Principle of Engineering ENG2301

Mechanics Section

Textbook: A Foundation Course in Statics and Dynamics Addison Wesley Longman 1997

Syllabus Overview

A Statics B Dynamics

Units

force Newton (N) stressNewton per metre squared (N/m2 ) or Pascal, 1 Pa = 1 N/m2 (Pa) pressure Newton per metre squared (N/m2 ) or bar, 1 bar = 1x105 N/m2 (bar) moment, torque, couple Newton . Metre (Nm)

Units

Most commonly used prefixes micro x 10-6 milli x 10-3m kilo x 103 k megax 106 M giga x 109 G * Note Capitals and lower case letters are

important

Scalars and Vector

Two kind of quantities: ScalarScalar VectorVector Scalar quantities Scalar quantities have magnitude have magnitude but but no no

directional propertiesdirectional properties can be handled by ordinary algebra, e.g. c= can be handled by ordinary algebra, e.g. c=

a+b, c= 8 if a=3, b= 5a+b, c= 8 if a=3, b= 5 e.g. time, mass, speed and energy etc. etc....e.g. time, mass, speed and energy etc. etc....

Vector Associated with directions and

magnitude e.g. Force, displacement,

acceleration and velocity Can be represented by a straight

line with arrowhead and the magnitude is shown by the length

l

Vector Addition and Subtraction By Triangle or Parallelogram laws

AdditionAddition V = V1 + V2

V is called the resultant vector

V1

1

V2

2

V1

V2

VV1

V2

V

(a) (c)(b)

Vector Addition and Subtraction

SubtractionSubtraction V’ = V1 - V2

can be regarded as V’ = V1 + (- V2) - V2 is drawn in the opposite direction

V’ is the resultant vector

V1

1

-V 2

2

(a) (b)

V'V1

-V 2 V1

V'

(c)

-V 2

V1

V2V3

V4V1 2+V

V1 2+V 3+V

V=V 1 2+V 3+V 4+V

V1V2

V3

V4

(b)(a)

Vector Addition and Subtraction

Adding more than two vectorsAdding more than two vectors V’ = V1 + V2 +V3 +V4

Resolution of Vectors Any vector can be resolved into components Commonly resolve into two components

perpendicular to each other V = Vx + Vy

Vx = V cos Vy = V sin magnitude V = Vx2 + Vy2) = tan-1 (Vy /Vx )

V

x

y

Vx

V y

o

Force and Newton’s First Law

First Law First Law - If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest), or will move with constant speed in a straight line (if originally in motion).

State of Equilibrium State of Equilibrium - Equilibrium exists when all the forces on a particle are in balance . The velocity of a particle does not change , if the particle is in Equilibrium .

Interpretation of First Law

A body is in Equilibrium if it moves with constant velocity. A body at rest is a special case of constant velocity i.e. v = 0 = constant.

For a body to be in Equilibrium the resultant force (meaning the vector addition of all the forces) acting on the body must be zero.

A Force can be defined as 'that which tends to cause a particle to accelerate', assuming that the force is not in Equilibrium with other forces acting on the body.

Force

A force cannot be seen, only the effect of a force on a body may be seen.

Force Units: S.I. Unit ,Newton, (N) or (kN)

Force is a vector quantity. It has both magnitude and direction.

Force Vectors

Polar and Rectangular Coordinates

Fx

Fy F

fx

fy

F = F F x y ;

;F.cos=Fx ;F.sinF y

2y

2x FF=F

x

y1-

F

Ftan

Example 1

Calculate the components in rectangular

coordinates of the 600 N force. Solution

Fx

Fy 600N

35

N 49135cos.600 xF

N 34435sin.600 yF

Example 2

A force vector has the components 600 kN and 300 kN in the x and y directions respectively, calculate the components in polar coordinates.

