Chapter 3 3 Linkages with Rolling and Sliding Contacts and Joints on moving Sliders ... P4/P3 |= 44.36 in/s2, //Q to O B (see next page) n Q P n O Q n P 4 /P 3 P 4 /O B

機構學

國立高雄應用科大機械系, C. F. Chang 1

KUAS ME, C. F. Chang 1

Chapter 3Chapter 3

Linkages with Rolling and SlidingContacts and Joints on moving Sliders


Velocity Relationship Between Coincident PointsVelocity Relationship Between Coincident Points(ref p. 105, Eq. 3.25)(ref p. 105, Eq. 3.25)

If points A and B are coincident but belong todifferent bodies, we have =0 and thus

J

I

Y

X

ro

i

yx

j

B

ρ

rp

ρr

A

rAB ρρωvv P相對於動座標系之速度

動座標系平移所造成之速度

動座標系旋轉所造成之速度,

ABArAB /vvρvv J

I

Y

X

B on link b

rA

ρr A fixed on link a

link a

vB/A must be tangent to the path that point Btraces on link a

General velocity relationship between any twopoints A and B :

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Acceleration Relationship Between Coincident PointsAcceleration Relationship Between Coincident Points(ref p. 105, Eq. 3.26)(ref p. 105, Eq. 3.26)

General velocity relationship between any twopoints A and B :

P在動座標系上運動之相對加速度

動座標系旋轉旋轉所造成之切線加速度

rrAB ρρωρωρωωaa 2)(

動座標系旋轉旋轉所造成之科氐速度

動座標系平移平移所造成之加速度

動座標系旋轉旋轉所造成之法向加速度

J

I

Y

X

B on link b

rA

ρr A fixed on link a

link ac

ABABArrAB //2 aaaρωρaa

If points A and B are coincident but belong todifferent bodies, we have =0 and thus


Example 3.1 Velocity and Acceleration Analysis of aExample 3.1 Velocity and Acceleration Analysis of aQuickQuick--Return Mechanism (pp. 107Return Mechanism (pp. 107--111)111)

Given:– Link 2 is driven with a constant angularconstant angular

velocityvelocity of 10 rpm CCW Find:

– The sliding velocity of the slider,– The angular accelerations of links 3 and 4– The acceleration of slider for the quick-

return mechanism Analysis of the problem :

2323 / AAAA vvv

3::3

AOBO BBAB vv

BCBC /vvv

A2, A3

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3232 / AAAA vvv

Velocity AnalysisVelocity Analysisstep 1: Loop Ostep 1: Loop OAA--AA22--AA33--OOBB

3::3

AOBO BBAB vv

A2, A3

vA2

vA2/A3

vA3

A2 is fixed on link 2A3 is fixed on link 3 but coincident with A2

vB

vA2/A3=點A2相對於桿3之相對速度=?

vA2/A3 // 滑軌OBB

|vA3|=(OBA3)3=?vA3 OBA3

|vA2|=(OAA2)2=10(10 2/60)=10.472 in/svA2 OAA2

2323 / AAAA vvv


Velocity AnalysisVelocity Analysisstep 2: Loop Ostep 2: Loop OBB--BB--CC

BCBC /vvv

vB

vC/B

vC

vA2

vA2/A3

vA3

vB

vC/B

vC

|vC/B|=(BC)4= ?VC/B BC

|vC| = ?VC //水平滑軌

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Velocity AnalysisVelocity Analysisstep 3: Calculate the Resultstep 3: Calculate the Result

vB

vC/B

vC

BCBC /vvv

|vC/B|=(BC)4= ?vC/B BC

|vC|= ?vC // fixed slot

Measuring the lengths on the velocity polygongives|vA3|=(oa3)Sv=9.07 in/s|vA3/A2|=(a2a3)Sv=5.09 in/s|vC/B| =(b3c4)Sv=2.823 in/s|vC| =(oc4)Sv=11.06 in/sANS

