Adam Paweł Zaborski. 8 Plasticity. reloading. 1. Bauschinger s effect. 2. unchanged yielding limit. 3. isotropic hardening

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1 8 lasticity Introduction Definitions loading/unloading/reloading words commonly used in lasticity lastic strain a ermanent strain that doesn t vanish after unloading, lastically assive rocess a rocess without the lastic strain change, there is no energy dissiation, lastically active rocess a rocess with ermanent strain change with the energy dissiation, Bauschinger s effect an effect of yielding limit change due to revious history of a rocess; a kinematic hardening which is in some sense oosite to the isotroic hardening rocess, Fig reloading 1. Bauschinger s effect. unchanged yielding limit. isotroic hardening Fig. 8.1 Bauschinger s effect and kinematic hardening Schematizations Fig. 8. Schematizations: Hooke s, Levy-Mises, and randtl In the following art of the course the randtl s schematization will be used. This is a schematization suitable for civil engineering where the lastic strains, if they exist, are the same range as the elastic ones. Elastic-lastic bending with tension of symmetric cross-section The assumtions of the engineering theory of bending are: uniaxial stress state E, for < Bernoulli s hyothesis. We will use the conditions of equivalence of external and internal force sets: da N, zda M. These conditions are the theorem of the theoretical mechanics and don t deend on the constitutive equations, so, they are always valid.

2 The following stages of the rocess we can distinguish of simultaneously acting bending moment and axial force on the cross-section, Fig. 8.: a) b) c) d) e) - - Fig. 8. Normal stress in the cross-section The first stage (a), is the stage of linear elasticity. The neutral axis osition remains unchanged. The stage ends when the yielding limit is attained at some fibers. The values of the cross-section forces at this oint of rocess (b) are called elastic load caacity. With the increasing cross-section forces a new lastic region is created. It is a locus where yielding occurs. The neutral axis during the stage (c) usually moves to another osition. In the next stage (d) the lastic regions occuy the extreme fibers from both sides of the cross-section. This corresonds to the sread of the yield from the outmost fibers inwards towards to neutral axis of the beam. This is known as the lastification or the contained lastic flow. The elastic region behavior is assumed to follow the Hooke s law and lastic regions yield. Two searate lastic zones have the elastic/lastic boundaries situated at equal distances from the neutral axis. Finally, at (e), the lastic zones almost adhere and a lastic hinge is created. The cross-section forces attain their extreme values that are called the lastic load caacity. When the lastification occurs the analysis of the stress-strain state becomes more comlicated. The suerosition rincile is no longer valid. The main roblem is that the existence and the ositions of the elastic/lastic boundaries are not known a riori. On the basis of the Bernoulli s hyothesis we can write the strain reartition as: 0 z where two indeendent arameters were used: the bar axis strain and its curvature. However, sometimes another choice of two indeendent arameters is more suitable, Fig There are: the neutral axis osition 0 ( z 0 ) 0 z0 the osition of the elastic/lastic boundaries 0 0 z E z E the extent of the elastic region E. E E central axis E z 0 z Fig. 8.4 arameters of stress/strain descrition

3 So, the arameters of stress-strain state descrition would be: the curvature of the bar axis,, the central axis strain, 0, the osition of the neutral axis, z 0, the osition of the elastic/lastic boundary, z, the extent of elastic region,. Only two arameters are indeendent. lastic section modulus In the case of the ure bending, M 0, N 0, the lastic cross-section modulus, analogical to the elastic section modulus, is a geometric quantity that can be calculated using a formula (sum of the static inertia moments): W S A 1 S A where the static moments may be calculated with resect to any axis arallel to the beam axis, in articular with resect to the neutral axis, which halves the cross-section: A1 A. If the central axis is taken as a reference, the formula for the lastic section modulus simlifies: but the osition of the centroid should be known. Examles Examle 8.1 W 0 S( A1, A ) A rectangular cross-section 5010 mm, Fig. 8.5, is subjected to a bending moment M =.8 knm. Assuming that the beam is made of an elastolastic material with a yielding strength of 40 Ma and an elasticity modulus of 00Ga, determine (a) the thickness of the elastic core, (b) the radius of curvature of the neutral axis, (c) the distribution of the residual stresses, (d) the radius of curvature after the bending moment has been decreased back to zero. z Fig. 8.5 lastification of cross-section Solution (a) thickness of elastic core First we determine the elastic and lastic load caacities of the section: M W knm M W knm 4 which means that the alied bending moment is the elastolastic one. From the equilibrium equation for external and internal forces we have: M.810 x z da A

4 z z z z z z z z 4010 zdz 0.0 z z 110 z z z 0.0 z z Then, the elastic core thickness is 0.08 m. (b) the radius of curvature We calculate the curvature on the basis of strain reartition in the elastic region: 4010 z, z z 0. 0, E z E so, the curvature radius is: 1 0. m 0.0 (c) the residual stress distribution The stress distribution, resulting from the alied bending moment is resented in Fig. 8..a. Because the unloading rocess is generally elastic, the unloading stresses may be calculated from usual elastic formula: M.810 max x 0.7 Ma W and from roortion for normal stresses: 0.04 x ( z 0.04) Ma. 0.0 The unloading stresses are resented in Fig. 8..b. Fig. 8. Stress distribution, unloading stress and residual stress Because of elastic unloading rocess we can recall the suerosition rincile and calculate the stresses as the sum of initial elastic-lastic stresses and the unloading ones. It gives us the stress redistribution after unloading, which means the residual stresses. We have: for z : Ma z x for z 0.1 : x Ma (and symmetrically in the bottom art of the section). The residual stress reartition is shown in Fig. 8..c. It is obvious that the axial force, resulting from the residual stress is zero. It may be found by simle calculations that the bending moment, resulting from the residual stress is also zero.

