<4D F736F F D20B9A4B3CCC1A6D1A7D7EED0C25FCEC4C7B05F2E646F63>

Transcription

1

2 Engineering Mechanics 工程力学 Kaifu WANG 王开福 著 北 京

3 Synopsis 内容简介 This book consists of statics, kinematics, kinetics, and mechanics of materials. The main contents of the book include: statics of a particle and of a rigid body, friction, kinematics of a particle and of a rigid body in plane motion, resultant motion of a particle, kinetics of a particle and of a rigid body in plane motion, mechanical properties of materials, aial tension and compression of bars, torsion of shafts, bending of beams, stress analysis and theories of strength, combined loadings, and stability of columns. The book can be used as an English, Chinese, or bilingual tetbook of engineering mechanics for the student majoring in aeronautical, mechanical, civil, and hydraulic engineering. 本书由静力学 运动学 动力学和材料力学组成 主要内容包括 : 质点静力学和刚体静力学 摩擦 质点运动学和刚体平面运动学 质点合成运动 质点动力学和刚体平面动力学 材料机械性能 杆的轴向拉伸与压缩 轴的扭转 梁的弯曲 应力分析与强度理论 组合载荷和压杆稳定 本书可作为高等院校航空 机械 土木和水利等学科专业学生的英文 中文或双语工程力学教材 图书在版编目 (CIP) 数据 工程力学 =Engineering Mechanics / 王开福著. 北京 : 科学出版社,2012 ISBN Ⅰ.1 工 Ⅱ.1 王 Ⅲ. 1 工程力学 双语教学 高等学校 教材 - 汉 英 IV. 1TB12 中国版本图书馆 CIP 数据核字 (2012) 第 号 责任编辑 : 尚雁罗吉 / 责任校对 : 黄海 责任印制 : 赵德静 / 封面设计 : 许瑞 科学出版社发行各地新华书店经销 * 2012 年 9 月第一版开本 :( ) 1/ 年 9 月第一次印刷印张 :33 1/2 字数 : 定价 :76.00 元 ( 如有印装质量问题, 我社负责调换 )

4 Preface 1. bjectives Engineering mechanics is a required subject for the student majoring in aeronautical, mechanical, civil, and hydraulic engineering and usually taught during the sophomore year. This book is intended to provide the student with the theory and application of engineering mechanics. It mainly includes theoretical mechanics and mechanics of materials; the former consisting of the first 10 chapters is devoted to the equilibrium and motion of bodies subjected to the action of forces, while the latter covering the other 10 chapters contributes to the analysis and design of structural members subjected to tension/compression, torsion, bending, and combined loadings. 2. Features The book is written by using English and Chinese, respectively; the English part of the book can be used as an English tetbook of engineering mechanics, while the Chinese part can be used as a Chinese tetbook of engineering mechanics. The study of engineering mechanics is based on the understanding of concepts and the use of principles. Eamples are thus presented in the tet in order to illustrate how these concepts and principles can be used in practical situation. Problem solving is of vital importance in the learning process of engineering mechanics. Hence problems for homework assignments are offered at the end of each chapter. The International System of Units is utilized in the entire book, and the National Standard of China is used as the design specification. 3. Contents Engineering mechanics consists of statics, kinematics, kinetics, and mechanics of materials; Statics is the study of bodies at rest or in equilibrium, kinematics deals with the geometry of the motion without regard to the forces acting on bodies, kinetics is with the relation between the motion of bodies and the forces acting on bodies, and mechanics of materials is to study the ability of structures and machines to resist failure. The book is organized into twenty chapters and five appendies. Chapter 1 is an introduction to the fundamental concepts and general principles of theoretical mechanics. Chapter 2 discusses the resultant and equilibrium of concurrent forces acting on a particle. In Chapter 3 the reduction and equivalence of a force system acting on a rigid body are discussed,

