MACHINES AND MECHANISMS APPLIED KINEMATIC ANALYSIS Fourth Edition David H. Myszka Solution Manual
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About This Presentation
MACHINES AND MECHANISMS
APPLIED KINEMATIC ANALYSIS
Fourth Edition
David H. Myszka
Solution Manual
Slide Content
r’s Solutions Manual
10 accompany
and Mechanism
PEARSON
Prentic
i
10 accompany
and Mechanism
PEARSON
Prentic
i
This work is protected by United States copyright laws and is provided
solely for the use of instructors in teaching their courses and assessing
student learning, Dissemination or sale of any part of this work (includ-
ing on the World Wide Web) will destroy the integrity of the work and
is not permitted. The work and materials from it should never be
made available to students except by instructors using the accom-
panying text in their classes. All recipients of this work are expected to
abide by these restrictions and to honor the intended pedagogical pur-
poses and the needs of other instructors who rely on these materials.
PEARSON
Prentice
Hall
solely for the use of instructors in teaching their courses and assessing
student learning, Dissemination or sale of any part of this work (includ-
ing on the World Wide Web) will destroy the integrity of the work and
is not permitted. The work and materials from it should never be
made available to students except by instructors using the accom-
panying text in their classes. All recipients of this work are expected to
abide by these restrictions and to honor the intended pedagogical pur-
poses and the needs of other instructors who rely on these materials.
PEARSON
Prentice
Hall
Solution Manual to Accompany:
Machines and Mechanisms
Applied Kinematic Analysis
4" Edition
David H. Myszka, Ph.D. P.E.
University of Dayton
300 College Park
Dayton OH, 45469
[email protected]
Machines and Mechanisms
Applied Kinematic Analysis
4" Edition
David H. Myszka, Ph.D. P.E.
University of Dayton
300 College Park
Dayton OH, 45469
[email protected]
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Carrer Crvele frame, Sued)
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DE
1-29 | Lmes-a dons 24
CERTES
OF: 20h yw
2b Ls <4 JomiseH
a y
L'ETAT à
121] Links dome 4
" an
por 304-1) U Ly
1-28 | Links = awe
Links =4 Lois Und
DE > 34-204 +17,
BE] Luca ents +
LS pend
Wore Blade yy
AL | Links =4 rats = À RT rte = a
EN N ES de tre ET)
DF BAND ly WF IN 20e
1S E] ET Oates o |
a“ Gens) x ES
ME > ey be 2d 21
1-35] Links = 5 dons = 5 db Links > A oats
Co em Teens, Van
due Bd -219= 27
wr AD -2-y
VST) E
pms, 2 shaia)
de=3(10 2(9=24
hE land
Dors ah ade y
EA] et re
(spss, Lau
DOF UN whey
EU
Uns E
outs 7
AS
WE 3-02 Vp
A
Links = {0 Spine 8
A
rise Eos 7
(0 pos, 3 shading)
wre 30d 240 > Y
(lepine, Verd)!
Dor: LAA Y
ae
ihe
Le
DE
1-29 | Lmes-a dons 24
CERTES
OF: 20h yw
2b Ls <4 JomiseH
a y
L'ETAT à
121] Links dome 4
" an
por 304-1) U Ly
1-28 | Links = awe
Links =4 Lois Und
DE > 34-204 +17,
BE] Luca ents +
LS pend
Wore Blade yy
AL | Links =4 rats = À RT rte = a
EN N ES de tre ET)
DF BAND ly WF IN 20e
1S E] ET Oates o |
a“ Gens) x ES
ME > ey be 2d 21
1-35] Links = 5 dons = 5 db Links > A oats
Co em Teens, Van
due Bd -219= 27
wr AD -2-y
VST) E
pms, 2 shaia)
de=3(10 2(9=24
hE land
Dors ah ade y
EA] et re
(spss, Lau
DOF UN whey
EU
Uns E
outs 7
AS
WE 3-02 Vp
A
Links = {0 Spine 8
A
rise Eos 7
(0 pos, 3 shading)
wre 30d 240 > Y
(lepine, Verd)!
Dor: LAA Y
dor + MAA 282 :
As Jon 12 TECH] tanks! Joel =1
Links 10, al Fa Lanks=4 us
Bor = 310) - 203 = y DoF 36D 20027
EST Let dot ET Loue ER
Links lay 5 darse
Dore BUN 20 yy Dore HUN ade y
[ATT Unks=8 20 [EB] LB Some To |
Aigher rt ey eres DoE ALBAN ZONE 1 |
BOF S@H-20-1 = Y . |
EN vil se 50 anks= os = 7
A Links rt DU 2 Mates ne bad)
ME 3-2)
Arge rocher
LS) s:15 19, p=,94 [ies2]s:4) 1212, p=12,9:5
S= 518. b-side
SAL dor Aigues
5.518 (je Wwe 11 Ged
=> crane rocker spcrank- rocker
153) 9231212, pe 8; qed MATS 3) =, pe 12, 425)
ss side. » Seside |
Berm 2 B+4 13+12< 1245
is S12 (nd 15417 ed
> crank rocker
As Jon 12 TECH] tanks! Joel =1
Links 10, al Fa Lanks=4 us
Bor = 310) - 203 = y DoF 36D 20027
EST Let dot ET Loue ER
Links lay 5 darse
Dore BUN 20 yy Dore HUN ade y
[ATT Unks=8 20 [EB] LB Some To |
Aigher rt ey eres DoE ALBAN ZONE 1 |
BOF S@H-20-1 = Y . |
EN vil se 50 anks= os = 7
A Links rt DU 2 Mates ne bad)
ME 3-2)
Arge rocher
LS) s:15 19, p=,94 [ies2]s:4) 1212, p=12,9:5
S= 518. b-side
SAL dor Aigues
5.518 (je Wwe 11 Ged
=> crane rocker spcrank- rocker
153) 9231212, pe 8; qed MATS 3) =, pe 12, 425)
ss side. » Seside |
Berm 2 B+4 13+12< 1245
is S12 (nd 15417 ed
> crank rocker
a Answers to the Chapter { Case Study Questions:
Case 11.
As link A rotates clockwise, 90, ide will move tothe lef
As link A rotates clockwise, 90%, the ball rapped in slide C wl drop down the lower, left
chute.
As link A continues ancther 90 clockwise link A wil be oriented st
will return tothe original poston shown inthe figure.
This device allons one feeder bowl to distribute balls to two separate tations.
‘The chamfers on slide C allow reli as a ball drops into the empty so as side C moves
under chute D.
This device would be useful in that only one Feder bowl will cad to be filled and
monitored.
Case12,
As handle A is rotated counterclockwise, flapper C rotates clockwise
‘As flapper € is raised, he water in thank is allowed to flow through the opening.
‘A buoyancy force will offset the water pressure, keeping the fapper in the upper poston
‘As the water leve lowers 1 the level ofthe Mepper, the Dapper vil lower wit the water
lie,
Item D floats on top ofthe water. As the water level lowers, float D also lowers.
As item D rotates counterclockwise, item F is also rotates counterclockwise.
Item Fis valve that controls water flow, nthe upper position, it file the tank. nthe
lower position, water flow seu off
‘These mechanisms allow a rapid flow of water frm the tank, slowly refill the tank, then
shutof? the water low.
“The water pressure in reidemial areas docs produce the required flow rate for a water
closet.
Case 13.
Inthe shown configuration, the wate level in bucket Bs rising, a the flow into the
bucket exceeds the amount that Is leaking fom the holes atthe boom ofthe bucket.
Inthe shown configuration, the water level in bucket Bs lowering, asthe water is leaking
Srom the oles atthe bottom ofthe bucket
If bucket B were forced upward, rocker arm C would rate clockwise
bucket B, were forced upward, ocker arm R would rotate counterclockwise.
Rocker arm R controls a diretional valve, channeling th water flow o either the upper
pipe or the lower pipe.
‘AS water drains fom one bucket, making it lighter, and fills the che, making it heavier,
the weight shift cases the rocker C to rotate and reverse the direction of the water flow.
‘The process repeats self and rotates rocker C back to the orignal position The continual
‘motion is escilation of rocker C.
‘As rocker C rotates, chanel $ moves between let and right postions, This allows the
ste! rod, which is constantly moving, 10 be ele ont a rel placed on the ef side or a
reel on placed om the right side
Since water is abundant and a common coling medium in most foundries, water flow can
be used to déve some machinery.
Case 11.
As link A rotates clockwise, 90, ide will move tothe lef
As link A rotates clockwise, 90%, the ball rapped in slide C wl drop down the lower, left
chute.
As link A continues ancther 90 clockwise link A wil be oriented st
will return tothe original poston shown inthe figure.
This device allons one feeder bowl to distribute balls to two separate tations.
‘The chamfers on slide C allow reli as a ball drops into the empty so as side C moves
under chute D.
This device would be useful in that only one Feder bowl will cad to be filled and
monitored.
Case12,
As handle A is rotated counterclockwise, flapper C rotates clockwise
‘As flapper € is raised, he water in thank is allowed to flow through the opening.
‘A buoyancy force will offset the water pressure, keeping the fapper in the upper poston
‘As the water leve lowers 1 the level ofthe Mepper, the Dapper vil lower wit the water
lie,
Item D floats on top ofthe water. As the water level lowers, float D also lowers.
As item D rotates counterclockwise, item F is also rotates counterclockwise.
Item Fis valve that controls water flow, nthe upper position, it file the tank. nthe
lower position, water flow seu off
‘These mechanisms allow a rapid flow of water frm the tank, slowly refill the tank, then
shutof? the water low.
“The water pressure in reidemial areas docs produce the required flow rate for a water
closet.
Case 13.
Inthe shown configuration, the wate level in bucket Bs rising, a the flow into the
bucket exceeds the amount that Is leaking fom the holes atthe boom ofthe bucket.
Inthe shown configuration, the water level in bucket Bs lowering, asthe water is leaking
Srom the oles atthe bottom ofthe bucket
If bucket B were forced upward, rocker arm C would rate clockwise
bucket B, were forced upward, ocker arm R would rotate counterclockwise.
Rocker arm R controls a diretional valve, channeling th water flow o either the upper
pipe or the lower pipe.
‘AS water drains fom one bucket, making it lighter, and fills the che, making it heavier,
the weight shift cases the rocker C to rotate and reverse the direction of the water flow.
‘The process repeats self and rotates rocker C back to the orignal position The continual
‘motion is escilation of rocker C.
‘As rocker C rotates, chanel $ moves between let and right postions, This allows the
ste! rod, which is constantly moving, 10 be ele ont a rel placed on the ef side or a
reel on placed om the right side
Since water is abundant and a common coling medium in most foundries, water flow can
be used to déve some machinery.
Answers to the Chapter 2 Case Study Questions:
Case 2-1.
L
‘As handle A is rotated, moving threaded rod B to the left, grp C also moves to the
Jet and slightly upward, Notice that inks E and F are pivoting in the middle, thus
‘pC is constrained to a swinging motion.
‘As handle A is rotated, moving threaded rod B to the lef, grip D moves tothe
Fight and slightly downward. Since links E and F are pivoting in the middle, grip D
il have motion opposing grip C
‘The purpose ofthis mechanism sto serve as a machining clamp for the workpiece.
‘The spring, G, pulling on link D would cause isto return to an upward and
rightward positon.
‘The purpose of spring G is, ultimately, o keep a positive contact between the
threaded rod and ink C.
Links E and F have a peculiar configuration to avoid interference with the
workpiece, throughout the range of motion ofthe clamp.
Such a device could be called a machining clamp.
Since link Cis moving in a swinging motion, the rounded end on the threaded rod,
‘assures a consistent point contact wit ink €.
Case 2-1.
L
‘As handle A is rotated, moving threaded rod B to the left, grp C also moves to the
Jet and slightly upward, Notice that inks E and F are pivoting in the middle, thus
‘pC is constrained to a swinging motion.
‘As handle A is rotated, moving threaded rod B to the lef, grip D moves tothe
Fight and slightly downward. Since links E and F are pivoting in the middle, grip D
il have motion opposing grip C
‘The purpose ofthis mechanism sto serve as a machining clamp for the workpiece.
‘The spring, G, pulling on link D would cause isto return to an upward and
rightward positon.
‘The purpose of spring G is, ultimately, o keep a positive contact between the
threaded rod and ink C.
Links E and F have a peculiar configuration to avoid interference with the
workpiece, throughout the range of motion ofthe clamp.
Such a device could be called a machining clamp.
Since link Cis moving in a swinging motion, the rounded end on the threaded rod,
‘assures a consistent point contact wit ink €.
Qupea |
an > o
ass da + 8 287 FE NES = 113 my
+
ES ui Eo x= an [EA] 8: 160-019: 307
CSD = E> Re ny
à tod: Go 8-97 AT, ae
RE ag eae rag
Do: 12ND > WH 25: UT Peels, 7
We pe BA ÿ
© tone 37 6 ;
ROE > 44 ay
ET
tam 35°F Hb X= y
MA ps: 4h ef
S+ 10 hypdanıse = 20:
dir 392 ¿> x= NS ag
cos 38" + Yao D Y= o
U ine 48x 337,
" 7
30°
A low ot cosines
80
4 CSN pS
2
_ s= VS my
ET wa Eur e
1.40 op al my“ Yn
LA ES
ES -8% 8
= A s A
eles IE ‘Ne ETES
anf "Hoh 22207 AN EE
El KR op Zah NY
Te
Em LA, some |
EN Rz De
'
sn" = Y,
u eye Wi
<i} Ze Pay
ton O= os 8 3812 -
TOS
cop EGY Ela,
Ya30 430, Cr RSA SC
an > o
ass da + 8 287 FE NES = 113 my
+
ES ui Eo x= an [EA] 8: 160-019: 307
CSD = E> Re ny
à tod: Go 8-97 AT, ae
RE ag eae rag
Do: 12ND > WH 25: UT Peels, 7
We pe BA ÿ
© tone 37 6 ;
ROE > 44 ay
ET
tam 35°F Hb X= y
MA ps: 4h ef
S+ 10 hypdanıse = 20:
dir 392 ¿> x= NS ag
cos 38" + Yao D Y= o
U ine 48x 337,
" 7
30°
A low ot cosines
80
4 CSN pS
2
_ s= VS my
ET wa Eur e
1.40 op al my“ Yn
LA ES
ES -8% 8
= A s A
eles IE ‘Ne ETES
anf "Hoh 22207 AN EE
El KR op Zah NY
Te
Em LA, some |
EN Rz De
'
sn" = Y,
u eye Wi
<i} Ze Pay
ton O= os 8 3812 -
TOS
cop EGY Ela,
Ya30 430, Cr RSA SC
p
E
©
ae nass
CRETE
I CS = 3BAin
WOE NOY
a TOR
af 2 sar
rire Ve 40 aviso)
= bb SC: 322,
3-20
reasons? 718 Algae
R004 CNY
Fa ai
B A
3 a
A
R- 43352869" R=20.13 37
[3723
A
R
R-22120 547
R
o PS
FT
Re [OF Sens = 24 jt
Best (BR st): 18.079
R218 Lis.o1
E
©
ae nass
CRETE
I CS = 3BAin
WOE NOY
a TOR
af 2 sar
rire Ve 40 aviso)
= bb SC: 322,
3-20
reasons? 718 Algae
R004 CNY
Fa ai
B A
3 a
A
R- 43352869" R=20.13 37
[3723
A
R
R-22120 547
R
o PS
FT
Re [OF Sens = 24 jt
Best (BR st): 18.079
R218 Lis.o1
>)
aaa]
1
À
PK
Re [aS I BADGE = 4335
Ba sir es] = 41.9
rhs: Bua Érugras
Rió
À ,
3
130
ns snl
245°
D
we Gus
Ran 37
ER 12.5 co04S su #15 corto
=-040
RS SAS + 1559010 155940 > 9.84
Rl Mosa as
mi 4
Ry AD -LO costo — Frage unse
RJ Eso. sus"
E ET eya
= 020
Ry = Me 4056 10= SL, |
Re as. Su
110.14 We
Bran Win og: LIT | 1
Re No. GTA
B-3 A
K=BPA
B
K= 8.074 ¿es 2uo
T:5587 Age
aaa]
1
À
PK
Re [aS I BADGE = 4335
Ba sir es] = 41.9
rhs: Bua Érugras
Rió
À ,
3
130
ns snl
245°
D
we Gus
Ran 37
ER 12.5 co04S su #15 corto
=-040
RS SAS + 1559010 155940 > 9.84
Rl Mosa as
mi 4
Ry AD -LO costo — Frage unse
RJ Eso. sus"
E ET eya
= 020
Ry = Me 4056 10= SL, |
Re as. Su
110.14 We
Bran Win og: LIT | 1
Re No. GTA
B-3 A
K=BPA
B
K= 8.074 ¿es 2uo
T:5587 Age
3-35
K & A
-B>
Measuring :
K=5581 RAG
336
J=A>Y
measuring
EAS ej
EEN
8 A
Measuring?
KA 65"
338) 5 EA
= 8.04
6 fat |.
E Bee > st
130-119 = 68,100
Te 800 T7
3398 x. [asas
= 8.04
9: coo (aot “eit a)
um”
BEYD- = 68.26
K=8.014 Bu
3-30 1ap-(45#39)= 105
RER OE
Eo) at
= 387 As
O
5-82 ne
K:5581 Ty
RENTE
Sal see
4
EAT
+8 > 5,581
A Core
Faser"
FA
ese (sn)
Peio- (ruse
TAN
DETAU
K & A
-B>
Measuring :
K=5581 RAG
336
J=A>Y
measuring
EAS ej
EEN
8 A
Measuring?
KA 65"
338) 5 EA
= 8.04
6 fat |.
E Bee > st
130-119 = 68,100
Te 800 T7
3398 x. [asas
= 8.04
9: coo (aot “eit a)
um”
BEYD- = 68.26
K=8.014 Bu
3-30 1ap-(45#39)= 105
RER OE
Eo) at
= 387 As
O
5-82 ne
K:5581 Ty
RENTE
Sal see
4
EAT
+8 > 5,581
A Core
Faser"
FA
ese (sn)
Peio- (ruse
TAN
DETAU
2
€
EGO
710+20=90
PR IST
150, =212.03
de Bes Gon
150
ge ms (91d eS"
Ranas UN
[44] 7-C+A=>B
A
B
J
Me i
Cou {As
| ST Boke
A
ce K
| Mason: 204.084 LON,
3-46] 32(+A>B+D
A
Meagyrin
T= 109 Atle AN
GA Ke Be DrA>C
asimag:
Kd. 924 SET
D B
Measur
de NE 5-C+*A+B+D>E
C
A
ay 19 18.287
SAV] EBS De AS CHE
Measuring *
Kz 107.869 (ade
38] RADAR Coe
AmB CHE
€
EGO
710+20=90
PR IST
150, =212.03
de Bes Gon
150
ge ms (91d eS"
Ranas UN
[44] 7-C+A=>B
A
B
J
Me i
Cou {As
| ST Boke
A
ce K
| Mason: 204.084 LON,
3-46] 32(+A>B+D
A
Meagyrin
T= 109 Atle AN
GA Ke Be DrA>C
asimag:
Kd. 924 SET
D B
Measur
de NE 5-C+*A+B+D>E
C
A
ay 19 18.287
SAV] EBS De AS CHE
Measuring *
Kz 107.869 (ade
38] RADAR Coe
AmB CHE
IS] A> cHe=DeR
AvE=Becad
[3-52] 1=C+A>B
NE ON (Ste) (Sc)
NE Con sd (iS sin)
goa ER
FE 260
Betas! (Sis ps 11:40
Ts 2.10 AS
33] K=B>A=>C
8
Ge (SSD) (1258049) CTSsinbd)
= Sle
ik Ky
Hero) (12 Scost)- (iSeastd)
A
a, ES
ù = 26.0
3-S4|T=CeASBSD
Tx SONS SAS ceed)
Cap coed) = -100,
I (won de D- {essia sd)
AD sn 10) = 44.8
ERP sms: y
ty
Ky = assiad)- (405,10) + (O)
A | Ban
Ke2bto navy Tea ZU
3-55] k= BOD PASC 3-50] 35 Co AB DIE
>
= (ids = 44.22
TESTA
Beta i(as So: NN
" Keaggı NET
Keen Sd «0s 30)
- (209090) + - 4045
[Sy = 0 + (Ad - Cube} (SosH30)
~ (20050) = DiR
DI TRAM
MY =101.08
Bota (ASYasS) = 18.3
J= 10.48 TRY
AvE=Becad
[3-52] 1=C+A>B
NE ON (Ste) (Sc)
NE Con sd (iS sin)
goa ER
FE 260
Betas! (Sis ps 11:40
Ts 2.10 AS
33] K=B>A=>C
8
Ge (SSD) (1258049) CTSsinbd)
= Sle
ik Ky
Hero) (12 Scost)- (iSeastd)
A
a, ES
ù = 26.0
3-S4|T=CeASBSD
Tx SONS SAS ceed)
Cap coed) = -100,
I (won de D- {essia sd)
AD sn 10) = 44.8
ERP sms: y
ty
Ky = assiad)- (405,10) + (O)
A | Ban
Ke2bto navy Tea ZU
3-55] k= BOD PASC 3-50] 35 Co AB DIE
>
= (ids = 44.22
TESTA
Beta i(as So: NN
" Keaggı NET
Keen Sd «0s 30)
- (209090) + - 4045
[Sy = 0 + (Ad - Cube} (SosH30)
~ (20050) = DiR
DI TRAM
MY =101.08
Bota (ASYasS) = 18.3
J= 10.48 TRY
[5S1]x=8-D>A>C> E
= (Lost) (50002) C3
+ (205m90)> 91.54
Ky = (-boenst) (-30 sind) d)-0}
+ (200050) = AR)
o Kx LEE
Ky
107.2k,
8: Tem Us!
