1 MPA104 Mechanical Design Engineering Statics Duncan Price IPTME, Loughborough University © Copyright: [email protected] (2006)

1

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 Statics – bodies at rest Dynamics – bodies 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 and structural 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 diagrams Resolution of forces (tutorial

sheet)Equilibrants and resultants of

forces (tutorial sheet)7 Class test (calculators needed)8-11 Levers, moments, reactions and

centre of gravity (tutorial sheet)12 Online test

5

Newton’s 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 don’t understand The only one that doesn’t have an equation

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Newton II

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

amFdt

vdmFvm

dt

dF ; ;

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 same

magnitude 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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Newton’s 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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Newton’s 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 L0 to L it exerts a force on the object to which it is attached:

F = K (L – L0)

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

K depends on spring material and design.

LO

L

FF

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Spring Forces (2)

F = K (L – L0) is also known as Hooke’s Law:

The force exerted by a spring is proportional to its extension.

Slope = F/ (L – L0) = 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 the body - 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 body diagram

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Rules: clear sketches draw in the correct orientation show all forces acting on the body don’t 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 is equivalent to a single force which is the vector sum of the forces.

FT is the resultant of the forces F1, F2 and F3

FT=F1+F2+F3

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Trigonometry (1)

A

B

C

Pythagoras Theorem:

A2 = B2 + C2

Internal angles of a triangle add up to 180°

90 -

90°

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Trigonometry (2)

A

(hypotenuse)B

(opposite)

C

(adjacent)

sine() = B/A

cosine() = C/A

tangent() = 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°) = 1√3 =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

√3 m

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 of component 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

30° Fx=500 cos 30°

= 433 N

Fy=500 sin 30°

= 250 N

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FTx= F1x+F2x+F3x

= 200 – 50 cos 30° + 80 cos 80°

= 171 N

FTy= F1y+F2y+F3y

= 0 – 50 sin 30° – 80 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+Fy

2)

= √(1712 + 1042)

= 200 N

= tan-1(Fy/Fx)

= tan-1(-104/171)

= -31°

FT

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30° 75°

100 N

A

B Chorizontal components:

FACcos75° = FABcos30°

FAC = FAB(cos30°/cos75°)

vertical components:

FACsin75° + FABsin30° = 100 N

FAB(cos30°/cos75°) sin75° + FABsin30° = 100

FAB = 26.8 N FAC = 89.7 N

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Resultant

If two or more forces at a point they can be replaced 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 straight line*, the forces are said to be in equilibrium.

* i.e. Newton’s 1st law

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Equilibrant

If two or more forces act at a point and are not in equilibrium a force equal in magnitude and opposite in direction to their resultant must be applied to restore equilibrium. Such a force is called the equilibrant.

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Equilbrant and resultant

ae

30 N

50 N

10 N

20 N

ae = resultant

ea = equilibrant

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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 perpendicular distance from the line of action of the force to the axis.

Moment about a = F d (units N m)

d Fa

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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 and b:

Moments about b: 49 1.25 = y 1.5

y = 61.25/1.5 = 40.83 N

Therefore x = 49 – 40.83 = 8.17 N

49 N1.75 m

x N

0.5 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

45


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 (1)

The centre of gravity of an object is a point at or near the object through which the resultant weight of the object acts.

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

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For a triangular section the centre of gravity occurs at a point 1/3 of the height from the base.

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For a circular section the centre of gravity occurs 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 of gravity 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

OTotal 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


0.625 X = 0.50.5 + 0.1250.25

X = 0.45 m

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0.5 m2

0.25 m

0.625 m

0.125 m2


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, where is the new centre of gravity?

0.4 m

0.25 m

0.25 m

Area 3 = 0.12564 m2

Total area = 0.49936 m2

57


Find new X and Y:

0.25 m 0.12564 m2

Total area = 0.49936 m2

0.49936 X = 0.50.5 + 0.1250.25 – 0.125640.25

X = 0.5 m

0.49936 Y = 0.50.25 + 0.1250.625 – 0.125640.25

Y = 0.344 m

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Another example:

59


Divide the section up into rectangles and triangles:

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Select the first solution (but either method would 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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Centre of Gravity (16)Select two axes A-A and B-B, at the extreme edge of the figure. The distance to the centre of gravity of 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 figure measured 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= 23.6X = 23.6/11.25 = 2.1 m

63


Taking moments about B-B:11.25Y = (2.25x6.0) + (6.0x3.0) + (3.0x0.5)= 13.5 + 18.0 + 1.5= 33.0Y = 33.0/11.25 = 2.9m

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Now let us support the section at the middle and far right…What is the ratio of forces on each support?

3 mX

Y

1.5 m

65


Moments about Y.

1.5 RX = 0.9

RX = 0.6

therefore RY = 0.43 mX

Y

1.5 m

Documents

1 MPA104 Mechanical Design Engineering Statics Duncan Price IPTME, Loughborough University © Copyright: [email protected] (2006)