1. CHAPTER ONE
General Principles

2. Statics, Dynamics and Mechanics of Materials
Statics deals with the state of rest or motion of bodies with constant velocities
In dynamics, the bodies are allowed to accelerate.
Two Branches of static equilibrium
Statics
Mechanics of Materials
Statics
Equilibrium of Bodies-material independent, uses only static equations
Mechanics of Materials
Relationship between the external loads, the intensity of internal forces and its deformation response

3. Basic Concepts Quantities Length (location, position, size)
Time (succession of events)
Force (Push, Pull)
Mass (Properties of Matter)
ONLY THREE OF FOUR
NEEDED (NEWTONS LAW)
Idealization
Particle (neglect, size, geometry)
Rigid Body (all points within remain in the same position, at fixed distances from each other)
Concentrated Force ( over a very small area, zero)

4. Physical Quantities Physical Qualities Mass, Volume Force area time
distance,
Density
Temperature
Volume
area
length
displacement
velocity
acceleration

5. Scalar and Vector Quantities
Scalar Quantities
Described by their magnitude, mass
(italic form) or lower case (a for A)
Vector Quantities
Described by a magnitude, a direction, and a point of application
(Bold Face) in the book
Bar or Arrow in handwritten work
Magnitude or A (italic) or a = R

6. Newton’s Laws Newton’s First Law
A body at rest tends to remain at rest & a body in motion at a constant velocity will tend to maintain the velocity.
Newton’s Second Law
Change of motion is proportional to the moving force impressed and takes place in the direction of the straight line in which such force is impressed.

7. Newton’s Laws Consequences of Newton’s Third Law
When two bodies interact, a pair of equal and opposite reaction forces will exist at their contact point.
This force pair will have the same magnitude and acts along the same direction, but have opposite sense.
The mutual force of action and reaction between two bodies are equal, opposite, and collinear.
Kg = mass=[m] Newton = Force=[F]
Slug= mass =[m] lbf = Force = [F]

8. Gravitational Law Gravitational Law
G = universal constant of gravitation
m1,m2 = mass of each of the two particles
r= distance

9. What is weight? Weight If = mass of the particle = mass of the earth
r = distance of the particle to the earth’s center
W = weight of the particle if

10. Units Units SI (International System of Units)
Length, time, mass, force -basic quantities
SI (International System of Units)
Meter (m)
Second (sec)
Kilogram (kg)
Newton (N)
(Note: we use bars to denote forces or vectors)
Ex : mass = 1kg
W = 9.81N

11. Conversion of Units Conversion of Units FPS SI Force 1 lb = 4.4482 N
Mass 1 slug = kg
Length 1 ft = m
Ex:

12. CHAPTER ONE.. concludes
General Principles

13. CHAPTER TWO
Force Vectors

14. 2.1 Scalars and Vectors Force Vectors
Scalars : A quantity represented be a number (positive or negative)
Ex: Mass, Volume, Length
(in the book scalars are represented by italics)
Vectors : A quantity which has both
A – magnitude (scalar)
B – direction (sense)
Ex: position, force, moment
Line of action
Magnitude
Sense

15. Forces Classification of Forces Contact
1 – Contacting or surface forces (mechanical)
2 – Non-Contacting or body forces (gravitational, weight)
Area
1 – Distributed Force, uniform and non-uniform
2 – Concentrated Force

16. Forces Classification of Forces Force System
1 – Concurrent : all forces pass through a point
2 – Coplanar : in the same plane
3 – Parallel : parallel line of action
4 – Collinear : common line of action
Three Types
1 – Free (direction, magnitude and sense)
2 – Sliding
3 – Fixed A O
Origin

17. 2.2 Basic Vector Operations
Properties of Vectors
1 – Vector Addition
2 – Vector Subtraction
3 – Vector Multiplication

18. Trigonometric Relations of a triangle
Phythagorean theorem is valid only for a right angled triangle. For any triangle (not necessarily right angled)
Sine Law
Cosine Law A B C

19. Example Ex: If the angle between F1 and F2 =60o and F1 = 54N
Find: the Resultant force and the angle b. R F1 F2
120o
60o

