Effect of Strain Rate on the Stress-Strain Behavior of Sand

Designation:D256–10Standard Test Methods forDetermining the Izod Pendulum Impact Resistance of Plastics1This standard is issued under thefixed designation D256;the number immediately following the designation indicates the year of original adoption or,in the case of revision,the year of last revision.A number in parentheses indicates the year of last reapproval.A superscript epsilon(´)indicates an editorial change since the last revision or reapproval.This standard has been approved for use by agencies of the Department of Defense.1.Scope*1.1These test methods cover the determination of the resistance of plastics to“standardized”(see Note1)pendulum-type hammers,mounted in“standardized”machines,in break-ing standard specimens with one pendulum swing(see Note2). The standard tests for these test methods require specimens made with a milled notch(see Note3).In Test Methods A,C, and D,the notch produces a stress concentration that increases the probability of a brittle,rather than a ductile,fracture.In Test Method E,the impact resistance is obtained by reversing the notched specimen180°in the clamping vise.The results of all test methods are reported in terms of energy absorbed per unit of specimen width or per unit of cross-sectional area under the notch.(See Note4.)N OTE1—The machines with their pendulum-type hammers have been “standardized”in that they must comply with certain requirements, including afixed height of hammer fall that results in a substantiallyfixed velocity of the hammer at the moment of impact.However,hammers of different initial energies(produced by varying their effective weights)are recommended for use with specimens of different impact resistance. Moreover,manufacturers of the equipment are permitted to use different lengths and constructions of pendulums with possible differences in pendulum rigidities resulting.(See Section5.)Be aware that other differences in machine design may exist.The specimens are“standard-ized”in that they are required to have onefixed length,onefixed depth, and one particular design of milled notch.The width of the specimens is permitted to vary between limits.N OTE2—Results generated using pendulums that utilize a load cell to record the impact force and thus impact energy,may not be equivalent to results that are generated using manually or digitally encoded testers that measure the energy remaining in the pendulum after impact.N OTE3—The notch in the Izod specimen serves to concentrate the stress,minimize plastic deformation,and direct the fracture to the part of the specimen behind the notch.Scatter in energy-to-break is thus reduced. However,because of differences in the elastic and viscoelastic properties of plastics,response to a given notch varies among materials.A measure of a plastic’s“notch sensitivity”may be obtained with Test Method D by comparing the energies to break specimens having different radii at the base of the notch.N OTE4—Caution must be exercised in interpreting the results of these standard test methods.The following testing parameters may affect test results significantly:Method of fabrication,including but not limited to processingtechnology,molding conditions,mold design,and thermaltreatments;Method of notching;Speed of notching tool;Design of notching apparatus;Quality of the notch;Time between notching and test;Test specimen thickness,Test specimen width under notch,andEnvironmental conditioning.1.2The values stated in SI units are to be regarded as standard.The values given in parentheses are for information only.1.3This standard does not purport to address all of the safety concerns,if any,associated with its use.It is the responsibility of the user of this standard to establish appro-priate safety and health practices and determine the applica-bility of regulatory limitations prior to use.N OTE5—These test methods resemble ISO180:1993in regard to title only.The contents are significantly different.2.Referenced Documents2.1ASTM Standards:2D618Practice for Conditioning Plastics for TestingD883Terminology Relating to PlasticsD3641Practice for Injection Molding Test Specimens of Thermoplastic Molding and Extrusion MaterialsD4066Classification System for Nylon Injection and Ex-trusion Materials(PA)D5947Test Methods for Physical Dimensions of Solid Plastics Specimens1These test methods are under the jurisdiction of ASTM Committee D20on Plastics and are the direct responsibility of Subcommittee D20.10on MechanicalProperties.Current edition approved May1,2010.Published June2010.Originally approved st previous edition approved in2006as D256-06a´1.DOI: 10.1520/D0256-10.2For referenced ASTM standards,visit the ASTM website,,or contact ASTM Customer Service at service@.For Annual Book of ASTM Standards volume information,refer to the standard’s Document Summary page on the ASTM website.*A Summary of Changes section appears at the end of this standard. Copyright©ASTM International,100Barr Harbor Drive,PO Box C700,West Conshohocken,PA19428-2959,United States.D6110Test Method for Determining the Charpy Impact Resistance of Notched Specimens of PlasticsE691Practice for Conducting an Interlaboratory Study to Determine the Precision of a Test Method2.2ISO Standard:ISO180:1993Plastics—Determination of Izod Impact Strength of Rigid Materials33.Terminology3.1Definitions—For definitions related to plastics see Terminology D883.3.2Definitions of Terms Specific to This Standard:3.2.1cantilever—a projecting beam clamped at only one end.3.2.2notch sensitivity—a measure of the variation of impact energy as a function of notch radius.4.Types of Tests4.1Four similar methods are presented in these test meth-ods.(See Note6.)All test methods use the same testing machine and specimen dimensions.There is no known means for correlating the results from the different test methods.N OTE6—Previous versions of this test method contained Test Method B for Charpy.It has been removed from this test method and has been published as D6110.4.1.1In Test Method A,the specimen is held as a vertical cantilever beam and is broken by a single swing of the pendulum.The line of initial contact is at afixed distance from the specimen clamp and from the centerline of the notch and on the same face as the notch.4.1.2Test Method C is similar to Test Method A,except for the addition of a procedure for determining the energy ex-pended in tossing a portion of the specimen.The value reported is called the“estimated net Izod impact resistance.”Test Method C is preferred over Test Method A for materials that have an Izod impact resistance of less than27J/m(0.5 ft·lbf/in.)under notch.(See Appendix X4for optional units.) The differences between Test Methods A and C become unimportant for materials that have an Izod impact resistance higher than this value.4.1.3Test Method D provides a measure of the notch sensitivity of a material.The stress-concentration at the notch increases with decreasing notch radius.4.1.3.1For a given system,greater stress concentration results in higher localized rates-of-strain.Since the effect of strain-rate on energy-to-break varies among materials,a mea-sure of this effect may be obtained by testing specimens with different notch radii.In the Izod-type test it has been demon-strated that the function,energy-to-break versus notch radius, is reasonably linear from a radius of0.03to2.5mm(0.001to 0.100in.),provided that all specimens have the same type of break.(See5.8and22.1.)4.1.3.2For the purpose of this test,the slope,b(see22.1), of the line between radii of0.25and1.0mm(0.010and0.040 in.)is used,unless tests with the1.0-mm radius give“non-break”results.In that case,0.25and0.50-mm(0.010and 0.020-in.)radii may be used.The effect of notch radius on the impact energy to break a specimen under the conditions of this test is measured by the value b.Materials with low values of b, whether high or low energy-to-break with the standard notch, are relatively insensitive to differences in notch radius;while the energy-to-break materials with high values of b is highly dependent on notch radius.The parameter b cannot be used in design calculations but may serve as a guide to the designer and in selection of materials.4.2Test Method E is similar to Test Method A,except that the specimen is reversed in the vise of the machine180°to the usual striking position,such that the striker of the apparatus impacts the specimen on the face opposite the notch.(See Fig. 1,Fig.2.)Test Method E is used to give an indication of the unnotched impact resistance of plastics;however,results ob-tained by the reversed notch method may not always agree with those obtained on a completely unnotched specimen.