course overview engineering mechanics statics (freshman fall) dynamics (freshman spring) strength of...
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Course Overview
Engineering Mechanics
Statics (Freshman Fall)
Dynamics (Freshman Spring)
Strength of Materials (Sophomore Fall)
Mechanism Kinematics and Dynamics (Sophomore Spring )
Aircraft structures (Sophomore Spring and Junior Fall)
Vibration(Senior)
Statics: force distribution on a system
Dynamics: x(t)= f(F(t)) displacement as a function of time and applied force
Strength of Materials: δ = f(F) deflection of deformable bodies subject to static applied force
Vibration: x(t) = f(F(t)) displacement on particles and rigid bodies as a function of time and frequency
0F
1
Engineering Mechanics: Statics
• Chapter Objectives
• Basic quantities and idealizations of mechanics
• Newton’s Laws of Motion and Gravitation
• Principles for applying the SI system of units
• General guide for solving problemsChapter Outline
• Mechanics
• Fundamental Concepts
• Units of Measurement
• The International System of Units
• Numerical Calculations
• General Procedure for Analysis
Chapter 1 General Principles
2
1.1 Mechanics
• Mechanics can be divided into 3 branches:
- Rigid-body Mechanics
- Deformable-body Mechanics
- Fluid Mechanics
• Rigid-body Mechanics deals with
- Statics – Equilibrium of bodies, at rest, and Moving with constant velocity
- Dynamics – Accelerated motion of bodies
3
1.2 Fundamentals Concepts
Basic Quantities
• Length - locate the position of a point in space
• Mass - measure of a quantity of matter
• Time - succession of events
• Force - a “push” or “pull” exerted by one body on another
Idealizations
• Particle - has a mass and size can be neglected
• Rigid Body - a combination of a large number of particles
• Concentrated Force - the effect of a loading
4
1.2 Newton’s Laws of Motion
• First Law - “A particle originally at rest, or moving in a straight line with constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force.”
• Second Law - “A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.”
• Third Law - “The mutual forces of action and reaction between two particles are equal and, opposite and collinear.”
maF
5
Chapter 2 Force Vector
• Scalars and Vectors
• Vector Operations
• Vector Addition of Forces
• Addition of a System of Coplanar Forces
• Cartesian Vectors
• Addition and Subtraction of Cartesian Vectors
• Position Vectors
• Force Vector Directed along a Line
• Dot Product
Chapter Outline
6
2.1 Scalar and Vector• Scalar
– A quantity characterized by a positive or negative number, indicated by letters in italic such as A, e.g. Mass, volume and length
• Vector – A quantity that has magnitude and direction, e.g. position, force and moment, presented as A and its magnitude (positive quantity) as A
• Vector Subtraction R’ = A – B = A + ( - B )
Finding a Resultant Force• Parallelogram law is carried out to find the resultant force FR = ( F1 + F2 )
7
2.4 Addition of a System of Coplanar ForcesScalar Notation
– Components of forces expressed as algebraic scalars
cos and sinx yF F F F
Cartesian Vector Notation • Cartesian unit vectors i and j are used to designate
the x and y directions with unit vectors i and jx yF F F i j
• Coplanar Force Resultants1 1 1
2 2 2
3 3 3
x y
x y
x y
F F
F F
F F
F i j
F i j
F i j
Coplanar Force Resultants
Scalar notation
1 2 3R Rx RyF F F F F F i j
yyyRy
xxxRx
FFFF
FFFF
321
321
8
Example 2.6
The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force.