20. Example… contd R F1 F2
120o
60o

21. Properties of Parallelogram
Parallelogram Law
Properties of Parallelogram
A – sum of the three angles in a triangle is 180o
B – sum of the interior angles is 360o
C- opposite sides are equal

22. Vector Addition R A B B C
A + B A R

23. Vector Subtraction
Vector Subtraction A
A - B -B B A

24. 2.3 Vector addition of forces
If we consider, only two forces at a time then the result
can be obtained using parallelogram law, and by using
law of sines and cosines of triangles.
Even if we have multiple (say 5) forces, we take two at
a time to resolve the resultant one by one.
Consider another example in Example 2.4 (a) page 24
Given
To find: Resultant
Procedure: Use law of cosines to find the magnitude, law of sines find the angles

25. 2.3 Vector addition of forces (example 2.4)
Given
To find:
Solution:
Draw the vector diagram
Use the sine law to find

26. 2.4 Cartesian Coordinate Systems
A much more logical way to add/subtract/manipulate vectors is to represent the vector in Cartesian coordinate system. Here we need to find the components of the vector in x,y and z directions.
Simplification of Vector Analysis x y A z n en i j k x y z

27. Vector Representation in terms unit vector
Suppose we know the magnitude of a vector in any
arbitrary orientation, how do we represent the vector?
Unit Vector : a vector with a unit magnitude A en ^

28. 2.5 Right-Handed Coordinate Systemx y z
Right-Handed System
If the thumb of the right hand points in the direction of the positive z-axis when the fingers are pointed in the x-direction & curled from the x-axis to the y-axis.

29. Components of a Vector y A Ay Ax Az z x In 3-D y A Ay Ax x In 2-D
Cartesian (Rectangular) Components of a Vector y A Ay Ax Az z x
In 3-D y A Ay Ax x
In 2-D

30. Cartesian Vectors z Az = Azk A = Aen en y Ax = Axi Ay = Ayj x
Cartesian Unit Vectors y
A = Aen
Ay = Ayj
Ax = Axi
Az = Azk z x n i j k en

31. Cartesian Vectors
Magnitude of a Cartesian Vector y A Ay Ax Az z x

32. 2.6. Addition and Subtraction of vectorsy F Fy Fx Fz z x

33. Force Analysis y F Fy Fx Fz z x

34. Force Analysis
Ex: y z x
10ft
6ft
8ft
F = 600

35. Summary of the Force Analysisz F y F x q =
cos - 1 y y F

36. 2.6 Addition/Subtraction of Cartesion vectors
Since any vector in 3-D can be expressed as components in x,y,z directions, we just need to add the corresponding components since the components are scalars.
Then the addition
Then the subtraction

37. Example 2-9, pg 41
Determine the magnitude and the coordinate direction angles of the resultant force on the ring
Solution:

38. 2.6 ---Example 2-9…2 Unit vector and direction cosines
From these components, we can
determine the angles

39. 2.4---Problem 2-27 (pg 35) Solution:
Four concurrent forces act on the plate. Determine the magnitude of the resultant force and its orientation measured counterclockwise from the positive x axis.
Solution:
Draw the free body diagram. Then resolve forces in x and y directions.

40. 2.4---Problem 2-27…..2
Find the magnitude and direction.

41. 2.7. Position Vectors z
(xb, yb, zb) r
( xa, ya, za) y x
Position vectors can be determined using the coordinates of the end and beginning of the vector

42. 2.8---Problem 2-51, page 56 Solution
Determine the length of the connecting rod AB by first formulating a Cartesian position vector form A to B and then determining its magnitude.
Solution

43. Force Vector Along a Line
A force may be represented by a magnitude & a position
Force is oriented along the vector AB (line AB) B z F
Unit vector along
the line AB A y x

44. 2.6---Problem 2-36 (pg 47)
Solution:

45. 2.6---Problem 2-36…..2

46. 2.6---Problem 2-36…..3

47. 2.9. Dot Product y x A
Dot Product (Scalar Product)

48. 2.9---Problem 2-76, pg 65
A force of F = 80 N is applied to the handle of the wrench. Determine the magnitudes of the components of the force acting along the axis AB of the wrench handle and perpendicular to it.
Hand first rotates 45 in the xy plane and 30 in that plane

