mpa104 statics

8/13/2019 MPA104 Statics

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MPA104

Mechanical Design

Engineering Statics

Duncan Price

IPTME, Loughborough University

Copyright: [email protected] (2006)


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Mechanics of Materials

Mechanics of Materials can be divided into

three categories:

Mechanics of Rigid Bodies Staticsbodies at rest

Dynamicsbodies in motion

Mechanics of Deformable Bodies

Mechanics of Fluids


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Statics

Statics is thoroughly used in the analysis of

structures, for instance in architectural andstructural engineering. Strength of materials

is a related field of mechanics that relies

heavily on the application of static

equilibrium.


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Course Content

12 Lectures:1-5 Forces, springs

Free body diagramsResolution of forces (tutorial sheet)

Equilibrants and resultants offorces (tutorial sheet)

7 Class test (calculators needed)

8-11 Levers, moments, reactions and

centre of gravity (tutorial sheet)12 Online test


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Newtons Laws

Sir Isaac Newton (1642-1729)

Principia 1687

Formulated three laws on which all

conventional motion is based


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Newton I

A particle remains at rest or continues to move

at a constant speed in a straight line unless

there is a constant force acting on it.

The most important law

The one that most people dont understand

The only one that doesnt have an equation


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Newton II

The resultant force on a particle is equal to therate of change of momentum of the particle.

amFdtvd

mFvmdt

d

F ;;

The form F=ma is only valid if the mass is

constant.


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Newton III

The forces of action and reaction between

interacting bodies are equal in magnitude and

opposite in direction.

BAAB FF

The force of the Earth on the Sun has the samemagnitude as the force of the sun on the earth

The force of a tennis ball on a racket has the same

magnitude as the force of the racket on the ball


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Newtons Law of Gravitation (1)

F = G M m / r2

Where:

M & m are particle masses

G is the universal constant of gravitation

(6.673 x 10-11 m3/kg s2)

r is the distance between the particles.

F in Newtons (kg m/s2)


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Newtons Law of Gravitation (2)

On Earth:

F = m g

Where:

m is the mass of the body in question (in kg)g = 9.81 m/s2


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Spring Forces (1)

If a spring is stretched from L0to L it exerts a force on

the object to which it is attached:

F = K (LL0)

K = spring constant = force required to stretch springby unit length (N/m).

K depends on spring material and design.

LO

L

FF


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Spring Forces (2)

F = K (LL0) is also known as Hookes Law:The force exerted by a spring is proportional to

its extension.

Slope = F/ (LL0) = KF

(L-Lo)


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Free body diagrams (1)

FBD is an essential step in the solution of all

problems involving forces on bodies

it is a diagram of the external surface of thebody - not interested in internal forces

all other bodies in contact with the one we are

interested in are replaced by vectors


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Sketch of person standing

mg

R1 R2

F=ma

R1+R2-mg=ma, but no acceleration so,

R1+R2=mg


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sketch

mg

T

free bodydiagram


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Rules:

clear sketches

draw in the correct orientation show all forces acting on the body

dont show any internal forces between

different parts of the body show the forces not the components


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Representation of a force by a vector

A force has the following characteristics:

Magnitude

Direction Point of application

A quantity which has magnitude and direction is

a vector.A quantity which has magnitude only is a scalar.


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Vector addition

F1F2

F3

F1

F2

F3FT

These three forces act together

on the particle. Their effect isequivalent to a single force

which is the vector sum of the

forces.

FTis the resultant of the

forces F1, F2and F3

FT=F1+F2+F3


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Trigonometry (1)

A

B

C

Pythagoras

Theorem:

A2= B2+ C2

Internal anglesof a triangle

add up to 180

90 -

90


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Trigonometry (2)

A

(hypotenuse) B(opposite)

