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.5.198

Type of Analysis : Plane Kinematics of Rigid Bodies, Linear Motion

Type of Element : Rigid Body (one part)

Comparison of results

Object Value Theory RecurDyn Error

𝜔𝐵𝐶 [rad/s] 0.576 0.577 0.001

𝑣𝐵 [m/s] 0.288 0.288 0.0

Note

1. Theoretical Solution

1) Basic Conditions

𝜔𝑂𝐴 = �̇� = 3 rad/s = const

θ = 30°

2) Geometrical Constraint

Second Law of Cosines

𝐶𝐴̅̅ ̅̅ 2 = 0.32 + 0.12 − 2×0.1×0.3× cos 120 = 0.13

𝐶𝐴̅̅ ̅̅ = 0.361

0.1

sin 𝛽=

𝐶𝐴̅̅ ̅̅

sin 120

β = 𝑠𝑖𝑛−1(0.1

𝐶𝐴̅̅ ̅̅sin 120) = 𝑠𝑖𝑛−1 (

0.1

0.361sin 120) = 13.88°

γ = 180 − 120 − 13.88 = 46.12°

3) Calculate a Velocity

𝑣𝐴 = 𝜔𝑂𝐴 ∙ 𝑂𝐴̅̅ ̅̅ = 3 ∙ 0.1 = 0.3 𝑟𝑎𝑑/𝑠

𝜔𝐶𝐴 =𝑣𝐴 cos 𝛾

𝐶𝐴̅̅ ̅̅=

0.3 ∙ cos 46.12

0.361= 0.576 𝑟𝑎𝑑/𝑠

Where, 𝜔𝐶𝐴 = 𝜔𝐵𝐶

𝑣𝐵 = 𝜔𝐵𝐶 ∙ 𝐵𝐶̅̅ ̅̅ = 0.576 ∙ 0.5 = 0.288 𝑚/𝑠

𝐴

𝐵

𝐷

𝐶

𝑂

𝑣𝐵

𝑣𝐴

𝜃

𝛾

0.3

0.1

𝜔𝑂𝐴

𝛽

2. Numerical Solution – Using RecurDyn

1) Create New Model

- Set the model name : P5_198

- Set the “Unit” to “MMKS”

- Set the “Gravity” to “-Y”

2) Create an Object Shape

(1) Measuring the locations and the angles of the connecting points for modeling

- Measure the locations and the angles of each connecting point using a CAD program to make

the modeling of the system easier.

(2) Adjust the “Icon Size” and the “Marker Size” by Using the “Icon Control” Tool

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

“Database” window.

Point : 0.0, 500.0, 0.0

Point : 0.0, 300.0, 0.0

- Create a “Box” in the “Ground Edit” mode.

Point1 : -60.0, 450.0, 0.0

Point2 : 60.0, 550.0, 0.0

Depth : 20

- Create another “Box” in the “Ground Edit” mode.

Point1 : -60.0, 480.0, 0.0

Point2 : 60.0, 520.0, 0.0

Depth : 20

- Use the “Boolean – Subtract” tool to remove “Box2” from “Box1”. After selecting

“Subtract”, select “Box1” and then select “Box2”.

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

(4) Create a “Link”, “Body1”

Point1 : 0.0, 0.0, 0.0

Point2 : 120.096, 485.363, 0.0

Depth : 20

- Double click “Body1” in the working window or select “Body1 – Edit” (Click the right mouse button

after selecting “Body1” 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 “Link1”.

Point : 120.096, 485.363, 0.0

(5) Create a “Link”, “Link2” in “Body1”

- Modify the properties of “Link2” as shown below.

- Use the “Boolean – Subtract” tool to remove “Link2” from “Link1”. After selecting

“Subtract”, select “Link1” and then select “Link2”.

- Create a marker on a point of “Link2”.

Point : 86.603, 350.0, 0.0

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

(6) Create a “Cylinder”, “Body2”

Point1 : 340.555, 0.0, 0.0

Point2 : 380.555, 0.0, 0.0

Radius : 10

(7) Rotate the Object

- Rotate “Body1” 76.102° to the left (counter clockwise) by using the “Basic Object Control”

with the “Inertial Marker” as the “Reference Frame” as shown below.

(8) 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.

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

(9) Create a “Cylinder”, “Body4”

Point1 : -20.0, 500.0, 0.0

Point2 : 20.0, 500.0, 0.0

Radius : 20

(10) Create a “Link”, “Body5”

- Double click “Body5” in the working window or select “Body5 – Edit” (Click the right mouse button

after selecting “Body5” 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.

(11) Change the names of the bodies : Rename

3) Create “Joint”

1) Revolute Joint

2) Translational Joint

4) Create “Motion”

- Select “Revjoint4” and click the right mouse button.

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

the “Name” and −3 [𝑟𝑎𝑑 𝑠⁄ ] as the “Expression”.

5) Create “Scope” : Create a scope of the velocity of point “B” of “BC”

- Enter the “Name” and click the “Et” button after select the “Entity” icon. And, drag

and drop the “CB-Marker3” in the “Database” window.

-Change the “Component” to the “Component – Vel _ TM”.

6) 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”.

7) Execute “Plot”

(1) The velocity of the point “B” of “CB” : 288.46 m/s

(2) The angular velocity in the Z direction of “CB” : -0.5769 rad/s

3. Problems to Consider

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

translational joints are created in the model, then it becomes a -5 degree of freedom system and

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

6 in total because of the driving constraint that exists in the model.

Calculate Degree of Freedom (D.O.F)

Count of Bodies 6 * 6 D.O.F (x, y, z, 𝜃𝑥 , 𝜃𝑦 , 𝜃𝑧 ) 36

Constraint

Ground 1 * 6 -6

Revolute Joint 5 * 5 -25

Translational Joint 2 * 5 -10

Driving Constraint 1 * 1 -1

D.O.F -6 D.O.F.

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

joints, and two spherical joints rather than five revolute joints and two translational joints 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 6 * 6 D.O.F (x, y, z, 𝜃𝑥 , 𝜃𝑦 , 𝜃𝑧 ) 36

Constraint

Ground 1 * 6 -6

Revolute Joint 1 * 5 -5

Translational Joint 2 * 5 -10

Universal Joint 2 * 4 -8

Spherical Joint 2 * 3 -6

Driving Constraint 1 * 1 -1

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?