Mechanical engineering - hand book

Frontmatter Mechanical Engineering Handbook Ed. Frank Kreith Boca Raton: CRC Press LLC, 1999

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1999 by CRC Press LLC

Contents

SECTION 1 Mechanics of Solids Bela I. Sandor

1.1 Introduction Bela I Sandor 1.2 Statics Bela I. Sandor 1.3 Dynamics Stephen M. Birn and Bela I. Sandor 1.4 Vibrations Bela I. Sandor 1.5 Mechanics of Materials Bela I. Sandor 1.6 Structural Integrity and Durability Bela I. Sandor 1.7 Comprehensive Example of Using Mechanics of Solids Methods Richard C. Duveneck, David A. Jahnke, Christopher J. Watson, and Bela I. Sandor

SECTION 2 Engineering Thermodynamics Michael J. Moran

2.1 Fundamentals Michael J.Moran 2.2 Control Volume Applications Michael J.Moran 2.3 Property Relations and Data Michael J.Moran 2.4 Combustion Michael J.Moran 2.5 Exergy Analysis Michael J.Moran 2.6 Vapor and Gas Power Cycles Michael J.Moran 2.7 Guidelines for Improving Thermodynamic Effectiveness Michael J.Moran

SECTION 3 Fluid Mechanics Frank Kreith

3.1 Fluid Statics Stanley A.Berger 3.2 Equations of Motion and Potential Flow Stanley A.Berger 3.3 Similitude: Dimensional Analysis and Data Correlation Suar W.Churchill 3.4 Hydraulics of Pipe Systems J.Paul Tullis 3.5 Open Channel Flow Frank M.White 3.6 External Incompressible Flow Alan T.McDonald 3.7 Compressible Flow Ajay Kumar 3.8 Multiphase Flow John C.Chen 3.9 Non-Newtonian Flow Thomas F.Irvine Jr. and Massimo Capobianchi 3.10 Tribology, Lubrication, and Bearing Design Francis E.Kennedy, E.Richard Booser, and Donald F.Wilcock 3.11 Pumps and Fans Rober F.Boehm 3.12 Liquid Atomization and Spraying Rolf D.Reitz 3.13 Flow Measurement Alan T.McDonald and Sherif A.Sherif 3.14 Micro/Nanotribology Bharat Bhushan

SECTION 4 Heat and Mass Transfer Frank Kreith

4.1 Conduction Heat Transfer Rober F.Boehm 4.2 Convection Heat Transfer George D.Raithby, K.G.Terry Hollands, and N.V.Suryanarayana 4.3 Radiation Michael F.Modest 4.4 Phase-Change Van P.Carey, John C.Chen and Noam Lior

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4.5 Heat Exchangers Ramesh K.Shah and Kenneth J.Bell 4.6 Temperature and Heat Transfer Measurements Robert J.Moffat 4.7 Mass Transfer Anthony F.Mills 4.8 Applications Arthur E.Bergles, Anthony F.Mills, Larry W.Swanson, and Vincent W.Antonetti 4.9 Non-Newtonian Fluids Heat Transfer Thomas F.Irvine,Jr. and Massimo Capobianchi

SECTION 5 Electrical Engineering Giorgio Rizzoni

5.1 Introduction Giorgio Rizzoni 5.2 Fundamentals of Electric Circuits Giorgio Rizzoni 5.3 Resistive Network Analysis Giorgio Rizzoni 5.4 AC Network Analysis Giorgio Rizzoni 5.5 AC Power Giorgio Rizzoni 5.6 Frequency Response,Filters,and Transient Analysis Giorgio Rizzoni 5.7 Electronics Giorgio Rizzoni 5.8 Power Electronics Giorgio Rizzoni 5.9 Operational Amplifiers Giorgio Rizzoni 5.10 Digital Circuits Giorgio Rizzoni 5.11 Measurements and Instrumentation Giorgio Rizzoni 5.12 Electromechanical Systems Giorgio Rizzoni

SECTION 6 Mechanical System Controls Jan F. Kreider

6.1 Human Machine Interaction Thomas B. Sheridan 6.2 The Need for Control of Mechanical Systems Peter S. Curtiss 6.3 Control System Analysis Peter S. Curtiss 6.4 Control System Design and Application Peter S. Curtiss 6.5 Advanced Control Topics Peter S. Curtiss, Jan Kreider, Ronald M.Nelson, and Shou-Heng Huang

SECTION 7 Energy Resources D. Yogi Goswami

7.1 Introduction D.Yogi Goswami 7.2 Types of Derived Energy D.Yogi Goswami 7.3 Fossil Fuels Robert Reuther, Richard Bajura, Larry Grayson, and Philip C. Crouse 7.4 Biomass Energy Michael C.Reed, Lynn L.Wright, Ralph P.Overend, and Carlton Wiles 7.5 Nuclear Resources James S. Tulenko 7.6 Solar Energy Resources D.Yogi Goswami 7.7 Wind Energy Resources Dale E.Berg 7.8 Geothermal Energy Joel L. Renner and Marshall J. Reed

SECTION 8 Energy Conversion D. Yogi Goswam

8.1 Steam Power Plant Lawrence Conway 8.2 Gas Turbines Steven I. Freedman 8.3 Internal Combustion Engines David E. Klett and Elsayed A.Adfify 8.4 Hydraulic Turbines Roger E.A. Arndt 8.5 Stirling Engines William B. Stine 8.6 Advanced Fossil Fuel Power Systems Anthony F. Armor 8.7 Energy Storage Chand K. Jotshi and D.Yogi Goswami 8.8 Nuclear Power Robert Pagano and James S. Tulenko

