[email protected] engr-36_lec-08_moments_math.ppt 1 bruce mayer, pe engineering-36:...

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

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

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 4: Moment

Mathematics



Moments are VECTORS

As Described Last Lecture a Moment is a measure of “Twisting Power”

A Moment has Both MAGNITUDE & Direction and can be Represented as a Vector, M, with Normal Vector properties

222zyx

zyx

MMMM

kMjMiM

M

ˆˆˆM



Moments are VECTORS

Describe M in terms of a unit vector, û, directed along the LoA for M

zyx kjiuMu cosˆcosˆcosˆMˆˆMM

Find the θm by Direction CoSines

Mcos

Mcos

Mcos z

zy

yx

x

MMM



M = r X F

1. Magnitude of M measures the tendency of a force to cause rotation of a body about an Axis thru the pivot-Pt O

sinFrFsinrM Fd

d



M = r X F

2. The sense of the moment may be determined by the right-hand rule

M

MomentDirection

• If the fingers of the RIGHT hand are curled from the direction of r toward the direction of F, then the THUMB points in the direction of the Moment



M = r X F

Combining (1) & (2) yields the Definition of the vector CROSS PRODUCT (c.f. MTH3)

bac

Engineering Mechanics uses the Cross Product to Define the Moment Vector

usinFrFrM • û is a unit vector directed by the Rt-Hand Rule• θ is the Angle Between the LoA’s for r & F



M = r X F → θ by Tail-toTail When Finding Moment

Magnitudes using:

The Angle θ MUST be determined by placing Vectors r & F in the TAIL-to-TAIL Orientation• See Diagram at Right

sin FrM

http://images.clipartof.com/small/17076-Two-Cats-One-White-One-Orange-Standing-Tail-To-Tail-And-Forming-A-Heart-Symbolizing-Love-On-Valentines-Day-Poster-Art-Print.jpg



Cross Product Math Properties

Recall Vector ADDITION Behaved As Algebraic Addition

– BOTH Commutative and Associative.

The Vector PRODUCT Math-Properties do NOT Match Algebra - Vector Products:• Are NOT Commutative• Are NOT Associative• ARE Distributive

veDistributi

tiveNONassocia

tiveNONcommuta

2121 QPQPQQP

SQPSQP

QPQPPQ

QP

PQ



Vector Prod: Rectangular Comps Vector Products Of Cartesian Unit

Vectors

0

0

0

kkikjjki

ijkjjkji

jikkijii

kQjQiQkPjPiPQPV zyxzyx

kQPQPjQPQPiQPQP xyyxzxxzyzzy

Vector Product In Terms Of Rectangular Coordinates



rxF in 3D Deteriminant Notation

Consider 3D versions of r & F Taking the Cross Product Yields M Determinant Notation provides a

convenient Tool For the Calculation

• Don’t Forget the MINUS sign in the Middle (j)Term– See also TextBook pg123

kFrFrjFrFriFrFr

FFF

rrr

kji

Fr xyyxxzzxyzzy

zyx

zyxˆˆˆ

ˆˆˆ



Varignon’s Theorem

The Moment About a Point OOf The Resultant Of SeveralConcurrent Forces Is Equal To The Sum Of The MomentsOf The Various Forces About The Same Point O• Stated Mathematically

2121 FrFrFFr

Varignon’s Theorem Makes It Possible To Replace The Direct Determination Of The Moment of a Force F By The Moments of Its Components (which are concurrent)



rxF in 3D Vector Properties

Cartesian CoOrds for a 3D M vector

xyyxz

xzzxy

yzzyx

FrFrM

FrFrM

FrFrM

The Magnitude of a 3D M vector

222xyyxxzzxyzzy FrFrFrFrFrFrM M

DirectionCoSines

M

M

M

xyyxzz

xzzxyy

yzzyxx

FrFr

M

M

FrFr

M

M

FrFr

M

M

cos

cos

cos Unit Vector

kM

Mj

M

Mi

M

Mu zyx ˆˆˆˆ



rxF in 2D r & F in XY Plane

If r & F Lie in the XY Plane, then rz = Fz = 0. Thus the rxF Determinant

kFrFrjrriFr

FF

rr

kji

Fr xyyxzxyy

yx

yxˆˆˆ

ˆˆˆ

0000

0

0

So in this case M is confined to the Z-Direction:

