清华弹性力学课件_Elasticity of Solids

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弹性力学绪论课件

课程安排
第三部分：弹性力学问题求解方法
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弹性力学问题的分类和特点
弹性力学问题的求解方法及其应用
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课程安排
第四部分：弹性力学在工程中的应用
弹性力学在材料科学中的应用
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弹性力学在结构分析中的应用
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弹性力学在其他领域中的应用
学习建议
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建立清晰的学习目标和方法，明确学习内容和重点
总结词
多物理场耦合下的弹性性质研究是当前弹性力学领域的另一个研究前沿，主要涉及多个物理场之间的相互作用对弹性性质的影响。
详细描述
多物理场耦合下的弹性性质研究主要关注多个物理场之间的相互作用对弹性性质的影响，
例如：热-力耦合、电磁-力耦合、化学-力耦合等。这些研究通常需要使用多物理场耦合理论和数值模拟方法来分析不同物理场之间的相互作用对弹性性质的影响，为多物理场耦合问题的解决提供理论支持和实践指导。
材料实例
介绍了一些具体的材料实例，如高强度轻质合金、纳米复合材料等，说明弹性力学在其中的应用和重要性。
弹性力学在生物医学工程中的应用
总结词
弹性力学在生物医学工程中应用日益广泛，为生物组织和器官的
力学特性研究提供有力工具。
详细描述
弹性力学的原理和公式可以用于研究生物组织和器官的力学特性，
如肌肉、骨骼、血管、心脏等组织的弹性、韧性和疲劳特性等。
弹性力学在工程中的应用
REPORTING
弹性力学在结构分析中的应用
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总结词
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详细描述
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工程实例
弹性力学在结构分析中应用广泛，为复杂结构分析提供理论支持。

弹性力学课件

研究对象
弹性力学的研究对象主要是弹性体，即在外力作用下能够发生变形，当外力去除后又能恢复到原来形状的物体。
弹性体基本假设与约束条件
基本假设
弹性体在变形过程中，其内部各点间距离的变化是微小的，且这种变化不影响物体的整体形状和大小。
约束条件
弹性体的变形受到外部约束条件的限制，如支撑、连接等，这些约束条件对弹性体的变形和内力分布产生影响。
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例题2
无限大平板受均布载荷作用下的应力分析。利用弹性力学理论求解无限大平板在均布载荷作用下的应力分布，并讨论平板厚度对应力分布的影响。
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例题3
圆柱体受内压作用下的应力分析。通过解析法或数值法求解圆柱体在内压作用下的应力分布，并讨论不同材料属性和几何参数对应力分布的影响。
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弹性体变形协调方程与几何方程
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讨论
通过对比各向同性和各向异性材料的力学行为，加深对材料本构关系的理解。
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平面问题求解方法与应用举例
平面问题定义及分类
平面应力问题
长柱形物体受平行于横截面的外力作用，横截面尺寸远小于轴向尺寸。
平面应变问题
平面或板状物体受平行于中面的外力作用，中面尺寸远大于厚度。
平面问题的简化
忽略体力，将空间问题简化为平面问题。
各向异性材料本构关系简介
各向异性假设
材料在各个方向上具有不同的力学性质。
本构关系特点
应力与应变之间的关系复杂，需要考虑材料的方向性。
典型各向异性材料
纤维增强复合材料、层合板等。
典型例题解析与讨论
1 2
例题一
求解各向同性材料在简单拉伸条件下的应力和应变。
例题二
分析各向异性材料在复杂应力状态下的力学行为。

