eng principles

8/6/2019 Eng Principles

1/22

Manchester Metropolitan University Simon Iwnicki

BSc(Hons) Mechanical Engineering Network Room E354

Unit 64ET4903 Mechanical Engineering Principles [email protected]

C:\teaching\bscep.doc

Mechanics


2/22





INTRODUCTIONBackground:

Engineers need to understand the effects that forces have on the bodies

which make up most mechanical systems. This field is called mechanics

and it is subdivided into two branches:

Statics

Where the forces acting on a body are balanced and the body is inequilibrium and:

Dynamics

Where the effects of forces on bodies and their subsequent motions are

studied.

Booklist:

Author Title Publisher CommentsHannah J. & Hillier M.J. Applied Mechanics Longman Still the best

introduction.

Johnson A. & Sherwin K. Foundations of Mechanical

Engineering

Nelson

Thornes

Statics, dynamics

and thermofluids.

Dyke P. & Whitworth R. Guide to Mechanics Palgrave

Stroud K.A. Engineering Mathematics Longman For maths

revisionBeer F.P & Johnston E.R. Vector Mechanics for Engineers McGraw Hill In 2 parts,

Good worked

examples


3/22





STATICS

Statics is the study of forces in equilibrium. A force in a balanced system

must be opposed by an equal and opposite force:

Multiple forces can also balance:

A force can be resloved into to components at right angles to each other:

To check if a system is in equilibrium, resolve all forces into x and y

components and check that balance:

F1x + F2x = 0 and F1y + F2y = 0

F1F2

F1

F1

F3

F4

F1x

F1

F2x

F2y F1


4/22


5/22





Free Body Diagrams

A diagram showing the particular body or bodies of interest in isolationfrom all other parts of the mechanism, with all the forces acting on that

body shown as vectors.

Rules:1. Draw the body of interest2. Mark the centre of gravity and the force due to gravity Mg3. Add normal forces at all points of contact4. Add friction forces at these points5. Add any other external forces

Example:


6/22





STRESS AND STRAIN

Stress

Is intensity of loading stress = load [N/m2]area

It may be tensile (+ve) or compressive (-ve)

Strain

Is distortion of the body caused by stress

Strain = change in length [-]

original length

Hookes law

States that strain is directly proportional to applied stress up to the elastic

limit. The constant slope is known as the modulus of elasticity or

Young's modulus E. The modulus is a measure of the stiffness or rigidityof the material.

(for steel E = 210 GN/m2

)

Example

Find the minimum diameter of a 1m long steel wire required to support a

mass of 50kg without extending by more than 1 mm.

Take Young's modulus for steel to be 210 GN/m2


7/22





Poisson's ratio

When a bar is loaded axially it stretches longitudinally aut it also

contracts laterally. Poisson's ration expresses the ratio of these twostrains:

Lateral strain [-]

Longitudinal strain

Shear stress

If two equal and opposite forces F, not in the same straight line, act on

parallel faces of a member , then it is said to be loaded in shear.

And the shear stress:

= F/A [N/m2]

Shear strain is the angle of deformation.

Shear stress = G ( the modulus of rigidity ) [N/m2]

Shear strain

(for steel G = 84 GN/m2

)

BENDING OF BEAMS

F

F

A


8/22





A beam under load usually experiences bending moments and shearforces which vary along and within the beam material.

There is a maximum bending moment and shear force that can be

sustained (different for each material) and understanding their

distribution can help an engineer to design efficient structures.

Both can be presented in diagrammatic form.

Shear force diagrams

w

w

w

+ve

w/2 w/2

w

w/2

-w/2

+ve


9/22





Bending moment diagrams

.

w

-wd

d

Fixing moment

w/2 w/2

w

Wd/4


10/22





Example

Freebody diagram

Must be in equilibrium therefore: RA+RB = 785 + 687

RA 785,2 + 687.5 = RB.4

so RB = 1250N

and RA = 785+687 - 1250

= 222N

2m

3m

4m 1m

George (80 kg) Vinny (70 kg)

BRIDGE

RA RB

80x9.81=785N 70x9.81=687N


11/22





Shear force diagram

Bending moment diagram

222 N

687 N

222.2 = 444 Nm

222.4-785.2 = -682 Nm


12/22





More examples draw shear force and bending moment diagrams

1m

200N

100N

200N

2m10m

3m

100N

226

4030

2000N

2.5m

6m

Cable

1m

Cable


13/22





Bending of beams

Bending equation for simple bending

( see Hannah + Hillier p 395 for derivation )

M = = EI Y R

Where M = bending moment [Nm]

I = 2nd

moment of area [m4]

= stress [N/m2]

y = distance from neutral axis [m]E = Youngs moduls [N/m

2]

R = radius of curvature [m]

For rectangular cross section

I = bd

3

12

y = d/2

For a circular cross section I = d4

64

y = d/2

So the stress at any point in a beam can be calculated or the radius of

curvature.

b

d

N A

N Ad


14/22





eg. For the practice example:

Find the maximum stress due to bending moment Mmax = 682 Nm

Need cross section

I = bd3

= 1.0.023

= 0.67.10-3

m4

12 12y = d/2 = 0.1m

M/I = /y

so = My = 682.0.1I 0.67.10

-3

= 102 kNm-2

1m

0.2m


15/22





DYNAMICSNewtons laws of motion

1st Law: The velocity of a body remains constant unless an external

force acts on it.

