Energy Concepts Brazil 08-21-12

7/23/2019 Energy Concepts Brazil 08-21-12

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ANINTRODUCTION TOENERGYCONCEPTS IN

TRAFFICCRASH

INVESTIGATION/RECONSTRUCTION

Presented for:

VIII National Seminar on Traffic Crash Accidents

Brasilia, Brasil

August 2012

Presented by:

John Daily

Jackson Hole Scientific Investigations, Inc.

Institute of Police Technology and Mangement


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INTRODUCTION

Many crash investigators/reconstruction do not

use, or shy away from using energy methods intheir analyses.

Wh ?? V ri r n

We will discuss the value and necessity of using

energy methods in our crash investigations.

We will explore various ideas and concepts inusing energy methods in traffic crash

investigations.

Copyright 2010Copyright 2010 -- 2012 by N. Shigemura & J. Daily2012 by N. Shigemura & J. Daily22


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WHYENERGYMETHODS?

Conservation of Linear Momentum (COLM) is

an excellent technique to use in analyzingcollisions.

,

momentum solution is not available or not

appropriate.

Energy methods may provide a solution or openother avenues to use in the analysis.



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WHAT IS ENERGY?

Ener is defined as the

Copyright 2010Copyright 2010 -- 2012 by N. Shigemura & J. Daily2012 by N. Shigemura & J. Daily

ability to do work.

44


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FORMS OFENERGY

Heat

Light

Sound

Mechanical

Kinetic

Potential

Electrical Chemical

Nuclear

we are most interested in

Kinetic Energy.



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ENERGY

Kinetic energyKinetic energyKinetic energyKinetic energy is the energy of motion.

21mvKe =

An object possesses kinetic energy by virtue of its

motion.

For an object to come to a stop all of its kinetic

energy must be dissipated.


2

66


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WORK

WorkWorkWorkWorkis defined the dot product of a force actingthrough a displacement.

In equation form:

If a force is applied to an object but there is nodisplacement, there is no work.

Notice the vector format for force anddisplacement and the scalar format for work (thedot product of two vectors).

=



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WORK

There are two kinds of work:

Reversible work, such as compressing an elastic spring Irreversible work, such as crushing a vehicle inelastically

In the same way, the force doing the work may be categorized:

Conservative, e.g. gravity Non-conservative or dissipative, e.g. friction

It is important to note that the work done by a dissipative force

is path dependent.


dsdFWs =


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WORK(CONT.)

If the applied force is in the same direction as

the displacement vector then the work done issimply product of the magnitude of the force Fand the distance d through which the object

moves.

dddd

FFFF

FdW =



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WORK(CONT.)

However, if the applied force is not in the same directionas the displacement vector, then the work done is theproduct of the magnitude of the force, F cos , in thedirection of the displacement and the distance dthrou h which the object moves.

F cos

( )dFW cos=

FFFF

dddd



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WORK ANDENERGY

WorkWorkWorkWork Energy TheoremEnergy TheoremEnergy TheoremEnergy Theorem

The work done by a force in displacing an object isequal to the change in kinetic energy of the object.

( )22

22

2

1

2

1

2

1

if

if

vvm

mvmv

=

=



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WORK ANDENERGY(CONT.)

The applied force that appears in the work

equation is the unbalanced force that is referred toin the definition of Newtons First Law of Motion.

If the force moves the object (causes

sp acement t w cause t e o ect to ave achange in kinetic energy equal to the amount ofwork accomplished in moving the object.

Now that the object possesses kinetic energy, ithas the ability to perform work.

Thus the Work-Energy Theorem.



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ENERGY

First Law of ThermodynamicsFirst Law of ThermodynamicsFirst Law of ThermodynamicsFirst Law of Thermodynamics

Energy cannot be created or destroyed. It is only transferred

from one form to another. Kinetic energy from a moving vehicle can be transformed

into other forms, primarily heat, in various ways.

Engine Service brakes

Collisions

For an object to come to a stop, allallallall of its kinetic energy must

be dissipated. If even an iota of kinetic energy remains, the object will still

be moving.



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DISSIPATINGKINETICENERGY

Skidding

FdW =mg

Ff =

mg=fmgF

Ffmg

=

=

Where: W= work

F= resistive force due to friction, Nf = drag factor

m = mass of the vehicle, kg

d = distance of the skid, m

g= gravitational acceleration, 9.81 M/sec2



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DISSIPATINGKINETICENERGY(CONT.)

Collision

Breaking/damaging objectsAttenuators

Poles

Guard rails, etc.