K 101.86 NE
3-50] APBrC-D»E
Measuring?
C= 14.01 607
ENR
ESSAIS EEE
AE CD SET
Faut
AB» B Fee,
Meagurins,
Bess a
AW 457
BUT arme C= bee
assume direckions 17 E>
O+ (cost) Ceosbd * es 30+
ver:
10+ (20siatS)s Cain bD= 15530 + 0
Co 42
Sdbshiuhna > E=11.5
um 797 + E
3-L2Ja-+ Bol p= ESE
assune directions DS, FA
Fez c si (1) «Des 19)
= CiScostn) —LEcos38
eo nes a) oor) ©
rene or) DS (NO)
= (sm CF an 0)
D Rae AD) US ©
using DIO anses
E-=23,89, de -3B.1L
Fama DL
= (Lost) (50002) C3
+ (205m90)> 91.54
Ky = (-boenst) (-30 sind) d)-0}
+ (200050) = AR)
o Kx LEE
Ky
107.2k,
8: Tem Us!
K 101.86 NE
3-50] APBrC-D»E
Measuring?
C= 14.01 607
ENR
ESSAIS EEE
AE CD SET
Faut
AB» B Fee,
Meagurins,
Bess a
AW 457
BUT arme C= bee
assume direckions 17 E>
O+ (cost) Ceosbd * es 30+
ver:
10+ (20siatS)s Cain bD= 15530 + 0
Co 42
Sdbshiuhna > E=11.5
um 797 + E
3-L2Ja-+ Bol p= ESE
assune directions DS, FA
Fez c si (1) «Des 19)
= CiScostn) —LEcos38
eo nes a) oor) ©
rene or) DS (NO)
= (sm CF an 0)
D Rae AD) US ©
using DIO anses
E-=23,89, de -3B.1L
Fama DL
BAS RCD: > EF
Assume dans BR, C7
Deos3)- (-Beos49)-(Cuos 43) +0
-45 cos 60) + (3084 30)
a O)
3 | Cooan8)-(Bont-(esnid) (43)
4 = -(4Ssin' 0D) +(2Dcos30)
BeCT.y-Comas venus O
Solving Ot Osimvitancovshy
Bra. ¿217
Beka aA C107 7
Assume dans BR, C7
Deos3)- (-Beos49)-(Cuos 43) +0
-45 cos 60) + (3084 30)
a O)
3 | Cooan8)-(Bont-(esnid) (43)
4 = -(4Ssin' 0D) +(2Dcos30)
BeCT.y-Comas venus O
Solving Ot Osimvitancovshy
Bra. ¿217
Beka aA C107 7
c ‚Answers to the Chapter 3 Case Study Questions:
Case3-
1, AS keys pressed, ockr plate A rotates back, o elckwise inthe e end-view.
2. Spring C proves ressaace 1 rotating the rocker pate, counterclockwise he LR end view.
5. Spring B provides resistance opresing the Kes, Land.
4 Ask 2 pressed, rocker plate A aes ack, and releases key 1 the Spring unde key 1
forces ey Vino an upward positon,
3. The purpose ofthis mechanism so hamper two Keys fom being inthe dowmwar position at the
6. Spring, acting on key Lin compression and forcing baton I upward,
7. Spring Csi tension, cg acer plate A 1 rote counterclockwise the ight end view
Stop prevents any further ration tan the postion show.
SA és tape player se similar mecanico) mean to preven the play, rewind or fs forward
buttons to be pressed at the same tne
9. As mentioned, pa Dseres sip fur rocker pat A. prvents Furl cockwis raton as
Seen Som the right nd view.
Case 3.2.
1. AS deivesha A tros, coll B als turns because ofa keyed coaneton.
2. The motion ofthe shaft and collar is transe 1 gear € because the proton Oak Dis
Seated nt the neh in al
ink D were freed upward ihe protrusion would come ot ofthe th colar Nothing
‘woud be Ie diving gear C, 6 I would step rotating.
Tink D would dislodge om the notch If gear C woud be exposed to significant rss to
€ The tendency o ik Dto ve upward motion mos oreome e spring sion.
This device intended oop the oto he ger, sigan rame encountered
Sicha dec bals sip euch,
‘Tis evce woud sop the wining mechanism wire Became je, us sopping
resisance on ear
Toro this dev, ik D mus be place under ik and ligne wih he noch in colar.
10. The spring mus Bein tension, Three, it pol ink F tthe wight and Ik Et ie ie
Case 33.
AS handle Ais rated, moving threaded rod B othe le, rip C also movesto the lef and
slighty upward, Notice that inks E and Far pvoig othe mide thus rip Cis constsined
{oa swinging mation
2. Ashandie A is route, moving threaded od Bi het, grip D moves othe right and slightly
‘downward Since inks E and F are pivoting in the mide grip D wil have mation oposing.
wine.
3. The purpose ofthis mechasim isto serve a machining clamp forthe wrkpiece
4, The preg G pling 0 ink D would ame to eur to an upward and rightward pesto,
Forma ls Gaia ep a pa ona wee he threaded oan
6, Links and F havea peculiar configuration 1 voi interference with he wrikpiee,
hroupkou ih ange af maton of the clamp.
7. Seca device could be call a machining lamp
& Since ink Cs moving la swinging mio, he rounded ed cn th headed rod, asus à
sise pole contac wit Ink C
Case3-
1, AS keys pressed, ockr plate A rotates back, o elckwise inthe e end-view.
2. Spring C proves ressaace 1 rotating the rocker pate, counterclockwise he LR end view.
5. Spring B provides resistance opresing the Kes, Land.
4 Ask 2 pressed, rocker plate A aes ack, and releases key 1 the Spring unde key 1
forces ey Vino an upward positon,
3. The purpose ofthis mechanism so hamper two Keys fom being inthe dowmwar position at the
6. Spring, acting on key Lin compression and forcing baton I upward,
7. Spring Csi tension, cg acer plate A 1 rote counterclockwise the ight end view
Stop prevents any further ration tan the postion show.
SA és tape player se similar mecanico) mean to preven the play, rewind or fs forward
buttons to be pressed at the same tne
9. As mentioned, pa Dseres sip fur rocker pat A. prvents Furl cockwis raton as
Seen Som the right nd view.
Case 3.2.
1. AS deivesha A tros, coll B als turns because ofa keyed coaneton.
2. The motion ofthe shaft and collar is transe 1 gear € because the proton Oak Dis
Seated nt the neh in al
ink D were freed upward ihe protrusion would come ot ofthe th colar Nothing
‘woud be Ie diving gear C, 6 I would step rotating.
Tink D would dislodge om the notch If gear C woud be exposed to significant rss to
€ The tendency o ik Dto ve upward motion mos oreome e spring sion.
This device intended oop the oto he ger, sigan rame encountered
Sicha dec bals sip euch,
‘Tis evce woud sop the wining mechanism wire Became je, us sopping
resisance on ear
Toro this dev, ik D mus be place under ik and ligne wih he noch in colar.
10. The spring mus Bein tension, Three, it pol ink F tthe wight and Ik Et ie ie
Case 33.
AS handle Ais rated, moving threaded rod B othe le, rip C also movesto the lef and
slighty upward, Notice that inks E and Far pvoig othe mide thus rip Cis constsined
{oa swinging mation
2. Ashandie A is route, moving threaded od Bi het, grip D moves othe right and slightly
‘downward Since inks E and F are pivoting in the mide grip D wil have mation oposing.
wine.
3. The purpose ofthis mechasim isto serve a machining clamp forthe wrkpiece
4, The preg G pling 0 ink D would ame to eur to an upward and rightward pesto,
Forma ls Gaia ep a pa ona wee he threaded oan
6, Links and F havea peculiar configuration 1 voi interference with he wrikpiee,
hroupkou ih ange af maton of the clamp.
7. Seca device could be call a machining lamp
& Since ink Cs moving la swinging mio, he rounded ed cn th headed rod, asus à
sise pole contac wit Ink C
u
€
HAT +=.1s
Tesco] =2 101
te LS
RNA CD EMETS
Axe (4.208)-(2.12)>-3.3812 3.
M2 423.2
of) = 1,02%
teils
23 cof SOLA DAO] = 26 in
Dx UNA = + 2.189 22.1
ay Tas:
te AR 6.060 137]
SEEN
TEE RR = LL HI mn —
7
measuring =
Meta À
Mantes 230", cows
| pomos
FT
€
HAT +=.1s
Tesco] =2 101
te LS
RNA CD EMETS
Axe (4.208)-(2.12)>-3.3812 3.
M2 423.2
of) = 1,02%
teils
23 cof SOLA DAO] = 26 in
Dx UNA = + 2.189 22.1
ay Tas:
te AR 6.060 137]
SEEN
TEE RR = LL HI mn —
7
measuring =
Meta À
Mantes 230", cows
| pomos
FT
9
fa)
AA
Aena = 3.038 “ETS
4-10)
Algen: «852"4
ass 2.00 SSH
us
et E
AZ,
AB ie 22.8, cu hr
(435
4-14
easing
SS
rel
Cay
Aena 10.254"
AU?
4-16! sl
Din,
Hos Dt
1435)
fa)
AA
Aena = 3.038 “ETS
4-10)
Algen: «852"4
ass 2.00 SSH
us
et E
AZ,
AB ie 22.8, cu hr
(435
4-14
easing
SS
rel
Cay
Aena 10.254"
AU?
4-16! sl
Din,
Hos Dt
1435)
9
‘iB
ABronile = AM cow
ad
AR
[4-20]
Ha
Ale 25-132: 83.48"
ASE) = 400 7m?
SADA Tee lo
Deed 52032907 TRAY
‘iB
ABronile = AM cow
ad
AR
[4-20]
Ha
Ale 25-132: 83.48"
ASE) = 400 7m?
SADA Tee lo
Deed 52032907 TRAY
lon
€
AL cctuste® 533 -32644 =206,007]
Rod? 82.4 437,50
ABpaña = 14,56, Cow ABieg = 2892", evs
Ae] + 1855"" 68.007 Bauer = 259-10 ZUSSH
2 EE
IT
ARE «TENA
Eva]
Mor AN" À
Ale 22,3507-20.184
= | Sbbl
longer
€
AL cctuste® 533 -32644 =206,007]
Rod? 82.4 437,50
ABpaña = 14,56, Cow ABieg = 2892", evs
Ae] + 1855"" 68.007 Bauer = 259-10 ZUSSH
2 EE
IT
ARE «TENA
Eva]
Mor AN" À
Ale 22,3507-20.184
= | Sbbl
longer
a]
us
a
Mos Remap atts
= 2.2514 sherk
4-32
iB
4
Ze, ig TU BIN
os RB
= ATT (ma)
us
a
Mos Remap atts
= 2.2514 sherk
4-32
iB
4
Ze, ig TU BIN
os RB
= ATT (ma)
434
Alim” 626, 0
1.4000
A 856
AB BAU, cow
4-31
Miu: 2038; ow
Alim” 626, 0
1.4000
A 856
AB BAU, cow
4-31
Miu: 2038; ow
at U sn 150°
Vs (6120-15: 10°
Les AT GL
ODA
= sn
Ciena D
RE EVA
N
EX Age, Ru
Bew “Cig: 10
10- (8-70
AR LA VIS
UE NEO
’ =9571"
Orsi ind) +70
a WE, 3
1-30] carat inte
ey BS sind) =F
age enden
Ro Re EURER
ES = [UH me
Bond) na
Em (oda
2 NA AS
Ares ASA IMAN 66.8 Lmm Y
ET
T eurer coberta:
SEN =]
R = Mar
pe
ARAM Big =
70"
Ro ISgesprdes WA |
Ep a a oie Es]
Ra
= 5545
R ACT I 4 sn 50
dspou eng: aD
y Opos]
|
=65.0
Rez bsial5~4si020
3.438"
Sr
ES
bee FAR.
4-3) D Comet epafigueadnen
A, Bes gt NS
Re ten 15
= 4.29"
dasplawd 5 Sin
ae
am = aar
Re
3.24 bú
Ae SU - Bat = 5.3231 4
Rz 1Seos Da Bsin 1D]
& Es se
Sen By Rs 3]
Vs (6120-15: 10°
Les AT GL
ODA
= sn
Ciena D
RE EVA
N
EX Age, Ru
Bew “Cig: 10
10- (8-70
AR LA VIS
UE NEO
’ =9571"
Orsi ind) +70
a WE, 3
1-30] carat inte
ey BS sind) =F
age enden
Ro Re EURER
ES = [UH me
Bond) na
Em (oda
2 NA AS
Ares ASA IMAN 66.8 Lmm Y
ET
T eurer coberta:
SEN =]
R = Mar
pe
ARAM Big =
70"
Ro ISgesprdes WA |
Ep a a oie Es]
Ra
= 5545
R ACT I 4 sn 50
dspou eng: aD
y Opos]
|
=65.0
Rez bsial5~4si020
3.438"
Sr
ES
bee FAR.
4-3) D Comet epafigueadnen
A, Bes gt NS
Re ten 15
= 4.29"
dasplawd 5 Sin
ae
am = aar
Re
3.24 bú
Ae SU - Bat = 5.3231 4
Rz 1Seos Da Bsin 1D]
& Es se
Sen By Rs 3]
L
=
€
Horse conFiguraiton?
Tas TANS
Azar msn gar
cos! AST Mo
pros | Nall pe
drug cria
[4-83] orginal confgrañon:
tere = 4
NOR Pe pe
Be PARE 30 el
do se
ES
fu os ats) | 4253
rsplased wa urahen: \
e Delia; e 5
és Les am
aa
cos! 103.17
bo fe re SEAN
SEE
a
4
TE ,
e
serine LT
pron ey. 200 sin 30] 59 06
EA cos [eat 435" E]: 9349°
AB te (ó-, à
= (389-40) -(253-57d "52.07,
ALT onagra configurer
Chr
C8
Bei 7 ul
SE
QA e
» % 3, 8 tas ES
Bons Ra
BD: AAT TENS ~44TIem
E SRE u
pe ws) o "| = 17.88"
637 ND ann nee 542
displaced, Contugoralin :
ar AT FE: 9. Be
ge sat Lt ssl = AE
Pos PR: a
[os 135~Low2b¥ SLA) = 1.40
Bb», FF TARO
a se [sms + 19.51
Al ur Ro
CRE» mer
re SD 8.88 = bey
Br =
Bt a ETRE
Ca) ur
= sfr. a
= ee | + 42.327
Oy GRAN tas 20.807
==
Ady= MA0+5,92 >16.82 Cw)
A0y= bb.14 -39:832 26318 cow
=
€
Horse conFiguraiton?
Tas TANS
Azar msn gar
cos! AST Mo
pros | Nall pe
drug cria
[4-83] orginal confgrañon:
tere = 4
NOR Pe pe
Be PARE 30 el
do se
ES
fu os ats) | 4253
rsplased wa urahen: \
e Delia; e 5
és Les am
aa
cos! 103.17
bo fe re SEAN
SEE
a
4
TE ,
e
serine LT
pron ey. 200 sin 30] 59 06
EA cos [eat 435" E]: 9349°
AB te (ó-, à
= (389-40) -(253-57d "52.07,
ALT onagra configurer
Chr
C8
Bei 7 ul
SE
QA e
» % 3, 8 tas ES
Bons Ra
BD: AAT TENS ~44TIem
E SRE u
pe ws) o "| = 17.88"
637 ND ann nee 542
displaced, Contugoralin :
ar AT FE: 9. Be
ge sat Lt ssl = AE
Pos PR: a
[os 135~Low2b¥ SLA) = 1.40
Bb», FF TARO
a se [sms + 19.51
Al ur Ro
CRE» mer
re SD 8.88 = bey
Br =
Bt a ETRE
Ca) ur
= sfr. a
= ee | + 42.327
Oy GRAN tas 20.807
==
Ady= MA0+5,92 >16.82 Cw)
A0y= bb.14 -39:832 26318 cow
Renga tance HAT ont enter:
A 0
we a ot be
on on PRO am
displaced tenfigorahion
:
ze Y o,
a = Ae
eee BES
AL+205- RATE L8.0Ímm
Ve E (ICO = 33,404 men?
= ABS €
50) oras entra:
¡VS de [ET eng”
A Yer 2247
Aspa) fan
wr 7 yee SIS al] E
ü Sa Y ET 2
Aye ste A dl
wea HSS
ET
DV Naher: zu
ADE > 24 - 20-184 > 3,581"
A 0
we a ot be
on on PRO am
displaced tenfigorahion
:
ze Y o,
a = Ae
eee BES
AL+205- RATE L8.0Ímm
Ve E (ICO = 33,404 men?
= ABS €
50) oras entra:
¡VS de [ET eng”
A Yer 2247
Aspa) fan
wr 7 yee SIS al] E
ü Sa Y ET 2
Aye ste A dl
wea HSS
ET
DV Naher: zu
ADE > 24 - 20-184 > 3,581"
1:52] orga. Confaoretton ? K-53 Dog cfa
E Pro [sera à OS
T Re ttes ie JR
Ay = os EN Aca i= we
pl I cepto
ebb | i. 3.8"
AA Beast 1 28% I
M Mosids AR sindye und $558 os ati aye wu
> |e grasa » 20.45"
ne teen Kae (A
&
displaced Sorhyyad isn >
el lad NR Mea
e ta a0 | A= sont gg) 49.32
(RE
EE) a
+ 288i (Br fs SA
cosdOrd gsnByr 28s (1-9) 472.
RO à Aus = 24 (8-H)=4] 47
[Matas
By x tows QE 8452 ERICH
G AR, 30. Imm Beso [Penta aaa: AU
& 6.5324 STA
q) = 25132]
Bem papa -9d)= 27.306"
are Tu: A ra = 65}
LESA) organ ondas: Ads (2300320 = 344.08 Be tan" (AM 35:11
os Res C0 Sm) 2343.08
ar E > E au
EACH
30+%30- 250 si (D,-$+8) = 21913
ROW - 250.00 (9-10 = 25.84
, Al
tan (3524/2599 = 21.68"
10 EU
> 2: <<): FRS
a (santa E
N > bar
230 +2390 -250 sin CAN an
20-180 -2% ens (Br +04) = VA
Me 20232 Bu FRS 30.02"
St A, (181+30.0))- (7340421. #553) cu)
he 2! RS
Bus pu DES
I
E Pro [sera à OS
T Re ttes ie JR
Ay = os EN Aca i= we
pl I cepto
ebb | i. 3.8"
AA Beast 1 28% I
M Mosids AR sindye und $558 os ati aye wu
> |e grasa » 20.45"
ne teen Kae (A
&
displaced Sorhyyad isn >
el lad NR Mea
e ta a0 | A= sont gg) 49.32
(RE
EE) a
+ 288i (Br fs SA
cosdOrd gsnByr 28s (1-9) 472.