清华大学弹性力学讲义chap2_Elasticity of Solids

2．Elasticity of SolidsReferencesJ.H.Weiner ，Statistical mechanics of elasticity, Wiley, 1981Green & Zerna ，Theoretical elasticity, 1968Ashby & Jones ，Engineering materials2.1 Definition of Elasticity ElasticityσFFigure 2.1 An elastic response.An elastic response of the material can be abstracted mathematically as()X F ,T σ= (2.1) where σ denotes the stress tensor, T the response function that depends only on the current values of the deformation gradient X x F ∂∂=, with X denoting the material coordinates of a point while x the spatial coordinates. If the material is homogeneous within the domain under consideration, the explicit dependence on X in (2.1) can be eliminated. Several remarks can be made to the definition in (2.1):(1) In the claim of ()()X t X,F ,T σ=, one pins down an elastic response as the one prtrayed by the current status of deformation, and henceforth irrelevant to thehistory or the process of deformation. The strain rate plays no role in the constitutive response. A hysteresis is ruled out, as shown in Fig. 2.1. Whenever the loading is removed, the original configuration (or the “non-distorted configuration”) is recovered.(2)For the special case of infinitesimal deformation, the response (2.1) is reduced to()X,εσ=, where εdenotes the strain tensor. The response function T is not Tnecessarily linear.(3)For homogeneous materials, one has ()Fσ=in finite deformation andT()εσ=for infinitesimal deformation.T(4)For the even special case of infinitesimal deformation, homogeneous material andlinear elasticity, the generalized Hooke’s law ε=is recovered, with Cσ:Cbeing the fourth-rank stiffness tensor. The notations of C as the stiffness tensor and S as the compliance tensor, not the otherwise according to their initials, unfortunately became the convention in the historic development of elasticity.The difference between the material responses at a solid state and a fluid state can be quoted as follows (Mechanics of Solids, The New Encyclopedia of Britannica, 15th edition, V ol. 23, pp. 734-747, 2002, written by J. R. Rice):“A material is called solid rather than fluid if it can also support a substantial shearing force over the time scale of some natural process or technological application of interest.”An elastic solid can resist the volume and shape changes, whereas a fluid can only resist the volume change but not the shape changes in a relatively long time scale. HyperelasticityHyperelasticity refers to an elastic response that can be defined by a potential. The basic assumptions are: (1) the response of the elastic body only depends on its current state, not the processes to achieve it; and (2) the current state of the elastic body can be described by a tensor, such as the strain tensor εfor the special case of infinitesimal deformation. The first assumption leads to the independence ofdeformation paths. It was the great mathematician Green who first exploited that condition to arrive the path independent condition of a multi-dimensional integration. The elasticity so-defined was called Green elasticity. As the readers may recall from their calculus course, such a path independency leads to the Green ’s conditions among partial derivatives of the integrands, and the existence of a potential function. We denote this elastic potential as W that has the physical significance of the elastic energy stored within the body. A hyperelastic constitutive response is thenij ij W εσ∂∂=. (2.2)That is also valid for non-linear elastic response. For a linear elastic material, one has kl ij ijkl C W εε21=. (2.3) The combination of (2.2) and (2.3) leads to the generalized Hooke ’s lawkl ijkl ij ij C W εεσ=∂∂= (2.4)2.2 Two Physical Origins of ElasticityThe elasticity of a solid may depend on other things beside the deformation state. For example, temperature and disorder status in the material structure may influence the elastic response. The influences from these factors are often material specific and they may get quite intricate. In terms of the temperature effect, a crystalline solid expands or shrinks when temperature rises or falls; while a polymeric solid, such as a rubber band, shrinks upon heating. These opposite behaviors come from different physical origins of elasticity that will be addressed briefly in this section. The book of J.H. Weiner is referred for the in-depth understanding.Energetic and Entropic StressesThe Helmholtz free energy H of a solid can be expressed asST U H -= (2.5) where U denotes the internal energy, S the entropy, and T the absolute temperature in Kelvin. The internal energy of an elastic material excluding the thermal fluctuation energy ST gives the Helmholtz energy that can be freed. Similar to the simplifiedversion of hyperelastic relation (2.2), the stress tensor can be deduced asij ij ij ij S T U H εεεσ∂∂-∂∂=∂∂=. (2.6)A Maxwell relation can be derived asT S ij ij ∂∂=∂∂σε. (2.7) Accordingly, (2.6) may have an alternative presentationT T U ij ij ij ∂∂-∂∂=σεσ (2.8) Figure 2.2 shows a graphical interpretation of (2.8). The total stress ij σ is composed of two terms: the energetic stress ijU ε∂∂ and the entropic stress T T ij ∂∂-σ.T -T Figure 2.2 Energetic and entropy stresses.Consider the stress versus the temperature curve above. At a prescribed temperature T , the corresponding stress is composed of two parts, the entropy stress denotes the interval between the horizontal line and the tangent extrapolated to the absolute zerotemperature, and the energetic stress for the remaining part. For a perfect crystal, the energetic stress dominates; for an amorphous polymer, the entropy stress dominates. These extreme cases will be discussed below to explore two physical origins of elasticity.CrystalsA crystal is built by periodic repetition of its unit cell. The primary bonding in the unit cell can be ionic, covalent or metallic. The bonding can be isotropic (such as metallic bonds) or polarized (such as the covalent and ionic bonds). Those bonds typical melt at a temperature (related to the kinetic energy of the atoms) range between 1000-5000K. The secondary bonding is composed by Van der Walls and hydrogen bonds, and they have a much lower range (100-500K) of melting point.The bonding between two atoms is described by an inter-atomic potential U (Fig. 2.3). The bonds can simplified as springs connecting the neighboring atoms, as shown in Fig. 2.3. The spring-like inter-atomic bonding provides a physical origin for the elasticity of a crystalline solid.bond energy UFigure 2.3 Spring-like inter-atomic bonding (above) and inter-atomic energy curve(below).The inter-atomic force F is given by the derivative of U with respect to the inter-atomic distance r . A plot of the inter-atomic force is given in Fig. 2.4. The location where d U /d r vanishes refers to an equilibrium location of the atom, marked by an