kFrFrkMFr xyyxzzXYˆˆM



Direction for r in rF Consider the CrowBar Below We Want to find the

Torque (Moment) About pt-B due to Pull, P, applied at pt-A using rP

We have Two Choices for r:• r points A→B

• r points B→A

Which is Correct?



Direction for r in rF We can find the Direction for r by

considering the SIGN of the Moment

In this case it’s obvious (to me, anyway) that P will cause CLOCKwise Rotation about Pt-B

In the x-y Plane ClockWise Rotation is defined as NEGATIVE

Test rP and rP

x

y



Direction for r in rF Put r and r into Component form

• Equal but Opposite

x

y

".cos"

".sin"

14235036

58275036

comp

comp

x

y

Then the two r’s

jir

jir

BAAB

ABBA

ˆ.ˆ.r

ˆ.ˆ.r

58271423

58271423

Now let

ilb 10P



Direction for r in rF then the rxP calculations noting

x

y

lb-in k8.2750

i10j58.27i14.23

^

^^^

Pr BA

Thus rB→A is the CORRECT position vector

lb-in k8.2750

i10j58.27i14.23

^

^^^

Pr AB

^^^^^

kij0ii



Direction for r in rF To Calc the Moment about pt-B use: The position Vector

points FROM the PIVOT-point TO the Force APPLICATION-point on the Force LoA

Summarize this as

FROM the PIVOT TO the FORCE



Unit Vector Notation: u ≡ λ

Our Text uses u to denote the unit vector

While u is quite popular as the unit vector notation, other symbols are often used (kind of like θ & φ for angles)

On Occasion I will use λ to represent the unit vector• This is usually apparent

from the problem or situation context ABAB

ABAB

u ˆ

λu



Example: 3D Moment A Rectangular Plate Is Supported

By The Brackets At A and B and By A Wire CD. Knowing That The Tension In The Wire is 200 N, Determine The Moment About A Of The Force Exerted By The Wire At connection-point C.

Solution Plan• The Moment MA Of The Force F

Exerted By The Wire Is Obtained By Evaluating The Vector Product

FrM ACA



Example 3D Moment - Solution Resolve Both F and rAC into

Cartesian Components Take Cross-Product Using Determinant

kjiM A

mN 8.82mN 8.82mN 687 .

kji

kji

CD

CDuFF

N 128N 69N 120

m 5.0

m 32.0m 0.24m 3.0N 200

N 200ˆ

kikkirrr ACAC m 080m 30m 320m 40m 30 .....

12896120

08.003.0

ˆˆˆ

ˆ

kji

uFACA rMWhich Moment will Most Likely Cause DEFORMATION?

ACr



Moment About an Axis (§4.5)

Moment MO Of A Force F ,Applied at The Point A, About a Point O, Recall

FrMO

FrMM OOL

Scalar Moment MOL About AnAXIS OL Is The Projection OfThe Moment Vector MO OntoThe OL Axis using the Dot Product

MOL it the tendency of the applied force to cause a rotation about the AXIS OL



Moment About an Axis – cont. Moments of F About The CoOrd Origin

xyz

zxy

yzx

yFxFM

xFzFM

kzjyixrzFyFM

as

BABA

BA

BBL

rrr

Fr

MM

Moment Of A Force AboutAn Arbitrary Axis BL

• Similar Analysis for CL, Starting With MC, Shows That MCL = MBL; i.e., the Result is Independent of the Location of the Point ON the Line

BABA rrr



Moment About an Axis – cont.

