validation...rigid body dynamics dynamics of rigid bodies.01 references: engineering mechanics:...
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1
Validation
Rigid Body Dynamics
Dynamics of Rigid Bodies.01
References: Engineering Mechanics: Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.31, P.2.13
Type of Analysis: Kinematics of particles, Linear Motion
Type of Element: Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Initial height ð 0 0 ð
Initial velocity ð£0 25 ð/ð
2
1. Theoretical Solution
ð =ðð
ðð¡ , â« ðð = â« ððð¡ , ð = ð0 + ðð¡ (ð = ðððð ð¡)
ð¡
ð¡0
ð
ð0
ð =ðð
ðð¡ , â« ðð = â« (ð0 + ðð¡)ðð¡ , ð = ð0 + ð0ð¡ +
1
2ðð¡2 (ð = ðððð ð¡)
ð¡
ð¡0
ð
ð0
ðð¡ =ðð
ð=
ðð
ð , â« ððð = â« ððð , ð2 â ð0
2 = 2ðð (ð = ðððð ð¡)ð
ð0
ð
ð0
2. Time Required to Reach Maximum Height
ð = ð0 + ðð¡
0 = 25 â 9.807 â t
t = 2.549 [s]
3. Maximum Height
ð = ð0 + ð0ð¡ +1
2ðð¡2
= 0+25 Ã 2.549 â1
2Ã 9.807 Ã 2.5492
= 31.865 m
4. Calculation of Height (h)
â = ð â 15 = 16.865 ð
5. Calculation of Velocity (ððµ) and Time Required to Drop to Point âBâ
ðð = 0
ððµ2 â ðð
2 = 2ðâ
ððµ2 â 0 = 2 Ã 9.807 Ã 16.865
ððµ = 18.188
ððµ = ðð + ðð¡
18.188 = 0 + 9.807 Ã ð¡
t = 1.855 s
3
6. Time Required to Reach Point âBâ after Launch
ð¡ðŽâðµ = 4.404 s
Numerical Solution - RecurDyn
1. Body modeling and Set initial velocity
2. Plot the results
- Measure the time when the ball reaches a height of 15 m
- Measure the velocity of the ball when the time is 4.404
4
Comparison of results
Object Value Theory RecurDyn Error(%)
ð [ð] 16.865 16.866 0.006
ð [ð¬] 4.404 4.404 0
ðð [ð/ð] 18.188 18.189 0.005
5
Dynamics of Rigid Bodies.02
Reference of Problem : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige),
7th Edition, pp.72, P.2.136
Type of Analysis : Kinematics of Particles, Polar Coordinates
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Initial length r 500 ðð
Decreasing rate ᅵᅵ -150 ðð/ð
Rotation rate ᅵᅵ 60 ððð/ð
1. Velocity of Point B
ð£ð = ᅵᅵ = â150 ðð/ð
ð£ð = ðᅵᅵ = 500 âð
3= 523.6 ðð/ð
ð£ = â(ð£ð2 + ð£ð
2) = 544.7 ðð/ð
6
2. Acceleration of the Point B
ðð = ᅵᅵ â ðᅵᅵ2 = â548.31 ðð/ð 2
ðð = ðᅵᅵ + 2ᅵᅵᅵᅵ = â314.16 ðð/ð 2
a = â(ðð2 + ðð
2) = 631.93 ðð/ð 2
â Note
ðᅵᅵ = ᅵᅵðð , ðᅵᅵ = âᅵᅵðð
ᅵᅵ = ᅵᅵ = ᅵᅵðð + ððᅵᅵ = ᅵᅵðð + ðᅵᅵðð
ᅵᅵ = ᅵᅵ = ᅵᅵðð + ᅵᅵðᅵᅵ + ᅵᅵᅵᅵðð + ðᅵᅵðð + ðᅵᅵðᅵᅵ
= ᅵᅵðð + ᅵᅵᅵᅵðð + ᅵᅵᅵᅵðð + ðᅵᅵðð â ðᅵᅵ2ðð
= (ᅵᅵ â ðᅵᅵ2)ðð + (2ᅵᅵᅵᅵ + ðᅵᅵ)ðð
Numerical Solution â RecurDyn
1. Modeling
2. Plot the results
- Velocity of Point âBâ
7
- Acceleration of Point âBâ
.
Comparison of results
Object Value Theory RecurDyn Error(%)
ð [ð/ð] 0.5447 0.5447 0
ð [ð/ðð] 0.63193 0.63193 0
8
Dynamics of Rigid Bodies.03
Reference : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.77, P.2.161
Type of Analysis : Kinematics of Particles, Polar Coordinates
Type of Element : Particle (One part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Initial angle ðœ 30 ððð
Angular velocity ᅵᅵ 60 ððð/ð
1. Geometrical Constraint
ð â ððð ð + 90 â cðð ðœ = 300
ð â ððð ð = 300 â 90 â ððð ðœ = 222.06
ð â ð ððð = 90 â sinβ = 45
ð â ððð ð = 300 â 90 â cosβ = 222.06
⎠ð¡ððð =45
222.06 â ð = 11.46°
9
⎠ð = 226.57
⎠ðŸ = ðœ + ð = 30 + 11.46 = 41.46°
2. Velocity and Acceleration of Point A of Link AC
ð£ = 90 â ᅵᅵ = 5400 ðð/ð
ð = 90 â ᅵᅵ2 = 324 â 103 ðð2/ð
3. Calculate ᅵᅵ, ᅵᅵ, ᅵᅵ, ᅵᅵ with the Velocity and Acceleration
ð£ð = ðᅵᅵ = ð£ â ððð ðŸ = 5400 â ððð 41.46 = 4046.86 ðð/ð
ᅵᅵ =4046.86
226.57= 17.86 ððð/ð
ð£ð = ᅵᅵ = ð£ â ð ðððŸ = 5400 â ð ðð41.46 = 3575.32 ðð/ð
ᅵᅵ = 3575.32 ðð/ð
ðð = ᅵᅵ â ðᅵᅵ2 = ð â ððð ðŸ
ᅵᅵ = ðᅵᅵ2 + ð â ððð ðŸ = 226.57 â 17.862 + 324 â 103 â ððð 41.46 = 315,082.68 ðð/ð 2
ðð = ðᅵᅵ + 2ᅵᅵᅵᅵ = âð â ð ðððŸ
ᅵᅵ =âð â ð ðððŸ â 2ᅵᅵᅵᅵ
ð=
(â324 â 103 â ð ðð41.46 â 2 â 3575.32 â 17.86)
226.57= â1,510.48 ððð/ð 2
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
10
- The relative velocity and acceleration
- The angular velocity of acceleration of the âRCaseâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ð« 3.58 3.58 0
ᅵᅵ 315 315.12 0.038
ᅵᅵ 17.88 17.86 0.112
ᅵᅵ -1510 -1510.43 0.028
11
Dynamics of Rigid Bodies.04
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.114, P.2.253
Type of Analysis : Kinematics of particles, Linear Motion
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Initial angle ð0 ð 3â ððð
âOAâ Length ð 0.8 ð
Where, ðð = âð
ð
ð = ð0 â ð ððððð¡
ᅵᅵ = ð0 â ðð â ððð ððð¡
ᅵᅵ = âð0 â ðð2 â ð ððððð¡
A tangential acceleration ðð¡ of the pendulum OA is,
12
ðð¡ = ð â ᅵᅵ = âð â ð0 â ðð2 â ð ððððð¡ = âð â ð0 â
ð
ðâ ð ððððð¡ = âð0 â ð â ð ððððð¡
A normal acceleration ðð is,
ðð = ð â ᅵᅵ2 = ð â ð02 â ðð
2 â ððð 2ððð¡ = ð â ð02 â
ð
ðâ ððð 2ððð¡ = ð0
2 â ð â ððð 2ððð¡
Thus, the acceleration of the pendulum OA is,
a = â(ðð¡2 + ðð
2) = ðð0 â â(ð ðð2ððð¡ + ð02 â ððð 4ððð¡)
A Period T of the pendulum OA is,
ð 4â = 0.4486, where T =2ð
ðð= 1.7943.
