Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-36_Lec-08_Moments_Equiv-Loads.ppt1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 5: Equivalent

Loads



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.



External vs. Internal Forces Two Classes of

Forces Act On Rigid Bodies:• External forces• Internal forces The Free-Body Diagram

Shows External Forces• UnOpposed External

Forces Can Impart Accelerations (Motion) – Translation– Rotation– Both



Transmissibility: Equivalent Forces Principle of

Transmissibility• Conditions 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 tothe rear bumper doesnot affect the motion or the other forcesacting on the truck



Transmissibility Limitations Principle of transmissibility may not

always apply in determining• Internal Forces• Deformations

Rigid Deformed

TENSION

COMPRESSION



Moment of a Couple COUPLE Two Forces F and −F With

Same • Magnitude • Parallel Lines Of Action• Distance separation• Opposite Direction

Moment of The Couple about O

distancethesin

dFdrFMFr

Frr

FrFrM

BA

BA



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 VECTOR

– i.e., It Can Be Applied At Any Point on a Body With The Same Effect

Two Couples Are Equal If• F1d1 = F2d2

• The Couples Lie In Parallel Planes• The Couples Have The Tendency

To Cause Rotation In The Same Direction



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



Couple Addition Consider Two Intersecting

Planes P1 and P2 WithEach Containing a Couple

222

111

planein

planein

PFrM

PFrM

21 FFrRrM

Resultants Of The Force Vectors Also Form a Couple



Couple Addition By Varignon’s Distributive Theorem

for Vectors

21

21

MM

FrFrM

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



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



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 Body

Attaching Equal & Opposite Force Vectors At O Produces NO Net Effect On The Body• But it DOES Produce a Couple

The Three Forces In The Middle Diagram May Be Replaced By An Equivalent Force Vector And Couple Vector; i.e., a FORCE-COUPLE SYSTEM



Force-Couple System at O’

FrMO

'

FsMM

FsFrFrsFrM

OO

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



Force-Couple System at O’

FsMM OO

'

Moving The Force-couple System From O to O’ Requires The Addition Of The Moment About O’ Generated by the Force At O



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



Example: Couples Attach Equal And Opposite 20 lb

Forces In the ±x Direction at A• No Net Change to the Structure

The Three Couples May Be Represented By 3 Vector Pairs

in.lb 540in. 18lb 30 xM

kji ˆˆˆM in.lb 180in.lb240in.lb 540

Mx

Mx

My

My

Mz

Mz

in.lb240in. 12lb 20 yM

in.lb 180in. 9lb 20 zM



Example: Couples Alternatively, Compute The Sum Of

The Moments Of The Four Forces About D

Only The Forces At C and E Contribute To The Moment About D• i.e., The Position vector, r, for the

Forces at D = 0

ikj

kjD

ˆˆˆ

ˆˆMM

lb 20in. 12in. 9

lb 30in. 18

rDErDC

kji ˆˆˆM in.lb 180in.lb240in.lb 540



Reduction to Force-Couple Sys

FrMFR RO

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



Reduction to a Force-Couple Sys

RsMM RO

RO

'

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:



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 Forcea) Concurrent Forces

– Generates NO Moment

b) Coplanar Forces (next slide)c) The Forces Are Parallel

– CoOrds for Vertical Forces

(a)

(b)

(c)

Rzy

Rxy

MxR

MzR



CoPlanar Force Systems System Of CoPlanar Forces Is

Reduced To A Force-couple System That Is Mutually Perpendicular

ROxy MyRxR

RMd RO

FrMFR RO

and

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 & Axes

0y 0x



Example: 2D Equiv. Sys.

.

For The Beam, Reduce The System Of Forces Shown To

a) An Equivalent Force-Couple System At A

b) An Equivalent Force-Couple System At B

c) A Single Force applied at the Correct Location

Solution Plana) Compute

– The Resultant Force – The Resultant Couple

About Ab) Find An Equivalent

Force-couple System at B Based On The Force-couple System At A

c) Determine The Point Of Application For The Resultant Force Such That Its Moment About A Is Equal To The Resultant Couple at A



Example: 2D Equiv. Sys. - Soln

jjjj

FRˆˆˆˆ N 250N 100N 600N 150

jR

N600 Now Calculate the Total Moment

About A as Generated by the Individual Forces.

ji

jiji

FrM RA

ˆˆ.

