overviews of statics

L E C T U R E R : P U I P U I L E E

B A S I C T E X T:1 . R . C . H I B B E L E R , E N G I N E E R I N G M E C H A N I C S : S TAT I C S,

1 2 T H E D. , N E W J E R S E Y, P R E N T I C E H A L L , 2 0 0 7 .

R E F E R E N C E S :1 . M E R I A M , J. L . , E N G I N E E R I N G M E C H A N I C S, VO L 1

S TAT I C S, 6 T H E D ( S I E D I T I O N ) . , J O H N W I L E Y & S O N S, 2 0 0 8 . 2 . A N D R E W P Y T E L A N D J A A N K I U S A L A A S. E N G I N E E R I N G

M E C H A N I C S : S TAT I C S 2 N D E D. T H O M S O N L E A R N I N G, 2 0 0 1 .

EGR1174- Engineering Mechanics:

Statics

Overview

1. General Principles & definitions

2. Fundamental concepts

3. Units of measurement, international systems of units

4. Scalars and Vectors

1. General Principles & Definition

Statics and Dynamics are introductory engineering mechanics courses, and they are among the first engineering courses encountered by most students.

Therefore it is appropriate that we begin with a brief exposition on the meaning of the term “engineering mechanics”


‘Physics’ – the science that relates the properties of matter and energy, excluding biological and chemical effects. Physics includes the study of mechanics, thermodynamics, electricity and magnetism, and nuclear physics.

Mechanics’ – the branch of physics that considers the action of forces or fluids that are at rest or in motion. Correspondingly, the primary topics of mechanics are statics and dynamics.

Physics

Thermodynamics

Mechanics Electricity

Deformable-Body Rigid-Body

Fluid

Statics Dynamics

Magnetism


‘Engineering’- the application of the mathematical and physical sciences (physics, chemistry and biology) to the design and manufacture of items that benefit humanity.

‘Engineering Mechanics’ – the branch of engineering that applies the principles of mechanics to mechanical design (i.e. any design that must take into account the effect of forces)

Introduction to statics

Mechanics is the oldest of the physical sciences.

The earliest recorded writings in mechanics are those of Archimedes (287-212 B.C.) on the principle of the lever and the principle of buoyancy.

Substantial progress came later with the formulation of the laws of vectors combination of forces by Stevinus (1548-1620), who also formulated most of the principles of statics.

Fundamental Concept

Four basic quantities are used throughout mechanics

1. Length- used to locate the position of a point in space and thereby describe the size of a physical system. It can be used to define distances and geometry

2. Time- the measure of the succession of events. Although the principles of statics are time independent, this quantity plays an important role in Dynamics

3. Mass- a measure of the inertia of body, which is its resistance to the change of velocity. Mass can also be thought of as the quantity of matter in a body. The mass of a body affects the gravitational attraction force between it and other bodies. This force appears in many applications in statics

4. Force- the action of one body on another. A force tends to move a body in the direction of its action. The action of a force is characterised by its magnitude, by the direction of its action, and by its point of application. Force is a vector quantities.

Fundamental Concepts

3 Idealisation are used in mechanics in order to simplify application of the theory.

1. Particle- a particle has a mass, but a size that can be neglected. 1. i.e. The size of the earth is insignificant compared to the size of its orbit, thus the size of the earth

can be modeled as a particle when studying its orbital motion. 2. When a body is idealised as a particle, the principles of mechanics reduce to a rather simplified form

since the geometry of the body will not be involved in the analysis of the problem

2. Rigid Body- a rigid body can be considered as a combination of a large number of particles in which all the particles remain at a fixed distance from one another, both before and after applying a load.

1. This is important because the material properties of any body that is assumed to be rigid will not have to be considered when studying the effects of forces acting on the body. In most cases, the actual deformations occurring in structures, machines, mechanisms are relatively small, thus the rigid-body assumption is suitable for analysis

3. Concentrated Force- it represents the effect of a loading which is assumed to act at a point on a body.

1. We can represent a load by a concentrated force, provided the area over which the load is applied is very small compared to the overall size of the body. I.e. the contact force between a wheel and the ground.

2.Fundamental Concepts

Engineering mechanics is formulated on the basis of Newton’s three laws of motion.

First law. Particle originally at rest, or moving in a straight line with constant velocity,

tends to remain in this state provided 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,

opposite, and collinear.

2.Fundamental concept

Shortly after formulating his 3 law, Newton postulated a law governing the gravitational attraction between any two particles.