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8.9 Nuclear Fusion Thomas E. Shannon 8.10 Solar Thermal Energy Conversion D.Yogi Goswami 8.11 Wind Energy Conversion Dale E. Berg 8.12 Energy Conversion of the Geothermal Resource Carl J. Bliem and Gregory L. Mines 8.13 Direct Energy Conversion Kitt C. Reinhardt, D.Yogi Goswami, Mysore L. Ramalingam , Jean-Pierre Fleurial, and William D. Jackson 8.14 Ocean Energy Technology Desikan Bharathan and Federica Zangrando 8.15 Combined Cycle Power Plants William W. Bathie 8.16 EMERGY Evaluation and Transformity Howard T.Odum 9.1 Introduction Shan K.Wang 9.2 Psychrometrics Shan K.Wang 9.3 Air Conditioning Processes and Cycles Shan K.Wang 9.4 Refrigerants and Refrigeration Cycles Shan K.Wang 9.5 Outdoor Design Conditions and Indoor Design Criteria Shan K.Wang 9.6 Load Calculations Shan K.Wang 9.7 Air Handling Units and Packaged Units Shan K.Wang 9.8 Refrigeration Components and Evaporative Coolers Shan K.Wang ' 9.9 Water Systems Shan K.Wang 9.10 Heating Systems Shan K.Wang 9.11 Refrigeration Systems Shan K.Wang 9.12 Thermal Storage Systems Shan K.Wang 9.13 Air Systems Shan K.Wang 9.14 Absorption Systems Shan K.Wang 9.15 Air Conditioning Systems and Selection Shan K.Wang 9.16 Desiccant Dehumidification and Air Conditioning Zalman Lavan

SECTION 9 Air Conditioning and Refrigeration Shan K. Wang

SECTION 10A Electronic Packaging

10A.1 Electronic Packaging Technologies Kevin D. Cluff and Michael G. Pecht 10A.2 Thermal Management in Electronic Packaging and Systems B.G. Sammakia and K. Ramakrishna 10A.3 Mechanical Design and Reliability of Electronic Systems Fred Barez 10A.4 Electronic Manufacturing: Processes, Optimization, and Control Roop L. Mahajan

SECTION 10 Transportation Frank Kreith

10.1 Transportation Planning Michael D.Meyer 10.2 Design of Transportation Facilities John Leonard II and Michael D.Meyer 10.3 Operations and Environmental Impact Paul W.Shuldiner and Kenneth B.Black 10.4 Transportation Systems Paul Schonfeld 10.5 Alternative Fuels for Motor Vehicles Paul Norton 10.6 Electric Vehicles Frank Kreith 10.7 Intelligent Transportation Systems James B. Reed

SECTION 11 Engineering Design Leonard D. Albano and Nam P. Suh

11.1 Introduction Nam P. Suh 11.2 Elements of the Design Process Nam P. Suh 11.3 Concept of Domains Nam P. Suh 11.4 The Axiomatic Approach to Design Nam P. Suh

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11.5 Algorithmic Approaches to Design Leonard D. Albano 11.6 Strategies for Product Design Michael Pecht 11.7 Design of Manufacturing Systems and Processes Leonard D. Albano 11.8 Precision Machine Design Alexander Slocum 11.9 Robotics Leonard D. Albano 11.10 Computer-Based Tools for Design Optimization Mark Jakiela, Kemper Lewis, Farrokh Mistree, and J.R. Jagannatha Rao

SECTION 12 Material Richard L. Lehman and Malcolm G. McLaren

12.1 Metals Victor A. Greenhut 12.2 Polymers James D. Idol and Richard L. Lehman 12.3 Adhesives Richard L. Lehman 12.4 Wood Daniel J. Strange 12.5 Portland Cement Concrete Steven H. Kosmatka 12.6 Composites Victor A. Greenhut 12.7 Ceramics and Glass Richard L.Lehman, Daniel J.Strange, and William F. Fischer III

SECTION 13 Modern Manufacturing Jay Lee and Robert E. Schafrik

13.1 Introduction Jay Lee and Robert E. Schafrik 13.2 Unit Manufacturing and Assembly Processes Robert E. Schafrik 13.3 Essential Elements in Manufacturing Processes and Equipment John Fildes, Yoram Koren, M. Tomizuka, Kam Lau, and Tai-Ran Hsu 13.4 Modern Design and Analysis Tools for Manufacturing David C .Anderson,Tien-Chien Chang,Hank Grant,Tien-I. Liu, J.M.A. Tanchoco,Andrew C. Lee,and Su-Hsia Yang 13.5 Rapid Prototyping Takeo Nakagawa 13.6 Underlying Paradigms in Manufacturing Systems and Enterprise for the 21st Century H.E.Cook, James J.Solberg, and Chris Wang

SECTION 14 Robotics Frank L. Lewis

14.1 Introduction Frank L.Lewis 14.2 Commercial Robot Manipulators John M.Fitzgerald 14.3 Robot Configurations Ian D.Walker 14.4 End Effectors and Tooling Mark R.Cutkosky and Peter McCormick 14.5 Sensors and Actuators Kok-Meng Lee 14.6 Robot Programming Languages Ron Bailey 14.7 Robot Dynamics and Control Frank L. Lewis 14.8 Planning and Intelligent Control Chen Zhou 14.9 Design of Robotic Systems Kok-Meng Lee 14.10 Robot Manufacturing Applications John W. Priest and G.T. Stevens, Jr. 14.11 Industrial Material Handling and Process Applications of Robots John M. Fitzgerald 14.12 Moblie, Flexible-Link, and Parallel-Link Robots Kai Liu

SECTION 15 Computer-Aided Engineering Kyran D. Mish

15.1 Introduction Kyran D. Mish 15.2 Computer Programming and Computer Architecture Kyran D. Mish 15.3 Computational Mechanics Kyran D. Mish

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15.4 Computer Intelligence Kyran D. Mish 15.5 Computer-Aided Design (CAD) Joseph Mello

SECTION 16 Environmental Engineering Jan F. Kreider

16.1 Introduction Ari Rabl and Jan F. Kreider 16.2 Benchmarks and Reference Conditions Ari Rabl, Nevis Cook, Ronald H. Hewitt Cohen, and Tissa Illangasekare ' 16.3 Sources of Pollution and Regulations Jan F.Kreider, Nevis Cook, Tissa Illangasekare, and Ronald H. Hewitt Cohen 16.4 Regulations and Emission Standards Nevis Cook and Ronald H Hewitt Cohen 16.5 Mitigation of Water and Air Pollution Jan F. Kreider, Nevis Cook, and Ronald H .Hewitt Cohen 16.6 Environmental Modeling Paolo Zannetti, Ronald H. Hewitt Cohen, Nevis Cook, Ari Rabl, and Peter S. Curtiss 16.7 Global Climate Change Frank Kreith

SECTION 17 Engineering Economics and Project Management Chan S. Park and Donald D. Tippett