5 ii Engineering Mechanics 工程力学 and in Chapter 4 the equilibrium of a rigid body, as well the internal force of a planar truss, is considered. The concepts of both sliding friction and rolling resistance are introduced in Chapter 5. The velocity and acceleration of a particle are analyzed in Chapter 6. Chapter 7 deals with the velocity and acceleration of a rigid body in translation, rotation, and general plane motion. The resultant motion of a particle is studied in Chapter 8. Chapter 9 and Chapter 10 are on the kinetics of a particle and of a rigid body in plane motion. The fundamental concepts of mechanics of materials are introduced in Chapter 11. Chapter 12 describes the mechanical properties of materials subjected to tension or compression. The stress and deformation are discussed in Chapter 13 for a bar subjected to tension and compression, and in Chapter 14 for a shaft subjected to torsion, respectively. The internal forces, stresses, and deformations of a beam in bending are, respectively, dealt with in Chapter 15, 16, and 17. The plane state of stresses and the failure criteria of materials are presented in Chapter 18. The stress analysis for a member subjected to combined loadings is considered in Chapter 19. In Chapter 20, the buckling of a column subjected to a centric compressive load is analyzed. The concepts of center of gravity, centroid, and mass moment of inertia are introduced in Appendi I and II, respectively. The geometrical properties of areas and of rolled-steel shapes are, respectively, presented in Appendi III and IV. The deflections and slopes of commonly-used beams are listed in Appendi V. The book can be used as an English, Chinese, or bilingual tetbook of engineering mechanics for the student majoring in aeronautical, mechanical, civil, and hydraulic engineering. Kaifu WANG Nanjing, June 2012

6 Contents 目录 iii 前 言 1. 目标工程力学是航空 机械 土木和水利工程等学科专业学生的必修课, 通常在大二学年讲授 本书旨在向学生传授工程力学的理论及其应用, 主要包括理论力学和材料力学 : 前者由前 10 章组成, 研究受力作用物体的平衡和运动 ; 后者包含其余 10 章, 研究受拉压 扭转 弯曲和组合载荷作用构件的分析和设计 2. 特点本书由英语和汉语分别撰写 : 英语部分可作为工程力学的英文教材, 而汉语部分则可作为工程力学的中文教材 工程力学的学习需要基于概念的理解和原理的应用, 因此书中给出了例题以说明怎样把这些概念和原理应用于实际情况 问题求解在工程力学的学习过程中显得非常重要, 因此在每章后面提供了课外习题以供学生练习 全书采用国际单位制, 并采用我国国标作为设计规范 3. 内容工程力学由静力学 运动学 动力学和材料力学组成 : 静力学研究物体的静止与平衡 ; 运动学在不涉及作用力的情况下研究物体的运动 ; 动力学研究物体的运动与作用力之间的关系 ; 材料力学研究结构或机械抵抗失效的能力 全书由 20 章正文和 5 个附录组成 第 1 章介绍理论力学的基本概念与普遍原理 第 2 章讨论作用于质点上的汇交力系的合成与平衡 第 3 章讨论作用于刚体上的力系的简化与等效 第 4 章考虑刚体的平衡以及平面桁架的内力 第 5 章介绍滑动摩擦与滚动摩阻的概念 第 6 章分析质点的速度与加速度 第 7 章涉及平移 转动和一般平面运动刚体的速度与加速度 第 8 章研究质点合成运动 第 9 章和第 10 章分别研究质点动力学和刚体平面动力学 第 11 章介绍材料力学的基本概念 第 12 章描述材料在拉压时的机械性能 第 13 章和第 14 章分别讨论拉压杆和扭转轴的应力与变形 第 15 章 16 章和 17 章分别涉及弯曲梁的内力 应力和变形 第 18 章介绍平面应力状态与材料失效准则 第 19 章考虑在组合载荷作用下构件的应力分析 第 20 章分析压杆的失稳 附录 I 和 II 分别介绍重心 形心和转动惯量的概念 附录 III 和 IV 分别介绍截面几何性质和型钢几何性质 附录 V 列出常用梁的挠度与转角 本书可作为高等院校航空 机械 土木和水利等学科专业学生的英文 中文或双语工程力学教材 王开福 2012 年 6 月于南京

7 Contents 目录 v Contents 目 录 Preface 前言 English Edition Chapter 1 Fundamental Concepts of Theoretical Mechanics What Is Theoretical Mechanics Basic Concepts General Principles...5 Chapter 2 Statics of Particle System of Concurrent Forces Resultant of Coplanar Concurrent Forces Equilibrium of Coplanar Concurrent Forces Resultant of Spatial Concurrent Forces Equilibrium of Spatial Concurrent Forces...19 Problems...20 Chapter 3 Reduction of Force System Moment of Force about Point Moment of Force about Given Ais Principle of Moments Components of Moment of Force about Point Moment of Couple Resultant of Couples Equivalence of Force Acting on Rigid Body Reduction of Force System...33 Problems...34 Chapter 4 Statics of Rigid Body Equilibrium of Rigid Body Equilibrium of Two-Dimensional Rigid Body Two-Force and Three-Force Bodies Planar Trusses Equilibrium of Three-Dimensional Rigid Body...47 Problems...48