Calculate the extreme value,
ðð
ðð¡= ðð0 â
2ððâð ððððð¡âððð ððð¡â4ð02âððâððð 3ððð¡âð ððððð¡
2â(ð ðð2ððð¡+ð02âððð 4ððð¡)
= 0
ðð0 â 2ðð â ð ððððð¡ â ððð ððð¡ â (1 â 2ð02 â ððð 2ððð¡) = 0
1 â 2ð02 â ððð 2ððð¡ = 0
ððð 2ððð¡ = 1
2ð02
ððð ððð¡ = ±â1
2ð02 = ±0.67524
⎠ð¡ = 0.23688 ðð 0.66026 ð
During the 1/4 period, calculate the âaâ value to find the maximum value and
the minimum value when the time is 0, 0.23688, and 0.4468,
1) t=0
ð = 0°
ð = 10.76 ð ð 2â : the maximum value
2) t=0.23688
ð = 0.77241 ððð = 44.26°
ð = 9.026 ð ð 2â : the minimum value
13
3) t=0.4468
ð = 1.04718 ððð = 60.0°
ð = 10.27 ð ð 2â
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- Measure the acceleration when t=0, tâ0.23688 and tâ0.4468
14
Comparison of results
Object Value Theory RecurDyn Error(%)
ðððð [ð ððâ ] 9.026 9.002 0.266
ð[ð], ðœ[ð ðð] 0.23688, 44.26° 0.23688, 44.26° 0
ðððð [ð ððâ ] 10.76 10.76 0
ð[ð], ðœ[ð ðð] 0.0, 0° 0.0, 0° 0
15
Dynamics of Rigid Bodies.05
References: Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.129, P.3.2
Type of Analysis : Kinetics of particles, Linear Motion
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Initial velocity ð£0 7 ð/ð
Initial angle ð 15, 30 ððð
Mass ð 50 ðð
Friction coefficient ð 0.40 -
âð¹ðŠ = ð â ð â ð â ððð ð = 0
ð = ð â ð â ððð ð
ð = ð â ð â ð â ððð ð
16
âð¹ð¥ = ð â ð â ð ððð â ð â ð â ð â ððð ð = ð â ð
⎠ð = ð â ð ððð â ð â ð â ððð ð
(1) ð = 15°
ð15 = 9.81 â (ð ðð15 â 0.4 â ððð 15) = â1.251 ð/ð 2
ð£2 â ð£02 = 2ðð
02 â 72 = â2 â 1.251 â ð
ð = 19.584ð
ð£ = ð£0 + ðð¡
ð¡ = âð£0/ð = â7/(â1.251) = 5.596 ð ðð
(2) ð = 30°
ð30 = 9.81 â (ð ðð30 â 0.4 â ððð 30) = 1.507 ð/ð 2
â Sliding occurs continuously because the acceleration has plus valu
e
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The displacement on the X axis of the block when the inclination of the
slope is 15° (The block slides down the slope as 19.063 m and then it stops
when the time is 5.593 ss)
17
Comparison of results
Object Value Theory RecurDyn Error(%)
ðœ = ðð° ð¡ [ð ] 5.59 5.593 0.054
ð¥ [ð] 19.58 19.069 2.61
ðœ = ðð° ð¡ [ð ] - - -
ð¥ [ð] - - -
18
Dynamics of Rigid Bodies.06
References: Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.129, P.3.2
Type of Analysis : Kinetics of particles, Linear Motion
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Units
Initial length ð¥ðŽ, ð¥ðµ 0.4, 0.3 ðð
Decreasing rate ð£ðŽ, ð£ðµ 0.9, 1.2 ðð/ð
Rotation rate ððŽðµ 3 ððð/ð
ððŽðµ =ð£ðŽ
0.3=
0.9
0.3= 3 =
ð£ðµ
0.4
ð£ðµ = ð¥ï¿œï¿œ = 1.2 m/s
Geometrical Constraint
ð¥ðŽ2 + ð¥ðµ
2 = 0.52
19
Differentiate this equation, then,
2ð¥ðŽð¥ï¿œï¿œ + 2ð¥ðµð¥ï¿œï¿œ = 0
Differentiate one more time, then,
ð¥ðŽ â ð¥ï¿œï¿œ + ð¥ï¿œï¿œ2 + ð¥ðµ â ð¥ï¿œï¿œ + ð¥ï¿œï¿œ
2 = 0
ð¥ï¿œï¿œ = â(ð¥ðŽ â ð¥ï¿œï¿œ + ð¥ï¿œï¿œ
2 + ð¥ï¿œï¿œ2)
ð¥ðµ= â
(ð¥ï¿œï¿œ2 + ð¥ï¿œï¿œ
2)
ð¥ðµâ
ð¥ðŽ â ð¥ï¿œï¿œ
ð¥ðµ= â
0.92 + 1.22
0.3â
0.4
0.3ð¥ï¿œï¿œ
ð¥ï¿œï¿œ = â7.5 â4
3ð¥ï¿œï¿œ (1)
F.B.D. of the Body A
âð¹ð¥ = ð â ð â ððð ð = ððŽð¥ï¿œï¿œ
40 â ð â4
5= 2 â ð¥ï¿œï¿œ
ð =5
4â (40 â 2 â ð¥ï¿œï¿œ) = 50 â
5
2â ð¥ï¿œï¿œ
F.B.D. of the Body B
âð¹ðŠ = âð â ð ððð = ððµð¥ï¿œï¿œ
âð â3
5= 3 â ð¥ï¿œï¿œ
ð = â5
3â 3 â ð¥ï¿œï¿œ = â5 â ð¥ï¿œï¿œ
P=40N
T
ð
ððŽ
ð¥ï¿œï¿œ
ððµ ð¥ï¿œï¿œ
ð
T
20
ð = 50 â5
2â ð¥ï¿œï¿œ = â5 â ð¥ï¿œï¿œ
ð¥ï¿œï¿œ = â10 +1
2â ð¥ï¿œï¿œ (2)
From equation (1) and (2),
ð¥ï¿œï¿œ = â7.5 â4
3ð¥ï¿œï¿œ = â10 +
1
2â ð¥ï¿œï¿œ
⎠ð¥ï¿œï¿œ = 1.364 ð/ð
⎠ð¥ï¿œï¿œ = â9.318 ð/ð
⎠ð = 46.59 ð
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The acceleration on the X axis of âBody Aâ and acceleration on the Y axis of
âBody Bâ
21
- The total reaction force of âRevJoint2â â âFM_Reaction_Forceâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ððš [ð/ðð] 1.36 1.361 0.074
ðð© [ð/ðð] -9.318 -9.315 0.032
ð» [ð] 46.59 46.592 0.004
22
Dynamics of Rigid Bodies.07
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.182, P.3.139
Type of Analysis : Kinetics of Particles, Work-energy Equation
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Spring coefficient ð 24 ð/ð
Initial length ð0 0.375 ð
Law of Energy Conservation
ð¥ðŽ = 0.4996 [ð]
âðµ = 0.250 [ð]
ð¥ðµ = 0.125 [ð]
ððâðŽ +1
2ðð£ðŽ
2 +1
2ðð¥ðŽ
2=ððâðµ +1
2ðð£ðµ
2 +1
2ðð¥ðµ
2
0 + 0 +1
2ðð¥ðŽ
2=ððâðµ +1
2ðð£ðµ
2 +1
2ðð¥ðµ
2
23
ð£ðµ = âð
ð(ð¥ðŽ
2 â ð¥ðµ2) â 2ðâðµ
ð£ðµ = 1.1558 ð/ð
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- Measure the velocity when the displacement on the X axis of the âSliderâ
is zero
Comparison of results
24
Object Value Theory RecurDyn Error(%)
ðð©[ð/ð] 1.1558 1.1556 0.017
25
Dynamics of Rigid Bodies.08
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.182, P.3.141
Type of Analysis : Kinetics of Particles, Work-energy Equation
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Units
Initial height âðŽ 4.5 ð
Initial velocity ð£ðŽ 0 ð/ð
Law of Energy Conservation
Calculate the velocity of the point âBâ
The point âBâ is the lowest point, thus âðµ = 0 ð, ð£ðµ =?
ððâðŽ +1
2ðð£ðŽ
2 = ððâðµ +1
2ðð£ðµ
2 = ððâð¹ +1
2ðð¥ð¹
2 +1
2ðð£ð¹
2
26
A-B : ððâðŽ +1
2ðð£ðŽ
2 = ððâðµ +1
2ðð£ðµ
2
ð£ðµ = â2ðâðŽ = 9.395ð/ð
Calculate the maximum deformation of the spring
A point of the spring that occurs the maximum deformation is called âFâ, âð¹ =
1.5ð, ð£ð¹ = 0, ð¥ð¹ = ð¿
A-F : ððâðŽ +1
2ðð£ðŽ
2 = ððâð¹ +1
2ðð¥ð¹
2 +1
2ðð£ð¹
2
ð¥ð¹ = ð¿ = â2ðð(âðŽ â âð¹)
ð= 54.24 ðð
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- Measure the maximum value from the velocity plot of the âSlider - Body
Aâ and the minimum deformation from the deformation plot of âSpringâ
27
Comparison of results
Object Value Theory RecurDyn Error(%)
ðœð© [ð/ð] 9.395 9.393 0.213
ð¹ [ðð] 54.24 54.08 0.295
28
Dynamics of Rigid Bodies.09
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.263, P.3.350
Type of Analysis : Kinetics of Particles, Work-energy Equation
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
- Calculate the time âtâ and the angle âΞâ when the particle is separated from
the tube with ð = 1[ð], ð0 = 0.5 [ððð/ð ] = ðððð ð¡, where ârâ is the length of the
tube and âð0â is the angular velocity of the tube.