F = Gm1m2/r² G= universal constant gravitation, according to experimental evidence,

G=66.73X10^-12 m³/(kgs²) m1 m2 – mass of each of the two particles r= distance between the two particles

W= Gm1Me/r² m1 = weight of a particle assume earth to have constant density and mass of m2=Me r = distance between the earth’s center and the particle

W=mg By comparison with F=ma, we can see that g is the acceleration due to gravity.

Since it depends on r, then the weight of a body is not an absolute quantity. Instead, its magnitude is determined from where the measurement was made. For most engineering calculations, however, g is determined at sea level and at a latitude of 45º, which is considered the ‘standard location’

3. Units of measurement

The four basic quantities are related by Newton’s second law of motion, F=ma

SI units. International Systems of Units defines length in Meters (m), time in Seconds (s), mass in Kilograms (kg), force in Newton (N).

F= ma, = kg (m/s²) = N

SI symbols and Prefixes: 1000 k, kilo 1000 000 M, mega 1000 000 000 G, giga 0.001 m, milli 0.000 001 µ, micro 0.000 000 001 n, nano

UNIT OF MEASUREMENTGRAVITATIONAL ATTRACTION

Exercise

Exercise

1. If a car is travelling at 88km/h, determine its speed in meter per seconds.

2. Evaluate each of the following and express with SI units having an appropriate prefixa. (50mN)(6GN)b. (400mm)(0.6 MN)²c. 45 MN³/900Gg

3. Two particles have a mass of 8kg and 12kg, respectively. If they are 800mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.

4. Use Newton’s law of universal gravitation to calculate the weight of a 70kg person standing on the surface of the earth. Then repeat the calculation by using W=mg and compare your two results. (Given R= 6371100, me= 5.976 X10^24)

4. Scalars & Vectors

2 kinds of quantities in mechanics- Scalars & Vectors

Scalar quantities are those with which only a magnitude is associated. I.e. time, volume, density, speed, energy, mass

Vector quantities possesses direction as well as magnitude, and must obey the parallelogram law of addition. I.e. displacement, velocity, acceleration, force, moment, and momentum

For ex: Speed is a scalar. It is the magnitude of velocity, which is a vector


Vectors representing physical quantities can be classified as;

Free vector is one whose action is not confined to a unique line in space. I.e. if a body moves without rotation, then the movement of any point in the body may be taken as vector. This vector describes equally well the direction and magnitude of the displacement of every point in the body. Thus, we may represent the displacement of such a body by a free vector

Sliding vector has a unique line of action is space but not a unique point of application. i.e. when an external force acts on a rigid body, the force can be applied at any point along its line of action without changing its effect on the body as a whole.

Fixed vector is one for which a unique point of application is specified. The action of a force on a deformable body must be specified by a fixed vector at the point of application of the force.


The length of the arrow represents the magnitude of the vector

The angle θ between the vector and a fixed axis defines the direction of its line of action.

The head of the arrow indicates the sense of direction of the vector

-V

V

θ

In scalar equation, the symbol will appear in lightface italic type, V.

Boldface type is used for vector quantities, V

When writing vector equation, always be certain to preserve the mathematical distinction between vectors and scalars. Use distinguishing mark for each vector quantity, I.e. V or V to take the place of boldface type in print.



Vector must obey the parallelogram law of addition.

V1 and V2 treated as free vectors, may be replaced by their equivalent vector V, which is diagonal of the parallelogram formed by V1 and V2 as its two sides.

The combination is called the vector sum V = V1 + V2 (vector) V ≠ V1 + V2 (scalar)

Parallelogram can then be reduced to a triangle, which represents the triangle rule


The two vectors V1 and V2 again treated as free vectors, may also be added head-to-tail by the triangle law to obtain the identical vector sum V

The difference V1 – V2 is easily obtained by adding –V2 to V1, where either the triangle or parallelogram

procedure may be used.V’ = V1 – V2, where the minus sign denotes vector

subtraction


Problems that involve the addition of two forces can be solved as follows:

Step 1. Parallelogram LawStep 2. TrigonometryCosine law:

C= √A² + B² - 2AB cos C

Sine Law: A/ sin a = B/sin b = c/sin c


When a force is resolved into two components along the x and y axes, the components are then called ‘Rectangular components’

Then,

Example 1

Figure below show two position vectors, the magnitudes of which are A=60m, and B=100m. A position vector is a vector drawn between two points in space. Determine the resultant R=A+B using the following methods

