刘刚-连续介质力学作业3
大连理工研究生连续介质力学作业题
f = xT Ax ， grad（f ）= ∂x T Ax = 2Ax ∂x
f ′ = (RT x′)T ART x′ = x′T RART x′
grad(f' ) = ∂x′TRART x′ = 2RART x′ ∂x′
= 2RA(RT x′) = 2RAx = R ⋅ 2Ax = R ⋅ grad(f)
（3） a1 = p, a 2 = q, a3 = r
⎡2 0 1⎤ 由[gij ] = ⎢⎢0 4 2⎥⎥
⎢⎣1 2 2⎥⎦ 及 ai = gij a j 得 a1 = 2 p+ r, a2 = 4 q+ 2 r, a3 = p+ 2 q+ 2 r
1
2. 已知笛卡尔坐标系 e1 , e3 , e3 ，一个新的坐标系定义为
2 3
x2'
+
2 3
x2'
−
4 3
x2'
+
2 3 x3' 4 3 x3' 2 3 x3'
⎫ ⎪ ⎪⎪ ⎬ ⎪ ⎪ ⎪⎭
= grad（f ′）
3
3.
二维情况下，一质点应力张量 σ 主值 λ1σ = 1.6 ， λσ2 = 2.3 。主方向 N1 =
3 2
e1
−
1 2
e2
，
N2
=
1 2 e1
+
3 2
e
（2） 求向量 x 参考新坐标系的表示形式 x = xi′ ei′ （3） 求函数 f 在新的坐标系下的表达形式 f ′(x1′ , x2′ , x3′ ) （4） 判断 grad（f ）的客观性。
¾ 解答：
（1） grad（f ）＝ (2 x1 , 0, − 2 x3 )T
连续介质力学基础第三章
速度定义
加速度
2 2 v u x w(a , t ) 2 2 t t t
6
物质导数:即随体导数,给定质点上函数对时间的变化率.
t F (a , t ) f ( x , t ) f ( x , t ) f x t a t a t x x t t a
9
例1: 已知位移场 u1 m2t 2a1 , u2 u3 0 (Lagrange)
m 2t 2 u1 x , u2 u3 0 2 2 1 1 m t
(Euler)
w 求速度场 v 和加速度场
解:
v1 u1 2m 2ta1 , v2 v3 0 t v w1 1 2m 2 a1 , w2 w3 0 t
1) 速度(Lagrange形式)
2) 速度(Euler形式) Du u v (v )u Dt t Dui ui u vi ( x , t ) vk i Dt t xk
v v ( x, t ) v ( x1 , x2 , x3 , t ) 为瞬时速度场. vi ( x, t ) vi ( x1 , x2 , x3 , t )
分量形式:
曲面
f1 ( x1, x2 , x3 , a) 0 f2 ( x1 , x2 , x3 , a) 0
a 迹线
15
消去时间 t
欧拉描述的迹线:
Dx x (a, t ) 轨迹与速度的联系: v ( x, t ) Dt t a
dx1 v1 ( x , t ) v1 ( x1 , x2 , x3 , t ) dt dx2 v2 ( x , t ) v2 ( x1 , x2 , x3 , t ) dt dx3 v3 ( x , t ) v3 ( x1 , x2 , x3 , t ) dt
《连续介质力学》期末复习提纲-总(2020年7月整理).pdf
<连续介质力学> QM 复习提纲(2010.12)一、基本要求1、掌握自由指标与哑指标的判别方法及表达式按指标展开；2、掌握ij 与ijk e 的定义、性质及相互关系；3、掌握二阶张量坐标转换的计算；4、掌握二阶张量特征值、特征向量与三个不变量的计算方法；5、掌握哈密顿微分算子及其基本计算；6、掌握小变形应变张量、转动张量及转动向量的计算；7、掌握正应变的计算；8、掌握正应力、剪应力及应力向量的计算； 9、掌握应力张量与应变张量的对称性； 10、掌握能量密度及能通量密度向量的计算；11、掌握各向同性线弹性体的广义胡克定律的两种形式； 12、掌握应力张量与体积膨胀率的关系； 13、掌握各向同性线弹性体的应变能密度函数； 14、会对材料的各个弹性参数之间的关系进行相互推导； 15、掌握从质点的运动方程推导Navier 方程的过程； 16、掌握从质点的运动方程出发推导纵横波的方程的过程； 17、掌握地震波速度与泊松比的关系； 18、掌握非均匀平面简谐波的传播特征；19、掌握P 波、SV 波入射到自由界面上的传播特征；20、掌握利用自由界面边界条件确定反射系数和反射波位移场的方法； 21、掌握Reilaygh 波和Stonely 波的传播特征；22、掌握P 波入射到两种弹性体接触面上的反射系数和透射系数的计算方法；二、复习题 简答论述题1、试解释“连续介质”所必须满足的条件。
2、简述弹性动力学基本假设。
3、说明应力、应变、正应力、正应变、剪应力及剪应变的含义。
4、说明杨氏模量、泊松比、体积模量与剪切模量的物理含义。
5、简述小变形应变张量的几何解释。
6、举例说明相容性条件的物理意义。
7、什么是应力主平面？什么是主应力与应力主方向？ 8、极端各向异性体有哪些特征？ 9、正交各向异性体有哪些特征？ 10、横向各向同性体有哪些特征？ 11、试说明Stoneley 波的传播特点？ 12、试说明Rayleigh 波的传播特点？13、以复数值形式表示的波向量所对应的位移为'''()i t A e e ω−−=k x kx u d其中的'k 及''k 满足式ωχ22⎫''''''⋅−⋅=⎪⎬⎪'''⋅=0⎭k k k k k k 试论述该平面波的传播特征。
3.连续介质力学
加速度
vX, t 2 uX, t aX, t v t t 2
2 变形和运动
运动描述：
独立变量是空间坐标 x 和时间 t，称为空间或Eulerian描述
v v( x, t ) vΦX, t , t
通过链规则得到材料时间导数 （全导数）
空间时间导数
2 变形和运动
运动条件