Given Symbol Value Unit
Tube length ð 1 ð
Angular velocity ð0 0.5 ððð/ð
Newtonâs Equation of Motion
âð¹ð = ððð
29
ððð ððð = ð(ᅵᅵ â ðᅵᅵ2)
ᅵᅵ â ðᅵᅵ2 = ðð ððð
Ξ = ð0 + ð0ð¡ +1
2ðŒð¡2, where ð0 = 0, ð0 = ðððð ð¡, and ðŒ = 0
Thus, ð = ð0ð¡
ᅵᅵ â ð02ð = ðð ððð0ð¡
That is the second order linear differential equation,
ð = ðð + ðð = ð¶1ðâð0ð¡ + ð¶2ð
ð0ð¡ â (ð
2ð02â ) ð ððð0ð¡
ᅵᅵ = âð¶1ð0ðâð0ð¡ + ð¶2ð0ð
ð0ð¡ â ð0 (ð
2ð02â ) ððð ð0ð¡
Where ð¡ = 0, and ð = 0, ᅵᅵ = 0
ð¶1 + ð¶2 = 0
âð¶1ð0 + ð¶2ð0 âð
2ð0â = 0
ð¶1 = âð
4ð02â
ð¶2 =ð
4ð02â
ð =ð
4ð02â (âðâð + ðð â 2ð ððð)
Where ð = 1[ð], ð0 = 0.5 [ððð/ð ]
Solve the nonlinear equation using numerical analysis which satisfies the
equation, âðâð + ðð â 2ð ððð â 0.10194 = 0
It is difficult to solve the nonlinear equation by hand. So the following solution
is calculated using numerical analysis on a computer (Newton-Rapson Method,
MatLab, Maple, Mathmatica, etc.)
ð = 0.5347 [ððð]
Where ð = ð0ð¡, thus ð¡ = ðð0
â = 1.07 [sec]
Numerical Solution - RecurDyn
30
1. Modeling
2. Plot the results
- Measure the time when the displacement of the âSliderâ is â1mâ
- A plot of the rotation angle of the âGuiderâ by the relative displacement
of âTraJoint1â
31
Comparison of results
Object Value Theory RecurDyn Error(%)
ð [ð] 1.07 1.068 0.187
ðœ [ð ðð] 30.653 30.681 0.009
32
Dynamics of Rigid Bodies.10
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.287, P.4.31
Type of Analysis : Kinetics of Particles, Work-energy Equation
Type of Element : Particle (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Mass ð 1 ðð
Length ð 0.5 ð
Law of Momentum Conservation
â«ð¹ðð¡ = âðº = ðºðŽâ² + ðºðµâ² â 0
0 = 2ð â ð£ð¥ + 2ð â (ð£ð¥ â ðᅵᅵ)
2ð£ð¥ = ðᅵᅵ (1)
33
For the entire system,
ð1â2 = âð + âðð
âð =1
2â 2ð â ð£ð¥
2 +1
2â 2ð â (ð£ð¥ â ðᅵᅵ)2 â 0
= ð â (2ð£ð¥2 â 2ðᅵᅵð£ð¥ + ð2ᅵᅵ2)
âðð = â2ðð â 2ð = â4ððð
ð1â2 = 0
ð â (2ð£ð¥2 â 2ðᅵᅵð£ð¥ + ð2ᅵᅵ2) â 4ððð = 0
2ð£ð¥2 â 2ðᅵᅵð£ð¥ + ð2ᅵᅵ2 â 4ðð = 0
Form equation (1), 2ð£ð¥ = ðᅵᅵ
2ð£ð¥2 â 4ð£ð¥
2 + 4ð£ð¥2 â 4ðð = 0
ð£ð¥ = â2ðð
ᅵᅵ = 2ð£ð¥
ð= 2â
2ð
ð (2)
ð = 1kg, ð = 0.5ð, ð = 9.81ð/ð 2
ð£ð¥ = â2ðð = â2 â 9.81 â 0.5 = 3.13 ð/ð
ᅵᅵ = 2ð£ð¥
ð= 2â
2ð
ð= 2â
2 â 9.81
0.5= 12.53 ððð/ð
Numerical Solution - RecurDyn
1. Modeling
34
2. Plot the results
- The angular velocity of the âPendulumâ
(X axis : rotation angle, Y axis : rotation velocity)
- The velocity of the âCartâ
(X axis : rotation angle of the âPendulumâ, Y axis : Velocity on X axis)
Comparison of results
Object Value Theory RecurDyn Error(%)
ᅵᅵ [ððð /ð] 12.53 12.53 0
ðð [ð/ð] 3.13 3.13 0
35
Dynamics of Rigid Bodies.11
Reference : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.353, E.5.9
Type of Analysis : Plane Kinematics of Rigid Bodies, Relative Velocity
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ðððµ 50 ð ððð/ð
Length ð1, ð2 0.25, 0.1 ð
âOBâ length ð 0.125 ð
âAOBâ angle Ξ 60 ððð
Geometrical Constraint
ðœ ð
B
A O
G
ð1 ð
ð2 ðððµ
ð£ðµ
ð£ðŽ
ð£ðŽ/ðµ
ðŸ
ððŽðµ
ððŽðµ
36
ð
ð ðððœ=
ð
ð ððð
ðœ = ð ððâ1 (ð
ðâ ð ððð) = ð ððâ1 (
0.125
0.35â ð ðð60) = 18.02°
⎠ðŸ = 71.98°
ð£ðµ = ð â ðððµ = 0.125 à 50ð = 19.63 ð/ð
The Equation of Relative Velocity
ð£ðŽ = ð£ðµ + ð£ðŽ/ðµ
ð£ðŽ
ð ðð78.02=
ð£ðµ
ð ðððŸ
ð£ðŽ =ð£ðµ
ð ðð71.98 â ð ðð78.02 = 20.19 ð/ð
ð£ðŽ/ðµ
ð ðð30=
ð£ðµ
ð ðððŸ
ð£ðŽ/ðµ =ð£ðµ
ð ðð71.98 â ð ðð30 = 10.32 ð/ð
ððŽðµ =ð£ðŽ/ðµ
ð = 29.49 ððð/ð
ð£ðº = ð£ðŽ + ð£ðº/ðŽ
= 20.19ð + 29.49ᅵᅵ à (0.25ððð 18.02ð + 0.25ð ðð18.02ð) = 20.19ð + (7.01ð â 2.28ð) =
17.91ð + 7.01ð
ð£ðº = â(17.912 + 7.012) = 19.23 ð/ð
ð£ðµ ð£ðŽ/ðµ
ð£ðŽ
ðŸ 30°
78.02°
37
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The velocity and angular velocity of the âCMâ of the âCRodâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ðð® [ð/ð] 19.23 19.24 0.27
ððšð© [ððð /ð] 29.5 29.53 0.346
38
Dynamics of Rigid Bodies.12
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.346, P.5.54
Type of Analysis : Plane Kinematics of Rigid Bodies, Absolute Motion
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ðððŽ 0.5 ððð/ð
âABCâ angle ð 45 ððð
âBCâ length ð 0.2 ð
âABâ length ð 0.1 ð
Geometrical Constraint
With the trigonometric function formula,
ð
ð ðððŒ=
ð
ð ðððœ
ð
sin (ð â (ð + ðœ))=
ð
ð ðððœ
39
ð
sin (ð + ðœ)=
ð
ð ðððœ
ð â ð ðð(ð + ðœ) = ð â ð ðððœ (1)
Differentiate both sides of this equation,
ð â (ᅵᅵ + ᅵᅵ) â ððð (ð + ðœ) = ð â ᅵᅵ â ððð ðœ
Where, ðððŽ = ᅵᅵ, ððŽðµ = ð = ᅵᅵ
(ð â ððð ðœ â ð â ððð (ð + ðœ)) ᅵᅵ = ð â ᅵᅵ â ððð (ð + ðœ)
ð = ᅵᅵ =ð â ᅵᅵ â ððð (ð + ðœ)
(ð â ððð ðœ â ð â ððð (ð + ðœ))
With the equation (1),
ð â ð ðð(ð + ðœ) = ð â ð ðððœ
ð â (ð ððð â ððð ðœ + ððð ð â ð ðððœ) = ð â ð ðððœ
Multiply both sides of this equation by 1
ððð ðœ,
ð â (ð ððð + ððð ð â ð¡ðððœ) = ð â ð¡ðððœ
(ð â ð â ððð ð) â ð¡ðððœ = ð â ð ððð
ð¡ðððœ =ð â ð ððð
(ð â ð â ððð ð)
ðœ = ð¡ððâ1ð â ð ððð
ð â ð â ððð ð
If, Ξ = 45°, ðððŽ = ᅵᅵ = 3 rad/s , ð = 0.2 ð , ð = 0.1 ð,
ðœ = ð¡ððâ10.1 â ð ðð45
0.2 â 0.1 â ððð 45= 28.68°
ð = ᅵᅵ =0.1 â 3 â ððð (45 + 28.68)
(0.2 â ððð 28.68 â 0.1 â ððð (45 + 28.68))= 0.572 ððð/ð
40
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The angular velocity on Z axis of the link âABâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ððšð© [ððð /ð] 0.572 0.572 0
41
Dynamics of Rigid Bodies.13
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.347, P.5.57
Type of Analysis : Plane Kinematics of Rigid Bodies, Absolute Motion
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ð0 3 ððð/ð
âABâ length ð 0.2 ð
âOBâ length ð 0.1 ð
âAOBâ angle ð 60 ððð
Geometrical Constraint
ð â ð ðððœ = ð â ð ððð (1)
ð ðððœ =ð
ðð ððð
42
ððð ðœ =âð2 â ð2ð ðð2ð
ð= â1 â (
ð
ðð ððð)2
Differentiate both sides of the equation (1),
ð â ᅵᅵ â ððð ðœ = ð â ᅵᅵ â ððð ð (2)
ððŽðµ = ᅵᅵ =ð â ᅵᅵ â ððð ð
ð â ððð ðœ=
ð
ðð0
ððð ð
â1 â (ððð ððð)2
Differentiate both sides of the equation (2),
ð â ᅵᅵ â ððð ðœ â ð â ᅵᅵ2 â ð ðððœ = ð â ᅵᅵ â ððð ð â ð â ᅵᅵ2 â ð ððð
Where ð0 = ᅵᅵ = ðððð ð¡, thus ᅵᅵ = 0
ðŒðŽðµ = ᅵᅵ =ð â ᅵᅵ2 â ð ðððœ â ð â ᅵᅵ2 â ð ððð
