RU2591294C1 - Method of constructing diagram of limit deformations of sheet material - Google Patents

Abstract

FIELD: technological processes.

SUBSTANCE: invention relates to stamping, in particular to investigation of mechanical properties of sheet materials for assessment of their ability, as well as for use in CAD/CAE-systems in computer simulation and sheet forming. Invention: on obtained from sheet material sample dividing screen is applied, sample is tested, dividing mesh is measured, minimum main deformation and maximum main deformation are calculated, and diagram of limit deformations is plotted using deformation values applied to coordinate grid on abscissa and on ordinate. Diagram of maximum deformations is built in relative uniform elongation and coefficient of anisotropy of driven certificate on sheet material, by which negative and positive abscissa ordinate of first point on left half of beam of maximum deformations are calculated. Straight line is drawn from first point at angle of 45° to coordinate axes to intersection with ordinate axis at second point, and left half of beam of maximum deformations is obtained. Third point on right half of beam of maximum deformations is built with abscissa and ordinate on condition that when testing sample for two-axis stretching with hemispherical punch or with liquid intensity of deformation of elements near rupture of sample is 2 times higher than intensity of deformations of elements of sample near break during testing on one-axis stretching, known for left half of beam of maximum deformations. Second and third points are connected with straight line, and right half of beam of maximum deformations is obtained.

EFFECT: reduced time and higher quality of design of technological processes and equipment, saving of sheet material, as well as significant simplification of selection of sheet material and equipment for sheet stamping of parts, for example, body parts of cars and other equipment.

1 cl, 2 dwg

Description

The invention relates to the field of sheet stamping, in particular to the study of the mechanical properties of sheet materials to assess their stampability as the possibility of obtaining plastic deformations without destroying the sheet billet obtained from the sheet material in forming sheet metal stamping operations, as well as for use in CAD / CAE systems (Computer-Aided-Design / Computer-Aided-Engineering-systems) in computer modeling and design of form-changing sheet metal stamping operations before their implementation in the automotive and other industries asl industry.

Known patent 2134872 G01N 3/28 from 08/20/1999 (A method of constructing a diagram of ultimate strains and a device for its implementation), in which a block assembled from a sample with a clamp and a matrix is installed in a container and deformed with a steel shot with a diameter of 0.5-1, 5 mm with a punch in the power plant. A disadvantage of the known analogue and prototype is that it requires the use of special expensive equipment and a long period of testing and building the diagram of ultimate deformation (DPD).

The objective of the invention is to reduce the complexity, timing and cost of building DPD sheet materials.

The problem is solved as follows. Two criteria are used to determine the likelihood of a sheet blank being destroyed in form-forming sheet stamping operations, such as drawing or molding complex parts such as box-shaped or body parts, stretching or tightening sheets:

1) destruction due to deformations: at each stage of deformation of the sheet stock, the points with the coordinates of the smallest principal strain ε ₂ and the largest principal strain ε ₁ for all elements of the sheet stock should be located below the DPD of the sheet material ε ₁ = f (ε ₂ ) with a certain margin P _d deformation ductility; for a given abscissa ε ₂ take the ordinate ε ₁ to DPD for 1;

.2) failure due to stresses: the points with the coordinates of the principal stresses σ ₁ and σ ₂ should be located below the diagram of ultimate stresses (DP) of the sheet material σ ₁ = f (σ ₂ ) with a certain margin of stress plasticity P _s ; DPN is built using DPD according to the equations of connection between strains and stresses; DPN corresponds to the ultimate plasticity ellipse $${\sigma }_{\mathrm{one}}^2-{\sigma }_{\mathrm{one}}{\mathrm{<figure-callout\; id="2"\; label="\sigma "\; filenames="00000023.png,00000024.png"\; state="\{\{state\}\}">\sigma </figure-callout>}}_2+{\mathrm{<figure-callout\; id="2"\; label="\sigma "\; filenames="00000023.png,00000024.png"\; state="\{\{state\}\}">\sigma </figure-callout>}}_2^2={\mathrm{<figure-callout\; id="2"\; label="\sigma \; s"\; filenames="00000023.png,00000024.png"\; state="\{\{state\}\}">\sigma \; </figure-callout>}}_s^{}$$

.