Cartesian Vector Notation
F1 = { 600cos30°i + 600sin30°j } N
F2 = { -400sin45°i + 400cos45°j } N
Thus,
FR = F1 + F2
= (600cos30º - 400sin45º)i + (600sin30º + 400cos45º)j
= {236.8i + 582.8j}N
The magnitude and direction of FR are determined in the same manner as before. 9
2.5 Cartesian Vectors• Right-Handed Coordinate System
A rectangular or Cartesian coordinate system is said to be right-handed provided:
• Rectangular Components of a Vector– A vector A may have one, two or three rectangular
components along the x, y and z axes, depending on orientation
– By two successive applications of the parallelogram law
A = A’ + Az
A’ = Ax + Ay
– Combing the equations, A can be expressed as
A = Ax + Ay + Az
AA A u 10
2.5 Direction Cosines of a Cartesian Vector– Orientation of A is defined as the coordinate direction
angles α, β and γ measured between A and the positive x, y and z axes, 0°≤ α, β and γ ≤ 180 °
– The direction cosines of A is
cos xA A
cos yA A
cos zA A
A / ( / ) ( ) ( / )x y zA A A u A A A / A Ai + j + k
A cos cos cos u i + j + k2 2 2cos cos cos = 1 + +
AA A u
cos cos cos A A Ai + j + k
x y zA A A i + j + k11
Example 2.8
Express the force F as Cartesian vector.
Since two angles are specified, the third angle is found by
( ) ( )
5.0707.05.01cos
145cos60coscos
1coscoscos
22
222
222
( ) 605.0cos 1 1205.0cos 1 orBy inspection, β= 60º since Fx is in the +x directionGiven F = 200N
Checking:
cos cos cos
(200cos60 ) (200cos60 ) (200cos 45 )
100.0 100.0 141.4 N
F F F
F i j k
i j k
= i j k
N
FFFF zyx
2004.1410.1000.100 222
222
12
2.7 Position Vector: Displacement and Force
Position Displacement Vector
– r = xi + yj + zk
F can be formulated as a Cartesian vector
( )F = u = r / rF F
13
Example 2.13The man pulls on the cord with a force of 350N. Represent this force acting on the support A as a Cartesian vector and determine its direction.
2 2 23 2 6 7m m m m r
Solution:
14
2.9 Dot ProductLaws of Operation
1. Commutative law
A·B = B·A or ATB=BTA
2. Multiplication by a scalar
a(A·B) = (aA)·B = A·(aB) = (A·B)a
3. Distribution law
A·(B + D) = (A·B) + (A·D)
4. Cartesian Vector Formulation
- Dot product of 2 vectors A and B
5. Applications
- The angle formed between two vectors
- The components of a vector parallel and perpendicular to a line
Cartesian Vector Formulation- Dot product of Cartesian unit vectors
i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 1 j·k = 1
1 0 0cos ( / ( )) 0 180 A B A B
A cosa A A u15
x x y y z zA B = A B A B A B
Example 2.17
The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB.
2 2 2
2 6 3
2 6 3
0.286 0.857 0.429
cos
. 300 0.286 0.857 0.429
(0)(0.286) (300)(0.857) (0)(0.429)
257.1
BB
B
AB
B
ru
r
F F
F u
N
i j k
i j k
j i j k
Since
Thus
16
Solution
Since result is a positive scalar, FAB has the same sense of direction as uB.
Perpendicular component
257.1 0.286 0.857 0.429
{73.5 220 110 }
300 (73.5 220 110 ) { 73.5 80 110 }
AB AB AB
AB
F F u
N
N
F F F N
i j k
i j k
j i j k i j k
Magnitude can be determined from F┴ or from Pythagorean Theorem
2 2 2 2
300 257.1 155ABF F F N N N
17
Chapter 3 Equilibrium of a Particle
• Concept of the free-body diagram for a particle
• Solve particle equilibrium problems using the equations of equilibrium
Chapter Outline• Condition for the Equilibrium of a Particle
• The Free-Body Diagram
• Coplanar Systems
• Three-Dimensional Force Systems
Chapter Objectives
18
3.2 The Free-Body Diagram
• Spring
– Linear elastic spring: with spring constant or stiffness k. F = ks
• Cables and Pulley– Cables (or cords) are assumed negligible weight and cannot
stretch– Tension always acts in the direction of the cable– Tension force must have a constant magnitude for
equilibrium– For any angle , the cable is subjected to a constant tension T
Procedure for Drawing a FBD
1. Draw outlined shape
2. Show all the forces Active forces: particle in motion and Reactive forces: constraints that prevent motion
3. Identify each of the forces 19
Example 3.1
The sphere has a mass of 6kg and is supported. Draw a free-body diagram of the sphere, the cord CE and the knot at C.
