MEM202 Engineering Mechanics - Statics 4.6 …cac542/L7.pdf4 MEM202 Engineering Mechanics - Statics...

1

MEM202 Engineering Mechanics - Statics MEM

4.6 Simplification of A Force System4-6.1 Coplanar (e.g., x-y plane) Force Systems

1Fr

1d 2d

3d

4d 2Fr

3Fr

4Fr

x

y

O

L1Fr

x

y

O1Fr

kdFMrr

111 =

1Fr

−

1d4Fr

x

y

O

4Fr

4Fr

−

4dkdFMrr

444 −=

∑∑=++=

=++=

iO

i

MMMM

FFFRrr

Lrr

rrL

rr

41

41x

y

O

Rr

OMRr

−

Rr

Rdx

y

O

OMRr

2



1Fr

1d 2d

3d

4d 2Fr

3Fr

4Fr

x

y

O

( ) kMkdFM

eRjRiRjFiFFR

OiiO

Ryxiyixirrr

rrrrrrr

==

=+=+==

∑∑∑∑

x

y

RrOM

Rr

Rd

Rx

Rr

−yRr

xRr

O

y

OR

OR

yxyxR

zyx

RMx

RMdR

jRR

iRRjieR

RRRRR

==

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛=+=

++=

or ,: of action of line theof Location

coscos: of Direction

: of Magnitude 222

r

rrrrrr

r

θθ

3



1Fr

1d 2d

3d

4d 2Fr

3Fr

4Fr

x

y

O

x

y

O

Rr

OMRr

−

Rr

Rd

x

y

O

OMCrr

=

A coplanar force system could be simplified to

mequilibriu in is system thei.e.,

0 and 0 if Zero ==== ∑ Oi MCFRrrrr

0

0

if couple A

≠=

== ∑O

i

MC

FRrr

rr

0

if force A

≠= ∑ iFRrr

4


4.6 Simplification of A Force SystemExamples of coplanar force systems

Problem 4-117 (p.168) Determine the resultant R of the three forces and the location of its line of action with respect to support A.

A B C D E

in14 in 16 in 10 in 02

lb 08 lb 57 lb 09

Resultant force and couple at A: Resultant force (couple = 0) at xR:

ARr

AMr R

r

Rx

lb 95907580 =+−=ARr

lb 95907580 =+−=ARr

lb-in 740,2409030751480

=×+×−×=AM

r

in 2695740,2

=== RR dx

5


4.6 Simplification of A Force SystemExamples of coplanar force systems

( ) ( ) 0541251312130220 =++−=xR

( ) ( ) 05312513513025 =−+=yR

0=Rr

Example Problem 4-21 (p.166)

( ) ( )( )( ) ( )( )( ) ( )( )( )lb-ft125

3125313021302220 53

135

1312

=

−+−=OM( ) ( )( ) ( ) ( )[ ]( ) ( ) ( )[ ]( )( ) ( )( ) ( )( )[ ]( )( )

lb-ft 125753

120250322021251253

13013023

2202

53

54

135

1312

=−+

−+=

−×+

+×++

−×=

jii

jiji

ijMO

rrr

rrrr

rr

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

( )( ) ( )( ) ( )( )( )lb-ft 125

213032522202125325

312521253130

1312

54

53

54

135

====

−−=

+−=

−+=

CBAO

C

B

A

MMMMMMM

6


4.6 Simplification of A Force System4-6.2 Non-coplanar Parallel (e.g., ⊥ x-y plane) Force Systems

( ) kRkFFFR i

rrrL

rr 51 ==++= ∑

jMiMjFxiFyFrM

jMiMjFxiFyFrM

yx

yx

rrrrrrrM

rrrrrrr

555555555

111111111

+=−=×=

+=−=×=

x

y

z1Fv

2Fv

3Fv

4Fv 5F

v

4y

4x4rr

jyixjrirr

jyixjrirr

yx

yx

rrrrrM

rrrrr

55555

11111

+=+=

+=+=

( )

MOyx

RRRRRO

eMjMiM

jRxiRyR

yxkji

RrM

rrr

rr

rrr

rrr

=+=

−==×= 00

0

( )? =+=+= jyixjrirr RRyx

rrrrr

x

y

z Rr

Ry

Rxrr

?? == rR rr

( )

jFxiFy

jMiMMM

iiii

iyixiFO irr

rrrrr

∑∑∑∑∑

−=

+==

kFkFF

kFkFF

z

z

rrrM

rrr

555

111

==

==

( ) ( )ROFO MMri

vvr= using determined is

7



x

y

z1Fv

2Fv

3Fv

4Fv 5F

v

4y

4x

x

y

z Rr

Ry

Rx4rr rr

?? == rR rr

( ) jyixrkRkFR RRi

rrrrrr+=== ∑

( ) jFxiFyjMiMM iiiiiyixFO i

rrrrrr ∑∑∑∑ −=+= ( ) jRxiRyjMiMM RRyxRO

rrrrr −=+=

( ) ( )