(See 28.1.)4,55.Significance and Use5.1Before proceeding with these test methods,reference should be made to the specification of the material being tested. Any test specimen preparation,conditioning,dimensions,and testing parameters covered in the materials specification shall take precedence over those mentioned in these test methods.If there is no material specification,then the default conditions apply.5.2The pendulum impact test indicates the energy to break standard test specimens of specified size under stipulated parameters of specimen mounting,notching,and pendulum velocity-at-impact.3Available from American National Standards Institute(ANSI),25W.43rd St., 4th Floor,New York,NY10036,.4Supporting data giving results of the interlaboratory tests are available from ASTM Headquarters.Request RR:D20-1021.5Supporting data giving results of the interlaboratory tests are available from ASTM Headquarters.RequestRR:D20-1026.FIG.1Relationship of Vise,Specimen,and Striking Edge to Each Other for Izod Test Methods A andC5.3The energy lost by the pendulum during the breakage of the specimen is the sum of the following:5.3.1Energy to initiate fracture of the specimen;5.3.2Energy to propagate the fracture across the specimen;5.3.3Energy to throw the free end (or ends)of the broken specimen (“toss correction”);5.3.4Energy to bend the specimen;5.3.5Energy to produce vibration in the pendulum arm;5.3.6Energy to produce vibration or horizontal movement of the machine frame or base;5.3.7Energy to overcome friction in the pendulum bearing and in the indicating mechanism,and to overcome windage (pendulum air drag);5.3.8Energy to indent or deform plastically the specimen at the line of impact;and5.3.9Energy to overcome the friction caused by the rubbing of the striker (or other part of the pendulum)over the face of the bent specimen.5.4For relatively brittle materials,for which fracture propa-gation energy is small in comparison with the fracture initiation energy,the indicated impact energy absorbed is,for all practical purposes,the sum of factors 5.3.1and 5.3.3.The toss correction (see 5.3.3)may represent a very large fraction of the total energy absorbed when testing relatively dense and brittle materials.Test Method C shall be used for materials that have an Izod impact resistance of less than 27J/m (0.5ft·lbf/in.).(See Appendix X4for optional units.)The toss correction obtained in Test Method C is only an approximation of the toss error,since the rotational and rectilinear velocities may not be the same during the re-toss of the specimen as for the original toss,and because stored stresses in the specimen may have been released as kinetic energy during the specimen fracture.5.5For tough,ductile,fiber filled,or cloth-laminated mate-rials,the fracture propagation energy (see 5.3.2)may be large compared to the fracture initiation energy (see 5.3.1).When testing these materials,factors (see 5.3.2,5.3.5,and 5.3.9)canbecome quite significant,even when the specimen is accurately machined and positioned and the machine is in good condition with adequate capacity.(See Note 7.)Bending (see 5.3.4)and indentation losses (see 5.3.8)may be appreciable when testing soft materials.N OTE 7—Although the frame and base of the machine should be sufficiently rigid and massive to handle the energies of tough specimens without motion or excessive vibration,the design must ensure that the center of percussion be at the center of strike.Locating the striker precisely at the center of percussion reduces vibration of the pendulum arm when used with brittle specimens.However,some losses due to pendulum arm vibration,the amount varying with the design of the pendulum,will occur with tough specimens,even when the striker is properly positioned.5.6In a well-designed machine of sufficient rigidity and mass,the losses due to factors 5.3.6and 5.3.7should be very small.Vibrational losses (see 5.3.6)can be quite large when wide specimens of tough materials are tested in machines of insufficient mass,not securely fastened to a heavy base.5.7With some materials,a critical width of specimen may be found below which specimens will appear ductile,as evidenced by considerable drawing or necking down in the region behind the notch and by a relatively high-energy absorption,and above which they will appear brittle as evidenced by little or no drawing down or necking and by a relatively low-energy absorption.Since these methods permit a variation in the width of the specimens,and since the width dictates,for many materials,whether a brittle,low-energy break or a ductile,high energy break will occur,it is necessary that the width be stated in the specification covering that material and that the width be reported along with the impact resistance.In view of the preceding,one should not make comparisons between data from specimens having widths that differ by more than a few mils.5.8The type of failure for each specimen shall be recorded as one of the four categories listed as follows:C =Complete Break —A break where the specimen separates into two or more pieces.H =Hinge Break —An incomplete break,such that one part of the specimen cannot support itself above the horizontal when the other part is held vertically (less than 90°included angle).P =Partial Break —An incomplete break that does not meet the definition for a hinge break but has frac-tured at least 90%of the distance between the vertex of the notch and the opposite side.NB =Non-Break —An incomplete break where the frac-ture extends less than 90%of the distance be-tween the vertex of the notch and the opposite side.For tough materials,the pendulum may not have the energy necessary to complete the breaking of the extreme fibers and toss the broken piece or pieces.Results obtained from “non-break”specimens shall be considered a departure from stan-dard and shall not be reported as a standard result.Impact resistance cannot be directly compared for any two materials that experience different types of failure as defined in the test method by this code.Averages reported must likewise be derived from specimens contained within a single failure category.This letter code shall suffix the reported impact identifying the types of failure associated with the reported value.If more than one type of failure is observed for asampleFIG.2Relationship of Vise,Specimen,and Striking Edge to EachOther for Test MethodEmaterial,then the report will indicate the average impact resistance for each type of failure,followed by the percent of the specimens failing in that manner and suffixed by the letter code.5.9The value of the impact methods lies mainly in the areas of quality control and materials specification.If two groups of specimens of supposedly the same material show significantly different energy absorptions,types of breaks,critical widths,or critical temperatures,it may be assumed that they were made of different materials or were exposed to different processing or conditioning environments.The fact that a material shows twice the energy absorption of another under these conditions of test does not indicate that this same relationship will exist under another set of test conditions.The order of toughness may even be reversed under different testing conditions.N OTE8—A documented discrepancy exists between manual and digital impact testers,primarily with thermoset materials,including phenolics, having an impact value of less than54J/m(1ft-lb/in.).Comparing data on the same material,tested on both manual and digital impact testers, may show the data from the digital tester to be significantly lower than data from a manual tester.In such cases a correlation study may be necessary to properly define the true relationship between the instruments.TEST METHOD A—CANTILEVER BEAM TEST6.Apparatus6.1The machine shall consist of a massive base on which is mounted a vise for holding the specimen and to which is connected,through a rigid frame and bearings,a pendulum-type hammer.(See 6.2.)The machine must also have a pendulum holding and releasing mechanism and a mechanism for indicating the breaking energy of the specimen.6.2A jig for positioning the specimen in the vise and graphs or tables to aid in the calculation of the correction for friction and windage also should be included.One type of machine is shown in Fig.3.One design of specimen-positioning jig is illustrated in Fig.4.Detailed requirements are given in subsequent paragraphs.General test methods for checking and calibrating the machine are given in Appendix X2.Additional instructions for adjusting a particular machine should be supplied by the manufacturer.6.3The pendulum shall consist of a single or multi-membered arm with a bearing on one end and a head, containing the striker,on the other.The arm must be suffi-ciently rigid to maintain the proper clearances and geometric relationships between the machine parts and the specimen and to minimize vibrational energy losses that are always includedin the measured impact resistance.Both simple and compound pendulum designs may comply with this test method.6.4The striker of the pendulum shall be hardened steel and shall be a cylindrical surface having a radius of curvature of 0.8060.20mm(0.03160.008in.)with its axis horizontal and perpendicular to the plane of swing of the pendulum.The line of contact of the striker shall be located at the center of percussion of the pendulum within62.54mm(60.100in.) (See Note9.)Those portions of the pendulum adjacent to the cylindrical striking edge shall be recessed or inclined at a suitable angle so that there will be no chance for other than this cylindrical surface coming in contact with the specimen during the break.N OTE9—The distance from the axis of support to the center of percussion may be determined experimentally from the period of small amplitude oscillations of the pendulum by means of the following equation:L5~g/4p2!p2FIG.3Cantilever Beam(Izod-Type)ImpactMachine FIG.4Jig for Positioning Specimen forClampingwhere:L=distance from the axis of support to the center of percussion,m or(ft),g=local gravitational acceleration(known to an accuracy of one part in one thousand),m/s2or(ft/s2),p= 3.1416(4p2=39.48),andp=period,s,of a single complete swing(to and fro)determined by averaging at least20consecutive and uninterrupted swings.Theangle of swing shall be less than5°each side of center.6.5The position of the pendulum holding and releasing mechanism shall be such that the vertical height of fall of the striker shall be61062mm(24.060.1in.).This will produce a velocity of the striker at the moment of impact of approxi-mately3.5m(11.4ft)/s.