弹性力学基础-中英.ppt

Elastic-plastic Mechanics of Materials Ming-an CHEN（陈明安） School of Materials Science and Engineering Central South University, Typical tensile specimen, Typical tensile test machine,gauge,length,(portion of sample with,reduced cross section),Chapt. 1 Introduction,1.1 Elasticity and plasticity（弹性与塑性） 1. STRESS-STRAIN TESTING（应力应变曲线拉伸试验）,gauge,length,=,2. ELASTIC DEFORMATION and ELASTICITY（弹性变形与弹性）,Elastic means reversible!,It is reversible and time independent. The deformation vanishes instantaneously as soon as forces are removed.,3. PLASTIC DEFORMATION (METALS) and PLASTICITY （塑性变形与塑性）,Plastic means permanent!,Plastic deformation-it is irreversible or permanent.,2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning is a trademark used herein under license.,The tension test is the standard test for determine E, the elastic or Youngs modulus. Test that load a cylindrical specimen in torsion are used to measure the shear modulus G. Knowing E and G, Poissons ratio may be obtained from the relationship we derived in the previous section.,杨氏（弹性）模量 E,Metals Alloys,Graphite Ceramics Semicond,Polymers,Composites /fibers,E(GPa),Based on data in Table B2, Callister 6e. Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers.,YOUNGS MODULI: COMPARISON, Plastic tensile strain at failure:, Another ductility measure:, Note: %AR and %EL are often comparable. -Reason: crystal slip does not change material volume. -%AR %EL possible if internal voids form in neck.,4. Ductile and brittle materials（韧性与脆性材料）, Energy to break a unit volume of material Approximate by the area under the stress-strain curve.,TOUGHNESS（韧性）,Low-carbon (mild) steel is different from most other metals in that there is a sudden small drop of load at the yield point followed by an extension at constant stress. The lower load is usually referred to as the yield point for mild steel.,The actual point of yield is often difficult to identify. A number of techniques are used to locatey. The tangent method (or knee method) locates the yield strength at the intersection of the elastic slope and the initial portion of the plastic region (not reliably). The preferred method is the percentage offset method where yield strength is obtained by drawing a line parallel to the initial elastic region data at 0.2% strain (0.002) offset. Where this line intersects the stress-strain curve then becomes known as the 0.2% yield strength.,5. Determination of yield strength 屈服强度,For most metals, loading beyond the yield point causes a permanent deformation. When a material is loaded to point B and then unloaded, it returns to a zero stress state along a line parallel to the initial elastic region but directly from B. The strain remaining in the material at point D is known as the plastic deformation.,On reloading from D there is a departure from linearity at C, slightly below B, and the stress-strain curve becomes the same as the original stress-strain curve (at E). Note that the point of departure from linearity on the reload curve (C) is slightly higher than for the first loading curve.,6. Unloading and reloading（卸载与再加载）,7. Idealizations of stress-strain curves 应力应变曲线简化,Ideal rigid-plastic,Rigid-plastic,Mechanics: branch of physics concerned with motion and body deformation created by mechanical disturbance or forces. Applied Mechanics: science of applying the principles of mechanics to design and analysis of mechanical system. Applied Mechanics Rigid Body Mechanics Statics Dynamics Kinematics, Kinetics Deformable Body Mechanics Elasticity Plasticity Viscoelasticity Fluid Mechanics Liquids Gases,1.2 Research objects and contents,Manufacturing processes that make use of cold working as well as hot working. Common metalworking methods,轧制、挤压、锻造、冲压、拉拔等,2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning is a trademark used herein under license.,Anisotropic behavior in a rolled aluminum-lithium sheet material used in aerospace applications. The sketch relates the position of tensile bars to the mechanical properties