C

(adjacent)

sine() = B/Acosine() = C/Atangent() = B/C


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Trigonometry (3)

sin(45) = 1/2 = 0.707

cos(45) = 1/2 = 0.707

tan(45) = 1/1 = 1

sin-1(0.707) = 45

cos-1(0.707) = 45

tan-1(1) = 45

2 m

1 m

1 m

45

45


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Trigonometry (4)sin(30) = 1/2 = 0.5

cos(30) = 3/2 = 0.866

tan(30) = 13 =0.577

sin(60) = 3/2 = cos(30)

cos(60) = 1/2 = sin(30)tan(60) = 3/1 = 1.73

Generally:

cos() = sin(90- )

tan() = sin()/cos()

sin2() + cos2() = 1

1 m

2 m

3m

30

60


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-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-360 -315 -270 -225 -180 -135 -90 -45 0 45 90 135 180 225 270 315 360

Sine and Cosine functions

sin()

cos()


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-10

-8

-6

-4

-2

0

2

4

6

8

10

-90 -45 0 45 90

Tangent function

tan()


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Resolving forces (1)

Forces can be broken down into any number ofcomponent forces

It is often convenient to choose two

perpendicular directions for resolution

FFy

F = Fx+Fy

F = (Fx2+Fy

2)1/2

Scalar magnitudesFX


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F

Fx

Fy

Fx=F cos

Fy=F sin

=tan-1(Fy/Fx)

If the components are

perpendicular, they may

be added independently

F1F2

F3

FT=F1+F2+F3 FTx=F1x+F2x+F3x FTy=F1y+F2y+F3y


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500 N

30Fx=500 cos 30

= 433 N

Fy=500 sin 30

= 250 N


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FTx= F1x+F2x+F3x

= 20050 cos 30 + 80 cos 80

= 171 N

FTy= F1y+F2y+F3y

= 050 sin 3080 sin 80

= - 104 N (i.e. downwards)

F1= 200 N

F2 = 50 N

F3= 80 N

30

80


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Resolving forces (5)Fx = 171 N

Fy= 104

N

FT= (Fx2+Fy2)

= (1712+ 1042)

= 200 N

= tan-1

(Fy/Fx)

= tan-1(-104/171)

= -31

FT


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30 75

100 N

A

B C horizontal components:

FACcos75 = FABcos30

FAC= FAB(cos30/cos75)

vertical components:

FACsin75 + FABsin30 = 100 N

FAB

(cos30/cos75)sin75 + FAB

sin30 = 100

FAB= 26.8 N FAC= 89.7 N


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Resultant

If two or more forces at a point they can bereplaced by a single force known as a

resultant.


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Equilibrium

When two or more forces act upon a body and

are so arranged that the body remains at rest

or moves at a constant speed in a straightline*, the forces are said to be in equilibrium.

* i.e. Newtons 1stlaw


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Y-hangs

What happens if we change the angle?

90

750 N

A B

C

Horizontally

ACsin45 = BCsin45

AC = BC

Vertically

ACcos45 + BCcos45=750 N

AC + BC =750 N/cos45

2 AC = 750 N/cos45

AC = 530 N = BC


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Trusses (1)

One of the basic methods to determine

loads in individual truss members is called

the Method of Joints. Each joint is treated as

a separate object and a free-body diagram

is constructed for the joint.


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Trusses (2)

30 45

A

B C

500 N

Horizontal forces

ABcos30 = ACcos45

AC = ABcos30/cos45

Vertical forces

ABsin30 + ACsin45 = 500 N

ABsin30 + AB(cos30/cos45)sin45 = 500 N

AB = 366 N

AC = 448 N


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Trusses (3)

30 45

A

B C

366 N

Horizontal forces

BC = ABcos30 = ACcos45

= 316 N

Vertical forces at B

RB= ABsin30

= 183 N

Vertical forces at C

RC= ACsin45

= 317 N

Check

RB+ RC= 183 N + 317 N = 500 N

448 N


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Moment of a force

The product of the force and the perpendiculardistance from the line of action of the force tothe axis.

Moment about a = F d (units N m)

d F

a


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Principle of Moments (1)

If a body is at rest under the action of several forces, the total

clockwise moment of the forces about any axis is equal to

the anticlockwise moment of the forces about the same axis.


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2 m 1.5 m

x N60 N

ab

y N

Moments about a: 60 2= x 1.5

x = 120/1.5 = 80 N

Moments about b: (2 + 1.5) 80 = 2 y

y = 280/2 = 140 N (ignoring mass of beam)

Or more simply: y = 60 + 80 = 140 N


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2 m 1.5 m

x N60 N

ab

y N

Now let the beam weigh 5 kg

Additional force of 49 N acting centrally.