49. 2.9---Problem 2-76…..2

50. Application of Dot Product
Component of a Vector along a line
A n A

51. Application of dot product
Angle between two vectors A B

52. CHAPTER TWO
Force Vectors

53. 2.1 Scalars and Vectors Force Vectors
Scalars : A quantity represented be a number (positive or negative)
Ex: Mass, Volume, Length
(in the book scalars are represented by italics)
Vectors : A quantity which has both
A – magnitude (scalar)
B – direction (sense)
Ex: position, force, moment
Line of action
Magnitude
Sense

54. Forces Classification of Forces Contact
1 – Contacting or surface forces (mechanical)
2 – Non-Contacting or body forces (gravitational, weight)
Area
1 – Distributed Force, uniform and non-uniform
2 – Concentrated Force

55. Forces Classification of Forces Force System
1 – Concurrent : all forces pass through a point
2 – Coplanar : in the same plane
3 – Parallel : parallel line of action
4 – Collinear : common line of action
Three Types
1 – Free (direction, magnitude and sense)
2 – Sliding
3 – Fixed A O
Origin

56. 2.2 Basic Vector Operations
Properties of Vectors
1 – Vector Addition
2 – Vector Subtraction
3 – Vector Multiplication

57. Trigonometric Relations of a triangle
Phythagorean theorem is valid only for a right angled triangle. For any triangle (not necessarily right angled)
Sine Law
Cosine Law A B C

58. Example Ex: If the angle between F1 and F2 =60o and F1 = 54N
Find: the Resultant force and the angle b. R F1 F2
120o
60o

59. Example… contd R F1 F2
120o
60o

60. Properties of Parallelogram
Parallelogram Law
Properties of Parallelogram
A – sum of the three angles in a triangle is 180o
B – sum of the interior angles is 360o
C- opposite sides are equal

61. Vector Addition R A B B C
A + B A R

62. Vector Subtraction
Vector Subtraction A
A - B -B B A

63. 2.3 Vector addition of forces
If we consider, only two forces at a time then the result
can be obtained using parallelogram law, and by using
law of sines and cosines of triangles.
Even if we have multiple (say 5) forces, we take two at
a time to resolve the resultant one by one.
Consider another example in Example 2.4 (a) page 24
Given
To find: Resultant
Procedure: Use law of cosines to find the magnitude, law of sines find the angles

64. 2.3 Vector addition of forces (example 2.4)
Given
To find:
Solution:
Draw the vector diagram
Use the sine law to find

65. 2.4 Cartesian Coordinate Systems
A much more logical way to add/subtract/manipulate vectors is to represent the vector in Cartesian coordinate system. Here we need to find the components of the vector in x,y and z directions.
Simplification of Vector Analysis x y A z n en i j k x y z

66. Vector Representation in terms unit vector
Suppose we know the magnitude of a vector in any
arbitrary orientation, how do we represent the vector?
Unit Vector : a vector with a unit magnitude A en ^

67. 2.5 Right-Handed Coordinate Systemx y z
Right-Handed System
If the thumb of the right hand points in the direction of the positive z-axis when the fingers are pointed in the x-direction & curled from the x-axis to the y-axis.

68. Components of a Vector y A Ay Ax Az z x In 3-D y A Ay Ax x In 2-D
Cartesian (Rectangular) Components of a Vector y A Ay Ax Az z x
In 3-D y A Ay Ax x
In 2-D

69. Cartesian Vectors z Az = Azk A = Aen en y Ax = Axi Ay = Ayj x
Cartesian Unit Vectors y
A = Aen
Ay = Ayj
Ax = Axi
Az = Azk z x n i j k en

70. Cartesian Vectors
Magnitude of a Cartesian Vector y A Ay Ax Az z x

71. 2.6. Addition and Subtraction of vectorsy F Fy Fx Fz z x

72. Force Analysis y F Fy Fx Fz z x

73. Force Analysis
Ex: y z x
10ft
6ft
8ft
F = 600

74. Summary of the Force Analysisz F y F x q =
cos - 1 y y F

75. 2.6 Addition/Subtraction of Cartesion vectors
Since any vector in 3-D can be expressed as components in x,y,z directions, we just need to add the corresponding components since the components are scalars.
Then the addition
Then the subtraction

76. Example 2-9, pg 41
Determine the magnitude and the coordinate direction angles of the resultant force on the ring
Solution:

77. 2.6 ---Example 2-9…2 Unit vector and direction cosines
From these components, we can
determine the angles

78. 2.4---Problem 2-27 (pg 35) Solution:
Four concurrent forces act on the plate. Determine the magnitude of the resultant force and its orientation measured counterclockwise from the positive x axis.
Solution:
Draw the free body diagram. Then resolve forces in x and y directions.

