Bruce Mayer, PE Licensed Electrical & Mechanical Engineer - PowerPoint Presentation

Slide1

Bruce Mayer, PELicensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

Engineering 36

Chp 5:

Equivalent Loads

Slide2

Introduction: Equivalent Loads

Any System Of Forces & Moments Acting On A Rigid Body Can Be Replaced By An Equivalent System Consisting of these “Intensities” acting at Single Point:One FORCE (a.k.a. a Resultant)One MOMENT (a.k.a. a Couple)

Equiv. Sys.

Slide3

External vs. Internal Forces

Two Classes of Forces Act On Rigid Bodies:External forcesInternal forces

The Free-Body Diagram Shows

External

Forces

UnOpposed

External Forces Can Impart Accelerations (Motion)

Translation

Rotation

Both

Slide4

Transmissibility: Equivalent Forces

Principle of TransmissibilityConditions Of Equilibrium Or Motion Are Not Affected By TRANSMITTING A Force Along Its LINE OF ACTION

Note:

F & F’ Are Equivalent Forces

Moving the point of application of

F

to

the rear bumper does

not affect the motion

or the other forces

acting on the truck

Slide5

Transmissibility Limitations

Principle of transmissibility may not always apply in determiningInternal ForcesDeformations





Rigid

Deformed

TENSION

COMPRESSION

Slide6

Moment of a Couple

COUPLE  Two Forces F and −F With Same Magnitude Parallel Lines Of ActionDistance SeparationOpposite DirectionMoment of The Couple about O

Slide7

M of a Couple → Free Vector

Thus The Moment Vector Of The Couple is INDEPENDENT Of The ORIGIN Of The Coord Axes Thus it is a FREE VECTORi.e., It Can Be Applied At Any Point on a Body With The Same Effect

Two Couples Are Equal If

F

1

d

1

=

F

2

d

2

(d

1

& d

2

are ┴ distances)

The Couples Lie In Parallel Planes

The Couples Have The Tendency To Cause Rotation In The

Same Direction

Slide8

Some Equivalent Couples

These Couples Exert Equal Twist on the Blk

For the Lug Wrench Twist

Shorter Wrench with greater Force

Would Have the Same Result

Moving Handles to Vertical, With

Same Push/Pull Has Same Result

Slide9

Couple Addition

Consider Two IntersectingPlanes P1 and P2 WithEach Containing a Couple

Resultants Of The Force Vectors

Also Form a Couple

Slide10

Couple Addition

By Varignon’s Distributive Theoremfor Vectors

Thus The Sum of Two Couples Is Also A Couple That Is Equal To The Vector Sum Of The

Two individual

Couples

i.e., Couples Add The Same as Force

Vectors

The resultant Couple is Also a FREE Vector

Slide11

Couples Are Vectors

Properties of Couples

A Couple Can Be Represented By A Vector With Magnitude & Direction Equal To The Couple-Moment

Couple Vectors Obey The Law Of Vector Addition

Couple Vectors Are

Free

Vectors

i.e., The Point Of

Application or

LoA

Is NOT Significant

Couple Vectors May Be Resolved Into

Component Vectors

Slide12

Resolution of a Force Into a Force at O and a Couple

Couple

r

x

F

Force Vector F Can NOT Be Simply Moved From A To O Without Modifying Its Action On The BodyAttaching Equal & Opposite Force Vectors At O Produces NO Net Effect On The BodyBut it DOES Produce a CoupleThe Three Forces In The Middle Diagram May Be Replaced By An Equivalent Force Vector And Couple Vector; i.e., a FORCE-COUPLE SYSTEM

Can be FREELY Moved

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Force-Couple System at O’

The Moments of

F

about

O

and

O’ are Related By The Vector s That Joins O and O’

Moving

F

from

A

To a Different Point

O

’ Requires Addition

of a

Different

Couple Vector

Slide14

Force-Couple System at O’



Moving The

Force-Couple

System From

O

to

O

’ Requires The Addition Of The Moment About

O

’ Generated by the Force At

O

Remember that

M

O

is a FREE Vector, and must be ADDED at

O

’ to obtain

M

O’

Slide15

Example: Couples

Determine The Components Of The Single Couple Equivalent To The Couples Shown

Solution Plan

Attach Equal And Opposite 20 Lb Forces In The ±x Direction At

A

, Thereby Producing 3 Couples For Which The Moment Components Are Easily Calculated

Alternatively, Compute The Sum Of The Moments Of The Four Forces About An Arbitrary Single Point.

The Point

D

Is A Good Choice As Only Two Of The Forces Will Produce Non-zero Moment Contributions

Slide16

Example: Couples

Attach Equal And Opposite 20 lb Forces In the ±x Direction at ANo Net Change to the StructureThe Three Couples May Be Represented By 3 Vector Pairs

M

x

M

x

M

y

M

y

M

z

M

z

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Example: Couples

Alternatively, Compute The Sum Of The Moments Of The Four Forces About DOnly The Forces At C and E Contribute To The Moment About Di.e., The Position vector, r, for the Forces at D = 0

r

DE

r

DC

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Reduction to Force-Couple Sys

A

SYSTEM

OF FORCES May Be REPLACED By A Collection Of FORCE-COUPLE SYSTEMS Acting at Given Point