RO à Aus = 24 (8-H)=4] 47
[Matas
By x tows QE 8452 ERICH
G AR, 30. Imm Beso [Penta aaa: AU
& 6.5324 STA
q) = 25132]
Bem papa -9d)= 27.306"
are Tu: A ra = 65}
LESA) organ ondas: Ads (2300320 = 344.08 Be tan" (AM 35:11
os Res C0 Sm) 2343.08
ar E > E au
EACH
30+%30- 250 si (D,-$+8) = 21913
ROW - 250.00 (9-10 = 25.84
, Al
tan (3524/2599 = 21.68"
10 EU
> 2: <<): FRS
a (santa E
N > bar
230 +2390 -250 sin CAN an
20-180 -2% ens (Br +04) = VA
Me 20232 Bu FRS 30.02"
St A, (181+30.0))- (7340421. #553) cu)
he 2! RS
Bus pu DES
I
(9)
4-55 | orignal conf! >
Seal tenfrayestion Gh AN 293.85"
Za as AAN] =
S en ] aa
IF Beer 2180 Bag = MiS"
ar
Bmr= ee FA
Bts Opos 4551
Oger = 180-B rey = 115.288"
A Fæ, >
SSE Ogee = Bier * Bag = 095°
N TER ES
Byegs ws pC st aay ea |: i CT EL Ore’ = 24.64
ee 29.19-39.00=-9.91> Gr cw/
ES An
el
(Cond, SIDI,
15123
4-55 | orignal conf! >
Seal tenfrayestion Gh AN 293.85"
Za as AAN] =
S en ] aa
IF Beer 2180 Bag = MiS"
ar
Bmr= ee FA
Bts Opos 4551
Oger = 180-B rey = 115.288"
A Fæ, >
SSE Ogee = Bier * Bag = 095°
N TER ES
Byegs ws pC st aay ea |: i CT EL Ore’ = 24.64
ee 29.19-39.00=-9.91> Gr cw/
ES An
el
(Cond, SIDI,
15123
\ Gus
| fra
& OL
(Paso 243)* Anm
rai cd vel: AIS SE q
Qe > [5,25 5% = 3.2"
Rad UTA-30M= su
| fra
& OL
(Paso 243)* Anm
rai cd vel: AIS SE q
Qe > [5,25 5% = 3.2"
Rad UTA-30M= su
Answers to the Chapter 4 Case Study Questions
Case 4-1.
1 As wheel Cis rotated clockwise, the motion of pin Dis circular, clocks, motion abou the center of
the whee
2. As wher Cis rotated cocos, the motion of pis Ps oscillating, slong an ar, centered about the
Joint onthe tapped sing ik
3. As cl Cs rated clockwise, be malen of pie K is osclating tension since item Us
ensraned to siding en ie fame
4, Turing he hand wheel wl eher as or lower the ped sido. As the lapped slides alse, fr
Je Ihe pl ofp wi be at Hier ee. The pal of ia K ll be pulled tthe le Pin.
(€)
and wil lo tend salt through siii gene placement
5. Tuning the hand-wel will move the osilating motion of slid U, ihr leward or rghiwar. The
range of motion wl az lr, gai.
6, This device ca De ud u rnning ajusment to machines thal require an oil
moon, such asa surface rider.
ing hing
ay modo ofthe Bow, coupled wih ton, brings the srew blanks tothe ges.
2. The lat havea head, which cn be ecient rented wth a parallel Rage. As he banks.
approach the fingers the Blanks that happen 10 randomly be aligned with the fingers wil end o ride
spite rack.
3. The rack intermitentes to aed poston, hen Rack down to he Bow
4. A comeing arm E, couples the motion of ik Di ak 8
5. Te mein fink D 1 rece wo pue ansia, and lak B sconstined to pure rotation
‘Therefore, a connecting ak, or a ltd jot, rquie io join the two components.
stack Bis in an elevated prison, the screw blanks are abl lie down towards C.
The tip ofthe fingers contacting the bow isthe less position. A tp os F would serve a an
fée method to preven the Mager ps o dap onthe ota ofthe bow
8. Asscrew blanks are congested inthe outlet wack, and Angers B, item F contacts the heeds ofthe
screw blanks. This cont probibis the al lowering ofthe Ginger.
9. Asscrew blanks av congested inthe ot wack, and agers By em E contacts the heads of the
Screw bans. Ths contac prob the finger B kom dipping int the bow. Therfre no other
Banks attempt eter he Gages
10. This device ia eer track, hat awonaticlly stos feeding srew blanks when the out ack Is
jammed.
11, Tis device must be even by a mechanis ha can compensat or a shorter displacement required
when the ack Is cese, À sideral mechanism, coupled wit a spring is ne opt.
Case 43.
12. BarB and ab, i rested to roma asian
3. Te op OF € can ony move borzotlly andthe boto can any move vertical. Sos cylinder
Lis shortened, the op ai C moves owar height, andthe Dtm of ink € moves downward
There, ak C moves tothe ight and roues clockwise.
3. Poin K moves othe righ, and douar,
4 Since the top of ink C is contained fom moving vera, andthe botom i constrained om
‘moving horizontal, the boto must be able to move vertically ¡any motion is o ocur Therefore,
plat mus ein lt,
3. ‘The purpose ofthis mechanism iso move table A KR and right. The motion ofthe cylinder is
amplified and the bot kage asias o suppor a load placed on tbl A.
6 Turning he headed ros end moves the "zone of ion of table A, pushing ler ear orto
ep.
Case 4-1.
1 As wheel Cis rotated clockwise, the motion of pin Dis circular, clocks, motion abou the center of
the whee
2. As wher Cis rotated cocos, the motion of pis Ps oscillating, slong an ar, centered about the
Joint onthe tapped sing ik
3. As cl Cs rated clockwise, be malen of pie K is osclating tension since item Us
ensraned to siding en ie fame
4, Turing he hand wheel wl eher as or lower the ped sido. As the lapped slides alse, fr
Je Ihe pl ofp wi be at Hier ee. The pal of ia K ll be pulled tthe le Pin.
(€)
and wil lo tend salt through siii gene placement
5. Tuning the hand-wel will move the osilating motion of slid U, ihr leward or rghiwar. The
range of motion wl az lr, gai.
6, This device ca De ud u rnning ajusment to machines thal require an oil
moon, such asa surface rider.
ing hing
ay modo ofthe Bow, coupled wih ton, brings the srew blanks tothe ges.
2. The lat havea head, which cn be ecient rented wth a parallel Rage. As he banks.
approach the fingers the Blanks that happen 10 randomly be aligned with the fingers wil end o ride
spite rack.
3. The rack intermitentes to aed poston, hen Rack down to he Bow
4. A comeing arm E, couples the motion of ik Di ak 8
5. Te mein fink D 1 rece wo pue ansia, and lak B sconstined to pure rotation
‘Therefore, a connecting ak, or a ltd jot, rquie io join the two components.
stack Bis in an elevated prison, the screw blanks are abl lie down towards C.
The tip ofthe fingers contacting the bow isthe less position. A tp os F would serve a an
fée method to preven the Mager ps o dap onthe ota ofthe bow
8. Asscrew blanks are congested inthe outlet wack, and Angers B, item F contacts the heeds ofthe
screw blanks. This cont probibis the al lowering ofthe Ginger.
9. Asscrew blanks av congested inthe ot wack, and agers By em E contacts the heads of the
Screw bans. Ths contac prob the finger B kom dipping int the bow. Therfre no other
Banks attempt eter he Gages
10. This device ia eer track, hat awonaticlly stos feeding srew blanks when the out ack Is
jammed.
11, Tis device must be even by a mechanis ha can compensat or a shorter displacement required
when the ack Is cese, À sideral mechanism, coupled wit a spring is ne opt.
Case 43.
12. BarB and ab, i rested to roma asian
3. Te op OF € can ony move borzotlly andthe boto can any move vertical. Sos cylinder
Lis shortened, the op ai C moves owar height, andthe Dtm of ink € moves downward
There, ak C moves tothe ight and roues clockwise.
3. Poin K moves othe righ, and douar,
4 Since the top of ink C is contained fom moving vera, andthe botom i constrained om
‘moving horizontal, the boto must be able to move vertically ¡any motion is o ocur Therefore,
plat mus ein lt,
3. ‘The purpose ofthis mechanism iso move table A KR and right. The motion ofthe cylinder is
amplified and the bot kage asias o suppor a load placed on tbl A.
6 Turning he headed ros end moves the "zone of ion of table A, pushing ler ear orto
ep.
|
a Qro pan [$21 .. By
© AED = | ee)
ws er DS =3i.beyod EN NEE = 109 rpm,
BAT. 180 rpm = 3 ee? 338
= RO 1322 E
= NS MARE = 882 rpm! A q
wee My
5-5] 2 15cpn = 1258 > 85%, [St lee s00rm sans ete
Qe Hoe = 1.517 =
ran, 09, ELA BPs OT see,
te ge by tags Discs
€
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SU a
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A
pacto IM e
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7 MATES es: Nes rem
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A = 49.0%,
Speaks “ts Yo
a Qro pan [$21 .. By
© AED = | ee)
ws er DS =3i.beyod EN NEE = 109 rpm,
BAT. 180 rpm = 3 ee? 338
= RO 1322 E
= NS MARE = 882 rpm! A q
wee My
5-5] 2 15cpn = 1258 > 85%, [St lee s00rm sans ete
Qe Hoe = 1.517 =
ran, 09, ELA BPs OT see,
te ge by tags Discs
€
(0)
SU a
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A
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very RS = us
AU) =
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7 MATES es: Nes rem
dis = BSH
A = 49.0%,
Speaks “ts Yo
|
ES
Qe EME
a SEE 242.8]
So 21€
pe = Tees 208
Teen (BE) = 102. 2rp)
e
Conueyor ? Conveyor
ak © 2 (OS eons Ups WANN 333%
ve Fe A Gas e d Ce 536 4%
Same ere um Po bay aux
m ek SW ES
> Su a. i EE m
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+
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Ñ
las ¡EL (BSE) > Soreny coe de = 150 can
SR] Qu) Gnine, teca) [SAT @125 ops ner
CES]
SUD +22.)
ke ge = ,45 mm,
sn la D ook D le
dor des (BEN = or)
ES
Qe EME
a SEE 242.8]
So 21€
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e
Conueyor ? Conveyor
ak © 2 (OS eons Ups WANN 333%
ve Fe A Gas e d Ce 536 4%
Same ere um Po bay aux
m ek SW ES
> Su a. i EE m
ee 168) = 26.1" =440/(nS+ 8337
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+
a | lm) 3
= |,
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5
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any ty wi work
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las ¡EL (BSE) > Soreny coe de = 150 can
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Answers to the Chapter 5 Case Study Questions:
Case 51.
1. As shaft A rotates 90”, clockwise, the eccentric of lobe B moves down to the 6
clock position.
2. Asa A rates 90°, clockwise, link C will move into a vertical posto
3. A slo is necessary at point E to accommodate for he vertical mation of ink C
4. As shall A rotates 90%, cloccvse, in H will move toward the le, and drop
douar
5. As shaft A rotates 90", lokos, pin will move toward the le.
‚6. The mechanism has 6 links, $ pin joints, 2 sliding joints, and one degree of
freedom.
7. Asthread G pulls roller E downward, the distance from the driveshaft o the
«tte pivot wil lengthen. As a esl he ange at which nk C wl oncles
reduced.
8. As thread G pulls roller E downward the sance from the dive shaft tothe
‘effective pivot will lengthen. As rs, the horizontal distance at which point H
‘will ose i reduced
9. The purpose ofthe mechanism isto move lock! back and forth. However, the
speed of rave i faster in one direction, yelling quick retum ation,
Case 5-2.
1. Asrod A moves to the right, sliding block B is also driven to the right.
. As rod A moves to the right, and the roller C reaches groove D, the roller falls into
the groove. Asa result, the belleank ink rotates clockwise, The sliding block B
‘Will sill move to the right, but at a slower speed.
3. Asrod A reverses and moves to the le, the bellerank link rotates
counterclockwise, pulling roller C from groove D. The sliding block B also moves.
to the lef, but ata relatively slow speed.
4. Asrod A oscillates horizontally, siding block B also oscilates horizontally.
However a the right extreme end of the stroke, sliding block B moves ata slower
speed.
5. The purpose ofthe mechanism isto take a singl-speed linear motion and create
two-speed linear motion.
6. Rod À can be driven with a sidercrank mechanism, a pneumatic or hydraulic
cylinder, a cam follower, ec.
7. The adjustment slots at E allow the user to modify the location where the speed of
sliding block B will change.
Answers to the Chapter 5 Case Study Questions:
Case 51.
1. As shaft A rotates 90”, clockwise, the eccentric of lobe B moves down to the 6
clock position.
2. Asa A rates 90°, clockwise, link C will move into a vertical posto
3. A slo is necessary at point E to accommodate for he vertical mation of ink C
4. As shall A rotates 90%, cloccvse, in H will move toward the le, and drop
douar
5. As shaft A rotates 90", lokos, pin will move toward the le.
‚6. The mechanism has 6 links, $ pin joints, 2 sliding joints, and one degree of
freedom.
7. Asthread G pulls roller E downward, the distance from the driveshaft o the
«tte pivot wil lengthen. As a esl he ange at which nk C wl oncles
reduced.
8. As thread G pulls roller E downward the sance from the dive shaft tothe
‘effective pivot will lengthen. As rs, the horizontal distance at which point H
‘will ose i reduced
9. The purpose ofthe mechanism isto move lock! back and forth. However, the
speed of rave i faster in one direction, yelling quick retum ation,
Case 5-2.
1. Asrod A moves to the right, sliding block B is also driven to the right.
. As rod A moves to the right, and the roller C reaches groove D, the roller falls into
the groove. Asa result, the belleank ink rotates clockwise, The sliding block B
‘Will sill move to the right, but at a slower speed.
3. Asrod A reverses and moves to the le, the bellerank link rotates
counterclockwise, pulling roller C from groove D. The sliding block B also moves.
to the lef, but ata relatively slow speed.
4. Asrod A oscillates horizontally, siding block B also oscilates horizontally.
However a the right extreme end of the stroke, sliding block B moves ata slower
speed.
5. The purpose ofthe mechanism isto take a singl-speed linear motion and create
two-speed linear motion.
6. Rod À can be driven with a sidercrank mechanism, a pneumatic or hydraulic
cylinder, a cam follower, ec.
7. The adjustment slots at E allow the user to modify the location where the speed of
sliding block B will change.
|
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eu [al coo cman) rer se
Cae ce (yt res ( Gene ÉS
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NEN Ve
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PACE) 31102 Sv
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w la omas
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Var (US) CT ER
©
Ve Ven a
>”
Ver Se
16:23] we 00) = (OATS ow
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ME Vo Ve
8
Ve
ye W
Vers
[6-24] 0,2 IE): 8.38% cow
Vs (8.28.73) = 6,2312 V3o
Me Ve Von
fe)
A
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NEN Ve
Me
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Ve Ven a
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ME Vo Ve
8
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Vs (8.28.73) = 6,2312 V3o
Me Ve Von
| |
[625 [0,4079 DH Eu
Ves 2B RTE
26.283)
= SL
158 Sut
SE
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ATL cus
LA
EN
6228 40,2 15: (5): 7.854 cu
Vo 80 (1890) = 628.3200
Me
DS
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USNS sages days 1/720
due 8U1.11/120 = 113245, Cows i:
ay 5 ; GE
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Ver 2 (523012108 % Sony Vos ZEUS) "TIMO ER
Vega,
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26.283)
= SL
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SE
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Me
DS
SE oe
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Volks
We
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VF15.99 ua 712.88"
a
a
we bn]
i
[Eat =
B
Sn
sor
EN = MARY cu
Ve (UNA = 198.40% Zr
Ver Vo? Veg, y A
Ve y
= | A ~~ 180
LA
3258 ae ns
ÉCHOS 150
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Ve= (BONES = 15.122 To
a «
| Vo NA so-n
NL vere, Te
390% ew
Ve” Ces « bot som
ADB = TA AT.
Ve Be se dol
45 se
aus an
2b. tA
Y
= HSM Mant woe Fam. NY
ENT
Dre
De w kr
WILDEN = 15.30"
sis. = SRP + ADB IT
OS
ANTES
\ ca a 242.57"
Bs Te ar
Len + 180-Fa2STH2L AS = 105°
t= 4Sepa(B) A WT, cows
NE) = 942% 33)
Ye 90 -[ede— ABE] = 93.5"
>\e0-(53.! ana
Ab CBD = 33.18
om ta
> \er 0.3% RS
Go 5545 =1880%, wr
we bn]
i
[Eat =
B
Sn
sor
EN = MARY cu
Ve (UNA = 198.40% Zr
Ver Vo? Veg, y A
Ve y
= | A ~~ 180
LA
3258 ae ns
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Ve= (BONES = 15.122 To
a «
| Vo NA so-n
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390% ew
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ADB = TA AT.
Ve Be se dol
45 se
aus an
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Y
= HSM Mant woe Fam. NY
ENT
Dre
De w kr
WILDEN = 15.30"
sis. = SRP + ADB IT
OS
ANTES
\ ca a 242.57"
Bs Te ar
Len + 180-Fa2STH2L AS = 105°
t= 4Sepa(B) A WT, cows
NE) = 942% 33)
Ye 90 -[ede— ABE] = 93.5"
>\e0-(53.! ana
Ab CBD = 33.18
om ta
> \er 0.3% RS
Go 5545 =1880%, wr
Nip ASUS
Ne
Lene
we Ya
E =53,18
no!
as I
Ad: RDA = 120,00 mm
PORE = tor (Yan) = SN
ZADE> Q0- ZDAE = 84.24%
LRAD = 180- (15+ LDAD)
80=/ nov +80 (mata
= 1221 Im
sinLABD. sin 0929, R s
Se = ar? LARD =7100
ZADB= 180-[2B AD +ZAQ5] = 3:11"
LOROS cos" EE Jus ay
AAA
oe fA
LRO: 129-(ZeBD+Z BOE!
Cos WepmB = 44274, cow
Mes DO =7154.0% 15 À
Me. MO im
FR
SIN
642 1348/10
= 1.00% , DS
LS
3 40
80/2100 os HO 18"
ESS HMO La 7.68"
ZABD= mdr Lape] = 32.32°
dog [LARES] 97°
Lens es a 627
rs ule
ES
2806 = 180-[ZeBD + RCD
ARE = LABD+ZERD = 58.20"
Ian ZADB + ¿ROL ='50.24
3,2 bOrpa( Sp) 6.283 %,cw
Ve (LD = 12.56% Var
CERTES
ARD ALU
LAR? SBI?
Ye
- 156
STE
& 199% Noe
das WEL 265 oy, wy
Ne
Lene
we Ya
E =53,18
no!
as I
Ad: RDA = 120,00 mm
PORE = tor (Yan) = SN
ZADE> Q0- ZDAE = 84.24%
LRAD = 180- (15+ LDAD)
80=/ nov +80 (mata
= 1221 Im
sinLABD. sin 0929, R s
Se = ar? LARD =7100
ZADB= 180-[2B AD +ZAQ5] = 3:11"
LOROS cos" EE Jus ay
AAA
oe fA
LRO: 129-(ZeBD+Z BOE!
Cos WepmB = 44274, cow
Mes DO =7154.0% 15 À
Me. MO im
FR
SIN
642 1348/10
= 1.00% , DS
LS
3 40
80/2100 os HO 18"
ESS HMO La 7.68"
ZABD= mdr Lape] = 32.32°
dog [LARES] 97°
Lens es a 627
rs ule
ES
2806 = 180-[ZeBD + RCD
ARE = LABD+ZERD = 58.20"
Ian ZADB + ¿ROL ='50.24
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Ve (LD = 12.56% Var
CERTES
ARD ALU
LAR? SBI?
Ye
- 156
STE
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em]
A, SUD Dons
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ip Brace ae Ae |
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a
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5
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A
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Ves Sa RSS
A, SUD Dons
hy y FIS UDcost|
el
t .