79. 2.4---Problem 2-27…..2
Find the magnitude and direction.

80. 2.7. Position Vectors z
(xb, yb, zb) r
( xa, ya, za) y x
Position vectors can be determined using the coordinates of the end and beginning of the vector

81. 2.8---Problem 2-51, page 56 Solution
Determine the length of the connecting rod AB by first formulating a Cartesian position vector form A to B and then determining its magnitude.
Solution

82. Force Vector Along a Line
A force may be represented by a magnitude & a position
Force is oriented along the vector AB (line AB) B z F
Unit vector along
the line AB A y x

83. 2.6---Problem 2-36 (pg 47)
Solution:

84. 2.6---Problem 2-36…..2

85. 2.6---Problem 2-36…..3

86. 2.9. Dot Product y x A
Dot Product (Scalar Product)

87. 2.9---Problem 2-76, pg 65
A force of F = 80 N is applied to the handle of the wrench. Determine the magnitudes of the components of the force acting along the axis AB of the wrench handle and perpendicular to it.
Hand first rotates 45 in the xy plane and 30 in that plane

88. 2.9---Problem 2-76…..2

89. Application of Dot Product
Component of a Vector along a line
A n A

90. Application of dot product
Angle between two vectors A B

91. Chapter 4: Equilibrium
Equilibrium means balance of forces to prevent body from translating, and balance of moments to prevent body from rotating.
Vector analysis in 3-D is the preferred method of solution.

92. 4.1 Conditions for Equilibrium
Equilibrium means that the object is at rest (if originally at rest), or in constant velocity (if originally moving).

93. 4.1 Conditions of Equilibrium Con’t
Moment about any point O,

94. 4.2 Free Body Diagrams
Need to know how to represent support and contact conditions.
If a support prevents translation in any direction, we have a reaction force in that direction.
If a support prevents rotation in any orientation, then we have a couple moment exerted.

95. 4.2 Free Body Diagrams Con’t
Weight always acts at the center of gravity.
W = m g
Consider the case of springs.

96. 4.2 Free Body Diagrams Con’t
Consider the cantilever beam supported by a fixed support at A.
Free body diagram

97. Free Body Diagrams of a Platform
Consider the platform
Exclude all other effects except the platform now!

98. 4.3 Equations of Equilibrium

99. 4.3 Supports and Reactions
Supports are idealized first. Reaction forces (magnitude and direction) and moments then depends on the type of support.
Roller allows motion along the plane. Reaction force is perpendicular to the surface.
Rocker allows rotation at that point. Reaction force is perpendicular to the surface.

100. 4.3 Supports and Reactions-2
Pin connected to a collar. Reaction force is perpendicular to the rod.
Hinge allows motion both in x and y direction but no rotation. Reaction force in x and y only. F not along member
Fixed allows no rotation and no translation. Reaction force vector and moment will result.

101. 4.3 Analysis Precedure
First draw the free body diagram for the loading shown to the right.
Apply equations of equilibrium through force and moment balance.

102. Problem 4-3 (page 138, Section 4.1-4.3)
4.3 Draw the free-body diagram of the automobile, which has a mass of 5 Mg and center of mass at G. The tires are free to roll, so rolling resistance can be neglected. Explain the significance of each force on the diagram.
Solution:

103. Problem 4-7 (page 139, Section 4.1-4.3)
4.7 Draw the free body diagram of the beam. The incline at B is smooth.
Solution:

104. Problem 4-17 (page 154, Section 4.4-4.5)
4.17 Determine the stretch of each spring for equilibrium of 20-Kg block. The springs are shown in their equilibrium position.
Solution:

105. 4.17 Determine the stretch of each spring for equilibrium of 20-Kg block. The springs are shown in their equilibrium position.
Solution-Con’t (slide 2)

106. 4.4 Two-Force Members
Two force members are trusses that have forces (tension or compression) but not couple moments.