连续可微，一对一（F可逆），J > 0 第一个条件，变形梯度通常在材料的界面上是非连续的。在 某些现象中，例如扩展裂纹，运动本身也是非连续的。要求在运 动及其导数中非连续的数量是有限的。实际发现，有些非线性解 答可能拥有无限数量的非连续。然而，这些解答非常罕见，不能 被有限元有效地处理，所以不关注这些解答。 第二个条件，即运动为一对一的，要求在参考构形上的每一 点，在当前构形上有唯一的点与之对应，反之亦然。这是 F 规则 的必要充分条件，即F是可逆的。当变形梯度F是正常的， 则 J 0 ，因为当且仅当 J 0 时F的逆才存在。因此，第二个和 第三个条件是有联系的，后者更强。 第三个条件，更强的条件， J 必须为正而不仅非零，在第 3.5.4节看到这遵循了质量守恒。这个条件在零尺度集合上也可能 违背，例如在一个裂纹的表面上，每一个点都成为了两个点。
如何建立x与X的关系？ 需要借助单元坐标和初始构形
2 变形和运动
解：三角形3节点线性位移单元的构形（见附录3）
1 2 3 1
xξ , t xI t I x1 t 1 x2 t 2 x3 t 3
运动条件
除了在有限数量的零度量集合上，假设描述运动和物体变形的映射
ΦX, t
满足以下连续性条件： 连续可微，一对一（F可逆），J > 0
连续介质力学1-3
3. 对称张量与反对称张量 命题一、对称性与反对称性与坐标系无关。 命题一、对称性与反对称性与坐标系无关。 证明： 证明：
Ti′j′ = β i′m β j′nTmn T j′i′ = β j′m β i′nTmn
命题二、 命题二、任意二阶张量可以唯一分解成一个对称 张量与一个反对称张量之和。 张量与一个反对称张量之和。 证明： 证明： 存在性 唯一性
证明： 证明：I 12 = (λ1 + λ2 + λ3 )
2
2 = λ1 + λ2 + λ2 + 2 λ1 λ 2 + λ 2 λ 3 + λ 3 λ1 2 3
(
) (
)
ˆ = T2
( )
kk
+ 2I2
§3-3 二阶实对称张量 ˆ的三个主值都是实数。 1. T的三个主值都是实数。 v v v 证明： 为主值， 是主方向， ˆ 证明：λ为主值，x是主方向，即T • x = λx v v ˆ • x # = λ# x # 则T v ˆ v# v 或x • T • x ˆ对称， ˆ v ˆ Q T对称，故T • x # = x # • T v# ˆ v = x •T • x v# ˆ # v# ∴ x •T = λ x v v v ˆ v x # • T • x = λ# x # • x
Tij a j = β ii ′ β jj′Ti ′j′ β jk ′ ak ′ = β ii ′δ k ′j′Ti ′j′ a k ′ = β ii ′Ti ′k ′ ak ′
λa i = λβ ik ′ a k ′ β ii ′Ti ′k ′ a k ′ = λβ ik ′ ak ′ β im′ β ii ′Ti ′k ′ ak ′ = λβ im′ β ik ′ ak ′
连续介质力学-例题与习题
《连续介质力学》例题和习题第一章 矢量和张量分析第一节 矢量与张量代数一、矢量代数令112233A A A =++A e e e ,112233B B B =++B e e e ,则有112233A A A αααα=++A e e e111222333()()()A B A B A B +=+++++A B e e e112233112233112233()()A A A B B B A B A B A B •=++•++=++A B e e e e e e112233112233111112121313212122222323313132323333()() A A A B B B A B A B A B A B A B A B A B A B A B ⨯=++⨯++=⨯+⨯+⨯+⨯+⨯+⨯+⨯+⨯+⨯A B e e e e e e e e e e e e e e e e e e e e e e e e又因为: 11⨯=e e 0;123⨯=e e e ;132⨯=-e e e ;213⨯=-e e e ;22⨯=e e 0;231⨯=e e e ; 312⨯=e e e ;321⨯=-e e e ;33⨯=e e 0则: 233213113212213(_)()()A B A B A B A B A B A B ⨯=+-+-A B e e e 习题:1、证明下列恒等式：1）[]2()()()()⨯•⨯⨯⨯=•⨯A B B C C A A B C2) [][]()()()()⨯•⨯=•⨯-•⨯A B C D A C D B B C D A2、请判断下列矢量是否线性无关？1232=-+A e e e 23=--B e e 12=-+C e e .其中i e 为单位正交基矢量。
3、试判断[]816549782-⎡⎤⎢⎥=⎢⎥⎢⎥--⎣⎦A 是否有逆矩阵；如有，请求出其逆阵[]1-A 。
二、张量代数例1：令T 是一个张量，其使得矢量a ，b 经其变换后变为2=+Ta a b ，=-Tb a b ，假定一个矢量2=+c a b ，求Tc 。
连续介质力学
R, ,V b ,a )b A ()a A ()b a (A ~~~~~~~~~∈βα∈∀⋅β+⋅α=β+α⋅~j~i ij ~i~i ij ~A,g g A g g A A ~⊗∂Π∂=⊗∂Π∂=∂Π∂=Π~i i mng g g g T TT ⊗⊗⊗∂=∂=Chapter 4Stress4.1 The Stress Vector and the Stress Principle of Euler and Cauchy1 External loads1) Body force: a distributed force per unit mass.in ΩThe body force is a long-range action.mFf m ΔΔ=→Δ~~lim2) Surface force: a force per unit area on surfaceon ∂ΩThe surface force is a short-range action, which manifests itself in the form of the contact forceSTp S ΔΔ=→Δ~~lim2 Interaction in the interior of continuum1) The Euler-Cauchy postulate(1) The interaction by a part applied on another part in the interior of a continuum is characterized by a stress vector field, which is defined upon an imagined interfacial surface S .