inter-atomic distance of 0r . The force is repulsive when 0r r < and attractive when 0r r >. The stiffness of the bonding spring equals to the second derivative ofthe inter-atomic potential with respect to the distance, 22drU d S =. The maximum inter-atomic force occurs at the location where S vanishes, as shown in Fig. 2.4. The value of S at the equilibrium location0220r r dr U d S =⎪⎪⎭⎫ ⎝⎛= (2.9)relates to the elasticity of the linear Hooke ’s law.FF Figure 2.4 Inter-atomic force.The elasticity tensor of a crystalline solid depends on the orientation, as the consequence of the atomic arrangement in the form of a lattice, which is spanned by the periodic repetition of a unit cell. For a simple cell, this repetitive construction of the crystal lattice is mathematically represented by the following specification on the atom sites332211321a a a r n n n n n n ++= (2.10) where i n (3,2,1=i ) are integers, and i a (3,2,1=i ) are lattice vectors that are not necessarily mutually perpendicular. The lattice formed by (2.10) has a translational symmetry.For a composite cell, the representation (2.10) is replaced byi i n n n n a r =321, ξa r +=i i n n n n 321' (2.11) Summation convention is implied for repeated indices. An example for a material of composite cell is grapheme and nanotubes.Aside from the translational symmetry, a crystal lattice can be categorized by various types. A generally accepted notion is the classification of Bravais lattices. The seven Bravais lattices are tri-clinic, mono-clinic, orthogonal, rhombohedral, tetragonal, hexagonal and cubic. The symmetry derived by these lattices can be characterized bythe group theory, via 32 point groups and 236 space groups. Their implications to the elasticity tensor will be explored in Section 2.4.Long Chain PolymersWe next turn to the elasticity of long chain polymers, a scientific wonder that was shaped in the first half of the 20th century by the famous scientists such as Mark, James, Guth and Flory.The long-chain polymers, such as rubbers, are highly elastic in terms of large extensibility and low elastic modulus. They are nearly incompressible. A strange feature of polymeric materials is the Gough-Joule effect: they shrink during heating and expand during cooling. Long chain polymers can be heated up rapidly during adiabatic deformation. You can test this phenomenon by fast pulling a rubber band and feeling it on your lip.These bizarre mechanical behaviors come from the unique microstructure of long chain polymers, and the latter manifests in terms of the entropy stress. Figure 2.5 depict the structure. A polymer is formed by the backbone connected by taunt covalent bonding attached with the tangling sidegroups formed by much weaker bonding. Neither the bonding length nor the bonding angle of the backbone can be changed appreciably. The sidegroups are used to form weak interaction, such as the van der Waals type, among the neighboring chains. A rotational degree of freedom, however, exists between the neighboring links in the backbone. That grants the polymer chain certain degree of flexibility for the actual shape of its backbone.(strong covalent bond)(weak bond)Figure 2.5 Long chain structure of polymeric solid.Consider a backbone connected by n links where n is a large number for the macromolecule. Two end points of the chain form a distance vector R . Under a fixed vector R , there are many ways for the polymer chain to arrange itself, as shown in Figure 2.6.Under a fixed number n for the links of the chain and a fixed end-to-end distance R , the number of possible arrangement, denoted by ()n W W ,R =, can be computed by the model of random walking. The same model was used by Albert Einstein in theoretical physics calculation. The number of possible arrangement of many polymer chains obeys a Gaussian distribution()⎪⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛=2223223exp 23,nb R nb n W πR , R =R (2.12) where b denotes the effective length of freely jointed links, or the step size of the random walk. For a freely jointed polymer chain, one has a b =; the length increase to a b 2= for freely rotating chain as shown in the lower graph of Fig. 2.5; and it becomes a b 7.6=for poly-ethelene.R ~Figure 2.6 Arrangement for a long chain polymeric to achieve a certain end to enddistance.Under a fixed n , the longer the distance R , the smaller the number of the possible arrangements. Subjected the polymer to a strain ε so that the vector r R →. That change provokes a related change of W . An elongation decreases the amount of W whilst a contraction increases it. The number of the possible arrangements W of n links in the chain and its stress response can be bridged by Boltzmann ’s formula for configuration entropy in statistical mechanics. That links the configuration entropy q S with W byW K S B q ln = (2.13) where B K is the famous Boltzmann constant. The total entropy of the system is composed of the configuration entropy q S and the vibration entropy S p , while the latter hardly changes during the deformation. The entropy stress, namely the second term in (2.6) can be computed asij Bij q ij W W T K S T S T εεε∂∂-=∂∂-≈∂∂- (2.13)The physical origin for rubber elasticity is revealed: the stress is caused by the reduction of the configuration entropy due to elongation.By means of the network model, James and Guth (1947) were able to derive the following elastic response for rubber-like polymers in uniaxial tension:()2--=λλσT NK B (2.14) where λ denotes the stretching ratio, and N the number of links per unit volume. The elastic law of (2.14) is referred as the Neo-Hookeian law in rubber elasticity.1234567812340λσFigure 2.7 Rubber elasticity: theory and experiment.Figure 2.7 plotted the experimental curve, decorated by circles, of rubber-like polymer undergoing stretch up to 8-folds of its original length. The Neo-Hookeian prediction (2.14) agrees with the experiment at the beginning, but two noticeable deviations occur in the later stage of elongation. For a stretching ratio between 2 and 5, the Neo-Hookeian law seems to overestimate the stress response. This effect was first corrected by Mooney, and called thereafter as the Mooney effect. The Mooney-Rivlin constitutive relation developed later can follow this deviation. For a stretching ratio higher than 6, the Neo-Hookeian prediction underestimate the stress response. Two explanations were offered for the upturn of the stress response: one due to the crystallization of polymers under large elongation, and the other due the non-Gaussian chain theory.2.3 Tensor Description of ElasticityVoigt SymmetryRecall the linear elastic response (2.4), namely kl ijkl ij C εσ=. All components of the fourth-rank elasticity tensor ijkl C can be measured from tests involving only homogeneous deformation. For a three-dimensional problem, all indices i , j , k and lmay have 3 