8 vi Engineering Mechanics 工程力学 Chapter 5 Friction Types of Friction Sliding Friction Angles of Friction Problems Involving Sliding Friction Rolling Resistance...59 Problems...60 Chapter 6 Kinematics of Particle Motion of Particle Motion of Particle Represented by Vector Motion of Particle Represented by Rectangular Coordinates Motion of Particle Represented by Natural Coordinates...65 Problems...67 Chapter 7 Kinematics of Rigid Body in Plane Motion Plane Motion of Rigid Body Translation Rotation about Fied Ais General Plane Motion...74 Problems...81 Chapter 8 Resultant Motion of Particle Motion of Particle Rates of Change of Vector Resultant of Velocities Resultant of Accelerations...87 Problems...91 Chapter 9 Kinetics of Particle Newton s Second Law of Motion Equations of Motion of Particle Method of Inertia Force for Particle in Motion Method of Work and Energy for Particle in Motion Method of Impulse and Momentum for Particle in Motion Problems Chapter 10 Kinetics of Rigid Body in Plane Motion Motion for System of Particles Motion of Mass Center of System of Particles Motion of System of Particles about Its Mass Center Equations of Motion for Rigid Body in Plane Motion Method of Inertia Force for Rigid Body in Plane Motion Method of Work and Energy for Rigid Body in Plane Motion

9 Contents 目录 vii 10.7 Method of Impulse and Momentum for Rigid Body in Plane Motion Problems Chapter 11 Fundamental Concepts of Mechanics of Materials What Is Mechanics of Materials Basic Assumptions of Materials Eternal Forces Internal Forces Stresses Strains Deformations of Members Problems Chapter 12 Mechanical Properties of Materials Tensile or Compressive Test Tension of Low-Carbon Steel Ductile and Brittle Materials Stress-Strain Curve of Ductile Materials without Distinct Yield Point Percent Elongation and Percent Reduction in Area Hooke s Law Mechanical Properties of Materials in Compression Chapter 13 Aial Tension and Compression of Bars Definition of Aial Tension and Compression Aial Force Normal Stress on Cross Section Saint-Venant s Principle Normal and Shearing Stresses on blique Section Normal Strain Deformation of Aially Loaded Bar Statically Indeterminate Aially Loaded Bar Design of Aially Loaded Bar Stress Concentrations Problems Chapter 14 Torsion of Shafts Definition of Torsion Twisting Moment Hooke s Law in Shear Shearing Stress on Cross Section of Circular Shaft Normal and Shearing Stresses on blique Section of Circular Shaft Angle of Twist Statically Indeterminate Circular Shaft...166

10 viii Engineering Mechanics 工程力学 14.8 Design of Circular Shaft Problems Chapter 15 Shearing Force and Bending Moment of Beams Definition of Bending Shearing-Force and Bending-Moment Diagrams Relations between Distributed Load, Shearing Force, and Bending Moment Relations between Concentrated Load, Shearing Force, and Bending Moment..178 Problems Chapter 16 ormal Stress and Shearing Stress in Beams Types of Bending Normal Stresses on Cross Section in Pure Bending Normal and Shearing Stresses on Cross Section in Transverse-Force Bending Design of Prismatic Beams in Bending Problems Chapter 17 Deflection and Slope of Beams Deformation of Beams Method of Integration Method of Superposition Statically Indeterminate Beams Problems Chapter 18 Stress Analysis and Theories of Strength State of Stress Transformation of Plane Stress Principal Stresses for Plane Stress Maimum Shearing Stress for Plane Stress Stresses in Pressure Vessels Generalized Hooke s Law Theories of Strength under Plane Stress Problems Chapter 19 Combined Loadings Definition of Combined Loadings Stress in Bar Subject to Eccentric Tension or Compression Stress in I-Section Beam Subject to Transverse-Force Bending Stress in Beam Subject to Bending and Aial Tension/Compression Stress in Shaft Subject to Torsion and Bending Problems Chapter 20 Stability of Columns Definition of Buckling Critical Load of Long Slender Columns under Centric Loadwith Pin