(2) The stress vector field applied on the material occupying the space interior to S is equipollent to theaction of the exterior material upon it.p nS(3) P n remains unchanged for all surfaces having at X the same normal vector n . This means that P n is independent of the surfaces chosen as long as they all have at X the same normal vector.XP nnComments:The Euler-Cauchy postulate implies, (1) no long-range interaction exists in the deformable continuum;(2) Interactions within the continuum are momentless.2) Stress vector (Traction)ST p nS n ΔΔ=→Δ~~limNote:The stress vector refers to adeformed surface element whose unit normal vector is n and represents a contact force per unit area of the deformed surface.Discuss:(1) What differences are there between the surface force and the stress vector? (2) The force state of a mass particle is determined by magnitude and direction of all forces applied on it. How should we describe the force state of a point within continuum?4.2 The Cauchy Stress FormulaCauchy tetrahedronwith inclined face having an arbitrary orientation n ,constructed about some material point and to be shrunk onto that point in the limit to be taken.The Cauchy stress tensorThe Cauchy stress formulaii g p ~~~⊗=σ~~~np n ⋅=σThe components of the Cauchy stress tensor and their meaningsThe component of σare denoted by σij , whose first index indicates the direction of the force of action and the second index indicates the direction of the normal to the surface on which σij acts.x 1x 2x 3σ11σ21σ31σ22σ12σ33σ13σ23Note:(1) The Cauchy stress tensor is an invariant independent of the choice of basis vectors.(2) The Cauchy stress tensor isdefined on the current configuration. (3) For non-polar materials, the Cauchy stress tensor is symmetric.Normal stress and shear stress~~~~~nn p n n n ⋅⋅=⋅=σσ2~~~~22~nnn n n n p σσσστ−⋅⋅⋅=−=n I σσ≤Chapter 5Balance Equations In this chapter we examine the local forms of the various conservations laws as expressed in the initial and current configuration we have introduced. To expedite our development, we first discuss how integral representations of equilibrium can be converted to pointwise conservation principles, a