(22) [N.sub.lim2b] 2nd limit axial force; Eqs. (10), (11) D 2nd stress-strain state; Eqs. (12)-(16) [N.sub.lim2c] 2nd limit axial force; Eqs. (10), (11) E 3rd stress-strain state; Eqs. (19)-(22) [N.sub.lim3c] 3rd limit axial force; Eqs. (17), (18) [N.sub.lim4] 4th limit axial force; Eq. (23) F 5th stress-strain state; Eqs. (28)-(30)

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Author: | Grigusevicius, Andrius; Blazevicius, Gediminas |
---|---|

Publication: | Engineering Structures and Technologies |

Article Type: | Report |

Date: | Sep 1, 2016 |

Words: | 3638 |

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