11 Contents 目录 i Supports Critical Load of Long Slender Columns under Centric Load with ther Supports Critical Stress of Long Slender Columns under Centric Load Critical Stress of Intermediate Length Columns under Centric Load Design of Columns under Centric Load Problems References Appendi I Centers of Gravity and Centroids I.1 Center of Gravity and Centroid of Plate I.2 Center of Gravity and Centroid of Composite Plate I.3 Center of Gravity and Centroid of 3D Body I.4 Center of Gravity and Centroid of 3D Composite Body Appendi II Mass Moments of Inertia II.1 Moment of Inertia and Radius of Gyration II.2 Parallel-Ais Theorem Appendi III Geometrical Properties of Areas III.1 First Moment and Centroid III.2 First Moment and Centroid of Composite Area III.3 Moment of Inertia and Polar Moment of Inertia III.4 Radius of Gyration and Polar Radius of Gyration III.5 Product of Inertia III.6 Parallel-Ais Theorem III.7 Moment of Inertia and Polar Moment of Inertia of Commonly-Used Areas Appendi IV Geometrical Properties of Rolled-Steel Shapes IV.1 I Steel IV.2 Channel Steel IV.3 Equal Angle Steel IV.4 Unequal Angle Steel Appendi V Deflections and Slopes of Beams 中文版 第 1 章理论力学的基本概念 什么是理论力学 基本概念 普遍原理 第 2 章质点静力学 汇交力系

12 Engineering Mechanics 工程力学 2.2 平面汇交力的合成 平面汇交力的平衡 空间汇交力的合成 空间汇交力的平衡 习题 第 3 章力系的简化 力对点之矩 力对轴之矩 力矩定理 力对点之矩的分量 力偶矩 力偶的合成 作用于刚体上力的等效 力系的简化 习题 第 4 章刚体静力学 刚体平衡 二维刚体的平衡 二力和三力物体 平面桁架 三维刚体的平衡 习题 第 5 章摩擦 摩擦分类 滑动摩擦 摩擦角 含有滑动摩擦的问题 滚动摩阻 习题 第 6 章质点运动学 质点的运动 质点运动的矢量表示 质点运动的直角坐标表示 质点运动的自然坐标表示 习题 第 7 章刚体平面运动学 刚体平面运动 平移 定轴转动

13 Contents 目录 i 7.4 一般平面运动 习题 第 8 章质点合成运动 质点的运动 矢量的变化率 速度的合成 加速度的合成 习题 第 9 章质点动力学 牛顿第二运动定律 质点运动方程 运动质点的惯性力法 运动质点的功 - 能法 运动质点的冲量 - 动量法 习题 第 10 章刚体平面动力学 质点系的运动 质点系质心的运动 质点系相对质心的运动 平面运动刚体的运动方程 平面运动刚体的惯性力法 平面运动刚体的功 - 能法 平面运动刚体的冲量 - 动量法 习题 第 11 章材料力学的基本概念 什么是材料力学 材料的基本假设 外力 内力 应力 应变 构件的变形 习题 第 12 章材料机械性能 拉伸或压缩试验 低碳钢拉伸 塑性和脆性材料 没有明显屈服点的塑性材料的应力 - 应变曲线 伸长率和断面收缩率

14 ii Engineering Mechanics 工程力学 12.6 胡克定律 材料压缩机械性能 第 13 章杆的轴向拉伸与压缩 轴向拉伸与压缩的定义 轴力 横截面上的正应力 圣维南原理 斜截面上的正应力和剪应力 线应变 轴向加载杆的变形 静不定轴向加载杆 轴向加载杆的设计 应力集中 习题 第 14 章轴的扭转 扭转的定义 扭矩 剪切胡克定律 圆轴横截面上的剪应力 圆轴斜截面上的正应力和剪应力 扭转角 静不定圆轴 圆轴的设计 习题 第 15 章梁的剪力与弯矩 弯曲的定义 剪力和弯矩图 分布载荷 剪力和弯矩之间的关系 集中载荷 剪力和弯矩之间的关系 习题 第 16 章梁的正应力与剪应力 弯曲的类型 纯弯曲梁横截面上的正应力 横力弯曲梁横截面上的正应力和剪应力 等截面弯曲梁的设计 习题 第 17 章梁的挠度与转角 梁的变形 积分法