process known as localization. The principle of energy conjugate and physical component are also given here.5.1 LocalizationSuppose we consider an arbitrary volume of material, V⊂Ω, in the reference configuration of a solid body, as depicted in Figure below: Suppose further that wecan establish the followinggeneral integral relationover this volume:The localized hypothesis asserts that (5-1) holds true for each and everysubvolume V of Ω,i.e.,)()(~~=∫ΩXdVXϕ)()(~~=∫V XdVXϕ(5-1)(5-2)According to the localization theorem, from Eq.(5-2) it yieldsϕ(X)= 0pointwise in Ω(5-3)Note: The procedure as same as the above can also be applied on a spatial domain.5.2 Balance of massConservation law of mass: The total mass of a closed continuum remains constant in motion.Integral representation:)(~=∫ΩxdvDtDρDifferential representation:)()(~~=∇⋅+⋅∇+∂∂v v tρρρ)(~=∇⋅+∂∂v tρρρρ=J 5.3 Conservation of momentumFor a given set of particles, the time rate ofchange of the total linear momentumequates to the vector sum of all the external forces acting on the particles of the set. This is expressed mathematically in the Equation below∫∫∫ΩΩΩ∂=+dv v DtDdv f ds p n ~~~ρρSubstituting the Cauchy formula into theEquation above yieldsApplying the divergence theorem and thetransport theorem gives∫∫∫ΩΩΩ∂=+⋅dv v DtDdv f ds n ~~~~ρρσ0)(~~~=−+∇⋅∫Ωdv DtvD f ρρσ By the localization theorem, it deduces toNote:1 The equilibrium equation of momentum isdefined in the current configuration.2 A nonlinearity is implicitly included in the equilibrium equation of momentum.DtvD f ~~~ρρσ=+∇⋅ Discuss: Can the equilibrium equation ofmomentum be represented by the second Piola-Kirchhoff stress?~0~0~~)(af F ρρ=+∇⋅Π⋅ 5.4 Balance of moment of momentumThe conservation of moment of momentumequates the time rate of change of the total moment of momentum for a set of particles to the vector sum of the moments of the external forces acting on the system. In the current configuration, we can write its balance of angular momentum via∫∫∫ΩΩ∂ΩΩ×=×+Ω×d v r Dt Dds p r d f r ~~~~~~ρρBy a series of operations and the localizedhypothesis, the equation above deduces to~~~~0:σσσ=⇔=∈TTThe conservation law of the moment ofmomentum guarantees that the stresstensor is symmetric under the condition that there does not exist the moment of body force and surface force. Therefore, there are only six independent components of stress—three normal components and three shear components.5.5 Physical ComponentsIn the polar coordinates, write out thebalance equation of two dimensional staticproblem