FBD at SphereTwo forces acting, weight 6kg (9.81m/s2) = 58.9N and the force on cord CE.
Cord CETwo forces acting: sphere and knotFor equilibrium, FCE = FEC
FBD at Knot3 forces acting: cord CBA, cord CE and spring CD
20
Example 3.7
Determine the force developed in each cable used to support the 40kN crate.
FBD at Point ATo expose all three unknown forces in the cables.Equations of EquilibriumExpressing each forces in Cartesian vectors, FB = FB(rB / rB) = -0.318FBi – 0.424FBj + 0.848FBk
FC = FC (rC / rC) = -0.318FCi – 0.424FCj + 0.848FCk
FD = FDi and W = -40k
22
3.4 Three-Dimensional Systems
Solution
For equilibrium,
∑F = 0; FB + FC + FD + W = 0
– 0.318FBi – 0.424FBj + 0.848FBk – 0.318FCi
– 0.424FCj + 0.848FCk + FDi - 40k = 0
∑Fx = 0; – 0.318FB – 0.318FC + FD = 0
∑Fy = 0; – 0.424FB – 0.424FC = 0
∑Fz = 0; 0.848FB + 0.848FC – 40 = 0
Solving,
FB = FC = 23.6kN
FD = 15.0kN 23
Chapter 4 Force System Resultants
Chapter Objectives•Concept of moment of a force in two and three dimensions
•Method for finding the moment of a force about a specified axis.
•Define the moment of a couple.
•Determine the resultants of non-concurrent force systems
•Reduce a simple distributed loading to a resultant force having a specified location
Chapter Outline• Moment of a Force – Scalar Formation
• Cross Product
• Moment of Force – Vector Formulation
• Principle of Moments
• Moment of a Force about a Specified Axis
• Moment of a Couple
• Simplification of a Force and Couple System
• Further Simplification of a Force and Couple System
• Reduction of a Simple Distributed Loading24
4.1 Moment of a Force – Scalar Formation• Moment of a force about a point or axis – a measure of the tendency of the force
to cause a body to rotate about the point or axis
• Torque – tendency of rotation caused by Fx or simple moment (Mo)z
Magnitude
• For magnitude of MO, MO = Fd (Nm)
where d = perpendicular distance from O to its line of action of force
Direction
• Direction using “right hand rule”
Resultant Moment MRo = ∑Fd
25
4.2 Cross Product• Cross product of two vectors A and B yields C, which is
written as C = A x B = (AB sinθ)uC
Laws of Operations
1. Commutative law is not valid
A x B ≠ B x A
Rather, A x B = - B x A
2. Multiplication by a Scalara( A x B ) = (aA) x B = A x (aB) = ( A x B )a
3. Distributive Law A x ( B + D ) = ( A x B ) + ( A x D )
Proper order of the cross product must be maintained since they are not commutative
26
4.2 Cross ProductCartesian Vector Formulation
• Use C = AB sinθ on a pair of Cartesian unit vectors
• A more compact determinant in the form as
x y z
x y z
i j k
A A A
B B B
A B
• Moment of force F about point O can be expressed using cross product
MO = r x F
• For magnitude of cross product,
MO = rF sinθ
• Treat r as a sliding vector. Since d = r sinθ,
MO = rF sinθ = F (rsinθ) = Fd 27
4.3 Moment of Force - Vector Formulation
O x y z
x y z
r r r
F F F
i j k
M r F
0 ( ) ( ) ( )
0
0
0
y z z y x z z x x y y x
z y x
z x y
y x z
r F r F r F r F r F r F
r r F
r r F
r r F
M i j k
For force expressed in Cartesian form,
with the determinant expended,
28
Example 4.4Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector.
5 m
4 5 2 m
A
B
j
i j k
r
r
The resultant moment about O is
0 5 0 4 5 2
60 40 20 80 40 30
0 0 5 60 0 2 5 80
0 0 0 40 2 0 4 40
5 0 0 20 5 4 0 30
30 40 60 kN m
O A B
i j k i j k
i j k
M r F r F r F
30
4.4 Principles of Moments
• Since F = F1 + F2,
MO = r x F = r x (F1 + F2) = r x F1 + r x F24.5 Moment of a Force about a Specified Axis
Vector Analysis• For magnitude of MA,
MA = MOcosθ = MO·ua
where ua = unit vector
• In determinant form,
( )ax ay az
a ax x y z
x y z
u u u
r r r
F F F
M u r F
31
Example 4.8
Determine the moment produced by the force F which tends to rotate the rod about the AB axis.