Since the moment, ML,about an arbitrary axisis INDEPENDENT ofposition vector, r, thatruns from ANY Pointon the axis to ANY pointon the LoA of the force we can choose the MOST CONVENIENT Points on the Axis and the Force LoA to determine ML

BAr

CAr



MOL Physical Significance

MOL Measures the Tendency of an Applied Force to Impart to a Rigid Body Rotation about a fixed Axis OL• i.e., How Much will the Applied Force

Cause The body to Rotate about an AXLE• MOL can be

Considered asthe Componentof M directed along “axis” OL



Example: MOL

A Cube With Side Length a is Acted On By a Force P as Shown

Determine The Moment Of P:a) About Pt A

b) About The Edge (Axis) AB

c) About The Diagonal (Axis) AG of The Cube

For Lines AG and FC d) Determine The

Perpendicular Distance Between them



Example MOL - Solutiona) Moment of P about A

kjPjiaM

kjPa

kjaP

aa

kajaP

FC

FCPP

jiajaiakajakaiarrr

PrM

A

AFAF

AFA

2

2222

kjiaPM A

2

kji

Pai

MiMkjiMM AAAABAB

2

001

aPaPM AB 707.02

aPaP

M A 225.12

3

b) Moment of P about AB

AFr



Example MOL - Solution

kaPjaPiaP

PP

aa

kji

PrAF ˆ2ˆ2ˆ2

220

0

ˆˆˆ

AFr

a) Alternative Moment of P about A

kjPa

kjaP

aa

kajaP

FC

FCPP

jiajaiakjaia

kzjyixrrr

PrM

AFAF

AFA

ˆˆ22

ˆˆˆˆ

ˆˆˆˆˆ0ˆˆ

ˆˆˆ

22

kjiaPM A

2



Example MOL - Solution

c) Moment of P About Diagonal AG

1116

23

12

3

1

3

aP

kjiaP

kjiM

kjiaP

M

kjia

kajaia

r

r

AG

AG

MM

AG

A

AG

AGAG

AAGAG

ˆ

ˆ

) toOPPOSITE is (dir.408.06

aPaP

M AG



Example MOL - Solution Perpendicular distance between AG and FC• Notice That Plane OFC Appears To

Be to Line AG, And FC Resides In this Plane– Since P Has Line-of-Action FC We Can

Test Perpendicularity with Dot Product

0

11063

1

2

P

kjikjP

P

PdaP

M AG 6

aa

d 408.06

• Then the Moment (or twist) Caused by P About AG = Pd; Thus



Mixed Triple Product

Do Find MOL we used the Qtyû•(r x F). Formalize thisOperation as the Mixed TripleProduct for vectors S, P, & Q resultscalar QPS

SPQQSPPQS

PSQSQPQPS

Associativity and Communtivity for the Mixed Triple Product Of Three Vectors



Evaluate the Mixed Triple Prod

Let V = PxQ, Then

zzyyxx VSVSVSVSQPS

zyyzzyx VVQPQPV andforSimilarly

zyx

zyx

zyx

xyyxz

xzzxyyzzyx

QQQ

PPP

SSS

QPS

QPQPS

QPQPSQPQPSQPS

And

Thus

• Determinant Notation Yet Again



Mixed Triple Product vs MOL

The Mixed Triple Product can be used to find the Magnitude of the Moment about an Axis.

zyx

zyx

zolyolxol

OL

olOL

FFF

rrr

uuu

u

,,, ˆˆˆ

M

FrˆM



WhiteBoard Work

Let’s WorkThis NiceProblem

Determine MA as caused by application ofthe 120 N force



Vector/Cross Product

TWISTING Power of a Force MOMENT of the Force• Quantify Using VECTOR

PRODUCT or CROSS PRODUCT Vector Product Of Two Vectors

P And Q Is Defined As TheVector V Which Satisfies:• Line of Action of V Is Perpendicular To Plane

Containing P and Q.– Rt Hand Rule Determines Direction for V

• |V| =|P|•|Q|•sin

[email protected] engr-36_lec-08_moments_math.ppt 1 bruce mayer, pe engineering-36:...

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