15 Contents 目录 iii 17.3 叠加法 静不定梁 习题 第 18 章应力分析与强度理论 应力状态 平面应力状态变换 平面应力状态的主应力 平面应力状态的最大剪应力 压力容器中的应力 广义胡克定律 平面应力状态强度理论 习题 第 19 章组合载荷 组合载荷的定义 偏心拉伸或压缩杆的应力 横力弯曲工字梁的应力 弯曲与拉压梁的应力 扭转与弯曲轴的应力 习题 第 20 章压杆稳定 失稳的定义 两端铰支中心加载细长压杆的临界载荷 其他支撑中心加载细长压杆的临界载荷 中心加载细长压杆的临界应力 中心加载中长压杆的临界应力 中心加载压杆的设计 习题 参考文献 附录 I 重心与形心 I.1 薄板的重心与形心 I.2 组合薄板的重心与形心 I.3 三维物体的重心与形心 I.4 三维组合物体的重心与形心 附录 II 转动惯量 II.1 转动惯量与回转半径 II.2 平行移轴定理 附录 III 截面几何性质 III.1 静矩与形心 III.2 组合截面的静矩与形心

16 iv Engineering Mechanics 工程力学 III.3 惯性矩与极惯性矩 III.4 惯性半径与极惯性半径 III.5 惯性积 III.6 平行移轴定理 III.7 常用截面的惯性矩与极惯性矩 附录 IV 型钢几何性质 IV.1 工字钢 IV.2 槽钢 IV.3 等边角钢 IV.4 不等边角钢 附录 V 常用梁的挠度与转角

17 English Edition 1

18 2 Engineering Mechanics

19 Chapter 1 Fundamental Concepts of Theoretical Mechanics 3 Chapter 1 Fundamental Concepts of Theoretical Mechanics 1.1 What Is Theoretical Mechanics Engineering mechanics is the science that applies the principles of mechanics to the analysis and design of engineering structures and machines. It usually includes theoretical mechanics and mechanics of materials. Theoretical mechanics is the study of equilibrium or motion of bodies subjected to the action of forces, and consists of statics, kinematics and dynamics. Statics is the study of bodies at rest or in equilibrium; kinematics treats the geometry of the motion without regard to the forces acting on bodies; and kinetics deals with the relation between the motion of bodies and the forces acting on bodies. In theoretical mechanics, bodies are assumed to be perfectly rigid. Though actual structures and machines are never absolutely rigid and deform under the action of forces, these deformations are usually small and do not affect the state of equilibrium or motion of the structures and machines under consideration. 1.2 Basic Concepts 1. Length Length is used to locate the position of a point in space. The position of a point can be defined by three lengths measured from a certain reference point in three given directions. 2. Time Time is used to represent a nonspatial continuum in which events occur in irreversible succession from the past through the present to the future. To define an event, it is not sufficient to indicate its position in space. The time of the event should be given. 3. Mass Mass is used to characterize the quantity of matter that a body contains. The mass of a body is not dependent on gravity and therefore is different from but proportional to its weight. Two bodies of the same mass, for eample, will be attracted by the earth in the same manner; they will also offer the same resistance to a change in velocity.

20 4 Engineering Mechanics 4. Force Force is used to represent the action of one body on another. A force tends to produce an acceleration of a body in the direction of its application. The effect of a force is completely characterized by its magnitude, direction, and point of application. 5. Particle If the size and shape of a body do not affect the solution of the specific problem under consideration, then this body can be idealized as a particle, i.e., a particle has a mass, but its size and shape can be neglected. For eample, the size and shape of the earth is insignificant compared to the size and shape of its orbit, and therefore the earth can be modeled as a particle when studying the orbital motion of the earth. 6. Rigid Body A rigid body can be considered as a combination of a large number of particles in which all the particles occupy fied positions with respect to each other within the body both before and after the action of forces, i.e., a rigid body is defined as one which does not deform when it is subjected to the action of forces. 7. Scalars Scalars possess only magnitude, e.g., length, time, mass, work, energy. Scalars are added by algebraic methods. 8. Vectors Vectors possess both magnitude and direction (direction is understood to includes both the inclination angle that the line of action makes with a given reference line and the sense of the vector along the line of action), e.g., force, displacement, impulse, momentum. Vectors are added by the parallelogram law. 9. Free Vectors A free vector can be moved anywhere in space provided it remains the same magnitude and direction. 10. Sliding or Slip Vectors A sliding or slip vector can be moved to any point along its line of action. 11. Fied or Bound Vectors A fied or bound vector must remain at the same point of application.