RFy

RMy

RFx

RM

x

RxFxMRyFyM

MM

iixR

iiyR

RiiyRiix

ROFO i

∑∑∑∑

===−=

−=−===

⇒= rrrr

8



x

y

z1Fv

2Fv

3Fv

4Fv 5F

v

4y

4x

x

y

z Rr

Ry

Rx

( ) jyixrkRkFR RRi

rrrrrr+=== ∑

4rr rr

RFy

RMy

RFx

RM

x iixR

iiyR

∑∑ ===−=

A non-coplanar parallel force system could be simplified tozero if R = 0 and MO = 0, i.e., the system is in equilibriuma couple C if R = 0 and MO ≠ 0a force R if R ≠ 0

9


4.6 Simplification of A Force SystemExamples of non-coplanar parallel force systems

x

y

z N 10N 15

N 20

m 14x

m2m2

m2

m 4m 3

( ) N 15201510 kkRrrr

−=−+−=

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

jMiM

ji

j

i

kji

kjikjiM

yx

O

rr

rr

r

r

rrr

rrrrrrr

+=

−−=

−++−−

−++−=

−×++

×++−×+=

m-N 1085

202156104

208157103

2082

15761034

m 67.51585m 667.0

1510

=−−

==−=−−

−=−=R

MyR

Mx x

Ry

R

( ) ( ) mN 6.851085 22 =−+−=OM

jijieM

rrrrr 117.0993.06.85

106.85

85−−=

−+

−=

10


4.6 Simplification of A Force SystemExamples of non-coplanar parallel force systems

( ) N 051510 kkRrrr

=−+−=

x

y

z N 10N 15

N5

m 14x

m2m2

m2

m 4m 3

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

jMiM

ji

j

i

kji

kjikjiM

yx

O

rr

rr

r

r

rrr

rrrrrrr

+=

−=

−++−−

−++−=

−×++

×++−×+=

m-N 4035

52156104

58157103

582

15761034

( ) ( ) mN 2.534035 22 =−+=OM

jijieM

rrrrr 753.0659.02.53

402.53

35−=

−+=

11


4.6 Simplification of A Force System4-6.3 General Force Systems

1Fr

Oy

x

z

2FrL

nFr

O yx

z

1Fr

1Cr

2Cr

nCr

2FrnF

r

nFFr

Lr

,,2

1Fr

1nr

O1rr

1Fr

1nr

O1rr

1Cr

O yx

z

Cr

Rr

( ) ( ) ( )R

zyx

eRkFjFiFR

r

rrrr

=

++= ∑∑∑

( ) ( ) ( )C

zyx

eCkCjCiCC

r

rrrr

=

++= ∑∑∑

0 if zero (iii) ,0,0 if couple A (ii) ,0 if force A (i) :outcomes possible Three ==≠=≠ CRCRRrrrrr

12



O yx

z

Cr

Rr

( )Ren r=

n′

n ′′

Plane containing R and C

n coincides with R

n´ on the plane and ⊥to n″ and n

n″ ⊥ to the plane

Rr

n

n′

n ′′ //Cr

⊥Cr

Cr

Rr

n

n′

n ′′

⊥Cr C

r

//Cr n

n′

( )//

//

//

CCC

eeCC

CCC

RRrrr

rrrr

rrr

−=

⋅=

+=

⊥

⊥

Rr n

n′

n ′′//Cr d

Rr

n

n′

n ′′

⊥Cr

RCd ⊥=Rr

Rr

−d

⇒⊥CRrr

and

Simplify

In the n - n´ - n" coordinate system

Resultant is a (positive) wrench

13



Determination of the location of the “Wrench”In the n-n′-n″ coordinate system, resultants R and C//, i.e., the wrench, are on the n-n″ plane and at a distance d = C⊥/R from the n–axis.

Rr

n

n′

n ′′ //Cr

⊥Cr

Rr n

n′

n ′′//Cr d

Cr

Rr

n

n′

n ′′x

y

z

O yx

z

//Cr R

r

n

rr

n′To find the location of R and C// in the x-y-zcoordinate system we draw a vector r from the origin to any point along the line of action of Rand C//. Vector r can be determined by using the relation r × R = C⊥ (why?)

14



Determination of the location of the “Wrench”For convenience, we usually use a vector on x-y, or y-z, or x-z plane.