ð â ððð ðœ=
ð âð2
ð2ð0
2 ððð 2ð
1 â (ððð ððð)2
ððð ððð â ð â ð0
2 â ð ððð
ð â â1 â (ððð ððð)2
=ð
ðð0
2 â ð ððð
ð2
ð2ððð 2ð
1 â (ððð ððð)2
â 1
â1 â (ððð ððð)2
=ð
ðð0
2 â ð ððð
ð2
ð2ððð 2ð â 1 +
ð2
ð2ð ðð2ð
(1 â (ððð ððð)2)
3/2=
=ð
ðð0
2 â ð ððð
ð2
ð2â 1
(1 â (ððð ððð)2)3/2
ðœ
ð ð â ð ððð
âð2 â ð2ð ðð2ð
43
The angular velocity and the angular acceleration of the link âABâ
Where ð = 0.2 ð , ð = 0.1 ð , ð0 = 3 ððð/ð , ð = 60°, thus
ððŽðµ =ð
ðð0
ððð ð
â1 â (ððð ððð)2
=0.1
0.2â 3 â
ððð 60
â1 â (0.10.2
ð ðð60)2
= 0.832 ððð/ð
ðŒðŽðµ =ð
ðð0
2 â ð ððð
ð2
ð2â 1
(1 â (ððð ððð)2)
3/2=
0.1
0.2â 32 â ð ðð60
0.12
0.22 â 1
(1 â (0.10.2
â ð ðð60)2)3/2
= â3.99 rad/ð 2
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The angular velocity and angular acceleration on Z axis of the link âABâ
44
Comparison of results
Object Value Theory RecurDyn Error(%)
ððšð© [ððð /ð] 0.832 0.832 0
ð¶ðšð© [ððð /ðð] -3.99 -3.99 0
45
Dynamics of Rigid Bodies.14
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.359, P.5.80
Type of Analysis : Plane Kinematics of Rigid Bodies, Relative Velocity
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ðððµ 0.8 ððð/ð
âABâ length ð 0.5 ð
âOBâ length ð 0.25 ð
âOA-OBâ angle β 20 ððð
46
The âOBâ rotates on the point of â0â, thus
ð£ðµ = ðððµ à ðððµ = â0.8ᅵᅵ à 0.25ð = â0.2ð
ððŽ = â(ð2 â ð2) = 0.433
Ξ = ððð â1(ð ðâ ) = 60°
ðŒðŒð¶ðŽ = ððŽ/ððð ðœ = 0.461
ððŽðµ =ð£ðµ
ðŒðŒð¶ðµâ = 0.20.408â = 0.49
ð£ðŽ = ððŽðµ à ðððŽ = â0.49ᅵᅵ à (0.158ð â 0.433ð) = â0.0774ð â 0.2122ð
⎠ð£ðŽ = 0.2259 [ð/ð ]
ð£ðµ
ð£ðŽ
r
ð
ðœ
ð O B
A
â(ð2 â ð2)
ðððµ
ð ð
C
47
ð£ð¶ = ð£ðµ + ð£ð¶ðµ
= ð£ðµ + ððŽðµ à ððµð¶ = â0.2ð â 0.49ᅵᅵ à (â0.25 â ððð 60ð â 0.25 â ð ðð60ð)
= â0.1061ð â 0.1388ð
⎠ð£ð¶ = 0.1747 [ð/ð ]
Numerical Solution - RecurDyn
1. Modeling
ð£ðµ
ð£ðŽ
r
ð
ðœ
ð
O B
A
â(ð2 â ð2)
ðððµ
C
IC
ðœ
â(ð2 â ð2) â ð¡ðððœ
â(ð2 â ð2)/ððð ðœ
ððŽðµ
48
2. Plot the results
- Measure the time when the displacement of the âSliderâ is â1mâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ððš [ð/ð] 0.2259 0.2261 0.089
ððª [ð/ð] 0.1747 0.1746 0.057
49
Dynamics of Rigid Bodies.15
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.361, P.5.89
Type of Analysis : Plane Kinematics of Rigid Bodies, Instantaneous Center of Zero
Velocity
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ðððµ 0.5 ððð/ð
Angle tan ð 4/3 -
50
The Velocities of Each End Point of the Link âABâ (âAâ and âBâ) and the
Relative Velocity of the âAâ with Respect to the âBâ
ð£ðµ = âðððµ ᅵᅵ à ðð = â0.5ᅵᅵ à 0.12ð = 0.06ð
ð£ðŽ = âððŽð¶ ᅵᅵ à (â0.12ð + 0.16ð) = 0.16 â ððŽð¶ð + 0.12 â ððŽð¶ð
ð£ðŽ/ðµ = ððŽðµ à ððµðŽ = âððŽðµ ᅵᅵ à (â0.24ð + 0.04ð) = 0.04 â ððŽðµð + 0.24 â ððŽðµð
Calculate the Accelerations of the Link âABâ and the Link âAC
ð£ðŽ = ð£ðµ + ð£ðŽ/ðµ
0.16 â ððŽð¶ð + 0.12 â ððŽð¶ð = 0.06ð + 0.04 â ððŽðµð + 0.24 â ððŽðµð
ð : 0.16 â ððŽð¶ = 0.06 + 0.04 â ððŽðµ
ð : 0.12 â ððŽð¶ = 0.24 â ððŽðµ
⎠ððŽðµ = 0.214 rad/s
⎠ððŽð¶ = 0.429 rad/s
A B
O C
ð
ð = 200 ð = 120
ðððµ
ððŽðµ
ððŽð¶
120
160
120
ð£ðµ
ð£ðŽ
ð¥
ðŠ
51
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The angular velocity on Z axis of the âABâ and âC
Comparison of results
Object Value Theory RecurDyn Error(%)
ððšð© [ððð /ð] 0.214 0.214 0
ððšðª [ððð /ð] 0.429 0.429 0
52
Dynamics of Rigid Bodies.16
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.405, P.5.195
Type of Analysis : Plane Kinematics of Rigid Bodies
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Initial velocity ð£ðŽ 1.2 ð/ð
Angular velocity ððð¶ 2 ððð/ð
The Relative Velocity Equation of the points âAâ, âBâ, and âCâ
ð£ðµ = ð£ðŽ + ð£ðµ/ðŽ = ð£ð¶ + ð£ðµ/ð¶
ð£ð¶ = âððð¶ ᅵᅵ à 0.3ð = 0.6ð
A Graph of the Velocity Relationship
53
ð£ðµ/ð¶ =ð£ðŽ â ð£ð¶
tan ð=
0.6
3/4= 0.8
ððµð¶ =ð£ðµ/ð¶
ðµð¶ =
0.8
0.4= 2 ððð/ð
ð£ðµ = ð£ðŽ + ð£ðµ/ðŽ = ð£ð¶ + ð£ðµ/ð¶ = 0.6ð + 0.8ð
The Angular Acceleration of Link âABâ
ð£ðµ/ðŽ =ð£ðŽ â ð£ð¶
sinð=
0.6
3/5= 1.0
ððŽðµ =ð£ðµ/ðŽ
ðŽðµ =
1.0
0.5= 2 ððð/ð
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The angular velocity on Z axis of the âBCâ
ð£ðµ/ðŽ
ð£ðŽ=1.2
ð£ðµ
ð£ð¶ = 0.6
ð£ðµ/ð¶
ð
54
Comparison of results
Object Value Theory RecurDyn Error(%)
ðð©ðª [ððð /ð] 2 2 0
55
Dynamics of Rigid Bodies.17
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.405, P.5.198
Type of Analysis : Plane Kinematics of Rigid Bodies, Linear Motion
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ðððŽ 3 ððð/ð
Angle ð 30 ððð
Geometrical Constraint - Second Law of Cosines
ð¶ðŽ 2 = 0.32 + 0.12 â 2 à 0.1 à 0.3 à cos 120 = 0.13
ð¶ðŽ = 0.361
0.1
sinðœ=
ð¶ðŽ
sin120
56
β = ð ððâ1(0.1
ð¶ðŽ sin120) = ð ððâ1 (
0.1
0.361sin120) = 13.88°
γ = 180 â 120 â 13.88 = 46.12°
Calculate a Velocity
ð£ðŽ = ðððŽ â ððŽ = 3 â 0.1 = 0.3 ððð/ð
ðð¶ðŽ =ð£ðŽ cos ðŸ
ð¶ðŽ =
0.3 â cos 46.12
0.361= 0.576 ððð/ð
Where, ðð¶ðŽ = ððµð¶
ð£ðµ = ððµð¶ â ðµð¶ = 0.576 â 0.5 = 0.288 ð/ð
ðŽ
ðµ
ð·
ð¶
ð
ð£ðµ
ð£ðŽ
ð
ðŸ
0.3
0.1
ðððŽ
ðœ
57
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The acceleration in the Z direction of âCBâ and the velocity of the point
âBâ of âBDâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ðð©ðª [ððð /ð] 0.576 0.577 0.174
ðð© [ð/ð] 0.288 0.286 0.694
58
Dynamics of Rigid Bodies.18
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.408, P.5.204
Type of Analysis : Plane Kinematics of Rigid Bodies
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ð0 5 ððð/ð
ð£ðµ = ð0 à ðððµ = â5ᅵᅵ à (0.25 â sin22.62 ð + 0.25 â cos22.62 ð) = 1.154ð â 0.481ð
600
250
ð0 650
ð£ðŽ
ðµ
ðŽ ð
ð£ðµ
ðœ
ð¥
ðŠ
59
β = ð¡ððâ1250
600= 22.62°
1. The Equation of Relative Velocity
If a point located on the point âCâ of the rod âBDâ is called âAâ,
ð£ðµ = ð0 à ðððµ = â5ᅵᅵ à (0.25 â sin 22.62 ð + 0.25 â cos 22.62 ð) = 1.154ð â 0.481ð
ð£ðŽ = ð£ðµ + ð£ðŽ/ðµ = ð£ðµ + ððŽðµ à ððŽ/ðµ
ð£ðŽ = ð£ðµ cosðœ = 1.25 â cos 22.62 = 1.154
ð£ðŽ/ðµ = ð£ðµ sinðœ = 1.25 â sin 22.62 = 0.481 = ððŽðµ â ððŽ/ðµ
ððŽ/ðµ = 0.65, then the angular velocity is ððŽðµ = 0.74 ððð/ð .