that are obtained,Rectangular Coordinates,The system of particles in the Figure is said to be in equilibrium if every one of its constitutive particles is in equilibrium. Consequently, the first condition for equilibrium: the vector sum of all the forces is zero.,where rn extends from point 0 to an arbitrary point on the line of action of force Fn.,If the surface and body forces are in balance, the body is in static equilibrium.,The second condition for equilibrium: the total moment of all the external forces about an arbitrary point 0 must be zero.,An Isolated System of Particles Showing External and Internal Forces,For an object to be at rest (ie static eq), net force and net moment must be zero.,Since forces and moments are vectors, with inherent directionality, it is frequently useful for decompose into individual components:,Shear Stress（剪应力） Shear stresses can also be generated by applied shear loads. Consider two equal and opposite shear forces V acting on a rectangular block as shown.,应力点的概念: 不同点处应力不同。 应力面的概念：同一点处不同截面上的应力不同。 应力必须指明是哪点、哪个截面上的应力。,A positive component of stress acts on a positive face in a positive coordinate direction or on a negative face in a negative coordinate direction.,Complementary Shear Stress（剪应力互等） Consider a rectangular block of unit thickness and suppose shear stresses 1, act on BC and AD. The forces1 *AD and1 * BC form a couple of magnitude1 * AD * AB and the block is not in equilibrium. There must bean equal and opposite couple formed by shear stresses on AB and CD.,Thus an applied shear stress is automatically accompanied by a shear stress of equal intensity at right angles (and causing an opposite turning moment) to the original shear stress. These are called complementary shear stresses.,The state of stress at a point can normally be determined by computing the stresses acting on certain conveniently oriented planes passing through the point of interest. Stresses acting on any other planes can then be determined by means of simple, standardized analytical or graphical methods. If so we can use the stresses, acting on these conveniently oriented planes passing through the point, for representing the stress state of the given point, and that the stress state at this point is known.,The selection of different cutting planes through a given point would, in general, result in stresses differing in both direction and magnitude. A complete description of the magnitudes and directions of stresses on all possible planes through the given point constitutes the state of stress at the given point.,Problem: The stress components, on which of and how much different planes, can be used for representing the stress state of the given point?,2.2.2 State of stress at a point（点的应力状态）,一点可以用无穷个微元表示，找出之间应力的关系，称为应力状态分析。,应力状态的概念： 过一点不同截面上应力的的集合，称为这一点的应力状态。,State of uniaxial stress: 单向应力状态,The stress normal to the cross-sectional surface:,Stresses on oblique planes:（斜面上的应力）,Stresses:,Forces:,Now suppose we cut the prismatic bar at an angle as shown below.,How do the normal and shear components of stress acting on a plane at a given point change as we change the orientation of the plane at the point.,2. General Stress Systems in 2-Dimensions （双向应力状态）,The stresses on the element ABCD in a component subjected to combined 2D loading (assuming no through thickness stresses, i.e. plane stress) are schematically shown in the Figure. The reference system of coordinate axes are as shown also.,What is the stress state on a chosen plane of interest?,Consider rotating the element ABCD by an angle to the x-axis so that it now has axes of x and y orientated at angleto the x and y axes. To determine the new stressesx,y andxy on the element in terms of the original stresses consider the free body diagram of a prismatic element ADE and the stresses acting on it are as shown in the Figure. The normal stress x and shear stress xy act on the plane AE and maintain the equilibrium of the prismatic element.,The stresses x andxy are obtained by resolution of forces in the