En =
ER Ve 24 y
1S" A
Me SAN]
=
ve
D | dou carer? Wear 37
8 sta]
13
ip Brace ae Ae |
= 2.0%
Vea Vaya woe
AS ay
Fa
a
ES
Neo __. No
CENTER
Vans ANA 707
5
4 doy B= 039% , cow
A
Ne
Va Ye
nas SE
Ne 24.0% Zs"
Vor 202.0°% WS"
Na Flow» (up) CaS] = 208.1%,
i)
SS
NS]
F Wir
sus
a Dame
op ab 10
en
su LADS
Satie
LABb= vel
Le
(Mas) = BAT,
nes Ds es
Io Neg
pre FN)
sit. sel ts
ME get
ve Uds se
RMS
Lee= ELA. Dest laces
Ves Sa RSS
UC
So
Os >
ee ve
ES
Io
postiar
6 e a
a SiN WS
a
Des A4
71 a
LABS cos nue
A as SPW
35%5-10)
LARGE cos* (ee estar
Lines Leg + ZBAC = 64.64
Voit Ver Vere
bs AU
so Tox
ibe nas, GA
paeesunes do
ellie = 3,37 Yn» CW
tr EA
= 32.30
We Gated)
DT autos
dons HA]
So
Os >
ee ve
ES
Io
postiar
6 e a
a SiN WS
a
Des A4
71 a
LABS cos nue
A as SPW
35%5-10)
LARGE cos* (ee estar
Lines Leg + ZBAC = 64.64
Voit Ver Vere
bs AU
so Tox
ibe nas, GA
paeesunes do
ellie = 3,37 Yn» CW
tr EA
= 32.30
We Gated)
DT autos
dons HA]
o
y
hee
+ >
husos lever =
Busen aan.
y
hee
+ >
husos lever =
Busen aan.
© 657. |:
instant Center | x Y
ua] 0.000 | 0500
(03) | 65028 | 22201
wa] soo] 6.000.
Wal ass] 2.2856.
ay 3.467] 000m
169) 15129] 34997
6558 Test Cone y
©
es
instant Center | x Y
ua] 0.000 | 0500
(03) | 65028 | 22201
wa] soo] 6.000.
Wal ass] 2.2856.
ay 3.467] 000m
169) 15129] 34997
6558 Test Cone y
©
es
e
ES
©
Answers to the Chapter 6 Case Study Questions:
Case 6-1.
1. As gear As forced to rotate counterclockwise, nating gear B wil rotate clockwise.
‘Mating, external, gears will always rotate in opposite dir
2.
3
4
8.
9.
As gear A rotates counterclockwise, stud pin C rotates clockwise.
‘As gear A rotates counterclockwise, lever D, which pivots about its right end,
‘oscillates rotationally.
Link D drives link E, in a rocker-rocker configuration. Since link E is shorter than
link D, link E wall have a larger throw angle and link D.
Link D will be placed ints lowest extreme position when the center ofthe slot is.
tangent to the bottom of the path of pin C.
Link D will be placed in ts upper extreme position when the center ofthe slot is
tangent to the top ofthe path of pin C.
It takes approximately 210 degrees of gear B to rotate link D (and E) upward and
about 150 degrees to retum downward.
With the imbalance observed in 7, it takes longer to rotate link E up than it takes.
to lower link E.
‘The motion of link E is “quick-retum.”
Case 6-2.
1.
2
3.
As wheel C is reed to rotate counterclockwise, pin D, wich is attached to wheel
, will rotate counterclockwise
As whee! C is forced to rotate counterclockwise ink G, which pivots at its
‘middle, oscillates rotationally
‘The left portion of ink G will be placed in its upper extreme position when the
center of the slot is tangent to the bottom of the path of pin D.
‘The let portion of ink D will be placed in its lowest extreme position when the
center of the slot is tangent to the top ofthe path of pin D.
Te takes approximately 200 degrees of wheel C 0 rotat link G counterclockwise
and about 160 degrees to return.
Te takes approximately 25% more time to rotate link G counterclockwise than it
takes to rotate link G clockwise.
The cyclical motion of link N is rotational oscilation. Et exhibits relatively slow
‘counterclockwise rotation, and a relatively quick clockwise rotation,
‘The cyclical motion of table Ris translational oscillation. It exhibits relatively slow
movement tothe left, and relatively quick movement 10 the right,
‘The mechanism creates quick return motion, moving the workpiece quickly
ough a grinding pass.
“The screw threads can be used to adjust the position of the translations oscillation
of the table
“The mechanism has eight inks, eight pin joints and two sliding joints, giving one
degree of freedom,
ES
©
Answers to the Chapter 6 Case Study Questions:
Case 6-1.
1. As gear As forced to rotate counterclockwise, nating gear B wil rotate clockwise.
‘Mating, external, gears will always rotate in opposite dir
2.
3
4
8.
9.
As gear A rotates counterclockwise, stud pin C rotates clockwise.
‘As gear A rotates counterclockwise, lever D, which pivots about its right end,
‘oscillates rotationally.
Link D drives link E, in a rocker-rocker configuration. Since link E is shorter than
link D, link E wall have a larger throw angle and link D.
Link D will be placed ints lowest extreme position when the center ofthe slot is.
tangent to the bottom of the path of pin C.
Link D will be placed in ts upper extreme position when the center ofthe slot is
tangent to the top ofthe path of pin C.
It takes approximately 210 degrees of gear B to rotate link D (and E) upward and
about 150 degrees to retum downward.
With the imbalance observed in 7, it takes longer to rotate link E up than it takes.
to lower link E.
‘The motion of link E is “quick-retum.”
Case 6-2.
1.
2
3.
As wheel C is reed to rotate counterclockwise, pin D, wich is attached to wheel
, will rotate counterclockwise
As whee! C is forced to rotate counterclockwise ink G, which pivots at its
‘middle, oscillates rotationally
‘The left portion of ink G will be placed in its upper extreme position when the
center of the slot is tangent to the bottom of the path of pin D.
‘The let portion of ink D will be placed in its lowest extreme position when the
center of the slot is tangent to the top ofthe path of pin D.
Te takes approximately 200 degrees of wheel C 0 rotat link G counterclockwise
and about 160 degrees to return.
Te takes approximately 25% more time to rotate link G counterclockwise than it
takes to rotate link G clockwise.
The cyclical motion of link N is rotational oscilation. Et exhibits relatively slow
‘counterclockwise rotation, and a relatively quick clockwise rotation,
‘The cyclical motion of table Ris translational oscillation. It exhibits relatively slow
movement tothe left, and relatively quick movement 10 the right,
‘The mechanism creates quick return motion, moving the workpiece quickly
ough a grinding pass.
“The screw threads can be used to adjust the position of the translations oscillation
of the table
“The mechanism has eight inks, eight pin joints and two sliding joints, giving one
degree of freedom,
ae Case 6:
Cc 1. Since link E is riding in slot J, disk F is driven counterclockwise.
2. At the instant shown, link E and disk Fare rotating counterclockwise. This motion
is pulling strap G towards the left.
3. At the instant shown, ink E and disk F are rotating counterclockwise. This motion
is pulling strap G and slide A towards the lef.
4, As lik E contacts the ramped pad M, it disengages from disk F. The compressed
spring N is allowed to expand
5. As ink E contacts the ramped pad M, it continues to rotate counterclockwise, but
is raised off of dik F.
6. Aslink E contacts the ramped pad M, it disengages from disk F. The energy stored
in the compressed spring N drives disk F quickly elockwise
7. As link E contacts the ramped pad M, it disengages from disk F. The energy stored
in the compressed spring N drives slide A quickly to the right
8. As link E continues to rotate, it falls of of ramped pad M but slides on top of disk
F. Disk F remains motionless once the energy in spring N is released.
9. Aslink E catches a second slot at K, it again engages disk F. Disk Fis again driven
counterclockwise.
10. The continual motion of side A is translational oscillation, It exhibits relatively
slow motion to the le, then rapid motion to the righ, followed by a short pause
before the cycle is repeated
Cc 1. Since link E is riding in slot J, disk F is driven counterclockwise.
2. At the instant shown, link E and disk Fare rotating counterclockwise. This motion
is pulling strap G towards the left.
3. At the instant shown, ink E and disk F are rotating counterclockwise. This motion
is pulling strap G and slide A towards the lef.
4, As lik E contacts the ramped pad M, it disengages from disk F. The compressed
spring N is allowed to expand
5. As ink E contacts the ramped pad M, it continues to rotate counterclockwise, but
is raised off of dik F.
6. Aslink E contacts the ramped pad M, it disengages from disk F. The energy stored
in the compressed spring N drives disk F quickly elockwise
7. As link E contacts the ramped pad M, it disengages from disk F. The energy stored
in the compressed spring N drives slide A quickly to the right
8. As link E continues to rotate, it falls of of ramped pad M but slides on top of disk
F. Disk F remains motionless once the energy in spring N is released.
9. Aslink E catches a second slot at K, it again engages disk F. Disk Fis again driven
counterclockwise.
10. The continual motion of side A is translational oscillation, It exhibits relatively
slow motion to the le, then rapid motion to the righ, followed by a short pause
before the cycle is repeated
a shat, $ Palo sure
Se Fons . obte
73) qu Zo AR [EST RL e
ON Lei 1%) AR esse “Soy (Seid
A TT RES os Y Et 49,0 192750 pol Th)
EY ICO) len
dre
da aw. 1g ss 4 Ys Hu a
Ello; ed 191% [alu [onan DELA
Les 0-104 BI Hr by, = 200m 261.80 %
TA E Fu EN
Rico AN C2
Fa] we oo Eee sr
to, 00e» (CH) = BRK ese ty
p= npr sn 333%)
PEN Gre I: sis Hobo aed
Gin - (1519 4 24 (e513. D
de = 1203 = lage » Cup
aj
Stage I’ 04t 42
0:5 veSt s=25t"
Sue ZELSS date
Sid
C10,
A E
sal ke
at 428, 5=700 Ko=-IL TD
Se Fons . obte
73) qu Zo AR [EST RL e
ON Lei 1%) AR esse “Soy (Seid
A TT RES os Y Et 49,0 192750 pol Th)
EY ICO) len
dre
da aw. 1g ss 4 Ys Hu a
Ello; ed 191% [alu [onan DELA
Les 0-104 BI Hr by, = 200m 261.80 %
TA E Fu EN
Rico AN C2
Fa] we oo Eee sr
to, 00e» (CH) = BRK ese ty
p= npr sn 333%)
PEN Gre I: sis Hobo aed
Gin - (1519 4 24 (e513. D
de = 1203 = lage » Cup
aj
Stage I’ 04t 42
0:5 veSt s=25t"
Sue ZELSS date
Sid
C10,
A E
sal ke
at 428, 5=700 Ko=-IL TD
mi
[A
me oH
ate v al ite e
EA
ARICA ey,
ak AA ARS A + 2533
UT ser:
Ver a Botto Bhs de,
au, Y
Be gana 12335
ES En A4, Mty Sb u
Arge 12330
Stage 2? Das AR ew +82,
er ie
24
[A
me oH
ate v al ite e
EA
ARICA ey,
ak AA ARS A + 2533
UT ser:
Ver a Botto Bhs de,
au, Y
Be gana 12335
ES En A4, Mty Sb u
Arge 12330
Stage 2? Das AR ew +82,
er ie
24
zum
©
e
EE A
lors, 50" arta At
ARE (OST 125 mm
[Stage 3:
GER Y SO, Mes
Ms 125 Am
Stage 2:
Suns DRA, AR
Be: SD
yess 2D
h re
¿50
ss
na]
a
|
UM] 007300020) 31.1278 Ka
EYE THe Yoo
=D
OS D
o5=D A
El t
[AT 2000 (o): 20.44 cas
DIESER Y 50
oi (D: 320% ATT |
ENESRET EN oe
= Bo
> tan (SSE) MUS
[I-18 ]o,°300pn GS) LATE ca
GANÓ: iR Ñ
ATOM nr
CAES &
Beten (LEP): 0.48"
Ge MY. STE
BAY UD: TOR 87
EC) (à = 14400 VES
CPE Re
Ben RIA - Suge EE
ART
0281." WOE
CET
las: (wa TEL 707
ke 98: 720074 20%
O, Cod 2
oven x
Bs tan (AUS ¿ÓN
©
e
EE A
lors, 50" arta At
ARE (OST 125 mm
[Stage 3:
GER Y SO, Mes
Ms 125 Am
Stage 2:
Suns DRA, AR
Be: SD
yess 2D
h re
¿50
ss
na]
a
|
UM] 007300020) 31.1278 Ka
EYE THe Yoo
=D
OS D
o5=D A
El t
[AT 2000 (o): 20.44 cas
DIESER Y 50
oi (D: 320% ATT |
ENESRET EN oe
= Bo
> tan (SSE) MUS
[I-18 ]o,°300pn GS) LATE ca
GANÓ: iR Ñ
ATOM nr
CAES &
Beten (LEP): 0.48"
Ge MY. STE
BAY UD: TOR 87
EC) (à = 14400 VES
CPE Re
Ben RIA - Suge EE
ART
0281." WOE
CET
las: (wa TEL 707
ke 98: 720074 20%
O, Cod 2
oven x
Bs tan (AUS ¿ÓN
|
ane |
OÙ w= 3000): 31.42%, cu] 2110 0= 300epm Ey: 31.4288 040
> DGSE sa WZ pOr 17
€ at: Goold): ‘SOD, m dés God (9: 1500% Sa
E Fe 2, Oe ESTE
= 5158 iD = ae.
! AS E Fig” Beta (Ya)
IES One Ben “aL
: 2] [Ea Ye Va
ME ves ce 53%
ER so
Menu ON “ re ste Ey}
An: ER
No: On Ag
un E Op, Ons (ad we
3 Oy ES AS
li ET TANT
_ L des ANA Ta
5 Pg [ETE = 3H,
ÿ We Bota Fae 35.0
+ 1.105%
Ga TS
Gp ge Ae 0
SN a) Ong [rast 2804
Cage 135282 ALI LEA ota (Es) +60
Cats as
FI
ane |
OÙ w= 3000): 31.42%, cu] 2110 0= 300epm Ey: 31.4288 040
> DGSE sa WZ pOr 17
€ at: Goold): ‘SOD, m dés God (9: 1500% Sa
E Fe 2, Oe ESTE
= 5158 iD = ae.
! AS E Fig” Beta (Ya)
IES One Ben “aL
: 2] [Ea Ye Va
ME ves ce 53%
ER so
Menu ON “ re ste Ey}
An: ER
No: On Ag
un E Op, Ons (ad we
3 Oy ES AS
li ET TANT
_ L des ANA Ta
5 Pg [ETE = 3H,
ÿ We Bota Fae 35.0
+ 1.105%
Ga TS
Gp ge Ae 0
SN a) Ong [rast 2804
Cage 135282 ALI LEA ota (Es) +60
Cats as
FI
mi
q
Lie
ce
o|
Fr ta = LA
Ken CA: Ver a 4
ara,
Sas db)
$
a
Neg APE
. Gig QAI
QE BBA y
1 Sd0qpa(E)- SLB ew
Mee (SON) Bay Zoo
Ye Ve ee,
VAR
Ve TRAE fe
%
Sa a ddr
E At)
& “ Su
a /\ &
E ams
= SpA |
DAY y
= Den CB) tel
Ves (ON YA = 240.8% Zur |
= 2800) % Ys
ON
20)
x af
y
Za rs, 20. Me
CAE = 410904 777
Au
Ent Vase
M DOH TA
LR Oye,
ij
UE
QE: (ood Ch =20000"% 107
O Thot ASA
> =B\4b1" a
FA ganas | LRM, cod
DICK DEE Terz
SS
Ve Var}
AA 57
$
de eat ae ot,
cfa. FT le
Qc Tag = ye
rg Gas, EN
q
Lie
ce
o|
Fr ta = LA
Ken CA: Ver a 4
ara,
Sas db)
$
a
Neg APE
. Gig QAI
QE BBA y
1 Sd0qpa(E)- SLB ew
Mee (SON) Bay Zoo
Ye Ve ee,
VAR
Ve TRAE fe
%
Sa a ddr
E At)
& “ Su
a /\ &
E ams
= SpA |
DAY y
= Den CB) tel
Ves (ON YA = 240.8% Zur |
= 2800) % Ys
ON
20)
x af
y
Za rs, 20. Me
CAE = 410904 777
Au
Ent Vase
M DOH TA
LR Oye,
ij
UE
QE: (ood Ch =20000"% 107
O Thot ASA
> =B\4b1" a
FA ganas | LRM, cod
DICK DEE Terz
SS
Ve Var}
AA 57
$
de eat ae ot,
cfa. FT le
Qc Tag = ye
rg Gas, EN
feminine
O
. helm.
Tab 04
e324:
ESTAN
AYNA
ALU ET
À de Aur
Va
A ET
Pl
a ozono
= 20.81% us
y VS 2
a LLO
Ne
fps VILA, 4
Aigo AREA
CARTE OY, 0%,
VS
Aya,
Ca
OS
se
737 Eu TS
oe 80 mE, bos fr A
A y
Aa
HD)
eR] Nele
AS di
BES
se
PHL
Vo
Vez 1040 —
Va ne
CNY TS
“MAYA
Re AV,
> 21007 NN
= 02835, co
" FLO TER
VE Vga
drt ®® Ve 2 GANT Ar
À UT
Ve RIS
Adra gra rock,
EAN 7
où (US)
“ ar
Be WR,
10 BA
Goes Oe? ER
O
. helm.
Tab 04
e324:
ESTAN
AYNA
ALU ET
À de Aur
Va
A ET
Pl
a ozono
= 20.81% us
y VS 2
a LLO
Ne
fps VILA, 4
Aigo AREA
CARTE OY, 0%,
VS
Aya,
Ca
OS
se
737 Eu TS
oe 80 mE, bos fr A
A y
Aa
HD)
eR] Nele
AS di
BES
se
PHL
Vo
Vez 1040 —
Va ne
CNY TS
“MAYA
Re AV,
> 21007 NN
= 02835, co
" FLO TER
VE Vga
drt ®® Ve 2 GANT Ar
À UT
Ve RIS
Adra gra rock,
EAN 7
où (US)
“ ar
Be WR,
10 BA
Goes Oe? ER
EE
far EE SEE
E
‘ave 154% Ayre
ARC VE
ee ete,
(9 a os
Ou
= NUT 7835 % reel
Oz AD VEAP
fon ae BR
+3 Pe VHB oo
A DTO
ME Ver
fonc: se
eS 40% ASS
p 535% Sat
b= TAY,
“2K ogy
psa
= DOC ANT
fon db 5372.12 (ES
oe ORY rk, 2
734 71
a Any Mg Sen aay
p \e28720% AA AO ITS
EU ven N agas)
GES N ler
yy betty CS Le e % N
Le eh „ARE
CAS y, = A
“pig hy OFF Lt De y le ù
E em
NEE “Gt an maa | |
NAME
ae: A mal
> UTR
EN
CSTRTE
ae ORT = LU
far EE SEE
E
‘ave 154% Ayre
ARC VE
ee ete,
(9 a os
Ou
= NUT 7835 % reel
Oz AD VEAP
fon ae BR
+3 Pe VHB oo
A DTO
ME Ver
fonc: se
eS 40% ASS
p 535% Sat
b= TAY,
“2K ogy
psa
= DOC ANT
fon db 5372.12 (ES
oe ORY rk, 2
734 71
a Any Mg Sen aay
p \e28720% AA AO ITS
EU ven N agas)
GES N ler
yy betty CS Le e % N
Le eh „ARE
CAS y, = A
“pig hy OFF Lt De y le ù
E em
NEE “Gt an maa | |
NAME
ae: A mal
> UTR
EN
CSTRTE
ae ORT = LU
B
€
Ver re Ve
Va Sy, 57
Va: DAME
RO a e
ie à
Ge o Zur
Mmeasorng?
21 m7
| lye MY, à
AO)
= (042% eeu
Ve (OO NET
No
x
Ne VASTES
L Ye ven sy
Deg Oe as =a} rag
a:
"A Gus
¿E ELISA
0% 08 (Md +300% aN
EEE Anar
Che (at: MALE VER
& ae 783.50"
A aan cc
PRES RES
1207
From CAD:
138 | dor AN IA
WLAN ay’
toy V4 = sag
Amok ar Hah va ot ES
Ce A = re
Ro
MOST = 47.12% 4
ie
la
A
Wes 2.11% ARIS
A '
MS
E Mar
Ot: bil «200127 Gon
Fo Doe a TEA A
Ge Gar = 2.028% cu
de Gor, 215.4 NH, ce
445 20rpm( Tas)
ay : PON,
SE
de Ve Veer,
e Bron CAS
aa Nas 384% 505%
MUSA age
ande RO
CN
ee 1S
> (AY ee
252% pus
LU
ES mes A
aay Ge
boxe (248
&u> LIV = SS]
N ac
‚Ast
on CA:
Red: Geddes
fr
wre, m
ES
€
Ver re Ve
Va Sy, 57
Va: DAME
RO a e
ie à
Ge o Zur
Mmeasorng?