possible numbers. That gives the maximum possible combinations of indices as 8134=.We now try to find ways to eliminate the unnecessary tests for a complete characterization of the stress tensor. First consider the symmetries of the stress and the strain tensors. For infinitesimal deformation, the strain tensor is given by a symmetric expression of ()ji i j j i ij u u εε=+=,,21. The anti-symmetric part, the rotation, bears no effect on the stress response. On the other hand, the reciprocal law of shear stresses (ji ij σσ=) derived by neglecting the couple acting on a body element gives rise of a symmetric stress tensor. We now use these symmetry requirements to derive the corresponding conditions for the elasticity tensor.First consider the stress symmetry ji ij σσ=. Substituting this condition to (2.4), one has kl jikl kl ijkl C C εε=, kl ε∀. Accordingly, one has the inter-changeability of the first two indices of the elasticity tensorjikl ijkl C C =Next consider the strain symmetry lk kl εε=. Equation (2.4) can be rearranged askl ijlk lk ijlk kl ijkl C C C εεε==, kl ε∀. That leads to the inter-changeability of the last two indices of the elasticity tensorijlk ijkl C C =Linear elasticity is a special case of hyperelasticity as described at the end of Section 2.1. An elastic potential kl ij ijkl C W εε21= exists as the strain energy density. Equation (2.2), ijij Wεσ∂∂=, leads to klij klij ijkl WC εεεσ∂∂∂=∂∂=2 In an Euclidian space, the order of differentiations bears no effect. One then concludes the interchangeability between the first two and the last two indices of the elasticitytensor:klij ijkl kl ij ijklC W W C =∂∂∂=∂∂∂=εεεε22 The above symmetry requirement can be summarized asklij ijlk jikl ijkl C C C C ===(2.15)They are called the V oigt relation of the elasticity tensor. Matrix RepresentationDue to the V oigt symmetry, it is frequently to represent the fourth rank of the elasticity tensor by a symmetric 66⨯ matrix, the elasticity matrix. The matrix is sequenced to put the 6 independent components of a symmetric second order tensor in the counter-clockwise manner described below.Therefore, the fourth rank symmetric elasticity tensor is transformed into the following symmetric elasticity matrix[]⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡=⇒665655464544363534332625242322161514131211C C C C C C C C C C C C C C C C C C C C C C ijklαβC (2.16)with j i --=9α and l k --=9β. The general anisotropy refers to the case where 21 elastic constants in (2.16) are totally unrelated. A material example for such a general anisotropy is the triclinic crystals whose three lattice vectors have different lengths, namely 321a a a ≠≠, and neither two of them is mutually perpendicular. The elastic strain energy for such a case could be in arbitrary quadratic form of the six⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡↑↑←342561⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡↑↑←)23()13(333231)12(232212131211symmetric components of strain, namely()231312332211,,,,,21εεεεεεεεW C W kl ij ijkl ==(2.17)Most natural or technical materials have a certain degree of symmetries. The implication of these symmetry to the reduction of the elastic constants will be revealed below. Coordinate Transform1XFigure 2.8 Coordinate transform.Consider an affine transformation from coordinates i x to coodinates i x , as shown in Fig. 2.8. The components of the second rank strain tensor and the fourth order elasicity tensor in the new coordinates with superimposed bars areij j s i r rs x x x x εε∂∂∂∂=, ijkl ltk p j s i r rspt C x x x x x x x x C ∂∂∂∂∂∂∂∂=(2.18)Reflective SymmetryA reflective symmetry refers to the invariance of material property with respect to a coordinate plane such as 03=x . It can be equally represented by rotating the x 3 axis of 180o . Consider the following coordinate transformationααx x =, 33x x -=, (α=1,2).(2.19)Denote ij δ as the Kronecker Delta, whose diagonal elements are equal to unity and the off-diagonal ones equal to zero. That givesr r x x ααδ=∂∂ and r r x x33δ-=∂∂. According to the first expression of (2.18), the strain components in the new coordinates areαβαβεε=，3333εε=，33ααεε-=(2.20)If the elastic properties of the material is invariant with respect to a mirror reflection of 03=x , then the strain energy cannot be any arbitrary functions of the strain components, but has to be restricted to the following form()231222321312332211,,,,,,εεεεεεεεW W =(2.21)That implies any quadratic terms related to 13ε and 23ε in the strain energyexpression may only take the form of 213ε, 223ε and 2312εε. Consequently, anyelastic constants related to the cross products between the components()12332211,,,εεεε and the components ()2313,εε should vanish. The elastic matrix isreduced to:[]⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡=6655454436332623221613121100000000C C C C C C C C C C C C C C (2.22)Only 13 elastic constants are present for a material with one reflective symmetry. If the material possesses a further reflective symmetry with respect to a second orthogonal coordinate plane, say the plane of 01=x , the response of the material would remain the same under the following coordinate transform:11x x -=，22x x =，33x x -=(2.23)It is a straightforward matter to deduce that the elastic strain energy would possess a more restricted form of()223213212332211,,,,,εεεεεεW W = (2.24)Its dependences with terms such as 2313εε, 1112εε, 2212εε and 3312εε are ruled out. The elasticity matrix for such a material has the following form[]⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡=665544332322131211000000000000C C C C C C C C C C (2.25)It has 9 independnt elastic constants, and is called orthotropic. Examples of such materials include crystals of orthogonal lattice, and cross-ply composite laminates. Rotation SymmetryThe elastic response for a certain material may remain invariant when rotating it about a symmetry axis by a specific angle. Such a symmetry is called a rotation symmetry. The reflective symmetry discussed above corresponds to two-fold rotation symmetry, or symmetry to a rotation of 180 degrees. Without loss of generality, we label the symmetry axis as 3x , an arbitrary rotation about that axis is given by the following coordinate transformation:ωωsin cos 211x x x +=, ωωcos sin 211x x x +-=, 33x x =(2.26)The cases of the rotation angels of 0180=ω, 0120=ω, 090=ω and 060=ω are termed rotation symmetries of 2-folds, 3-folds, 4-folds and 6-folds, respectively. Transverse IsotropyIf a solid maintains its elastic response by rotation an arbitrary angle with respect to an aixs, the solid is termed transversely isotropic. In the coordinate transformation stated in (2.26), the following relations always hold for an arbitrary value of ω:22112211εεεε+=+, 21222112122211εεεεεε-=-, 223213223213εεεε+=+, ij ij εε= (2.27) where ij ε denotes the determinant of