Chapter 4 Rigid Bodies Equivalent Force/Moment …cac542/L4.pdfChapter 4 Rigid Bodies Equivalent Force/Moment Systems. 2 ... Equilibrium for non-concurrent force ... the individual

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MEM202 Engineering Mechanics - Statics MEM

Chapter 4Rigid Bodies

Equivalent Force/Moment Systems

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Equilibrium of Rigid Bodies

0=−=∑ PPFrrr

Pr

Pr

0=∑Fr

Pr

Pr

0≠∑Mr

0 0 == ∑∑ MFrr

Equilibrium for non-concurrent force systems

Force systems to be studied:1. Coplanar, parallel force systems2. Coplanar, non parallel, non-concurrent force systems3. Non-coplanar, parallel force systems4. Non-coplanar, nonparallel, non-concurrent force systems

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4.2 Moments and Their Characteristics

OMr

O

FrMFdM OO

rrr×== or

AAF −about ofMoment r

Fr

A

A

Orr

Fr

O

drr

• The magnitude of a moment is the product of the magnitude of theforce (F) and the perpendicular distance (d) from the line of action of the force to the axis of rotation (A-A).

• The units for moments are in⋅lb (U.S.) and N⋅m (SI).• A moment is a vector. Thus, parallelogram law of addition applies.

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4.2 Moments and Their CharacteristicsSign Convention for Moments

Right-hand Rule :When fingers of right hand curl from positive r to positive F, the thumb is pointing in the direction of positive moment.

In 2-D, a counter-clockwise moments ( ) is positive and a clockwise moments ( ) is negative.

x

y

x

y

+ −

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4.2 Moments and Their CharacteristicsExamples

BCd

Moment of FA about point E( ) (ccw) lbin 100010100 ⋅=== AEAE dFM

r

Moment of FE about point A( ) (ccw) lbin 240012200 ⋅=== EAEA dFM

r

Moment of FD about point B( ) (cw) lbin 420014300 ⋅=== DBDB dFM

r

Moment of FC about point B( ) (ccw) mN 2.2512.016.090 ⋅=+== CBCB dFM

r

Moment of FD about point A( ) (ccw) mN 0.6620.035.0120 ⋅=+== DADA dFM

r

Moment of FB about point C( )[ ] (cw) mN 4.3630cos12.016.0150 ⋅=+== o

rBCBC dFM

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Moment of FB about point C( ) (cw) lbin 60015400 ⋅=== BCBC dFM

r

Moment of FC about point B( ) (ccw) lbin 360012300 ⋅=== CBCB dFM

r

Moment of FD about point B( ) (ccw) lbin 7503250 ⋅=== DBDB dFM

r

o13.531520tan 1 == −φ

in. 251520 22 =+=ACL

in. 0.313.53sin1515

in. 0.1213.53sin15

in. 0.1510

=−=

==

=−=

o

o

DB

CB

ACBC

d

d

Ld

4.2 Moments and Their CharacteristicsExamples

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Varignon’s Theorem: The moment of the resultant of a system of forces with respect to any axis or point is equal to the vector sum of the moments of the individual forces of the system with respect to the same axis or point.

4.2 Moments and Their Characteristics

2α

1α

2αα

Rr

1Fr

2Fr

1α

α

1d

d2d

( )αcos :about ofMoment hRRdMOR O ==r

21

2211

2211

2211

coscoscoscoscoscos

MMMdFdFRd

hFhFRhFFR

R +=⇒+=⇒

+=⇒+=

αααααα

nnO

n

dFdFdFRdMFFFR+++==⇒

+++=L

rL

rrr

2211

21 general In11 cosαF22 cosαF

αcosR

O

( )111111 cos :about ofMoment αhFdFMOF ==r

h

( )222222 cos :about ofMoment αhFdFMOF ==r

21 FFRrrr

+=

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4.2 Moments and Their CharacteristicsExample of Varignon’s Theorem

ft 83.230sin330cos5 =−= ood lb 0.43330cos50030cos === ooFFx

lb 0.25030sin50030sin === ooFFylbft 1415)83.2(500 ⋅−=−=−= FdMO

( ) ( ) lbft 1415534 ⋅−=−= xO FFM

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lb 8.25930cos30030cos === ooFFx

lb 0.15030sin30030sin === ooFFy

4.2 Moments and Their CharacteristicsExample of Varignon’s Theorem and Principle of Transmissibility

( ) ( ) mN 0.9520.025.0 ⋅−=−−= yxB FFM

( ) m 366.030tan2.025.02 =+= od( ) mN095366.082592 ⋅−=−=−= ..dFM xB

( ) m 633.030cot25.020.03 =+= od( ) mN095633.00.1503 ⋅−=−=−= .dFM yB

Move F to D (aligned horizontally with B)