21 m7
| lye MY, à
AO)
= (042% eeu
Ve (OO NET
No
x
Ne VASTES
L Ye ven sy
Deg Oe as =a} rag
a:
"A Gus
¿E ELISA
0% 08 (Md +300% aN
EEE Anar
Che (at: MALE VER
& ae 783.50"
A aan cc
PRES RES
1207
From CAD:
138 | dor AN IA
WLAN ay’
toy V4 = sag
Amok ar Hah va ot ES
Ce A = re
Ro
MOST = 47.12% 4
ie
la
A
Wes 2.11% ARIS
A '
MS
E Mar
Ot: bil «200127 Gon
Fo Doe a TEA A
Ge Gar = 2.028% cu
de Gor, 215.4 NH, ce
445 20rpm( Tas)
ay : PON,
SE
de Ve Veer,
e Bron CAS
aa Nas 384% 505%
MUSA age
ande RO
CN
ee 1S
> (AY ee
252% pus
LU
ES mes A
aay Ge
boxe (248
&u> LIV = SS]
N ac
‚Ast
on CA:
Red: Geddes
fr
wre, m
ES
‘a
AAN
STR cos
DN le
Ve
DES & Con CR
ea
Ene Gr a gs
SUS foam niet
OS =e Ase
Re ( Vos ae Ass"
ERST LUN
Dm QD: OF MH kur
Le RSS = 400", c
Aus ya, = 2.0814, Coe
Ar: Varga (7/30)
Las cous
(1.288)
ei
tn me ler
MN
AA Vedder
% Be
AE y Ayes
A, EH mes
Fm bib: OF WS ~ 807
FR = BBM, cu
dus CRAVE = ho, CO
Taq 3
2.
= A
SP: SLR ge se
= 180000 (TA 218849 0% tw
MENO = Hea leo?
VERS Ve » = Vo
A i
> ASH
ake AN
RL
Ve 23310 —7
Ma + Va
Sia TBM Sm LD
Va 33.20% ST
a. a Va
mo Les.
ER 4000"
20 ZA
>
Del 20330 +24000coskO
LOMA OCs TAL
= OL] sa 10 + WOO sind.
° AD sinh 54-08, S14 18
Oke —10S1l,
SOS TEMA
de: se
AAN
STR cos
DN le
Ve
DES & Con CR
ea
Ene Gr a gs
SUS foam niet
OS =e Ase
Re ( Vos ae Ass"
ERST LUN
Dm QD: OF MH kur
Le RSS = 400", c
Aus ya, = 2.0814, Coe
Ar: Varga (7/30)
Las cous
(1.288)
ei
tn me ler
MN
AA Vedder
% Be
AE y Ayes
A, EH mes
Fm bib: OF WS ~ 807
FR = BBM, cu
dus CRAVE = ho, CO
Taq 3
2.
= A
SP: SLR ge se
= 180000 (TA 218849 0% tw
MENO = Hea leo?
VERS Ve » = Vo
A i
> ASH
ake AN
RL
Ve 23310 —7
Ma + Va
Sia TBM Sm LD
Va 33.20% ST
a. a Va
mo Les.
ER 4000"
20 ZA
>
Del 20330 +24000coskO
LOMA OCs TAL
= OL] sa 10 + WOO sind.
° AD sinh 54-08, S14 18
Oke —10S1l,
SOS TEMA
de: se
ie oe ORAL ane Zar ~
AS OE = Gog(= 97 ZEN
© % Ss. gh 30 SN
Ti o SET» ns TES
| eu «sn:
le DEM Of, Zur og 4
FAR CS ot.
Ve Ruda ha CeO Ol, Cg
Ve Van Yay STE ER Pe 102Messto ~Aeos30
RS
Vig PASS ag:
16 ARE
Ve. 5 > Ve HAB 518 ST His MM
TON E Aa Ke Y
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=180- (154d) = qu.
PEACE
4
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Eu
uf
|
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LESE ie
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Our ZADB BIG = bb,
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mst qe ELISE
13
Assume
ors Wt
ON ak = ays + OO,
heer $02 eosubid-+ of eos 23.4 = “1295 LOS TS
AB BES ART 05 34:94 = af, Cs 5.07
Vert > -8.0L sib. WA QE Sie DLL 1890 Sm AS
+2bsn!S=,11151n8 464 Of sin $5.04
= 20m) = ORY, Cu
Ve=@ 044 (3.29) = ROG NIP
OST
Ven a
Ne WS SLA
N 3.00% BER
Vee 1,20% W108
La es (RE) ss eg Ba ing AZ :
: aN gm Tae Zeit
=180- (154d) = qu.
PEACE
4
RD
|LAD8= 188 =(£4,3149b.
162: LCBD+ AABD 19"
Eu
uf
|
41 Ab: [aris = 322887 LETTER CHa) = 125TH, Cow
LESE ie
LABD= sun (EEE sinat.d)=
Our ZADB BIG = bb,
OST ATED 4.
OE= HD ALIN. £238
SD 15,080 7 ZU
Ve Vas aye
due Ke 550 cou
1S Bb By,
ORD PFR- SEN)
. Ya. Mi F |
2 Sap? REST CES
eos 2<35714" is
4131" eh nary ree Sener aL
mst qe ELISE
13
Assume
ors Wt
ON ak = ays + OO,
heer $02 eosubid-+ of eos 23.4 = “1295 LOS TS
AB BES ART 05 34:94 = af, Cs 5.07
Vert > -8.0L sib. WA QE Sie DLL 1890 Sm AS
+2bsn!S=,11151n8 464 Of sin $5.04
= 20m) = ORY, Cu
Ve=@ 044 (3.29) = ROG NIP
OST
Ven a
Ne WS SLA
N 3.00% BER
Vee 1,20% W108
OS
ate Gode Bed Bo Mi,
08 WEY 180% 37 dO
ere DANE WA, of OMY, says
Pu: OCIA EZ Cin? ORAR
SH (GA, Of INTA LOS, oe SULLY
PAM TZ, Ope TUE DES»
dar ao md co
ate Gode Bed Bo Mi,
08 WEY 180% 37 dO
ere DANE WA, of OMY, says
Pu: OCIA EZ Cin? ORAR
SH (GA, Of INTA LOS, oe SULLY
PAM TZ, Ope TUE DES»
dar ao md co
O
Engineers Computaion Pad
No. 997 011€
2 STAEDTLER”
(8)
7-5) As BUS LAM, cw
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le Me Ve = a
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AN
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Engineers Computaion Pad
No. 997 011€
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le Me Ve = a
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%
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Answers to the Chapter 7 Case Study Questions:
Case 7-1.
1. As crankshaft I rotates 30 degrees from the postion shown, slide J will move upward.
2. As crankshaft I rotates a few more than 30 degrees, he crank will lose contact with
tab K, and J wil fall
3. Component C serves to catch tab G and prevents slide J from falling toa lowest
position.
4. After approximately 270 degrees of crankshaft rotation, crank D will contact fever C
1 release slide J. Side J will encounter a fre-fall.
5. Pin B serves asa stop, determining a lowest position for slide J
6. The mechanism raises slide Jin a controlled manner. Slide J will ten fll slightly and
remain an upward position for a set length of time. Eventually slide J will be released
and free-fall o a lower position and remain there fr a shor time.
Case 72,
1. As the rotating machine table tums counterclockwise, component G pushes roller F
causing a clockwise rotation of ink E.
2. Spring K provides resistance for clockwise rotation of ink E, main
between Fand 6.
3. As he rotating machine table tums counterlockwise an link E rotates clockwise,
Link D is pled towards the right
4. Spring L create a force that tends to reduce the angle betwen link D and E.
5. The hood on D and groove on is called a ratchet and pawl arrangement. As the
ratchet wheel tums, the paw als into the groove allowing rotation is single
direction (lockwis in this cas),
6. Item His a ball detent mechanism. I is used to hold wheel Bin fixed positon, until
sufficient force is applied to overcome the spring loaded ball.
7. Asthe rotary machine table rotates, bel B advanes clockwise one position,
releasing a rive int an assembly machine.
contact
Case 73.
1. As link J rotates 90 degrees clockwise from the position shown, bellerank H will
rotate counterclockwise.
2. As link J rotates 90 degrees clockwise from the position shown, pusher E and plate C
move toward the lft
3. As link J rotates 90 degrees clockwise from the position shown, pin S will also move
toward the left.
4. As link J rotates 90 degrees clockwise from the position shown, guide pin R will
move along the groove, left and upward.
5. As link] rotates 90 degrees clockwise from the position shown, bellerank P moves.
toward the left and rotates counterclockwise.
Case 7-1.
1. As crankshaft I rotates 30 degrees from the postion shown, slide J will move upward.
2. As crankshaft I rotates a few more than 30 degrees, he crank will lose contact with
tab K, and J wil fall
3. Component C serves to catch tab G and prevents slide J from falling toa lowest
position.
4. After approximately 270 degrees of crankshaft rotation, crank D will contact fever C
1 release slide J. Side J will encounter a fre-fall.
5. Pin B serves asa stop, determining a lowest position for slide J
6. The mechanism raises slide Jin a controlled manner. Slide J will ten fll slightly and
remain an upward position for a set length of time. Eventually slide J will be released
and free-fall o a lower position and remain there fr a shor time.
Case 72,
1. As the rotating machine table tums counterclockwise, component G pushes roller F
causing a clockwise rotation of ink E.
2. Spring K provides resistance for clockwise rotation of ink E, main
between Fand 6.
3. As he rotating machine table tums counterlockwise an link E rotates clockwise,
Link D is pled towards the right
4. Spring L create a force that tends to reduce the angle betwen link D and E.
5. The hood on D and groove on is called a ratchet and pawl arrangement. As the
ratchet wheel tums, the paw als into the groove allowing rotation is single
direction (lockwis in this cas),
6. Item His a ball detent mechanism. I is used to hold wheel Bin fixed positon, until
sufficient force is applied to overcome the spring loaded ball.
7. Asthe rotary machine table rotates, bel B advanes clockwise one position,
releasing a rive int an assembly machine.
contact
Case 73.
1. As link J rotates 90 degrees clockwise from the position shown, bellerank H will
rotate counterclockwise.
2. As link J rotates 90 degrees clockwise from the position shown, pusher E and plate C
move toward the lft
3. As link J rotates 90 degrees clockwise from the position shown, pin S will also move
toward the left.
4. As link J rotates 90 degrees clockwise from the position shown, guide pin R will
move along the groove, left and upward.
5. As link] rotates 90 degrees clockwise from the position shown, bellerank P moves.
toward the left and rotates counterclockwise.
3. As link J rotates 90 degrees clockwise from the position shown, slide Land plate M
‘will stay nearly stationary asthe motion due 1 the translation ofS is counterbalanced
by the rotation of P. Further rotation of the link J will move he slide L toward the
left as the profile ofthe groove changes and imparts a clockwise rotation of bellerank
P.
‘A connecting link N is required as link P rotates and drives slide L, which is
constrained to horizontal sliding.
As link J rotates 90 degrees clockwise from the position shown, slide M and C will be
nearly parallel.
‘The continual motion ofthe device will move plates M and C together on the left side
of channel B, and move plate C alone on the right side of the channel while pte M
remains nearly stationary.
‘will stay nearly stationary asthe motion due 1 the translation ofS is counterbalanced
by the rotation of P. Further rotation of the link J will move he slide L toward the
left as the profile ofthe groove changes and imparts a clockwise rotation of bellerank
P.
‘A connecting link N is required as link P rotates and drives slide L, which is
constrained to horizontal sliding.
As link J rotates 90 degrees clockwise from the position shown, slide M and C will be
nearly parallel.
‘The continual motion ofthe device will move plates M and C together on the left side
of channel B, and move plate C alone on the right side of the channel while pte M
remains nearly stationary.
Chapter 8 | |
8-1 Crank Ang] Theias] Gamma] _ Sider DES Diepi | Som
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SLIDER-CRANK KINEMATICS (length, de, engtvsec, radisec)
von crak oot + Coupler * + er ses
Angle Speed Accel. Angle Speed Accel’ Lengh Speed
0 40 00 32 057 05 9 17
30 40 00 73 050 LE SRG 16
G 40 où 1042 02% 20 SSA 22
% 40 00 NS 000 1% 514 300
m 40 00 mn 0% 20 479 ar
10 40 00 73 0 LR 456 164
10 400 On 328 037 om 49 17
20 4m O0 0% 04 415 460 154
200 400 00 SM 02% 1 46 270
m 4m 00 4% 00 323 300
wo 40 00 38 02 #1 230
30 40 00 om 04 50 M46 18
360 40 00 328 057 yo 17 A
Maximum 00 115 06 23 59 300 1028
Minimum 00 49 06 23 M9 300 BBS
8410.
SLIDER-CRANK KINEMATICS (length, de, lengiísec, rad/see)
Crank se 1 Couple 8 Pr
Angle Speed Accel. Angie Speed Al Leng Spoul Ati
O 1000 (D TS TE
30 1051 100 954 165 832 91 3502 “11653,
WO 100 m 0 WG STR 724 0
D 1146 100 479 000 26 S076 -10776 I
m HO 100 HT LI DM 88 24
130 DM 100 JS I 4163 90 8588
mo 127 109 a 22% 220 223 17 Im
20 BIS 0 076 20-40 MIS 630 za
20 DS 100 A 122 M 4764 140 705
m 94 100 50 522 1301 948
mu 100432 os 116 «11038
300 1 100 0 ms 6735-19011
30 1502 100437 ns 1079 “MR
Maximum 100 MA 23 M3 175 Ina 124698
Minimum 100 59 (23 350 4295 -11028 19108
Transnission
Angle
167
1326
1096
785
#6
26
33
ya
362
51
1162
1492
Asa
167
62
Angie
1356
103
1065
752
405
ns
SLIDER-CRANK KINEMATICS (length, de, engtvsec, radisec)
von crak oot + Coupler * + er ses
Angle Speed Accel. Angle Speed Accel’ Lengh Speed
0 40 00 32 057 05 9 17
30 40 00 73 050 LE SRG 16
G 40 où 1042 02% 20 SSA 22
% 40 00 NS 000 1% 514 300
m 40 00 mn 0% 20 479 ar
10 40 00 73 0 LR 456 164
10 400 On 328 037 om 49 17
20 4m O0 0% 04 415 460 154
200 400 00 SM 02% 1 46 270
m 4m 00 4% 00 323 300
wo 40 00 38 02 #1 230
30 40 00 om 04 50 M46 18
360 40 00 328 057 yo 17 A
Maximum 00 115 06 23 59 300 1028
Minimum 00 49 06 23 M9 300 BBS
8410.
SLIDER-CRANK KINEMATICS (length, de, lengiísec, rad/see)
Crank se 1 Couple 8 Pr
Angle Speed Accel. Angie Speed Al Leng Spoul Ati
O 1000 (D TS TE
30 1051 100 954 165 832 91 3502 “11653,
WO 100 m 0 WG STR 724 0
D 1146 100 479 000 26 S076 -10776 I
m HO 100 HT LI DM 88 24
130 DM 100 JS I 4163 90 8588
mo 127 109 a 22% 220 223 17 Im
20 BIS 0 076 20-40 MIS 630 za
20 DS 100 A 122 M 4764 140 705
m 94 100 50 522 1301 948
mu 100432 os 116 «11038
300 1 100 0 ms 6735-19011
30 1502 100437 ns 1079 “MR
Maximum 100 MA 23 M3 175 Ina 124698
Minimum 100 59 (23 350 4295 -11028 19108
Transnission
Angle
167
1326
1096
785
#6
26
33
ya
362
51
1162
1492
Asa
167
62
Angie
1356
103
1065
752
405
ns
CM
eue Crank
Angle Speed
o 1800
20 1500
@ 1500
91500
mo 1500
150 1500
10 1500
210 1500
240 1500
2m 1500
50 100
50 1500
56 1500
Maximum
Minima
en.
Maximum
Minimum,
FOUR BAR KINEMATICS (deg,radsec)
e copier oo
Angle
1:
25
256
267
257
m2
202
ma
153
ui
10
154
155
267
Mo
Spent
ss
29
us
vo
va
19
157
won Follower +
Angie Spar
251 08
ar a
ar 32
ns 04
0. 49
CRE
5 150
#7 96
199 317
459 61
#1 498
a 04
ast cas
2. us
27 ps
FOUR BAR KINEMATICS (seg, adísec
200
ns
ss
mo
#70
Palm aus
Ange Spa
150 15420
ma 1604
bo a
204 16510
no mn
55 mm
ne 62
59 4195
ws 15007
768 m
m 20026
0 25534
180 su
CET
HA 2605
Transmision
“Angle
164
wi
559
ous
m3
ns
ns
ns
ms
on
359
on
4
ms
14
Transmission
Angle
as
302
466
er
on
383
20
ws
a
os
466
302
8
920
as
eue Crank
Angle Speed
o 1800
20 1500
@ 1500
91500
mo 1500
150 1500
10 1500
210 1500
240 1500
2m 1500
50 100
50 1500
56 1500
Maximum
Minima
en.
Maximum
Minimum,
FOUR BAR KINEMATICS (deg,radsec)
e copier oo
Angle
1:
25
256
267
257
m2
202
ma
153
ui
10
154
155
267
Mo
Spent
ss
29
us
vo
va
19
157
won Follower +
Angie Spar
251 08
ar a
ar 32
ns 04
0. 49
CRE
5 150
#7 96
199 317
459 61
#1 498
a 04
ast cas
2. us
27 ps
FOUR BAR KINEMATICS (seg, adísec
200
ns
ss
mo
#70
Palm aus
Ange Spa
150 15420
ma 1604
bo a
204 16510
no mn
55 mm
ne 62
59 4195
ws 15007
768 m
m 20026
0 25534
180 su
CET
HA 2605
Transmision
“Angle
164
wi
559
ous
m3
ns
ns
ns
ms
on
359
on
4
ms
14
Transmission
Angle
as
302
466
er
on
383
20
ws
a
os
466
302
8
920
as
Answers fo the Chapter 8 Case Study Questions:
Case 8-1
1. As slide bar Eis pulled to the lef, link D rotates clockwise.
2. As slide bar E is pulled tothe left, side block I moves upward
3. As pulley J is rotated, being eccentric from the center of rotates crank pin C will move
along a circular path
‘As pulley J rotates, slide bar E will rotate about it's about axis of shaft A.
Moving slide bar E tothe left will increase the eccentricity of rank pin C. Crank pin
€ will move along a circular path of a larger radius.
If sleeve Fis keyed to the housing, it remains stationary encasing shaft A.
Item G is an enveloping gear that mates with a worm gear.
As item G rotates, the threads thrust sleeve F either left or right.
As item G rotate, slide bar Eis moved either left or right, depending on the direction
of rotation of G.
10. This mechanism can be used to adjust the erank length, which the shafts spinning.
Case 8-1
1. As slide bar Eis pulled to the lef, link D rotates clockwise.
2. As slide bar E is pulled tothe left, side block I moves upward
3. As pulley J is rotated, being eccentric from the center of rotates crank pin C will move
along a circular path
‘As pulley J rotates, slide bar E will rotate about it's about axis of shaft A.
Moving slide bar E tothe left will increase the eccentricity of rank pin C. Crank pin
€ will move along a circular path of a larger radius.
If sleeve Fis keyed to the housing, it remains stationary encasing shaft A.
Item G is an enveloping gear that mates with a worm gear.
As item G rotates, the threads thrust sleeve F either left or right.
As item G rotate, slide bar Eis moved either left or right, depending on the direction
of rotation of G.