21 Chapter 1 Fundamental Concepts of Theoretical Mechanics 5 1. Parallelogram Law 1.3 General Principles This law states that two forces acting on a particle can be replaced by a single resultant force obtained by drawing the diagonal of the parallelogram which has sides equal to the given forces. For eample, two forces and F 2 acting on a particle, Fig. 1.1a, can be replaced by a single force R, Fig. 1.1b, which has the same effect on the particle and is called the resultant force of the forces and F 2. The resultant force R can be obtained by drawing a parallelogram using and F 2 as two adjacent sides of the parallelogram. The diagonal that passes through represents the resultant force R, i.e., R= F1+ F 2. This method for finding the resultant force of two forces is known as the parallelogram law. F 2 F 2 R (a) (b) Fig. 1.1 From the parallelogram law, an alternative method for determining the resultant force of two forces by drawing a triangle, Fig. 1.2b, can be obtained. The resultant force R of the forces and F 2 can be found by arranging and F 2 in tip-to-tail fashion and then connecting the tail of with the tip of F 2, i.e., R= F1+ F 2. This is known as the triangle rule. F 2 R F 2 (a) (b) Fig Principle of Transmissibility This principle states that the state of equilibrium or motion of a rigid body will remain unchanged if one force acting at a given point of the rigid body is replaced by another force of the same magnitude and same direction, but acting at a different point, provided that the two forces have the same line of action. For eample, a force F, Fig. 1.3a, acting on a given point of a rigid body can be

22 6 Engineering Mechanics replaced by a force F, Fig. 1.3b, of the same magnitude and same direction, but acting at a different point on the same line of action. The two forces F and F have the same effect on the rigid body and are said to be equivalent. This principle shows that the effect of a force on a rigid body remains unchanged provided the force acting on the rigid body is moved along its line of action. Thus forces acting on a rigid body are sliding vectors. F F = (a) (b) Fig Newton s First Law This law states that if the resultant force acting on a particle is zero, then the particle will remain at rest (if originally at rest) or will move with constant velocity in a straight line (if originally in motion). 4. Newton s Second Law This law states that if the resultant force acting on a particle is not zero, then the particle will have an acceleration proportional to the magnitude of the resultant force and in the direction of this resultant force. This law can be epressed mathematically as F = ma (1.1) where F, m, and a are, respectively, the resultant force acting on the particle, the mass of the particle, and the acceleration of the particle. 5. Newton s Third Law This law states that the forces of action and reaction between two bodies in contact have the same magnitude (equal), same line of action (collinear), and opposite sense (direction). 6. Newton s Law of Gravitation This law states that two particles are mutually attracted by equal and opposite forces. The magnitude of the two forces can be given by mm 1 2 F = G (1.2) r 2

23 Chapter 1 Fundamental Concepts of Theoretical Mechanics 7 where F is the force of gravitation between the two particles, G is the universal constant of gravitation, m 1 and m 2 are, respectively, the mass of each of the two particles, and r is the distance between the two particles. When a particle is located on or near the surface of the earth, the force eerted by the earth on the particle is defined as the weight of the particle. Taking m 1 equal to the mass M of the earth, m 2 equal to the mass m of the particle, and r equal to the radius R of the earth, and letting M g = G (1.3) R 2 where g is the acceleration of gravity, then the magnitude of the weight of the particle can be given by W = mg (1.4) The value of g is approimately equal to 9.81 m/s 2 in SI units, as long as the particle is located on or near the surface of the earth.

