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ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY
Dynamic Force Analysis II
Dr. Sadettin KAPUCU
© 2007 Sadettin Kapucu
2Gaziantep University
ExampleExample
The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.
Freebody Diagram
Equations Of Motion
GamF
IM
3Gaziantep University
ExampleExample
The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.
Kinematics of the slender rod
?? yxG aaa
?
Freebody DiagramEquations Of Motion
GamF
IM
xx maF yy maF AA IM
4Gaziantep University
ExampleExample
The slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.
Kinematics of the slender rod
?? yxG aaa
?
)x(x rax
ray
x
r
At the instant bar is released, its angular velocity 0 0xa
jlilkay
2
1
2
1x
0xa
lay 2
1
5Gaziantep University
ExampleExampleThe slender bar of mass m is released form rest in the horizontal position as shown. At that instant, determine the force exerted on the bar by the support A.
Freebody DiagramEquations Of Motion
AA IM
00
xx
xx
AmA
maF
lmmgAmamgA
maF
yyy
yy
2
14
mgAy
3212
222 mll
mml
mdII GA l
gmllmg
2
3
32
1 2
0xa
lay 2
1
6Gaziantep University
D’Alembert’s PrincipleD’Alembert’s Principle
D’Alembert’s principle permits the reduction of a problem in dynamics to one in statics. This is accomplished by introducing a fictitious force equal in magnitude to the product of the mass of the body and its acceleration, and directed opposite to the acceleration. The result is a condition of kinetic equilibrium.
fictitious force and torque
0 aaaF
mmm
0 ααατIII
The meaning of the equation; i.e. indication of a dynamic case still holds true, but equation, having zero on right hand side becomes very easy to solve, like that in a “static force analysis” problem.
CGF
m, Ia
CGF
m, Ia
-ma
7Gaziantep University
Solution of a Solution of a DDynamic ynamic PProblem roblem UUsing sing D’Alembert’s D’Alembert’s PPrinciplerinciple
1. Do an acceleration analysis and calculate the linear acceleration of the mass centers of each moving link. Also calculate the angular acceleration of each moving link.
2. Masses and centroidal inertias of each moving link must be known beforehand.
3. Add one fictitious force on each moving body equal to the mass of that body times the acceleration of its mass center, direction opposite to its acceleration, applied directly onto the center of gravity, apart from the already existing real forces.
4. Add fictitious torque on each moving body equal to the centroidal inertia of that body times its angular acceleration, direction or sense opposite to that of acceleration apart from the already existing real torques.
5. Solve statically.
8Gaziantep University
Example 1Example 1
AB=10 cm, AG3=BG3=5 cm, =60o m2=m4=0.5 kg, m3=0.8 kg, I3=0.01 kg.m2
In the figure, a double- slider mechanism working in horizontal plane is shown. The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate the amount of force on this mechanism in the given kinematic state.
4
B
2A
3
G3
x
BABA VVV
?AV
smVB /1
ABtoVB
A ?
BV
AV
BAV
smVA /5774.0
smVB
A /1547.1
9Gaziantep University
Example 1 contExample 1 contIn the figure, a double- slider mechanism working in horizontal plane is shown. The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate the amount of force on this mechanism in the given kinematic state.
4
B
2A
3
G3
x
BABA VVV
BtoAfromsmAB
Va B
An
BA
22
2
/33.131.0
1547.1
?Aa
BABA aaa
0
BAA aa
t
BA
n
BA aa
ABtoa t
BA
?
10Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3
G3
x
BtoAfromsmAB
Va B
An
BA
22
2
/33.131.0
1547.1
?Aa
BAA aa
t
BA
n
BA aa
ABtoa t
BA
?
2/698,7 sma t
BA
t
BA
a
Aa
2/396.15 smaA
G3
3Ga
n
BA
a
2/698.72/3
smaa AG
3ABa t
BA
CCWsradAB
a t
BA
23 /98.78
1.0
698.7
11Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3
G3
x
t
BA
a
Aa
2/396.15 smaA
G3
3Ga
n
BA
a
2/698.72/3
smaa AG
CCWsradAB
a t
BA
23 /98.78
1.0
698.7
3Gma
Ama
3I
D’Alembert forces and moments
090698.7396.15*5.0 NmaA
0901584.6698.7*8.03
NmaG
CWNmI 7689.098.76*01.03
12Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3
G3
x
N698.7
N1584.6
Nm7689.0
BF4
B
2A
3
G3
N698.7
N1584.6
Nm7689.0
BF
12F
14F
13Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3
G3
N698.7
N1584.6
Nm7689.0
BF
12F
14F
1212 0;0 FFFFF BBx
NF
FFy
86.13
01584.6698.7;0
14
14
NF
F
M B
11.150867.0
3087.1
060sin*1.0*60cos*05.0*1584.6
60cos*1.0*698.77698.0
;0
12
12
NFB 11.15
+
x
y
14Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3
G3
x
3Gma
Ama
3I
090698.7396.15*5.0 NmaA
0901584.6698.7*8.03
NmaG
CWNmI 7689.098.76*01.03
3Gma
3I
3Gma
3Gma
h
hmaI G *33
3
3
Gma
Ih
mh 125.01589.6
7689.0
h3G
ma
3Gma
15Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3
G3
x
h
N1584.6
N698.7
BF4
B
2A
3G3
N698.7
BF
12F
14F
mh 125.01589.6
7689.0
h
N1584.6
16Gaziantep University
Example 1 contExample 1 cont
4
B
2A
3G3
N698.7
BF
12F
14F
h
N1584.6
1212 0;0 FFFFF BBx
NF
FFy
86.13
01584.6698.7;0
14
14
NF
F
M B
11.150867.0
3087.1
060sin*1.0*60cos*1.0*698.7
60cos*)125.005.0(*1584.6;0
12
12
NFB 11.15
+
x
y
17Gaziantep University
Example 2Example 2Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.