in terms of the tensorial laws.Take the balance equation of momentum as example. If the body force is not concerned, the it is represented as;=βαβσ In order to calculate the Cristtoffelsymbol,we firstly give the transformation between the polar and Cartesian coordinates as follows:x =r cos θ, y =r sin θ.,=Γ+Γ+αγβγβγβαγββαβσσσ Therefore, we get g r =cos θe 1+sin θe 2g θ=–r sin θe 1+r cos θe 2g rr =1, g θθ=r 2, g r θ= g θr =0;g rr =1, g θθ=1/r 2, g r θ= g θr =0;Γθr θ=Γr θθ=–Γθθr =g θθ,r /2=r ;Γθr θ=Γr θθ=1/r, Γθθr =–r.Note: The Base vector has dimensions.As thus,However, in the standard textbook ofelasticity, the balance equation ofmomentum has the form below:030=+∂∂+∂∂=−+∂∂+∂∂rr r r r r r rr r θθθθθσθσσσσθσσ02101=+∂∂+∂∂=−+∂∂+∂∂rr r r r r r r r r r θθθθθσθσσσσθσσWhy ?This is because the physical dimensions ofbase vectors.Let us define the base vectors without dimensions as follows:.ˆ,ˆ~~~~><><==ααααααααgggg g gSo we haveβααββαββαααββααββααββαββαααββααβσσσσσσσ~~~~~~~~~~~~~ˆˆˆˆˆˆˆˆˆˆg g g gggg g g g g gg g g g ⊗=⊗=⊗=⊗=⊗=⊗=><><><>< Cleverly,These are the so-called physicalcomponents of tensor. By this concept, wecan obtain the standard balance equation of momentum..ˆ ,ˆ.ˆ ,ˆ><><><><><><><><====ββαααβαβββαααβαβββαααβαβββαααβαβσσσσσσσσggg g g g g g 5.6 Balance of energy1 Balance equation of energyThe global form of the equation of energyconservation may be written as∫∫∫∫∫ΩΩ∂ΩΩΩ∂Ω+⋅=⋅−Ω+Ω⋅+⋅d e v v Dt D sd h rd d f v ds p v )21(~~~~~~~~ρρρρDue to the fact that∫∫∫∫∫∫ΩΩΩΩΩ∂Ω∂Ω∇⋅σ⋅+σ=Ω∇⋅σ⋅+σ=Ω∇⋅σ⋅+σ∇⊗=Ω∇⋅σ⋅=⋅σ⋅=⋅d )](v d :[d )](v :L [d )](v :)v [(d )v (ds n v ds p v ~~~~~~~~~~~~~~~~~~~∫∫ΩΩ∂Ω∇⋅=⋅d h s d h ~~~∫∫∫ΩΩΩΩ+⋅=Ω+⋅=Ω+⋅d ea v d e v v Dt D d e v v Dt D][])(21[)21(~~~~~~&&ρρρρρρthe equation of energy conservation reducesto∫∫ΩΩΩρ=Ωσ+ρ+∇⋅−d ed )d :r h (~~~& By the localized hypothesis, one hased :r h ~~~&ρ=σ+ρ+∇⋅−2 Energy conjugateProblem:1 When a strain measure is given, is thechoice of stress arbitrary?Select a pairs of strain and stress according to what criterion?~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~:1)(:1)(1)(1:)(1:1)(1)(1)(1:)(1::E JF d F J F L F tr J F F tr J F F J FJ F tr J F L tr J L F tr J L F J L d eTT T T T TT &&&&&&Σ=⋅⋅Σ=⋅⋅⋅Σ=⋅⋅Σ=Σ⋅==⋅=⋅⋅=⋅⋅=⋅===πππππσσρ Different strain measure needs differentstress to be associated with it in order toenable to uniquely determine the internal energy of body. This is the principle of energy conjugate.。
连续介质力学习题二
连续介质力学习题二二．变形与运动2-1 如果物体在运动过程中保持任意两点间的距离不变，则称这样的运动为刚体运动，试证：物体的运动若为刚体运动，则参考构形中的物质点X变换到当前构形中的空间位置x时，必满足：)()()(t a A X t Q x +-⋅=，其中)(t Q 为正常正交仿射量。