17.1 Engineering Economic Decisions Chan S. Park 17.2 Establishing Economic Equivalence Chan S. Park 17.3 Measures of Project Worth Chan S. Park 17.4 Cash Flow Projections Chan S. Park 17.5 Sensitivity and Risk Analysis Chan S. Park 17.6 Design Economics Chan S. Park 17.7 Project Management Donald D. Tippett

SECTION 18 Communications and Information Systems Lloyd W. Taylor

18.1 Introduction Lloyd W. Taylor 18.2 Network Components and Systems Lloyd W. Taylor and Daniel F. DiFonzo 18.3 Communications and Information Theory A. Britton Cooper III 18.4 Applications Lloyd W. Taylor, Dhammika Kurumbalapitiya, and S.Ratnajeevan H.Hoole

SECTION 19 Mathematics William F. Ames and George Cain

19.1 Tables William F.Ames 19.2 Linear Algebra and Matrices George Cain 19.3 Vector Algebra and Calculus George Cain 19.4 Difference Equations William F. Ames 19.5 Differential Equations William F. Ames 19.6 Integral Equations William F. Ames 19.7 Approximation Methods William F. Ames 19.8 Integral Transforms William F. Ames 19.9 Calculus of Variations Approximation William F. Ames 19.10 Optimization Methods George Cain 19.11 Engineering and Statistics Y.L. Tong 19.12 Numerical Methods William F. Ames 19.13 Experimental Uncertainty Analysis W.G. Steele and H.W. Coleman 19.14 Chaos R.L. Kautz 19.15 Fuzzy Sets and Fuzzy Logic Dan M. Frangopol

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SECTION 20 Patent Law and Miscellaneous Topics Frank Kreith

20.1 Patents and Other Intellectual Property Thomas H. Young 20.2 Product Liability and Safety George A. Peters 20.3 Bioengineering Jeff R. Crandall, Gregory W. Hall, and Walter D. Pilkey 20.4 Mechanical Engineering Codes and Standard Michael Merker 20.5 Optics Roland Winston and Walter T. Welford 20.6 Water Desalination Noam Lior 20.7 Noise Control Malcolm J. Crocker 20.8 Lighting Technology Barbara Atkinson, Andrea Denver, James E. McMahon, Leslie Shown, Robert Clear, and Craig B Smith 20.9 New Product Development Philip R. Teakle and Duncan B. Gilmore

APPENDICES Paul Norton

A. Properties of Gases and Vapors B. Properties of Liquids C. Properties of Solids D. SI Units E. Miscellaneous

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Sandor, B.I.; Roloff, R; et. al. Mechanics of Solids Mechanical Engineering Handbook Ed. Frank Kreith Boca Raton: CRC Press LLC, 1999

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1999 by CRC Press LLC

Mechanics of Solids

Bela I. Sandor

University of Wisconsin-Madison

1.1 1.2

Introduction .1-1 Statics .1-3

Vectors. Equilibrium of Particles. Free-body Diagrams Forces on Rigid Bodies Equilibrium of Rigid Bodies Forces and Moments in Beams Simple Structures and Machines Distributed Forces Friction Work and Potential Energy Moments of Inertia

Ryan Roloff

Allied Signal Aerospace

Stephen M. Birn

Allied Signal Aerospace

1.3

Dynamics.1-31

Kinematics of Particles Kinetics of Particles Kinetics of Systems of Particles Kinematics of Rigid Bodies Kinetics of Rigid Bodies in Plane Motion Energy and Momentum Methods for Rigid Bodies in Plane Motion Kinetics of Rigid Bodies in Three Dimensions

Maan H. Jawad

Nooter Consulting Services

Michael L. Brown

A.O. Smith Corp.

1.4

Vibrations .1-57

Undamped Free and Forced Vibrations Damped Free and Forced Vibrations Vibration Control Random Vibrations. Shock Excitations Multiple-Degree-of-Freedom Systems. Modal Analysis Vibration-Measuring Instruments

1.5

Mechanics of Materials.1-67

Stress Strain Mechanical Behaviors and Properties of Materials Uniaxial Elastic Deformations Stresses in Beams De ections of Beams Torsion Statically Indeterminate Members Buckling Impact Loading Combined Stresses Pressure Vessels Experimental Stress Analysis and Mechanical Testing

1.6 1.7

Structural Integrity and Durability.1-104

Finite Element Analysis. Stress Concentrations Fracture Mechanics Creep and Stress Relaxation Fatigue

Comprehensive Example of Using Mechanics of Solids Methods .1-125

The Project Concepts and Methods

1.1 Introduction

Bela I. Sandor

Engineers use the concepts and methods of mechanics of solids in designing and evaluating tools, machines, and structures, ranging from wrenches to cars to spacecraft. The required educational background for these includes courses in statics, dynamics, mechanics of materials, and related subjects. For example, dynamics of rigid bodies is needed in generalizing the spectrum of service loads on a car, which is essential in de ning the vehicle s deformations and long-term durability. In regard to structural

© 1999 by CRC Press LLC

1-1

1-2

Section 1

FIGURE 1.1.1 Artist s concept of a moving stainless steel roadway to drive the suspension system through a spinning, articulated wheel, simulating three-dimensional motions and forces. (MTS Systems Corp., Minneapolis, MN. With permission.) Notes: Flat-Trac Roadway Simulator, R&D100 Award-winning system in 1993. See also Color Plate 1.*

integrity and durability, the designer should think not only about preventing the catastrophic failures of products, but also of customer satisfaction. For example, a car with gradually loosening bolts (which is dif cult to prevent in a corrosive and thermal and mechanical cyclic loading environment) is a poor product because of safety, vibration, and noise problems. There are sophisticated methods to assure a product s performance and reliability, as exempli ed in Figure 1.1.1. A similar but even more realistic test setup is shown in Color Plate 1.* It is common experience among engineers that they have to review some old knowledge or learn something new, but what is needed at the moment is not at their ngertips. This chapter may help the reader in such a situation. Within the constraints of a single book on mechanical engineering, it provides overviews of topics with modern perspectives, illustrations of typical applications, modeling to solve problems quantitatively with realistic simpli cations, equations and procedures, useful hints and reminders of common errors, trends of relevant material and mechanical system behaviors, and references to additional information. The chapter is like an emergency toolbox. It includes a coherent assortment of basic tools, such as vector expressions useful for calculating bending stresses caused by a three-dimensional force system on a shaft, and sophisticated methods, such as life prediction of components using fracture mechanics and modern measurement techniques. In many cases much more information should be considered than is covered in this chapter.

