31ConceptsofStressandStrain.pptx

第第6次课次课Concepts of Stress and Strain College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAWhy Study Mechanical Properties?College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAContentw3.1 Introductionw3.2 Concepts of Stress(应力应力)and Strain(应变应变)College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINA The mechanical behavior of a material reflects the relationship between its response of deformation to an applied load or force.1 IntroductionwExamples of designing materials based on their characteristics:the aluminum alloy from which an airplane wing(机翼机翼)is constructed and the steel in an automobile axle(车轴车轴).College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINA Factors to be considered as design of materials:the nature of the applied load and its duration,as well as the environmental conditions.Examples:Examples:wIt is possible for the load to be tensile(拉伸),compressive(压缩),or shear(剪切),and its magnitude may be constant with time or it may fluctuate continuously.wApplication time may be only a fraction of a second,or it may extend over a period of many years.wService temperature may be an important factor.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAMechanical propertiesStrength(强度强度)Hardness(硬度硬度)Ductility(塑性塑性)Stiffness(韧度韧度)The testing techniques of these mechanical properties have been standardized in order to have consistency for a variety of parts.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAStrain:当材料受到外力作用而又不产生惯性移动时，其几何形当材料受到外力作用而又不产生惯性移动时，其几何形状和尺寸会发生变化，这种变化称为形变。状和尺寸会发生变化，这种变化称为形变。Stress:材料宏观形变时，其内部分子及原子间发生材料宏观形变时，其内部分子及原子间发生 相对位移，相对位移，产生分子间及原子间对抗外力的附加内力，达到平衡时，产生分子间及原子间对抗外力的附加内力，达到平衡时，附加内力与外力大小相等，方向相反，定义单位面积上附加内力与外力大小相等，方向相反，定义单位面积上的内力为应力，其值与外加的力相等。的内力为应力，其值与外加的力相等。2 Concepts of Stress(应力应力)and Strain(应变应变)College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINA(1)A load is static or changes slowly with time and is applied uniformly over a cross section or surface of a member.(2)It is commonly conducted at room temperature.There are three principal ways in which a load may be applied:namely,tension,compression,and shearSimple Stress-Strain Test:Simple Stress-Strain Test:College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINA Schematic illustration of how a tensiletensile load produces an elongation(延长)and positive linear strain.Dashed lines represent the shape before deformation;solid lines,after deformationTensionCollege of Materials Science&EngineeringNORTH UNIVERSITY OF CHINATension tests A specimen is deformed,usually to fracture,with a gradually increasing tensile load that is applied uniaxially along the long axis of a specimen.Standard tensile specimen:Standard tensile specimen:wNormally,the cross section is circular,but rectangular specimens are also used.During testing,deformation is confined to the narrow center region.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAwThe requirements of a standard tensile specimen:The requirements of a standard tensile specimen:1.the standard diameter is approximately 12.8 mm,where the reduced section length should be at least four times this diameter;60 mm is common.2.Gauge length is used in ductility computations;the standard value is 50 mm.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINATesting:Testing:wThe specimen is mounted by its ends into the holding grips of the testing apparatus,as shown in following figure:Schematic representation of the apparatus used to conduct tensile stress-strain tests.The specimen is elongated by the moving crosshead;load cell and extensometer measure,respectively,the magnitude of the applied load and the elongation.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINARequirements in testing:1.The tensile testing machine is designed to elongate the specimen at a constant rate,and continuously and simultaneously measure the instantaneous applied load(瞬时外加负载),and to the resulting elongations(发生长度).2.A stress-strain test typically takes several minutes to perform and is destructive.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAEngineering stressEngineering stress(MPa):F:the instantaneous load applied perpendicular to the specimen cross section(N);A0:the original crossectional area before any load is applied(m2).Engineering strain Engineering strain(untiless):l0:original length before any load is applied,li:the instantaneous length,and l=li-l0.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAEg a cylindrical brass rod that has a diameter of 10 mm(0.4 in.)is pulled in tension with 5600N force,producing only elastic deformation.Calculate the resulting strain.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINASOLUTIONThis deformation situation is represented in the accompanying drawing.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAWhen the force F is applied,the specimen will elongate in the z direction and at the same time experience a reduction in diameter,d,of 2.510-3 mm in the x direction.For the strain in the x directionFinally,from Equation 7.1,the applied force may be determined asCollege of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAProblem A specimen of aluminum having a rectangular cross section 10 mm12.7 mm(0.4 in0.5 in.)is pulled in tension with 35,500 N(8000 lbf)force,producing only elastic deformation.Calculate the resulting strain.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINA Schematic illustration of how a compressivecompressive load produces contraction(收缩)and negative linear strain.CompressionCollege of Materials Science&EngineeringNORTH UNIVERSITY OF CHINACompression testsw A compression test is conducted in a manner similar to the tensile test,except that the force is compressive and the specimen contracts along the direction of the stress.wPrevious equations are utilized to compute compressive stress and strain,respectively.College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINA LeftLeft:Schematic representation of shear strain,where=tan;RightRight:Schematic representation of torsional deformation(i.e.angle of twist)produce by an applied torque T,which is applied in engineering practice.ShearCollege of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAShear and torsional testsThe shear stress:wWhere F is the load or force imposed parallel to the upper and lower faces,each of which has area of A0The shear stain:=tanCollege of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAeg A piece of copper originally 305 mm(12 in.)long is pulled in tension with a stress of 276 MPa(40,000 psi).If the deformation is entirely elastic,what will be the resultant elongation?College of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAwSOLUTIONSince the deformation is elastic,strain is dependent on stress according to Equation 7.5.Furthermore,the elongation l is related to the original length l0 through Equation 7.2.Combining these two expressions and solving for l yieldsCollege of Materials Science&EngineeringNORTH UNIVERSITY OF CHINAThe values of and l0 are given as 276 MPa and 305 mm,respectively,and the magnitude of E for copper from Table 7.1 is 110 GPa(16 106 psi).Elongation is obtained by substitution into the expression above as