These velocities as a vector are,
ð£ðŽ = 1.154ð
ð£ðŽ/ðµ = 0.481ð (1)
ððŽ/ðµ = â0.65ð
If the angular velocity is assumed to be ððŽðµ = 0.74ᅵᅵ, then
ð£ðŽ/ðµ = ððŽðµ à ððŽ/ðµ = 0.74ᅵᅵ à (â0.65ð) = â0.481ð (2)
Since the signs of (1) and (2) are different, ððŽðµ = â0.74ᅵᅵ
Finally, the angular velocity is ððŽðµ = â0.74ᅵᅵ because the equation (1) and the
equation (2) have different signs.
ð£ðŽ
ð£ðµ
ð£ðŽ/ðµ = ððŽðµ à ððŽ/ðµ
ðœ
60
2. The Equation of Relative Acceleration
ððŽ = ððµ + ððŽðµ à ððŽ/ðµ + ððŽðµ à (ððŽðµ à ððŽ/ðµ ) + 2ððŽðµ à ð£ððð + ðððð
ððµ = ð0 à (ð0 à ðððµ ) = â5ᅵᅵ à (â5ᅵᅵ à (0.25 â sin 22.62 ð + 0.25 â cos 22.62 ð))
= â2.404ð â 5.769ð
ððŽðµ à ððŽ/ðµ = ððŽðµ ᅵᅵ à (â0.65ð) = â0.65 â ððŽðµ ð
ððŽðµ à (ððŽðµ à ððŽ/ðµ ) = â0.74ᅵᅵ à (â0.74ᅵᅵ à (â0.65ð)) = 0.356ð
ð£ððð = 0
ðððð = 0
ððŽ = ððŽð¥ð + ððŽðŠð = â2.404ð â 5.769ð â 0.65 â ððŽðµ ð + 0.356ð
ð : ððŽð¥ = â2.404 + 0.356 = â2.048
ð : ððŽðŠ = â5.769 â 0.65 â ððŽðµ
Consider Coriolis acceleration of the pivoted collar that rotates about a fixed
axis through the point âCâ,
ððŽðŠ = 2ðð¶ à ð£ðŽ = â2 â 0.74ᅵᅵ à 1.154ð = â1.708ð
ððŽðŠ = â5.769 â 0.65 â ððŽðµ = â1.708
⎠ððŽðµ = â6.248 rad/s
⎠ᅵᅵ = ððŽðµ = â6.248ᅵᅵrad
s
61
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The angular acceleration about the Z axis of âBDâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ð¶ð©ð« [ððð /ðð] -6.25 -6.24 0.16
62
Dynamics of Rigid Bodies.19
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.408, P.5.212
Type of Analysis : Plane Kinematics of Rigid Bodies
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ð 157.08 ððð/ð
âABâ length ð 0.35 ð
âOBâ length ð 0.125 ð
63
Geometrical Constraint
ð¥ = ð â ððð ðœ + ð â ððð ð
ð£ðŽ = âᅵᅵ = ð â ᅵᅵ â ð ðð ðœ + ð â ᅵᅵ â ð ðð ð
ð â ð ðð ðœ = ð â ð ðð ð
Calculate a Velocity
Differentiate both sides of the equation,
ð â ᅵᅵ â ððð ðœ = ð â ᅵᅵ â ððð ð
ᅵᅵ =ð
ðâððð ð
ððð ðœâ ᅵᅵ =
ð
ðâ ᅵᅵ â
ððð ð
â1 â ð ðð2ðœ=
ð
ðâ ᅵᅵ â
ððð ð
â1 â (ððsinð)
2=
ð â ððð ð
â(ðð)2
â ð ðð2ð
⎠ð£ðŽ = ð âð â ððð ð
â(ðð)
2
â ð ðð2ð
âð
ðsinð + ð â ð â ð ðð ð = ð â ð â ð ðð ð â
[
1 +ððð ð
â(ðð)
2
â ð ðð2ð]
Where, ð = 0.35 ð , ð = 0.125 ð , ð = 1500 ððð£/ð = 1500 â 2ð/60 = 157.08 ððð/ð
ð£ðŽ = 19.635 â ð ðð ð â [1 +ððð ð
â7.84 â ð ðð2ð]
ð£ðŽ O A
B
ðœ ð
ð
ð ð
ð¥
ðŠ
ð¥
64
⎠ð£ðŽððð¥ = 20.865 ð/ð ðð¡ ð = 1.2566 ððð = 72°
Calculate the Acceleration
Differentiate the equation to calculate the acceleration,
ððŽ =ðð£ðŽ
ðð¡= ð â ð â
[
ᅵᅵ â ððð ð +1
2
2ᅵᅵ cos2ð â â(ðð)
2
â ð ðð2ð â sin2ðâ2 sinð â cos ð â ᅵᅵ
2â(ðð)
2
â ð ðð2ð
(ðð)
2
â ð ðð2ð
]
= ð â ð2 â
[
ððð ð +
cos 2ð â â(ðð)2
â ð ðð2ð â sin 2ðâ
12
sin2ð
2â(ðð)
2
â ð ðð2ð
(ðð)
2
â ð ðð2ð
]
= ð â ð2 â
[
ððð ð +
cos 2ð â [(ðð)
2
â ð ðð2ð] +14 ð ðð22ð
[(ðð)
2
â ð ðð2ð]
3/2
]
= ð â ð2 â
[
ððð ð +ð
ðâ(ððð 2Ξ â ð ðð2Ξ)â [1 â (
ðð)2ð ðð2ð] +
14 (
ðð)2(4 â ð ðð2Ξ â ððð 2Ξ)
[1 â (ðð)2ð ðð2ð]
3/2
]
65
⎠ððŽ = ð â ð2 â
[
ððð ð +ð
ðâ[1 â 2ð ðð2Ξ + (
ðð)2ð ðð4ð]
[1 â (ðð)2ð ðð2ð]
3/2
]
Where, ð = 0.35 ð , ð = 0.125 ð , ð = 1500 ððð£/ð = 1500 â 2ð/60 = 157.08 ððð/ð
ððŽ = 3084.27 â [ððð ð + 0.357 â[1 â 2ð ðð2Ξ + 0.128 â ð ðð4ð]
[1 â 0.128 â ð ðð2ð]3/2]
ððŽððð¥ = 4172.97 ð/ð 2 ðð¡ ð = 0°
ððŽ = â1983.19 ð/ð 2 ðð¡ ð = 180°
ððŽ = 0 ð/ð 2 ðð¡ ð = 1.26125ððð = 72.26°
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
66
- The velocity on X axis of the âPistonâ
(X axis : the angle of the âRevJoint1â,Y axis : The velocity on X axis
of the âPistonâ)
- Calculate of the âðâ value when ððŽ = 0 ð/ð 2
Comparison of results
Object Value Theory RecurDyn Error(%)
ðœðš,ððð [ð/ð] 20.865 20.875 0.048
ðœ [ððð ] 1.2566 1.2566 0
67
Dynamics of Rigid Bodies.20
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.433, E.6.4
Type of Analysis : Plane Kinematics of Rigid Bodies, Rotation about a Fixed Axis
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Mass ð 7.5 ðð
âOGâ length ðð 0.295 ð
Initial angle ð 60 degree
âðð = ðŒðα = ððᅵᅵ cos ð
mðð2α = 7.5 à 0.2952 à α = 7.5 à 9.81 à 0.25 à cos ð
α = 28.18 cos ð
⎠α = 14.091 rad/ð 2
68
â« ðððð
0
= â« ðŒððð
0
= â« 28.18 â cos ð ððð
0
1
2ð2 = 28.18 â sin ð
âŽ Ï = 6.99 rad/s
âð¹ð = ðð â ðð sin ð = ððð = ðᅵᅵð2
ðð â 7.5 Ã 9.81 Ã sin60 = 7.5 Ã 0.25 Ã 6.992
⎠ðð = 155.33 [N]
âð¹ð¡ = âðð¡ + ðð cos ð = ððð¡ = ðᅵᅵ α
âðð¡ + 7.5 Ã 9.81 Ã cos 60 = 7.5 Ã 0.25 Ã 14.091
⎠ðð¡ = 10.37 [N]
⎠O = âðð¡2 + ðð
2 = 155.68 [N]
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The total reaction force when the angle is â60.06°â
ð
ðð¡
ðð
ð
ðº
ðð
ᅵᅵ
ð
ð
ðŒ
69
Comparison of results
Object Value Theory RecurDyn Error(%)
ð¹ð¶ [ðµ] 155.65 155.67(60.06°) 0.013
70
Dynamics of Rigid Bodies.21
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.428, P.6.25
Type of Analysis : Plane Kinematics of Rigid Bodies, Translational Motion