2323 / AAAA vvv

|vA2|=(OAA2)2=10(10 2/60)=10.472 in/svA2 OAA2

|vA3/A2|=?vA3/A2 // OBB

|vA3|= (OBA3)3=?vA3 OBA3

3::3

AOBO BBAB vv

3=|vA3|/(OBA3) =0.344 rad/s, CCWANS3=|vA3|/(OBA3) =0.344 rad/s, CCWANS

4=|vC/B|/(BC) =0.608 rad/s, CCWANS4=|vC/B|/(BC) =0.608 rad/s, CCWANS


Acceleration AnalysisAcceleration Analysisstep 1: Loop Ostep 1: Loop OAA--AA22--AA33--OOBB

cAAAAAA 323232 // aaaa

32323322 /3/ 2 AAAAtA

nA

tA

nA vaaaaa

|anA3|= (OBA3)(3) 2 = (26.37)(0.344)2= 3.12in/s2

anA3//OBA3 and point to OB

|anA2|= (OAA2)(2) 2= 10.97in/s2

anA2//OAA2 and point to OA

|atA3|= (OBA3)3=?

atA3OBA3

|aA2/A3|= ?aA2/A3 //moving slot OBB

|acA2/A3|=23(vA2/A3)=2(5.09)(0.344)=3.5in/s2

acor =rotate vA2/A3 by 90according to 3

Then, DetermineThen, Determine aaBB =4.12=4.12 in/s2 by using acceleration image theoremby using acceleration image theorem

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Acceleration AnalysisAcceleration Analysisstep 2: Loop Ostep 2: Loop OBB--BB--CC (a(aBB isis knownknown))

|anC/B|= (BC)(4)2= (5)(0.608)2 =1.85in/s2, //BC

|atC/B|= (BC)(3)=? in/s2, BC

aC=? in/s2, // horizontal fixed slot

BCBC /aaa )( //t

BCn

BCBC aaaa

By measurement in figure, we getaC =0.7986 in/s2, to the rightANS3 =|at

C/B|/(BC)=0.071in/s2


Basic Relationships for Rolling ContactBasic Relationships for Rolling Contact

rolling)pure(24BB vv

B2, B4

nBO

nOO

nOBBB 44422242 //// aaaa

042 / BBv

|anB2/O2|= (B2O2)(2)2=|vB2/O2|2/(B2O2)

B2 to O2

|anO2/O4|= (O2O4)(x)2=|vO2/O4|2/ (O2O4)

O2 to O4

|anO4/B4|= (O4B4)(4)2 =|vO4/B4|2/ (O4B4)

O4 to B4

aB2/B4| // C-C (O2O4)

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Example 3.2 Analysis of Linage with aExample 3.2 Analysis of Linage with a Rolling Contact JointRolling Contact JointGivienGivien:: 22=10=10 radrad/s,/s, 22 = 0= 0Find: the angular acceleration,Find: the angular acceleration, 44 , of gear 4, of gear 4

Note: A, Q, and PNote: A, Q, and P33, are on the same link 3, are on the same link 3

ChoosingChoosing SSvv=2.5 to draw the velocity polygon yields=2.5 to draw the velocity polygon yields

APAPP /34 3vvvv

|vP3/A|=(AP3)3=?vP3/A AP3

|vA|= (OAA)2= (1 in)(10 rad/s) =10 in/svA OAA

|vP4|= (OBP4)4=?vP4 OBP4

|vP/A| =(ap)Sv=6.85 in/s|vP| =(op)Sv=38.32 in/s

3=|vP/A|/(AP3) =6.85/3.258=2.10 rad/s, CW4=|vP|/(OBP4) =38.32/1.125=7.39 rad/s, CCW

※※TriangleTriangle apqapq in velocity polygon can be determined by using velocity image tin velocity polygon can be determined by using velocity image theoremheoremvvQQ/OB/OB==vvQQ -- vvOBOB==vvQQ -- 0=6.540=6.54