T
0.6 0
0 , 0
0.3 300
0 0.3 0 0 0
0.3 0 0.6 0 180
0 0.6 0 300 0
0.4 0.41
0.2 , 0.20.2
0 0
01
M 0.4 0.2 0 180 80.40.2
0
c
B B
AB B
r F
M r F
r u
u M
32
T
T
0.4 0.4 0.2
300
0.1 0.4 0.3
0 0.3 0.4 200 100
0.3 0 0.1 200 50
0.4 0.1 0 100 100
1001
M 0.6 0.8 0 50 100
100
CD
CD
CD
OC
OC
OA OAOA
M
M
r
rF
r
r
r F
rr
Solution:
33
4.6 Moment of a Couple• Couple
– two parallel forces of the same magnitude but opposite direction separated by perpendicular distance d
• Resultant force = 0
Scalar Formulation
• Magnitude of couple moment M = Fd
• M acts perpendicular to plane containing the forces
Vector Formulation
• For couple moment, M = r x F
Equivalent Couples
• 2 couples are equivalent if they produce the same moment
• Forces of equal couples lie on the same plane or plane parallel to one another
34
Example 4.12
Determine the couple moment acting on the pipe. Segment
AB is directed 30° below the x–y plane.
Take moment about point O,
M = rA x (-250k) + rB x (250k)
= (0.8j) X (-250k) + (0.66cos30ºi + 0.8j – 0.6sin30ºk) x (250k)
= {-130j}N.cm
Take moment about point A
M = rAB x (250k)
= (0.6cos30°i – 0.6sin30°k) x (250k)
= {-130j}N.cm
Take moment about point A or B,
M = Fd = 250N(0.5196m) = 129.9N.cm
M acts in the –j direction M = {-130j}N.cm 35
4.7 Simplification of a Force and Couple System
• Equivalent resultant force acting at point O and a resultant couple moment is expressed as
• If force system lies in the x–y plane and couple moments are perpendicular to this plane,
R
R OO
F F
M M M
R xx
R yy
R OO
F F
F F
M M M36
4.8 Simplification of a Force and Couple SystemConcurrent Force System
• A concurrent force system is where lines of action of all the forces intersect at a common point O
R F F
Coplanar Force System
• Lines of action of all the forces lie in the same plane
• Resultant force of this system also lies in this plane
37
4.8 Simplification of a Force and Couple System• Consists of forces that are all parallel to the z axis
38
4.9 Distributed Loading• Large surface area of a body may be subjected to distributed loadings,
often defined as pressure measured in Pascal (Pa): 1 Pa = 1N/m2
39
Magnitude of Resultant Force• Magnitude of dF is determined from differential area dA under the
loading curve. • For length L, AdAdxxwF
AL
R
Location of Resultant Force
• dF produces a moment of xdF = x w(x) dx about O
• Solving for
( )R
L
xF xw x dxx
( )
( )L
L
xw x dx
xw x dx
Example 4.21
Determine the magnitude and location of the equivalent resultant force acting on the shaft.
Solution
For the colored differential area element,
For resultant force
For location of line of action,
dxxwdxdA 260
22
0
23 3 3
0
;
60
2 060 60 160
3 3 3
R
R
A
F F
F dA x dx
xN
22 4 4 42
0 0
2 060(60 ) 60
4 4 41.5
160 160 160A
A
xx x dxxdA
x mdA
40
Chapter 5 Equilibrium of a Rigid BodyObjectives
• Develop the equations of equilibrium for a rigid body
• Concept of the free-body diagram for a rigid body
• Solve rigid-body equilibrium problems using the equations of equilibrium
Outline• Conditions for Rigid Equilibrium• Free-Body Diagrams• Equations of Equilibrium• Two and Three-Force Members• Free Body Diagrams• Equations of Equilibrium• Constraints and Statical Determinacy 41
5.1 Conditions for Rigid-Body Equilibrium
• The equilibrium of a body is expressed as
• Consider summing moments about some other point, such as point A, we require
0
0
R
R OO
F F
M M
0A R R O M r F M
42
5.2 Free Body Diagrams
Support Reactions
• If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction.