O

The Force And Couple Vectors

May then Be

Combined Into

a

single

Resultant

Force-Vector and a

Resultant Couple-Vector

Slide19

Reduction to a Force-Couple Sys

Two Systems Of Forces Are EQUIVALENT If They Can Be Reduced To The SAME

Force-Couple

System

The Force-Couple System at

O

May Be Moved To

O

’ With The

Addition

Of The Moment Of

R

About

O

’ as before:

Slide20

More Reduction of Force Systems

If the Resultant Force & Couple At O Are Perpendicular, They Can Be Replaced By A Single Force Acting With A New Line Of Action.Force Systems That Can be Reduced to a Single ForceConcurrent ForcesGenerates NO MomentCoPlanar Forces (next slide)The Forces Are ParallelCoOrds for Vertical Forces

(a)

(b)

(c)

Slide21

CoPlanar Force Systems

System Of CoPlanar Forces Is Reduced To A Force-couple System That Is Mutually Perpendicular

System Can Be Reduced To a Single Force By Moving The Line Of Action

R

To Point-A Such That d:

In Cartesian

Coordinates use transmissibility to slide the Force

PoA

to Points on the X & Y Axes

Slide22

Parallel Force Systems

Proper Placement of the Resultant (at in this case Produces the SAME MOMENT about the & Axes as did the 3 the VectorsExample of this methodology on next Slide

 

 

 

 

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Example: 2D Equiv. Sys.

.

For The Beam, Reduce The System Of Forces Shown To

An Equivalent Force-Couple System At A An Equivalent Force-Couple System At BA Single Force applied at the Correct Location

Solution Plan

Compute

The Resultant Force

The Resultant Couple About

A

Find An Equivalent Force-couple System at

B

Based On The

Force-Couple

System At

A

Determine The

Point Of Application

For The Resultant Force Such That Its Moment About

A

Is Equal To The Resultant Couple at

A

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Example: 2D Equiv. Sys. - Soln

Now Calculate the Total Moment About

A

as Generated by the Individual Forces.

Find the resultant force and the resultant couple at

A

.

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Example: 2D Equiv. Sys. - Soln

Find An Equivalent Force-couple System At B Based On The Force-couple System at AThe Force Is Unchanged By The Movement Of The Force-Couple System From A to B

The Couple At

B

Is Equal To The Moment About

B

Of The Force-couple System Found At

A

r

BA

Slide27

Example: 2D Equiv. Sys. - Soln

Determine a SINGLE Resultant Force (NO Couple)The Force Resultant Remains UNCHANGED from parts a) & b)The Single Force Must Generate the Same Moment About A (or B) as Caused by the Original Force System

Then the Single-Force Resultant

Chk 1000 Nm at B



Slide28

Example: 3D Equiv. Sys.

3 Cables Are Attached To The Bracket As Shown. Replace The Forces With An Equivalent Force-Couple System at A

Solution Plan:Determine The Relative Position Vectors For The Points Of Application Of The Cable Forces With Respect To A.Resolve The Forces Into Rectangular ComponentsCompute The Equivalent Force

Calculate The Equivalent Couple

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Example Equiv. Sys. - Solution

Determine The Relative Position Vectors w.r.t. A

Resolve The Forces Into Rectangular Components

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Example Equiv. Sys. - Solution

Compute Equivalent Force

Compute Equivalent Couple

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Distributed Loads

The Load on an Object may be Spread out, or Distributed over the surface.

Load Profile, w(x)

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Distributed Loads

If the Load Profile, w(x), is known then the distributed load can be replaced with at POINT Load at a SPECIFIC LocationMagnitude of thePoint Load, W, is Determined by Area Under the Profile Curve

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Distributed Loads

To Determine the Point Load Location employ Moments

Recall: Moment = [LeverArm]•[Intensity]

In This Case

LeverArm = The distance from

the Baseline Origin, x

n

Intensity = The Increment of Load, dW

n

, which is that load, w(x

n

) covering a

distance dx located at x

n

That is: dW

n

= w(x

n

)•dx

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Distributed Loads

Now Use Centroidal Methodology

And

also:

Equating the

Ω

Expressions

find

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Distributed Loads on Beams

A distributed load is represented by plotting the load per unit length,

w

(

N/m

). The total load is equal to the area under the load curve.

A distributed load can be REPLACED by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the areal centroid.

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Integration Not Always Needed

The Areas & Centroids of Common Shapes Can be found on the Inside Back-Cover of the Text BookStd Areas can be added & subtracted directlyStd Centroids can be combined using [LeverArm]∙[Intensity] methods

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Scalene TriAngle Centroid

Slide39

Example:Trapezoidal Load Profile

A beam supports a distributed load as shown. Determine the equivalent concentrated load and its Location on the Beam

Solution PlanThe magnitude of the concentrated load is equal to the total load (the area under the curve)The line of action of the concentrated load passes through the centroid of the area under the Load curve.The Equivalent Causes the SAME Moment about the beam-ends as does the Concentrated Loads

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Example:Trapezoidal Load Profile

SOLUTION:

The magnitude of the concentrated load is equal to the total load, or the area under the curve.

The line of action of the concentrated load passes through the area centroid of the curve.

Slide41

WhiteBoard Work

Let’s Work

This NiceProblem

For the Loading & Geometry shown Find:

The Equivalent Loading

HINT: Consider the Importance of the Pivot Point

The Scalar component of the Equivalent Moment about line

OA

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Bruce Mayer, PELicensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

Engineering 36

Appendix

WhtBd Soln to Channel Bracke

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