3.258 in P3, P4


Example 3.2 : Acceleration AnalysisExample 3.2 : Acceleration Analysis

ChoosingChoosing SSaa=25 to drawing the acceleration polygon gives=25 to drawing the acceleration polygon gives

|anP/3A|= (AP)(3)2= (3.258)(2.1)2 =45.4 in/s2, //AP

|atP3/A|= (AP)(3)= (3.258) 3 =? in/s2, AP

|anA|= (OAA)(2)2= 1(10)2 =100 in/s2, //OAA

|anP4|= (OBP)(4)2= (1.125)(7.39)2 =61.4 in/s2, //OBP

|atP4|= (OBP)(4)= (4.0) 4 =? in/s2, OBP

|atP| =(op)Sa=(1.816)(50)=90.8 in/s2 =74.25 in/s2

4=|atP|/(OBP) =74.25/1.125=66.0 rad/s2ANS

3433434 /// )( ppApAPPpP aaaaaa

nPP

tAP

nAP

tA

nA

tP

nP 3433 /// )()()( aaaaaaa

3=|vP/A|/(AP) =2.10 rad/s, CW4= |vP|/(OBP) =7.39 rad/s, CCW

3.258 in2=const

|aP4/P3|= 44.36 in/s2, //Q to OB (see next page)

nPQ

nQO

nOPPP BB 3434 //// aaaa

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Remark onRemark on aP4/P3

|aP4/P3|=61.531-21.386+4.213= 44.36 in/s2, //QOB|aP4/P3|=61.531-21.386+4.213= 44.36 in/s2, //QOB

nPQ

nQO

nOPPP BB 3434 //// aaaa

531.61125.132.8

438.6139.7125.1

2

4

2

//

2244/

4

4

4

PO

v

PO

B

OPnOP

Bn

OP

B

B

B

a

a

386.21254.6 2

2

/

2

//

B

OQ

B

QOnQO QO

v

QO

vBB

Ba

213.4875.092.1

859.31.2875.0

2

3

2

//

2233/

3

3

3

QP

v

QP

PQnPQ

nPQ

a

a

計算加速度時, 由於製圖之誤差以及計算方法的不同, 可能出現不同之結果, 例如:


Another Method: Using a Virtual LinkageAnother Method: Using a Virtual Linkage

)()()( //t

AQn

AQnA

tQ

nQ aaaaa

Determine aQ by solving

|atQ| =(OBQ) 4

4=|atQ|/(OBQ)=66 rad/s2ANS

A

OA

Q

OB

2

3

4

aQ

aA

a’

o’

q’

aQ/A

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Comparing the ResultsComparing the Results

aQ

aA

a’

o’

q’

aQ/A

aQ

aA

a’

o’

q’

aQ/A

aa’’pp33’’qq’’is similar to that of APQ;is similar to that of APQ;

that is, satisfying the acceleration imagethat is, satisfying the acceleration imagetheoremtheorem


Example 3.3 Analysis of a Geared LinkageExample 3.3 Analysis of a Geared Linkage——RollingRollingContact (pp. 119Contact (pp. 119--120)120)

Given:– Cam 2 is rotated CCW with constant angular velocityconstant angular velocity of 1000 rpm– OAOB=4.0, OBB=4.25, r2=0.5, r3=2.5, OAA=1.153, AB=0.901 (unit: in)

Find:– The angular velocity and– The angular acceleration of the arm (link4)

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Equivalent LinkagesEquivalent Linkages

When motion analysis is to be made for a direct-contactmechanism, the problem can be simplified by replacing themechanism by an equivalent four-bar linkage.

An equivalent four-bar linkage is one whose driving link anddriven link have angular velocities and accelerations which areidentical at the instant to those of the original linkage.

C2

O2 O4

C4

4’2’

3


Examples for Determining Equivalent Linkages (cont.)Examples for Determining Equivalent Linkages (cont.)

C2

O2O4

C4

4’2’

3

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Examples for Determining Equivalent LinkagesExamples for Determining Equivalent Linkages

2‘4‘

Path of C2 on link 4

2‘4‘

Path of C2 on link 4

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Motion Analysis Using the concept of Equivalent LinkageMotion Analysis Using the concept of Equivalent LinkageEx 3.4 & Ex 3.5 (pp. 122Ex 3.4 & Ex 3.5 (pp. 122--128)128)

Find the velocity and acceleration of the cam follower (link 3) ifthe cam is rotating at a constant angular velocity of 100 rad/sCCW

Find the sliding velocity at the point of contact, C.