The yield stress σ _s depending on the strain intensity ε _i = ln (l + δ _p ) is calculated taking into account the strengthening of the workpiece by the formula (Zharkov VA Modeling in the Marc system of material processing in mechanical engineering. Part 7. Tensile testing and editing. - Engineering Bulletin, 2013, No. 3, pp. 43-48):

, n=ln{l-σ_т/[σ_в(1+δ_p)]}/ln[ln(1+δ_p)], $${\sigma }_s={\sigma }_{0.2}+{\sigma }_{\mathrm{at}}(\mathrm{one}+{\delta }_p){\epsilon }_i^n$$

, n = ln {l-σ _t / [σ _in (1 + δ _p )]} / ln [ln (1 + δ _p )],

where the yield strength σ _0.2 , the tensile strength σ _in and the relative uniform elongation δ _p to start the formation of the neck on the sample is determined according to GOST 11701-84 “Metals. Tensile test methods for thin sheets and ribbons. ”

The base point on the DPD is determined by testing according to GOST 11701-84, in which the sample undergoes three main stages of extension: a - uniform elongation, when at each stage of extension the decreasing thickness and width of the sample are the same along the calculated length; b - uneven diffuse elongation with the formation of an ever increasing neck by reducing the thickness and width of the sample in the neck zone; c - local thinning in the form of a local decrease in the thickness, width and cross-sectional area of the neck and rupture of the sample in this place of the neck. In FIG. Figure 1 shows the estimated length of the sample at the end of each of these three stages: a — end of uniform elongation; b - end of neck formation; c is the end of the gap; d is the design diagram of the neck after rupture in the form of two polyhedrons-obelisks; e is the design diagram of the neck after rupture in the form of two wedge polyhedra.

The calculation of a uniform decrease in the width of the sample at the first stage of stretching of the sample is performed according to the following scheme. Metallurgical plants deliver to the machine-building enterprises a sheet, as a rule, with only one plasticity characteristic of the sheet material — relative elongation after rupture δ _f according to GOST 11701-84, and only according to additional technical conditions in the certificate the sheet gives relative uniform elongation δ _p , otherwise δ _{l, p} in the direction of the length l of the sample, and the anisotropy coefficient R _a according to GOST 11701-84. If in the certificate for the selected sheet only δ _f is known and δ _{p is} not known, and there is no machine for tensile testing of a sheet material sample to determine δ _p , then first for a sheet material similar in mechanical properties with known values of δ _p and δ _f the coefficient C = δ _f / δ _{p is} calculated, and then δ _p = δ _f / C is calculated for the selected sheet material. Based on these two properties, δ _{l, p,} and R _a are used to calculate a uniform decrease in the width b of the sample ε _{b, p} , which is required for the construction of the DPD. In a tensile test of a specimen with the initial dimensions s ₀ , b ₀ , l ₀ (s ₀ is the thickness, b ₀ is the width, l ₀ is the initial calculated length) at the moment of uniform elongation and the beginning of neck formation (when the tensile force is maximum), the dimensions the samples are equal to s _p , b _p , l _p and δ _{l, p} = (l _p -l ₀ ) / l ₀ , ε _{l, p} = ln (l _p / l ₀ ) = ln (1 + δ _{l, p} ) . At this moment, the anisotropy coefficient is equal to R _a = ln (b _p / b ₀ ) / ln (s _p / s ₀ ) = ln (b _p / b ₀ ) / ln [l ₀ b ₀ / (l _p b _p )] , whence we find the logarithmic deformation ε _{b, p} = ln (b _p / b ₀ ) = - R _a / (1 + R _a ) ln (1 + δ _{l, p} ) and from the equations ε _{b, p} = ln (1+ δ _{b, p} ), δ _{b, p} = (b _p -b ₀ ) / b ₀ - relative deformation δ _{b, p} = exp (ε _{b, p} ) -1 in width. From the condition of incompressibility of the sheet material ε _{l, p} + ε _{b, p} + ε _{s, p} = 0, which is strictly valid only for logarithmic strains ε and is valid only approximately for relative strains δ, a uniform decrease in the sample thickness is determined: ε _{s, p} = -ε _{l, p} -ε _{b, p} , δ _{s, p} = (s _p -s ₀ ) / s ₀ = exp (ε _{s, p} ) -1.