5 (d) the radius of curvature after unloading Because the elastic core was unaffected by lasticity during the whole loading-unloading rocess, we can aly the Hooke s law at any oint of the core. For the elastic/lastic boundary we have: x x E and because ( ) 4 x z x x z m. z z 0.04 Examle 8. Determine the ratio of the lastic load caacity to the elastic load caacity for the cross-section in Fig Solution the cross-section area: A = 15.5 cm the centroid osition: z c =.0 cm rincial central inertia moment: y 1 the elastic section modulus: Fig. 8.7 Cross-section 4 J cm 4 J y 55.5 W 1.7 cm z max.4 the osition of neutral axis in limit lastic state (the axis asses through lower flange): z0. 0.5cm 4 the lastic section modulus: W S y cm the ratio is: W 5.8 s W 1.7 The ratio of the lastic to the elastic load caacities is calculated by the rogram rzekrój ( A. Zaborski) view roblems roblem 8.1 Determine the ratio of the lastic load caacity to the elastic load caacity of the beam resented in Fig

6 1.1 0 cm 8 cm 1 m m m 5 cm 8 cm Fig. 8.8 Beam with loading roblem 8. Straight rods of -mm diameter and 0-m length are stored by coiling the rods inside a drum of 1.5 m inside diameter. Assuming the elastic behavior of the rods, determine (a) the stress distribution in the coiled rod cross-section, (b) the residual curvature with the bending moment decrease back to zero. Use E = 00 Ga and = 40 Ma.. roblem 8. A cantilever is loaded by a single force at its free end, Fig Assuming, that at the fixed end the lastic load caacity is reached, determine the length of that art of the cantilever which sustains artial lastification. m 5 5 cm Fig. 8.9 Cantilever with load Addendum Statically admissible stress field A statically admissible stress field is any stress field which is comatible with the static boundary conditions and inside the boundary the stress values don t exceed admissible values anywhere, x. Lower bound theorem If a statically admissible stress field can equilibrate alied loading, the lastic load caacity is not less than the alied loading. Kinematically admissible dislacement field A kinematically admissible dislacement field is any dislacements field which is comatible with the kinematic boundary conditions. Uer bound theorem If, on the kinematically admissible dislacement field, the virtual work of external and internal forces are equal, the lastic load caacity is not greater than the alied loading.

7 Examles of limit analysis use 1. Truss Determine the lastic load caacity of the truss in Fig. 8.10, assuming A 1 = cm, A = cm, A = 4 cm and = 00 Ma. 1 / / Solution Statically admissible stress field (SASF) The load caacities of the truss bars are: Fig N A, so N 1 40 kn, N 0kN, N 80 kn (a) We assume lastification of the bars 1 and : N1 40kN and N 0 kn. From the node equilibrium we have: N N1 40 kn (admissible, < N ) and 100 kn (b) We assume lastification of the bars 1 and : N1 40kN and N 80 kn. From the node equilibrium we find it is not admissible (no equilibrium of rojection onto horizontal axis) (c) We assume lastification of the bars and : N 0 kn and N 80 kn. From the node equilibrium we find it is not admissible, the force in bar 1 exceeds its load caacity Finally, the answer is: 100 kn Kinematically admissible dislacement field (KADF) We assume lastification of the bars 1 and. In this case the instantaneous rotation centre is situated at the uer hinge of the bar and dislacements field would be as in Fig π/ π/ Fig Dislacement field From the comarison of the external and internal forces work, we have: 4 A1 A, ( A1 A ) R e kN The solution is the same as in the static aroach, so it is the exact solution and calculations for other KADF are not necessary.. Beam Determine the lastic load caacity for the beam in Fig Solution KADF 1 1 Fig. 8.1 Beam with loading

8 We suose three KADFs, Fig. 8.1: 4 Fig. 8.1 Kinematic failure schemes and calculate the load caacity for each scheme, resectively: M M 5M 0. 5M M 1. 5M We choose the minimal value, so 0. 5M. SASF We verify the solution admitting the lastic hinges like in nd kinematic scheme, i.e. at the fixed end and under force, Fig M M M 1 1 Fig Beam with loading and hinges From the beam equilibrium we finally get 0. 5M, which confirms the solution. Glossary loading/unloading/reloading obciążanie/odciążanie/onowne obciążanie active rocess roces aktywny (lastyczny) assive rocess roces bierny (srężysty) Bauschinger s effect efekt Bauschingera kinematic hardening wzmocnienie kinematyczne isotroic hardening wzmocnienie izotroowe elastic load caacity nośność srężysta lastic load caacity nośność lastyczna lastification, contained lastic flow ulastycznienie elastic/lastic boundaries front lastyczny statically admissible stress field statycznie douszczalne ole narężenia kinematically admissible dislacement field kinematycznie douszczalne ole rzemieszczeń kinematic failure scheme kinematyczny mechanizm zniszczenia limit analysis teoria stanów granicznych

FE FORMULATIONS FOR PLASTICITY

FE FORMULATIONS FOR PLASTICITY G These slides are designed based on the book: Finite Elements in Plasticity Theory and Practice, D.R.J. Owen and E. Hinton, 1970, Pineridge Press Ltd., Swansea, UK. 1 Course Content: A INTRODUCTION AND