Moments about a: (60 2) + (49 0.25)= x 1.5

x = 144.5/1.5 = 88.2 N

Reaction of support y = 60 + 49 + 88.2 = 197.2 N

49 N1.75 m


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1.5 m 1.5 mab

y N

Now support the unloaded beam at a, b and c:

Moments about b: 49 1.25 = y 1.5 + z 3

Moments about a: 49 0.25 = x 1.5 - z 1.5

Moments about c: 49 1.75 = x 3 + y 1.5

49 N1.75 m

x N

0.5 m c

z N


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1.75 m

Now support the unloaded beam at c only

49 N1.75 m

c

z N


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Centre of Gravity (2)If a vertical line through the centre of gravity falls outside the

base upon which the body relies for stability, overturning will

result, unless precautions, such as tying the base down, are

taken.


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Centre of Gravity (3)

For a rectangle or square, the centre ofgravity occurs at the centre of the section.


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For a triangular section the centre ofgravity occurs at a point 1/3 of the height

from the base.


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For a circular section the centre of gravityoccurs at the centre of the circle.

r


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Centre of Gravity (6)To determine the centre of gravity for a compound section:

1. Divide the section into several parts (i.e. rectangles and

triangles, circles) so that the centre of gravity of each

individual part is known.

2. Determine the area and position of centre of gravity of each

part.

3. Take moments about a convenient axis to determine the

centre of gravity of the whole body.

This is based on the principle that, along any one axis (or in any

one direction):

when the distance is measured from the same point in each case.


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Determine the position of the centre ofgravity of the L-section shown below:

0.5 m 0.5 m

1 m

0.75 m


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Divide up section into two rectangles,identify c.o.g of each relative to O and

calculate area.

0.5 m

0.25 m

Area 1 = 0.5 m2

0.25 m

0.625 m Area 2 = 0.125 m2

O

Total area = 0.625 m2


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Let X = location in X axis of c.o.g0.5 m

0.25 m

0.5 m2

0.125 m2

O


0.625 X = 0.50.5 + 0.1250.25X = 0.45 m


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0.5 m2

0.25 m

0.625 m

0.125 m2

O


Let Y = location in Y axis of c.o.g

0.625 Y = 0.125 0.625 + 0.5 0.25

Y = 0.325 m


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Now let us drill a hole in the object, whereis the new centre of gravity?

0.4 m

0.25 m

0.25 m

Area 3 = 0.12564 m2



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Find new X and Y:

0.25 m 0.12564 m2


0.49936 X = 0.50.5 + 0.1250.250.125640.25X = 0.5 m

0.49936 Y = 0.50.25 + 0.1250.6250.125640.25

Y = 0.344 m


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Divide the section up into rectangles andtriangles:


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Select the first solution (but either methodwould give the same answer).

Area 1 3x1.5/2 = 2.25m2

Area 2 4x1.5 = 6.00m2

Area 3 3x1.0 = 3.00m2

Total = 11.25m2


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Select two axes A-A and B-B, at the extreme edge

of the figure. The distance to the centre of gravityof each section can then be calculated from these

axes.


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Let X be the horizontal distance to the

centre of gravity of the whole figuremeasured from A-A and Y be the vertical

distance to the centre of gravity of the

whole figure measured from B-B.

Taking moments about A-A:

11.25X = (2.25x2.25)+6.0x2.25)+(3.0x1.5)

= 5.6+13.5+4.5

= 23.6

X = 23.6/11.25 = 2.1 m


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Taking moments about B-B:

11.25Y = (2.25x6.0) + (6.0x3.0) + (3.0x0.5)= 13.5 + 18.0 + 1.5

= 33.0

Y = 33.0/11.25 = 2.9m


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Now let us support the section at the middle

and far rightWhat is the ratio of forces on each support?

3 mX

Y

1.5 m


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Moments about Y.

1.5 RX = 0.9

RX= 0.6

therefore RY= 0.4

3 mX

Y

1.5 m

mpa104 statics

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