Type of Element : Rigid Body (three parts)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Momentum ð 200 ð â ð
Angular velocity ð 5 ððð/ð
71
In the case of the link âCDâ,
ðð = rð2 = 0.6 Ã 52 = 15 ð/ð 2
In the case of the link âBDâ,
âððµ = ð·ð Ã 0.8 = ðððd
ð·ð Ã 0.8 = 25 Ã 0.6 Ã 52 Ã 0.5 = 187.5
⎠ð·ð = 235.375 [ð]
It is assumed that the mass of the link âABâ is ignored,
âððŽ = ð â ðµð¡ à 0.6 = ððŽðµðð¡ð1 â 0
ðµð¡ = 200/0.6 = 333.33
In the case of the link âBDâ,
âð¹ð¡ = ððð¡ = ðððŒ
ðµð¡ â ðð = ðððŒ
⎠α =(ðµð¡ â ðð)
(ðð)â = 5.872 [r/ð 2]
72
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The size of the reaction force of the Joint âDâ and the angular acceleration
of the link âCDâ
Comparison of results
Object Value Theory RecurDyn Error(%)
ð¹ð« [ð/ðð] 235.375 234.375 0.425
73
ð¶ [ððð /ðð] 5.871 5.872 0.017
74
Dynamics of Rigid Bodies.22
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.429, P.6.29
Type of Analysis : Plane Kinematics of Rigid Bodies, Translational Motion
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
âCEâ length ðð¶ðž 600 ðð
âCDâ length ðð¶ð· 1200 ðð
The Driving Constraint
Ξ =ð
6(1 â ððð
ðð¡
2)
Ξ =ð
6âð
2â ð ðð
ðð¡
2=
ð2
12â ð ðð
ðð¡
2
Ξ =ð2
12âð
2â ððð
ðð¡
2=
ð3
24â ððð
ðð¡
2
75
(a) Ξ = 0°, t = 0
Ξ = 0
Ξ =ð3
24
ðð¡ = ðᅵᅵ = 1.2 Ãð3
24= 1.55
âðð¹ = (ðð â ð¹ð·) â 0.6 = ððð¡ â 0.48
⎠ð¹ð· = 1714 [ð]
(b) t = 1 sec
Ξ =ð
6(1 â ððð
ð
2) =
ð
6
Ξ =ð2
12â ð ðð
ð
2=
ð2
12
Ξ =ð3
24â ððð
ð
2= 0
480 ðð
ð¹ð· ð¹ð¹ð
ð¹ð¹ð¡
ððð¡ ðº
ð· ð¹ 600
76
ðð¡ = 0
ðð = ðΞ2 = 1.2 à [ð2
12]
2
= 0.812
âðð¹ = ðð â 0.6 â ð¹ð· â 0.6 â cos ð = ððð â (0.6 â cos ð â 0.48 â sinð)
⎠ð¹ð· = 2178.13 [ð]
Numerical Solution - RecurDyn
1. Modeling
480 ðð
ð¹ð· ð¹ð¹ð
ð¹ð¹ð¡ ððð
ðº
ð· ð¹
ð ð ð
ð
ððð¡
77
2. Plot the results
- The reaction force of the revolute joint âDâ when the time is 0 and 1
Comparison of results
Object Value Theory RecurDyn Error(%)
ð«ð=ð [ðµ] 1714 1714 0
ð«ð=ð [ðµ] 2178 2178 0
78
Dynamics of Rigid Bodies.23
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.439, P.6.57
Type of Analysis : Plane Kinematics of Rigid Bodies, Rotation about a Fixed Axis
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Mass ð 6000 ðð
Torque ð 30 ð â ð
79
ðŒðº =1
126000 â [1.52 + 2.52] = 4,250 kg â ð2
ðŒð = ðŒðº + ðð2 = 4250 + 6000 â 2.052 = 29,465 kg â ð2
âðð = ðŒðα
ð = 30 = 29465 à ðŒ
ðŒ = 1.018 à 10â3 ððð/ð 2
Ξ = ð0 + ð0ð¡ +1
2ðŒð¡2
Calculate the time when the rotation angle, âΞ", is 45°,
Ξ0 = 0 , ð0 = 0 , Ξ = 45° =ð
4
ð
4=
1
2Ã 1.018 Ã 10â3 Ã ð¡1
2
ð¡1 = 39.28 ð
⎠ð¡ = 78.56 ð
Numerical Solution - RecurDyn
1. Modeling
ðº 2.05
0.8
1.5
1.25
1.25
ð T
O
80
2. Plot the results
- Measure the time when the rotation angle is 90°
Comparison of results
Object Value Theory RecurDyn Error(%)
ð [ð] 78.56 78.6 0.051
81
Dynamics of Rigid Bodies.24
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.454, P.6.92
Type of Analysis : Plane Kinematics of Rigid Bodies, General Plane Motion
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Mass ð 10 ðð
Angular velocity ð0 4.5 ððð/ð
Angular velocity ððŽðµ 3 ððð/ð
82
The Equation of Relative Velocity
ððµ = ððŽ + (ððµ/ðŽ)ð + (ððµ/ðŽ)ð¡
ððŽ = ððŽ â ð02 = 0.4 à 4.52 = 8.1 ð/ð 2
ð
O
B
A 0.4
1.0
ð0
ð£ðµ
0.6
0.8
ððµ
ð£ðŽ
ð
O
B
A
0.5
0.6
0.8
ððµ
ðŽð¥
ðŽðŠ
ð
ðð
ð(ððº/ðŽ)ð¡
ð(ððº/ðŽ)ð
0.5
ððŽðµ
G
ð(ððµ/ðŽ)ð¡
ð(ððµ/ðŽ)ð ðŒðŽðµ
(ððµ/ðŽ)ð¡ (ððµ/ðŽ)ð
ððŽ ððµ ð
83
(ððµ/ðŽ)ð = ðŽðµ â ððŽðµ2 = 1.0 à 32 = 9 ð/ð 2
tan ð =(ððµ/ðŽ)ð
(ððµ/ðŽ)ð¡
(ððµ/ðŽ)ð¡ = ðŽðµ â ðŒðŽðµ =(ððµ/ðŽ)ð
tan ð=
9
3/4= 12 ð/ð 2
ðŒðŽðµ =(ððµ/ðŽ)ð¡
ðŽðµ = 12 rad/ð 2
âððŽ = ðŒï¿œï¿œ + ðᅵᅵð = ðŒðºðŒ + ðððºð
ððº = ððŽ + (ððº/ðŽ)ð + (ððº/ðŽ)ð¡
(ððº/ðŽ)ð¡ = ððŒðŽðµ = 0.5 à 12 = 6 rad/ð 2
⎠âððŽ = ðð â 0.3 â ððµ â 0.6 =1
12ðð2 â ðŒðŽðµ + ð(ððº/ðŽ)
ð¡â 0.5 â ðððŽ â 0.4
10 Ã 9.81 Ã 0.3 â ððµ Ã 0.6 =1
12Ã 10 Ã 1.02 Ã 12 + 10 Ã 6 Ã 0.5 â 10 Ã 8.1 Ã 0.4
⎠ððµ = 36.38 [ð]
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The reaction force acting on the end of the roller âBâ, which is attached at
one end of the link âABâ
84
Comparison of results
Object Value Theory RecurDyn Error(%)
ðµð© [ðµ] 36.38 36.38 0
85
Dynamics of Rigid Bodies.25
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.457, P.6.105
Type of Analysis : Plane Kinematics of Rigid Bodies, General Plane Motion
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity ðððµ 100ð ððð/ð
Angular acceleration ðŒððµ 0 ððð/ð 2
ððŽðµ = 0.6 ðð, ðð = 0.82 ðð
ð1 = 0.075 ð, ð2 = 0.0325 ð
ð = 0.0425 ð, ððº = 0.028 ð
Calculate the velocity and the acceleration
ð£ðµ = ðððµ à 0.0425 = 13.35 m/s
A
O B
(ððŽ/ðµ)ð¡
ð
ð ðŒ = 0
(ððµ)ð ð£ðµ
(ððŽ/ðµ)ð ð£ðŽ
ð
G
ð1
ð2
86
ð£ðŽ = ð£ðµ = 13.35 m/s
⎠ððŽðµ = 0 ððð/ð
ððµ = (ððµ)ð=ðððµ â ðððµ2 = 0.0425 Ã (100ð)2 = 4.195 Ã 103 ð/ð 2
Ξ = ððð â1ðððµ
ð1 + ð2= ððð â1
0.0425
0.075 + 0.0325= 66.71°
The Equation of Relative Velocity
ððŽ = ððµ + (ððŽ/ðµ)ð + (ððŽ/ðµ)ð¡