*

Color Plates 1 to 16 follow page 1-131.

Mechanics of Solids

1-3

1.2 Statics

Bela I. Sandor

Vectors. Equilibrium of Particles. Free-Body Diagrams

Two kinds of quantities are used in engineering mechanics. A scalar quantity has only magnitude (mass, time, temperature, ). A vector quantity has magnitude and direction (force, velocity, .). Vectors are represented here by arrows and bold-face symbols, and are used in analysis according to universally applicable rules that facilitate calculations in a variety of problems. The vector methods are indispensable in three-dimensional mechanics analyses, but in simple cases equivalent scalar calculations are suf cient. Vector Components and Resultants. Parallelogram Law A given vector F may be replaced by two or three other vectors that have the same net effect and representation. This is illustrated for the chosen directions m and n for the components of F in two dimensions (Figure 1.2.1). Conversely, two concurrent vectors F and P of the same units may be combined to get a resultant R (Figure 1.2.2).

FIGURE 1.2.1 Addition of concurrent vectors F and P.

FIGURE 1.2.2 Addition of concurrent, coplanar vectors A, B, and C.

Any set of components of a vector F must satisfy the parallelogram law. According to Figure 1.2.1, the law of sines and law of cosines may be useful. Fn F F = m = sin sin sin 180 ( + )

[

] ]

(1.2.1)

2 F 2 = Fn2 + Fm 2 Fn Fm cos 180 ( + )

[

Any number of concurrent vectors may be summed, mathematically or graphically, and in any order, using the above concepts as illustrated in Figure 1.2.3.

FIGURE 1.2.3 Addition of concurrent, coplanar vectors A, B, and C.

1-4

Section 1

Unit Vectors Mathematical manipulations of vectors are greatly facilitated by the use of unit vectors. A unit vector n has a magnitude of unity and a de ned direction. The most useful of these are the unit coordinate vectors i, j, and k as shown in Figure 1.2.4.

FIGURE 1.2.4 Unit vectors in Cartesian coordinates (the same i, j, and k set applies in a parallel x y z system of axes).

The three-dimensional components and associated quantities of a vector F are shown in Figure 1.2.5. The unit vector n is collinear with F.

FIGURE 1.2.5 Three-dimensional components of a vector F.

The vector F is written in terms of its scalar components and the unit coordinate vectors, F = Fx i + Fy j + Fz k = Fn where Fx = F cos x Fy = F cos y Fz = F cos z

(1.2.2)

F = Fx2 + Fy2 + Fz2 n x = cos x n y = cos y nz = cos z

2 2 2 n x + n y + nz = 1

n x n y nz 1 = = = Fx Fy Fz F The unit vector notation is convenient for the summation of concurrent vectors in terms of scalar or vector components: Scalar components of the resultant R: Rx =

F

x

Ry =

F

y

Rz =

F

z

(1.2.3)

Mechanics of Solids

1-5

Vector components: Rx =

F = F i

x x

Ry =

F = F j

y y

Rz =

F = F k

z z

(1.2.4)

Vector Determination from Scalar Information A force, for example, may be given in terms of its magnitude F, its sense of direction, and its line of action. Such a force can be expressed in vector form using the coordinates of any two points on its line of action. The vector sought is F = Fx i + Fy j + Fz k = Fn The method is to nd n on the line of points A(x1, y1, z1) and B(x2, y2, z2): n= d x i + d y j + dz k vector A to B = 2 2 distance A to B d x + d y + d z2

where dx = x2 x1, dy = y2 y1, dz = z2 z1. Scalar Product of Two Vectors. Angles and Projections of Vectors The scalar product, or dot product, of two concurrent vectors A and B is de ned by A B = ABcos

(1.2.5)

where A and B are the magnitudes of the vectors and is the angle between them. Some useful expressions are A B = B A = Ax Bx + Ay By + Az Bz = arccos Ax Bx + Ay By + Az Bz AB

The projection F of a vector F on an arbitrary line of interest is determined by placing a unit vector n on that line of interest, so that F = F n = Fx n x + Fy n y + Fz nz Equilibrium of a Particle A particle is in equilibrium when the resultant of all forces acting on it is zero. In such cases the algebraic summation of rectangular scalar components of forces is valid and convenient:

F = 0 F = 0 F = 0

x y z

(1.2.6)

Free-Body Diagrams Unknown forces may be determined readily if a body is in equilibrium and can be modeled as a particle. The method involves free-body diagrams, which are simple representations of the actual bodies. The appropriate model is imagined to be isolated from all other bodies, with the signi cant effects of other bodies shown as force vectors on the free-body diagram.

1-6

Section 1

Example 1 A mast has three guy wires. The initial tension in each wire is planned to be 200 lb. Determine whether this is feasible to hold the mast vertical (Figure 1.2.6).

FIGURE 1.2.6 A mast with guy wires.

Solution. R = TAB + TAC + TAD The three tensions of known magnitude (200 lb) must be written as vectors.

TAB = (tension AB)( unit vector A to B) = 200 lb n AB = 200 lb = 200 lb 5 + 10 + 4

2 2 2

( d i + d j + d k)

x y z

d

( 5i 10 j + 4k)

ft = 84.2 lb i 168.4 lb j + 67.4 lb k ft

TAC =

200 lb (5i 10 j + 4k) ft = 84.2 lb i + 168.4 lb j + 67.4 lb k 11.87 ft 200 lb (0i 10 j + 6k) ft = 171.5 lb j 102.9 lb k 11.66 ft

TAD =

The resultant of the three tensions is R=

F i + F j + F k = ( 84.2 + 84.2 + 0) lb i + ( 168.4 168.4 171.5) lb j

x y z

+ (67.4 + 67.4 102.9) lb k = 0 lb i 508 lb j + 31.9 lb k There is a horizontal resultant of 31.9 lb at A, so the mast would not remain vertical.

Forces on Rigid Bodies

All solid materials deform when forces are applied to them, but often it is reasonable to model components and structures as rigid bodies, at least in the early part of the analysis. The forces on a rigid body are generally not concurrent at the center of mass of the body, which cannot be modeled as a particle if the force system tends to cause a rotation of the body.

