Level 3

Mechanical Engineering Principles

Energy

1

Tim Coley for Coleg Llandrillo Cymru

ENERGY FORMS

A mass may possess several forms of energy. We are only concerned with mechanical energy so the energy due to temperature is not covered here.

GRAVITATIONAL or POTENTIAL ENERGY

In order to raise a mass m kg a height z metres, a lifting force is required which must be at least equal to the weight mg.

The work done in raising the mass is as always, force x distance moved so

Work = mgz

Since energy has been used to do this work and energy cannot be destroyed, it follows that theenergy must be stored in the mass and we call this gravitational energy or potential energy P.E.

There are many examples showing how this energy may be got back, e.g. a hydro-electric powerstation.

P.E. = mgz

(Note z is the SI symbol for altitude but many other symbols are used for height such as h)

LINEAR KINETIC ENERGY

When a mass m kg is accelerated at‘a’ m/s2 from rest to a velocity ofv m/s, a force ‘F’ is needed toaccelerate it.

This is given by the formula derived from Newton's 2nd Lawof Motion

F= m a.

After time t seconds the masstravels x metres and reaches avelocity v m/s. The laws relatingthese quantities are

a = v/t

andx = vt/2

The work done is

W = Fx = max = mv2/2

Energy has been used to do this work and this must be stored in the mass and carried along with it.

This is KINETIC ENERGY.

K.E. = mv2/2

ANGULAR KINETIC ENERGY

Rotating objects also have kinetic energy. Consider a simple wheelrotating about its centre. Every particle in the wheel ‘δm’ has a velocity‘v’ along a circular path at a radius ‘r’.

The kinetic energy of the particle is

δm v2/2

The linear velocity is linked to the angular velocity by

v = ω r

so thekinetic energy of the particle isδm (ωr)2/2

If we sum the energy of all the particles in the wheel then the total K.E. is:

K.E. =Σδmω2r2/2

By definitionΣδmr2 is the moment of inertia I

so the K.E.of any rotating body is

K.E. = Iω2/2

STRAIN ENERGY

When an elastic body such as a spring is deformed, work is done. The energy used up is stored inthe body as strain energy and it may be regained by allowing the body to relax. The best example ofthis is a clockwork device which stores strain energy and then gives it up.

Springs take many forms. Any elastic material may be stretched, compressed, twisted, sheared orbent. In all cases work is done to deform the material so strain energy is stored in it. Strain energy isusually given the symbol U. Consider the two simple cases shown below of linear springs that obey

Hooke’s law.

F = k x

k is the stiffness in N/m and x is the deflection in m. If we start from zero and gradually increase theforce to Fmax, the F – x graph produced is a straight line with a gradient k. The work done is the areaunder the graph.

W = F x/2 so U = F x/2

F and x are the final or maximum values. If we substitute F = kx then

U = kx2/2

ENERGY TRANSFORMATION IN MECHANICAL SYSTEMS

Many mechanical systems involve the exchange of energy from one form to another. We shouldalways consider the Law of Conservation of Energy and remember if energy is not transferred with100% efficiency then energy is lost as other unspecified forms such as heat and sound. Frictionalways produces a loss of energy. Let’s start with simple cases that you should be familiar with.

FALLING BODY

Suppose a ball of mass M kg is allowed tofall from rest to the ground a distance zmetres below. The body will acceleratedown under the action of gravity and justbefore it hits the ground the ball it will havea velocity v m/s. The velocity is simplysolved by considering the energy changes.

The ball initially has potential energy Mg z and just before it hits the ground it has kinetic energyMv2/2. If no energy is lost then we may say the

PE lost = KE gained

so equating :

Mgz = Mv2/2

v =

Note that the result is the same whatever the mass.

WORKED EXAMPLE No. 1

Calculate the velocity of ball just before it hits the ground when dropped from a height of 6 m.

SOLUTION

v == =10.85 m/s

FALLING AND ROLLING

Consider a cylindrical body that starts at rest at the top of a ramp and then rolls down hill without slipping or sliding. As it rolls down the ramp it will roll and obtain both linear and angular kinetic energy so if energy is conserved the potential energy is converted into both.

APPLICATION TO IMPACT LOADS

Consider a mass M that is dropped a height z onto a springof stiffness k. When the mass hits the spring, thespring will deflect a distance x before the mass stopsmoving down.

At the moment the spring is compressed to its maximumthe force in the spring is F and the strain energy is

U = kx2/2

The potential energy given up by the falling mass is

P.E. = Mg(z + x)

SIMPLIFIED SOLUTION

If x is small compared to the distance z then we may say

P.E. = Mgz

Equating the energy lost to the strain energy gained we have

Mgz = kx2/2

Hence

x =

EXACT SOLUTION

Equating P.E and Strain energy we have

Mg{z + x) = kx2/2

Rearrange into a quadratic equation

kx2 - 2Mgx – 2Mgz = 0

Solving with the quadratic equation we find

x =2Mg±()/2k

There are two solutions

x =2Mg+()/2k

and

x =2Mg-()/2k

Discuss the relevance of 2 answers.

SUDDENLY APPLIED LOADS

A suddenly applied load occurs when z = 0. This is not the same as a static load. Putting z = 0yields the result:

x =2Mg+ )/ 2k

= 2Mg/ k

The static deflection of the spring when the mass just rests on it is

xs= Weight/k = Mg/k

fromwhich it follows that

x = 2 xs

The deflection is double that of the static load.

It also follows that the instantaneous force in the spring is double the static weight.

This theory also applies to loads dropped on beams.

WORKED EXAMPLE No. 3

A mass of 1600 kg rolls onto the end of a simple cantilever bridge as shown. When the massrests on the end of the cantilever, the deflection is 50 mm. What is the maximum deflectionwhen the ball first rolls onto the end?

SOLUTION

This is a suddenly applied load so the deflection is twice the static deflection and is 100 mm.

WORKED EXAMPLE No. 4

A mass of 5 kg is slides on a rod suspended from a spring as shown. The spring stiffness is4000 N/m. Calculate the maximum deflection of the spring when mass is dropped from aheight of 0.3 m onto collar at the end. Calculate the deflection of the spring when the masscomes to a rest.

SOLUTION

x =2Mg+()/2k

x =2 x 5 x 9.81 +()/2 x 4000

x = 0.099 m or 99 mm

Static deflection xs= Mg/k = 5 x 9.81/4000 = 0.012 m or 12 mm.

SELF ASSESSMENT EXERCISE No. 3

  1. A spring loaded platform supports a mass of 150 kg as shown and the platform deflects 20 mm from its normal position. If the same mass is dropped onto the platform from a height of 200 mm what will be the maximum deflection of the platform? (Answers 131 mm)
  1. A beam is placed across a span as shown. When a force of 20 kN is applied at the centre, it deflects 3 mm. Calculate the maximum deflection when a mass of 5000 kg is dropped from a height of 10 mm onto the middle and the deflection when the mass rests on the beam.

(Answers 22 mm and 7.36 mm)

1

Tim Coley for Coleg Llandrillo Cymru