the strain tensor. Relations in (2.27) are called the invariance relations under the rotation transform (2.26). It can also show that the relations in (2.27) are the only ones invariant at any values of ω.If the material under consideration possesses the transverse isotropy, its elastic strain energy can only be the function of these strain invariants. Consequently, it has to be ina form of ()3322321321222112211,,,,εεεεεεεεεij W W +-+=. The only possible quadratic terms in strain components are ()22211εε+, 2122211εεε-, 223213εε+, 233ε and ()332211εεε+. Therefore, only 5 independent elastic constants exist. The elastic matrixis reduced to:[]()⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡-=221144443313111312112100000000000C C C C C C C C C C C (2.28)Examples for this class of materials include the wood and bamboo in nature, the pole vault in sport, and the man made composite materials reinforced by uni-directional fibers. IsotropyWe finally examine the most special materials that are isotropic in their elastic response. A typical example of such materials is a polycrystal composed of grains with complete random orientations. Consider an arbitrary rotation ij R on the coordinates such as j ij i x R x =. The rotation tensor obeys the relation of1R R =⋅T or ik jk ij R R δ=(2.29)Due to the second expression of (2.18), the elastic tensor in an arbitrarily rotated coordinates is ijkl tl pk sj ri rspt C R R R R C =. The only possibility for rspt C so defined to be invariant under arbitrary rotation is the following formjk il jl ik kl ij ijkl C δδμδμδδλδ'++=.Furthermore, the symmetry with respect to indices i and j implies 'μμ=. One then arrives at the elastic tensor for an isotropic material()jk il jl ik kl ij ijkl C δδδδμδλδ++=(2.30)Only two independent elastic constants λ and μ are encountered in (2.30) for anisotropic material. They are termed the Lamè constants. Substituting (2.30) into (2.4), one obtains the following generalized Hooke ’s law:ij kk ij ij μεελδσ2+=(2.31)Simple ShearWe now consider the generalized Hooke ’s law for three particular cases: simple shear, radial dilation, and uniaxial tension. The case of simple shear will be discussed first, as shown in Fig. 2.9.The kinematics given in Fig. 2.9 can be described by 21x u γ= and 032==u u . Consequently, all strain components are zero except 2/12γε=, where γ is the shear amount. The generalized Hooke ’s law leads to a shear stress of μγσ=12, while the other stress components are absent. The physical meaning of the second Lamè constant μ now becomes transparent: it is the shear modulus of the material.Figure 2.9 Simple shear.DilationNext consider dilatation in radial direction. The radial dilation adopts a kinematical field of i i x u 3θ=. That delivers a spherical strain tensor of ij ij δθε3=, where kk εθ=symbolizes the volume dilation. The generalized Hooke ’s law (2.31) becomes ij ij K θδσ=(2.32)whereμλ32+=K (2.33)denotes the bulk modulus of the material by (2.32). The stress tensor presented in (2.32) is spherical. In the general case, we can always decompose the stress and strain tensors into the spherical and deviatoric (traceless) parts as follows:ij kkij ij S δσσ3+=, ij kkij ij e δεε3+=.(2.34)It is a straightforward matter to show that 0=kk S and 0=kk e . The decomposition (2.34) suggest another representation of the generalized Hooke ’s law asij ij e S μ2=, kk kk K εσ3=(2.35)The strain energy density can be explicitly expressed as221814121kk ij ij kk ij ij KS S K e e W σμεμ+=+= (2.36)Uniaxial TensionOur last example concerns uniaxial tension where the only non-trivial stress component is σσ=11. The generalized Hooke ’s law (2.31) can be inverted as⎪⎭⎫⎝⎛-=ij kk ij ij K δσλσμε321 (2.37)by utilizing the relation (2.33). Take the “11”, “22”, and “33” components of the above equation, one hasσσλμεEK 1312111≡⎪⎭⎫ ⎝⎛-=, σσμλεεE v K -≡-==63322, (2.38)where the Young ’s modulus and the Poisson ’s ratio are given byμμ+=K K E 39, μμ+-=K K v 32321,(2.39)Among 5 symbols for elastic response, namely the Young ’s modulus E , the Poisson ’s ratio v , the shear modulus μ, the bulk modulus K and the first Lamè constant λ, any one of them can be expressed by any two others.2.4 Physical Foundation of Elastic SymmetryCrystals and Bravais LatticesThe elastic symmetry in a perfect crystal is motivated by its lattice structure. Table 2.1 lists 7 Bravais lattices, along with their corresponding symmetry groups and the number of independent elastic constants.Table 2.1 Seven Bravais LatticesLattice type Symmetry group Elastic constantstriclinic null (center symmetry)21 mono-clinic 2-fold rotation13 orthogonal two orthogonal 2-fold rotations9 tetragonal 4-fold rotation 7 rhombohedral 3-fold rotation 9 hexagonal 6-fold rotationcubic four 3-fold rotations about the cube diagonalsCauchy ’s RelationThe great mathematician Cauchy explored the physical foundation of a elastic crystal 180 years ago. His formulation led to the so-called Cauchy ’s relation for elastic constants. For the elasticity tensor ijkl C , the V oigt symmetry (2.15) indicates the interchangeabilities of the following index groups(i ,j )↔(j ,i ), (k ,l )↔(l ,k ), (i ,j )↔(k ,l )while the Cauchy ’s symmetry adds the interchangeability of(j ,l )↔(l ,j ).That grants the complete symmetry of an elastic tensor. For the general anisotropic materials (such as tri-clinic crystal), the Cauchy ’s symmetry adds 6 additional relations for the components of elastic matrix5614C C =, 6612C C =, 5513C C =, 4536C C =, 4423C C =, 4625C C = (2.40)Therefore, only 15 independent elastic constants are allowed for a generallyanisotropic material provided the Cauchy ’s symmetry holds.Central Force Assumption of CauchyCauchy derived his relation among elastic constants based on an assumption that the interaction among the atoms were in the form of central forces of any atom pairs. As shown in Fig. 2.10, a vector linking atom l and atom l ’ is denoted as ()()()l x l x l l x i i i -=∆'',, whose length can be computed as ()()()',',',2l l x l l x l l R i i ∆∆=. The change in that distance by imposing a strain tensor ij ε is()()()()()',',2',',22l l x l l x l l R l l r j i ij ij ∆∆+=-εδ (2.41) Cauchy assumed that the inter-atomic potential U depended only on the distances of various pairs of atoms in the aggregate, namely()()∑=','',l l ll l l r U ϕ(2.42)where 'll ϕ denotes the interaction energy between atom l and atom l ’, with l and 'l taking values in all atoms within the lattice. Generalizing (2.9) for the three-dimensional case, one can compute the elastic constants from the second derivatives of the inter-atomic potential as ()()()()()()()',',',',',',41',2'22l l x l l x l l x l l x l l r l l r V U V C l k j i l l oll okl ij ijkl ∆∆∆∆∂∂=∂∂∂=∑ϕεε. (2.43) l'lFigure 2.10 Interaction between two atoms.Expression (2.43) indicates that indices l k j i ,,, can switch their orders arbitrarily。