In production, to improve accuracy and quality, as well as to assess the formability of the part, a dividing grid is applied to the workpiece, after stamping in hazardous places, the parts are calculated from the nets by deformations, compared with DPD, determining the ductility margin before failure, and, if necessary, measures are prescribed to reduce deformations in hazardous locations. Often, calculation of workpiece deformations over grids is replaced or combined with CAD / CAE modeling, which also requires DPD.

The DPD of a sheet material in the form of a functional dependence ε ₁ = f (ε ₂ ) is constructed from the points obtained from uniaxial and biaxial tensile tests of samples of different shapes and sizes that are cut from this sheet material, giving different points on the DPD. A sample thickness s ₀ is applied to the cell separating grid, typically in the form of circles with a diameter l _0. The diameter of the cells is selected so that after the test near the place of discontinuity of the sample, the circles turn into ovals or ellipses with a small axis of symmetry of length l _min and a large axis of symmetry of length l _max , and the thickness s _{f of the} sample gradually increases in the direction from the point of rupture to the grip of the test machine . In this case, the shear deformations and tangential stresses in the direction of the minor and major axes of the oval are equal to zero, as a result of which the linear strains ε ₁ and ε ₂ and stresses σ ₁ and σ _2, respectively, in the direction of the major and minor axes of the oval are major. The third principal stress σ ₃ in the thickness direction of the sheet material is zero. The axis of the ovals l _min and l _max measure and calculate ε ₁ = ln (l _max / l ₀ ) and ε ₂ = ln (l _min / l ₀ ) in the center of the cell. The third main strain ε ₃ = ln (s _f / s ₀ ) is calculated either from the results of measurements of the thickness s _f in the center of the cell, or from the condition ε ₁ + ε ₂ + ε ₃ = 0 of the incompressibility of the sheet material: ε ₃ = -ε ₁ - ε ₂ . If all three strains ε ₁ , ε ₂ and ε ₃ are measured, then the incompressibility condition is used to evaluate the accuracy of the measurements.

To build a DPD on a grid of a rectangular coordinate system, they postpone: in the positive and negative directions of the horizontal abscissa axis, the smallest strain ε ₂ = ln (l _min / l ₀ ); in the positive direction of the vertical ordinate axis, the greatest strain ε ₁ = ln (l _max / l ₀ ), and from the condition ε ₁ + ε ₂ + ε ₃ = 0 of the incompressibility of the sheet material, it follows that the three strains ε ₁ , ε ₂ and ε ₃ , at least one deformation during plastic deformation of the sheet material is of positive value. Since the destruction of the sample during the test or billet of sheet material in the process of stamping parts can only occur due to thinning, always s _f <s ₀ , and deformation δ _{s, f} = (s _f -s ₀ ) / s ₀ , ε ₃ = ln (s _f / s ₀ ) = ln (1 + δ _{s, f} ) cells near the place of rupture of the sample or workpiece will always have negative values.

The left half of the DPD for ε ₂ <0 corresponds to uniaxial tension with compression of the sheet material elements, the axis ε ₂ = 0 to plane deformation, the right half of the DPD for ε ₂ > 0 to biaxial tension of the sheet material elements.

Designations are introduced for the base point 1 (ε _2.1 ; ε _1.1 ), the second digit in the indices means the number of the point on the DPD. The designations of the remaining points of the DPD are similar, only the second digit in the indices changes.

As an example in FIG. 2 in the upper half-plane (ε ₂ , ε ₁ ) the solid line shows the experimentally constructed DPD ε ₁ = f (ε ₂ ) in logarithmic deformations (ε ₂ , ε ₁ ) for sheet steel widely used in industry: galvanized low-carbon steel grade 01YUT VOSV GV GTS according to TU 14-105-685-2002 with a nominal thickness s ₀ = 0.7 mm, where the WWTP category is a very particularly difficult hood, and hot-dip galvanized steel. In the calculations of the technology for manufacturing parts from sheet material, it is necessary to use the properties of samples cut at the angle θ _r relative to the direction of rolling of the sheet material, where the plasticity characteristics are worse, so that there is an additional margin of plasticity when stamping parts from sheet material in production. For sheet steel products θ _r = 90 °. When tested in accordance with GOST 11701-84 obtained and calculated using the formulas derived above mechanical properties of sample width b ₀ = 12.5 mm from this steel: yield point σ _0,2 = 171 N / mm ^2, tensile strength σ _B = 304 H / mm ² , relative and logarithmic uniform elongations δ _p = δ _{l, p} = 0.24, ε _{l, p} = 0.215; elongation after rupture δ _f = δ ₈₀ = 0.408 for samples with an initial calculated length of 80 mm, anisotropy coefficient R _a = 2.722; relative and logarithmic uniform decreases in the width of the sample δ _b = δ _{b, p} = -0.145, ε _{b, p} = -0.157; relative and logarithmic uniform decreases in sample thickness δ _{s, p} = -0.0563, ε _{s, p} = -0.058.