(See Note10.)The mechanism shall be so constructed and operated that it will release the pendulum without imparting acceleration or vibration to it.N OTE10—V5~2gh!0.5where:V=velocity of the striker at the moment of impact(m/s),g=local gravitational acceleration(m/s2),andh=vertical height of fall of the striker(m).This assumes no windage or friction.6.6The effective length of the pendulum shall be between 0.33and0.40m(12.8and16.0in.)so that the required elevation of the striker may be obtained by raising the pendulum to an angle between60and30°above the horizontal.6.7The machine shall be provided with a basic pendulum capable of delivering an energy of2.760.14J(2.0060.10 ft·lbf).This pendulum shall be used with all specimens that extract less than85%of this energy.Heavier pendulums shall be provided for specimens that require more energy to break. These may be separate interchangeable pendulums or one basic pendulum to which extra pairs of equal calibrated weights may be rigidly attached to opposite sides of the pendulum.It is imperative that the extra weights shall not significantly change the position of the center of percussion or the free-hanging rest point of the pendulum(that would consequently take the machine outside of the allowable calibration tolerances).A range of pendulums having energies from2.7to21.7J(2to16 ft·lbf)has been found to be sufficient for use with most plastic specimens and may be used with most machines.A series of pendulums such that each has twice the energy of the next will be found convenient.Each pendulum shall have an energy within60.5%of its nominal capacity.6.8A vise shall be provided for clamping the specimen rigidly in position so that the long axis of the specimen is vertical and at right angles to the top plane of the vise.(See Fig.1.)This top plane shall bisect the angle of the notch with a tolerance of0.12mm(0.005in.).Correct positioning of the specimen is generally done with a jig furnished with the machine.The top edges of thefixed and moveable jaws shall have a radius of0.2560.12mm(0.01060.005in.).For specimens whose thickness approaches the lower limiting value of3.00mm(0.118in.),means shall be provided to prevent the lower half of the specimen from moving during the clamping or testing operations(see Fig.4and Note11.)N OTE11—Some plastics are sensitive to clamping pressure;therefore,cooperating laboratories should agree upon some means of standardizing the clamping force.One method is using a torque wrench on the screw of the specimen vise.If the faces of the vise or specimen are notflat and parallel,a greater sensitivity to clamping pressure may be evident.See the calibration procedure in Appendix X2for adjustment and correction instructions for faulty instruments.6.9When the pendulum is free hanging,the striking surface shall come within0.2%of scale of touching the front face of a standard specimen.During an actual swing this element shall make initial contact with the specimen on a line22.0060.05 mm(0.8760.002in.)above the top surface of the vise. 6.10Means shall be provided for determining the energy expended by the pendulum in breaking the specimen.This is accomplished using either a pointer and dial mechanism or an electronic system consisting of a digital indicator and sensor (typically an encoder or resolver).In either case,the indicated breaking energy is determined by detecting the height of rise of the pendulum beyond the point of impact in terms of energy removed from that specific pendulum.Since the indicated energy must be corrected for pendulum-bearing friction, pointer friction,pointer inertia,and pendulum windage,in-structions for making these corrections are included in10.3and Annex A1and Annex A2.If the electronic display does not automatically correct for windage and friction,it shall be incumbent for the operator to determine the energy loss manually.(See Note12.)N OTE12—Many digital indicating systems automatically correct for windage and friction.The equipment manufacturer may be consulted for details concerning how this is performed,or if it is necessary to determine the means for manually calculating the energy loss due to windage and friction.6.11The vise,pendulum,and frame shall be sufficiently rigid to maintain correct alignment of the hammer and speci-men,both at the moment of impact and during the propagation of the fracture,and to minimize energy losses due to vibration. The base shall be sufficiently massive that the impact will not cause it to move.The machine shall be so designed,con-structed,and maintained that energy losses due to pendulum air drag(windage),friction in the pendulum bearings,and friction and inertia in the indicating mechanism are held to a minimum.6.12A check of the calibration of an impact machine is difficult to make under dynamic conditions.The basic param-eters are normally checked under static conditions;if the machine passes the static tests,then it is assumed to be accurate.The calibration procedure in Appendix X2should be used to establish the accuracy of the equipment.However,for some machine designs it might be necessary to change the recommended method of obtaining the required calibration measurements.Other methods of performing the required checks may be substituted,provided that they can be shown to result in an equivalent accuracy.Appendix X1also describes a dynamic test for checking certain features of the machine and specimen.6.13Micrometers—Apparatus for measurement of the width of the specimen shall comply with the requirements of Test Methods D5947.Apparatus for the measurement of the depth of plastic material remaining in the specimen under the notch shall comply with requirements of Test Methods D5947, provided however that the one anvil or presser foot shall beatapered blade conforming to the dimensions given in Fig.5.The opposing anvil or presser foot shall be flat and conforming to Test Methods D5947.7.Test Specimens7.1The test specimens shall conform to the dimensions and geometry of Fig.6,except as modified in accordance with 7.2,7.3,7.4,and 7.5.To ensure the correct contour and conditions of the specified notch,all specimens shall be notched as directed in Section 8.7.1.1Studies have shown that,for some materials,the location of the notch on the specimen and the length of the impacted end may have a slight effect on the measured impact resistance.Therefore,unless otherwise specified,care must be taken to ensure that the specimen conforms to the dimensions shown in Fig.6and that it is positioned as shown in Fig.1or Fig.2.7.2Molded specimens shall have a width between 3.0and 12.7mm (0.118and 0.500in.).Use the specimen width as specified in the material specification or as agreeduponN OTE 1—These views not to scale.N OTE 2—Micrometer to be satin-chrome finished with friction thimble.N OTE 3—Special anvil for micrometer caliper 0to 25.4mm range (50.8mm frame)(0to 1in.range (2-in.frame)).N OTE 4—Anvil to be oriented with respect to frame as shown.N OTE 5—Anvil and spindle to have hardened surfaces.N OTE 6—Range:0to 25.4mm (0to 1in.in thousandths of an inch).N OTE 7—Adjustment must be at zero when spindle and anvil are in contact.FIG.5Early (ca.1970)Version of a Notch-DepthMicrometerbetween the supplier and the customer.All specimens having one dimension less than 12.7mm (0.500in.)shall have the notch cut on the shorter side.Otherwise,all compression-molded specimens shall be notched on the side parallel to the direction of application of molding pressure.(See Fig.6.)N OTE 13—While subsection 7.5requires perpendicular pairs of plane parallel surfaces,the common practice has been to accept the non-parallel drafted surfaces formed when directly injection molding specimens for Izod ers must be aware that employing a trapezoidal section rather than a rectangular section may lead to data shifts and scatter.Unequal stress,created by clamping in the fracture region and dynamic twisting,caused by uneven striking of the specimen are prone to occur when the faces of the specimen are not parallel.Interlaboratory compari-sons must clearly spell out the specimen preparation conditions.7.2.1Extreme care must be used in handling specimens less than 6.35mm (0.250in.)wide.Such specimens must be accurately positioned and supported to prevent twist or lateral buckling during the test.Some materials,furthermore,are very sensitive to clamping pressure (see Note 11).7.2.2A critical investigation of the mechanics of impact testing has shown that tests made upon specimens under 6.35mm (0.250in.)wide absorb more energy due to crushing,bending,and twisting than do wider specimens.Therefore,specimens 6.35mm (0.250in.)or over in width are recom-mended.The responsibility for determining the minimumspecimen width shall be the investigator’s,with due reference to the specification for that material.7.2.3Material specification should be consulted for pre-ferred molding conditions.The type of mold and molding machine used and the flow behavior in the mold cavity will influence the impact resistance obtained.A specimen taken from one end of a molded plaque may give different results than a specimen taken from the other end.Cooperating laboratories should therefore agree on standard molds con-forming to the material specification.Practice D3641can be used as a guide for general molding tolerances,but refer to the material specification for specific molding conditions.7.2.4The impact resistance of a plastic material may be different if the notch is perpendicular to,rather than parallel to,the direction of molding.The same is true for specimens cut with or across the grain of an anisotropic sheet or plate.7.3For sheet materials,the specimens shall be cut from the sheet in both the lengthwise and crosswise directions unless otherwise specified.The width of the specimen shall be the thickness of the sheet if the sheet thickness is between 3.0and 12.7mm (0.118and 0.500in.).Sheet material thicker than 12.7mm shall be machined down to 12.7mm.Specimens with a 12.7-mm square cross section may be tested either edgewise or flatwise as cut from the sheet.When specimens are tested flatwise,the notch shall be made on the machined surface ifthemmin.A 10.1660.050.40060.002B 31.861.0 1.2560.04C 63.562.0 2.5060.08D 0.25R 60.050.010R 60.002E12.7060.200.50060.008FIG.6Dimensions of Izod-Type TestSpecimen。