Solution

kN 8.670600300 22 F

56.26600

300tan 1

Resultant Force

Parallelogram Method

ForceF2

F1

Resultant R

Force

Particle

means "equivalent to"Note

F1

F2

R

Fy

Fx

Resultant Force

Algebraic Method

F x2 ,

F y2 ,

F2

2

F x1,

F y1,F1

1

2211,2,1 coscos FFFFR xxx

2211,2,1 sinsin FFFFR yyy

22yx RRR

x

y

R

R1tan

Resultant Force

Triangle of Forces Method

R

F1

F2

Fx

Fy

RF1

F2

Fx

Fy

Order is not important

Example 3

Find the magnitude and direction of the resultant (i.e. in polar coordinates) of the two forces shown in the diagram,

a) Using the Parallelogram Method b) Using the Triangle of Forces Method c) Using the algebraic calculation method Solution

Example 3 (Solution)6kN

4kN

15 30

kN 332.115cos430cos6 xR

kN 035.415sin430sin6 yR

kN 249.4)035.4()332.1( 22 R

73.71332.1

035.4tan 1

Or -108.260 from +ve x axis

Equilibrium of Concurrent forces

Force

F1

F2 Resultant R

Equilibrant EEquilibrant E

E

F1

F2

Fx

Fy R Equilibrant E are equal and opposite to Resultant R

E = -R

Conditions for Equilibrium

Coplanar: all forces being in the same plane (e.g.only x-y plane, no forces in z direction)

Concurrent: all forces acting at the same point (particle)

;0F and ;0For ;0Fn=i

0=ii y,

n=i

0=ii x,

n=i

0=ii

;0FFF x,3x,2x,1 For three forces acting on a particle

;0FFF y,3y,2y,1

Some Definitions Particle is a material body whose linear

dimensions are small enough to be irrelevant

Rigid Body is a body that does not deform (change shape) as a result of the forces acting on it .

Polygon of Forces

Equilibrium under multiple forces

Rigid body under concurrent forces

F1

F2 F3

F4

F5F1

F2F3

F4

F5

Forces acting on particle

Resultant and Equilibrant

F1F2

F3

F5

F4

Fy

Fx

F1F2

F3

F4

y

x

R

Resultant = - Equilibrant

R = - F5

Example 4 The diagram shows three forces acting on a The diagram shows three forces acting on a

particle .particle . Find the equilibrant by drawing the polygon

of forces.

20

50

45 kN

122 kN

84 kN

30

Newton’s Third Law

The forces of action and reaction between bodies in contact have the same magnitude, but opposite in direction.

BANG!

Solid Surface

Hammer

Action and Reaction

Solid Surface

Hammer

Force on Surface caused by Hammer

Force on Hammer caused by Surface

Free Body Diagram of Hammer

Free Body diagram of Surface

Free BodyBoundary

Free Body Diagram

Free Body Diagram Free Body Diagram - used to describe the system of forces acting on a body when considered in isolation

R

mgR

R

R

Free Body Diagram

W

W

Wr Reaction

Wa Action

Free Body

Boundary

System of Particles or BodiesTwo or more bodies or particles connected together are referred to as a system of bodies

or particles.

External Force External forces are all the forces acting on a body defined as a free body or free system of bodies, including the actions due to other

bodies and the reactions due to supports.

Transmissibility of Force

FF F

Load and Reaction

Loads are forces that are applied to bodies or systems of bodies.

Reactions at points supporting bodies are a consequence of the loads applied to a body and the equilibrium of a body.

Tensile and Compressive Forces

Action Reaction

Compressive Force

Action Reaction

Tensile Force

Push on the body which is called a compressive force

Pull on a body which is called a tensile force

Procedure for drawing a free body diagram

Step 1: Imagine the particle to be isolated or cut “free from its surroundings. Draw or sketch its outlined shape.

Step 2: Indicate on this sketch all the forces that act on the particle. These forces can be applied surface forces, reaction forces and/or force of attraction.

Procedure for drawing a free body diagram

Supportedpulley

PullingForce

Mass

Spring

PullingForce

Mass

Spring

mg

Fp

F1

F1

Fs

Fs

Procedure for drawing a free body diagram

Step 3: The forces that known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that unknown.

