Chapter 1 Statics of Particles
　　If the size of a body does not affect the solution to the problem under consideration， then this body can be idealized as a particle. A particle has a finite mass， but a negligible size. All the forces acting on a particle can be assumed to be applied at a common point in space， and will thus form a concurrent force system.
　　1.1 Resultant of Concurrent Coplanar Force System
　　A concurrent coplanar force system is one in which the lines of action of all the forces lie in a common plane and intersect at a common point.
　　1. Graphical Solution
　　Two forces acting on a particle can be replaced by a resultant force obtained by drawing the diagonal， passing through the intersection point of the two forces， of the parallelogram which has sides equal to the given forces. For example， two forces F1 and F2 acting on particle O， Fig. 1.1a， can be replaced by a single force R ， Fig. 1.1b， which is called the resultant force of the forces F1 and F2 . The resultant force R can be obtained by drawing a parallelogram using F1 and F2 as two adjacent sides of the parallelogram. The diagonal that passes through particle O represents the resultant force R ， i.e.， R . F1 . F2 . This is known as the parallelogram law.
　　Fig. 1.1
　　When only half of the parallelogogram is considered， an alternative method， Fig. 1.2b， can be obtained by drawing a triangle. The resultant force R of the forces F1 and F2 can be found by arranging F1 and F2 tip-to-tail (or head-to-tail) and then connecting the tail of F1 with the tip of F2 ， i.e.， R . F1 . F2 . This is known as the triangle rule (or trigonometry).
　　Fig. 1.2
　　If a particle is acted upon by three or more concurrent coplanar forces， the resultant force can be obtained by repeated applications of the triangle rule. Considering that particle O is acted upon by concurrent coplanar forces F1， F2 ， and F3 ， Fig. 1.3a， the resultant force R can be obtained graphically by arranging all the given forces tip-to-tail and connecting the tail of the first force with the tip of the last one， Fig. 1.3b， i.e.， R .F .F .F . This is known as the polygon rule.
　　Fig. 1.3
　　We thus conclude that a concurrent coplanar system of forces acting on a particle can be replaced by a resultant force through the intersection point， and that the resultant force is equal to the vector sum of the given concurrent coplanar forces， i.e.，
　　(1.1)
　　Example 1.1 Two rods AC and AD are attached at A to column AB， Fig. E1.1a. Knowing that F .150 N ， θ1=30.， andθ2=15.， determine (a) F2 if the resultant force is directed vertically upward， (b) the magnitude of the resultant force.
　　Solution F1 and F2 can be replaced by R from the parallelogram law， Fig. E1.1b. Considering the shaded triangle shown in Fig. E1.1b and using the law of sines， we have
　　Substituting F1 .150 N ， θ1=30.， andθ2=15. into the above equations， we get
　　Fig. E1.1
　　Example 1.2 Two rods AC and AD are attached at A to column AB， Fig. E1.2a. Knowing that F . 120 N ， F . 100 N ，θ1=35. ， andθ2=20. ， determine the resultant force.
　　Fig. E1.2
　　Solution The force triangle is shown in Fig. E1.2b. Using the laws of cosines and of sines， we have
　　Substituting F1 =120 N ， F2=100 N ， θ1=35. ， and θ2=20. into the above equations
　　2. Analytical Solution
　　Two or more forces acting on a particle can be replaced by a resultant force. Conversely， one force acting on a particle can also be replaced by two or more component forces which， together， have the same effect on the particle. For example， F can be replaced by F1 and F2 ， Fig. 1.4a， where F1 and F2 are the vector components of F . Substituting F1 and F2 for F is called the resolution of a force into components. Clearly， for F there exist infinite sets of vector components， Fig. 1.4b.
　　Fig. 1.4
　　It is often convenient to resolve a force into components perpendicular to each other. For example， F can be resolved into two vector components Fx and Fy， Fig. 1.5a.
　　Fig. 1.5
　　By introducing two unit vectors i and j ， Fig. 1.5b， F can also be expressed as F=F i=F j ， where F and F are the scalar components of F .
　　Denoting by F the magnitude of F and by . the angle of F from the positive x axis， Fig. 1.5c， we can express the scalar components Fx and Fy as follows: Fx =F cos .，Fy= F sin θ.
　　Using the graphical solution to determine the resultant force often requires extensive geometric or trigonometric calculation， especially for finding the resultant force of three or more forces. Instead， problems of this type are easily solved by using the analytical solution.