弹性力学第03章精品PPT课件

在主要边界上： y h ,
2
s y 0,
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前的坐标描述的，因此只有在小变形的条件下才可以使用叠加原理。即变形对外力作用点位置的改变可以忽略不计。
圣维南原理及应用
➢对于工程实际问题，构件表面面力或者位移很难满足
严格的边界条件。这使得弹性力学解的应用将受到极大的限制。为了扩大弹性力学解的适用范围，放宽这种限制，圣维南提出了局部影响原理。
➢圣维南原理主要内容：物体表面某一小面积上作用的
外力力系，如果被一个静力等效的力系所替带，那么物体内部只能导致局部应力的改变。而在距离力的作用点较远处，其影响可以忽略不计。
圣维南原理及应用
➢根据圣维南局部影响原理，假如我们用一静力等效力系取
代弹性体上作用的原外力，则其影响仅在力的作用区域附近。离此区域较远处，几乎不受影响。
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(3)在确定的坐标系下，考察具有确定的几何尺寸和形状的弹性体，根据主要边界上的面力边界条件(2-15) 或次要边界上的积分边界条件, 分析这些应力分量对应于边界上什么样的面力，从而得知所选取的应力函数可以解决什么样的问题。（或者根据已知面力确定应力函数或应力分量表达式中的待定系数）

第4章弹性力学ppt课件

4.4 梁的整体稳定
Ø梁的临界弯矩用稳定理论求解最简单的工况：纯弯曲的简支梁，截面双轴对称。
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此式含有侧向弯曲刚度E I y ，两个扭转刚度 G I t 和E I ，和失稳现象完全符合。复杂的工况：承受任意横向荷载的简支梁，截面单轴对称。
1 为单肢对其平行虚轴的形心轴的长细比。
截面选择：先根据绕实轴稳定要求选出单肢截面，再按照等稳要求确定两肢之间的距离。
计算时可先取缀条尺寸，以后再验算。
4.3 实腹式和格构式压杆的截面选择
Ø 对单肢长细比的要求：不是和杆件长细相等，而是更严格。
原因：杆件的初弯曲使凹侧肢的压应力大于杆件的平均值。

式(4-25b)的 2 3 相当于式(4-20)的 1 0 。因此， 1 3 2 相当于无量纲化的综合初曲挠度。它包含了几何缺陷和残余应力两种因素的效应，并且用于计算极限荷载而不是边缘屈服荷载。系数 1 , 2 , 3 对a，b，c，d 四类截面各不相同。详见GB50017规范。稳定系数由正则化来表达，计算公式可以通用于各种强度等级的
(2) 失稳是构件的整体行为。
由第一点，可以认为失稳是Pδ效应(即荷载位移效应)累积的结果。由第二点，可以领会杆件失稳和截面强度破环的差别。
4.1 稳定问题的一般特点
Ø 杆件稳定的极限承载力
欧拉临界力不能直接用于钢结构设计。
原因：现实构件都存在缺陷：几何缺陷——几何非线性
力学缺陷(残余应力)——材料非线性
当于兼承P和αP的理想直杆。4.1.1节的计算都适用。

弹性力学及有限元基础全套PPT课件 431页

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14
工科弹性力学教学
面向工程师的能力培养
知微观、重宏观，知数学、重力学，知计算、重概念。
教材：《弹性力学简明教程》（第四/三版）徐芝纶
参考书：
《Theory of Elasticity》
中文译本
S.Timoshenko
16
课堂要求：
认真听课，积极思考，踊跃讨论；独立完成作业，认真思考思考题。
x
y
y
(等厚薄板 t 很小）

弹性力学Chap8

σ ij = λε kk δ ij + 2 µε ij = −2vω,3δ ij − (1 − 2v )(δ i 3ω, j + δ j 3ω,i ) + x3ω, ij
x3ω,11 − 2vω,3 = x3ω,12 x3ω, 22 − 2vω,3 x3ω,12 − (1 − 2v )ω,1 x3ω, 23 − (1 − 2v )ω, 2 x3ω,33 − 2(1 − v )ω,3
Chapter 8.1
− Fk , k = xkψ k + φ 2µu i = −4(1 − v ) i + ( x kψ k + φ ),i ψ
6
Solutions by Displacement Potentials
Papkovich－Neuber Solution
The Papkovich-Neuber solution is complete. Papkovichcomplete. R.D. Mindlin, Note on the Galerkin and Papkovichi Stress Function, Bull. Am. Math. Soc 42(1936), 373-376 The uniqueness of the Papkovich-Neuber Papkovichsolution is not a trivial matter
Boussinesq Solution
Solution E Rotation
x ⋅ψ
is a harmonic function
ψ = −2(1 − v )χ
2µu i = −4(1 − v )eij 3 χ , j ≡ 2eij 3ψ , j