Damage energy

To the bullet vehicle

To the target vehicle



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DISSIPATINGKINETICENERGY(CONT.) Post impact movement

Slide to a stop

Spinning during translation to final positionfwdWork=

fadj= fr+ (- fr)|sin avg| + m eta = 0.80

Rotational kinetic energy, spinning more or less in one placerather than translating


2

2

1IKE=

1998 Ford Taurus

Interval d (ft) i i+1 avg fr sin m fadj S (mph)

1 20.20 0 27 13.5 0.70 0.06 0.233 -0.035 0.174 10.28

2 19.41 27 56.5 41.75 0.70 0.06 0.666 -0.035 0.451 16.21 S = 39.16 mph

3 24.54 56.5 87 71.75 0.70 0.06 0.950 -0.035 0.633 21.58

4 18.30 87 109 98 0.70 0.06 0.990 -0.035 0.659 19.02 f_overall = 0.50

5 19.72 109 134 121.5 0.70 0.06 0.853 -0.035 0.571 18.37

102.17 f_ref = 0.70

eta = 0.71

1616


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ENERGYANALYSISTECHNIQUES

Bridge abutment

Crash attenuators

Skid

Example 1



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ENERGYANALYSISTECHNIQUES(CONT.)

Determining speed at the start of the skid in

Example 1 is not a momentum problem. Can be solved using energy techniques.

e erm ne a e ma or ener y osses. Add the energy losses together to get energy

at the start of the skid.

Determine speed at the start of the skid from

the total energy.



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Four energy losses in Example 1.

1. Skid to impact.

2. Crushing/destroying crash attenuators.

. .

4. Damage (crush) sustained by the vehicle.



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Skid

1

2

Example 1

3

4



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1

2

weight of vehicle, w= 1364 kg (13,377N)

52,240 J

J

mgfdw

706,142

24.15)70.0(377,13

=

=

=1.

3

4

d 15.24 mf= 0.70

d = 25 m

f= 0.40

284,760 J

2. 52,240 J

J

mgfdw

770,133

25)40.0(377,13

=

=

=3.

4. 284,760 J


(e.g. from specs)

(from damage analysis)

2121


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Calculate the total energy at the start of the

skid by addingaddingaddingadding the individual energies together.

J

EEEEET

476,613

760,284770,133240,52706,142

4321

=

+++=+++=



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ENERGYANALYSISTECHNIQUES(CONT.) Now can calculate speed at the start of the skid

from the total energy value.

KeS

26=

( )

hkm

m

/108

8.693,11

1364

950376,15

1364

476,61326

=

=

=

=



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Example 2 Consider an

impact with a wooden polewhere the pole breaks.

The impact event can have

our energy osses

1. Pole fracture energy

2. Pole stump movement in

the ground

3. Damage to the vehicle

4. Post impact movement of

the vehicle

Copyright 2010Copyright 2010 -- 2012 by N. Shigemura & J.2012 by N. Shigemura & J.

DailyDaily 2424


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ANALOGY Energy can be thought of like money.

The quantity of money that a person hadat the start of a shopping trip can be

determined by adding all the receipts

oge er.

By determining each energy

expenditures and adding up all the

energy receipts, the total energy at thestart of the event can be determined, and

subsequently the speed at the start of

the event.Copyright 2010Copyright 2010 -- 2012 by N. Shigemura & J. Daily2012 by N. Shigemura & J. Daily

2525


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ANALOGY(CONT.)

Thinking of kinetic energy dissipation as

individual expenditures can help theinvestigator answer the question should I take



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WHATABOUTSPEEDEQUATIONS?

Basic speed equation:

dfS 254=

Its an energy equation!



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WHATABOUTSPEEDEQUATIONS?

The basic

speedequation Fdmv

2

1WorkEnergyKinetic

2=

=

fdv

gfdv

)81.9(2

2

=

=

Continue to convert to speed:

the work-energy

theorem.


gfdv

gfdv

fdg

v

dmgfmv

2

2

2

2

2

2

2

=

=

=

=

fdS

fdS

fdS

fdS

Sv

254

)62.19(6.3

62.196.3

62.196.3

6.3

.

2

=

=

=

=

=

Velocity in fundamental units

2828


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WHATABOUTSPEEDEQUATIONS? (CONT.)

Combined speed equation:

23

3

2

2

2

1 nSSSSS ++++= K

These are energy equations!


( ) 2332211254 fnn SdfdfdfdfS +++++= K

The speeds in these equations are Kinetic Energy Equivalent (KEES)

speeds because each represents a discreet kinetic energy change.

2929


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( )22

321

fo

finalninitial

mSmS

KEWWWWKE +++++= K

For multiple surfaces with an event at the end:

( )

( )

( ) 2332211

2

332211

2

2

332211

2

332211

254

)81.9(92.25

92.2592.25

92.2592.25

fnno

fnno

f

nno

nn

SdfdfdfdfS

SdfdfdfdfS

Sgdfgdfgdfgdf

S

+++++=

+++++=

+++++=

K

K

K

K



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If we continue with the last equation:

( ) 2332211254 fnno SdfdfdfdfS +++++= K


22

3

2

2

2

1

2

332211 254254254254

no

fnno

SSSSS

SdfdfdfdfS

++++=

+++++=

K

K

Generic equation for

multiple events ( is

just another event so it

is encompassed by ).

2

fS

2

nS

3131


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Thus:

Speed equations come from energy (the Work-Energy Theorem).