FdM B = But d is not easy to determine

d

Determine Moment of F about B

Move F to C (aligned vertically with B)Find the components of F first

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4.3 Vector Representation of a MomentDot and Cross Products of Vectors (Appendix A)

( ) ( ) ( )kBABAjBABAiBABA

BBBAAAkji

eABBAC

xyyxzxxzyzzy

zyx

zyx

rrr

rrr

rrrr

−+−+−=

=

=×= sinθ

kAjAiAA zyx

rrrr++=

kBjBiBB zyx

rrrr++=

zzyyxx BABABAABBA ++==⋅ θcos

BAe

BeAe

C

CCrrr

rrrr

and containing plane

,

⊥

⊥⊥Czyx eCkCjCiCC rrrrr

=++=

y

z

Ar

BrC

rPlane containing A and B

θ

x OCer

Dot (Scalar) product of two vectors yields a scalarrr

Cross (Vector) product of two vectors yields a vector

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4.3 Vector Representation of a MomentCross Product of Two Vectors

kBjBiBBkAjAiAA zyxzyx

rrrrrrrr++=++= ,

kCjCiCBBBAAAkji

BAC zyx

zyx

zyx

rrr

rrr

rrr++==×=

yxzyx

yxzyx

zyx

zyx

BBBBBAAAAAjikji

BBBAAAkji

rrrrrrrr

=

+ + +− − −( ) ( ) ( )kBABAjBABAiBABA xyyxzxxzyzzy

rrr −+−+−=

CkCjCiC

eCCCC zyxCzyx

rrrr ++

=++= ,222

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4.3 Vector Representation of a MomentCross Product of Two Vectors

kjiBkjiArrrrrrrr

352 ,243 :Example ++−=+−=

( ) ( ) ( )( ) ( )( ) ( ) ( )( )

kji

kji

kBABAjBABAiBABA

jikjikji

BBBAAAkji

BAC

xyyxzxxzyzzy

zyx

zyx

rrr

rrr

rrr

rrrrrrrrrrr

rrr

7 13 22

2453 3322 5234

5235243243

352243

+−−=

−×−−×+×−−×+×−×−=

−+−+−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−−−=

−−==×=

( ) ( )( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( ) 0264.03491.05830.02

0264.02491.04830.03

264.0491.0830.071322

71322 :NOTE222

=+−+−−=⋅

=+−−+−=⋅

+−−=+−+−

+−−=

C

C

C

eB

eA

kjikjie

rr

rr

rrrrrr

r

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4.3 Vector Representation of a MomentMoment of a Force About A Point

FdMO =

2-D 3-D

Fr

d x

y

O

FdMO =

x

y

z Fr

rr

Od

α

α

rF

erF

eFrFrMO

rr

r

rr

rrrr

and containing plane the to

larperpendicur unit vecto A : and between angle The :

sin

α

α=×=

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αsinrd =

eMeFd

eFrFrM

O

Orr

rrrr

==

=×=

sinα

4.3 Vector Representation of a MomentMoment of a Force About A Point

Frrr

O

α

1α2α

1rr

2rr

rF rr and

containing Plane

d

Moment of a force about O is equal to the cross product of a vector drawn from O to any point along the force and the force itself.22

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sin

sin

α

α

r

rd

=

=

eMeFd

eFrFr

eFrFrM

O

O

rr

rrr

rrrr

==

=×=

=×=

sin

sin

222

111

α

α

Direction of M is determined by using the right-hand rule

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4.3 Vector Representation of a MomentBF point about of Moment :Example

r

x y

zFr

O

B

ABArr rr

=

Brr

Arr

• Select any point (A) on the line of action of F.

• Determine vector r using the coordinates of points A and B:r = rA/B = rA − rB

• Compute the moment of F about B:MB = r × F = rA/B × F

If the coordinates of points A and B are (xA, yA, zA) and (xB, yB, zB), then ( ) ( ) ( )kzzjyyixxrrrr BABABABABA

rrrrrrr −+−+−=−==

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4.3 Vector Representation of a Moment2-D Cases

Fr

rr

x

y

O d

αFr

rr x

y

Oyr

xr xF

yF

kFr

kFd

kMM OO

r

r

rr

sin

α=

=

=

( )kFrFr

FrM

xyyx

Or

rrr

−=

×=

Fr

rr

x

y

O

α

rθ

Fθ

rθ

( )rFrF

rF

θθθθθθα

sincoscossinsinsin

−=−=

( )( )kFrFr

kFrFr

kFrM

xyyx

rFrF

O

r

r

rr

sincoscossin

sin

−=

−=

=

θθθθ

α

ry

rx

Fy

Fx

rF

rrrrFFFF

θθθθ

θθα

sincos

sincos

====

−=

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4.3 Vector Representation of a Moment2-D Cases - Example