10. This mechanism can be used to adjust the erank length, which the shafts spinning.
| |
Liam Se Soar = EBA DATE
Jm 63
I
| 9
30. 35 ung )
ı ı | y (do)
ptt me
y EJ Gan = BE TT SUR = Weer
4 TS
[E
o
ES
J
ra
=
E
Liam Se Soar = EBA DATE
Jm 63
I
| 9
30. 35 ung )
ı ı | y (do)
ptt me
y EJ Gan = BE TT SUR = Weer
4 TS
[E
o
ES
J
ra
=
E
® = ime
FAY Sms = TRY 227%), = 13.0 rpm |
AT Gens DEE: ones Alarm
FAY Sms = TRY 227%), = 13.0 rpm |
AT Gens DEE: ones Alarm
za
ra
W306 = Hho EE = ER
Zep om
ra
W306 = Hho EE = ER
Zep om
a]
FEE ous
Tess “tive, mx Vel,
wit oveor then:
Noa Yr = Y
Oman > 00
5%
ar
im
0
0000
0467
033
0500
0887
0833
1000
1000
1000
1000
1000
1.000
0750
‘0.500
0.250
000
0.000
Since Fall OR va
less tame, mar vel
will occur Shen
Viney Mrz 18/2
YA z
Follows Diplcamet a)
ar
(i)
0
0000
0168
0375
0563
0750
0750
0750
0750
0750
0750
0563
0375
0.188
0000 *
0000
0000
0000
FEE ous
Tess “tive, mx Vel,
wit oveor then:
Noa Yr = Y
Oman > 00
5%
ar
im
0
0000
0467
033
0500
0887
0833
1000
1000
1000
1000
1000
1.000
0750
‘0.500
0.250
000
0.000
Since Fall OR va
less tame, mar vel
will occur Shen
Viney Mrz 18/2
YA z
Follows Diplcamet a)
ar
(i)
0
0000
0168
0375
0563
0750
0750
0750
0750
0750
0750
0563
0375
0.188
0000 *
0000
0000
0000
O
9-16
Since Lol Orcos ín less time,
man Vel 4 aceel Wil occur then
bey? 245 Y ge AL
un AY Ae
ET
Both Lol sequences ace.
identico) + exar mox
vel + gore)
roy rs ad
Ae
Oman? RENTE
ENS
188
Bash
9-16
Since Lol Orcos ín less time,
man Vel 4 aceel Wil occur then
bey? 245 Y ge AL
un AY Ae
ET
Both Lol sequences ace.
identico) + exar mox
vel + gore)
roy rs ad
Ae
Oman? RENTE
ENS
188
Bash
y
JL
z
|
9-14
Max Vel taceel ur
In the Final Gall sequence.
Mana y
nas SET
or ipisament
dit)
JL
z
|
9-14
Max Vel taceel ur
In the Final Gall sequence.
Mana y
nas SET
or ipisament
dit)
920
Re a een
Van ETES
23.1
Quay W* Narr = y
24.81%
Moe POY aus
9-21
pe eos in = =
Vase" Yor = HY (ands ==.
= 128599 Soto
Re a een
Van ETES
23.1
Quay W* Narr = y
24.81%
Moe POY aus
9-21
pe eos in = =
Vase" Yor = HY (ands ==.
= 128599 Soto
|
422
Mex veloc 15 dunng
@ | Leah sequence
Vinay 2 UY see
=2%%
Omas 27s = 2 Hu
= LR
a ee
Cy
Max velocity 1s doing
[rise sequence
Vinay = Y > Ue
=1.43 1%
nos 2e Yu
= Ae
Follow pace
422
Mex veloc 15 dunng
@ | Leah sequence
Vinay 2 UY see
=2%%
Omas 27s = 2 Hu
= LR
a ee
Cy
Max velocity 1s doing
[rise sequence
Vinay = Y > Ue
=1.43 1%
nos 2e Yu
= Ae
Follow pace
Ce
8
8
3)
ha
ha
ES
=
ie
Mie
a
IS)"
N |
RE
N
=: =
=
ie
Mie
a
IS)"
N |
RE
N
=: =
pe]
E
ES
E
ES
4
1-20
1-20
9-62
Re e.
zu
a (sin bd = 38,
P= 180- Was 2
SFL Do
85 mm
ET B= sist [BER sin Le 204
born) "028%; ia + Ton (E) 21. 33°64,
E
w
7 SS AA El Dean ue HUE
] BD o nn tai dr ERY Seas ud LESC
SOBA Year” = rea lo BUN EN
Ey, 05
©
was rl): gan
eet (au) cos (ase-nd= same
barsa (Sas
5492-389)
* sa
Re e.
zu
a (sin bd = 38,
P= 180- Was 2
SFL Do
85 mm
ET B= sist [BER sin Le 204
born) "028%; ia + Ton (E) 21. 33°64,
E
w
7 SS AA El Dean ue HUE
] BD o nn tai dr ERY Seas ud LESC
SOBA Year” = rea lo BUN EN
Ey, 05
©
was rl): gan
eet (au) cos (ase-nd= same
barsa (Sas
5492-389)
* sa
Answers to the Chapter 9 Case Study Questions:
Case 9-1.
1. As shaft G turns clockwise, item E moves upward,
2. Connection between item E and Fis a rack and pinion gear joint.
3. Counterweight K balances the attack of papers sitting on item J.
4. As paper is used, the weight of the stack decreased. The radius of item H changes to
reduce the moment, and hence the force countering the weight ofthe papers.
5. A pin is used to join shaft G with cam H, constraining both to rotate together.
6. For smaller stacks of paper, springs are used as counterbalances. They share the
‘common feature of reducing the supporting force at lower deflections.
Case 92.
LAS cam D rotates clockwise, the immediate motion of link B will be clockwise
rotation, followed by a dwell,
‘Cam D isa plate cam.
Item C is a roller mounted on a pivoted follower.
lem F isa spring.
Spring F that attempts to maintain alignment of links A and B.
Item Bis a screw.
‘Screw E serves as a stop to limit the motion of item A.
Item Bis a pivoted follower. It moves together with item A, until stop E is reached,
‘When A is in contact with the stop, the pivot point for link B changes and follower C
traces a shortened radius
9. Lengthening serew E will cause the pivot point to move from the bottom of link A to
the top of link A for a greater portion ofthe cycle.
10, Shortening screw E will cause the pivot point to move from the bottom of link A to
the top of link A for a reduced portion ofthe cycle.
Case 9-3.
1. As rod C stats to slide downward, cam E rotates counterclockwise.
2. As rod C starts to side downward and cam E rotates counterclockwise, plunger H is
forced downward.
3. The strip of sheet metal clamped at W is contacted by plunger H and formed into an
inverted channel shape (LJ )
4. As rod C continues to slide downward, follower proceeds P past the lobe of eam E,
‘and plunger H raises,
5. As rod C continues to slide downward, the angled surfaces D contact slides Land
force them inward,
6. Asslides I move inward, they contact the sides of the steel stip (already formed as an
inverted channel) and bend them over (€)
7. As rod C starts to move upward, he slides retum tothe position shown, and follower
P once again rides over the lobe of cam E, The plunger H again strikes the top ofthe
steel, forming an open box profile over (==).
Case 9-1.
1. As shaft G turns clockwise, item E moves upward,
2. Connection between item E and Fis a rack and pinion gear joint.
3. Counterweight K balances the attack of papers sitting on item J.
4. As paper is used, the weight of the stack decreased. The radius of item H changes to
reduce the moment, and hence the force countering the weight ofthe papers.
5. A pin is used to join shaft G with cam H, constraining both to rotate together.
6. For smaller stacks of paper, springs are used as counterbalances. They share the
‘common feature of reducing the supporting force at lower deflections.
Case 92.
LAS cam D rotates clockwise, the immediate motion of link B will be clockwise
rotation, followed by a dwell,
‘Cam D isa plate cam.
Item C is a roller mounted on a pivoted follower.
lem F isa spring.
Spring F that attempts to maintain alignment of links A and B.
Item Bis a screw.
‘Screw E serves as a stop to limit the motion of item A.
Item Bis a pivoted follower. It moves together with item A, until stop E is reached,
‘When A is in contact with the stop, the pivot point for link B changes and follower C
traces a shortened radius
9. Lengthening serew E will cause the pivot point to move from the bottom of link A to
the top of link A for a greater portion ofthe cycle.
10, Shortening screw E will cause the pivot point to move from the bottom of link A to
the top of link A for a reduced portion ofthe cycle.
Case 9-3.
1. As rod C stats to slide downward, cam E rotates counterclockwise.
2. As rod C starts to side downward and cam E rotates counterclockwise, plunger H is
forced downward.
3. The strip of sheet metal clamped at W is contacted by plunger H and formed into an
inverted channel shape (LJ )
4. As rod C continues to slide downward, follower proceeds P past the lobe of eam E,
‘and plunger H raises,
5. As rod C continues to slide downward, the angled surfaces D contact slides Land
force them inward,
6. Asslides I move inward, they contact the sides of the steel stip (already formed as an
inverted channel) and bend them over (€)
7. As rod C starts to move upward, he slides retum tothe position shown, and follower
P once again rides over the lobe of cam E, The plunger H again strikes the top ofthe
steel, forming an open box profile over (==).
8. The mechanism converts a strip of steel ino a box profile with a downward and
upward stroke of rod C.
9. The springs contacting slides ae return springs. They return the slides to the
position shown when not being contacted by the angled surfaces D.
10. Spring under item K give the base some compliance. As the strip steel W is fist
‘contacted, a single material thickness is tapped between Hand K. On the second.
strike stroke, two material thicknesses are trapped between Hand K.
11. The linear reciprocation of C can be actuated by a slider-crank mechanism, another
cam system, a hydraulic eylinder, along with a variety of other linkages and
mechanisms.
Case 94,
1. As gear K rotates clockwise, lnk F (which sa its limiting rightward posstion)
rotates clockwise.
inks J and F re configured as shape-crank mechanism and exhibit qiek-retum
motion.
3. A gear K rotates elockwise, he motion of slide D is slowly to the right and a quick
return to the ef
4. As gear K rotates clockwise, gearN rotates counterclockwise,
5. As pear K rotates clockwise link Q does not move fr the next 180 degrees of
xotaion, then rotates counterclockwise.
Component Pisa plate cam.
Item V is constrained to vertical translation.
Link Q and link V are coupled with a connecting link,
‘As gear K rotates clockwise the workpiece X is sid 1 the righ, then lifted upward.
‘Timing is controlled since all motion is achieved with a single actuator.
10, Pick and place mechanisms are used throughout a manufacturing facility.
upward stroke of rod C.
9. The springs contacting slides ae return springs. They return the slides to the
position shown when not being contacted by the angled surfaces D.
10. Spring under item K give the base some compliance. As the strip steel W is fist
‘contacted, a single material thickness is tapped between Hand K. On the second.
strike stroke, two material thicknesses are trapped between Hand K.
11. The linear reciprocation of C can be actuated by a slider-crank mechanism, another
cam system, a hydraulic eylinder, along with a variety of other linkages and
mechanisms.
Case 94,
1. As gear K rotates clockwise, lnk F (which sa its limiting rightward posstion)
rotates clockwise.
inks J and F re configured as shape-crank mechanism and exhibit qiek-retum
motion.
3. A gear K rotates elockwise, he motion of slide D is slowly to the right and a quick
return to the ef
4. As gear K rotates clockwise, gearN rotates counterclockwise,
5. As pear K rotates clockwise link Q does not move fr the next 180 degrees of
xotaion, then rotates counterclockwise.
Component Pisa plate cam.
Item V is constrained to vertical translation.
Link Q and link V are coupled with a connecting link,
‘As gear K rotates clockwise the workpiece X is sid 1 the righ, then lifted upward.
‘Timing is controlled since all motion is achieved with a single actuator.
10, Pick and place mechanisms are used throughout a manufacturing facility.
= dub sd = Min
a as A dea Nun
ee drets» Va
de ve née any
wi ed du d-2he 5.86%
pe Tan Mp= 0393: /
103] a+ Nas Vive 280
durdente 25 «HS 24205,
SOL + dar dr2a= 2.628 spe
LOT ge min == ams |
Le ADO Coded ho =
ey
ABE ro ahs on
Be md =D
NTE ton make wth an oe sd)
LE: ym > Aud dor 2444
ACTE = =D
losJa- E dress ME den a Bets
AS 215 Ce}, E sat
tye TUD, à 7 Pye TD cos aa = 0198
Le 128 > Cod Raho ge; 29/4
Ze rue gone ob an gear
ABE ye Se 015%
BEY Brandt = = 036
CG img a
Be FO) C0825 /ag = 05m
Ls 0, > Comélhos De. 1964
18 ton eames ge (os dy
ren
Cady: Corde = 2406 à Car are = Any
DAT gs ras de Bei AA
ya: 2105
Art Dennis ann
La BULA 3 cb Cube» El 31 87/7
Notes? Bt pri thon q
Abs à . IB:
ETA 001 der
Gaye COCs bla Gest
BA a ss, a: any
ACTOS 125157
10-10
204 de
EE 3G IAN
+ Sige
a as A dea Nun
ee drets» Va
de ve née any
wi ed du d-2he 5.86%
pe Tan Mp= 0393: /
103] a+ Nas Vive 280
durdente 25 «HS 24205,
SOL + dar dr2a= 2.628 spe
LOT ge min == ams |
Le ADO Coded ho =
ey
ABE ro ahs on
Be md =D
NTE ton make wth an oe sd)
LE: ym > Aud dor 2444
ACTE = =D
losJa- E dress ME den a Bets
AS 215 Ce}, E sat
tye TUD, à 7 Pye TD cos aa = 0198
Le 128 > Cod Raho ge; 29/4
Ze rue gone ob an gear
ABE ye Se 015%
BEY Brandt = = 036
CG img a
Be FO) C0825 /ag = 05m
Ls 0, > Comélhos De. 1964
18 ton eames ge (os dy
ren
Cady: Corde = 2406 à Car are = Any
DAT gs ras de Bei AA
ya: 2105
Art Dennis ann
La BULA 3 cb Cube» El 31 87/7
Notes? Bt pri thon q
Abs à . IB:
ETA 001 der
Gaye COCs bla Gest
BA a ss, a: any
ACTOS 125157
10-10
204 de
EE 3G IAN
+ Sige
10-11
pense
SRW diet
D a: zas a Woes
IC 285
JO 4-28: 297 dr Beas
CCD EE "2
FON} cogs Al EC) JDA vo (re) HBB rp coy
3234 Spm eu = ® be
= INT med (M) 155
El Syn Ve uo “sad 26128,
We Gide TE = 138.5 5
E ed. ala cu | Et Loge Bt AM) = 2875 pc
$ ‘23 0.1
wa (as 38 ; a
Ne SSSR)» 15.8 By We NA 90.33% /
MA) vee 5: + 9: Sa,
Cia (ar o aa y,
No Ra N= HD ay das 20.07
Nas 200 Hey
1920| ve= 4: > des 40 dy
C2 395= Edad) > dé
NY
Nae Aie
Ba Ram a, Ba,
1922] yrsd =r a= 44,
vost kr examined
C= 2.086% (9,+84) 242 S ES e de 24
Ne Pad Nils y Az | me: NA
x N wiesen
Ny Bl Naot un nz,
must Toe, examin
BES] ve= 6 > dee bd, loza es > 4.30
C2382 4 (div ba de boy | cede EG dd Pas ys
Could vse Gly | ee die Sige
R= le Would avoid mél “Py =
Neste Nas bk ower topaulf | > 20 eco
fm
pense
SRW diet
D a: zas a Woes
IC 285
JO 4-28: 297 dr Beas
CCD EE "2
FON} cogs Al EC) JDA vo (re) HBB rp coy
3234 Spm eu = ® be
= INT med (M) 155
El Syn Ve uo “sad 26128,
We Gide TE = 138.5 5
E ed. ala cu | Et Loge Bt AM) = 2875 pc
$ ‘23 0.1
wa (as 38 ; a
Ne SSSR)» 15.8 By We NA 90.33% /
MA) vee 5: + 9: Sa,
Cia (ar o aa y,
No Ra N= HD ay das 20.07
Nas 200 Hey
1920| ve= 4: > des 40 dy
C2 395= Edad) > dé
NY
Nae Aie
Ba Ram a, Ba,
1922] yrsd =r a= 44,
vost kr examined
C= 2.086% (9,+84) 242 S ES e de 24
Ne Pad Nils y Az | me: NA
x N wiesen
Ny Bl Naot un nz,
must Toe, examin
BES] ve= 6 > dee bd, loza es > 4.30
C2382 4 (div ba de boy | cede EG dd Pas ys
Could vse Gly | ee die Sige
R= le Would avoid mél “Py =
Neste Nas bk ower topaulf | > 20 eco
fm
|
CANA
ep AN 2 1SDBrpm—
EXA
10-21] From table 15
Bhp Ai Dom Pa > By
16:28] Fron ae 1.3
Bip À Bree Randy
Me Ges Bes VR 983 = Aa
À Een Na= 120 “ N=8/ Nes 807.
[OS rr tah | x 18 uam -
ae 18 Ne 27 Nas | BOD) = 180 som
CES 453.0, mp de 13.309
Ea ST 10-30] From tbe 95
Une + MSrpn 7 P4=20 0 wee LUS com > he
Ves FB = 3.08 ye: MB. SM
e Nelos Nin ihe ENE NEBR
Po HEC) = 5150 Send] we aye BLM = 22 opm
Ae OBS d24% | de WY dr
AIT From tie 93
EN 10 hp 4 WIS rpm > BB
10-32] From table 9.5
ERA > GID
VR= WS Tre, wee ID 20.03% bar dur bd
cn aa. ÓN) de ECS = (46d)
ae = dee Ei
Nae W 3
re ren n AR arm,
Load bey Stoner Seal are R=
ALA as
N,=20/ Nas 107
Dur 1015/7 > 1.8 om
em Frum table 9.5
A Brom > 4:5
vee EN
C=20= LG al)
Fer das
ND Nites
Nee 8%, = 206 rpm
[OBST as
ESA
dex 24. 2GY ys
> 1303 radios (Loney,
= 2.0 coins
CANA
ep AN 2 1SDBrpm—
EXA
10-21] From table 15
Bhp Ai Dom Pa > By
16:28] Fron ae 1.3
Bip À Bree Randy
Me Ges Bes VR 983 = Aa
À Een Na= 120 “ N=8/ Nes 807.
[OS rr tah | x 18 uam -
ae 18 Ne 27 Nas | BOD) = 180 som
CES 453.0, mp de 13.309
Ea ST 10-30] From tbe 95
Une + MSrpn 7 P4=20 0 wee LUS com > he
Ves FB = 3.08 ye: MB. SM
e Nelos Nin ihe ENE NEBR
Po HEC) = 5150 Send] we aye BLM = 22 opm
Ae OBS d24% | de WY dr
AIT From tie 93
EN 10 hp 4 WIS rpm > BB
10-32] From table 9.5
ERA > GID
VR= WS Tre, wee ID 20.03% bar dur bd
cn aa. ÓN) de ECS = (46d)
ae = dee Ei
Nae W 3
re ren n AR arm,
Load bey Stoner Seal are R=
ALA as
N,=20/ Nas 107
Dur 1015/7 > 1.8 om
em Frum table 9.5
A Brom > 4:5
vee EN
C=20= LG al)
Fer das
ND Nites
Nee 8%, = 206 rpm
[OBST as
ESA
dex 24. 2GY ys
> 1303 radios (Loney,
= 2.0 coins
9-33) qe ras Lol ass
Ela td Lee Abd | As Adj» MENA U
AD «Boa (AER) = Th rave SA y,
CNEA 0-38) 4-8: 15.
ase A = 314 ny
ve oda 002 2g: 26088)
Loa RAS 10,6 ven
ar 10.6 rp
03%) a= 38-2.
Go = Born Gath) = 502 A
Veud= sds
ffo-40 ae he aye
ee
We Weose = Beasts 11.27
Pre post Td cose
ATA costs Me
m werden, nes Ser Has
OAT a EEE
€ a 4
Re eso = 18.87
Rare Enid D
vite ag GR
[ln Stes
REITER
30m
a es
MY
TR
cost = Wen
Nei Nagy Des de
RaQ este Toy
pu ee $2 3.0
Ce U6 aan
Hana gears
SES
tos y = Ma
Con Use
Ne 0 / Naty Pr
VEUT Pye resid
10.607
[sa] yp. By
td, 9.40, 3:15)
he May
fond, acordo
% 15.017
Ela td Lee Abd | As Adj» MENA U
AD «Boa (AER) = Th rave SA y,
CNEA 0-38) 4-8: 15.
ase A = 314 ny
ve oda 002 2g: 26088)
Loa RAS 10,6 ven
ar 10.6 rp
03%) a= 38-2.