14
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=4 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
18Gaziantep University
Example 2 cont.Example 2 cont.Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.
14
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=4 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
CBCB VVV
ABtoVB ?
sec/5 mVC
BCtoVC
B ?
smVC /5
5 m/s
smVC
B /05.5smVB /85.3
19Gaziantep University
Example 2 cont.Example 2 cont.Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.
14
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
CBCB aaa
0Can
CB
n
CB
tB
nB aaaa
AtoBfromsmAB
Va BnB
222
/5.192077.0
85.3
BCtoa t
CB ?
CtoBfromsmBC
Va C
Bn
CB
22
2
/6.63704.0
05.5
ABtoa tB ?
2/5.192 smanB
2/6.637 sman
CB
2/483 sma t
CB
2/776 sma tB
20Gaziantep University
Example 2 cont.Example 2 cont.Crank AB of the mechanism shown is balanced such that the mass center is at A. Mass center of the link CD is at its mid point. At the given instant, link 4 is translating rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required on crank AB to keep at the given kinematics state.
14
2
3
A
C
B
D
=45
F
AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
??,?, 32 Ba
2/5.192 smanB
2/6.637 sman
CB
2/483 sma t
CB
2/776 sma tB
CCWsradABa tB
222 /10078
077.0
776*
CWsradBCa t
CB
233 /20000
04.0
800*
Ba
262/800 2 smaB
B
21Gaziantep University
Example 2 contExample 2 contD’Alembert forces and moments
14
2
3
A
C
B
D
=45
F
2/5.192 smanB
2/6.637 sman
CB
2/483 sma t
CB
2/776 sma tB
CCWsrad 22 /10078
CWsrad 23 /20000
Ba
262/800 2 smaB
B
82400080053 N*am B
CWNm*.I 5041007805022
CCWNm*.I 10002000005033
m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2
-m3aB
22Gaziantep University
Example 2 contExample 2 cont
14
2
3
A
C
B
D
=45
F
-m3aB
4
23
A
C
B
D
=45
F
C
B
A22°
504 Nm
4000 N
1000 Nm
CyF
CxF
BxFByF
ByF
BxF
CyFCxF
AxFAyF
4AF
A4
23Gaziantep University
Example 2 contExample 2 cont
4
23
A
C
B
D
=45
F
C
B
A22°
504 Nm
4000 N
1000 Nm
CyF
CxF
BxFByF
ByF
BxF
CyFCxF
AxFAyF
4AF
A4
BxAxBxAxx FFFF;F 00
ByAyByAyy FFFF;F 00
007140028280504
022225040
ByBx
ByBxA
F*.F*.
cos*AB*Fsin*ABFT;M
24Gaziantep University
Example 2 contExample 2 cont
4
23
A
C
B
D
=45
F
C
B
A22°
504 Nm
4000 N
1000 Nm
CyF
CxF
BxFByF
ByF
BxF
CyFCxF
AxFAyF
4AF
A4
069.556;082cos*4000;0 BxCxBxCxx FFFFF
007.3961;082sin*4000;0 ByCyByCyy FFFFF
0*02828.0*02828.01000;045cos**45sin*1000;0 CyCxCyCxB FFBCFBCFM
25Gaziantep University
Example 2 contExample 2 cont
4
23
A
C
B
D
=45
F
C
B
A22°
504 Nm
4000 N
1000 Nm
CyF
CxF
BxFByF
ByF
BxF
CyFCxF
AxFAyF
4AF
A4
00 Cxx F;F
00 4AyCyy FF;F
00 4 CA*FT;M CyAA
26Gaziantep University
Example 2 contExample 2 cont
4
23
A
C
B
D
=45
F
C
B
A504 Nm
4000 N
1000 Nm
CyF
CxF
BxFByF
ByF
BxF
CyFCxF
AxFAyF
4AF
A4
BxAxBxAxx FFFF;F 00
ByAyByAyy FFFF;F 00
0*0714.0*02828.0504
022cos**22sin*504;0
ByBx
ByBxA
FFT
ABFABFTM
069.556
;082cos*4000;0
BxCx
BxCxx
FF
FFF
007.3961
;082sin*4000;0
ByCy
ByCyy
FF
FFF
0*02828.0*02828.01000
;045cos**45sin*1000;0
CyCx
CyCxB
FF
BCFBCFM
0;0 Cxx FF
00 4AyCyy FF;F
00 4 CA*FT;M CyAA
N.FBx 69556 N..
FCy 3435355028280
1000
N...FBy 41393160739613435355
CCWNm.T
.*.).(*.
933326
041393160714069556028280504
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