Mechanics of Solids

1-7

Moment of a Force The turning effect of a force on a body is called the moment of the force, or torque. The moment MA of a force F about a point A is de ned as a scalar quantity M A = Fd

(1.2.7)

where d (the moment arm or lever arm) is the nearest distance from A to the line of action of F. This nearest distance may be dif cult to determine in a three-dimensional scalar analysis; a vector method is needed in that case. Equivalent Forces Sometimes the equivalence of two forces must be established for simplifying the solution of a problem. The necessary and suf cient conditions for the equivalence of forces F and F are that they have the same magnitude, direction, line of action, and moment on a given rigid body in static equilibrium. Thus, F = F and MA = MA

For example, the ball joint A in Figure 1.2.7 experiences the same moment whether the vertical force is pushing or pulling downward on the yoke pin.

FIGURE 1.2.7 Schematic of testing a ball joint of a car.

Vector Product of Two Vectors A powerful method of vector mechanics is available for solving complex problems, such as the moment of a force in three dimensions. The vector product (or cross product) of two concurrent vectors A and B is de ned as the vector V = A B with the following properties: 1. 2. 3. 4. 5. V is perpendicular to the plane of vectors A and B. The sense of V is given by the right-hand rule (Figure 1.2.8). The magnitude of V is V = AB sin , where is the angle between A and B. A B B A, but A B = (B A). For three vectors, A (B + C) = A B + A C.

FIGURE 1.2.8 Right-hand rule for vector products.

1-8

Section 1

The vector product is calculated using a determinant, i V = Ax Bx j Ay By k Az = Ay Bz i + Az Bx j + Ax By k Ay Bx k Ax Bz j Az By i Bz

(1.2.8)

Moment of a Force about a Point The vector product is very useful in determining the moment of a force F about an arbitrary point O. The vector de nition of moment is MO = r F

(1.2.9)

where r is the position vector from point O to any point on the line of action of F. A double arrow is often used to denote a moment vector in graphics. The moment MO may have three scalar components, Mx, My , Mz, which represent the turning effect of the force F about the corresponding coordinate axes. In other words, a single force has only one moment about a given point, but this moment may have up to three components with respect to a coordinate system, M O = M x i + M y j + Mz k Triple Products of Three Vectors Two kinds of products of three vectors are used in engineering mechanics. The mixed triple product (or scalar product) is used in calculating moments. It is the dot product of vector A with the vector product of vectors B and C, Ax A (B C) = Bx Cx Ay By Cy Az Bz = Ax By Cz Bz Cy + Ay ( Bz Cx Bx Cz ) + Az Bx Cy By Cx Cz

(

)

(

)

(1.2.10)

The vector triple product (A B) C = V C is easily calculated (for use in dynamics), but note that

(A B) C A (B C)

Moment of a Force about a Line It is common that a body rotates about an axis. In that case the moment M of a force F about the axis, say line , is usefully expressed as nx M l = n M O = n (r F) = rx Fx ny ry Fy nz rz Fz

(1.2.11)

where n is a unit vector along the line , and r is a position vector from point O on line of action of F. Note that M is the projection of MO on line .

to a point on the

Mechanics of Solids

1-9

Special Cases 1. The moment about a line zero). 2. The moment about a line MO on is zero). Moment of a Couple A pair of forces equal in magnitude, parallel in lines of action, and opposite in direction is called a couple. The magnitude of the moment of a couple is M = Fd where d is the distance between the lines of action of the forces of magnitude F. The moment of a couple is a free vector M that can be applied anywhere to a rigid body with the same turning effect, as long as the direction and magnitude of M are the same. In other words, a couple vector can be moved to any other location on a given rigid body if it remains parallel to its original position (equivalent couples). Sometimes a curled arrow in the plane of the two forces is used to denote a couple, instead of the couple vector M, which is perpendicular to the plane of the two forces. Force-Couple Transformations Sometimes it is advantageous to transform a force to a force system acting at another point, or vice versa. The method is illustrated in Figure 1.2.9. is zero when the line of action of F intersects is zero when the line of action of F is parallel to (the moment arm is (the projection of

FIGURE 1.2.9 Force-couple transformations.

1. A force F acting at B on a rigid body can be replaced by the same force F acting at A and a moment MA = r F about A. 2. A force F and moment MA acting at A can be replaced by a force F acting at B for the same total effect on the rigid body. Simpli cation of Force Systems Any force system on a rigid body can be reduced to an equivalent system of a resultant force R and a resultant moment MR. The equivalent force-couple system is formally stated as R=

F

i =1

n

i

and M R =

M = (r F )

i i i i =1 i =1

n

n

(1.2.12)

where MR depends on the chosen reference point. Common Cases 1. 2. 3. 4. The resultant force is zero, but there is a resultant moment: R = 0, MR 0. Concurrent forces (all forces act at one point): R 0, MR = 0. Coplanar forces: R 0, MR 0. MR is perpendicular to the plane of the forces. Parallel forces: R 0, MR 0. MR is perpendicular to R.

1-10

Section 1

Example 2 The torque wrench in Figure 1.2.10 has an arm of constant length L but a variable socket length d = OA because of interchangeable tool sizes. Determine how the moment applied at point O depends on the length d for a constant force F from the hand.

FIGURE 1.2.10 Model of a torque wrench.

Solution. Using MO = r F with r = Li + dj and F = Fk in Figure 1.2.10, M O = ( Li + dj) Fk = Fdi FLj Judgment of the Result According to a visual analysis the wrench should turn clockwise, so the j component of the moment is justi ed. Looking at the wrench from the positive x direction, point A has a tendency to rotate counterclockwise. Thus, the i component is correct using the right-hand rule.