　　Considering F1， F2 ， and F3 acting on particle O， Fig. 1.6， then the resultant force R of these forces can be expressed， using the graphical solution， as R=F1+F2+F3.Resolving each force， including the resultant force， into its rectangular components， we write
　　(1.2)
　　from which it follows that

Contents
Chapter 1 Statics of Particles 1
1.1 Resultant of Concurrent Coplanar Force System 1
1.2 Equilibrium of Concurrent Coplanar Force System 6
1.3 Resultant of Concurrent Noncoplanar Force System 8
1.4 Equilibrium of Concurrent Noncoplanar Force System 10
Problems 11
Chapter 2 Reduction of Force System 15
2.1 Moment of Force about Point 15
2.2 Moment of Force about Axis 16
2.3 Principle of Moments 17
2.4 Moment of Couple 18
2.5 Equivalence of Force 19
2.6 Reduction of Force System to Force-Couple System 21
2.7 Reduction of Force-Couple System 21
Problems 23
Chapter 3 Statics of Rigid Bodies 25
3.1 Equilibrium of Nonconcurrent Coplanar Force System 25
3.2 Two-Force and Three-Force Members 27
3.3 Planar Trusses 28
3.4 Equilibrium of Nonconcurrent Noncoplanar Force System 32
Problems 32
Chapter 4 Friction 37
4.1 Laws of Friction 37
4.2 Angle of Static Friction 38
4.3 Problems Involving Friction 38
Problems 41
Chapter 5 Kinematics of Particles 43
5.1 Vector Representation 43
5.2 Rectangular Components 44
5.3 Tangential and Normal Components 45
Problems 47
Chapter 6 Kinematics of Rigid Bodies in Plane Motion 49
6.1 Translation 49
6.2 Rotation about Fixed Axis 50
6.3 General Plane Motion 53
Problems 59
Chapter 7 Resultant Motion of Particle 61
7.1 Time Derivatives of Vector 61
7.2 Resultant of Velocities 62
7.3 Resultant of Accelerations 64
Problems 66
Chapter 8 Kinetics of Particles 69
8.1 Equations of Motion 69
8.2 Method of Inertia Force 71
8.3 Method of Work and Energy 72
8.4 Method of Impulse and Momentum 74
8.5 Equations of Motion for Particle System 75
8.6 Equations of Motion for Mass Center of Particle System 76
8.7 Equations of Motion for Particle System about Mass Center 77
Problems 78
Chapter 9 Kinetics of Rigid Bodies in Plane Motion 81
9.1 Equations of Motion 81
9.2 Method of Inertia Force 83
9.3 Method of Work and Energy 84
9.4 Method of Impulse and Momentum 86
9.5 Impact of Rigid Body in Plane Motion 88
Problems 90
Chapter 10 Analytical Mechanics 95
10.1 Constraints and Virtual Work 95
10.2 Principle of Virtual Work 95
10.3 Generalized Coordinates and Generalized Forces 97
10.4 Lagrange’s Equations 98
Problems 101
Chapter 11 Stresses and Strains 105
11.1 External Forces 105
11.2 Internal Forces 105
11.3 Stresses 107
11.4 Strains 109
11.5 Hooke’s Law 111
11.6 Tensile Properties of Low-Carbon Steel 111
11.7 Stress-Strain Curve of Ductile Materials without Distinct Yield Point 114
11.8 Ductile and Brittle Materials 114
11.9 Properties of Materials in Compression 115
Problems 115
Chapter 12 Tension and Compression 117
12.1 Axial Force 117
12.2 Normal Stress on Cross Section 117
12.3 Normal and Shearing Stresses on Oblique Section 119
12.4 Normal Strain 120
12.5 Axial Deformation 122
12.6 Statically Indeterminate Axially-Loaded Bar 123
12.7 Design of Axially-Loaded Bar 124
Problems 125
Chapter 13 Torsion 129
13.1 Torsional Moment 129
13.2 Hooke’s Law in Shear 129
13.3 Shearing Stress on Cross Section 130
13.4 Normal and Shearing Stresses on Oblique Section 133
13.5 Angle of Twist 134
13.6 Statically Indeterminate Shaft 135
13.7 Design of Torsional Shaft 136
Problems 137
Chapter 14 Bending Internal Forces 141
14.1 Shearing-Force and Bending-Moment Diagrams 141
14.2 Relations between Distributed Load， Shearing Force， and Bending Moment 144
14.3 Relations between Concentrated Load， Shearing Force， and Bending Moment 147
Problems 149
Chapter 15 Bending Stresses 151
15.1 Normal Stresses on Cross Section in Pure Bending 151
15.2 Normal and Shearing Stresses on Cross Section in Transverse-Force Bending 155
15.3 Design of Bending Beam 159
Problems 161
Chapter 16 Bending Deformation 163
16.1 Method of Integration 164
16.2 Method of Superposition 165
16.3 Statically Indeterminate Beam 167
Problems 169
Chapter 17 Stress Analysis and Strength Theories 171
17.1 Stress Transformation 171
17.2 Principal Stresses 173
17.3 Maximum Shearing Stress 174
17.4 Pressure Vessels 176
17.5 Generalized Hooke’s Law 178
17.6 Strength Theories 180
Problems 185
Chapter 18 Combined Loadings 187
18.1 Eccentric Tension or Compression 187
18.2 Transverse-Force Bending of I-Beam 189
18.3 Axial Loading and Bending 192
18.4 Torsion and Bending 193
Problems 196
Chapter 19 Stability of Columns 199
19.1 Critical Load of Long Column with Pin Supports 199
19.2 Critical Load of Long Column with Other Supports 200
19.3 Critical Stress of Long Column 201
19.4 Critical Stress of Intermediate Column 203
19.5 Design of Column 204
Problems 206
Chapter 20 Energy Methods 209
20.1 External Work 209
20.2 Stain-Energy Density 211
20.3 Strain Energy 212
20.4 Pri