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材料力学双语教学学习资料3
主讲教师：陈晓峰
第七章应力和应变分析 强度理论
Chapter SevenStre and Strainprincipal strees are not equal to zero.二向应力状态：一个主应力为零的应力状AnalysisStrength Theories
§7–1应力状态概述
§7–1Concepts of the State of Stre
1．一点的应力状态：过一点有无数的截面，这一点的各个截面上应力情况的集合，称为这点的应力状态。
state of stre at a point: There are countle sections through a point.The gathering of
strees in all sections is called the state of stre at this point.2．单元体：构件内的点的代表物，是包围被研究点的无限小的几何体，常用的是正六面体。
Element: Delegate of a point in the member.It is an infinitesimalgeometricbody envelopingthe studied point.In common use it is a correctitudecubicbody.3．主单元体：各侧面上剪应力均为零的单元体。
principalelement :The element in which the shearing strees in side planes are all zero.4．主平面：剪应力为零的截面。
Principal Planes:The planes on which the shearing strees are zero.5．主应力：主平面上的正应力。
principal strees: Normal strees acting on the principle planes.6．主应力排列规定；按代数值大小 123Convention of the order for three principal strees: In magnitudeof the algebraicvalue.7．三向应力状态：三个主应力都不为零的应力状态。
state of the triaxial stre（three dimensional state of stre）:State of stre that all the three
态。
state of the biaxial stre（plane state of stre）: state of stre that one principal stre is equal to zero.单向应力状态：一个主应力不为零的应力状态
state of the uniaxial stre（unidirectional state of stre）:state of stre that one principal stre is not equal to zero.§7–2 二向应力状态分析——解析法 §7–2 Analysis of the State of Plane Stre—
AnalyticalMethod
1.任意斜截面上的应力：strees acting in arbitraryinclined plane
2.正负规定： 截面外法线同向为正；
绕研究对象顺时针转为正；
逆时针为正。
Sign Stipulate： is positive if its direction is the same with one of the external normal line of the section；is positive if it make the element rotate clockwise； A
counterclockwise angle is considered to be positive.§7–3 二向应力状态分析——图解法 §7–3 Analysis of the State of Plane Stre—
GraphicalMethod
1.应力圆 ：stre circle
2.单元体与应力圆的对应关系：
Corresponding relation between the element and stre circle
(1)面上的应力( ， )应力圆上一点( ， )
(2)面的法线 应力圆的半径
(3)两面夹角 两半径夹角2 ；且转向一致。
(1)stre( ， )in planea point( ， )on the stre circumference
(2)normal line of planeradius of the stre circle(3)angle between two sectionsangle 2 between two radiuses；And the direction of rotation is the same.§7–10强度理论概述
§7–10Concepts of Strength Theories1．材料的破坏形式：⑴ 屈服 ⑵ 断裂 ：types of failure of materials：⑴ yield⑵ rupture
2．强度理论：是关于“构件发生强度失效起因”的假说。
theories of strength：some aumptions about the cause of the strength failure of materials.§7–11四种常用强度理论
§7–11 Four Common Used Strength Theories 1.最大拉应力（第一强度）理论:
认为构件的断裂是由最大拉应力引起的。当最大拉应力达到单向拉伸的强度极限时，构件就断了。
theory of the maximum tensile stre（the first strength theory）:
This theory considers the main cause of rupture to be the maximum tensile stre.The member will rupture as the maximum tensile stre reaches the strength limit in axial tension.2.最大伸长线应变（第二强度）理论：
认为构件的断裂是由最大伸长线应变引起的。当最大伸长线应变达到单向拉伸试验下的极限应变时，构件就断了
Theory of the maximum tensile strain（the second strength theory）:
This theory considers the main cause of rupture to be the maximum tensile strain.The member will rupture as the maximum tensile strain reaches the limit strain in axial tension3.最大剪应力（第三强度）理论：
认为构件的屈服是由最大剪应力引起的。当最大剪应力达到单向拉伸试验的极限剪应力时，构件就破坏了。
Theory of the maximum shearing stre（the third strength theory）:
This theory considers the main cause of rupture to be the maximum shearing stre.The member will rupture as the maximum shearing stre reaches the limit shearing stre in axial tension.4．畸变能密度（第四强度）理论：
认为构件的屈服是由形状改变比能引起的。当形状改变比能达到单向拉伸试验屈服时的形状改变比能时，构件就破坏了。
Theory of the distortionalenergy density(the fourth strength theory）:
This theory considers the main cause of yield to be the distortional strain energy.The member will rupture as the distortional strain energy reaches the distortional strain energy of yield in axial tension
5.相当应力：equivalentstre 6.断裂准则：criterionof rupture屈服准则:criterion of yield
第八章组合变形
Chapter Eight Composite Deformation
§8–1组合变形和叠加原理 §8–1Composite Deformation and
Superposition Principle
1.叠加原理的步骤：
The steps for principle of superposition
①外力分析：外力向形心简化并沿形心主惯性轴分解
Analysis of external forces：External forces are reduced along the centroidof section and resolved along principal axes of inertia.②内力分析：求每个外力分量对应的内力方程和内力图，确定危险面。