2024版弹性力学5PPT课件

2024/1/25
5
边界条件与约束类型
边界条件
位移边界条件、应力边界条件、混合边界条件。
约束类型
几何约束、运动约束、动力约束。
2024/1/25
பைடு நூலகம்
6
应力、应变及位移关系
2024/1/25
应力
单位面积上的内力，包括正应力和剪应力。
应变
物体在外力作用下形状和尺寸的改变，包括线应变和角应变。
位移
物体在外力作用下某点位置的改变，包括线位移和角位移。
广义平面应力问题与广义平面应变问题的定义
阐述广义平面应力问题和广义平面应变问题的基本概念和定义。
广义平面应力问题与广义平面应变问题的求解方法
介绍如何利用弹性力学的基本方程和边界条件，求解广义平面应力问题和广义平面应变问题。
广义平面应力问题与广义平面应变问题的实例分析
通过具体实例，展示广义平面应力问题和广义平面应变问题求解方法的实际应用。
10
功的互等定理与卡氏定理
01
功的互等定理的基本内容
在弹性力学中，如果两个载荷系统在相同的物体上分别作用并产生相同
的位移场，则这两个载荷系统所做的功相等。
2024/1/25
02 03
卡氏定理的基本内容
在弹性力学中，如果物体在某一载荷作用下处于平衡状态，那么在该载荷作用下物体内部任意点的应力分量与另一与之平衡的载荷在该点所引起的位移分量成正比。
2024/1/25
03
平面问题求解方法
13
平面应力问题与平面应变问题
平面应力问题
分析薄板在面内荷载作用下的应力、变形和稳定性。
2024/1/25
平面应变问题
研究长柱体或深埋在地下的结构物，在垂直于轴线或地面的荷载作用下，其横截面内的应力和变形。

清华大学研究生弹塑性力学讲义 5弹塑性_弹性力学的基本方程与解法

弹塑性力学第四章弹性力学的基本方程与解法一、线性弹性理论适定问题的基本方程和边界条件对于在空间占有体积域V 的线弹性体在外加恒定载荷和固定几何约束条件下引起的小变形问题，若以, ,u εσ作为求解变量，则可以建立如下偏微分方程边值问题：几何方程()1,,2ij i j j i u u ε=+ ()12∇+∇u u ε= (1a)广义胡克定律 ij ijkl kl E σε= :E σ=ε(1b)平衡方程 ,0ij j i f σ+= ∇⋅+=f 0σ V∀∈x (1c)以上方程均要求在域内各点均满足。

边界条件 u u i i = ∀∈x S ui (2a)n t j ji i σ= ∀∈x S ti(2b)对于适定问题，即不仅要求保证解存在唯一，而且有较好的稳定性。

当载荷或边界条件给定值有微小摄动时，应能保证问题解的变化也是微小的。

对于边界条件的提法就有严格的要求。

即要求：S S S S S ui ti ui ti U I ==∅(2c)对于各向同性材料，其广义胡克定律可具体写成 σλεδεij kk ij ij G =+2 ()tr 2G λ+I σ=εε (3a)()11ij ij kk ij E ενσνσδ⎡⎤=+−⎣⎦ ()()1tr Eνν=⎡⎤⎣⎦I ε1+σ−σ (3b)以上就域内方程来说，一共是对于u ,,σ ε的15个独立分量u i ij ij ,, σε的15个方程。

对于边界条件来说，三维问题每点有三个边界条件，而且是在三个正交方向上每个方向有一个边界条件，这个边界条件或者给定位移、或者给定面力。

这三个正交第四章弹性力学的基本方程与解法方向可以是整体笛卡儿坐标系的三个方向，也可以是边界自然坐标系的三个方向（即法向和两个切向）。

从更一般来说，除去给定位移或面力外，还有另一种线性的边界条件t K u c i ij j i +=(4)这是一种弹性约束条件。

用这个条件可以取代给定位移或给定面力的条件。

弹性力学基础教学课件PPT

弹性力学基础教学课件
目录
• 引言 • 弹性力学基本概念 • 弹性力学基本方程 • 弹性力学问题解法 • 弹性力学应用实例 • 总结与展望
01
引言
课程简介
弹性力学基础是一门介绍弹性力学基本原理和方法的课程，旨在为学生提供解决工程问题中弹性力学问题的能力。
本课程将介绍弹性力学的基本概念、基本原理、基本方法以及在工程实践中的应用，帮助学生建立对弹性力学的基本认识，培养其解决实际问题的能力。
弹性力学基本方程
平衡方程
静力平衡方程
描述了弹性体在力的作用下保持平衡的状态，表达了物体内部各点的应力与外力之间的关系。
运动平衡方程
在考虑了物体运动的情况下，描述了弹性体在力的作用下保持运动的平衡状态，涉及到速度和加速度。
几何方程
应变与位移关系
描述了物体在受力变形过程中，位移与应变之间的关系。
应变与速度关系
描述了物体在受力变形过程中，速度与应变之间的关系。
本构方程
弹性本构方程
描述了弹性体在受力变形过程中，应力与应变之间的关系，涉及到弹性模量和泊松比等参数。
塑性本构方程
描述了塑性体在受力变形过程中，应力与应变之间的关系，涉及到屈服准则和流动法则等参数。
04
弹性力学问题解法
总结词
弹性梁的弯曲问题
总结词
实际工程应用
详细描述
在建筑工程、机械工程和航空航天工程等领域，弹性梁的弯曲问题具有广泛的应用。例如，在桥梁和建筑结构中，梁是主要的承载构件，其弯曲变形会影响结构的稳定性和安全性。通过掌握弹性力学的基本原理和方法，可以更加准确地分析梁的弯曲问题，优化梁的设计和计算。
弹性薄板的弯曲问题
越广泛。未来可以进一步研究和发展更加高效、精确的数值计算方法，