If a point with coordinates for the end of uniform elongation and the beginning of neck formation on the sample is applied to the experimentally constructed DPD ε ₁ = f (ε ₂ ), then this point will be much lower than the DPD. Such a big difference is explained by the fact that the DPD is built according to the change in the dividing grid applied to the sample near the place of rupture of the neck of the sample (Fig. 1, c), and not at the moment the neck formation begins (Fig. 1, a). For sheets with high characteristics of plastic properties, the beginning of the formation of a neck is not yet considered the beginning of a rupture of a sample on a testing machine or a workpiece during stamping of a part in production. However, GOST 11701-84 does not provide for the calculation of ε ₂ and ε ₁ near the gap, although the point with these coordinates is the base from which the construction of the DPD should begin. This gap and fill in this method.

The coordinates of the base point 1 ″ (ε _{2.1 ″} ; ε _{1.1 ″} ) on the DPD are determined by the mechanical properties of the certificate for sheet material according to GOST 11701-84 according to the following scheme.

, имеющих расположенные в параллельных плоскостях большие и малые основания-прямоугольники и соединенных по малым основаниям; противоположные грани одинаково наклонены к каждому основанию (фиг. 1, d). При испытании образца по ГОСТ 11701-84 начальная расчетная длина l₀ в конце равномерного удлинения и начала образования шейки увеличивается до значения l_p=(1+δ_p)l₀=(1+0,24)80=99,2 мм, а после диффузного образования шейки, локального утонения и разрыва образца дополнительно увеличивается до значения l_f=(1+δ_f)l₀=(1+0,408)80=112,64 мм при дополнительном ходе захвата испытательной машины h_n=(l_f-l_p)=112,64-99,2=13,44 мм. Так как в образовании шейки участвуют сопряженные с верхней и нижней частями шейки участки образца, то длина шейки l_n≥h_n. На верхнем и нижнем концах шейки большие основания двух многогранников имеют размеры, соответствующие концу равномерного удлинения образца: ширина b_p=(1+δ_b,p)b₀=(1-0,145)·12,5=10,687 мм, толщина s_p=(1+δ_s,p)s₀=(1-0,0563)·0,7=0,661. При ходе h_n объем V этого основания равен: V=s_pb_ph_n. Экспериментально установлено, что объем V₂ образца, идущий на образование шейки, равен (40…50)%, в среднем, 45% от V: V₂=0,45s_pb_ph_n=0,45·0,661·10,687·13,44=42,724 мм³. Объем V₁ образца, идущий на образование одной нижней или верхней половины шейки: V₁=V₂/2=21,362 мм³. При малой номинальной толщине s₀ образца, как в данном примере, толщина s_c шейки в месте разрыва s_c≈0, и два многогранника-обелиска шейки превращаются в два многогранника-клина (фиг. 1, e), каждый из которых имеет объем V₁ и высоту $$l_n{}^{\prime }\approx h_n/2=\mathrm{13,44}/2=\mathrm{6,72}$$

мм. Теперь можно определить ширину b_c шейки в месте разрыва: $$b_c=(6V_1/l_n{}^{\prime }-2s_p{b}_p)/s_p=\mathrm{7,481}$$