[the concept of inertia]

2nd Law: When a force acts on a body its momentum changes at a rate

proportional to the magnitude of the force and in the direction

of the force.

[the concept of acceleration]

3rd Law: To every action there is an equal and opposite reaction.

[the concept of equilibrium]

NOTES

Weight and Mass:

Mass is a fixed property of a body (kg)

Weight is the force on a body due to a gravitational field (N)

Momentum = mass x velocity (kgm/s)

Translation and Rotation:

Force (N) = Mass (kg) * Acceleration (m/s2)

Torque (Nm) = Moment of inertia (kgm2) * Angular acceleration (rad/s

2)

Friction

friction force f = N (where N = normal force

= coefficient of friction )

always opposes the motion

Impulse = force x time acting (Ns)


16/22





Equations governing motion with constant acceleration

V = U + a x t

s =1

2(U+V) x t

s = U x t +1

2a x t

2

V2

= U2

+ 2 x a x s

where V = final velocity

U = initial velocity

a = acceleration

t = time

s = displacement

(for derivations see e.g. Hannah and Hillier page 73)

Example 1: Projectiles

A projectile is a moving body that is acted on only by gravity. The

motion of a projectile can be fully described if its initial speed (Sp) and

angle to the horizontal () are known.

Sp


17/22





It is convenient to separate vertical and horizontal motions for all

calculations. e.g.:

1. Time to maximum height

use V = u + a x t on vertical motion:

t1 =

2. Maximum heightuse s = U x t +

1

2a x t

2on vertical motion:

h =

3. Overall time (actually 2.t1 but need to proove)

use s = U x t + 12

a x t2 on vertical motion from top to bottom:

t2 =

4. Distance travelled

use s = U x t +1

2a x t

2on horizontal motion:

R =

(Now calculate the velocity at impact)


18/22





Motion in a circle

From Newtons first law we can see that a body moving in a circular path

must have a force acting on it. The general term for this force is the

centripetal force and it acts towards the centre of the circle.

For a body of mass M [kg] moving in a circular path of radius r [m] and

with constant velocity v[m/s]

the centripetal force required is F =Mv

r

2

Example 2: A cyclist in a curve

Draw a free body diagram of the cyclist:


19/22





Calculate:

1. The angle at which the cyclist must lean:

tan =v

rg

2

2. The maximum velocity the cyclist can travel without sliding:

v rg=


20/22





ENERGY

Background:

Principle of conservation of energy:

Energy cannot be lost or gained in a closed system but

can be interchanged in form (at specific rates of

conversion).

There are many different type of stored energy (chemical, electrical, heat

etc. ) but in mechanical systems we are mainly interested in the

interchange between energy stored in a mass in a gravitational field

(potential energy) and energy associated with a moving body (kinetic

energy).

Equations governing energy

Potential energy: P.E. = mgh

Kinetic energy: K.E. =1

2

2mv (for translational motion)

K.E. =1

2

2I (for rotational motion)

Where: m = mass [kg]v = velocity [m/s]

I = moment of inertia [kgm2]

= angular velocity [rad/s]


21/22





A note on I

If we consider a generalbody made up of n particles

each with mass mn and that

body rotates about a pivot at

0 with an angular velocity

:

The kinetic energy of each particle isK.E. =

1

2

2m vi i

and therefore the kinetic energy of the whole body is

K.E. =1

2

2

1

m vi ii

n

=

K.E. =1

21

2 2

i

n

i i im r

=

K.E. =1

2

2

1

2

i

n

i im r

=

and the boxed part is I.

I can be calculated by integrating mr2 over the total cross sectional area

of the body concerned (perpendicular to the axis of rotation).

Or it can be looked up in a table!........


22/22





Some common values of I:

Cylinder I =1

2

2Mr

Sphere I =2

5

2Mr

Thin rod I =1

12

2ML

I =1

3

2ML

Example

A 25 mm dimeter cylinder is allowed to roll from the top of a plane 1.25

m long inclined at an angle of of 20o

to the horizontal. Neglecting friction

and air resistance and assuming that the cylinder does not slip determine

the linear velocity of the cylinder on reaching the bottom of the plane.

[2.3 m/s]

eng principles

Documents