O

yx

z

//Cr R

r

nrr

Ry Rx

x-y plane

O

yx

z

//Cr R

r

n

rr

Ry

Rz

y-z plane

//Cr

O

yx

z

Rr

n

rrRz Rx

x-z plane

( )kRyRx

jRxiRy

RRRyx

kjiRrC

xRyR

yRzR

zyx

RR

r

vr

rrr

rrr

−+

−=

=×=⊥ 0

( )kRyjRz

iRzRy

RRRzykji

RrC

xRxR

yRzR

zyx

RR

rv

r

rrr

rrr

−+

−=

=×=⊥ 0

( )kRx

jRxRziRz

RRRzxkji

RrC

yR

yRxRyR

zyx

RR

r

vr

rrr

rrr

+

−+−=

=×=⊥ 0

15


4.6 Simplification of A Force SystemExamples of General Force Systems

( )

kji

kjiRRe

R

kji

kjiR

R

rrr

rrrrr

rrr

rrrr

507.0 406.0761.062.98

50 4075

lb 62.98504075

lb 50 4075

3080 4075

222

++=

++==

=++=

++=

−++=(P4-139)

( ) ( ) ( )

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

kji

kjikjjiji

FrFrFrC COCBOBAOA

rrr

rrr

rrrrrrrvr

rrrrrrr

751880980183018

lb-in 350,1720 900

80189304018 75189

−+++−=

−+=

×+−+−×+×+=

×+×+×= ∑∑∑

16



( )( ) ( )

( ) ( )( )

( ) ( )

kji

kjiRRe

R

kji

k

jiR

kji

kjieFF

R

BBB

rrr

rrrrr

rrr

r

rrr

rrr

rrrrrr

775.0 629.0066.01.311

0.241 7.1956.20

N 1.3110.2417.1956.20

N 0.241 7.1956.20

1200.161200

257.120506.10080

N 0.1617.1206.100

865

865225

222

222

−+−=

−+−==

=−++−=

−+−=

+−−+

+++−=

−+−=

−++−

−+−==

( ) ( )( )

( ) ( )

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

ji

kji

kjji

kjij

kjiji

FrFrFrC COBBOBAOA

r

rr

rrr

rrrr

rrrr

rrrvr

rrrrrrr

255.26.100750505.2806

1205.22005.212061615.72006

m-N 7.330.80 8.168

1202565.2

0.1617.1206.1005.7

20050 0865.2

−++−+

+++−−=

++−=

+×+−+

−+−×+

−+×+=

×+×+×= ∑∑∑

(P4-140)

17



(P4-141)( ) ( )

kjkjRRe

R

kjkjR

R

rrrrr

r

rrrrr

771.0 637.02.149

lb 115 95

lb 2.14911595

lb 115 95 4075 3012522

+=+

==

=+=

+=++−=

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

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

ji

kji

kkjijji

kjkji

FrFrFrC DADCACBAB

r

rr

rrr

rrrrrrr

rrrvr

rrrrrrr

1256308

75640875164016950

lb-in 990130 570,1

758166125166

40309168

−−+

+−+++−=

−+=

×++−+×+−+

+−×−+=

×+×+×= ∑∑∑

18



( ) ( )

( )

kji

kjiRRe

R

kji

kjiR

R

rrr

rrrrr

rrr

rrrr

248.0 866.0433.0lb 9.403

lb 100 350175

lb 9.403100350175

lb 100 350175

125225 150200175

222

−+=

−+==

=−++=

−+=

+−+++=

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

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

j

i

kji

kjikji

jkjikki

FrFrFrC COCBOBAOA

r

r

r

rrr

rrrrrr

rrrrrrr

rrrrrrr

175221502020020

125201753222520

125221503220022

lb-in 150,3600,7 450,6

125150175322220

2002222202252220

−++

−++

+−−=

++−=

++×+++

×+++−×+=

×+×+×= ∑∑∑

(P4-147)

19


4.6 Simplification of A Force SystemExamples of General Force Systems – A Wrench

( )

kjkjRRe

R

kjkjR

R

rrrrr

r

rrrrr

952.0 307.015550

155 50

lb 8.16215550

lb 155 50 7580 50

22

22

+=+

+==

=+=

+=++=

( )( ) ( ) ( ) lb-in 450 90050 9 7512 kikiCrrrrr

+=+=

( ) ( ) lb-in 8.407 5.131 4.428 RR// kjeeeCC R

rrrrrrr+==⋅=

lb-in 2.42 5.131900// kjiCCCrrrrrr

+−=−=⊥

lb-in 2.425.131900

lb-in 50155155

1555000

kji

kxjxiy

yxkji

RrC

RRR

RR

rrr

rrr

rrr

rrr

+−=

+−=

=×=⊥

lb-in 2.425.131900

lb-in 5015550

1555000

kji

kxjxiz

zxkji

RrC

RRR

RR

rrr

rrr

rrr

rrr

+−=

+−−=

=×=⊥

NOTE: For this particular example, you cannot use an rvector on y-z plane to determine the location of resultants. Why?