мм.The lower and upper parts of the neck after rupture of the sample are approximated by a design scheme in the form of two identical polyhedra of the obelisk type $$l_n^{\text{'}\text{'}}=l_n/2$$

having large and small rectangle bases located in parallel planes and connected along small bases; opposite faces are equally inclined to each base (Fig. 1, d). When testing a sample in accordance with GOST 11701-84, the initial calculated length l ₀ at the end of uniform elongation and the beginning of neck formation increases to a value l _p = (1 + δ _p ) l ₀ = (1 + 0.24) 80 = 99.2 mm, and after diffuse neck formation, local thinning and rupture of the sample, it additionally increases to a value of l _f = (1 + δ _f ) l ₀ = (1 + 0.408) 80 = 112.64 mm with an additional grip of the testing machine h _n = (l _f -l _p ) = 112.64-99.2 = 13.44 mm. Since the sections of the sample conjugated with the upper and lower parts of the neck participate in the formation of the neck, the neck length l _n ≥h _n . At the upper and lower ends of the neck, the large bases of the two polyhedra have dimensions corresponding to the end of the uniform elongation of the sample: width b _p = (1 + δ _{b, p} ) b ₀ = (1-0.145) · 12.5 = 10.687 mm, thickness s _p = (1 + δ _{s, p} ) s ₀ = (1-0.0563) 0.7 = 0.661. During the course of h _{n, the} volume V of this base is: V = s _p b _p h _n . It was experimentally established that the volume V _{2 of the} sample, which goes to the formation of the neck, is (40 ... 50)%, on average, 45% of V: V ₂ = 0.45 s _p b _p h _n = 0.45 · 0.661 · 10.687 · 13.44 = 42.724 mm ³ . The volume V ₁ of the sample, extending the formation of a lower or upper half of the neck: V ₁ = V _2/2 = 21.362 mm ^3. For a small nominal thickness s _{0 of the} sample, as in this example, the thickness s _{c of the} neck at the point of rupture s _c ≈ 0, and two polyhedrons-obelisks of the neck turn into two polyhedrons-wedges (Fig. 1, e), each of which has a volume V ₁ and height $$l_n{}^{\prime }\approx h_n/2=13.44/2=6.72$$

mm Now you can determine the width b _{c of the} neck at the gap: $$b_c=(6V_{\mathrm{one}}/l_n{}^{\prime }-2s_p{b}_p)/s_p=\mathrm{7,481}$$

mm

и $$l_n{}^{″}$$

поперечными сечениями, перпендикулярными оси Y, и принимаем, что на длине $$l_n/2=l_n{}^{\prime }/2+l_n{}^{″}/2$$

соединенных в одном общем малом основании половин многогранников происходило локальное утонение и разрушение образца, где ячейки делительной сетки на передних гранях многогранников также были разрушены, а на прилегающей к большому основанию передней грани каждой из половин многогранников ячейки делительной сетки не были разрушены, превратились в овалы, большие оси которых равны половине передней грани, были измерены и рассчитаны деформации вблизи места разрыва. Так как толщина образца по длине $$l_n{}^{\prime }$$

каждого многогранника распределена по линейному закону от s_p до s_c≈0 и так как деформации рассчитывают в центре ячейки посередине большой оси овала, расположенной вдоль образца, то для определения толщины s_f образца в центре овальной ячейки следует в сечении образца плоскостью YZ длину по грани каждого многогранника и соответственно (из принципа подобия) длину $$l_n{}^{\prime }$$

по оси Y и толщину s_p основания многогранника разделить на 4 части и умножить на 3: s_f=3/4·s_p=3/4·0,661=0,495 мм.Divide the upper and lower polyhedra in half by length $$l_n{}^{\prime }$$

and $$l_n{}^{″}$$

cross-sections perpendicular to the Y axis, and assume that the length $$l_n/2=l_n{}^{\prime }/2+l_n{}^{″}/2$$

localized in one common small base of the polyhedra, local thinning and destruction of the sample took place, where the cells of the dividing grid on the front faces of the polyhedra were also destroyed, and on the adjacent front edge of each of the halves of the polyhedra the cells of the dividing grid were not destroyed, turned into ovals, the major axes of which are equal to half of the front face were measured and calculated strains near the rupture site. Since the thickness of the sample along the length $$l_n{}^{\prime }$$

of each polyhedron is linearly distributed from s _p to s _c ≈ 0 and since the strains are calculated in the center of the cell in the middle of the major axis of the oval located along the sample, to determine the thickness s _{f of the} sample in the center of the oval cell, the length along the faces of each polyhedron and accordingly (from the principle of similarity) the length $$l_n{}^{\prime }$$

divide along the Y axis and thickness s _{p of the} base of the polyhedron into 4 parts and multiply by 3: s _f = 3/4 · s _p = 3/4 · 0.661 = 0.495 mm.