ˆˆ.ˆˆ.

25084

1008260061

kM RA

ˆmN 1880

a) Find the resultant force and the resultant couple at A.



Example: 2D Equiv. Sys. - Solnb) Find An Equivalent Force-

couple System At B Based On The Force-couple System at A• The Force Is Unchanged By The

Movement Of The Force-Couple System From A to B

jR ˆN 600

kk

jik

RrMM BARA

RB

ˆˆ

ˆˆ.ˆ

mN 2880mN 1880

N 600m 84mN 1880

kM RB

mN 1000

• The Couple At B Is Equal To The Moment About B Of The Force-couple System Found At A

rBA



Example: 2D Equiv. Sys. - Solnc) 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

m 133N 600 .ˆ xjR

kxk

jixk

RrM AxRA

ˆN 600ˆmN 1880

ˆN 600ˆˆmN 1880

Then the Single-Force Resultant

kji

ji

RrM BxRB

ˆmN 1002

ˆN 600ˆ67.1

ˆN 600ˆ8.413.3

Chk 1000 Nm at B



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 Components

• Compute The Equivalent Force

FR

FrM RA

• Calculate The Equivalent Couple



Example Equiv. Sys. - Solution

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

m 10001000

m 05000750

m 05000750

jir

kir

kir

AD

AC

AB

..

..

..

N ˆ200ˆ600ˆ300

ˆ289.0ˆ857.0ˆ429.0

1755015075ˆ

ˆN 700

kjiF

kji

kjirru

uF

B

BE

BE

B

N ˆ707ˆ707

ˆ45cosˆ45cosN 1000

ki

kiFC

N ˆ1039ˆ600

ˆ30cosˆ60cosN 1200

ji

jiFD

Resolve The Forces Into Rectangular Components



Example Equiv. Sys. - Solution Compute Equivalent Force

k

j

i

FR

ˆ

ˆ

ˆ

707200

1039600

600707300

N 5074391607 kjiR ˆˆˆ

kkji

Fr

jkji

Fr

kikji

Fr

FrM

DAD

cAC

BAB

RA

ˆ...

ˆˆˆ

ˆ...

ˆˆˆ

ˆˆ..

ˆˆˆ

916301039600010001000

68177070707050000750

4530200600300050000750

kjiM RA

9.11868.1730

Compute Equivalent Couple



Distributed Loads The Load on an Object may be Spread

out, or Distributed over the surface.

Load Profile, w(x)



Distributed Loads If the Load Profile, w(x), is known then

the distributed load can be replaced with at POINT Load at a SPECIFIC Location

Magnitude of thePoint Load, W, is Determined by Area Under the Profile Curve

span

dxxwW

N 3

100W



Distributed Loads To Determine the Point Load Location

employ Moments Recall: Moment = [LeverArm]•[Intensity] In This Case

• LeverArm = The distance from the Baseline Origin, xn

• Intensity = The Increment of Load, dWn, which is that load, w(xn) covering a distance dx located at xn

– That is: dWn = w(xn)•dx



Distributed Loads Now Use Centroidal Methodology

span

nnspan

x dxxwxIntensityLeverArm

And also: Location Centroid theis xWxx

Equating the Ω Expressionsfind

W

dxxwxx span

nn



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.

AdAdxwWL

0

AxdAxAOP

dWxWOPL

0

• 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.





Integration Not Always Needed The Areas & Centroids of Common

Shapes Can be found on Inside Back-Cover of the Text Book

Std Areas can be added & subtracted directly

Std Centroids can be combined using [LeverArm]∙[Intensity] methods



Example:Trapezoidal Load Profile

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

Solution Plan• The 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



Example:Trapezoidal Load ProfileSOLUTION:• The magnitude of the concentrated load is

equal to the total load, or the area under the curve.

kN 0.18F

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

kN 18mkN 63

X m5.3X

m6mN

245001500

F



WhiteBoard Work

Let’s WorkThis 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



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

Documents

pe engineering

forces f

equivalent system

couple free vectorthus

couple resultants

force vectors

n s j o h n s o n c

f2d2the couples