(ððŽ/ðµ)ð¡ =(ððµ)ð
sinð=
4.195 Ã 103
sin 66.71= 4.567 Ã 103 ð/ð 2
ðŒðŽðµ =(ððµ/ðŽ)ð¡
ðŽðµ = 42483.7 rad/ð 2
ððŽ =(ððµ)ð
tan ð=
4.195 Ã 103
tan 66.71= 1.806 Ã 103 ð/ð 2
(1) Piston
âð¹ð¥ = ððððð¥
ðŽð¥ = 0
âð¹ðŠ = ðððððŠ
ðŽðŠ â ððð = ððððŽ
ððð â 0 : Ignore
ðŽðŠ = 0.82 à 1.806 à 103 = 1480.7 [ð]
ððð â 0 : Consider
ðŽðŠ = 0.82 à (9.81 + 1.806 à 103) = 1488.8 [ð]
ð
(ððµ)ð
ððŽ
(ððŽ/ðµ)ð¡
ðŽðŠ
ððð ðŽð¥
87
(2) The link âABâ
âððµ = ðŒðµðŒ + ᅵᅵ à ðððµ
ðŒðµ = ðŒðº + ðᅵᅵ2 = ðððº2 + ðᅵᅵ2 = ð(ððº
2 + ᅵᅵ2) = 0.6 â (0.0282 + 0.03252) = 0.00110415
ððµ = 4.195 Ã 103 ð/ð 2
ðŽð¥ â ð â sinð â ðŽðŠ â ðððµ â ððŽðµð â ᅵᅵ â cos ð = ðŒðµ â ðŒðŽðµ â ððŽðµ â ððµ â ᅵᅵ â sin ð
ððŽðµð â 0 : Ignore
ðŽð¥ à 0.1075 à sin 66.71 â 1488.8 à 0.0425
= 0.0011 Ã 42483.7 â 0.6 Ã 4195 Ã 0.0325 Ã sin 66.71
ðŽð¥ = 353.1 [ð]
⎠ðŽ = âðŽð¥2 + ðŽðŠ
2 = 1530.1 [ð]
ððŽðµð â 0 : Consider
ðŽð¥ = 353.9 [ð]
⎠ðŽ = âðŽð¥2 + ðŽðŠ
2 = 1530.3 [ð]
âððµ = ðŒï¿œï¿œ + ðᅵᅵð
ððº = ððµ + (ððº/ðŽ)ð + (ððº/ðŽ)ð¡
ðŒðº = ðððº2 = 0.6 à 0.0282 = 0.0004704
ððŽðµð â 0 : Ignore
ðŽð¥ à 0.1075 à sin 66.71 â 1488.8 à 0.0425
= 0.0004704 Ã 42483.7 + 0.6 Ã (â4195 Ã 0.0325 Ã sin 66.71
+ 0.03252 Ã 42483.7)
ðŽð¥ = 354.9 [ð]
⎠ðŽ = âðŽð¥2 + ðŽðŠ
2 = 1530.5 [ð]
ððŽðµð â 0 : Consider
ðµð¥
G
ð1
ðµðŠ
ðŽð¥
ðŽðŠ
ððŽðµð
ð
ðŒðŽðµ
ᅵᅵ
88
ðŽð¥ = 355.69 [ð]
⎠ðŽ = âðŽð¥2 + ðŽðŠ
2 = 1530.7 [ð]
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The reaction force of the joint when ð = 90°
(X axis : the angle of Revolute Joint âOâ, Y axis : the reaction force of
Revolute Joint âAâ)
89
Comparison of results
Object Value Theory RecurDyn Error(%)
ð¹ðš [ðµ] 1530.5 1531.3 0.052
90
Dynamics of Rigid Bodies.26
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.474, P.6.139
Type of Analysis : Plane Kinematics of Rigid Bodies, Work-energy Equation
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
0.4
0.8
G
0.18
1.0
0.4
0.8
G
0.18
0.2
ð
ð£ðµ
ð£ðŽ 0.4
0.4
91
Given Symbol Value Unit
Mass ðððŽ,ððŽðµ 8, 12 ðð
Gyration radius ððº 0.22 ð
ð£ðŽ = 0.4Ï m/s
ð£ðŽ = ð£ðµ = ð£
ððŽðµ = 0
Work-energy Equation
The link âOAâ
âðð = âðððŽðâ1 = â8 à 9.81 à 0.18 = â14.13
âT =1
2ðŒðð2 =
1
2à ðððŽðð
2 à [ð£ðŽ
0.4]2
=1
2Ã 8 Ã 0.222 Ã
ð£ðŽ2
0.42= 1.21 â ð£2
The link âABâ
âðð = âððŽðµðâ2 = â12 à 9.81 à 0.2 = â23.54
âT =1
2ððŽðµð£ðµ
2 =1
2à 12 à ð£ðµ
2 =1
2à 12 à ð£ðµ
2 = 6 â ð£2
The total system
ð1â2 = âT + âðð
0 = â14.13 + 1.21 â ð£2 â 23.54 + 6 â ð£2
⎠ð£ = 2.29 ð/ð
Numerical Solution - RecurDyn
1. Modeling
92
2. Plot the results
- The velocity of end B (X axis : the rotation angle of Revolute Joint âOâ, Y
axis : the velocity on X axis of âBâ)
Comparison of results
Object Value Theory RecurDyn Error(%)
ðð© [ð/ð] 2.29 2.29 0
93
Dynamics of Rigid Bodies.27
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.509, P.6.221
Type of Analysis : Plane Kinematics of Rigid Bodies
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Mass ð 1.2 ðð
Spring coefficient ð 100 ð/ð
ð1 = 0.6 m, ð2 = 0.2 m
ð = 0° â ð£0 = 0, ÎŽ0 = 0
The length of the spring without deformation, ð0 = 0.6â2 m
94
Work-energy Equation
The length of the spring with deformation when the angle is ð,
ð2 = 0.62 + 0.62 â 2 Ã 0.6 Ã 0.6 Ã cos(90 + ð) = 2 Ã 0.62 Ã (1 + sinð)
Thus, the equation of deformation of the spring is, ð¿ = ð â ð0 =
â2 Ã 0.62 Ã (1 + sin ð) â 0.6â2 = 0.6â2(â1 + sin ð â 1)
ð1â2 = âT + âðð = 0
ðððð¥ â ð£ = 0
ððâ1 +1
2ðð¥1
2 +1
2ðŒðð1
2 = ððâ2 +1
2ðð¥2
2 +1
2ðŒðð2
2
ð¥1 = 0 , ð1 = 0 , â2 = 0 , ð2 = 0
1.2 Ã 9.81 Ã 0.4 sinð =1
2Ã 100 Ã [0.6â2(â1 + sin ð â 1)]
2
4.7088 â sin ð = 36 â (2 + sinð â 2â1 + sinð)
31.2912 â sinð â 72â1 + sin ð + 72 = 0
If sinð = ð, then,
31.2912 â ð â 72â1 + X + 72 = 0
31.2912 â ð + 72 = 72â1 + X
Squares both sides of the equation,
979.14 â ð2 + 4505.93 â ð + 5184 = 5184(1 + X)
979.14 â ð2 â 678.07 â ð = 0
0.6
ð
0.6
0.2
0.6â2
0.6âsin ð
B
O
A G
95
ð â (979.14 â ð â 678.07) = 0
ð = 0 ðð ð = 0.6925
⎠ð = 0° ðð ð = 43.83°
⎠ðððð¥ = 43.83°
When ð is an arbitrary angle,
ð1â2 = âT + âðð = 0
ððâ1 +1
2ðð¥1
2 +1
2ðŒðð1
2 = ððâ2 +1
2ðð¥2
2 +1
2ðŒðð2
2
ð¥1 = 0 , ð1 = 0 , â2 = 0
1.2 Ã 9.81 Ã 0.4 sinð =1
2Ã 100 Ã [0.6â2(â1 + sin ð â 1)]
2+
1
2Ã
1
3Ã 1.2 Ã 0.82 Ã ð2
4.7088 â sin ð = 36 â (2 + sin ð â 2â1 + sinð) + 0.128 â ð2
0.128 â ð2 = â31.2912 â sinð + 72â1 + sin ð â 72
⎠ð = ââ244.4625 â sinð + 562.5 â â1 + sinð â 562.5
If, ð = ââ244.4625 â sin ð + 562.5 â â1 + sin ð â 562.5 = âð
ðð
ðð=
â244.4625 â cos ð + 562.5 âcos ð
2â1 + sinð
2âð= 0
â244.4625 â cos ð â 2 â â1 + sinð + 562.5 â cos ð = 0
96
cos ð â (â488.925 â â1 + sin ð + 562.5) = 0
cos ð = 0 ðð â1 + sinð = 1.1505
⎠ð = 90° ðð ð = 18.88°
⎠ð = 18.88° (0.3295 ððð) â ðððð¥ = 2.353 ððð/ð
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The angular velocity on Z axis of the âABâ and âCAâ
97
Comparison of results