As a result, the abscissa ε _{2.1 ″} base point 1 ″ (ε _{2.1 ″} ; ε _{1.1 ″} ) on the DPD is calculated for linear stretching of the sample under the condition ε _{2.1 ″} = ε _{1.1 ″} = -2ε _{2, 1 ″} : ε _{2.1 ″} = ε _{3.1 ″} = ln (s _f / s ₀ ) = ln (0.495 / 0.7) = - 0.3465, and the ordinate is ε _{1.1 ″} = -2ε _{2 , 1 ″} = -2 (-0.3465) = 0.693.

On the experimentally constructed DPD (Fig. 2) in the form of a solid line of logarithmic strains ε ₁ = f (ε ₂ ) from the obtained point 1 ″ (ε _{2.1 ″} ; ε _{1.1 ″} ), a dashed line is drawn to the intersection with the axis ε ₁ at point 3 ″ (ε _{2.3 ″} ; ε _{1.3 ″} ) with an abscissa ε _{2.3 ″} = 0. This dashed straight line is drawn at an angle γ to the ordinate axis, which is 45 ° for sheet materials widely used in the automotive and other industries. The ordinate is ε _{1.3 ″} point 3 ″ (ε _{2.3 ″} ; ε _{1.3 ″} ): ε _{1.3 ″} = ε _{1.1 ″} + ε _{2.1 ″} and the equation ε ₁ = f (ε ₂ ) left straight 1 ″ -3 ″ DPD:

ε ₁ = −ε ₂ + ε _{1.3 ″} with ε _{2.1 ″} ≤ ε ₂ ≤0.

For steel 01YUT VOSV GTS ε _{1.3 ″} = 0.3465.

. Например, из точки 1″(ε_2,1″; ε_1,1″) до новой точки 0″(ε_2,0″; ε_1,1″) при ε_2,0″=ε_2,min, $${\epsilon }_{\mathrm{1,}{0}^{″}}={\epsilon }_1{{}^{\prime }}_{,\max }$$

проводят отрезок прямой с уравнением ε₁=-ε₂+ε_1,3″ при ε_2,min≤ε₂≤ε_2,1″.If required, then the straight line segment 1 ″ -3 ″ DPD is extended to the left to the point with coordinates ε _{2, min} , $${\epsilon }_{\mathrm{one}}{{}^{\prime }}_{,\max }$$

. For example, from point 1 ″ (ε _{2.1 ″} ; ε _{1.1 ″} ) to the new point 0 ″ (ε _{2.0 ″} ; ε _{1.1 ″} ) with ε _{2.0 ″} = ε _{2, min} , $${\epsilon }_{\mathrm{one,}{0}^{″}}={\epsilon }_{\mathrm{one}}{{}^{\prime }}_{,\max }$$

carried straight line with equation ₁ ε ₂ = -ε ε + _{1.3 "when} ε _{2, min} ≤ε ₂ ≤ε _2,1".

равна $${\epsilon }_i{}^{\prime }=\mathrm{0,693}$$

.To plot the right half of the DPD using two points 3 ″ and 6 ″, the coordinates of the first point 3 ″ (ε _{2.3 ″} ; ε _{1.3 ″} ) on the axis ε _{1 are} already known, and the coordinates of the second point 6 ″ (ε _{2.6 ″} ; ε _{1.6 ″} ) is determined as follows. For point 1 ″ (ε _{2.1 ″} ; ε _{1.1 ″} ) with abscissa ε _{2.1 ″} = -0.3465 and ordinate ε _{1.1 ″} = 0.693 under uniaxial tension of the sample, the strain rate

is equal to $${\epsilon }_i{}^{\prime }=0.693$$

.

или $${\epsilon }_i{}^{″}=\ln (s_0/s)$$

.When testing the biaxial tension of samples with a hemispherical punch using lubricants in combination with anti-friction films, for example, Teflon, to reduce friction between the sample and the punch, or when testing biaxial tension of samples with a strain fluid ε ₁ and ε ₂ along two mutually perpendicular tangents to the surface molded by the punch or liquid, the poles of the sample are equal to each other: ε ₁ = ε ₂ = -0.5 ε ₃ , and the formula for the strain intensity ε _i takes the form: $${\epsilon }_i{}^{″}=\left|{\mathrm{<figure-callout\; id="3"\; label="\epsilon "\; filenames="00000023.png,00000024.png"\; state="\{\{state\}\}">\epsilon </figure-callout>}}_3\right|=\left|\ln (s/s_0)\right|$$

or $${\epsilon }_i{}^{″}=\ln (s_0/s)$$

.