第42卷第4期2021年4月哈㊀尔㊀滨㊀工㊀程㊀大㊀学㊀学㊀报Journal of Harbin Engineering UniversityVol.42ɴ.4Apr.2021非对称循环载荷下Q235钢力学响应特性分析张庆玲1,2,金淼1,2,李群1,2,郭宝峰1,2(1.燕山大学先进锻压成形技术与科学教育部重点实验室,河北秦皇岛066004;2.燕山大学机械工程学院,河北秦皇岛066004)摘㊀要:为了准确判断Q235钢在非对称应力循环载荷作用下产生的棘轮效应㊁包申格效应及循环软/硬化特性对材料性能的影响,本文进行了多种非对称应力条件下的循环加载试验㊂采用数据分析的方法,研究了Q235钢力学变形行为与加载工况之间的关系㊂试验表明:Q235钢棘轮应变和棘轮应变率的正负与平均应力符号相同;平均应力为负值时,表现为循环硬化特性,压缩屈服应力大于拉伸屈服应力;平均应力为正值时,表现为循环软化特性,拉伸屈服应力大于压缩屈服应力;在平均应力符号相同条件下,第1周的屈服应力也基本相同㊂研究结果为结构设计和建立精确的循环本构模型提供了理论依据㊂关键词:非对称循环载荷;包申格效应;循环软硬化;棘轮效应;屈服应力;Q235钢;平均应力;应力幅值DOI :10.11990/jheu.201905041网络出版地址:http :// /kcms /detail /23.1390.u.20210302.1345.014.html 中图分类号:TB31㊀文献标志码:A㊀文章编号:1006-7043(2021)04-0581-07Analysis of the mechanical response characteristics of Q 235steel under asymmetrical cyclic loadingZHANG Qingling 1,2,JIN Miao 1,2,LI Qun 1,2,GUO Baofeng 1,2(1.Key Laboratory of Advanced Forging &Stamping Technology and Science (Yanshan University),Ministry of Education of China,Qinhuangdao 066004,China;2.School of Mechanical Engineering,Yanshan University,Qinhuangdao 066004,China)Abstract :To accurately determine the effects of the ratcheting effect,Bauschinger effect,and cyclic softening and hardening characteristics of Q235steel on the material properties under asymmetric cyclic stress loading,we con-duct various cyclic loading tests under asymmetric stress conditions.The relationships between the mechanical de-formation behaviors of Q235steel and loading conditions are discussed by means of data analysis.The test shows that the positive and negative signs of ratcheting strain and ratcheting strain rate of Q235steel are the same as those of the mean stress.When the mean stress is negative,it shows a cyclic hardening characteristic,and the compres-sive yield stress is greater than the tensile yield stress.When the mean stress is positive,it shows a cyclic softening characteristic,and the tensile yield stress is greater than the compression yield stress.Under the condition of the same mean stress sign,the yield stress of the first cycle is basically the same.Results provide theoretical data for the structural design and the establishment of an accurate cyclic constitutive model.Keywords :asymmetric cyclic loading;Bauschinger effect;cyclic softening and hardening;ratcheting;yield stress;Q235steel;mean stress;stress amplitude收稿日期:2019-05-13.网络出版日期:2021-03-02.基金项目:国家自然科学基金项目(52075474);河北省自然科学基金项目(E2019203560);河北省高校创新团队领军人才培育计划(LJRC012).作者简介:张庆玲,女,高级实验师,博士研究生;金淼,男,教授,博士生导师.通信作者:金淼,E-mail:jmiao@.㊀㊀工程中的承载结构在工作中常由于载荷或几何形状不连续而产生应力集中,甚至会在局部出现塑性变形㊂此时,对于受循环载荷作用的结构件则可能出现棘轮效应㊁包申格效应㊁循环软/硬化等变形行为,从而影响结构性能㊂因此,深入了解材料在循环载荷作用下的力学相应特性对于结构设计及安全评定具有十分重要的意义㊂棘轮应变会随着循环次数的增加而增加,会严重恶化部件的性能[1]㊂棘轮应变的累积取决于载荷中平均应力和应力幅值的组合[2-3]㊂在恒平均应力时,无论应力速率如何变化,随应力幅值的增加,棘轮寿命都会下降,棘轮应变累积率增加[4-6]㊂这些研究使人们对金属材料棘轮效应的基本特性有了较为深入的了解㊂哈㊀尔㊀滨㊀工㊀程㊀大㊀学㊀学㊀报第42卷包申格效应是指金属材料在经历了一定量的单向拉伸或压缩塑性变形之后再反向加载,其屈服应力会低于连续正向变形的屈服应力,这是造成金属材料力学方向性的重要原因之一[7-9]㊂盛光敏等[10]通过对AZ31进行拉压和压拉循环试验,得出其包申格效应比反包申格效应明显;文献[11-12]分析了不同应变历史㊁预应变量㊁应变速率和循环周次对铝合金7A04和高强钢10CrNi5MoV 包申格效应的影响;文献[13-14]对Q345㊁Q460和Q235进行循环加载试验,指出3种钢材均存在包申格效应㊂当外加循环载荷使得材料进入塑性变形后,反复变形会令金属的塑性流动特性发生变化,造成材料抵抗变形的能力增强或减弱,这种现象称为循环硬化或循环软化㊂文献[15]探讨了不同加载条件下不锈钢316L 的循环软硬化行为;文献[16-18]指出低碳钢S355循环硬/化行为随塑性应变范围的增大而增大,循环软化行为随塑性应变范围的减小而减小;文献[19-21]研究发现Q235钢在不同应变幅值和平均应变组合下表现为循环硬化,循环硬化指数随平均应变水平的增加而增大㊂学者们对不同材料在不同条件下的循环变形特性进行了深入探讨,但对于焊接结构件最为常用的Q235钢在非对称应力控制下产生棘轮效应㊁包申格效应及循环软硬化特性的系统研究却鲜见报道㊂本文以Q235钢为研究对象,进行了多种条件下的循环加载试验,运用数据分析的方法,并结合唯象理论,对试验结果进行系统分析,深入研究了此材料的力学响应特性㊂1㊀应力循环加载试验方案试验所用原材料为20mm 厚Q235钢板,测得其弹性模量为210GPa,上㊁下屈服极限分别为310MPa 和243MPa㊂沿轧制方向取样,按照GB /T 3075-2008‘金属材料疲劳试验轴向力控制方法“加工成圆形截面循环加载试样,其平行段直径9mm,平行段长度27mm,过渡圆弧半径25mm㊂在精度为0.2kN 的Instron8801型电液伺服疲劳试验机上进行循环加载试验,并通过精度为0.1μm,标距为25mm 的接触式引伸计采集轴向应变㊂试验加载波形如图1所示,加载应力率为40MPa /s,循环周次为30周,具体试验方案如表1所示㊂图1㊀应力循环加载曲线Fig.1㊀Single stage stress cyclic loading curves表1㊀应力循环试验加载工况Table 1㊀Loading conditions of stress cycle test试样编号平均应力/MPa应力幅值/MPaFYL-01-20300FYL-02-20320FYL-03-20340FYL-04-40300FYL-05-60300ZYL-0920300ZYL-1020320ZYL-1120340ZYL-1240300ZYL-13603002㊀棘轮效应分析本文材料在非对称应力循环载荷作用下产生的棘轮应变εr 为:εr =(εT max +εT min )/2(1)式中εT max ㊁εTmin 分别表示某一个循环周次的最大真应变和最小真应变㊂如无特殊说明,本文所涉及应变均为真应变㊂由于Q235钢存在屈服平台,使得第1周的应变值较大,棘轮应变的计算均从第2周开始㊂将相邻2个循环周次内棘轮应变的变化量定义为棘轮应变率Δεr ,其反映了循环加载过程中棘轮应变累积的快慢程度㊂应力幅值为300MPa,平均应力分别为-20MPa 和20MPa 时,循环加载过程中Q235钢的应力应变曲线如图2所示㊂可以看出,正㊁负平均应力时均产生了棘轮效应㊂当平均应力为负值时,随循环周次的增加,滞回曲线向负应变方向移动;当平均应力为正值时,滞回曲线则向正应变方向移动㊂应力幅值为300MPa,不同平均应力条件下棘轮应变随循环周次的变化如图3所示㊂由图可知,平均应力为正值时棘轮应变为正值,平均应力为负值时棘轮应变为负值,棘轮应变的绝对值均随循环周次增加而增大㊂不同正负平均应力条件下棘轮应变的变化趋势不同,且棘轮应变率也不相同,但棘轮应变率均在第20周后趋于稳定㊂平均应力为正值时,棘轮应变率为正值,平均应力越大,棘轮应变率越高;平均应力为负值时,棘轮应变率为负值,平均应力绝对值越大,棘轮应变率也越高㊂应力幅值为300MPa 时,棘轮应变率稳定值㊃285㊃第4期张庆玲,等:非对称循环载荷下Q235钢力学响应特性分析与平均应力之间的关系曲线如图4所示,两者之间呈指数函数关系㊂图2㊀循环应力应变曲线Fig.2㊀Cyclic stress-straincurve图3㊀应力幅值为300MPa 时的棘轮应变Fig.3㊀Ratcheting strain at stress amplitude 300MPa图4㊀应力幅值为300MPa 时不同平均应力下的棘轮应变率Fig.4㊀Ratcheting strain rate under different mean stressat stress amplitude 300MPa㊀㊀图5所示为不同应力幅值下Q235钢的棘轮应变曲线㊂无论平均应力为正值还是负值,棘轮应变及棘轮应变速率均随应力幅值的增加而增大㊂由上述分析可知,Q235钢在循环过程中产生的棘轮应变与平均应力㊁应力幅值和循环周次有关,根据唯象理论得出棘轮应变的预测模型:εr =10-4λησm σᶄs0-σs0exp(η)N +εᶄη,σm >010-4λσm(σᶄs0-σs0)ηexp(η)N -εᶄη,σm <0ìîíïïïïïï(2)式中:λ为材料参数,可通过试验数据获取,此处取λ=3.5;N 为循环周次;σm 为平均应力;σa 为应力幅值;σᶄs0㊁σs0分别为单向拉伸时的上㊁下屈服极限;η=σa /σs ;εᶄ为循环加载试验时应力峰值在单向拉伸试验曲线中所对应的应变值㊂图5㊀不同应力幅值条件下的棘轮应变Fig.5㊀Ratcheting strain at different stress amplitude如图6所示,将计算得到的棘轮应变与试验数据进行对比,两者吻合良好,说明该公式可以在非对称应力控制的循环加载试验中,较好的表征Q235钢的棘轮效应㊂图6㊀棘轮应变试验值与拟合值比较Fig.6㊀Comparison of test and fitting data of ratchetingstress㊃385㊃哈㊀尔㊀滨㊀工㊀程㊀大㊀学㊀学㊀报第42卷3㊀包申格效应分析包申格效应在金属材料构件中一般扮演着负面的角色,会影响到材料的抗疲劳性能,造成工件不能满足正常的服役条件㊂试验发现,Q235钢在循环加载过程中产生了明显的包申格效应㊂为使计算结果更有可比性,在循环第1周取屈服平台的值作为屈服应力,其他不产生屈服平台且无明显屈服点的循环周次,取相应周次发生0.2%相对塑性变形时对应的应力值,即取该周应力应变曲线起始部分的斜率,然后偏移0.2%应变量对应得到的应力值㊂图7给出了应力幅值为300MPa,不同平均应力条件下的屈服应力随循环周次的变化曲线㊂图中显示,当平均应力为负值时,压缩屈服应力大于拉伸屈服应力,平均应力绝对值越大,拉伸屈服应力越小,压缩屈服应力越大;而当平均应力为正值时,拉伸屈服应力大于压缩屈服应力,平均应力越大,拉伸屈服应力越大,压缩屈服应力越小㊂且随循环周次的增加无论平均应力为正值还是负值,拉伸和压缩方向的屈服应力均呈现下降趋势㊂图7㊀不同平均应力条件下屈服应力随循环周次变化曲线Fig.7㊀Changing curves of yield stress with cycle numbers at different mean stresses㊀㊀图8为相同平均应力不同幅值条件下各循环周次的屈服应力㊂结合图7的结论,可以得出,在应力载荷控制下,在拉伸和压缩2个方向哪个方向载荷大,对应方向的屈服应力相对较高㊂平均应力一定时,随应力幅值的增大,拉伸屈服应力和压缩屈服应力均略有增大,在循环到第30周时相邻载荷条件下的屈服应力相对变化量不超过2%㊂对比图7和图8,可以看出,平均应力和应力幅值分别增大所得到的屈服应力变化规律并不一致,相同峰值条件下,平均应力比应力幅值对屈服应力的影响更为明显㊂这是因为,平均应力为负值时,平均应力增大,应力峰值减小,应力谷值绝对值增大;平均应力为正值时,平均应力增大,应力峰值增大,应力谷值绝对值减小;而应力幅值增大,正负平均应力下的应力峰值和应力谷值的绝对值均增大㊂图8㊀不同应力幅值下的屈服应力变化曲线Fig.8㊀Changing curve of yield stress at different stress amplitude and same mean stresses㊀㊀从上述对屈服应力的数据分析可以得出,屈服应力是关于循环周次㊁平均应力和应力幅值的函数㊂当外加载荷大于上屈服极限时,屈服应力σs 随循环周次的演化规律为:σs =σi ,N =1(μ+λi )σi ηMiN,N >1{(3)式中:i =1,2分别代表拉伸和压缩;σi 为第1周屈服应力值,平均应力为正值时,拉伸屈服应力为289MPa,㊃485㊃第4期张庆玲,等:非对称循环载荷下Q235钢力学响应特性分析压缩屈服应力为189MPa;平均应力为负值时,拉伸屈服应力为191MPa,压缩屈服应力为320MPa;λi 为材料参数,通过试验获取,平均应力为正值时,λ1取0.78,λ2取1.10;平均应力为负值时,λ1取0.87,λ2取1.15;μ=σmσᶄs0;M i 为指数指标,M 1=10σm ησP,M 2=10σmσP,σp 为应力峰值㊂将计算得到的屈服应力与试验值进行比较,如图9所示㊂不同正负平均应力条件下,两者在初始几个循环周次相差较多,最大相对误差为5%,但在10周之后最大相对误差仅为2%,说明拟合效果良好㊂图9㊀屈服应力试验值与拟合值比较Fig.9㊀Comparison of test data and fitting data about yield stress㊀㊀包申格系数B 与循环载荷作用下预拉伸/压缩变形后屈服应力的变化直接相关,可对包申格效应进行定量表征和描述,其表达式为:B =σ-0.2-σ+0.2σ-0.2,σm <0σ+0.2-σ-0.2σ+0.2,σm >0ìîíïïïïïï(4)式中:σ+0.2㊁σ-0.2分别为材料的拉伸和压缩屈服应力㊂屈服应力下降越大,包申格系数越高,包申格效应越明显㊂Q235钢在幅值为300MPa,不同平均应力下包申格系数的变化如图10所示㊂图10㊀应力幅值为300MPa 时包申格系数随循环周次的变化Fig.10㊀The variation of Bauschinger coefficient with cyclenumbers at stress amplitude 300MPa可以看出,不同应力状态第1周时的包申格系㊀㊀数基本相同,约为0.346,说明Q235在经历一定的预拉伸/压缩变形后再反向压缩/拉伸时均表现出明显的包申格效应㊂从第2周开始,包申格系数随平均应力的变化表现出较大差异,但相同应力条件下则变化很小㊂因此,可以用第2周的包申格系数来描述材料的包申格效应㊂平均应力绝对值越大,包申格效应越显著㊂平均应力绝对值相同时,Q235钢在正平均应力条件下表现出的包申格效应更明显㊂4㊀循环软/硬化特性分析Q235钢在不同应力组合条件下表现出不同的循环软/硬化行为㊂不同平均应力和应力幅值条件下的响应应变幅值随着循环周次的变化曲线如图11所示㊂可以看出,当平均应力为负值时,表现出轻微的硬化特性;当平均应力为正值时,则表现出明显的软化特性,但2种工况下,应力幅值越大应变幅值越大,而平均应力绝对值越大应变幅值却越小㊂随循环周次的增加,Q235钢循环软/硬化速率在第20周之后趋于稳定㊂将不同应力条件下的循环软/硬化速率稳定值列于表2,负值表示循环硬化,正值表示循环软化㊂比较发现,负平均应力时,循环硬化速率随平均应力绝对值的增大逐渐减小,即硬化程度减弱;正平均应力时,循环软化速率随平均应力的增大呈增长趋势,即软化程度增强;应力幅值增大使循环软硬化速率均增加,软硬化程度更显著㊂㊃585㊃哈㊀尔㊀滨㊀工㊀程㊀大㊀学㊀学㊀报第42卷图11㊀不同条件下的应变幅值变化曲线Fig.11㊀Changing curve of strain amplitude at different conditions表2㊀不同条件下的循环软/硬化速率Table 2㊀Cyclic softening /hardening rate at different condi-tions应力/MPa幅值均值速率/10-6300-60-1.46010.4-40-3.2408.5-20-3.420 6.0320-20-4.420 6.3340-20-14.32010.95㊀结论1)在非对称应力循环载荷作用下,负平均应力产生负棘轮应变,正平均应力产生正棘轮应变㊂2)外加载荷平均应力相同,应力幅值增大,或外加载荷应力幅值相同,平均应力绝对值增大,均会造成棘轮应变绝对值增大,棘轮应变率升高㊂3)在非对称应力循环载荷作用下,Q235钢表现出明显的包申格效应㊂当平均应力为负值时,压缩屈服应力大于拉伸屈服应力;当平均应力为正值时,拉伸屈服应力大于压缩屈服应力㊂在循环的第1周屈服应力值基本相同,可以用第2周的包申格系数来表征材料的包申格效应㊂4)当外加载荷平均应力为负值时,Q235钢表现出循环硬化特性,平均应力绝对值越小或应力幅值越大,循环硬化现象越明显;当外加载荷平均应力为正值时,Q235钢表现出循环软化特性,平均应力或应力幅值越大,循环软化现象越明显㊂参考文献:[1]陈旭,焦荣,田涛.棘轮效应预测及其循环本构模型研究进展[J].力学进展,2003,33(4):461-470.CHEN Xu,JIAO Rong,TIAN Tao.Research advances of ratcheting effects and cyclic constitutive models [J].Ad-vances in mechanics,2003,33(4):461-470.[2]LIN Y C,LIU Zhenghua,CHEN Xiaomin,et al.Uniaxialratcheting and fatigue failure behaviors of hot-rolled AZ31B magnesium alloy under asymmetrical cyclic stress-controlled loadings[J].Materials science and engineering:A,2013,573:234-244.[3]赵路远,黄俊,陈涛,等.7075铝合金的单轴棘轮行为[J].