1. Moment of F about O2. Perpendicular distance d from O to F

( ) N 60080060.080.01000 jijiFrrrrr

+=+=

( )m 20.010.0 jirr OA

rrrr+==

( ) mN 100

00

⋅−==−=

=×=

kkMkFrFr

FFrr

kjiFrM

zxyyx

yx

yxO

rrr

rrr

rr

1. Moment of F about O

d

BAr

2. Determine the distance d:

m 22.0==F

Md B

r

r

jir BA

rr35.010.0 += mN 220 ⋅−=×= kFrM BAB

rrrr

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4.3 Vector Representation of a Moment3-D Cases

kFjFiFF

krjrirr

zyx

zyxrrrr

rrrr

++=

++=

( ) ( ) ( )

MO

zyx

xyyxzxxzyzzy

zyx

zyxO

eMkMjMiM

kFrFrjFrFriFrFr

FFFrrrkji

FrM

r

rrr

rrr

rrr

rrr

=

++=

−+−+−=

=×=

xyyxzzxxzyyzzyx FrFrMFrFrMFrFrM −=−=−=

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4.3 Vector Representation of a MomentMagnitude and Direction of the Resulting Moment

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

++=

++=

−−−

O

zz

O

yy

O

xx

zyxM

zyxO

MM

MM

MM

kjie

MMMM

111

222

coscoscos

coscoscos

θθθ

θθθrrrr

MOzyxO eMkMjMiMM rrrrr =++=

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4.3 Vector Representation of a MomentForce as a Sliding Vector (Principle of Transmissibility)

CACBABA rrrrr rrrrr+=+=

( )( ) Frr

Frr

FrM

CAC

BAB

AO

rrr

rrr

rrr

×+=

×+=

×=FrAr

r

O

Brr

Crr

AB

C0//

0//

=×⇒

=×⇒

FrFr

FrFr

CACA

BABArrrr

Q

rrrrQ

FrFrFrM CBAO

rrrrrrr×=×=×=∴

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4.3 Vector Representation of a Moment

( ) ( ) ( )N 7.5608.6004.300

14015075

14015075875222

kji

kjiF

rrr

rrrr

++=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

++=

kjir OA

rrrr 15.025.020.0 ++=

mN 1.451.671.50

7.5608.6004.30015.025.020.0

⋅+−=

=×=

kji

kjiFrM OAO

rrr

rrr

rrr

1. Moment of F about O2. Perpendicular distance d from O to F

( ) ( ) ( ) mN 95.11.451.671.50 222 ⋅=+−+== OO MMr

mm 108.6m 0.108687595.1

====F

Md O

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4.3 Vector Representation of a MomentMoment of a Force About A Line (Axis)

MOO eMFrM rrrr=×= ( )

( )[ ] nOBnn

nnOOB

eMeeFr

eeMMrrrrr

rrrr

=⋅×=

⋅=

knjnine zyxn

rrrr++=

( ) ( ) ( ) ( )[ ] [ ]

( ) ( ) ( ) zxyyxyzxxzxyzzy

zyx

zyx

zyx

n

zyx

zyx

zyxxyyxzxxzyzzynnOOB

nFrFrnFrFrnFrFrFFFrrrnnn

eFFFrrrkji

knjninkFrFrjFrFriFrFreFreMM

−+−+−==⋅=

++⋅−+−+−=⋅×=⋅=

r

rrr

rrrrrrrrrrr

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4.3 Vector Representation of a Moment

( ) ( ) ( )N 9.1829.3189.338

8.458800850

8.458800850500222

kji

kjiF

rrr

rrrr

+−−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−+−

+−−=

kir OB

rrr 75.085.0 +−=

mN 1.2717.982.239

9.1829.3189.33875.0085.0

⋅+−=

−−−=×=

kji

kjiFrM OBO

rrr

rrr

rrr

mN 2.2399.1829.3189.338

75.0085.0000.1

mN 2.239

⋅−=−−

−−

=⋅×=

⋅−=⋅=

OCOB

OCOOC

eFr

eMM

rrr

rrieOC

rr 0.1−=

Determine moment of F about line OC

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Chapter 4 Rigid Bodies Equivalent Force/Moment …cac542/L4.pdfChapter 4 Rigid Bodies Equivalent Force/Moment Systems. 2 ... Equilibrium for non-concurrent force ... the individual