dynamics of rigid bodies 5. plane kinematics of rigid...

Dynamics of Rigid Bodies

5. Plane Kinematics of Rigid Bodies

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)

Comparison of results

Object Value Theory RecurDyn Error

𝜔𝐵𝐶 [rad/s] 2 2 0.0

Note

1. Theoretical Solution

1) Basic Conditions

𝑣𝐴 = 1.2 m/s = const

𝜔𝑂𝐶 = 2 𝑟𝑎𝑑/𝑠

2) The Relative Velocity Equation of the points “A”, “B”, and “C”

𝑣𝐵⃗⃗ ⃗⃗ = 𝑣𝐴⃗⃗⃗⃗ + 𝑣𝐵/𝐴⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ = 𝑣𝐶⃗⃗⃗⃗ + 𝑣𝐵/𝐶⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗

𝑣𝐶⃗⃗⃗⃗ = −𝜔𝑂𝐶�⃗� ×0.3𝑗 = 0.6𝑖

3) A Graph of the Velocity Relationship

𝑣𝐵/𝐶 =𝑣𝐴 − 𝑣𝐶

tan 𝜃=

0.6

3/4= 0.8

𝜔𝐵𝐶 =𝑣𝐵/𝐶

𝐵𝐶̅̅ ̅̅=

0.8

0.4= 2 𝑟𝑎𝑑/𝑠

𝑣𝐵⃗⃗ ⃗⃗ = 𝑣𝐴⃗⃗⃗⃗ + 𝑣𝐵/𝐴⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ = 𝑣𝐶⃗⃗⃗⃗ + 𝑣𝐵/𝐶⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ = 0.6𝑖 + 0.8𝑗

4) The Angular Acceleration of Link “AB”

𝑣𝐵/𝐴 =𝑣𝐴 − 𝑣𝐶

sin 𝜃=

0.6

3/5= 1.0

𝜔𝐴𝐵 =𝑣𝐵/𝐴

𝐴𝐵̅̅ ̅̅=

1.0

0.5= 2 𝑟𝑎𝑑/𝑠

𝑣𝐵/𝐴

𝑣𝐴=1.2

𝑣𝐵

𝑣𝐶 = 0.6

𝑣𝐵/𝐶

𝜃

2. Numerical Solution – Using RecurDyn

1) Create New Model

- Set the model name : P5_195

- Set the “Unit” to “MMKS”

- Set the “Gravity” to “-Y”

2) Create an Object Shape

(1) Create a marker after clicking the “Ground” or selecting the “Ground – Edit” in the

“Database” window.

Point : -400.0, 0.0, 0.0

(2) Create two “Cylinder” bodies in the “Ground Edit” mode.

(3) Use the “Boolean – Subtract” tool to remove “Cylinder2” from “Cylinder1”. After

selecting the “Subtract” tool, select “Cylinder1” and then select “Cylinder2”.

- Click the “Exit” icon to exit from the “Ground Edit” mode.

(4) Create a “Cylinder”, “Body1”

- Click the point (-400, 0, 0) and the point (-800, 0, 0) to create a cylinder with radius of 15mm as

shown below.

Point1 : -400.0, 0.0, 0.0

Point2 : -800.0, 0.0, 0.0

Radius : 15

- Create a marker on a point of “Body1”.

Point : -800.0, 0.0, 0.0

(5) Create a “Link”, “Body2”

- Double click “Body2” in the working window or select “Body2 – Edit” (Click the right mouse button

after selecting “Body2” in the “Database” window, then, select “Edit” in the pop up box.) in the

“Database” window on the right to enter the “Body Edit” mode.

- Modify the properties of “Link1” as shown below.

- Create a marker on a point of “Body2”.

Point : 0.0, 300.0, 0.0

- Click the “Exit” icon to exit from the “Body Edit” mode.

(6) Create a “Link”, “Body3”

- Double click “Body3” in the working window or select “Body3 – Edit” (Click the right mouse button

after selecting “Body3” in the “Database” window, then, select “Edit” in the pop up box.) in the

“Database” window on the right to enter the “Body Edit” mode.

- Modify the properties of “Link1” as shown below.

- Create a marker on a point of “Body3”.

Point : -400.0, 300.0, 0.0

- Click the “Exit” icon to exit from the “Body Edit” mode.

(7) Create a “Link”, “Body4”

- Double click “Body4” in the working window or select “Body4 – Edit” (Click the right mouse button

after selecting “Body4” in the “Database” window, then, select “Edit” in the pop up box.) in the

“Database” window on the right to enter the “Body Edit” mode.

- Modify the properties of “Link1” as shown below.

- Click the “Exit” icon to exit from the “Body Edit” mode.

(8) Change the names of the bodies : Rename

3) Create “Joint”

1) Revolute Joint

2) Translational Joint

4) Create “Motion”

- Select “Trajoint1” and click the right mouse button.

- Select “Property” – “Include Motion” – “Velocity” (angular) – “EL” button – “Create” – Enter

the “Name” and 1200 [𝑚𝑚 𝑠⁄ ] as the “Expression”.

- Select “RevJoint1” and click the right mouse button.

- Select “Property” – “Include Motion” – “Velocity” (angular) – “EL” button – “Create” – Enter

the “Name” and −2 [𝑟𝑎𝑑 𝑠⁄ ] as the “Expression”.

5) Analysis

- Execute the Dynamic/Kinematic icon to simulate (this example problem is a kinematic

problem because the number of degrees of freedom is “0”).

- Set the “End Time” and the “Step” as shown below.

- Set the “Maximum Time Step” to “0.001” and the “Integrator Type” to “DDASSL”.

6) Execute the “Plot”

- The angular velocity on Z axis of the “BC” (Vel_RZ): -2 rad/s ([View] – [TraceData])

3. Problems to Consider

1) The model is a zero degree of freedom system in reality. However, if four revolute joints and a

translational joint are created in the model, then it becomes a -1 degree of freedom system and is

an over-constrained mechanism. Moreover, the number of degrees of freedom in the model is -3

in total because two driving constraints exist in the model.

Calculate Degree of Freedom (D.O.F)

Count of Bodies 5 * 6 D.O.F (x, y, z, 𝜃𝑥 , 𝜃𝑦 , 𝜃𝑧 ) 30

Constraint

Ground 1 * 6 -6

Revolute Joint 4 * 5 -20

Translational Joint 1 * 5 -5

Driving Constraint 2 * 1 -2

D.O.F -3 D.O.F.

The constraints in the model should consist of two revolute joints, a spherical joint, a universal joint,

and a translational joint rather than four revolute joints and a translational joint to make a zero

degree of freedom system, which is equivalent to the system in reality. RecurDyn has an algorithm

to eliminate this sort of redundancy, automatically. This is a very powerful and convenient function

of RecurDyn. However, keep in mind that this algorithm can eliminate ‘meaningful constraints’ in

the case of a complicated model, which can cause unintended motion.

Calculate Degree of Freedom (D.O.F)

Count of Bodies 5 * 6 D.O.F (x, y, z, 𝜃𝑥 , 𝜃𝑦 , 𝜃𝑧 ) 30

Constraint

Ground 1 * 6 -6

Revolute Joint 2 * 5 -10

Translational Joint 1 * 5 -5

Universal Joint 1 * 4 -4

Spherical Joint 1 * 3 -3

Driving Constraint 2 * 1 -2

D.O.F 0 D.O.F.

2) How many degrees of freedom are in this system? In other words, is this problem a “Dynamic”

problem or a “Kinematic” problem?