(1) analytically, using triangle law(2) graphically, using the triangle law

30 º 70 º

AB

Example 2

The vertical force P of magnitude 100kN is applied to the frame shown in figure below. Resolve P into components that are parallel to the members AB and AC of the truss

A

P

B

C

70º

35º

Example 3

4.2 Scalar notation

The rectangular components of force V shown in figure below are found using the parallelogram law, so that V= Vx + Vy

Because these components form a right triangle, their magnitudes can be determined from

Vx = Vcos θVy = V sin θ

4.3 Cartesian Vector Notation

It is also possible to represent the x and y components of a force in terms of Cartesian unit vectors i and j

Each of these unit vectors has a dimensionless magnitude of one, and so they can be used to designate the directions of the x and y axes

V = Vx i + Vy j

4.4 Coplanar Force Resultants

We can use either Cartesian Vector Notation or the Scalar notation to determine the resultant of several coplanar forces

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4.4 Coplanar Forces

Scalar Notation x and y axes are designated positive and

negative Components of forces expressed as

algebraic scalars

sin and cos FFFF

FFF

yx

yx


4.4 Coplanar Forces

Cartesian Vector Notation Cartesian unit vectors i and j are used to

designate the x and y directions Unit vectors i and j have dimensionless

magnitude of unity ( = 1 ) Magnitude is always a positive

quantity, represented by scalars Fx and Fy

jFiFF yx


4.4 Coplanar Forces

Coplanar Force ResultantsTo determine resultant of several coplanar

forces: Resolve force into x and y components Addition of the respective components

using scalar algebra Resultant force is found using the

parallelogram law Cartesian vector notation:

jFiFF

jFiFF

jFiFF

yx

yx

yx

333

222

111


4.4 Coplanar Forces

Coplanar Force Resultants Vector resultant is therefore

If scalar notation are used

jFiF

FFFF

RyRx

R

321

yyyRy

xxxRx

FFFF

FFFF

321

321


4.4 Coplanar Forces

Coplanar Force Resultants In all cases we have

Magnitude of FR can be found by Pythagorean Theorem

yRy

xRx

FF

FF

Rx

RyRyRxR F

FFFF 1-22 tan and

* Take note of sign conventions

Example 4

Using method (1)Scalar notation(2)Cartesian Vector Notation

Example 5

Combine the two forces P and T, which act on the fixed structure at B, into a single equivalent force RP=800 N

T= 600 N

α

A C D

B

60°

6m

3m

Using - Graphical Method, - Geometric method, - Algebraic method (xy coordinate)

Example 6

Using method (1)Scalar notation(2)Cartesian Vector Notation

Example 7

5. Cartesian Vectors

In particularly 3-D problems, it is convenient to express the rectangular components of V in terms of unit vectors i , j , k, which are vector in x-, y- and z-direction.

Cartesian Vector Separating magnitude and direction Simplify vector algebra

V = Vx + Vy + Vz


Magnitude of a Cartesian Vector

Direction of a Cartesian Vector Cos α = Vx/ V, Cos β = Vy/ V Cosγ = Vz/ V

An easy way of obtaining these direction cosines is to form a unit vector nv in the direction of V. nv will have a magnitude of one and be dimensionless provided V is divided by its magnitude.


A vector V may be expressed mathematically by multiplying its magnitude V by a vector n whose magnitude is one and whose direction coincides with that of V. The n is called a ‘unit vector’.

nv Represent the direction of cosines of V nv = Cos α i + Cos β j + Cosγ k Cos ²α + Cos ²β + Cos²γ = 1

n = V/ V

Example 8

Objective 2

To express force and position in Cartesian vector form and explain how to determine the vector’s magnitude and direction.

2 D 3 D

Example 9


Solution I

Scalar Notation:

N

NNF

FF

N

NNF

FF

Ry

yRy

Rx

xRx

8.582

45cos40030sin600

:

8.236

45sin40030cos600

:


Solution I

Resultant Force

From vector addition, direction angle θ is

N

NNFR629

8.5828.236 22

9.67

8.236

8.582tan 1

N

N


Solution II

Cartesian Vector NotationF1 = { 600cos30°i + 600sin30°j } N

F2 = { -400sin45°i + 400cos45°j } N

Thus, FR = F1 + F2

= (600cos30ºN - 400sin45ºN)i + (600sin30ºN + 400cos45ºN)j = {236.8i + 582.8j}NThe magnitude and direction of FR are determined in the same manner as before.