24 8 Engineering Mechanics Chapter 2 Statics of Particle 2.1 System of Concurrent Forces A body under consideration can be idealized as a particle if its size and shape are able to be neglected. All the forces acting on this particle can be assumed to be applied at the same point and will thus form a system of concurrent forces. 2.2 Resultant of Coplanar Concurrent Forces A coplanar system of concurrent forces consists of concurrent forces that lie in one plane. 1. Graphical Method for Resultant of Forces The resultant force of a coplanar system of concurrent forces acting on a particle can be obtained by using the graphical method. If a particle is acted upon by three or more coplanar concurrent forces, the resultant force can be obtained by the repeated applications of the triangle rule. Considering that a particle is acted upon by coplanar concurrent forces, F 2, and F 3, Fig. 2.1a, the resultant force R of these forces can be obtained graphically by arranging all the given forces in tip-to-tail fashion and connecting the tail of the first force with the tip of the last one, Fig. 2.1b. This method is known as the polygon rule. R F 3 F 3 F 2 F 2 (a) (b) Fig. 2.1 We thus conclude that a coplanar system of concurrent forces acting on a particle can be replaced by a resultant force through the concurrence, and that the resultant force is equal to the vector sum of the given coplanar concurrent forces, i.e., R F F F F (2.1) = = Eample 2.1 Two rods, AC and AD, are attached at A to column AB, Fig. E2.1a. Knowing that the

25 Chapter 2 Statics of Particle 9 force in the left-hand rod is = 150 N, and that the inclination angles of the rods are θ 1 = 30 and θ 2 = 15, using the graphical method determine (a) the force F 2 in the right-hand rod if the resultant of the forces eerted by the rods on the column is to be vertical, (b) the corresponding magnitude of the resultant. R C θ 1 D C F θ F 2 2 θ1 2 2 D θ A A B B (a) (b) Fig. E2.1 Solution The forces and F 2 acting at A can be replaced by a resultant force R from the parallelogram law, Fig. E2.1b. Considering the shaded triangle shown in Fig. E2.1b and using the law of sines, we have 2 = F = R sin(90 θ ) sin(90 θ ) sin( θ + θ ) Using = 150 N, θ 1 = 30, and θ 2 = 15, we obtain sin(90 θ ) sin( θ + θ ) F = F = N, R= F = N sin(90 θ2) sin(90 θ2) Eample 2.2 Two rods, AC and AD, are attached at A to column AB, Fig. E2.2. Knowing that the forces in the rods are = 120 N and F 2 = 100 N, and that the inclination angles of the rods are θ 1 = 35 and θ 2 = 20, using the graphical method determine the resultant force. Solution The force triangle drawn based on the triangle rule is shown in Fig. E2.2b. Using the law of cosines and the law of sines, we have = θ1+ θ2 R F F 2FF cos[180 ( )] F2 R = sin( θ θ) sin[180 ( θ + θ )] 1 1 2

26 10 Engineering Mechanics C θ 2 F 2 θ 1 D A θ 2 R F 2 θ C θ 1 B (a) (b) Fig. E2.2 Using = 120 N, F 2 = 100 N, θ 1 = 35, and θ 2 = 20, we can obtain R= F1 + F2 + 2FF 1 2cos( θ1+ θ2) = N, θ = θ1 arcsin[ F sin( θ1+ θ2)] = R 2. Components of Force Two or more forces acting on a particle can be replaced by a single force which has the same effect on the particle. Conversely, one force acting on a particle can also be replaced by two or more forces which, together, have the same effect on the particle. For eample, a force F acting on a particle, Fig. 2.2, can be replaced by and F 2. and F 2 are called the vector components of F, and the process of substituting and F 2 for F is called the resolution of a force into components. Clearly, for a force F there eist an infinite number of possible sets of vector components, Fig. 2.2b. F 2 F F 2 F (a) (b) 3. Rectangular Components of Force Fig. 2.2 It is often convenient to resolve a force into components which are perpendicular to each other. For eample, a force F acting on a particle, Fig. 2.3, can be resolved into two vector components F and F y, respectively along the and y aes, where F and F y are called the rectangular components of the force F. Thus we have F = F + F (2.2) y

27 Chapter 2 Statics of Particle 11 y F y F F Fig. 2.3 By introducing two unit vectors i, and j, directed respectively along the positive, and y aes, Fig. 2.4, F can also be epressed as y F j y F j i F i Fig. 2.4 F = F i+ F j (2.3) where F and F y are called the scalar components of the force F respectively along the and y aes. F and F y may be positive or negative, respectively depending upon the sense of F and F y. F is positive when F has the same sense as the positive ais and is negative when F has the opposite sense. A similar conclusion can be drawn regarding the signs of F y. Denoting by F the magnitude of the force F and by θ the angle of F from the positive ais, Fig. 2.5, we can epress the scalar components F and F y of F as follows: y y F j y F θ F i Fig. 2.5 F = Fcos θ, F = Fsinθ (2.4) y 4. Analytical Method for Resultant of Forces Using the graphical method to determine the resultant force of coplanar concurrent forces