机械工程材料,2018,42(9):21-25.ZHAO Luyuan,HUANG Jun,CHEN Tao,et al.Uniaxial ratcheting behavior of 7075aluminum alloy[J].Materials for mechanical 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英文原版教材班“材料科学基础”考试试题试卷一Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. Glossary (2 points for each)1) crystal structure:2) basis (or motif):3) packing fractor:4) slip system:5) critical size:6) homogeneous nucleation:7) coherent precipitate:8) precipitation hardening:9) diffusion coefficient:10) uphill diffusion:2. Determine the indices for the planes in the cubic unit cell shown in Figure 1. (5 points)Fig. 13. Determine the crystal structure for the following: (a) a metal with a0 =4.9489 Å, r = 1.75 Å and one atom per lattice point; (b) a metal with a0 = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. (10 points)4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (4 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (4 points)(3)Plot the isothermal section at 150o C. (7 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(1)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(2)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(1)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(2)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibi ts a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题答案Solution s of the course of “fundament of materials science”1. Glossary (2 points for each)1) The arrangement of the atoms in a material into a repeatable lattice.2) A group of atoms associated with a lattice.3) The fraction of space in a unit cell occupied by atoms.4) The combination of the slip plane and the slip direction.5) The minimum size that must be formed by atoms clustering together in the liquid before thesolid particle is stable and begins to grow.6) Formation of a critically sized solid from the liquid by the clustering together of a largenumber of atoms at a high undercooling (without an external interface).7) A precipitate whose crystal structure and atomic arrangement have a continuousrelationship with matrix from which precipitate is formed.8) A strengthening mechanism that relies on a sequence of solid state phase transformationsin a dispersion of ultrafine precipitates of a 2nd phase. This is same as age hardening. It is a form of dispersion strengthening.9) A temperature-dependent coefficient related to the rate at which atom, ion, or otherspecies diffusion. The DC depends on temperature, the composition and microstructure of the host material and also concentration of the diffusion species.10) A diffusion process in which species move from regions of lower concentration to that ofhigher concentration.2. Solution: A(-364), B(-340), C(346).3. Solution: (a)fcc; (b) bcc.4-1. What is the characteristic of brinell hardness test, rockwell hardness test and Vickers hardness test? What are the effects of strain rate and temperature on the mechanical properties of metallic materials? (15 points)4-2. What are the effects of cold-work on metallic materials? How to eliminate those effects? And what is micro-mechanism for the eliminating cold-work effects? (15 points)5-1. Based on the Pb-Sn-Bi ternary diagram as shown in Fig. 2, try to(1)Show the vertical section of 40wt.%Sn; (5 points)(2) Describe the solidification process of the alloy 2# with very low cooling speed (includingphase and microstructure changes); (5 points)(3)Plot the isothermal section at 150o C. (5 points)Fig. 25-2. A 1mm sheet of FCC iron is used to contain N2in a heated exchanger at 1200o C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. At 1000 o C, if same N concentration is demanded at the second surface and the flux of N becomes to half of that at 1200o C, then what is the thickness of sheet?(15 points)6-1. Supposed that a certain liquid metal is undercooled until homogeneous nucleation occurs. (15 points)(3)How to calculate the critical radius of the nucleus required? Please give the deductionprocess.(4)For the Metal Ni, the Freezing Temperature is 1453︒C, the Latent Heat of Fusion is 2756J/cm3, and the Solid-liquid Interfacial Energy is 255⨯10-7 J/cm2. Please calculate the critical radius at 1353︒C. (Assume that the liquid Ni is not solidified.)6-2. Fig.3 is a portion of the Mg-Al phase diagram. (15 points)(3)If the solidification is too rapid, please describe the solidification process of Mg-10wt%Alalloy.(4)Please describe the equilibrium solidification process of Mg-20wt%Al alloy, and calculate theamount of each phase at 300︒C.Fig. 37-1. Figure 4 shows us the Al-Cu binary diagram and some microstructures found in a cooling process for an Al-4%Cu alloy. Please answer following questions according to this figure. (20 points)Fig. 4(1)What are precipitate, matrix and microconstituent? Please point them out in the in the figure and explain.(2)Why is need-like precipitate not good for dispersion strengthening? The typical microstructure shown in the figure is good or not? why?(3)Please tell us how to obtain the ideal microstructure shown in this figure.(4)Can dispersion strengthened materials be used at high temperature? Please give the reasons (comparing with cold working strengthening)7-2. Please answer following questions according to the time-temperature-transformation (TTT) diagram as shown in Fig. 5. (20 points)(1)What steel is this TTT diagram for? And what means P, B, and M in the figure? (2)Why dose the TTT diagram exhibits a ‘C’ shape?(3)Point out what microconstituent will be obtained after austenite is cooled according to the curves I, II, III and IV .(4)What is microstructural difference between the curve I and the curve II? (5)How to obtain the steel with the structure of(a) P+B(b) P+M+A (residual) (c) P+B+M+A (residual)(d) Full tempered martensiteIf you can, please draw the relative cooling curve or the flow chart of heat treatment.Fig. 5III III IV英文原版教材班“材料科学基础”考试试题试卷二Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to physically separate different materials in a scrap recycling plant. Describe some possible methods that might be used to separate materials such as polymers, aluminum alloys, and steels from one another. (5 points)2. Plot the melting temperature of the elements in the 1A column of the periodic table versus atomic number (i.e., plot melting temperatures of Li through Cs). Discuss this relationship, based on atomic bonding and binding energy. (10 points)3.Above 882℃, titanium has a BCC crystal structure, with a = 0.332 nm. Below this temperature, titanium has a HCP structure, with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC titanium transforms to HCP titanium. Is this a contraction or expansion? (10 points)4. The density of BCC iron is 7.882 g/cm3and the lattice parameter is 0.2866 nm whenhydrogen atoms are introduced at interstitial positions. Calculate (a) the atomic fraction of hydrogen atoms and (b) the number of unit cells required on average to contain one hydrogen atom. (15 points)5. A carburizing process is carried out on a 0.10% C steel by introducing 1.0% C at the surface at 980℃, where the iron is FCC. Calculate the carbon content at 0.01 cm, 0.05 cm, and 0.10 cm beneath the surface after 1 h. (15 points)6. The following data were collected from a standard 0.505-in.-diameter test specimen of acopper alloy (initial length (t o) = 2.0 in.):Load Gage Length Stress Strain(lb) (in.) (psi) (in/in.)0 2.00000 0 0.03,000 2.00167 15,000 0.0008356,000 2.00333 30,000 0.0016657,500 2.00417 37,500 0.0020859,000 2.0090 45,000 0.004510,500 2.040 52,500 0.0212,000 2.26 60,000 0.1312,400 2.50 (max load) 62,000 0.2511,400 3.02 (fracture) 57,000 0.51After fracture, the gage length is 3.014 in. and the diameter is 0.374 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A 1.5-em-diameter metal bar with a 3-cm gage length is subjected to a tensile test. Thefollowing measurements are made.Change in Force (N) Gage length (cm) Diameter (cm)16,240 0.6642 1.202819,066 1.4754 1.088419,273 2.4663 0.9848Determine the strain hardening coefficient for the metal. Is the metal most likely to be FCC, BCC, or HCP? Explain.(15 points)8. Based on Hume-Rothery’s conditions, which of the following systems would be expected todisplay unlimited solid solubility? Explain. (15 points)(a) Au-Ag (b) Al-Cu (c) Al-Au (d)U-W(e) Mo-Ta (f) Nb-W (g) Mg-Zn (h) Mg-Cd英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2.T (o C)L i–180.7N a– 97.8K – 63.2R b– 38.9As the atomic number increases, the melting temperature decreases, (10 points)3. We can find the volume of each unit cell. Two atoms are present in both BCC and HCP titanium unit cells, so the volumes of the unit cells can be directly compared.V BCC = (0.332 nm)3 = 0.03659 nm3V HCP= (0.2978 nm)2(0.4735 nm)cos30 = 0.03637 nm3△V=x 100 =×100= -0.6%Therefore titanium contracts 0.6% during cooling. (10 points)4. (a) 7.882 g/cm3 =x = 0.0081 H atoms/cellThe total atoms per cell include 2 Fe atoms and 0.0081 H atoms. Thus:(10 points)(b) Since there is 0.0081 H/cell, then the number of cells containing H atoms is:cells = 1/0.0081 = 123.5 or 1 H in 123.5 cells (5 points)5. D = 0.23 exp[-32,900/(1.987)(1253)] = 42 × 10-8 cm2/sC x= 0.87% CC x= 0.43% CC x= 0.18% C(15 points)6. σ=FI (π/4)(0.505)2 = F/0.2ε = (l-2)/2(a) 0.2% offset yield strength = 45,000 psi(b)tensile strength = 62,000 psi(c) E = (30,000 - 0) / (0.001665 - 0) = 18 x 106 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 57,000 psi(g)true stress at fracture = 11,400 lb / (TC/4)(0.374)2= 103,770 psi (h) From the graph, yielding begins at about 37,500 psi. Thus:(15 points)7.Force(lb) Gage length(in.) Diameter(in.) True stress(psi) True strain(psi)16,240 3.6642 12.028 143 0.20019,066 4.4754 10.884 205 0.40019,273 5.4663 9.848 249 0.600σt=Kεt2or ln143=ln K + n ln0.2ln 249 = ln K + nln 0.6n=0.51A strain hardening coefficient of 0.51 is typical of FCC metals.