Object Value Theory RecurDyn Error(%)
ðððð [ððð /ð] 2.353 2.353 0
ð [ð ðð] 18.88 19.07 1.001
98
Dynamics of Rigid Bodies.28
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.511, P.6.229
Type of Analysis : Plane Kinematics of Rigid Bodies
Type of Element : Rigid Body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Mass ð 30 ðð
Angular velocity ð0 4 ððð/ð
ð0 = 4 ððð/ð â ÎŽ0 = 0
ð = 2.4 m
ð2 = 0.2 m
ð = 3 kN/m
ðŒð =1
3ðð2 =
1
3Ã 30 Ã 2.42 = 57.6 m
99
Work-energy Equation
ð0 = 1.2â2 m
The length of the spring with deformation when the angle is ð,
ð2 = 1.22 + 1.22 â 2 Ã 1.2 Ã 1.2 Ã cos(90 + ð) = 2 Ã 1.22 Ã (1 + sinð)
Thus, the equation of deformation of the spring is, ð¿ = ð â ð0 =
â2 Ã 1.22 Ã (1 + sinð) â 1.2â2 = 1.2â2(â1 + sin ð â 1)
ð1â2 = âT + âðð = 0
ððâ1 +1
2ðð¥1
2 +1
2ðŒðð1
2 = ððâ2 +1
2ðð¥2
2 +1
2ðŒðð2
2
ð¥1 = 0 , â2 = 0
30 Ã 9.81 Ã 1.2 Ã (1 â cos ð) +1
2Ã 57.6 Ã 42
=1
2Ã 3000 Ã [1.2â2(â1 + sinð â 1)]
2+
1
2Ã 57.6 Ã ð2
813.96 â 353.16 â cos ð = 4320 â (2 + sin ð â 2â1 + sinð) + 28.8 â ð2
28.8 â ð2 = â4320 â sin ð + 8640â1 + sin ð â 7826.04 â 353.16 â cosð
⎠ð = ââ150 â sin ð + 300 â â1 + sinð â 271.74 â 12.263 â cosð
ð
ð
ð0
O
A
1.2
1.2
1.2
1.2
â1
100
⎠ðððð = 0.909 ððð/ð ðð¡ ð = 1.2879 ððð = 73.79°
ð = 90° â ð = 1.589 ððð/ð
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- The angular velocity on the Z axis of âAâ
101
(X axis : the rotation angle of revolute joint âKâ, Y axis : the angular
velocity on the Z axis of âAâ)
Comparison of results
Object Value Theory RecurDyn Error(%)
ðððð [ððð /ð] 0.909 0.901 0.88
ðœ [ð ðð] 73.79 73.99 0.271
ððšð©,ð=ð [ððð /ð] 1.589 1.599 0.629
102
Dynamics of Rigid Bodies.29
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.536, P.7.46
Type of Analysis : 3-Dimensional Rigid Body Dynamics, Rotational Motion about a
Fixed Point
Type of Element : Rigid body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Angular velocity
of disk ð 10ð ððð/ð
Angular velocity
of frame Ω 4ð ððð ð â
ð = 10ðð [ððð ð â ] = ðð [ððð ð â ] = const
Ω = 4ðᅵᅵ [ððð ð â ] = Ωᅵᅵ [ððð ð â ] = const
ᅵᅵð = 0.3ð + 0.1ᅵᅵ
ððŽ/ð = 0.1ᅵᅵ
103
Interaction Formula of Acceleration
ᅵᅵ = ᅵᅵ = ᅵᅵð + ðð = Ω à ð = Ωᅵᅵ à ðð = âðΩð = â40ð2ð
ᅵᅵðŽ = ᅵᅵð + Ωà ððŽ/ð + Ω à (Ω à ððŽ/ð) + 2Ω à ᅵᅵððð + ᅵᅵððð
ᅵᅵð = Ω à (Ω à ᅵᅵð) = 4ðᅵᅵ à (4ðᅵᅵ à (0.3ð + 0.1ᅵᅵ)) = 4ðᅵᅵ à 1.2ðð = â4.8ð2ð
Ω = 0
Ω à ððŽ/ð = 4ðᅵᅵ à 0.1ᅵᅵ = 0
ᅵᅵððð = ð à ððŽ/ð = 10ðð à 0.1ᅵᅵ = ðð
ᅵᅵððð = ð à (ð à ððŽ ðâ ) = 10ðð à (10ðð à 0.1ᅵᅵ) = 10ðð à ðð = â10ð2ᅵᅵ
⎠ᅵᅵðŽ = â4.8ð2ð + 0 + 0 + 8ðᅵᅵ à ðð â 10ð2ᅵᅵ = â4.8ð2ð + 8ð2ð â 10ð2ᅵᅵ
= 2ð2(â2.4ð + 4ð â 5ᅵᅵ) ð/ð 2
⎠ᅵᅵðŽ = â47.37ð + 78.96ð â 98.70ᅵᅵ ð/ð 2
⎠|ᅵᅵðŽ| = â(47.372 + 78.962 + 98.702 +) = 134.98 ð/ð 2
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
104
1-1) Create a scope of the point âAâ which is on the disk
1. Enter the âNameâ and click the âEtâ button after select the âEntityâ
icon. And, drag and drop âDisk-Marker2â in the âDatabaseâ window
2. Change the âComponentâ to âComponent â Acc_TXâ
3. Change the âReference Frameâ to âGround-InertiaMarkerâ
4. Click the âAdd to Plotâ icon to add these scopes to the âPlotâ
1-2) âAdd to Plotâ for the acceleration on the X,Y,Z axis of the âDiskâ
2) Draw a plot of an angular acceleration on the X axis of the âDiskâ
105
Comparison of results
Object Value Theory RecurDyn Error(%)
ð¶ [ððð ððâ ] -394.78ð â394.78ð 0
(ððš)ð [ð ððâ ] -47.37 -47.37 0
(ððš)ð [ð ððâ ] 78.96 78.96 0
(ððš)ð [ð ððâ ] -98.70 -98.70 0
106
Dynamics of Rigid Bodies.30
References : Engineering Mechanics : Dynamics (J.L.Meriam, L.G.Kraige), 7th Edition,
pp.598, P.8.36
Type of Analysis : Vibration and Time Response, Free Vibration of Particles
Type of Element : Rigid body (one part)
Theoretical Solution
Basic Conditions
Given Symbol Value Unit
Mass ð 3 ðð
Spring Coefficient ð 108 ð ðâ
Damping Coefficient ð 18 ð â ð /ð
The Equation of Motion of Free Damped Oscillation
ðᅵᅵ + ðᅵᅵ + ðð¥ = 0
ᅵᅵ +ð
ðᅵᅵ +
ð
ðð¥ = 0
107
ðð = âð
ð= â
108
3= 6 ððð/ð
ζ =ð
2âðð=
18
2â3 â 108= 0.5
ðð = ððâ1 â ζ2 = 6 â â1 â 0.52 = 5.196 ððð/ð
The Solution of the Second Order Differential Equation
ᅵᅵ + 2ζððᅵᅵ + ðð2ð¥ = 0
ð¥ = (ðŽ1 cosððð¡ + ðŽ2 sinððð¡)ðâζððð¡
If, ð¡ = 0 â ð¥ = ð¥0
⎠ðŽ1 = ð¥0
If, ð¡ = 0 â ᅵᅵ = 0
ᅵᅵ = ðð(âðŽ1 sinððð¡ + ðŽ2 cosððð¡)ðâζððð¡ â ζðð(ðŽ1 cosððð¡ + ðŽ2 sinððð¡)ðâζððð¡
0 = ðððŽ2 â ζðððŽ1
⎠ðŽ2 =ζðð
ðððŽ1 =
0.5 Ã 6
5.196ð¥0 = 0.577ð¥0
⎠ð¥ = ð¥0(cos 5.196ð¡ + 0.577 sin 5.196ð¡)ðâ3ð¡
ðð =5.196
2ð ð»ð§ = 0.827 ð»ð§
ð =1
ðð= 1.209 ð /ððŠððð
ð¡ =ð
2= 0.605 ð ðð
⎠ð¥ = ð¥1 = â0.163ð¥0
ð¥0 = 0.3 ð
ð¥1 = â0.0489
108
Numerical Solution - RecurDyn
1. Modeling
2. Plot the results
- Draw the âDEFL_TSDAâ of Spring1 (the deformation of the spring) from the
database
109
Comparison of results
Object Value Theory RecurDyn Error(%)
ðð [ð] -0.0489 -0.0489 0