Example 5

Weight = 10 N

Ceiling Support

Ring

Weight = 10 N

Ceiling Support

Ring

Gravity Load =10 N

Reaction Force = 10 N

Action Load = 10 N

Free Body boundary

Example 6

Tug No.1

Tug No.2

Barge

60

25

Ring.

tow rope

tow rope

Example 6 (Solution)

resultant R of the two forces in tow ropes No.1 & No. 2 from the components in the x and y directions:

kN 06.2460cos3025cos10

xR

kN 75.2160sin3025sin10

yR

Tug No.1

Tug No.2

Barge

60

25

Ring.

tow rope

tow rope

Example 6 (Solution)

kN 43.3275.2106.24 22 R

1.4206.24

75.21tan 1

R = 32.43 kN

42.1

E = 32.43 kN

42.1

Equilibrant E = - R

Example 6 (Solution)

Tug No.1

Tug No.2

Barge

60

25

Ring.

tow rope

tow rope

Resultant R is the sum of the actions of the tow ropes on the barge

Equilibrant E is the reaction of the barge to the ropes

E = - R

Moment and Couple Moment of Force Moment M of the force

F about the point O is defined as:

M = F dwhere d is the perpendicular distance from O to F

Moment is directional

dF

oM

Moment and Couple

d

F

r

M = F.d = F.r.cos

A

B

Moment = Force x Perpendicular Distance

Resultant of A System of Forces

An arbitrary body subjected to a number of forces F1, F2 & F3.

Resultant R = F1 + F2 + F3

ComponentsRx = F1x + F2x + F3x

Ry = F1y + F2y + F3y

O

R

F 3

F 2

F 1

d 3

d 2

d 1

d

R

R x

R yF 3

F 2

F 1

F 1x

F 3y

F 3x

F 1yF 2x

F 2y

Resultant of A System of Forces

Resultant moment Mo= Sum of Moments

Mo = F1 d1 + F2 d2 + F3 d3

Mo = R d

O

R

F 3

F 2

F 1

d 3

d 2

d 1

d

R

R x

R yF 3

F 2

F 1

F 1x

F 3y

F 3x

F 1yF 2x

F 2y

Couple

For a Couple R =F = 0 But Mo 0 Mo = F(d+l) - Fl = Fd Moment of couple is the

same about every point in its plane

d

F

O

F

l

Example 7

Calculate the total (resultant) moment on the body.

15 N

15 N300 mm

30 N

30 N

170 mm

100 mm50 mm

A

Example 7 (Solution)

Taking moments about the corner A

Note that the forces form two couples or pure moments 3.6 Nm and 3.0 Nm (resultant force =0, moment is the same about any point).

1.01505.0303.01517.030 M

Nm 6.62.015120.030

Equilibrium of Moments

The sum of all the moments is zero when the body is in moment equilibrium. or

If the body is in equilibrium the sum of the moments of all the forces on acting on a rigid body is the same for all points on the body. It does not matter at which point on a rigid body you choose for taking moments about

n

ii

M1

0 0.......21

MM

Example 8

Calculate the resultant moment and the equilibrant moment.

3.0 m

2.5 m

0.5 m

1.0 m 1.25 m

10 N

15 N

5 N

A

B

Example 8 (Solution)

3.0 m

2.5 m

0.5 m

1.0 m 1.25 m

10 N

15 N

5 N

A

B

Nm 5.225.250.1155.010 R

M

Nm 75.185.0525.0155.210 R

M

Take moment about A

Take moment about B

Example 8 (Solution)

Note that the body is not in vertical and horizontal equilibrium.

There is no unique value for the resultant moment. The value depends on where the resultant force

acts, ie., depends on the perpendicular distance between the resultant force and the point for taking moment.

Therefore, the moments about A and B are different.

Example 9

Cantilever beam

W

d

Beam

Point Load of weight WBuilt in Endor Fixed EndWall

Find the reaction force and moment at the built in end

Example 9 (Solution)

Taking moment about A

W

dReaction V

Moment Reaction MA

Free Body Diagram of Beam

A B

C

0.dWMMA

dWMA

.

0WVFy

WV

General Equations of Equilibrium of a Plane (Two Dimensional) Rigid Body

(Non-concurrent forces)

;0F and ;0F n=i

0=ii y,

n=i

0=ii x,

n

ii

M1

0

;0...FFFx,3x,2x,1

;0...FFFy,3y,2y,1

0.......