2.5 Cartesian Vectors

Right-Handed Coordinate SystemA rectangular or Cartesian coordinate system is said to be right-handed provided: Thumb of right hand points in the direction

of the positive z axis z-axis for the 2D problem would be

perpendicular, directed out of the page.



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 application of the parallelogram lawA = A’ + Az

A’ = Ax + Ay

Combing the equations, A can be expressed asA = Ax + Ay + Az



Unit Vector Direction of A can be specified using a unit

vector Unit vector has a magnitude of 1 If A is a vector having a magnitude of A ≠ 0,

unit vector having the same direction as A is expressed by uA = A / A. So that

A = A uA



Cartesian Vector Representations 3 components of A act in the positive i, j and

k directions

A = Axi + Ayj + AZk

*Note the magnitude and direction of each components are separated, easing vector algebraic operations.



Magnitude of a Cartesian Vector From the colored triangle,

From the shaded triangle,

Combining the equations gives magnitude of A

222zyx AAAA ++=

22' yx AAA +=

22' zAAA +=



Direction of a Cartesian Vector Orientation of A is defined as the coordinate

direction angles α, β and γ measured between the tail of A and the positive x, y and z axes

0° ≤ α, β and γ ≤ 180 ° The direction cosines of A is

A

Axcos

A

Aycos

A

Azcos



Direction of a Cartesian Vector Angles α, β and γ can be determined by the

inverse cosinesGiven A = Axi + Ayj + AZk

then, uA = A /A = (Ax/A)i + (Ay/A)j + (AZ/A)k

where 222zyx AAAA



Direction of a Cartesian Vector uA can also be expressed as

uA = cosαi + cosβj + cosγk

Since and uA = 1, we have

A as expressed in Cartesian vector form isA = AuA = Acosαi + Acosβj + Acosγk = Axi + Ayj + AZk

222zyx AAAA

1coscoscos 222


2.6 Addition and Subtraction of Cartesian Vectors

Concurrent Force Systems Force resultant is the vector sum of all the

forces in the system

FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk


Example 10

Express the force F as Cartesian vector.


Solution

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

Two possibilities exit, namely

1205.0cos 1

( ) o605.0cos 1 == -a

( ) ( )

oo

5.0707.05.01cos

145cos60coscos

1coscoscos

22

222

222

±=--=

=++

=++

a

a

gba


Solution

By inspection, α = 60º since Fx is in the +x directionGiven F = 200N

F = Fcosαi + Fcosβj + Fcosγk = (200cos60ºN)i + (200cos60ºN)j

+ (200cos45ºN)k = {100.0i + 100.0j + 141.4k}N

Checking:

N

FFFF zyx

2004.1410.1000.100 222

222

Example 10

Note: ‘+ve’ or ‘-ve’ sign MUST take into account when

substituting into Cos α, cos β and cos γ

Example 11


2.7 Position Vectors

x,y,z Coordinates Right-handed coordinate system Positive z axis points upwards, measuring the

height of an object or the altitude of a point Points are measured relative

to the origin, O.



Position Vector Position vector r is defined as a fixed vector

which locates a point in space relative to another point.

E.g. r = xi + yj + zk



Position Vector Vector addition gives rA + r = rB

Solving r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)kor r = (xB – xA)i + (yB – yA)j + (zB –zA)k



Length and direction of cable AB can be found by measuring A and B using the x, y, z axes

Position vector r can be established Magnitude r represent the length of cable Angles, α, β and γ represent the direction

of the cable Unit vector, u = r/r

Example 12


Solution

Position vectorr = [-2m – 1m]i + [2m – 0]j + [3m – (-

3m)]k = {-3i + 2j + 6k}m

Magnitude = length of the rubber band

Unit vector in the director of ru = r /r = -3/7i + 2/7j + 6/7k

mr 7623 222


Solution

α = cos-1(-3/7) = 115°

β = cos-1(2/7) = 73.4°

γ = cos-1(6/7) = 31.0°


2.8 Force Vector Directed along a Line

In 3D problems, direction of F is specified by 2 points, through which its line of action lies

F can be formulated as a Cartesian vector

F = F u = F (r/r)

Note that F has units of forces (N) unlike r, with units of length (m)


2.8 Force Vector Directed along a Line

Force F acting along the chain can be presented as a Cartesian vector by- Establish x, y, z axes- Form a position vector r along length of chain

Unit vector, u = r/r that defines the direction of both the chain and the force

We get F = Fu

Example 13

Example 14

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