2-2 现取物质坐标系}{A X 和空间坐标系}{i x 为同一个直角坐标系，其单位基向量为),,(321e e e，有一物体的变形为：33222011,,X x X x X k X x ==+=，试写出以下各量：1）变形梯度张量F 和变形梯度张量之逆1-F ；2）右，左Cauchy-Green 张量B C ,；并计算C 和B的三个主不变量；3）写出C 和B的特征方程，并求出三个特征值αη和相应的特征方向αL 和αl)3,2,1(=α。
4）试给出极分解R V F ⋅=中的左伸长张量V 和正交张量R的矩阵表示。
2-3 现取物质坐标系}{A X 为直角坐标系{X ，Y ，Z}，空间坐标系}{i x 为圆柱坐标系},,{z r θ，令z 轴与Z 轴重合，0=θ与X 轴重合，图示长方体发生纯弯曲，题2-3图变形满足),(X r r =),(Y θθ=)(Z z z =，且存在逆关系：),(r X X =，),(θY Y =)(z Z Z =，试写出以下各量：1）变形梯度张量F 和变形梯度张量之逆1-F ；2）右，左Cauchy-Green 张量B C ,；并计算C 和B的三个主不变量；3）写出C 和B的特征方程，并求出三个特征值αη和相应的特征方向αL 和αl)3,2,1(=α。
2-4 现取空间坐标系}{i x 为直角坐标系，其单位基向量为),,(321e e e，有一物体的小变形位移场为3212323213131))(())((e x x e x x x x e x x x x u-+++--=，试求：（1）P （0，2，-1）点的小应变张量e ，小转动张量Ω 及其反偶矢量ω； （2）求P 点在9/)48(321e e e+-=ν方向上的线应变；（3）求P 点在9/)48(321e e e +-=ν和9/)744(321e e e-+=μ二方向上的直角的变化量。
连续介质力学-第4章-四川大学
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4.2 基本定律的积分形式
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2017力学复习的策略(南陵中学刘刚)
2017力学复习的策略南陵中学刘刚在2017年的高考大纲中，相比2016年大纲的调整将选修3-5的动量从选修内容调整到了必考的95分中，同时将“牛顿运动定律””调整为“牛顿运动定律及其应用”。
从物理学科的构造来看力学复习，基本上说就是高考总复习，因为力学的概念、规律和方法涵盖或联系了高中物理的全部内容。
加上新大纲为了考虑力学部分内容的完整性将选修3-5纳为必修内容更加体现了力学复习的效果如何，决定了对高中物理知识体系的整体理解水平及应用能力水平，也决定了应试者最终能否高考成功。
2017年的高考力学可从以下几个方面进行把握：一.紧扣考纲，弄清力学地位力学的学习内容，是高中物理教学的核心，主要的概念、规律和方法，在教材中被集中安排在必修一、必修二，选修3-4和选修3-5四本教材中。
高中物理力学的学习内容包括：建立较完整的运动学、动力学的基本概念；研究对物体进行受力分析的方法；对物体的直线运动和曲线运动进行描述；从动力学的角度揭示运动和力的关系；从能量的角度研究物体的运动规律；从动量守恒的角度研究物体的运动规律。
因此，较熟练掌握高中物理必修一、必修二(人教版)中力学的概念、规律和方法，即奠定了高中物理知识和方法的基础，获得了高中物理知识体系的主干。
在高三物理总复习课的教学中，把力学的规律和方法归纳、理解、练习到位，虽功在力学，却能在最大程度上有利于高中物理其它各模块知识的熟练掌握，把物理学各部分规律通过相互作用联系在一起。
因此，提醒大家对力学内容的复习要有全局意识，多投入精力，计划充足的时间，夯实基础，稳步提高。
再一次在学、思、辨、悟中提升对力学的概念、规律和方法的整体理解及应用水平。
二．串起解题的知识脉络对于高中力学的基本思想应该是从三个方面考虑，其一是对象，其二是选择适当的物理规律解题。
第一．对象可以分为是系统还是单体。
然后研究的是过程还是状态。
从而对应了不同的规律，我们的力学规律总而言之就六类，1，运动学方程（恒力，单体，定空间，时间过程）2，动能定理（恒力，单体，定空间过程）3，动量定理（恒力，单体，定时间过程）4，牛顿运动定律（一般单体，定状态）。
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May 16th, 2013 Computer Exercise of Continuum Mechanics IIIMajor: Mechanical EngineeringCourse: Continuum MechanicsStudent name: LiuGangStudent ID: 12S057007Director: Prof. Tongtong GuoBasic method:Applications/ScalesBasic conditionsAssumptionsAnalysisSummeryExample 1: A cantilever beam with its dimensions shown in the figure, bearing a centroid force and a moment on the right side of the beam. The force can be denoted by F and the moment by FLM .Solution:1.Application: balcony, book shelves, diving platform, tower crane, dinner table...2. Basic Conditions : A cantilever beam with left-fixed and right side free. The free end loaded a