21 MM

For complete equilibrium, all 3 equations must be satisfied

Types of Beam Supports

A beam simply supported on points or knife-edges

A side view or elevation of a simply supported beam

A diagrammatic view of a simply supported beam

Simply supported beam

Types of Beam Supports

Pin Joint supported by a roller

Pin Joint Fixed to the ground

Fix Support

Force perpendicular

to surface only.

Both parallel and

perpendicular forces

Two components of force and

a moment

Ry

Rx

Ry

Ry

Rx

M

Types of Supports and Connections

A beam simply supported on points or knife-edges

A side view or elevation of a simply supported beam

A diagrammatic view of a simply supported beam

Simply supported beam

Types of Loading on Beams

Point load

Diagrammatic representation

Distrubuted load (uniform)

Diagrammatic representation

a)

b)

W

W

W / unit length

W / unit length

Another way of representingUniformly distributed loads

W / unit length

Types of Loading on Beams

Non-uniformly distributed loadsc)

W

Types of Loading on Beams

RA

RB

A B

C

W

WA

C

B

a b

Diagrammatic view of simply

supported beam with a

Free body diagram replacing

loads and supports by forces

concentrated load W

Types of Loading on Beams

(a+b)/2

F= w(b-a)

A w N/m B

a b

Diagrammatic view of simply

supported beam with uniformly

Free body diagram replacing

loads and supports by forces

distributed load w

Example 10

Find the reactions at the supports for the beam shown in the diagram.

20 kN5 kN/m

3.5m 3.5m 7 m

Example 10(Solution)

20 kN

3.5m 3.5m7 m

35 kN

R RA B

kN 357 0005 W

05.3205.103514 B

R

kN 25.31B

R

2035BA

RR

kN 75.23A

R

Example 11

Express F in terms of m, a and b.

a bFW

Example 11(Solution)a bF

W

R

0=W-F-R ;+ 0;=F

0FaWbM

Wa

bF Ratio a/b is called

Mechanical Advantage

Example 12

Find reaction forces at supports A and B.

30 o 200N

B

350N

A

0.15m

0.30 m

0.40 m

Example 12 (Solution)

Consider the sum of vertical forces and horizontal forces are zero, and since RAx = 0 as point B can only take up vertical force.

RAx = 200 x sin30o = 100 N

RAy + RBy = 350 + 200 cos30o

= 350 + 173.2 N = 523.2 N

Example 12 (Solution)

Taking moment about A, RBy x 0.3 = 350 x 0.15 + 200 cos30o x 0.4

RBy x 0.3 = 121.8 N

RBy = 405.9 N

Therefore the reaction at point B is 405.9N upward.

RAy = 523.2 - 405.9 = 117.3 N

Example 12 (Solution)

The reaction at point A is 117.3 N upward and 100 N to the left.

Resultant at A is: RA = (117.32 + 1002)

= 154.1 N

angle = 49.55o

100

117.3

Example 13

Find reaction forces at supports A and B.

600 mm

750 mm

55 kN

A

B

Example 13 (Solution)

600 mm

750 mm

55 kN

HA

VA

HB

055Ay

VF

0BAx

HHF

06.05575.0 AHM

kN 55A

V

kN 44A

H kN 44B

H

Example 14

250 mm

75 N

100 mm

T

O

20 o

Example 14 (Solution)

75 N

T

O R

R

R

x

y

Example 14 (Solution)

Fy = 0: Ry - 75 = 0 Ry = 75 N Fx = 0: T - Rx = 0T = Rx MO = 0: 75 x 250 - T x 100 x cos 200 = 0

Therefore Rx = T = 200 N R = (Rx2 + Ry2) = 214 N = tan-1 (Ry/Rx) = 20o 33’

Example 14 (Solution)

Reaction R is that exerted by the frame on the bellcrank, which is equal and opposite to that on the chassis.

R

20 33'o

T = 200 N

R = 214 N75 N

Example 14 (Graphical Solution)

Having determined the line of reaction R, a scaled force polygon can be drawn.

By measurement, T = 200 N, R = 214 N

20 33'o

T = 200 N

R = 214 N75 N