respective directions.,Transformation of Stresses（应力变换）,x-y,xp-yp,These results clearly illustrate how the values for the normal and shear stress components of a force distributed over a plane inside of an object depends upon how you look at the point inside the object in the sense that the values of the shear and normal stresses at a point within a continuum depend upon the orientation of the plane you have chosen to view.,n = Sxl + Sym + Szn = xl2 + ym2 + zn2 + 2(xylm + yzmn + zxnl),l = cos(n, x), m = cos(n, y) and n = cos(n,z),n2 = S2 - n2,The above mentioned showes that if we know the 9 stresses components on the three mutually perpendicular planes as faces of a cube of infinitesimal size (element) which surround the given point we can determine the sterss components acting on any plane through the point. So these 9 stresses components can be used to represent the stress state of a point.,一点应力状态可表示为：,2.3 Stress tensor and principal stresses （应力张量与主应力） 2.3.1 Stress tensor（应力张量）,Tensor is the generalised term for a vector. Its full mathematical definition is: A mathematical entity specifiable by a set of components with respect to a system of co-ordinates and such that the transformation that has to be applied to the components to obtain components with respect to a new system of co-ordinates is related in a certain way to the transformation that has to be applied to the system of coordinates. The components of a vector change when the co-ordinate system is rotated. However, the vector still has the same magnitude and direction as it did before the co-ordinate system was rotated. Second rank tensors (e.g. stress, inertia) see their components change when a co-ordinate system is rotated and unlike vectors the magnitude and orientation of the tensor may also change., Theorem of conjugate shearing stresses,Therefore only 6 independent stress components,Stress tensor is symmetric.,Could you write out the stress tensors corresponding to the following figures ？,Could you show the stress tensor in a corresponding element ？,三向应力状态下的应力变换,2.3.2 Principal stresses（主应力）,The actual values of the 6 stress components in the stress matrix for a given body subjected to loading will depend on the orientation of the cube in the body itself. If we rotate the cube, it should be possible to find the directions in which the normal stress components take on maximum and minimum values. It is found that in these directions the shear components on all faces of the cube become zero. The principal stresses are defined as those normal components of stress that act on planes that have shear stress components with zero magnitude.,一点处一般有三个主平面，互相垂直。,假设该斜微分面即为待求的主平面，面上0，正应力全应力S。全应力S在3个坐标轴上的投影为：,以l、m、n为未知数的齐次线性方程组，其解就是应力主轴的方向。显然lmn0是一组解，但l2m2n21，故应求其非0解。,Stress invariants (应力张量不变量),第一、第二、第三应力不变量,1. 可以证明，在应力空间，主应力平面是存在的； 2. 三个主平面是相互正交的； 3. 三个主应力均为实根，不可能为虚根； 4. 应力特征方程的解是唯一的； 5. 对于给定的应力状态，应力不变量也具有唯一性； 6. 应力第一不变量I1反映变形体体积变形的剧烈程度，与塑性变形无关； I3也与塑性变形无关；I2与塑性变形有关; 7. 应力不变量不随坐标而改变，是点的确定性的判据。,用主应力表示的各种应力状态的图示：,2.3.3 Principal shear stresses（主剪应力）,剪应力取极值的平面上的剪应力-主剪应力；主剪应力所在的平面-主剪应力平面； 主剪应力平面的法线方向-主剪应力方向。,n = Sxl + Sym + Szn = xl2 + ym2 + zn2 + 2(xylm + yzmn + zxnl),n2 = S2 - n2,S,l2 + m2 +n2 =1,现考虑主应力空间下主剪应力、主剪应力平面的求解：,2 = l 212 + m 222 + n 232- 1l 2 + 2m 2 + 3n 2 2,将n2 =1- l2 - m2 代入上式，取2 对l 和m的偏导数并令其为零，可解出对应的l、m、n和极值剪应力。,n =0，l=1/(2) ，m=1/,m =0，l=1/(2) ，n=1/,l =0，m=1/(2) ，n=1/,三组（6个）主剪应力平面分别与一个主应力平面垂直，与另两个主应力平面呈45。,最大剪应力(maximun shear stress),2.3.4 Decomposition of stress tensor （应力张量分解）,Deviatoric stress components（偏应力分量）,(i, j=x, y, z or 1,2,3),The spherical or hydrostatic stress tensor (球应力张量),The deviatoric stress tensor （偏应力张量）,1、分解的依据：静水压力实验证实，静水压力不会引起变形体形状的改变，只会引起体积改变，即对塑性条件无影响。 2、为引起形状改变的偏应力张量(deviatoric stress tensor)，为引起体积改变的球张量(spherical stress tensor)（静水压力）。 3、与应力张量类似，偏应力张量也存在相应的不变量：,One Dimensional State of Stresses,Shearing State of Stresses,2.4 The Mohr circle of stress（应力莫尔圆）,应力极值,哪个几何图形可代表该点的应力状态？,画出球应力状态的Mohr圆？,The Figure illustrates the orientation of one of the eight octahedral planes which are a