force F and moment M.4. Analysis : a : Consider the whole beam as a free body. Study the fixed end: 0=∑Yand=∑oMWe can get0=+F F y0=+-FL FL M o 0=o Mb : Model parameters: concrete, EX=2GPa. u=0.3, L=5m, h=0.6m,b=0.4m, F=10000N, M z =50000N·mc : Algorithms:/PREP7 ET,1,PLANE82 MPTEMP,,,,,,,, MPTEMP,1,0 MPDA TA,EX,1,,2e9 MPDA TA,PRXY ,1,,0.3 RECTNG ,0,5,0,0.6, AESIZE,ALL,0.1, MSHKEY ,0 CM,_Y ,AREA ASEL, , , , 1 CM,_Y1,AREA CHKMSH,'AREA' CMSEL,S,_Y AMESH,_Y1 CMDELE,_Y CMDELE,_Y1 CMDELE,_Y2 MSHKEY ,0 /SOL/STATUS,SOLU FLST,2,1,1,ORDE,1 FITEM,2,108 F,P51X,FY ,10000 FLST,2,1,4,ORDE,1 FITEM,2,4DL,P51X, ,ALL,SA VEFLST,2,1,1,ORDE,1FITEM,2,2FK,P51X,MZ,-50000And I captured some diagrams when modeling5:Summery: In this example, I learned how to apply a moment to a structure.We can see that when F=10000N, M=50000N.m, the maximum deflection ( U y=0.01172m ) occurs on the free end of the cantilever beam. While the largest stress(=xσ515858pa) and the largest strain(=ε0.458mm) appear on the fixed end of the beam. So in actual engineering, the contact or the fixed end is the most dangerous positions.Example 2: A long slender block with its dimensions shown in the figure subjected to a uniformed-pressure q on the lateral side. The bottom is fixed and the top is free.Solution1.Application: running train, transforming bends...2.Basic conditions: bottom area fixed, top area freeand lateral loaded by uniformed-pressure.3.Assumptions: simplified to a plane stressproblem and ignore weight and obey small displacement, continuum, isotropic assumptions.4.Analysis: a:Model parameters: concrete, EX=200MPa. u=0.3, b=10m, h=100m, q=1000N/m2b: Algorithms:/PREP7ET,1,PLANE82MPTEMP,,,,,,,,MPTEMP,1,0MPDA TA,EX,1,,2e8MPDA TA,PRXY,1,,0.3RECTNG,0,10,0,100,AESIZE,ALL,1,MSHKEY,0CM,_Y,AREAASEL, , , , 1CM,_Y1,AREACHKMSH,'AREA'CMSEL,S,_YAMESH,_Y1CMDELE,_YCMDELE,_Y1CMDELE,_Y2MSHKEY,0FINISH/SOLFLST,2,1,4,ORDE,1FITEM,2,1DL,P51X, ,ALL,SA VEFLST,2,201,1,ORDE,4FITEM,2,1FITEM,2,222FITEM,2,242FITEM,2,-440F,P51X,FY,-1000FLST,2,201,1,ORDE,3FITEM,2,2FITEM,2,22FITEM,2,-221F,P51X,FY,1000FINISH/POST1FINISHAnd I captured some diagrams when modeling:5. Summery: (1) Though hand calculation, we ge0=xσ,xybq y212=σand)121(222by q xy -=τ. (2) We can see that when q=1000N/m 2, the maximumdeflection ( U x =0.402444m ) occurs on top free area of the block. While the largest stress(=σ143308pa) and the largest strain(=ε0.00392754m) all appear on the fixed bottom of the block. So in actual engineering, the bottom area of the block is the most dangerous area. Because the block is rather