Path of B on 3


Ex 3.4 & Ex 3.5 Velocity AnalysisEx 3.4 & Ex 3.5 Velocity Analysis

Define:Define:– B2 on link 2– B3 on link 3– B2 and B3 are coincident with joint B

3232 / BBBB vvv

vB2/B3=點B2相對於桿3之相對速度=?vB2/B3 // moving slide

|vB3|=?vB3 //fixed slide

|vB2|=(AB2)2=0.5100=50 in/svB2 AB2

AnsAns::–vB3=0.7x50=35 in/s = the velocity of

link 3

0.7 in

vB2/B3

v3 ,vB3

B2, B3

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Ex 3.4 & Ex 3.5 Acceleration AnalysisEx 3.4 & Ex 3.5 Acceleration Analysis

ChoosingChoosing SSaa=5000 to drawing the acceleration polygon gives=5000 to drawing the acceleration polygon gives

|atB3|= ? in/s2, // fixed slide

|anB2|= (AB2)(2)2= 0.5(100)2 =5000 in/s2, //AB2

|atB2/B3|=? in/s2, // moving slide

|atB3| =(o’b’3)Sa=3250 in/s2ANS

cBB

tBB

nBB

tB

nB

tB

nB 3232323322 /// )()()( aaaaaaa

cBBBBBB 323232 // aaaa

3=0

B2, B3


Ex 3.4 & Ex 3.5 Sliding Velocity at Contact Point CEx 3.4 & Ex 3.5 Sliding Velocity at Contact Point C

vC3

vC3/C2

vC3 ,vB3

C2, C3

2323 / CCCC vvv

|vC3/C2|=|vC2/C3|=?vC3/C2=-vC2/C3 //the common tangent at C

|vC2|= (AC2)2vC2 AC2

vC3=vB3 is known (Link 3 translates)

|vC3/C2 |=148 in/s

vC2

vC3/C2

vC2

c

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Ex 3.7 (pp. 132Ex 3.7 (pp. 132--136)136)

Given: 2=2 rad/s CCW, 2=3 rad/s2 CCW Find:

–3, 3=3, vD=3, and aD

– The center of curvature of the path that B3 traces on link 2

3232 / BBBB vvv


323232 /3/ 2 BBBBBB vωaaa

B2 on link 2B3 on link 3


Ex 3.7 : Velocity AnalysisEx 3.7 : Velocity Analysis

3232 / BBBB vvv

|vB2/B3|=|vC2/C3|=?, tangent to the slot

|vB2|=(AB2)2=(5)(2)=10 m/s, AB2

From the velocity polygon, we get|vB3/B2 |=8.7 m/s|vD |=7 m/sANS

|vB3|= (CB3)3=? CB3

3=|vB3|/(CB3)=5/5=1 rad/sANS

vB2 ,AB2

vB3/B2 ,EB3

vB3 ,CB3

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anB3 //BC

anB2/B3+ac

B2/B3//EB

atB2/B3 EB

atB3CB

atB2 AB an

B2//AB

Ex 3.7 : Acceleration AnalysisEx 3.7 : Acceleration Analysis

|anB2/B3|=|vB2/B3|2/(EB)=8.72/3=25.23 m/s2, from B to E

|atB2/B3|=?, EB3

|anB2|=(AB2)(2)2=(5)22=20 m/s2, from B2 to A

|atB2|=(AB2)2=(5)(3)=15 m/s2, AB2

|anB3|=(CB3)(3)2=(5)(1)=5 m/s2?, from B3 to C

|atB3|=(CB3)3=?, CB3


|acB2/B3|=23vB2/B3=2(1)(8.7)=17.4 m/s2, from E to B


Ex 3.7 : Acceleration Analysis (cont)Ex 3.7 : Acceleration Analysis (cont)

anB3 //BC

//EB

atB2/B3 EB

atB3CB

atB2AB

anB2//AB

From the acceleration polygon, we get|at

B3 |=32.2 m/s2

|aD |=45.9 m/s2ANS

3=|atB3|/(CB3)=32.2/5=11.2 rad/sANS

b'3 can be determined fromaB3=an

B3+atB3

Then, d'3 is determined byusing the concept ofacceleration imagec’-b’-d’is similar to C-B-D


atB3

Documents

Chapter 3 3 Linkages with Rolling and Sliding Contacts and Joints on moving Sliders ... P4/P3 |= 44.36 in/s2, //Q to O B (see next page) n Q P n O Q n P 4 /P 3 P 4 /O B