Engineering Mechanics: Statics 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

• Basic quantities and idealizations of mechanics

• Newton’s Laws of Motion and gravitation

• Principles for applying the SI system of units

Chapter Outline

• Mechanics

• Fundamental Concepts

• The International System of Units

• Numerical Calculations

• General Procedure for Analysis

Chapter 1 General Principles

2

1.1 Mechanics

• Mechanics can be divided into:

- Rigid-body Mechanics

- Deformable-body Mechanics

- Fluid Mechanics

• Rigid-body Mechanics deals with

- Statics – Equilibrium of bodies; at rest or 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

• Addition of a System of Coplanar Forces

• 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 Forces

Scalar Notation: Components of forces expressed as algebraic scalars

cos and sinx yF F F F

Cartesian Vector Notation in unit vectors i and j

x yF F F i j

• Coplanar Force Resultants

1 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

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

9

2.5 Cartesian Vectors

• Right-Handed Coordinate System

A right-handed rectangular or Cartesian

coordinate system.

• 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

A = A’ + Az

A’ = Ax + Ay

A = Ax + Ay + Az

AA A u10

2.5 Direction Cosines of a Cartesian Vector

cosxA

A

cosyA

A

coszA

A

A / ( / ) ( ) ( / )x y zA A A u A A A / A Ai + j + k

A cos cos cos u i + j + k

2 2 2cos cos cos =1 + +

AA A u

cos cos cos A A Ai + j + k

x y zA A A i + j + k

11

Example 2.8

Express the force F as Cartesian vector.

Since two angles are specified, the third angle is found by

( ) o 60 5 . 0 cos

1 o1205.0cos 1 or

By inspection, = 60º since Fx is in the +x direction

Given F = 200N

Checking:

2 2 22 2 2 100.0 100.0 141.4 200x y zF F F F N

12

5.0cos

145cos60coscos

1coscoscos

222

222

oo

cos i cos j cos k

(200cos60 )i (200cos60 ) j (200cos 45 )k

100 0i 100 0j 141 4k

F F F

. . . N

F

2.7 Position Vector: Displacement and Force

Position Displacement Vector

r = xi + yj + zk

F can be formulated as a Cartesian vector

( )F F F= u = r / r

13

Example 2.13

The 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 2

3 2 6 7m m m m r

14

= 3/7i – 2/7j -6/7k

r

F

F

2.9 Dot Product Laws 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 cos Ta A A u A u

15

x x y y z z

A 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

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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

3. Identify each of the forces

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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 Sphere

Cord CE

FBD at Knot

20

21

Example 3.7 Determine the force developed in each

cable used to support the 40kN crate.

FBD at Point A

To expose all three unknown forces in the cables.

Equations of Equilibrium

Expressing 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

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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

→FB = FC = 23.6kN

FD = 15.0kN

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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

• Moment of Force – Vector Formulation

• Moment of a Force about a Specified Axis

• Moment of a Couple

• Simplification of a Force and Couple System

• Reduction of a Simple Distributed Loading

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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

A x B = - B x A

2. Multiplication by a Scalar

a( 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.

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4.2 Cross Product Cartesian 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 4.4 Two 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

4.5 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

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

0.6 0.8 0 50 100

100

CD

CD

CD

OC

OC

OA OA

OA

M

M

T

TM

r

rF

r

r

r F

rr

33

4.6 Moment of a Couple

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 ,then the

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 System

Concurrent 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

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,

AdAdxxwFAL

R

Location of Resultant Force

• dF produces a moment of xdF = x w(x) dx about O

• Solving for

( )R

L

xF xw x dx

x

( )

( )

L

L

xw x dx

xw x dx

Determine the magnitude and location of the

equivalent resultant force acting on the shaft.

For the differential area element,

For resultant force

For location of line of action,

2

2

0

23 3 3

0

60

2 060 60 160

3 3 3

R

A

F dA x dx

xN

Example 4.21

dxxwdxdA 260

22 4 4 42

0 0

2 060(60 ) 60

4 4 41.5

160 160 160

A

A

xx x dxxdA

x mdA

40

Chapter 5 Equilibrium of a Rigid Body

Objectives

• Equations of equilibrium for a rigid body

• Concept of the free-body diagram for a rigid body

Outline

• Conditions for Rigid Body Equilibrium

• Free-Body Diagrams

• Two and Three-Force Members

• 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 RO

M r F M

42

5.2 Free Body Diagrams

• 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 Diagrams

44

5.2 Free Body Diagrams

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.

Free-Body Diagram

• Support at A is a fixed wall

• Two forces acting on the beam at A denoted as Ax, Ay, with moment MA

• For uniform beam,

Weight, W = 100(9.81) = 981N

acting through beam’s center of gravity

47

5.4 Two- and Three-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 Members

When subjected to three forces,

the forces are concurrent or parallel

48

5.5 3D Free-Body Diagrams

Types 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

5.7 Constraints for a Rigid Body

Redundant Constraints

• More support than needed for equilibrium

• Statically indeterminate: more unknown

loadings than equations of equilibrium

50

5.7 Constraints for a Rigid Body Improper Constraints

• Instability caused by the improper constraining by the supports

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

improperly constrained

51