.


23

3

2

2

2

1 nSSSSS ++++= K( ) finalninitial KEWWWWKE ++++= K321

3232


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WHATELSECANENERGYANALYSES BEUSEDFOR?

Calculating impact speeds in an inline

(collinear) collision. The Conservation of Linear Momentum (COLM) and

Conservation of Ener COE e uations are solved

simultaneously.



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(CONT.)

Calculating the speed of a vehicle that collideswith a barrier and comes to a stop.

Damage energy is used to calculate the speed.

If the barrier doesnt move and doesnt sustain

damage, the speed calculated will be both a delta-V and an impact speed.

If the barrier moves or sustains damage or if there

is any post impact movement of the vehicle, thespeed calculated will be an EBS (EquivalentBarrier Speed).



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(CONT.)

Calculating delta-Vs

=

11

11

2

m

Ev crush

1

2

12 v

m

mv =

22

1

m

Where: Ecrush = E1 + E2 (damage energy of Unit 1 plus the damage energy of Unit 2), J

m1 = mass of Unit 1, kgm2 = mass of Unit 2, kg

= effective mass ratio



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SOMEDEFINITIONS Damage Energy is based on the assumption the vehicle deforms according

to Hookes Law.

3636Copyright 2010Copyright 2010 -- 2012 by N. Shigemura & J. Daily2012 by N. Shigemura & J. Daily


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SOMEDEFINITIONS In the previous slide, A & B are the Stiffness

Coefficients of the damaged vehicle

The A stiffness value represents the force the

vehicle can sustain without manifesting residual

crush.

The B coefficient is the slope of the force-deflection

line.


B

AG 2

2

=


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SOMEDEFINITIONS What is , the effective mass ratio?

Gamma is a ratio of the acceleration of the vehicle

center of mass to the acceleration at the centroid

of the damage area.


22

2

hkk+

=

k = radius of gyrationh = lever arm from collision (impulse) force line to CM


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SOMEDEFINITIONS

The area of the force-deflection triangle

represents the work done to crush the vehicle.Because damage areas are irregular in most cases,

a series of trian les all with the same stiffness

values, are summed up across the width of thedamage on the vehicle.


dxdLBxAE )()tan1( 2

++=


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BASICDAMAGEENERGYEQUATION

( ) ( )2

2

tan12

+

++=

BAAxBAE D


Where: E = total damage energy; irreversible work done to crush vehicle

A, B = stiffness coefficients

AD = area of damage

= depth of the damage centroid from the damage face

= angle the collision force makes with respect to a

line perpendicular to the damage face.

(1+tan2) = Magnification factor

x

4040


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WHAT ISNEEDED TOCALCULATEDAMAGEENERGY?

Four things are needed to calculate damage

energy:1. Stiffness coefficients

.

3. Depth of the damage centroid from the damage

face

4. The angle the collision force makes with respect

to a line normal (perpendicular) to the damage

face.



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DAMAGEENERGY

Determining damage energy allows us to:

Calculate the amount of kinetic energy/workexpended to crush the car.

-

determined.

Will allow progression to further analysis.

Delta-V

Impact speeds via a simultaneous equation solution



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DETERMININGSTIFFNESSCOEFFICIENTS National Highway Traffic Safety Administration website

http://www-nrd.nhtsa.dot.gov/database/veh/veh.htm

Copyright 2010Copyright 2010 -- 2012 by N. Shigemura & J.2012 by N. Shigemura & J.

DailyDaily 4343


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DETERMINING

STIFFNESS

COEFFICIENTS

(CONT

) Commercial software packages



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Copyright 2010 - 2012 by N. Shigemura & J. Daily4545


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DETERMININGDAMAGE AREA

Use scale diagram and geometric shapes.

Measure damage area using Tumbas andSmith protocol, SAE 880072.



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DETERMINING

DAMAGE AREA


1 2 3

4848


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DETERMININGDEPTH OFCENTROID OF DAMAGEAREA

The centroidcentroidcentroidcentroid is the center of a volume or a two

dimensional area defined by a boundary. Can be determined:

,

Paper cutouts,

Calculations.


DETERMININGANGLE OFCOLLISIONFORCE WITH


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RESPECT TO A LINENORMAL TODAMAGEFACE

Equations and measurements are based onNHTSA crash tests where the vehicles crashperpendicularly into barriers.

If the collision force is not perpendicular to the

damage face, a geometric/trigonometricadjustment needs to be made to take intoaccount the angle.

This adjustment is called the magnificationmagnificationmagnificationmagnificationfactor.factor.factor.factor.


)2tan1+5050


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CONCLUSION

Understanding energy concepts and energyanalysis techniques puts more tools in thecrash investigator/reconstructionists tool box.

By looking at the movements of vehicles from

an energy point of view helps clarify manyissues and questions.

Utilizing energy analysis techniques can assist

the investigator in solving crashes that mayhave previously gone unsolved.


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Energy Concepts Brazil 08-21-12

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