Go = Born Gath) = 502 A
Veud= sds
ffo-40 ae he aye
ee
We Weose = Beasts 11.27
Pre post Td cose
ATA costs Me
m werden, nes Ser Has
OAT a EEE
€ a 4
Re eso = 18.87
Rare Enid D
vite ag GR
[ln Stes
REITER
30m
a es
MY
TR
cost = Wen
Nei Nagy Des de
RaQ este Toy
pu ee $2 3.0
Ce U6 aan
Hana gears
SES
tos y = Ma
Con Use
Ne 0 / Naty Pr
VEUT Pye resid
10.607
[sa] yp. By
td, 9.40, 3:15)
he May
fond, acordo
% 15.017
| |
[0-455 yg $-3157 lO=4b) ve- 2-50
Ur or A CUEE)
doz \ 2 ee 10,47
dandy: Yen wen) fonds cts
be 4857 Yer Woy
DAT ve: SR = Lo O48) ye= $800 = mw
Single thread: Er
Nas MBs Some ter u
dl ar + 5258
Ide Be 3295 1 e cdo Alita > en AS ons
ora} ae 20 o o. ¿A (uo) 100 me
[nie hr „82 (eed)
Sein Ardo anal LEE + Za "85 in
varo der ES =D
durées ce 2051.05 ee
E “repas
ke ae. à SE LE eg
CNET E Ey GUY rad)
EL mu 048 à
alo. eN c= ga Se
= 112597
[0-455 yg $-3157 lO=4b) ve- 2-50
Ur or A CUEE)
doz \ 2 ee 10,47
dandy: Yen wen) fonds cts
be 4857 Yer Woy
DAT ve: SR = Lo O48) ye= $800 = mw
Single thread: Er
Nas MBs Some ter u
dl ar + 5258
Ide Be 3295 1 e cdo Alita > en AS ons
ora} ae 20 o o. ¿A (uo) 100 me
[nie hr „82 (eed)
Sein Ardo anal LEE + Za "85 in
varo der ES =D
durées ce 2051.05 ee
E “repas
ke ae. à SE LE eg
CNET E Ey GUY rad)
EL mu 048 à
alo. eN c= ga Se
= 112597
10-56
en
= 244 ran] di
po gan
= 35 Th rpm WA
Los? gale AR ea
= 2520 wy
CS "x. as Ex. oS).
ay
© DU ge Bara a
400 + 8 NN 2 55 Ne |
o ER
recone A Mccces |
In Bee
DADAAEROY th 30 Nilda
"80 oi À AQ (LAGO = Das
\ + Lo' cu
FSSC
Matos" 8 coy
en
= 244 ran] di
po gan
= 35 Th rpm WA
Los? gale AR ea
= 2520 wy
CS "x. as Ex. oS).
ay
© DU ge Bara a
400 + 8 NN 2 55 Ne |
o ER
recone A Mccces |
In Bee
DADAAEROY th 30 Nilda
"80 oi À AQ (LAGO = Das
\ + Lo' cu
FSSC
Matos" 8 coy
Du Na re
r N een
rt À EA
Landes fio Eo
4
BB 0 25
Loan LITE TC) Le »
on pa
dau à
Seat: Rea pur
Mana 251-3.
CH rk
TS
N,
ns
vera
pd NS
AED ce
Ts, Eos 20
Ce Los TOTS
can
sda
A Burg! 251-2009: 0 Miss
ws
LD Wale Walia Ny SBD]
sel ay
sur à
5
woe
ax ou» ren cone
ON = 2200 cpm u
toy (Den) = ZED em u
=
Cd > Bb Cys
lun]
A O]
su, ot = — NER Nee IA
u 35 u x, 2-4-3 &
o Le, 0 AO A a AR
A wy ya
(60) > Born cee a
Ot AM 048 -08 211
SO = 33 rpm of
Brom Cuy
CaN ie co 7
r N een
rt À EA
Landes fio Eo
4
BB 0 25
Loan LITE TC) Le »
on pa
dau à
Seat: Rea pur
Mana 251-3.
CH rk
TS
N,
ns
vera
pd NS
AED ce
Ts, Eos 20
Ce Los TOTS
can
sda
A Burg! 251-2009: 0 Miss
ws
LD Wale Walia Ny SBD]
sel ay
sur à
5
woe
ax ou» ren cone
ON = 2200 cpm u
toy (Den) = ZED em u
=
Cd > Bb Cys
lun]
A O]
su, ot = — NER Nee IA
u 35 u x, 2-4-3 &
o Le, 0 AO A a AR
A wy ya
(60) > Born cee a
Ot AM 048 -08 211
SO = 33 rpm of
Brom Cuy
CaN ie co 7
10-11] Net@}20 NS Wee POM) Nez Hew Was
Ns: 3d N=150
loas] EN
SUR OED BE 1 ein o in
Ml AA AA Meta A A
4 a RBA 4 0 = 6 4 5 =
Une 9.28) 24500 rpm cuyo
oa O = 15 qe ug
(to BRE. (003) > 176g Cu
À (D) 20 ep te
(nd 00 cpm coy
toys Syed} 490 ren ey de Elda > Biome 6
UP Bye NEDO Rm cy (400) = 318 cpm camp
to. OF
ET N ea
Nas 25(B=20 Nyz32
1 2 tm 3 4
MERE E
MH 1 1 4 +
ALO 1 2 ©
Outpst hable wil ot ratte!
Ns: 3d N=150
loas] EN
SUR OED BE 1 ein o in
Ml AA AA Meta A A
4 a RBA 4 0 = 6 4 5 =
Une 9.28) 24500 rpm cuyo
oa O = 15 qe ug
(to BRE. (003) > 176g Cu
À (D) 20 ep te
(nd 00 cpm coy
toys Syed} 490 ren ey de Elda > Biome 6
UP Bye NEDO Rm cy (400) = 318 cpm camp
to. OF
ET N ea
Nas 25(B=20 Nyz32
1 2 tm 3 4
MERE E
MH 1 1 4 +
ALO 1 2 ©
Outpst hable wil ot ratte!
Answers to the Chapter 10 Case Study Quest
Case 10-1.
1. As segment gear A rotates counterclockwise, rack C is pushed to the left
2. As segment gear A rotates counterclockwise pushing rack C tothe left, gear B rotates
counterclockwise.
3. Once tooth E disengages with the rack, the leading tooth on segment gear A will
engage gear B, driving gear D clockwise.
4, For continuous counterclockwise motion of segment gear A, segment gear B
reciprocates counter and clockwise.
5. When engaged with gear A, rack C is pushed let. When disengaged from A, but
engaged with B, rack C is pushed back to the right
6. This mechanism creates reciprocating linear (rack C) and rotation (gear B) with a
uniform continuously rotary input.
7. The constant engaging and disengaging can cause shock and vibration problems, if
the mechanism is driven at high speed or with substantial loads.
Case 10:2.
‘As link A moves t the left, link B rotates counterclockwise.
As link A moves tothe le, gear C rotates counterclockwise.
As link A moves tothe left, gear D rotates clockwise.
{As ink A moves tothe left, link E rotates clockwise
‘As link A moves to the left, link F rotates clockwise. Note that the configuration of
links E and F form a shaper-crank mechanism and exhibit quick-return motion. It the
position shown, the mechanism is within the slower portion of its stroke.
6. As link A moves to the le, link G moves to the le. However, due to the quick
return nature ofthe arrangement, link G moves slower than link A.
7. As link A reciprocates back and forth link G also reciprocates but with a smaller
stoke,
8, Iflink E were rotated 90 degrees, then link G would reciprocate with a larger stoke
than the stroke of link A.
Case 10.3.
1. As gear A rotates clockwise, gear B rotates counterclockwise.
2. As gear A rotates clockwise and gear B rotates counterclockwise, gear C rotates
clockwise.
3. As gear A rotates clockwise and gear B rotates counterclockwise, gear D rotates
clockwise.
4. Alter the handle is rotated upward, gear B disengages with gear A but gear C engages
with gear À.
5. As gear A continues to rotate clockwise, gear C rotates counterclockwise, gear B
rotates clockwise and gear D rotates counterclockwise.
6. The mechanism produces forward and reverse rotation of an output gear (D) from a
constant direction of an input gear (A).
Case 10-1.
1. As segment gear A rotates counterclockwise, rack C is pushed to the left
2. As segment gear A rotates counterclockwise pushing rack C tothe left, gear B rotates
counterclockwise.
3. Once tooth E disengages with the rack, the leading tooth on segment gear A will
engage gear B, driving gear D clockwise.
4, For continuous counterclockwise motion of segment gear A, segment gear B
reciprocates counter and clockwise.
5. When engaged with gear A, rack C is pushed let. When disengaged from A, but
engaged with B, rack C is pushed back to the right
6. This mechanism creates reciprocating linear (rack C) and rotation (gear B) with a
uniform continuously rotary input.
7. The constant engaging and disengaging can cause shock and vibration problems, if
the mechanism is driven at high speed or with substantial loads.
Case 10:2.
‘As link A moves t the left, link B rotates counterclockwise.
As link A moves tothe le, gear C rotates counterclockwise.
As link A moves tothe left, gear D rotates clockwise.
{As ink A moves tothe left, link E rotates clockwise
‘As link A moves to the left, link F rotates clockwise. Note that the configuration of
links E and F form a shaper-crank mechanism and exhibit quick-return motion. It the
position shown, the mechanism is within the slower portion of its stroke.
6. As link A moves to the le, link G moves to the le. However, due to the quick
return nature ofthe arrangement, link G moves slower than link A.
7. As link A reciprocates back and forth link G also reciprocates but with a smaller
stoke,
8, Iflink E were rotated 90 degrees, then link G would reciprocate with a larger stoke
than the stroke of link A.
Case 10.3.
1. As gear A rotates clockwise, gear B rotates counterclockwise.
2. As gear A rotates clockwise and gear B rotates counterclockwise, gear C rotates
clockwise.
3. As gear A rotates clockwise and gear B rotates counterclockwise, gear D rotates
clockwise.
4. Alter the handle is rotated upward, gear B disengages with gear A but gear C engages
with gear À.
5. As gear A continues to rotate clockwise, gear C rotates counterclockwise, gear B
rotates clockwise and gear D rotates counterclockwise.
6. The mechanism produces forward and reverse rotation of an output gear (D) from a
constant direction of an input gear (A).
7. Engaging and disengaging gears while transferring power causes abrupt shock loading
on the gear teeth
Case 10-4,
1. As gear B rotates clockwise, gear C rotates counterclockwise.
2. As gear B rotates clockwise, gear D rotates clockwise.
3. Link J, gear C and link E (along with the frame) create a four-bar mechanism, The
rotation of gear C actuates the mechanism and causes link J to oscillate, pivoting at A.
4. The center of gear D will trace an arc, centered at A.
5. Piston G is constrained to horizontal siding.
6. The mechanism creates a large stroke for piston G.
Case 10.5.
1. As shaft G rotates as shown, gear H rotates counterclockwise, when viewing the
mechanism from a right side view.
2. Gears F, J and H are bevel gear.
3. tem A is a nut that is pushed leftward withthe counterclockwise (as viewed from a
right side view) rotation of serew C.
4. As item A contacts color L, the serew is pushed to the right. tem K disengages with
gear H, and contacts gear J. This action results in the reversal of serew C and nut A.
5. Item O is a spring-loaded ball detent, which holds screw C from axial translation,
until enough force is created to overcome the spring force and move the screw.
‘Setscrews are located on collars L and Q to alter the stroke of nut A.
‘The mechanism creates reciprocating, constant velocity, motion of nut A between
collars L and Q. This type of mechanism is common in winding machinery (such as
cable on a spool).
on the gear teeth
Case 10-4,
1. As gear B rotates clockwise, gear C rotates counterclockwise.
2. As gear B rotates clockwise, gear D rotates clockwise.
3. Link J, gear C and link E (along with the frame) create a four-bar mechanism, The
rotation of gear C actuates the mechanism and causes link J to oscillate, pivoting at A.
4. The center of gear D will trace an arc, centered at A.
5. Piston G is constrained to horizontal siding.
6. The mechanism creates a large stroke for piston G.
Case 10.5.
1. As shaft G rotates as shown, gear H rotates counterclockwise, when viewing the
mechanism from a right side view.
2. Gears F, J and H are bevel gear.
3. tem A is a nut that is pushed leftward withthe counterclockwise (as viewed from a
right side view) rotation of serew C.
4. As item A contacts color L, the serew is pushed to the right. tem K disengages with
gear H, and contacts gear J. This action results in the reversal of serew C and nut A.
5. Item O is a spring-loaded ball detent, which holds screw C from axial translation,
until enough force is created to overcome the spring force and move the screw.
‘Setscrews are located on collars L and Q to alter the stroke of nut A.
‘The mechanism creates reciprocating, constant velocity, motion of nut A between
collars L and Q. This type of mechanism is common in winding machinery (such as
cable on a spool).
HAT o> Boon
TER
ae 195738) = 180.64%%
YY (N)
11-2] o, EN [N
eg]
be W5QCTAD) = 120.420
Ne (o 5/0 = 301 % (4943)
lbo,= 1200 (Ad) = 125 lolo Ys
N= (rs y)= 628% (O79)
= 3142 Ams
= 158) Pin = 1505 mia
EN E we 1-9] dog Reo
. .
= 1D (1200) 3000 tw = BS (109312590,
Cor IS) = 117.81 M/s
Vein. 2-5/2) = 41% (602)
= 13. m7
lo o,
= BS (De 10m,
toys V1S0 (He) = 1832, Ys
Ke amas): 508 0% (690)
= 2978 ma
[Ue dos Retos
= AU =2100rg0, can
[o OCT = 188.0 0%
= GS DER) = 848.2% (61D
bel = Bie Sina?
NT
=24
Bl 12% Cua 3G 251
Cos 23 Curtin range)
L2@-Eends Sas
SAR
as wer Bug
1409-27 6410 = 181.19
lox Ba (a I”
i
= 2.315 lun range)
9, 180-20 Sa) 162.07
TER
ae 195738) = 180.64%%
YY (N)
11-2] o, EN [N
eg]
be W5QCTAD) = 120.420
Ne (o 5/0 = 301 % (4943)
lbo,= 1200 (Ad) = 125 lolo Ys
N= (rs y)= 628% (O79)
= 3142 Ams
= 158) Pin = 1505 mia
EN E we 1-9] dog Reo
. .
= 1D (1200) 3000 tw = BS (109312590,
Cor IS) = 117.81 M/s
Vein. 2-5/2) = 41% (602)
= 13. m7
lo o,
= BS (De 10m,
toys V1S0 (He) = 1832, Ys
Ke amas): 508 0% (690)
= 2978 ma
[Ue dos Retos
= AU =2100rg0, can
[o OCT = 188.0 0%
= GS DER) = 848.2% (61D
bel = Bie Sina?
NT
=24
Bl 12% Cua 3G 251
Cos 23 Curtin range)
L2@-Eends Sas
SAR
as wer Bug
1409-27 6410 = 181.19
lox Ba (a I”
i
= 2.315 lun range)
9, 180-20 Sa) 162.07
[1-9] 1240 3809 = LO
B=468) -21(2:19) = 224.33
[EU
244 Caux LAB + Vo
_fB=4G)- 27 (Bed = 214.94
RE, À CEST
ts BHA QR ICH
Ve
3 28.224 ui rang |
e 1D- ee my
=25, 807 (oithin rango)
Dy=120-2sis* Ear 143. Y!
EUT 10hp+ 35 Drgm > 3V belt
one possible design :
ve di:25 6,27
10. tea LAS = 39.3
try CE Sin Sri ya
Le AT CS) + Que
MUSA
Use ndo Thay
BeACD 25 (25+10.D) = 201.49
a BUA rent 3204-25)
WG
= ABLA
(9,2 35509) = 371.1
22323 USA
25 (887400) = 9.80, use Dy? 1D.)
VEN GHA As A)= MA 1 |
N12] 25hp4 Asp > SV
B=468) -21(2:19) = 224.33
[EU
244 Caux LAB + Vo
_fB=4G)- 27 (Bed = 214.94
RE, À CEST
ts BHA QR ICH
Ve
3 28.224 ui rang |
e 1D- ee my
=25, 807 (oithin rango)
Dy=120-2sis* Ear 143. Y!
EUT 10hp+ 35 Drgm > 3V belt
one possible design :
ve di:25 6,27
10. tea LAS = 39.3
try CE Sin Sri ya
Le AT CS) + Que
MUSA
Use ndo Thay
BeACD 25 (25+10.D) = 201.49
a BUA rent 3204-25)
WG
= ABLA
(9,2 35509) = 371.1
22323 USA
25 (887400) = 9.80, use Dy? 1D.)
VEN GHA As A)= MA 1 |
N12] 25hp4 Asp > SV
1-3] 5hpa Drm > BV keit [HAT 10hp + 20007pm-> SV tt
si
I
TAS] Ihpt Dion > SVB [IL] 2hp 1 2ddrpm> SV bar
©
si
I
TAS] Ihpt Dion > SVB [IL] 2hp 1 2ddrpm> SV bar
©
im
los WOUYad) = 12.84 MY
Dat Tin (896A) = 20.088 in
Nex (18. nam: au
29441 Kms
vx kon veu
NE] cs Matos I PE
RA ES re an Las) = 83 Lam ep)
NI Lot 23.09 09%
De SAin (PA) =3.037 15
Ms (23.0% & De 34 aa x (à
0 mot
LS ai Fe colon
j (tee) wo, Be bo,
= EE (ae 1.47 0 WY
axe HT) 18.84 44
= Yan (BA = EC
Y (1.8.0 = 31.14% (2/0)
= 40517 Mn
Us manval rae
IN do, Nun
= FE (es)= 160 opm, cw,
25 yn ts
ne 2500): 26.18 "Ys
Das 129/5:(180/50) = 30,25 1
Me BA) = 288.0% (98)
= 197
ini
je ot ar NA
pea] lem YA) “0
va Batty
= Æfro- QD + —
Let —
(1-22.
D,= ER) LOW /
de m
I fro trad} EN]
RS
B= 180-200" > Wey
By sores Ga) 2
ac
= BIG
UBUD. bep
al spy
a CAMA .
ANRT ES |
los WOUYad) = 12.84 MY
Dat Tin (896A) = 20.088 in
Nex (18. nam: au
29441 Kms
vx kon veu
NE] cs Matos I PE
RA ES re an Las) = 83 Lam ep)
NI Lot 23.09 09%
De SAin (PA) =3.037 15
Ms (23.0% & De 34 aa x (à
0 mot
LS ai Fe colon
j (tee) wo, Be bo,
= EE (ae 1.47 0 WY
axe HT) 18.84 44
= Yan (BA = EC
Y (1.8.0 = 31.14% (2/0)
= 40517 Mn
Us manval rae
IN do, Nun
= FE (es)= 160 opm, cw,
25 yn ts
ne 2500): 26.18 "Ys
Das 129/5:(180/50) = 30,25 1
Me BA) = 288.0% (98)
= 197
ini
je ot ar NA
pea] lem YA) “0
va Batty
= Æfro- QD + —
Let —
(1-22.
D,= ER) LOW /
de m
I fro trad} EN]
RS
B= 180-200" > Wey
By sores Ga) 2
ac
= BIG
UBUD. bep
al spy
a CAMA .
ANRT ES |
Waals nal "
Dessen "33957 Dee Nana
Í 3 : =
© A Ba De y RA
C= any a x L-15 fren- Ge,
i
an | | f20- Eo att GEN 3 Seat: TS |
a } 2 52424
ma nz | Or wa sy
| ; Ev) Be 25 [REAR | us)
N-2S| 90 he + Hail ii
EE A
One fan désun:
Neu
Ne SD ÓN = LOS 2 OR
©
o
Be AE [00.88
del» 385%
us Wis
ea sae ED 5 _
ENG Di eme
1 and - ana
ve eM lt
ut thn
E]
ANN
Dessen "33957 Dee Nana
Í 3 : =
© A Ba De y RA
C= any a x L-15 fren- Ge,
i
an | | f20- Eo att GEN 3 Seat: TS |
a } 2 52424
ma nz | Or wa sy
| ; Ev) Be 25 [REAR | us)
N-2S| 90 he + Hail ii
EE A
One fan désun:
Neu
Ne SD ÓN = LOS 2 OR
©
o
Be AE [00.88
del» 385%
us Wis
ea sae ED 5 _
ENG Di eme
1 and - ana
ve eM lt
ut thn
E]
ANN
(1241
120hga rpm? No HO cha
11-28 |50hp+ Om» No 100 chain
8 hp à Dey” No. LD chain
11301 10hp4-425rpnw1> No ED chain
120hga rpm? No HO cha
11-28 |50hp+ Om» No 100 chain
8 hp à Dey” No. LD chain
11301 10hp4-425rpnw1> No ED chain
Answers to the Chapter 11 Case Study Quest
Case 1-1,
I. As sprocket A drives clockwise, sprocket B is driven clockwise.
2. As sprocket A drives clockwise, he instantaneous motion of pin C (which is attached
to the chain) is moving left and upward toward sprocket A.