элементов вблизи места разрыва образца в 2 раза больше интенсивности деформаций $${\epsilon }_i{}^{\prime }$$

элементов образца вблизи места разрыва при испытании на одноосное растяжение: $${\epsilon }_i{}^{″}=2{\epsilon }_i{}^{\prime }$$

, в данном примере, $${\epsilon }_i{}^{″}=2\cdot \mathrm{0,693}=\mathrm{1,386}$$

.It was experimentally established that, when tested for biaxial tension of a sample with a hemispherical punch or liquid, the strain rate $${\epsilon }_i{}^{″}$$

elements near the point of rupture of the sample is 2 times greater than the strain intensity $${\epsilon }_i{}^{\prime }$$

elements of the sample near the rupture point during uniaxial tensile testing: $${\epsilon }_i{}^{″}=2{\epsilon }_i{}^{\prime }$$

, in this example, $${\epsilon }_i{}^{″}=2\cdot 0.693=\mathrm{1,386}$$

.

для двухосного растяжения строят точку 6″ (ε_2,6″; ε_1,6″) выше оси абсцисс ε₂ с абсциссой ε_2,6″=0,693 и ординатой ε_1,6″=0,693, проводят пунктирную линию (фиг. 2) с уравнением ε₁=f(ε₂) правой половины ДПД в виде прямой через две известные точки 3″(ε_2,3″; ε_1,3″) и 6″(ε_2,6″; ε_1,6″):From the condition $$\epsilon \mathrm{one}=\mathrm{<figure-callout\; id="2"\; label="\epsilon "\; filenames="00000023.png,00000024.png"\; state="\{\{state\}\}">\epsilon </figure-callout>}2=0.5{\epsilon }_i{}^{″}=0.693$$

for biaxial tension, a point 6 ″ (ε _{2.6 ″} ; ε _{1.6 ″} ) is built above the abscissa axis ε ₂ with abscissa ε _{2.6 ″} = 0.693 and ordinate ε _{1.6 ″} = 0.693, a dashed line is drawn (FIG. 2) with the equation ε ₁ = f (ε ₂ ) of the right half of the DPD in the form of a straight line through two known points 3 ″ (ε _{2.3 ″} ; ε _{1.3 ″} ) and 6 ″ (ε _{2.6 ″} ; ε _{1, 6 ″} ):

(ε ₁ -ε _{1.3 ″} ) / (ε _{1.6 ″} -ε _{1.3 ″} ) = (ε ₂ -ε _{2.3 ″} ) / (ε _{2.6 ″} -ε _{2.3 ″} ) at 0≤ε ₂ <ε _{2.6 ″} .

, то из условия, что ε₁ является наибольшей деформацией и что всегда ε₁≥ε₂, следует, что продленная часть ДПД должна быть выше, касаться или совпадать с прямой ε₁=ε₂. Например, из точки 6″(ε_2,6″; ε_1,6″) до новой точки 7″(ε_2,7″; ε_1,7″). При ε_2,7″=ε_2,max, $${\epsilon }_{\mathrm{1,}{7}^{″}}={\epsilon }_1{{}^{″}}_{,\max }$$

проводят отрезок прямой с уравнением ε₁=f(ε₂): ε₁=ε₂ при ε_2,6″≤ε₂≤ε_2,max.If you want to extend the right half of the DPD to the right to the point with coordinates ε _{2, max} , $${\epsilon }_{\mathrm{one}}{{}^{″}}_{,\max }$$

, then from the condition that ε ₁ is the largest strain and that always ε ₁ ≥ ε ₂ , it follows that the extended part of the DPD should be higher, tangent to or coincide with the straight line ε ₁ = ε ₂ . For example, from point 6 ″ (ε _{2.6 ″} ; ε _{1.6 ″} ) to the new point 7 ″ (ε _{2.7 ″} ; ε _{1.7 ″} ). When ε _{2.7 ″} = ε _{2, max} , $${\epsilon }_{\mathrm{one,}{7}^{″}}={\epsilon }_{\mathrm{one}}{{}^{″}}_{,\max }$$

draw a straight line with the equation ε ₁ = f (ε ₂ ): ε ₁ = ε ₂ with ε _{2.6 ″} ≤ ε ₂ ≤ ε _{2, max} .