(15 points)8.The Au–Ag, Mo–Ta, and Mg–Cd systems have the required radius ratio, the same crystal structures, and the same valences. Each of these might be expected to display complete solid solubility. [The Au –Ag and Mo –T a d o have isomorphous phase diagrams. In addition, the Mg–Cd alloys all solidify like isomorphous alloys; however a number of solid state phase transformations complicate the diagram.] (15 points)英文原版教材班“材料科学基础”考试试题试卷三Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to be able to identify different materials without resorting to chemical analysis or lengthy testing procedures. Describe some possible testing and sorting techniques you might be able to use based on the physical properties of materials. (5 points)2. Plot the melting temperatures of elements in the 4A to 8-10 columns of the periodic table versus atomic number (i.e., plot melting temperatures of Ti through Ni, Zr through Pd, and Hf through Pt). Discuss these relationships, based on atomic bonding and binding energy, (a) as the atomic number increases in each row of the periodic table and (b) as the atomic number increases in each column of the periodic table. (10 points)3. Beryllium has a hexagonal crystal structure, with a o= 0.22858 nm and c o= 0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm3, and the atomic weight is 9.01 g/mol. Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell.(10 points)4. Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find (a) the density and (b) the packing factor. (15 points)5. Iron containing 0.05% C is heated to 912oC in an atmosphere that produces 1.20% C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if (a) the iron is BCC and (b) the iron is FCC. Explain the difference. (15 points)6. The following data were collected from a 0.4-in. diameter test specimen of poly vinyl chloride(l0 = 2.0 in):Load(lb) Gage Length(in.) Stress(psi) Strain(in/in.)0 2.00000 0 0.0300 2.00746 2,387 0.00373600 2.01496 4,773 0.00748900 2.02374 7,160 0.011871200 2.032 9,547 0.0161500 2.046 11,933 0.0231660 2.070 (max load) 13,206 0.0351600 2.094 12,729 0.0471420 2.12 (fracture) 11,297 0.06After fracture, the gage length is 2.09 in. and the diameter is 0.393 in. Plot the data and calculate (a) the 0.2% offset yield strength, (b) the tensile strength, (c) the modulus of elasticity, (d) the %Elongation, (e) the %Reduction in area, (f) the engineering stress at fracture, (g) the true stress at fracture, and (h) the modulus of resilience. (15 points)7. A titanium alloy contains a very fine dispersion of tiny Er203 particles. What will be the effectof these particles on the grain growth temperature and the size of the grains at any particular annealing temperature? Explain. (15 points)8. Suppose 1 at% of the following elements is added to copper (forming a separate alloy witheach element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? Is any of the alloying elements expected to have unlimited solid solubility in copper?(a) Au (b) Mn (c) Sr (d) Si (e) Co (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1.Steels can be magnetically separated from the other materials; steel (or carbon-containing iron alloys) are ferromagnetic and will be attracted by magnets. Density differences could be used—polymers have a density near that of water; the specific gravity of aluminum alloys is around2.7;that of steels is between 7.5 and 8. Electrical conductivity measurements could be used—polymers are insulators, aluminum has a particularly high electrical conductivity.(5 points)2. Ti –1668 Zr – 1852 Hf – 2227V –1900 Nb –2468 Ta – 2996Cr –1875 Mo–2610 W–3410Mn–1244 Tc –2200 Re–3180Fe –1538 Ru –2310 Os–2700Co –1495 Rh –1963 Ir –2447Ni –1453 Pd –1552 Pt –1769For each row, the melting temperature is highest when the outer “d” energy level is partly full. In Cr, there are 5 electrons in the 3d shell; in Mo, there are 5 electrons in the 4d shell; in W there are 4 electrons in the 5d shell. In each column, the melting temperature increases as the atomic number increases—the atom cores contain a larger number of tightly held electrons, making the metals more stable. (10 points)3.V= (0.22858 nm)2(0.35842 nm)cos 30 = 0.01622 nm3 = 16.22 × 10-24 cm3(a)From the density equation:1.848 g/cm3 =x = 2 atoms/cell(b)The packing factor (PF) is:PF == 0.77 (10 points)4. There is one carbon atom per 100 iron atoms, or 1 C/50 unit cells, or 1/50 C per unit cell:(a)(b)(15 points)5. t= (24 h)(3600 s/h) = 86,400 sD BCC = 0.011 exp[-20,900/(1.9871185)] = 1.54 × 10-6 cm2/sD FCC = 0.23 exp[-32,900/(1.987)(1185)] = 1.97×10-7 cm2/sBCC: = erf[0.0685] = 0.077C x= 1.11% CFCC: = erf[0.192] = 0.2139C x = 0.95% CFaster diffusion occurs in the looser packed BCC structure, leading to the higher carbon content at point “x”. (15 points)6. σ=F /(π/4)(0.4)2 = F/0.1257ε = (l-2)/2(a)0.2% offset yield strength = 11,600 psi(b) tensile strength = 12,729 psi(c) E= (7160 - 0) / (0.01187 - 0) = 603,000 psi(d)%Elongation =(e) %Reduction in area =(f) engineering stress at fracture = 11,297 psi(g)true stress at fracture = 1420 lb / (TC/4)(0.393)2= 11,706 psi (h) From the figure, yielding begins near 9550 psi. Thus:(15 points)7. These particles, by helping pin die grain boundaries, will increase the grain growth temperature and decrease the grain size. (15 points)8.The Cu-Sr alloy would be expected to be strongest (largest size difference). The Cu-Au alloy satisfies Hume-Rothery ’s conditions and might be expected to display complete solid solubility—in fact it freezes like an isomorphous series of alloys, but a number of solid state transformations occur at lower temperatures.(15 points)英文原版教材班“材料科学基础”考试试题试卷四Examination problems of the course of “fundament of materials science”姓名：班级：记分：1.Aluminum has a density of2.7 g/cm3. Suppose you would like to produce a compositematerial based on aluminum having a density of 1.5 g/cm3. Design a material that would have this density. Would introducing beads of polyethylene, with a density of 0.95 g/cm3, into the aluminum be a likely possibility? Explain. (5 points)2. (a) Aluminum foil used for storing food weighs about 0.3 g per square inch. How many atomsof aluminum are contained in this sample of foil?(b) Using the densities and atomic weights given in Appendix A, calculate and compare thenumber of atoms per cubic centimeter in (i) lead and (ii) lithium. (10 points)3. The density of potassium, which has the BCC structure and one atom per lattice point, is0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the latticeparameter; and (b) the atomic radius of potassium. (10 points)4. The density of a sample of HCP beryllium is 1.844 g/cm3 and the lattice parameters are a0=0.22858 nm and c0= 0.35842 nm. Calculate (a) the fraction of the lattice points that containvacancies and (b) the total number of vacancies in a cubic centimeter. (15 points)5. A ceramic part made of MgO is sintered successfully at 1700℃in 90 minutes. To minimizethermal stresses during the process, we plan to reduce the temperature to 1500℃. Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions? What time will be required at the lower temperature? (15 points)6. (a) A thermosetting polymer containing glass beads is required to deflect 0.5 mm when aforce of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports. Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.(b) The flexural modulus of alumina is 45 x 106 psi and its flexural strength is 46,000 psi. Abar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart.Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs. (15 points)7. Based on the following observations, construct a phase diagram. Element A melts at 850°Cand element B melts at 1200°C. Element B has a maximum solubility of 5% in element A, and element A has a maximum solubility of 15% in element B. The number of degrees of freedom from the phase rule is zero when the temperature is 725°C and there is 35% B present. At room temperature 1% B is soluble in A and 7% A is soluble in B. (15 points)8.Suppose that age hardening is possible in the Al-Mg system (see Figure 10-11). (a)Recommend an artificial age-hardening heat treatment for each of the following alloys, and(b) compare the amount of the precipitate that forms from your treatment of each alloy. (i)Al-4% Mg (ii) Al-6% Mg (iii) Al-12% Mg (c) Testing of the alloys after the heat treatment reveals that little strengthening occurs as a result of the heat treatment. Which of the requirements for age hardening is likely not satisfied? (15 points)英文原版教材班“材料科学基础”考试试题答案Solutions of the course of “fundament of materials science”1. In order to produce an aluminum-matrix composite material with a density of 1.5 g/cm 3, we wouldneed to select a material having a density considerably less than 1.5 g/cm 3. While polyethylene’s density would make it a possibility, the polyethylene has a very low melting point compared to aluminum; this would make it very difficult to introduce the polyethylene into a solid aluminum matrix —processes such as casting or powder metallurgy would destroy the polyethylene .Therefore polyethylene would NOT be a likely possibility.One approach, however, might be to introduce hollow glass beads .Although ceramic glasses have densities comparable to that of aluminum, a hollow bead will have a very low density. The glass also has a high melting temperature and could be introduced into liquid aluminum for processing as a casting. (5 points)2. (a) In a one square inch sample:number ==6.69 × 1021 atoms(b) (i) In lead:= 3.3 × 1022 atoms/cm 3(ii) In lithium:= 4.63 × 1022 atoms/cm 3 (10 points)3. (a) Using Equation 3-5:0.855 g/cm 3 =a o 3 = 1.5189 × 10-22 cm 3 or a o = 5.3355 × 10-8 cm(b) From the relationship between atomic radius and lattice parameter:r == 2.3103 × 10-8cm (10 points)4. V u.c.