Equilibrium of Rigid Bodies

The concept of equilibrium is used for determining unknown forces and moments of forces that act on or within a rigid body or system of rigid bodies. The equations of equilibrium are the most useful equations in the area of statics, and they are also important in dynamics and mechanics of materials. The drawing of appropriate free-body diagrams is essential for the application of these equations. Conditions of Equilibrium A rigid body is in static equilibrium when the equivalent force-couple system of the external forces acting on it is zero. In vector notation, this condition is expressed as

F = 0 M = (r F) = 0

O

(1.2.13)

where O is an arbitrary point of reference. In practice it is often most convenient to write Equation 1.2.13 in terms of rectangular scalar components,

F = 0 M F = 0 M F = 0 M

x y z

x

=0 =0 =0

y

z

Mechanics of Solids

1-11

Maximum Number of Independent Equations for One Body 1. One-dimensional problem: F = 0 2. Two-dimensional problem:

or or

F = 0 F = 0 M = 0 F = 0 M = 0 M = 0 M = 0 M = 0 M = 0

x y A x A B A B C x y z

( x axis not AB) ( AB not BC)

3. Three-dimensional problem:

F = 0 F = 0 F = 0 M = 0 M = 0 M = 0

x y z

where xyz are orthogonal coordinate axes, and A, B, C are particular points of reference. Calculation of Unknown Forces and Moments In solving for unknown forces and moments, always draw the free-body diagram rst. Unknown external forces and moments must be shown at the appropriate places of action on the diagram. The directions of unknowns may be assumed arbitrarily, but should be done consistently for systems of rigid bodies. A negative answer indicates that the initial assumption of the direction was opposite to the actual direction. Modeling for problem solving is illustrated in Figures 1.2.11 and 1.2.12.

FIGURE 1.2.11 Example of two-dimensional modeling.

FIGURE 1.2.12 Example of three-dimensional modeling.

Notes on Three-Dimensional Forces and Supports Each case should be analyzed carefully. Sometimes a particular force or moment is possible in a device, but it must be neglected for most practical purposes. For example, a very short sleeve bearing cannot

1-12

Section 1

support signi cant moments. A roller bearing may be designed to carry much larger loads perpendicular to the shaft than along the shaft. Related Free-Body Diagrams When two or more bodies are in contact, separate free-body diagrams may be drawn for each body. The mutual forces and moments between the bodies are related according to Newton s third law (action and reaction). The directions of unknown forces and moments may be arbitrarily assumed in one diagram, but these initial choices affect the directions of unknowns in all other related diagrams. The number of unknowns and of usable equilibrium equations both increase with the number of related free-body diagrams. Schematic Example in Two Dimensions (Figure 1.2.13)

FIGURE 1.2.13 Free-body diagram.

Given: F1, F2, F3, M Unknowns: P1, P2, P3, and forces and moments at joint A (rigid connection) Equilibrium Equations

F = F + P = 0 F = P + P F F = 0 M = P c + P (c + d + e) + M F a F ( a + b ) = 0

x 1 3 y 1 2 2 3 O 1 2 2 3

Three unknowns (P1, P2, P3) are in three equations. Related Free-Body Diagrams (Figure 1.2.14)

FIGURE 1.2.14 Related free-body diagrams.

Dimensions a, b, c, d, and e of Figure 1.2.13 are also valid here.

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1-13

New Set of Equilibrium Equations Left part: (OA)

F = F + A

x 1 y 1 y

x

=0

F = P + A F = 0

2

M

Right side: ( AB)

O

= P c + Ay (c + d ) + M A F2 a = 0 1 + P3 = 0

F = A

x y 2 A

x

F = P A F = 0 M = M + P e + M F f = 0

y 3 A 2 3

Six unknowns (P1, P2, P3, Ax, Ay , MA) are in six equations. Note: In the rst diagram (Figure 1.2.13) the couple M may be moved anywhere from O to B. M is not shown in the second diagram (O to A) because it is shown in the third diagram (in which it may be moved anywhere from A to B). Example 3 The arm of a factory robot is modeled as three bars (Figure 1.2.15) with coordinates A: (0.6, 0.3, 0.4) m; B: (1, 0.2, 0) m; and C: (0.9, 0.1, 0.25) m. The weight of the arm is represented by WA = 60 Nj at A, and WB = 40 Nj at B. A moment MC = (100i 20j + 50k) N m is applied to the arm at C. Determine the force and moment reactions at O, assuming that all joints are temporarily xed.

FIGURE 1.2.15 Model of a factory robot.

Solution. The free-body diagram is drawn in Figure 1.2.15b, showing the unknown force and moment reactions at O. From Equation 1.2.13,

F = 0

FO + WA + WB = 0 FO 60 N j 40 N j = 0

1-14

Section 1

FO = 100 N j

M

O

=0

M O + M C + (rOA WA ) + (rOB WB ) = 0 M O + (100i 20 j + 50k) N m + (0.6i 0.3 j + 0.4k) m ( 60 N j) + (i 0.2 j) m ( 40 N j) = 0 M O + 100 N m i 20 N m j + 50 N m k 36 N m k + 24 N m i 40 N m k = 0 M O = ( 124i + 20 j + 26k) N m Example 4 A load of 7 kN may be placed anywhere within A and B in the trailer of negligible weight. Determine the reactions at the wheels at D, E, and F, and the force on the hitch H that is mounted on the car, for the extreme positions A and B of the load. The mass of the car is 1500 kg, and its weight is acting at C (see Figure 1.2.16).

FIGURE 1.2.16 Analysis of a car with trailer.

Solution. The scalar method is best here.

Put the load at position A rst For the trailer alone, with y as the vertical axis MF = 7(1) Hy(3) = 0, Hy = 2.33 kN On the car Hy = 2.33 kN Ans. Fy = 2.33 7 + Fy = 0, Fy = 4.67 kN Ans. For the car alone ME = 2.33(1.2) Dy(4) + 14.72(1.8) = 0 Dy = 5.93 kN Ans. Fy = 5.93 + Ey 14.72 2.33 = 0 Ey = 11.12 kN Ans. Put the load at position B next For the trailer alone MF = 0.8(7) Hy(3) = 0, Hy = 1.87 kN On the car Hy = 1.87 kN Ans. Fy = 1.87 7 + Ey = 0 Ey = 8.87 kN Ans. For the car alone ME = (1.87)(1.2) Dy(4) + 14.72(1.8) = 0 Dy = 7.19 kN Ans. Fy = 7.19 + Ey 14.72 ( 1.87) = 0 Ey = 5.66 kN Ans.

Forces and Moments in Beams

Beams are common structural members whose main function is to resist bending. The geometric changes and safety aspects of beams are analyzed by rst assuming that they are rigid. The preceding sections enable one to determine (1) the external (supporting) reactions acting on a statically determinate beam, and (2) the internal forces and moments at any cross section in a beam.

