Course Overview Engineering Mechanics Statics (Freshman Fall) Dynamics (Freshman Spring) Strength of...

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

4 12 12 AB

AB

r

r

0 12 0 4

or = 12 0 0 12

0 0 0 12

AB

AB

r

r

29

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.


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 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.



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.




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

5.7 Constraints for a Rigid BodyImproper Constraints

• Instability caused by the improper constraining by the supports

• When all reactive forces are concurrent at this point, the body is improperly constrained

52

Course Overview Engineering Mechanics Statics (Freshman Fall) Dynamics (Freshman Spring) Strength of...

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Transcript of Course Overview Engineering Mechanics Statics (Freshman Fall) Dynamics (Freshman Spring) Strength of...