= (0.22858 nm)2(0.35842 nm)cos30 = 0.01622 nm 3= 1.622 x 10~23 cm 3 (a) From the density equation:x = 1.9984fraction =29984.12 = 0.0008(b) number == 0.986 x 1020 vacancies/cm 3 (15 points)5. Diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.D 1700= 0.000043 exp[-82,100/(1.987)(1973)] = 3.455 × 10-14 cm 2/sD1500= 0.000043 exp[-82,100/(1.987)(1773)] = 3.255 × 10-15 cm2/st1500 = D1700 t1700/D1500== 955 min = 15.9 h (15 points)6. (a) Solution:The minimum distance L between the supports can be calculated from the flexural modulus.L3 = 4w/z3δ(flexural modulus)/3FL3 = (4)(20 mm)(5 mm)3(0.5 mm)(6.9 GPA)(1000 MPa/GPa) / 500 NL3 = 69,000 mm3 or L = 41 mmThe stress acting on the bar when a deflection of 0.5 mm is obtained isσ= WL/2wh2 = (3)(500 N)(41 mm) / (2)(20 mm)(5 mm)2 = 61.5 MPaThe applied stress is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.(b) Solution:The force required to break the bar isF = 2w/z2(flexural strength)/3LF= (2)(1 in)(0.3 in)2(46,000 psi / (3)(7 in.) = 394 lbThe deflection just prior to fracture is8 = FZ3/4wh3(flexural modulus)8 = (394 lb)(7 in)3/(4)(l in)(0.3 in)3(45 x 106 psi) = 0.0278 in. (15 points)7.(15 points)8. (a) The heat treatments for each alloy might be:Al-4% Mg Al-6% Mg Al-12% MgT Eutectic451°C 451°C 451°CT Solvs210°C 280°C 390°CSolutionTreat at: 210-451°C 280-451°C 390-451°CQuench Quench QuenchAge at: <210°C <280°C <390°C(b) Answers will vary depending on aging temperature selected. If all threeare aged at 200°C, as an example, the tie line goes from about 3.8 to 35% Mg:A1-4% Mg: %β = (4− 3.82)/(35 − 3.82) X 100 = 0.6%Al-6% Mg: %β = (6 − 3.82)/(35 − 3.82) X 100 = 7.1%Al-12% Mg: %β = (12 −3.82)/(35− 3.82) X 100 = 26.8%(c) Most likely, a coherent precipitate is not formed; simple dispersionstrengthening, rather than age hardening, occurs. (15 points)英文原版教材班“材料科学基础”考试试题试卷五Examination problems of the course of “fundament of materials science”姓名：班级：记分：1. You would like to design an aircraft that can be flown by human power nonstop for adistance of 30 km. What types of material properties would you recommend? What materials might be appropriate? (5 points)2. Boron has a much lower coefficient of thermal expansion than aluminum, even though bothare in the 3B column of the periodic table. Explain, based on binding energy, atomic size, and the energy well, why this difference is expected. (10 points)3. Determine the ASTM grain size number if 20 grains/square inch are observed at amagnification of 400. (10 points)4. We currently can successfully perform a carburizing heat treatment at 1200o C in 1 h. In aneffort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950℃. What time will be required to give us a similar carburizing treatment? (15 points)5.The data below were obtained from a series of Charpy impact tests performed on foursteels, each having a different manganese content. Plot the data and determine (a) the transition temperature (defined by the mean of the absorbed energies in theductile and brittle regions) and (b) the transition temperature (defined as the temperature that provides 50 J absorbed energy). Plot the transition temperature versus manganese content and discuss the effect of manganese on the toughness of steel. What would be the minimum manganese allowed in the steel if a part is to be used at 0°C?Test temperature°C Impact snergy (J)0.30% Mn 0.39% Mn 1.01% Mn 1.55% Mn-100 2 5 5 15-75 2 5 7 25-50 2 12 20 45-25 10 25 40 700 30 55 75 11025 60 100 110 13550 105 125 130 14075 130 135 135 140100 130 135 135 140(15 points)。

elasticitytheory of elasticity homogeneous state ofstressstress invariant strain invariant strain ellipsoid homogeneous state ofstrainequation of strain compatibilityLame constants isotropic elasticityrotating circular diskwedgeKelvin problemBoussinesq problemAiry stress functionKolosoff-Muskhelishvili methodKirchhoff hypothesisPlateRectangular plate Circular plate Annular plate Corrugated plate Stiffened plate,reinforced弹性力学 弹性理论 均匀应力状态 应力不变量 应变不变量 应变椭球 均匀应变状态应变协调方程拉梅常量各向同性弹性旋转圆盘楔开尔文问题布西内斯克问题 艾里应力函数 克罗索夫―穆斯赫利什维 利法基尔霍夫假设板 矩形板 圆板 环板 波纹板 加劲板PlatePlate of moderate thickness Stress function of bendingShell Shallow shell Revolutionary shell Spherical shell Cylindrical shell Conical shell Toroidal shell Closed shell Corrugated shell Stress function of torsionWarping function semi-inverse method Rayleigh-Ritz method Relaxation methodLevy method Relaxation Dimensional analysis self-similarity Influence surface Contact stress Hertz theory Conforming contact Sliding contact Rolling contact中厚板 弯［曲］应力函数壳 扁壳 旋转壳 球壳 ［圆］柱壳锥壳 环壳 封闭壳 波纹壳 扭［转］应力函数翘曲函数 半逆解法 瑞利―里茨法松弛法 莱维法 松弛 量纲分析 自相似［性］影响面 接触应力 赫兹理论 协调接触压入Indentation各向异性弹性Anisotropic elasticity 颗粒材料Granular material散体力学Mechanics of granular media 热弹性Thermoelasticity超弹性Hyperelasticity粘弹性Viscoelasticity对应原理Correspondence principle 褶皱Wrinkle塑性全量理论Total theory of plasticity 滑动Sliding微滑Microslip粗糙度Roughness非线性弹性Nonlinear elasticity 大挠度Large deflection突弹跳变snap-through有限变形Finite deformation格林应变Green strain阿尔曼西应变Almansi strain弹性动力学Dynamic elasticity运动方程Equation of motion准静态的Quasi-static气动弹性Aeroelasticity水弹性Hydroelasticity颤振Flutter弹性波Elastic wave简单波Simple wave柱面波Cylindrical wave水平剪切波Horizontal shear wave 竖直剪切波Vertical shear wave 体波body wave无旋波Irrotational wave 畸变波Distortion wave膨胀波Dilatation wave瑞利波Rayleigh wave等容波Equivoluminal wave 勒夫波Love wave界面波Interfacial wave 边缘效应edge effect塑性力学Plasticity可成形性Formability金属成形Metal forming耐撞性Crashworthiness结构抗撞毁性Structural crashworthiness 拉拔Drawing破坏机构Collapse mechanism回弹Springback挤压Extrusion冲压Stamping穿透Perforation层裂Spalling塑性理论Theory of plasticity 安定［性］理论Shake-down theory 运动安定定理kinematic shake-downtheoremStatic shake-down theorem rate dependent theoremload factor Loading criterion Loading function Loading surface Plastic loading Plastic loading waveSimple loading Proportional loadingUnloading Unloading wave Impulsive load step load pulse load limit load nentral loading instability in tension acceleration wave constitutive equation complete solution nominal stress over-stress true stress equivalent stressflow stress stress discontinuity静力安定定理 率相关理论 载荷因子 加载准则 加载函数 加载面 塑性加载 塑性加载波 简单加载 比例加载 卸载 卸载波 冲击载荷 阶跃载荷 脉冲载荷 极限载荷 中性变载 拉抻失稳 加速度波 本构方程 完全解 名义应力 过应力 真应力 等效应力 流动应力 应力间断stress space principal stress space hydrostatic state of stresslogarithmic strain engineering strain equivalent strain strain localizationstrain ratestrain rate sensitivitystrain space finite strain plastic strain incrementaccumulated plastic strainpermanent deformationinternal variable strain-softening rigid-perfectly plasticMaterialrigid-plastic materialperfectl plastic material stability of material deviatoric tensor of strain deviatori tensor of stress spherical tensor of strain spherical tensor of stresspath-dependency linear strain-hardening应力空间 主应力空间 静水应力状态 对数应变 工程应变 等效应变 应变局部化 应变率 应变率敏感性 应变空间 有限应变塑性应变增量累积塑性应变永久变形 内变量 应变软化 理想刚塑性材料刚塑性材料 理想塑性材料 材料稳定性 应变偏张量 应力偏张量 应变球张量 应力球张量 路径相关性strain-hardening kinematic hardening isotropic hardening strain-hardening moduluspower hardening plastic limit bendingMomentplastic limit torque elastic-plastic bending elastic-plastic interface elastic-plastic torsionViscoplasticityInelasticityelastic-perfectly plasticMaterial limit analysislimit design limit surface upper bound theorem upper yield point lower bound theorem lower yield point bound theorem initial yield surface subsequent yield surface convexity of yield surface shape factor of cross-section应变强化 随动强化 各向同性强化 强化模量 幕强化 塑性极限弯矩塑性极限扭矩 弹塑性弯曲 弹塑性交界面 弹塑性扭转粘塑性非弹性理想弹塑性材料极限分析 极限设计 极限面 上限定理 上屈服点 下限定理 下屈服点 界限定理 初始屈服面 后继屈服面 屈服面［的］外沙堆比拟屈服屈服条件屈服准则屈服函数屈服面塑性势能量吸收装置能量耗散率塑性动力学塑性动力屈曲塑性动力响应塑性波运动容许场静力容许场流动法则速度间断滑移线滑移线场移行塑性铰塑性增量理论米泽斯屈服准则普朗特―罗伊斯关系特雷斯卡屈服准则sand heap analogyYieldyield conditionyield criterionyield functionyield surfaceplastic potential energy absorbing device energy absorbing device dynamic plasticity dynamic plastic buckling dynamic plastic response plastic wave kinematically admissibleFieldstatically admissibleFieldflow rule velocity discontinuityslip-linesslip-lines field travelling plastic hinge incremental theory ofPlasticityMises yield criterion prandtl- Reuss relation Tresca yield criterion洛德应力参数莱维―米泽斯关系亨基应力方程赫艾一韦斯特加德应力空间洛德应变参数德鲁克公设盖林格速度方程结构力学结构分析结构动力学拱三铰拱抛物线拱圆拱穹顶空间结构空间桁架雪载［荷］风载［荷］土压力地震载荷弹簧支座支座位移支座沉降Lode stress parameterLevy-Mises relation Hencky stress equation Haigh-Westergaardstress space Lode strain parameter Drucker postulateGeiringer velocityEquation structural mechanics structural analysis structural dynamicsArchthree-hinged archparabolic archcircular archDomespace structurespace trusssnow loadwind loadearth pressureearthquake loadingspring support support displacementsupport settlementdegree of indeterminacy kinematic analysis method of joints method of sectionsjoint forces conjugate displacementinfluence line three-moment equation unit virtual force stiffness coefficient flexibility coefficientmoment distributionmoment distribution methodmoment redistribution distribution factor matri displacement method element stiffness matrix element strain matrix global coordinates Betti theorem Gauss-Jordan eliminationMethod buckling mode mechanics of compositescomposite materialfibrous composite unidirectional composite超静定次数 机动分析 结点法 截面法 结点力 共轭位移 影响线 三弯矩方程 单位虚力 刚度系数柔度系数力矩分配力矩分配法 力矩再分配 分配系数 矩阵位移法 单元刚度矩阵 单元应变矩阵 总体坐标 贝蒂定理 高斯一若尔当消去法屈曲模态复合材料力学 复合材料foamed composite particulate compositeLaminate sandwich panel cross-ply laminate angle-ply laminatePlycellular solid ExpansionDebulk Degradation DelaminationDebond fiber stress ply stress ply strain interlaminar stress specific strength strength reduction factor strength -stress ratio transverse shear modulustransverse isotropyOrthotropyshear lag analysis