in 815in 848.0 .yx RR == in 0.18in 848.0 −== RR zx

20


4.6 Simplification of A Force SystemExamples of General Force Systems – A Wrench

( )kjiRRe

R

kjiR

R

rrrrr

rrrr

939.0 332.0084.0

N 6.1994.1873.6675.16

N 4.187 3.6675.16222

++−==

=++−=

++−=

( )m-N 8.994.838.166

160118186

kji

FFrFrC OBOArrr

rrr

+−=

+×+×=

( ) m-N 84.48 27.17369.4 R// kjieeCC R

rrrrrrr++−=⋅=

m-N 96.50 7.1002.171// kjiCCCrrrrrr

+−=−=⊥

( )m-N 96.507.1002.171

m-N 75.163.664.1874.187

4.1873.6675.160

kji

kyxjxiy

yxkji

RrC

RRRR

RR

rrr

rrr

rrr

rrr

+−=

++−=

−=×=⊥

N 9.12495.99

N 5.621.100

N 4.1662.83

160

118

186

kiF

kjF

jiF

rr

rr

rr

+−=

+−=

+=

m 914.0m 537.0 == RR yx

21


ReviewParallelograms and Laws of Sine and Cosine

2112 :1 Step FFRrrr

+=

lb 954cos2 121

22

2112

=

++= φFFFFR

o0.27sinsin12

1111 =⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

RF φβ

ooo 6030301 =+=φ

312 :2 Step FRRrrr

+=

lb 386,1cos2 23122

3212 =++= φFRFRR

o3.39sinsin 21212 =⎟

⎠⎞

⎜⎝⎛= −

RR φβ

ooo 6740272 =+=φ

ooo 3.59203.39 =+=θ

Determine the resultant of F1, F2, and F3forces

22


ReviewResultants by Rectangular Components

kip 628.22284.28972.16

kip 299.16955.20971.13

kip 348.13534.9442.11

333

222

111

kjieFF

kjieFF

kjieFF

rrrrr

rrrrr

rrrrr

++==

+−−==

+−==

( ) ( ) ( )kip 275.52205.2443.14 kji

kFjFiF

kRjRiRR

iziyix

zyx

rrr

rrr

rrrr

+−=

++=

++=

∑∑∑

kip 3.54222 =++= zyx RRRR

R

ooo 6.15cos 3.92cos 6.74cos 111 ====== −−−

RR

RR

RR z

zy

yx

x θθθ

23


ReviewSimplification of A General Force System

( ) ( )

( )

kji

kjiRRe

R

kji

kjiR

R

rrr

rrrrr

rrr

rrrr

248.0 866.0433.0lb 9.403

lb 100 350175

lb 9.403100350175

lb 100 350175

125225 150200175

222

−+=

−+==

=−++=

−+=

+−+++=

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

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

j

i

kji

kjikji

jkjikki

FrFrFrC COCBOBAOA

r

r

r

rrr

rrrrrr

rrrrrrr

rrrrrrr

175221502020020

125201753222520

125221503220022

lb-in 150,3600,7 450,6

125150175322220

2002222202252220

−++

−++

+−−=

++−=

++×+++

×+++−×+=

×+×+×= ∑∑∑

(P4-147)

24


Review - Moment about an axis

( ) ( ) ( )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

25


ReviewSimplification of A General Force Systems - Wrench

( )kjiRRe

R

kjiR

R

rrrrr

rrrr

939.0 332.0084.0

N 6.1994.1873.6675.16

N 4.187 3.6675.16222

++−==

=++−=

++−=

( )m-N 8.994.838.166

160118186

kji

FFrFrC OBOArrr

rrr

+−=

+×+×=

( ) m-N 84.48 27.17369.4 R// kjieeCC R

rrrrrrr++−=⋅=

m-N 96.50 7.1002.171// kjiCCCrrrrrr

+−=−=⊥

( )m-N 96.507.1002.171

m-N 75.163.664.1874.187

4.1873.6675.160

kji

kyxjxiy

yxkji

RrC

RRRR

RR

rrr

rrr

rrr

rrr

+−=

++−=

−=×=⊥

N 9.12495.99

N 5.621.100

N 4.1662.83

160

118

186

kiF

kjF

jiF

rr

rr

rr

+−=

+−=

+=

m 914.0m 537.0 == RR yx

MEM202 Engineering Mechanics - Statics 4.6 …cac542/L7.pdf4 MEM202 Engineering Mechanics - Statics...

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Transcript of MEM202 Engineering Mechanics - Statics 4.6 …cac542/L7.pdf4 MEM202 Engineering Mechanics - Statics...