In FIG. 2 it can be seen that the DPD constructed by this method (dashed line) near the ε ₁ axis, where the points with the coordinates of dangerous places on the workpiece made of this sheet material are located when drawing complex body-type parts, corresponds exactly to the experimental DPD (solid line).

The DPD constructed according to this method is used in production to determine the risk of damage to the workpiece during stamping of a part or is introduced into a CAD / CAE system, for example, into Marc system of MSC Software Corporation (USA), which combines a CAD program and a CAE program, or to the AutoForm program of AutoForm Engineering (Switzerland), in the form of the above equations or in the form of a large number of point coordinates obtained by these equations. Based on these data, the CAD / CAE system constructs the DPS in a rectangular coordinate system with a given scale of the coordinate grid and, after computer simulation of the stamping process, displays an array of points with the coordinates ε ₁ and ε _{2 of} all the blank elements from this sheet material at all stamping stages on this coordinate grid blanks. By the number and color of the dots, they determine those elements on the stamped workpiece that are closely located (taking into account the plasticity margin) or are beyond the DPD and where the probability of failure of the workpiece is high, and develop measures to reduce the likelihood of destruction of the workpiece during processing in production.

The diagram of ultimate strains of isotropic (DPDI) sheet material constructed by this method is a first approximation. Over time, as additional tests are carried out, a diagram of the ultimate strains of anisotropic (DPDA) sheet material is imposed on the DPDI as subsequent approximations for the real anisotropic material in the production, the DPDA is introduced into the CAD / CAE system and more accurately than with the DPDI, it is modeled and designed sheet stamping process, as well as using the anisotropy coefficient A _θ = (ε ₃ -ε ₂ ) / (0,5ε ₁ ), ε ₁ = ln (1 + δ _p ), ε ₂ = ln (1 + δ _b ), ε ₃ = -ε ₁ -ε ₂ (Zharkov V.A. Modeling in the Marc system of material processing in mechanical engineering. Part 6. Extraction of a complex part - Vestnik Mashinostroeniya, 2012, No. 9, pp. 67-73) for angles θ of cutting samples relative to the direction of rolling of the sheet material, a qualitative and quantitative characterization of the effect of anisotropy on the ultimate shape change of the sheet material is obtained.

The technical result consists in the fact that, without waiting for complex and expensive tests of sheet material on a testing machine and in a stamp device, it is possible by this method to quickly build a DPD based on only two standard mechanical properties given in the certificate for sheet material in accordance with GOST 11701- 84: relative uniform elongation and anisotropy coefficient, and use this DPD both in the CAD / CAE system and in production to calculate the probability of destruction of the sheet stock during sheet head operations povki. This method reduces time and improves the quality of the design of technological processes and equipment, saves sheet material by reducing the percentage of rejects during debugging of technological processes, and also greatly simplifies the choice of sheet material and equipment for sheet metal stamping of parts, for example, body parts of cars and other equipment.

Claims (1)

A method of constructing a diagram of ultimate strains of sheet material, namely, that a dividing grid is applied to the sample obtained from the sheet material, the sample is tested, the dividing grid is measured, the smallest principal strain ε ₂ and the largest principal strain ε _{1 are} calculated and the strain values applied to the coordinate grid ε ₂ along the abscissa axis and ε ₁ along the ordinate axis build a diagram of ultimate strains ε ₁ = f (ε ₂ ), characterized in that the diagram of ultimate strains is built according to the relative uniform and the anisotropy coefficient given in the certificate for sheet material, according to which, respectively, the negative abscissa and the positive ordinate of the first point on the left half of the limit strain diagram are calculated, from the first point constructed, draw a straight line at an angle of 45 ° to the coordinate axes until it intersects with the ordinate axis at the second point and get the left half of the diagram of ultimate strains, the third point on the right half of the diagram of ultimate strains is built with an abscissa and ordinate from the condition that when tested on water drain extension of the sample with a hemispherical punch or liquid the strain rate of the elements near the rupture point of the sample is 2 times greater than the strain rate of the sample elements near the rupture when tested for uniaxial tension, known for the left half of the ultimate strain diagram, connect the second and third points of the straight line and get the right half of the diagram ultimate strains.

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