Mechanics of Solids

1-15

Classi cation of Supports Common supports and external reactions for two-dimensional loading of beams are shown in Figure 1.2.17.

FIGURE 1.2.17 Common beam supports.

Internal Forces and Moments The internal force and moment reactions in a beam caused by external loading must be determined for evaluating the strength of the beam. If there is no torsion of the beam, three kinds of internal reactions are possible: a horizontal normal force H on a cross section, vertical (transverse) shear force V, and bending moment M. These reactions are calculated from the equilibrium equations applied to the left or right part of the beam from the cross section considered. The process involves free-body diagrams of the beam and a consistently applied system of signs. The modeling is illustrated for a cantilever beam in Figure 1.2.18.

FIGURE 1.2.18 Internal forces and moments in a cantilever beam.

Sign Conventions. Consistent sign conventions should be used in any given problem. These could be arbitrarily set up, but the following is slightly advantageous. It makes the signs of the answers to the equilibrium equations correct for the directions of the shear force and bending moment. A moment that makes a beam concave upward is taken as positive. Thus, a clockwise moment is positive on the left side of a section, and a counterclockwise moment is positive on the right side. A

1-16

Section 1

shear force that acts upward on the left side of a section, or downward on the right side, is positive (Figure 1.2.19).

FIGURE 1.2.19 Preferred sign conventions.

Shear Force and Bending Moment Diagrams The critical locations in a beam are determined from shear force and bending moment diagrams for the whole length of the beam. The construction of these diagrams is facilitated by following the steps illustrated for a cantilever beam in Figure 1.2.20.

FIGURE 1.2.20 Construction of shear force and bending moment diagrams.

1. Draw the free-body diagram of the whole beam and determine all reactions at the supports. 2. Draw the coordinate axes for the shear force (V) and bending moment (M) diagrams directly below the free-body diagram. 3. Immediately plot those values of V and M that can be determined by inspection (especially where they are zero), observing the sign conventions. 4. Calculate and plot as many additional values of V and M as are necessary for drawing reasonably accurate curves through the plotted points, or do it all by computer. Example 5 A construction crane is modeled as a rigid bar AC which supports the boom by a pin at B and wire CD. The dimensions are AB = 10 , BC = 2 , BD = DE = 4 . Draw the shear force and bending moment diagrams for bar AC (Figure 1.2.21). Solution. From the free-body diagram of the entire crane,

F = 0 F = 0

x y

M

A

=0

Ax = 0

P + Ay = 0 Ay = P

P(8l) + M A = 0 M A = 8Pl

Mechanics of Solids

1-17

FIGURE 1.2.21 Shear force and bending moment diagrams of a component in a structure.

Now separate bar AC and determine the forces at B and C.

F = 0

x

F = 0

y

M

1 T 5 CD

A

=0

Bx + TCDx = 0 (a ) Bx = 2 T 5 CD

P By TCDy = 0 ( b) By = P

2 T (12l) + Bx (10l) + M A = 0 5 CD 24l 20l T + T = 8Pl 5 CD 5 CD (c) TCD = 8 5 P = 2 5P 4

From (a) and (c), Bx = 4P and TCDx = 4P. From (b) and (c), By = P 2P = P and TCDy = 2P. Draw the free-body diagram of bar AC horizontally, with the shear force and bending moment diagram axes below it. Measure x from end C for convenience and analyze sections 0 x 2 and 2 x 12 (Figures 1.2.21b to 1.2.21f). 1. 0 x 2

F = 0

y

M

K

=0

4 P + VK1 = 0 VK1 = 4 P 2. 2 x 12

M K1 + 4 P( x ) = 0 M K1 = 4 Px

1-18

Section 1

F = 0

y

M

K

=0

4 P 4 P + VK2 = 0 VK2 = 0

M K2 4 P( x 2l) + 4 P( x ) = 0 M K2 = 8Pl

At point B, x = 2 , M K1 = 4P(2 ) = 8P = M K2 = MA. The results for section AB, 2 x 12 , show that the combined effect of the forces at B and C is to produce a couple of magnitude 8P on the beam. Hence, the shear force is zero and the moment is constant in this section. These results are plotted on the axes below the free-body diagram of bar A-B-C.

Simple Structures and Machines

Ryan Roloff and Bela I. Sandor

Equilibrium equations are used to determine forces and moments acting on statically determinate simple structures and machines. A simple structure is composed solely of two-force members. A machine is composed of multiforce members. The method of joints and the method of sections are commonly used in such analysis. Trusses Trusses consist of straight, slender members whose ends are connected at joints. Two-dimensional plane trusses carry loads acting in their planes and are often connected to form three-dimensional space trusses. Two typical trusses are shown in Figure 1.2.22.

FIGURE 1.2.22 Schematic examples of trusses.

To simplify the analysis of trusses, assume frictionless pin connections at the joints. Thus, all members are two-force members with forces (and no moments) acting at the joints. Members may be assumed weightless or may have their weights evenly divided to the joints. Method of Joints Equilibrium equations based on the entire truss and its joints allow for determination of all internal forces and external reactions at the joints using the following procedure. 1. Determine the support reactions of the truss. This is done using force and moment equilibrium equations and a free-body diagram of the entire truss. 2. Select any arbitrary joint where only one or two unknown forces act. Draw the free-body diagram of the joint assuming unknown forces are tensions (arrows directed away from the joint). 3. Draw free-body diagrams for the other joints to be analyzed, using Newton s third law consistently with respect to the rst diagram. 4. Write the equations of equilibrium, Fx = 0 and Fy = 0, for the forces acting at the joints and solve them. To simplify calculations, attempt to progress from joint to joint in such a way that each equation contains only one unknown. Positive answers indicate that the assumed directions of unknown forces were correct, and vice versa. Example 6 Use the method of joints to determine the forces acting at A, B, C, H, and I of the truss in Figure 1.2.23a. The angles are = 56.3 , = 38.7 , = 39.8 , and = 36.9 .

Mechanics of Solids

1-19

FIGURE 1.2.23 Method of joints in analyzing a truss.