3. As sprocket A drives clockwise, he instantaneous motion of yoke D (which is
‘constrained to horizontal sliding) is toward the left at constant velocity.
4. Inroughly 1/2 revolution, the portion of the chain that contains pin C will engage
with sprocket A.
5. As the portion ofthe chain hat contains pin C engages on sprocket A, rod E will
decelerate
6. AS the portion of the chain that contains pin C moves onto the top of sprocket A, rod
E will change directions then accelerate toward the right. Upon further rotation and
disengagement of the portion of chain that contains pin C, rod E will move toward the
right with constant velocity.
7. The mechanism creates reciprocating, constant velocity, motion of rod E.
Case 11-2.
1. As handle H rotates, shaft [rotates with the handle,
2. As handle H rotates, he right half sheave rotates threading itself onto (or off the left
half
3. As handle H rotates, two halfsheaves are brought either together or apar.
4. Item J is a key and transfers rotational power from the sheave to shaft A.
5. tem guides shaft.
6. Item G isa spring, which keeps engagement of shaft I with half-sheave C.
7. This devise isa variable radius sheave, commonly used in CVTs (continuously
variable transmissions).
Case 11-3.
1. As tab B is forced upward ino lever C, item D is forced downward,
2. As tab Bis forced upward into lever C, item D will move downward into a slot
integral with sheave G, The resulting action is that sheave G will no longer be able to
rotate.
3. As tab B is forced upward and sheave G is no longer abe to rotate, the paint shaking.
‘machine will stop operation.
4. The purpose of this device is stop operation of the machine when the door is raised.
5. Sheave E has many notches for more immediate response to opening the door,
stopping the shaking machine more quickly.
6. Spring His a return spring to keep D in an upward position until forced downward,
7. This mechanism is a safety device.
Case 1-1,
I. As sprocket A drives clockwise, sprocket B is driven clockwise.
2. As sprocket A drives clockwise, he instantaneous motion of pin C (which is attached
to the chain) is moving left and upward toward sprocket A.
3. As sprocket A drives clockwise, he instantaneous motion of yoke D (which is
‘constrained to horizontal sliding) is toward the left at constant velocity.
4. Inroughly 1/2 revolution, the portion of the chain that contains pin C will engage
with sprocket A.
5. As the portion ofthe chain hat contains pin C engages on sprocket A, rod E will
decelerate
6. AS the portion of the chain that contains pin C moves onto the top of sprocket A, rod
E will change directions then accelerate toward the right. Upon further rotation and
disengagement of the portion of chain that contains pin C, rod E will move toward the
right with constant velocity.
7. The mechanism creates reciprocating, constant velocity, motion of rod E.
Case 11-2.
1. As handle H rotates, shaft [rotates with the handle,
2. As handle H rotates, he right half sheave rotates threading itself onto (or off the left
half
3. As handle H rotates, two halfsheaves are brought either together or apar.
4. Item J is a key and transfers rotational power from the sheave to shaft A.
5. tem guides shaft.
6. Item G isa spring, which keeps engagement of shaft I with half-sheave C.
7. This devise isa variable radius sheave, commonly used in CVTs (continuously
variable transmissions).
Case 11-3.
1. As tab B is forced upward ino lever C, item D is forced downward,
2. As tab Bis forced upward into lever C, item D will move downward into a slot
integral with sheave G, The resulting action is that sheave G will no longer be able to
rotate.
3. As tab B is forced upward and sheave G is no longer abe to rotate, the paint shaking.
‘machine will stop operation.
4. The purpose of this device is stop operation of the machine when the door is raised.
5. Sheave E has many notches for more immediate response to opening the door,
stopping the shaking machine more quickly.
6. Spring His a return spring to keep D in an upward position until forced downward,
7. This mechanism is a safety device.
e E
12-1 | Ya-20 > p=dy= .05m
AN A
et
Ale
tan he .O1<.2
Breads are self lodeing
[122] 4280 p=7g 0887
= tar KOST .
it) Aree pe
tan =.05 4.2
Ehreods are self locking
1273| Mlbx2.0 > P= Zn
COS
tan 2=.093 4.2
Birds One self-locking
ANR Sp EDR
tar aus
tan A=-0644 2.2
threads are self - looking,
eS) Ao > e357 .0Si0
ARE Ne p AB=()(05 Mere)
215i
[FeblACME 2-4 ph 25m
AR= Wip AB= (OCASO
=25 “Y
121] Arme 1-9%p=¥=.20in
AR Ne pb (MO
=2Sù 4
MST HN >p=2=.0 Tn
Amr NEDP ia > EYbre)
= Ure’
ARENepAB= OTAN
=A in
12-1 | Ya-20 > p=dy= .05m
AN A
et
Ale
tan he .O1<.2
Breads are self lodeing
[122] 4280 p=7g 0887
= tar KOST .
it) Aree pe
tan =.05 4.2
Ehreods are self locking
1273| Mlbx2.0 > P= Zn
COS
tan 2=.093 4.2
Birds One self-locking
ANR Sp EDR
tar aus
tan A=-0644 2.2
threads are self - looking,
eS) Ao > e357 .0Si0
ARE Ne p AB=()(05 Mere)
215i
[FeblACME 2-4 ph 25m
AR= Wip AB= (OCASO
=25 “Y
121] Arme 1-9%p=¥=.20in
AR Ne pb (MO
=2Sù 4
MST HN >p=2=.0 Tn
Amr NEDP ia > EYbre)
= Ure’
ARENepAB= OTAN
=A in
12-9] Ame 10 © pe Lie
= BREN PB GER Or)
= Zin
[12-10] Acme Yo-1D > p=45= in
AR D (mes rel
Ray bAB2i0 À
12-11] Acme J2-10 => Pty,
30ce
Are Ne PAB = Na
= 3dın
A Ryan: 2.750 de
Na] 2-0 > pr Blin
[ANT CET OA)
Lin
Reken: 2 Sin ANZ
= BREN PB GER Or)
= Zin
[12-10] Acme Yo-1D > p=45= in
AR D (mes rel
Ray bAB2i0 À
12-11] Acme J2-10 => Pty,
30ce
Are Ne PAB = Na
= 3dın
A Ryan: 2.750 de
Na] 2-0 > pr Blin
[ANT CET OA)
Lin
Reken: 2 Sin ANZ
N
1213| 94-10 » p= Je = Due
AR GYCAMAL Brae „Fin
Raden? Luzern IT
RAT 344-10 person
Rus Ls) + 2 Sin
Aute: ZT,
Abend = lotir 2283"
MAS] 94-10 psdozonin
BEN =; 13
Actor » LIE, saw
an Harpe QE
AR = (NON YS re)» 38 Tin
OS
BR]
12-07-19 > p
FAR: ONL NS
Fa
[218 TAcmE Five hr Sn
Vars Ne po NS Yo Ya
= 10% EUR
NES ”
1213| 94-10 » p= Je = Due
AR GYCAMAL Brae „Fin
Raden? Luzern IT
RAT 344-10 person
Rus Ls) + 2 Sin
Aute: ZT,
Abend = lotir 2283"
MAS] 94-10 psdozonin
BEN =; 13
Actor » LIE, saw
an Harpe QE
AR = (NON YS re)» 38 Tin
OS
BR]
12-07-19 > p
FAR: ONL NS
Fa
[218 TAcmE Five hr Sn
Vars Ne po NS Yo Ya
= 10% EUR
NES ”
N
JA] ACME 1-5 > pr$=.201n
Ve Ne pw
= (20/40 )(200 m)
= 240'nm (RE)
= 4 Mer
Vins baton = 41Ys À
|
12201 = LDO rpm
Bun: (DO) = 12D 09
Nbr PBUNC p> À 3.01
Vases Na ps
= OS OS]
= 4.23 Yan (LE)
= ASA Mec À
12-22]
3-10» p= 15>
Major (INEA)
Inia
A |
10 ph in
Zee
Vals Yer Vo Vert re Ve Vers Varo.
Nee" eee Yet Seemed?
Ve 1925072 À MET Yan AQIS?
12-23) 12-24
JA] ACME 1-5 > pr$=.201n
Ve Ne pw
= (20/40 )(200 m)
= 240'nm (RE)
= 4 Mer
Vins baton = 41Ys À
|
12201 = LDO rpm
Bun: (DO) = 12D 09
Nbr PBUNC p> À 3.01
Vases Na ps
= OS OS]
= 4.23 Yan (LE)
= ASA Mec À
12-22]
3-10» p= 15>
Major (INEA)
Inia
A |
10 ph in
Zee
Vals Yer Vo Vert re Ve Vers Varo.
Nee" eee Yet Seemed?
Ve 1925072 À MET Yan AQIS?
12-23) 12-24
(dar (RO) = 240rgm
ey
Avs:
P= 152.0164 in
Ne DA" PNL
0307 Mu + AA“
12:25 AUME 2-43 p= ay =RSm fe eurem: 15» -3=20 in
5% “bron Gd = “Le =
22D ie nella: Wer
pe ©- a IV Le: Go- Ach 228.51 Yoo?
NE [rt NA on Vee)
AR a À Conste vel) cna Rare dir as vel
ea de 6
on 1200cpm
=
ey
Avs:
P= 152.0164 in
Ne DA" PNL
0307 Mu + AA“
12:25 AUME 2-43 p= ay =RSm fe eurem: 15» -3=20 in
5% “bron Gd = “Le =
22D ie nella: Wer
pe ©- a IV Le: Go- Ach 228.51 Yoo?
NE [rt NA on Vee)
AR a À Conste vel) cna Rare dir as vel
ea de 6
on 1200cpm
=
(2-29 Acme 2x10
F mE 0 ‘J
= 24.0 in LA
OL
4 306
12-30
ack xl (si)! a) A A)
EN mn ga
re een Nowy
= Be Roberts 1
TO
=.259 = Wey
ISS) we vce sag. bs E ICA cee ea
tanker Se few E ut ong
EEE nn,
PS A 248 oy
We HI 7118 Ay
E EN
ALAN
EN ña EN
44: a
BT Dao om
À Lo Fume MUT
A ee ee"
a (tes 30.15 (07
Semen) 245
we-10 pen
Tube we 20 p 2.08
=24,5%,
12-35 136
An Sat (LL) Bis
Sur Henn
Sat SE Lata Baro e OTe
Thnod B ve MID x 150 ar vs M108 180
The ue Mido 1.00 Trad Ce Mbs —
237
Sra > (LL) Bruns
Bes Late
Tiresa D
F mE 0 ‘J
= 24.0 in LA
OL
4 306
12-30
ack xl (si)! a) A A)
EN mn ga
re een Nowy
= Be Roberts 1
TO
=.259 = Wey
ISS) we vce sag. bs E ICA cee ea
tanker Se few E ut ong
EEE nn,
PS A 248 oy
We HI 7118 Ay
E EN
ALAN
EN ña EN
44: a
BT Dao om
À Lo Fume MUT
A ee ee"
a (tes 30.15 (07
Semen) 245
we-10 pen
Tube we 20 p 2.08
=24,5%,
12-35 136
An Sat (LL) Bis
Sur Henn
Sat SE Lata Baro e OTe
Thnod B ve MID x 150 ar vs M108 180
The ue Mido 1.00 Trad Ce Mbs —
237
Sra > (LL) Bruns
Bes Late
Tiresa D
Chapter 13 |
I
À
BT
hore: 160 - 140¢0s 25 +33," = By
100 + HD si 253 159,28 Ry
PR RedRe*Ry MZA
Lo ate) 1627
13-2
Morte: 10-190 coslS= 100.8" By
vert: rie 226 A8 Ry
y RER = 208.38
H a: tas (RQ) = bb.0°
E RL art ree
33 EN as
TOP 160-140 eos 105= Nr Re
Vert: 1004 Ds 106 + =By A MA
1D BRE ES
Ba tas (By) = SD"
= 200i ows
ai Re 306.3" £501"
[13-5] Ber EN EU
ral TR, Le Y _
TT
:
[KM re > en Pp
e
DEMO (ei an RÔ1S - wold
ee hero 5
We 3132104 Compression
cr PI
‘Dama=o:
xo] rode = BC eos (i) + Ta 20
To Tye + 28020 À me
= 29.0 Nm Y
I
À
BT
hore: 160 - 140¢0s 25 +33," = By
100 + HD si 253 159,28 Ry
PR RedRe*Ry MZA
Lo ate) 1627
13-2
Morte: 10-190 coslS= 100.8" By
vert: rie 226 A8 Ry
y RER = 208.38
H a: tas (RQ) = bb.0°
E RL art ree
33 EN as
TOP 160-140 eos 105= Nr Re
Vert: 1004 Ds 106 + =By A MA
1D BRE ES
Ba tas (By) = SD"
= 200i ows
ai Re 306.3" £501"
[13-5] Ber EN EU
ral TR, Le Y _
TT
:
[KM re > en Pp
e
DEMO (ei an RÔ1S - wold
ee hero 5
We 3132104 Compression
cr PI
‘Dama=o:
xo] rode = BC eos (i) + Ta 20
To Tye + 28020 À me
= 29.0 Nm Y
GS Br ER]
0 0: Ñ E DEMO
: Bean
el“ oe sti sdb ne
Py fotos’ (WA) soa CR
E > A (eee
ae (adi rita
a ATACA Op un
Sa [Es RER EU = a =
a ann D se
oe een
gas) ans a
vr va ut Damen er
Dinos as f ad io ~~ Said
Midas “ÉD do ter
= AS
See Ea tar
al es Tes
C js colon Beant (ng
wren PET BU Se tye Gena
fy soe Fe am
nl CUS fea weet
= er.
= a
% ern
sé ” ES
Ses O)
SANS
ROSES
Bus BED ler en by
Er ASS Genser)
por dard Me Resinthse0
PATTES des 3
EN
Damgeo:
Wale -0
Tee BA int ew
0 0: Ñ E DEMO
: Bean
el“ oe sti sdb ne
Py fotos’ (WA) soa CR
E > A (eee
ae (adi rita
a ATACA Op un
Sa [Es RER EU = a =
a ann D se
oe een
gas) ans a
vr va ut Damen er
Dinos as f ad io ~~ Said
Midas “ÉD do ter
= AS
See Ea tar
al es Tes
C js colon Beant (ng
wren PET BU Se tye Gena
fy soe Fe am
nl CUS fea weet
= er.
= a
% ern
sé ” ES
Ses O)
SANS
ROSES
Bus BED ler en by
Er ASS Genser)
por dard Me Resinthse0
PATTES des 3
EN
Damgeo:
Wale -0
Tee BA int ew
Chapter 14 |
Wenger nee v
a od ana
N FR
run az
DO EE We th SH
[Bis 0% yr
ua
[43] T= mits ge BAT mE
LS 1) i ds
Rox 024 loft EIERN
20.24 Les mg = 135000 kq ma = 0,135 at
[AT re mr“
RE EN
2328000 la mas 8125 kam
ke (eee. Bs Som = Sk nny
(Te Emer = gn
En DNS LAS
À
Lin (Gr SI
| EICHE
1,99 te &
MA
Ten
O
> 085 th vw st ENS
Ty Tare 0 08S+
ben
DL
Ahlers Ga] 08 kan?
Tes Temas 089 GU
= Die ka my
132 Vo in ss
[EE
wel Va ESS
as E
sp po
ese (13)
Mao = 08 la
DES
AA
IS
+ mont - into", BES 918]
=.000133 tb LA
JNE
le IT za
HS a wk ABD
Ma pa Vas: EAN
\ a all
245,
soa LOGE (193.289 = COOL oo
ern] uns
Hanan GS] 682107 Mb me
Lol +000 + Sees: N
HRS OS CL
DH nl ET
Wenger nee v
a od ana
N FR
run az
DO EE We th SH
[Bis 0% yr
ua
[43] T= mits ge BAT mE
LS 1) i ds
Rox 024 loft EIERN
20.24 Les mg = 135000 kq ma = 0,135 at
[AT re mr“
RE EN
2328000 la mas 8125 kam
ke (eee. Bs Som = Sk nny
(Te Emer = gn
En DNS LAS
À
Lin (Gr SI
| EICHE
1,99 te &
MA
Ten
O
> 085 th vw st ENS
Ty Tare 0 08S+
ben
DL
Ahlers Ga] 08 kan?
Tes Temas 089 GU
= Die ka my
132 Vo in ss
[EE
wel Va ESS
as E
sp po
ese (13)
Mao = 08 la
DES
AA
IS
+ mont - into", BES 918]
=.000133 tb LA
JNE
le IT za
HS a wk ABD
Ma pa Vas: EAN
\ a all
245,
soa LOGE (193.289 = COOL oo
ern] uns
Hanan GS] 682107 Mb me
Lol +000 + Sees: N
HRS OS CL
DH nl ET
ETS
is
— fes SR T7
o
ap ES
> ea STAR BS
ne LE
¡Es From CAB
NEE
an HZ”
ld Bar
Reiko ma
ri Ti
A y
Seen Cd
Nas TP
Vos 118.0% —
qu Ls 128078 Sis"
Get GaAs 5H pa
& Sa nv Zur
aE ag PA Op,
om CAD
AS —
hans Rand à Na EU) oy ñ
5 issus LE Rte
ES à run aa -ageo
Eng SUS NEAR IS +
ES Be $ > Y: 148,30
NEN E Nind> ©
A The 211300 ls ES ¡EA Y
Op o nn Es woe Se
Ms mam» RUES pe Gay, 2 sine ur
ao
08 = (185000) 8150" 287
Com CAB
a gt WG
Fe mela sake
ental O et
nee RESTO)
uo = 23 be
> Re: Al
EMO
ONE
D
Ay
Tr ASA à lb 4
= dede Sie Od,
is
— fes SR T7
o
ap ES
> ea STAR BS
ne LE
¡Es From CAB
NEE
an HZ”
ld Bar
Reiko ma
ri Ti
A y
Seen Cd
Nas TP
Vos 118.0% —
qu Ls 128078 Sis"
Get GaAs 5H pa
& Sa nv Zur
aE ag PA Op,
om CAD
AS —
hans Rand à Na EU) oy ñ
5 issus LE Rte
ES à run aa -ageo
Eng SUS NEAR IS +
ES Be $ > Y: 148,30
NEN E Nind> ©
A The 211300 ls ES ¡EA Y
Op o nn Es woe Se
Ms mam» RUES pe Gay, 2 sine ur
ao
08 = (185000) 8150" 287
Com CAB
a gt WG
Fe mela sake
ental O et
nee RESTO)
uo = 23 be
> Re: Al
EMO
ONE
D
Ay
Tr ASA à lb 4
= dede Sie Od,
CA
E 4 E
Gant Ob + 9h78 ody Eu
fon CAD
AS
lp 95176 VER
VE 1ER
AE (57
38% BAY
DE = 02 e, as
dd m cart SE
=
A Os LS: page VS
R E Y
5
CRE
Using FRD'S Bom Melk
KERN, R
NAN
Guanes
105.38,
a i ee iy
ER hemos so
¿a Soe
peal as, eae em $ ad e
A
Tay 17.81 NY
LA
MAT Using analysis From IL, witht Je a=
PBT Vers ane fe tow
> Candy 1074 OF
se
Fe ea.
Ding FED'S from Walk:
WEN, Fag? EU N
IT ET RTS
Grat e dh,
QE 9 0 wa
E 4 E
Gant Ob + 9h78 ody Eu
fon CAD
AS
lp 95176 VER
VE 1ER
AE (57
38% BAY
DE = 02 e, as
dd m cart SE
=
A Os LS: page VS
R E Y
5
CRE
Using FRD'S Bom Melk
KERN, R
NAN
Guanes
105.38,
a i ee iy
ER hemos so
¿a Soe
peal as, eae em $ ad e
A
Tay 17.81 NY
LA
MAT Using analysis From IL, witht Je a=
PBT Vers ane fe tow
> Candy 1074 OF
se
Fe ea.
Ding FED'S from Walk:
WEN, Fag? EU N
IT ET RTS
Grat e dh,
QE 9 0 wa
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