107. 4.4 Two and Three-Force Members
Hydraulic cylinder is a two-force member
If a member is subjected to three coplanar forces, then the forces should either be concurrent or coplanar for equilibrium.

108. Problem 4-26 (page 155, Section 4.4-4.5)
4.26 Determine the horizontal and vertical components of reaction at the pin A and the force in the short link BD.
Solution:

109. 4.5 Equilibrium in 3-D
The concept of equilibrium in 3-D is similar. Here we need to solve all the known and unknown in 3-D. Once again we need to know the reaction forces and moments for each type of support, see Table 4-2
Ball- Only 3 forces
Pin- All except 1 moment
Bearing- 2 forces+2 moments
Fixed- All 6 forces & moments

110. 4.6 Equations of Equilibrium
Procedure for Analysis:
Draw Free body diagram for the body under analysis
Mark all the reaction and external forces/moments.
Use the above equations to solve.

111. Problem 4-68 (page 174, Section 4.6-4.7)
4.68 Member AB is supported by a cable BC and at A by a smooth fixed square rod which fits loosely through the square hole of the collar. If the force lb, determine the tension in cable BC and the x, y, z components of reaction at A.
Solution:

112. 4.68 Member AB is supported by a cable BC and at A by a smooth fixed square rod which fits loosely through the square hole of the collar. If the force lb, determine the tension in cable BC and the x, y, z components of reaction at A.
Solution-Con’t (slide 2)

113. 4.68 Member AB is supported by a cable BC and at A by a smooth fixed square rod which fits loosely through the square hole of the collar. If the force lb, determine the tension in cable BC and the x, y, z components of reaction at A.
Solution-Con’t (slide 3)

114. 4.7 Friction
Friction is the force of resistance offered by a body that prevents or retards a body from motion relative to the first.
Friction always acts tangent to the surface and opposing any possible motion.
Friction is caused by small asperities as shown here.
We should consider all the minor surface asperities to get a distributed load .

115. 4.7 Friction-2
Frictional coefficient changes from static to kinetic when the value reduces..
Consider the motion of the following structure.

116. 4.7 Friction-3

117. 4.7 Tipping or impending motion
Tipping during motion or sliding depends if the clockwise moment at the bottom corner is CW or CCW.
Evaluate the location of N with respect to W.

118. Example of pipes stacked
The concrete pipes are stacked. Determine the minimum coefficient of static friction so that the pile does not collapse.
Solution:
Draw the Free body diagrams first.
Top pipe

119. Example of pipes stacked
Solution:
Bottom pipe

120. Example of man on a plank
Solution:

121. Example of man on a plank-2
Solution continued
If the plank is on the verge of moving, slipping would occur at A.

122. Problem 4-100 (page 192, Section 4.6-4.7)
Two boys, each weighing 60 lb, sit at the ends of a uniform board, which has a weight of 30 lb. If the board rests at its center on a post having a coefficient of static friction of with the board, determine the greatest angle of tilt before slipping occurs. Neglect the size of post and the thickness of the board in the calculations.
Solution:

123. Problem 4-112 (page 195, Review Problems)
The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is , determine the required stiffness of the spring at B so that if the beam is loaded with the 800-N force it remains in the horizontal position both before and after loading.
Solution:

124. 4. 112 The horizontal beam is supported by springs at its ends
The horizontal beam is supported by springs at its ends. If the stiffness of the spring at A is , determine the required stiffness of the spring at B so that if the beam is loaded with the 800-N force it remains in the horizontal position both before and after loading.
Solution-Con’t (slide 2)

125. Chapter 4: Equilibrium.. concludes

126. Chapter 6: Center of Gravity and Centroid

127. Chapter 6.1 C.G and Center of Mass

128. Chapter 6.1 C.G and Center of Mass..2
The same way you can find the centroid of the line and the volume.

129. Example 6-2 (pg.251, sections 6.1-6.3)
Locate the centroid of the area shown in figure
Solution I
Differential Element. A differential element of thickness dx is shown in the Figure. The element intersects the curve at the arbitrary point (x,y), and so it has a height of y.
Area and Moment Arms. The area of the element is dA=y dx, and its centroid is located at

130. Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)
Integrations. Applying Equations 6-5 and integrating with respect to x yields

131. Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)
Solution II
Differential Element. The differential element of thickness dy is shown in Figure. The element intersects the curve at the arbitrary point (x,y) and so it has a length (1-x).
Area and Moment Arms. The area of the element is dA = (1-x) dy, and its centroid is located at

132. Con’t Example 6-2 (pg. 251, Sections 6.1-6.3)
Integrations. Applying Equations 6-5 and integrating with respect to y, we obtain

133. Chapter 6: In-class exercise

134. Example 6-6 (pg. 261, Sections 6.1-6.3)
Locate the centroid C of the cross-sectional area for the T-beam
Solution I The y axis is placed along the axis of symmetry so that To obtain we will establish the x axis (reference axis) thought the base of the area.l The area is segmented into two rectangles and the centroidal location for each is established.

135. Con’t Example 6-6 (pg. 261, Sections 6.1-6.3)
Solution II
Using the same two segments, the x axis can be located at the top of the area. Here
The negative sign indicates that C is located below the origin, which is to be expected. Also note that from the two answers in in =13.0 in., which is the depth of the beam as expected

136. Con’t Example 6-6 (pg. 261, Sections 6.1-6.3)
Solution III
It is also possible to consider the cross-sectional area to be one large rectangle less two small rectangles. Hence we have

137. Problem 6-30 (pg. 264, Sections 6.1-6.3)
Determine the distance to the centroid of the shaded area.
Solution

138. 6.6 Moments of Inertia For Areas
By definition moments of inertia with respect to any axis (i.e. x and y) are
Polar moment of inertia
Always positive value
Units:

139. Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)

140. Geometric Properties of An Area and Volume Geometric Properties of An Area and Volume (page 786-787)

141. Con’t 6.6 Moments of Inertia For Areas
Example 6.14
Lets proof for rectangular area
Solution
Part (a)

142. Example 6-15 (pg. 287 Sections 6.8-6.9)
Determine the moment of inertia of the shaded area about the x axis.
Solution I (CASE I)
A differential element of area that is parallel to the x axis is chose for integration.
dA =(100-x) dy.
Limits of integration wrt y, y=0 to y=200 mm

143. 6-7 Parallel-Axis Theorem
If we know Moment of Inertia of a given axis, we can compute M.I about another parallel axis

144. Con’t 6-7 Parallel-Axis Theorem
As a result:

145. Con’t 6-7 Parallel-Axis Theorem

146. Con’t Example 6-15 (pg. 287 Sections 6.8-6.9)
Solution II (CASE 2)
A differential element parallel to the y axis is chosen for integration.
Use the parallel-axis theorem to determine the moment of inertia of the element with respect to this axis.
For a rectangle having a base b and height h, the moment of inertia about its centroidal axis is

147. Con’t Example 6-15 (pg. 287 Sections 6.8-6.9)
Solution II (CASE 2)
For the differential element , b = dx and h = y, and thus
Since the centroid of the element is at from the x axis, the moment of inertia of the element about this axis is

148. Example 6-18

149. Con’t Example 6-18

150. Example 6-17 (pg. 291, Sections 6.8-6.9)
Determine the moment of inertia of the cross-sectional area of the T-beam about the centroidal axis.
In class workout

151. Problem 6-87 (pg. 294, Sections 6.8-6.9)
Determine the moment of inertia of the shaded area with respect to a horizontal axis passing through the centroid of the section
Solution:

152. Problem 6-92 (pg. 294, Sections 6.8-6.9)
Determine the moment of inertia of the beam’s cross-sectional area about the y axis
Solution:

153. Problem 6-8 (pg. 257, Sections 6.1-6.3)
Determine the location of the centroid of the quarter elliptical plate.
Solution

154. Con’t Problem 6-8 (pg. 257, Sections 6.1-6.3)
Con’t Solution

155. Chapter 6: Concludes….