huge, so the maximum displacement is quite small comparing to the whole block The biggest strain and stress comparing to the extreme strain and stress of is quite small, which can reach 100GPa.Thanks to anti-symmetry, there are no shear force. When I work on this problem, I still don ’t know how to apply a uniform pressure on an area. In future study, I will look though more books and papers to slove a series problems.Example 3: An annular disc with its dimension on the following figureapplied with a moment M along its outer boundary.Solution:1. Application : Flywheels, discs , wheels, bearings, pulleys...2. Basic Conditions : Fixed around the interSurface area loaded a moment M along the outer-surface area.3. Assumption : Homogeneous and isotropic. Considering weight, obey small displacement, continuum, isotropic assumptions.4. Analysis : a:By the given formulation:⎪⎩⎪⎨⎧=-=2)(ρρρc E k w B dtd where σε21=wWe set a series of parameters like E 1,,1ρσε,; E 2,2ρσε, .... Consider whenmaxρρ>n or minρρ<nthe value of the εand σ.b: Model parameters: soft-steel, EX=2GPa, u=0.3, r 1=100m, r 2=60m. M=5000N.m. c: Algorithms:/PREP7 ET,1,SOLID95MPTEMP,1,0MPDA TA,EX,1,,2e9 MPDA TA,PRXY,1,,0.3 MPTEMP,,,,,,,, MPTEMP,1,0MPDA TA,DENS,1,,2500 CYLIND,100,60,0,10,0,360, SMRT,6SMRT,5SMRT,4SMRT,3MSHAPE,1,3D MSHKEY,0CM,_Y,VOLUVSEL, , , , 1CM,_Y1,VOLU CHKMSH,'VOLU' CMSEL,S,_YVMESH,_Y1 CMDELE,_Y CMDELE,_Y1 CMDELE,_Y2APLOTET,2,SURF154 KEYOPT,2,2,0 KEYOPT,2,4,0 KEYOPT,2,6,0 KEYOPT,2,7,0 KEYOPT,2,11,0 KEYOPT,2,12,0ASEL,S,,,3,4NSLA,,1NPLOTTYPE, 2MAT, 1REAL,ESYS, 0 SECNUM,FLST,5,208,1,ORDE,10 FITEM,5,1FITEM,5,-4FITEM,5,9FITEM,5,-12FITEM,5,-76FITEM,5,121FITEM,5,-180FITEM,5,1803FITEM,5,-1882CM,_Y,NODENSEL, , , ,P51XCM,_Y1,NODE CMSEL,,_Y CMSEL,,_Y1ESURF,0CMSEL,,_Y CMDELE,_Y CMDELE,_Y1 LOCAL,11,1,0,0,0 FLST,2,2734,2,ORDE,2 FITEM,2,1FITEM,2,-2734 EMODIF,P51X,ESYS,11, FINISH/SOLFLST,5,2734,2,ORDE,2 FITEM,5,1FITEM,5,-2734CM,_Y,ELEMESEL, , , ,P51XCM,_Y1,ELEM CMSEL,S,_Y CMDELE,_YSFE,_Y1,2,PRES, ,5000, , , /PSF,PRES,TANX,2 EPLOTALLSELEPOLTFLST,2,2,5,ORDE,2 FITEM,2,5FITEM,2,-6DA,P51X,ALL, EQSLV,PCGSOLVEFINISHAnd I captured some diagrams when modeling:5. Summery: We can see that the maximum deformation U x=0.067926m which is very small compared to the dimension of the disc, the maximum stress occurs on a small zone of the inter area of the disc , which is well coincided with the actual situation of hard computing, From this example, I’ve learned how to apply an moment on a curved surface and how to define several element types and mesh on a volume.。