28 12 Engineering Mechanics often requires etensive geometric or trigonometric calculation, especially for finding the resultant force of three or more coplanar concurrent forces. Instead, problems of this type are easily solved by using the analytical method. Considering coplanar concurrent forces, F 2, and F 3 acting on a particle, Fig. 2.6, then the resultant force R of these forces can be epressed, using the graphical method, as y F 2 F 3 Fig. 2.6 R= F1+ F2 + F 3 (2.5) Resolving each force, including the resultant force, into its rectangular components, we write from which it follows that R i+ R j = ( F + F + F ) i+ ( F + F + F ) j (2.6) y y 2y 3y R = F + F + F, R = F + F + F (2.7) y 1y 2y 3y We thus conclude that the scalar components along the and y aes of the resultant force of coplanar concurrent forces acting on a particle are respectively equal to the algebraic sums of the scalar components on the same ais of the given forces, i.e., (2.8) R = F, R = F y y The magnitude R of the resultant force and the angle θ that the resultant force forms with the positive ais can be written by 2 2 R R= R, arctan y + Ry θ = (2.9) R Eample 2.3 Two rods, AC and AD, are attached at A to column AB, Fig. E2.3a. Knowing that the force in the left-hand rod is = 150 N, and that the inclination angles of the rods are θ 1 = 30 and θ 2 = 15, using the analytical method determine (a) the force F 2 in the right-hand rod if the resultant of the forces eerted by the rods on the column is to be vertical, (b) the corresponding magnitude of the resultant. Solution Establishing the system of reference as shown in Fig. E2.3b, then the scalar components of the resultant force can be epressed as

29 Chapter 2 Statics of Particle 13 y R C C D θ F θ D θ F θ A A B B (a) (b) Fig. E2.3 R = F cosθ F cos θ, R = F sinθ + F sinθ y Since the resultant of the forces eerted by the rods on the column is vertical, i.e., R = 0, we have cosθ F = F = N, R= R = F sin + F sin = N y 2 θ2 1 θ1 cosθ2 Eample 2.4 A block subjected to three forces is located on a surface of inclination angle α = 25, Fig. E.2.4a. Assuming that θ = 40, and that = 150 N, F 2 = 250 N, and F 3 = 200 N, using the analytical method determine the resultant of the forces acting on the block. R F 2 F 2 F 3 θ θ F 3 θ y θ R θ α α (a) (b) Fig. E2.4 Solution Establishing the system of reference as shown in Fig. E2.4b, then the scalar components of the resultant force can be epressed as R = F + F cosθ F sinθ = N, R = F sinθ + F cosθ = N y 2 3 Using the scalar components above, we can obtain the magnitude and direction of the resultant

30 14 Engineering Mechanics as follows: 2 2 Ry R= R + Ry = N, θr = arctan = R Thus the resultant force is N in magnitude and in inclination. 2.3 Equilibrium of Coplanar Concurrent Forces 1. Free-Body Diagram In solving a problem concerning the equilibrium of a particle, it is essential to consider all the forces acting on the particle. This can be done by choosing the particle under consideration and drawing a separate diagram showing this particle and all the forces acting on it. Such a diagram is called a free-body diagram. 2. Graphical Solution for Equilibrium of Forces A particle is said to be in equilibrium if the resultant force of forces acting on the particle is zero. Thus the necessary and sufficient condition for equilibrium of a particle subjected to a coplanar system of concurrent forces can be epressed as = 0 F (2.10) It can be seen from Eq. (2.10) that a particle is in equilibrium if the given forces acting on the particle form a closed polygon. Considering that a particle is acted upon by forces, F 2, and F 3, Fig. 2.7a, the resultant force of the given forces can be obtained by the polygon rule. Starting from point with and arranging the forces in tip-to-tail fashion, we find that the tip of F 3 coincides with the starting point, Fig. 2.7b. Thus the resultant of the given forces is zero, and the particle is in equilibrium. F 2 F 2 F 3 F 3 (a) (b) Fig. 2.7 Eample 2.5 Three cables are tied together at A and are loaded as shown in Fig. E2.5a. Using the graphical method determine the tension (a) in cable AB, (b) in cable AC.