Analysis of internal forces：Determine the internal force equation and its diagram corresponding to each external force component and the critical section.③应力分析：画危险面应力分布图，叠加，建立危险点的强度条件。Analysis of strees：Plot the distribution diagram of the stre in the critical section，do the superposition of the strees and establish the strength condition of the critical point.平均应力。
Critical stre : average stre in the cro section §8–2拉伸(或压缩)与弯曲的组合of the compreive column in the critical state.§8–2Composite Deformation of Tension
（or Compreion）and Bending 2.柔度（或长细比）：flexibility
(or slenderne ratio)§8–4扭转与弯曲的组合§8–4 Combination of Torsion and Bending大柔度杆的临界应力由欧拉公式来求;小 柔度杆的临界应力就是它的屈服极限;中柔度
杆的临界应力由经验公式来求第九章压杆稳定
The critical stre of the large flexibility column Chapter NineStabilization
is calculated by Euler’s formula.of Compreive Columns
The critical stre of small flexibility column is
its yield limit.§9–1压杆稳定性的概念
The critical stre of the middle flexibility §9–1 Concepts of Stability of
column may be determined by the empirical Compreed Columns
formula.1.失稳：lo of stability
2.稳定与不稳定平衡:
stable and instable equilibrium§9–5压杆的稳定校核3.临界压力： critical preure§9–5Stability Check of Compreed Column1.稳定安全因数：safety coefficient of stability §9–2两端铰支细长压杆的临界压力 2.稳定条件：stability condition§9–2 Critical Preure of The SlenderCompreed Column With Two Hinged Ends §9–6提高压杆稳定性的措施1.欧拉公式:Euler’s formula §9–6 Method to Improve Stability of 2.理想压杆：材料绝对理想；轴线绝对直；压Compreed Column力绝对沿轴线作用。1.选择合理的截面形状:choose reasonable Ideal compreive column: the material is section of the column: absolutelyideal；the axis is 2.改变压杆的约束条件:change constraint
condition of the column absolutely straight；the compreive force is
absolutely along the axis of the column.3.合理选择材料: choose reasonable material.§9–3其它支座条件下细长压杆的临界压力 第十章 动载荷 §9–3Critical Preure of The Slender Compreed Chapter TenDynamic Load
Column With Other End Conditions
§10–1 概述 1.长度系数（约束系数）：length coefficient
（or constraint coefficient）§10–1Introduction
2.两端铰支: two hinged ends 1.静载荷：static loads
The loads don’t change with time(or change 3.一端固定另端自由:
one fixed end and one free end very stably and slowly)and acceleration of each
member is zero or may be neglected4.两端固定: two fixed ends
5.一端固定另端铰支:2.动载荷：dynamic loadsone fixed end and one hinged end The loads change sharply with time and thevelocity of the member changes obviously
§9–4欧拉公式的适用范围经验公式
§9–4Application Range of Euler’s Formula §9–2动静法的应用Empirical Formula§9–2 Application of The Method of 1.临界应力: 压杆处于临界状态时横截面上的Kinetic Statics
1.惯性力：inertiaforce
2.动荷系数：dynamic load coefficient
3.达朗伯原理：处于不平衡状态的物体，存在惯性力，惯性力的方向与加速度方向相反，惯性力的数值等于加速度与质量的乘积。只要在物体上加上惯性力，就可以把动力学问题在形式上作为静力学问题来处理，这就是动静法。D’Alembert’s principle: There is inertial force on the body in unequilibrium.The direction of the inertial force is opposite to the acceleration of the body and the magnitude of the inertial force is the product of the ma and the acceleration of the body.After the inertial force is applied on the body the dynamic problem may be dealt with the static problem in form, which is called the method of kinetic statics.Exercise 1: The state of stre at a point as shown.(unit:Mpa),try to determine the three principal strees.(either by analytical method or by graphical method).And compute the
equivalent stre of the fourth strength theory.Exercise 4:A hollow circular shaft is shown in the figure.Its inside diameter is d=24mm and its outside diameter is D=30mm.The diameters of pulley B and D are respectively D1=400mm and D2=600mm，P1=600N，[]=100MPa.Try to check the strength of the shaft with the third strength theory.Exercise 5: A compreed rod produces the bending deformation due to the lo of stability.A beam produces the bending deformation due to the action of transverse forces.What are the differences of the two in nature?
Exercise 2: A circular rod made of cast iron is subjected to the loads T=7kNm, P=50kN as shown in the figure.Its diameter is d=0.1m, []=40MPa.Try to check the strength of the rod according to the theory of the first strength.Exercise 3: The cro-section area of the
square-section rod is reduced half at the section mn.Try to determine the maximum tensile stre at the section mn due to the axial force P.
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