弹性力学ppt课件

应变定义
物体在外力作用下产生的形变，表示物体尺寸和形状的变化。
应力与应变关系
应力与应变之间存在一一对应关系，通过本构方程来描述。
广义胡克定律及应用
1 2
广义胡克定律又称作弹性本构关系，表示应力与应变之间的线性关系。
广义胡克定律的应用用于计算弹性体在复杂应力状态下的应力和应变，是弹性力学中的重要基础。
弹性力学ppt课件
contents
目录
• 弹性力学概述 • 弹性力学基本原理 • 线性弹性力学问题求解方法 • 非线性弹性力学问题简介 • 弹性力学实验方法与技术应用 • 弹性力学在相关领域拓展应用
01 弹性力学概述
弹性力学定义与研究对象
弹性力学定义
弹性力学是研究弹性体在外力和其他外界因素作用下产生的变形和内力，从而在变形与外力之间建立一定关系的科学。
有限元法在弹性力学中应用
有限元法基本原理
将连续体离散化为有限个单元，每个单元用简单的函数近似表示，通过变分原理得到有限元方程。
有限元法求解过程
包括网格划分、单元分析、整体分析、边界条件处理和求解有限元方程等步骤。
有限元法的优缺点
有限元法可以求解复杂几何形状、非均质材料和非线性问题，但存在网格划分和计算精度等问题。
布。
弹性模量和泊松比测定实验
拉伸法
通过对标准试件进行拉伸实验，测量试件的应力和应变，从而计算得到弹性模量和泊松比。
压缩法
通过对标准试件进行压缩实验，测量试件的应力和应变，进而计算弹性模量和泊松比，适用于脆性材料的测量。
弯曲法
通过对梁式试件进行三点或四点弯曲实验，测量试件的挠度和应力，从而推算出弹性模量，特别适用于细长构件的测量。

弹性力学课件Chapter3

y 1
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（3-2）
5
应用弹塑性力学
APPLIED ELASTO-PLASTICITY OF SOLIDS
3-3、应变分量与位移分量之间的微分关系—几何方程:
应变是与位移有关的。物体中各质点相对位置的改变，则产生变形。因此要分析物体的变形，就要研究各点的位置的变化。
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APPLIED ELASTO-PLASTICITY OF SOLIDS
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w' w( x, y, z) w dx w dy w dz
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应用弹塑性力学
APPLIED ELASTO-PLASTICITY OF SOLIDS
为了表示成应变分量的关系，对 u, v, w 作如下改写：
x
v v(x, y, z) v' v(x dx, y dy, z dz)
位移分量是坐标的函数 w w(x, y, z) w' w(x dx, y dy, z dz)

弹性力学ppt课件(2024)

建立一维拉伸或压缩问题的数学模型
通过受力分析，确定物体在拉伸或压缩过程中的内力分布和变形情况。
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求解一维拉伸或压缩问题的基本方法
运用弹性力学的基本原理和公式，如胡克定律、应力-应变关系等，对一维拉伸或压缩问题进行求解。
一维拉伸或压缩问题的有限元分析
介绍有限元方法在一维拉伸或压缩问题中的应用，包括网格划分、单元刚度矩阵和总体刚度矩阵的建立、边界条件的处理等。
适用范围
适用于大多数金属材料在常温、静载条件下的力学行为。对于非金属材料、高温或动载条件下的情况，需考虑其他因素或修正虎克定律。
2024/1/25
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弹性力学分析方法与技巧
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解析法求解思路及步骤
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建立弹性力学基本方程
选择适当的坐标系和坐标…
求解基本方程
件和载荷。
平面应变问题建模
02
探讨平面应变问题的特性，构建适当的力学模型，并确定边界
条件和载荷。
求解方法
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介绍适用于平面应力和平面应变问题的求解方法，如有限元法
、有限差分法等，并讨论各种方法的优缺点和适用范围。
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极坐标下二维问题处理方法
极坐标系的引入
阐述极坐标系的定义和性质，以及与直角坐标系的关系。
根据问题的实际情况，确定位移边界条件、应力边界条件以及初始条件。
通过与其他方法（如数值法、实验法）的结果进行比较，验证解析解的正确性和有效性。
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数值法（有限元法）在弹性力学中应用
有限元法基本原理
有限元模型建立

弹性力学第一章课件

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yx xz
y yz x
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xy yx y
zx
O
y z
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xy
第1个下标 x 表示τ所在面的法线方向；第2个下标 y 表示τ的方向.
应力正负号的规定：
正应力—— 拉为正，压为负。剪应力—— 坐标正面上，与坐标正向一致时为正；
坐标负面弹性上力学，第一与章坐标正向相反时为正。
数学弹性力学；弹性力学
应用弹性力学。
弹性力学是塑性力学、断裂力学、岩石力学、振动理论、有限单元法等课程的基础。
弹性力学第一章
小结：
弹性力学是固体力学的一个分支，研究弹性体由于外力作用或温度改变等原因而发生的应力、形变和位移。
本课程较为完整的表现了力学问题的数学建模过程，建立了弹性力学的基本方程和边值条件，并对一些问题进行了求解。弹性力学基本方程的建立为进一步的数值方法奠定了基础。
弹性力学是学习塑性力学、断裂力学、有限元方法等课程的基础。
弹性力学第一章
§1-2 弹性力学中的几个基本概念
基本概念：外力、应力、形变、位移。
1. 外力
体力、面力（材力：集中力、分布力。）
(1) 体力 —— 弹性体内单位体积上所受的外力
lim F
Q —— 体力分布集度
V 0 V
（矢量）
F Xi Yj Zk
弹性力学以微元体为研究对象，建立方程求解，得到弹性体变形的一般规律。所得结果更符合实际。
弹性力学第一章
（3）数学理论基础材力、结力 —— 常微分方程（4阶，一个变量）。弹力 —— 偏微分方程（高阶，二、三个变量）。数值解法：能量法（变分法）、差分法、有限单元法等。