chopped fiber continuous fiber fiber direction泡沫复合材料 颗粒复合材料层板 夹层板 正交层板 斜交层板 层片 多胞固体 膨胀 压实 劣化 脱层 脱粘 纤维应力 层应力 层应变层间应力比强度强度折减系数 强度应力比 横向剪切模量 横观各向同性 正交各向异 剪滞分析 短纤维 长纤维fiber break fiber pull-out fiber reinforcementDensification optimum weight design netting analysis rule of mixture failure criterion Tsai-W u failure criterionDugdale model fracture mechanics probabilistic fractureMechanicsGriffith theory linear elastic fracturemechanics, LEFMelastic-plastic fracturemecha-nics, EPFMFracture brittle fracturecleavage fracture creep fracture ductile fracture inter-granular fracture quasi-cleavage fracture trans-granular fractureCrack纤维断裂 纤维拔脱 纤维增强 致密化 最小重量设计 网格分析法 混合律 失效准则 蔡一吴失效准则 达格代尔模型断裂力学概率断裂力学格里菲思理论线弹性断裂力学弹塑性断裂力学断裂 脆性断裂 解理断裂 蠕变断裂 延性断裂 晶间断裂 准解理断裂 裂纹Flaw Defect Slit MicrocrackKinkelliptical crack embedded crack penny-shape crackPrecrack short crack surface crack crack blunting crack branching crack closure crack front crack mouthcrack opening angle,COAcrack opening displacement,CODcrack resistancecrack surfacecrack tipcrack tip opening angle,CTOAcrack tip openingdisplacement, CTOD crack tip singularity裂缝 缺陷 割缝 微裂纹 折裂 椭圆裂纹 深埋裂纹 ［钱］币状裂纹预制裂纹 短裂纹 表面裂纹 裂纹钝化 裂纹分叉 裂纹闭合 裂纹前缘 裂纹嘴 裂纹张开角 裂纹张开位移裂纹阻力裂纹面裂纹尖端 裂尖张角裂尖张开位移Fieldcrack growth rate stable crack growth steady crack growth subcritical crack growthcrack retardation crack arrest arrest toughness fracture mode sliding mode opening mode tearing mode mixed mode Tearingtearing modulus fracture criterionJ-integral J-resistance curve fracture toughness stress intensity factor Hutchinson-Rice-RosengrenFieldconservation integraleffective stress tensor strain energy density energy release ratecohesive zone裂纹扩展速率 稳定裂纹扩展 定常裂纹扩展 亚临界裂纹扩展 裂纹［扩展］减速 止裂 止裂韧度 断裂类型 滑开型 张开型 撕开型 复合型 撕裂 撕裂模量 断裂准则 J 积分 J 阻力曲线 断裂韧度 应力强度因子HRR 场守恒积分 有效应力张量 应变能密度 能量释放率塑性区plastic zone张拉区stretched zone热影响区heat affected zone, HAZ延脆转变温度brittle-ductile transitiontempe- rature剪切带shear band 剪切唇shear lip无损检测non-destructive inspection双边缺口试件double edge notchedspecimen, DEN specimen 单边缺口试件single edge notchedspecimen, SEN specimen 三点弯曲试件three point bendingspecimen, TPB specimen 中心裂纹拉伸试件center cracked tensionspecimen, CCT specimen 中心裂纹板试件center cracked panelspecimen, CCP specimen 紧凑拉伸试件compact tension specimen,CT specimen 大范围屈服large scale yielding 小范围攻屈服small scale yielding 韦布尔分布Weibull distribution 帕里斯公式paris formula空穴化Cavitation应力腐蚀stress corrosion概率风险判定probabilistic riskassessment, PRAdamage mechanicsDamagecontinuum damage mechanics microscopic damage mechanicsaccumulated damage brittle damage ductile damage macroscopic damage microscopic damage microscopic damagedamage criteriondamage evolution equationdamage softeningdamage strengtheningdamage tensor damage threshold damage variable damage vector damage zone Fatigue low cycle fatigue stress fatigue random fatigue creep fatigue corrosion fatigue fatigue damage 损伤力学 损伤 连续介质损伤力学 细观损伤力学 累积损伤 脆性损伤 延性损伤 宏观损伤 细观损伤 微观损伤损伤准则损伤演化方程损伤软化 损伤强化 损伤张量 损伤阈值 损伤变量 损伤矢量 损伤区 疲劳 低周疲劳 应力疲劳 随机疲劳 蠕变疲劳 腐蚀疲劳fatigue failure fatigue fracture fatigue crack fatigue life fatigue rupture fatigue strength fatigue striations fatigue threshold alternating load alternating stress stress amplitudestrain fatiguestress cyclestress ratio safe life overloading effect cyclic hardening cyclic softening environmental effectcrack gage crack growth, crackPropagation crack initiationcycle ratio experimental stressAnalysisactive[strain] gage疲劳失效 疲劳断裂 疲劳裂纹 疲劳寿命 疲劳破坏 疲劳强度 疲劳辉纹 疲劳阈值 交变载荷 交变应力应力幅值应变疲劳应力循环 应力比 安全寿命 过载效应 循环硬化 循环软化 环境效应 裂纹片 裂纹扩展裂纹萌生 循环比工作［应变］片backing material stress gage zero shift, zero drift strain measurementstrain gage strain indicator strain rosette strain sensitivity mechanical strain gage rectangular rosetteExtensometertelemetering of strain transverse gage factor transverse sensitivity weldable strain gage balanced bridge bonded strain gage bonded foiled gage bonded wire gage bridge balancing capacitance strain gage compensation technique compensation techniquereference bridge resistance strain gageself-temperature compensating gage基底材料 应力计 零［点］飘移 应变测量 应变计 应变指示器 应变花 应变灵敏度 机械式应变仪 直角应变花弓I 伸仪 应变遥测 横向灵敏系数 横向灵敏度 焊接式应变计 平衡电桥 粘贴式应变计 粘贴箔式应变计 粘贴丝式应变计桥路平衡 电容应变计 补偿片 补偿技术 基准电桥 电阻应变计semiconductor strainGageslip ring strain amplifier fatigue life gage inductance [strain] gagePhotomechanics Photoelasticity Photoplasticity Young fringe birefrigent effect contour of equal Displacement dark fringefringe multiplication interference fringeIsochromatic Isoclinic isopachic stress- optic lawIsostatic light fringe optical path differencephoto-thermo -elasticityphotoelastic coatingMethodphotoelastic sandwich半导体应变计集流器 应变放大镜 疲劳寿命计 电感应变计 光［测］力学光弹性 光塑性 杨氏条纹 双折射效应 等位移线暗条纹 条纹倍增 干涉条纹 等差线 等倾线 等和线 应力光学定律 主应力迹线亮条纹光程差热光弹性 光弹性贴片法Methoddynamic photo-elasticityspatial filtering spatial frequencyPolarizerreflection polariscope residual birefringentEffectstrain fringe valuestrain-optic sensitivitystress freezing effectstress fringe valuestress-optic pattern temporary birefringentEffect pulsed holographytransmission polariscope real-time holographic interfero - metrygrid methodholo-photoelasticityHologram Holographholographic interferometry holographic moire techniqueHolography whole-field analysis动态光弹性 空间滤波 空间频率 起偏镜 反射式光弹性仪 残余双折射效应应变条纹值应变光学灵敏度应力冻结效应应力条纹值 应力光图 暂时双折射效应脉冲全息法 透射式光弹性仪 实时全息干涉法网格法 全息光弹性法全息图 全息照相 全息干涉法 全息云纹法 全息术散斑干涉法speckle interferometry 散斑Speckle错位散斑干涉法speckle-shearinginterferometry,shearography散斑图Specklegram 白光散斑法white-light speckle method 云纹干涉法moire interferometry ［叠栅］云纹moire fringe［叠栅］云纹法moire method 云纹图moire pattern离面云纹法off-plane moire method 参考栅reference grating试件栅specimen grating分析栅analyzer grating面内云纹法in-plane moire method脆性涂层法brittle-coating method 条带法strip coating method坐标变换transformation ofCoordinates计算结构力学computational structuralmecha-nics加权残量法weighted residual method 有限差分法finite difference method 有限［单］元法finite element method 配点法point collocation里茨法Ritz method广义变分原理generalized variationalPrinciple 最小二乘法least square method胡［海昌］一鹫津原理Hu-Washizu principle赫林格-赖斯纳原理Hellinger-ReissnerPrinciple修正变分原理modified variationalPrinciple约束变分原理constrained variationalPrinciple混合法mixed method杂交法hybrid method边界解法boundary solution method有限条法finite strip method半解析法semi-analytical method协调兀conforming element非协调兀non-conforming element混合元mixed element杂交元hybrid element边界元boundary element强迫边界条件forced boundary condition自然边界条件natural boundary condition离散化Discretization离散系统discrete system连续问题continuous problem广义位移generalized displacement广义载荷generalized load广义应变generalized straingeneralized stress interface variable node, nodal pointElement corner node mid-side node internal node nodeless variablebar element truss element beam elementtwo-dimensional elementone-dimensional elementthree-dimensional element axisymmetric elementplate element shell elementthick plate element triangular element quadrilateral element tetrahedral element curved element quadratic element linear element cubic element quartic element isoparametric element广义应力 界面变量 节点 ［单］元 角节点 边节点 内节点 无节点变量杆元 桁架杆元梁元二维元一维元 三维元 轴对称元厚板元 三角形元 四边形元 四面体元 曲线元 二次元 线性元 三次元 四次元 等参［数］super-parametric element sub-parametric element variable-number-nodeelement Lagrange element Lagrange family serendipity element serendipity family infinite element element analysis element characteristicsstiffness matrixgeometric matrixequivalent nodal forcenodal displacementnodal load displacement vectorload vector mass matrix lumped mass matrix consistent mass matrixdamping matrix Rayleigh damping assembly of stiffnessMatricesconsistent mass matrix assembly of mass matrices assembly of elements超参数元 亚参数元 节点数可变元 拉格朗日元 拉格朗日族 巧凑边点元 巧凑边点族 无限元 单元分析 单元特性刚度矩阵几何矩阵等效节点力节点位移 节点载荷 位移矢量 载荷矢量 质量矩阵 集总质量矩阵 相容质量矩阵 阻尼矩阵 瑞利阻尼 刚度矩阵的组集载荷矢量的组集 质量矩阵的组集local coordinate systemlocal coordinate area coordinates volume coordinates curvilinear coordinates static condensation contragradienttransformation shape function trial function test function weight function spline function substitute function reduced integration zero-energy mode p-convergenceh-convergenceblended interpolation isoparametric mapping bilinear interpolationpatch test incompatible modenode number element number band width banded matrix profile matrix局部坐标系 局部坐标 面积坐标 体积坐标 曲线坐标 静凝聚合同变换 形状函数 试探函数 检验函数 权函数 样条函数 代用函数 降阶积分 零能模式P 收敛H 收敛 掺混插值 等参数映射 双线性插值 小块检验 非协调模式 节点号 M 二 口. 单兀号minimization of band widthfrontal method subspace iteration method determinant search methodstep-by-step methodNewmark Wilsonquasi-Newton method Newton-Raphson method incremental method initial straininitial stresstangent stiffness matrixsecant stiffness matrix mode superposition method equilibrium iterationSubstructure substructure techniquesuper-element mesh generationstructural analysis programpre-processing post-processing mesh refinement stress smoothing composite structure带宽最小化 波前法 子空间迭代法 行列式搜索法逐步法 纽马克法 威尔逊法 拟牛顿法 牛顿-拉弗森法增量法初应变初应力切线刚度矩阵 割线刚度矩阵 模态叠加法 平衡迭代 子结构 子结构法 超单元 网格生成 结构分析程序前处理 后处理 网格细化 应力光顺。

这是同济大学结构研一硕士上的《高等混凝土结构理论》期末考试的复习要点，希望对考博选考《混凝土结构基本理论》这门课的同学有所帮助。

1．Stress-strain curves of concrete under monotonic, repeated and cyclic uniaxial loadings. 单轴受力时混凝土在单调、重复、反复加载时的应力应变曲线。

2．Creep of concrete (linear and nonlinear) 混凝土的徐变（线性、非线性徐变）3．Components of deformation of concrete 混凝土变形的多元组成4．Process of failure of concrete under uniaxial compression 混凝土在单向受压时破坏的过程。

5．Strength indices of concrete and the relations among them 混凝土的强度指标及其之间关系6．Features of stress-strain envelope curve of concrete under repeated compressive loading. 混凝土单向受压重复加载时的应力应变关系的包络线的特征。

7．The crack contact effect of concrete and its representation in stress-strain diagram. 混凝土的裂面效应及其在应力应变关系图上的表示。

8．The multi-level two-phase system of concrete. 混凝土的多层次二相体系。

9．The rheological model of concrete. 混凝土的流变学模型。

10．Influence of stress gradient on strength of concrete. 应力梯度对混凝土强度的影响。