• If rotation is prevented, a couple moment is exerted on the body.
43
5.2 Free Body DiagramsTypes of Connection Reaction Number of Unknowns
One unknown. The reaction is a tension force which acts away from the member in the direction of the cable
One unknown. The reaction is a force which acts along the axis of the link.
One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.
One unknown. The reaction is a force which acts perpendicular to the slot.
One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.
θ θ
F
θ θ
Fθ
F
θF
θ
θ θF
θF
θ θF
(1)
(2)
(3)
(4)
(5)
cable
weightless link
roller
roller or pin in confined smooth slot
rocker
44
5.2 Free Body Diagrams
One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.
One unknown. The reaction is a force which acts perpendicular to the rod.
Two unknowns. The reactions are two components of force, or the magnitude and direction φ of the resultant force. Note that φ and θ are not necessarily equal [usually not, unless the rod shown is a link as in (2)].
Two unknowns. The reaction are the couple moment and the force which acts perpendicular to the rod.
Three unknowns. The reaction are the couple moment and the two force components, or the couple moment and the magnitude and direction φ of the resultant force.
θ θsmooth contacting surface
F
θ
member pin connected θ
Fθ
F
θ
smooth pin or hinge
Fx
Fy
F
φ
Member fixed connected to collar on smooth rod
F M
fixed support
Fx
Fy
M
φ
F
(6)
(7)
(8)
(9)
(10)
45
5.2 Free Body Diagram
Weight and Center of Gravity• Each particle has a specified weight
• System can be represented by a single resultant force, known as weight W of the body
• Location of the force application is known as the center of gravity
46
Example 5.1
Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg.
SolutionFree-Body Diagram• Support at A is a fixed wall• Two forces acting on the beam at A denoted as Ax, Ay, with moment MA
• Unknown magnitudes of these vectors• For uniform beam,
Weight, W = 100(9.81) = 981N acting through beam’s center of gravity, 3m from A 47
5.4 Two- and Three-Force Members
Two-Force Members
• When forces are applied at only two points on a member, the member is called a two-force member
• Only force magnitude must be determined
Three-Force MembersWhen subjected to three forces, the forces are concurrent or parallel
48
5.5 3D Free-Body DiagramsTypes of Connection Reaction Number of Unknowns
One unknown. The reaction is a force which acts away from the member in the known direction of the cable.
One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.
One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact.
Three unknown. The reaction are three rectangular force components.
Four unknown. The reaction are two force and two couple moment components which acts perpendicular to the shaft. Note: The couple moments are generally not applied of the body is supported elsewhere. See the example.
F
F
F
Fx
Fy
Fz
Fx
Fz
Mz
Mx
(1)
(2)
(3)
(4)
(5)
cable
smooth surface support
roller
ball and socked
single journal bearing
49
5.5 Free-Body Diagrams
Five unknowns. The reaction are two force and three couple moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the example.
Five unknowns. The reaction are two force and three couple moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the example.
Five unknowns. The reaction are two force and three couple moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the example.
Five unknowns. The reaction are two force and three couple moment components. Note: The couple moments are generally not applied if the body is supported elsewhere. See the example.
Six unknowns. The reactions are three force and three couple moment components.
(6)
(7)
(8)
(9)
(10)
fixed support
single hinge
single smooth pin
single thrust bearing
single journal bearing with square shaft
Fx
Fz
Mz
Mx
Fz
Mz
Mx
My
Fy
Mz
My
Fz
Fx Fy
Fx
Fy
Fz
Mz
My
Mx
Fx
Mx
Fy My
Fz
Mz
50
5.7 Constraints for a Rigid Body
Redundant Constraints• More support than needed for equilibrium
• Statically indeterminate: more unknown loadings than equations of equilibrium
51