Solution. First the reactions at the supports are determined and are shown in Figure 1.2.23b. A joint at which only two unknown forces act is the best starting point for the solution. Choosing joint A, the solution is progressively developed, always seeking the next joint with only two unknowns. In each diagram circles indicate the quantities that are known from the preceding analysis. Sample calculations show the approach and some of the results. Joint A:

F = 0

x

F = 0

y

FAI = 0

FAB Ay = 0 FAB 50 kips = 0 FAB = 50 kips (tension)

Joint H:

F = 0

x

F = 0

y

FGH sin FCH cos FBH = 0 FGH (0.625) + (60.1 kips) (0.555) 0 = 0 FGH = 53.4 kips (compression) Method of Sections

FCH sin + FDH + FGH cos FHI = 0 (60.1 kips) (0.832) + FDH (53.4 kips) (0.780) + 70 kips = 0 FDH = 21.7 kips (tension)

The method of sections is useful when only a few forces in truss members need to be determined regardless of the size and complexity of the entire truss structure. This method employs any section of the truss as a free body in equilibrium. The chosen section may have any number of joints and members in it, but the number of unknown forces should not exceed three in most cases. Only three equations of equilibrium can be written for each section of a plane truss. The following procedure is recommended. 1. Determine the support reactions if the section used in the analysis includes the joints supported.

1-20

Section 1

2. Section the truss by making an imaginary cut through the members of interest, preferably through only three members in which the forces are unknowns (assume tensions). The cut need not be a straight line. The sectioning is illustrated by lines l-l, m-m, and n-n in Figure 1.2.24. 3. Write equations of equilibrium. Choose a convenient point of reference for moments to simplify calculations such as the point of intersection of the lines of action for two or more of the unknown forces. If two unknown forces are parallel, sum the forces perpendicular to their lines of action. 4. Solve the equations. If necessary, use more than one cut in the vicinity of interest to allow writing more equilibrium equations. Positive answers indicate assumed directions of unknown forces were correct, and vice versa.

FIGURE 1.2.24 Method of sections in analyzing a truss.

Space Trusses A space truss can be analyzed with the method of joints or with the method of sections. For each joint, there are three scalar equilibrium equations, Fx = 0, Fy = 0, and Fz = 0. The analysis must begin at a joint where there are at least one known force and no more than three unknown forces. The solution must progress to other joints in a similar fashion. There are six scalar equilibrium equations available when the method of sections is used: Fx = 0, Fy = 0, Fz = 0, Mx = 0, My = 0, and Mz = 0. Frames and Machines Multiforce members (with three or more forces acting on each member) are common in structures. In these cases the forces are not directed along the members, so they are a little more complex to analyze than the two-force members in simple trusses. Multiforce members are used in two kinds of structure. Frames are usually stationary and fully constrained. Machines have moving parts, so the forces acting on a member depend on the location and orientation of the member. The analysis of multiforce members is based on the consistent use of related free-body diagrams. The solution is often facilitated by representing forces by their rectangular components. Scalar equilibrium equations are the most convenient for two-dimensional problems, and vector notation is advantageous in three-dimensional situations. Often, an applied force acts at a pin joining two or more members, or a support or connection may exist at a joint between two or more members. In these cases, a choice should be made of a single member at the joint on which to assume the external force to be acting. This decision should be stated in the analysis. The following comprehensive procedure is recommended. Three independent equations of equilibrium are available for each member or combination of members in two-dimensional loading; for example, Fx = 0, Fy = 0, MA = 0, where A is an arbitrary point of reference. 1. Determine the support reactions if necessary. 2. Determine all two-force members.

Mechanics of Solids

1-21

3. Draw the free-body diagram of the rst member on which the unknown forces act assuming that the unknown forces are tensions. 4. Draw the free-body diagrams of the other members or groups of members using Newton s third law (action and reaction) consistently with respect to the rst diagram. Proceed until the number of equilibrium equations available is no longer exceeded by the total number of unknowns. 5. Write the equilibrium equations for the members or combinations of members and solve them. Positive answers indicate that the assumed directions for unknown forces were correct, and vice versa.

Distributed Forces

The most common distributed forces acting on a body are parallel force systems, such as the force of gravity. These can be represented by one or more concentrated forces to facilitate the required analysis. Several basic cases of distributed forces are presented here. The important topic of stress analysis is covered in mechanics of materials. Center of Gravity The center of gravity of a body is the point where the equivalent resultant force caused by gravity is acting. Its coordinates are de ned for an arbitrary set of axes as

x=

x dW

W

y=

y dW

W

z=

z dW

W

(1.2.14)

where x, y, z are the coordinates of an element of weight dW, and W is the total weight of the body. In the general case dW = dV, and W = dV, where = speci c weight of the material and dV = elemental volume. Centroids If is a constant, the center of gravity coincides with the centroid, which is a geometrical property of a body. Centroids of lines L, areas A, and volumes V are de ned analogously to the coordinates of the center of gravity,

Lines:

x=

x dL

W

y=

y dL

L

z=

z dL

L

(1.2.15)

Areas:

x=

x dA

A

y=

y dA

A

z=

z dA

A

(1.2.16)

Volumes:

x=

x dV

V

y=

y dV

V

z=

z dV

V

(1.2.17)

For example, an area A consists of discrete parts A1, A2, A3, where the centroids x1, x2, x3 of the three parts are located by inspection. The x coordinate of the centroid of the whole area A is x obtained from Ax = A1x1 + A2x2 + A3x3.

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Section 1

Surfaces of Revolution. The surface areas and volumes of bodies of revolution can be calculated using the concepts of centroids by the theorems of Pappus (see texts on Statics). Distributed Loads on Beams The distributed load on a member may be its own weight and/or some other loading such as from ice or wind. The external and internal reactions to the loading may be determined using the condition of equilibrium. External Reactions. Replace the whole distributed load with a concentrated force equal in magnitude to the area under the load distribution curve and applied at the centroid of that area parallel to the original force system. Internal Reactions. For a beam under a distributed load w(x), where x is distance along the beam, the shear force V and bending moment M are related according to Figure 1.2.25 as

FIGURE 1.2.25 Internal reactions in a beam under distributed loading.

w( x ) =

dV dx

V=

dM dx

(1.2.18)

Other useful expressions for any two cross sections A and B of a beam are VA VB = MB MA = Example 7 (Figure 1.2.26)

xB

w( x ) dx = area under w( x )