weeks 2-3 dynamics of machinerykisi.deu.edu.tr/hasan.ozturk/makina dinamigi/makina... ·...

1

WEEKS 2-3 Dynamics of Machinery

• References • Theory of Machines and Mechanisms, J.J.

Uicker, G.R.Pennock ve J.E. Shigley, 2003 • Makine Dinamiği, Prof. Dr. Eres SÖYLEMEZ,

2013 • Uygulamalı Makine Dinamiği, Jeremy

Hirschhorn, Çeviri: Prof.Dr. Mustafa SABUNCU, 2014

Prof.Dr.Hasan ÖZTÜRK

Example:This slider-crank mechanism is in static equilibrium in the shown configuration. A known force F acts on the slider block in the direction shown. An unknown torque acts on the crank. Our objective is to determine the magnitude and the direction of this torque in order to keep the system in static equilibrium.

F12x

F12y

F32y F32x

F23y

F23x

F43y

F43x

F34x

F34y

F14=F14yT12 T14

12 32

12 32

32 32 12

2

00

0

(Sum of moments about O )

x x

y y

x y

F FF F

aF bF T

+ =+ =

− + + =

23 43

23 43

43 43

00

0

(Sum of moments about A )

x x

y y

x y

F FF F

cF dF

− + =− + =

+ =

34

34 14

14

00

0(Sum of moments about B )

x

y

F FF F

T

− + =− + =

=

2 3

4


2

3

Numerical values for the link lengths are L2 = 2 m and L3 = 4 m. From the figures we extract the following measurements: a = 1.8 m , b = 1 m , c = 2 m , d = 3.6 m. Assume the applied force is given to be F = 10 N in the negative direction.

12

12

32

32

12

105.56

105.56

23.55

x

y

x

y

F NF NF NFT N

= −

=

=

= −

=

43

43

14

14

105.56

5.560

x

y

F NF NF NT

== −

= −=

2T2

AF32

F12x

F12y

Simplified FBD method: The connecting rod of this mechanism is a two-force member. The reaction forces at A and B must be equal but in opposite directions. These reaction forces are named F2 3 and F 43 , and given arbitrary directions

B

3

F23

A

F43

4

1

F

F14

F34 B


4

Graphical:

00

0 == =

∑∑ ∑x

y

FF

F

14 34 0F F F+ + =

23 43F F= −

0F = ⇒∑32 12F F= −

2 2 32

2 32

0 . 0.

OM T h FT h F

= ⇒ − =

=∑


5

Coulomb Friction: Coulomb friction can be included between two contacting surfaces in a static force analysis. Given the static coefficient of friction, μ (s) , the friction force can be described as the product of the coefficient of friction and the reaction force normal to the contacting surfaces. The friction force must act in the opposite direction of the tendency of any motion.

VB

Ffriction=µ F14

12 32

12 32

32 32 12

2

00

0

(Sum of moments about O )

x x

y y

x y

F FF F

aF bF T

+ =+ =

− + + =

23 43

23 43

43 43

00

0

(Sum of moments about A )

x x

y y

x y

F FF F

cF dF

− + =− + =

+ =

34 .

34 14

0

0x fric

y

F F FF F

− + + =

− + =


6

Example: An external force of 10 N is acting horizontally on the rocker link, 30 mm from the point D. Find the amount of torque to be applied to the crank AB to keep the mechanism in static equilibrium. [ME 302 DYNAMICS OF MACHINERY, Prof. Dr. Sadettin KAPUCU]

0 0 060 , 20.02 , 89.86= = =bθ φ


7 Prof.Dr.Hasan ÖZTÜRK

8


9

Graphical method:

F14 & F34 are measured directly from the scaled force polygon. 10 N stands for 50 mm


10

Example: Onto link 6 of the mechanism given, a 100 N vertical force acting. Calculate the amount of the torque required on the crank AB to keep the mechanism in static equilibrium, using the graphical approach. [ME 302 DYNAMICS OF MACHINERY, Prof. Dr. Sadettin KAPUCU]


Prof.Dr.Hasan ÖZTÜRK 13

The kinematic function of gears is to transfer rotational motion from one shaft to another. Since these shafts may be parallel, perpendicular, or at any other angle with respect to each other, gears designed for any of these cases take different forms and have different names: spur, helical, bevel, worm, etc.

GEAR KINEMATICS

Kinematics of meshing gears.

( ) ( )

1 2

1 1 2 2

1 1 2 2

1 1 2 2

1 1 1 2 2 2

1 2 1 1 2 2

260

t tV V.r ( ) .r.D ( ) .D

N .D ( )N .D , N : rpmN ,rad / sn

N . m .z ( )N . m .z , D m.z*****m m N .z ( )N .z

=ω = − ωω = − ω

= −π

ω =

= − =

= ⇒ = −

1 2

1 1 2 2

t ta a.r ( ) .r=

α = − α

tangential acceleration

pitch diameter

number of teeth

module

14

Spur Gear Force Analysis (Static):

The reaction forces between the teeth occur along the pressure line AB, tipped by the pressure angle, tipped by the pressure angle φ from the common tangent to the pitch circles.

32 23

32 23

radial force component

transmitted force component

r r

t t

F FF F

= − =

= − =

( )32 32r tF F tan= φ

In applications involving gears, the power transmitted and the shaft speeds are often specified. remembering that power is the product of force times velocity or torque times angular velocity, we can find the relation between power and the transmitted force. Using the symbol P to denote power, we obtain


Spur Gear Force Analysis (Dynamic):

The reaction forces between the teeth occur along the pressure line AB, tipped by the pressure angle, tipped by the pressure angle φ from the common tangent to the pitch circles.

32 23

32 23

radial force component

transmitted force component

r r

t t

F FF F

= − =

= − =

( )32 32r tF F tan= φ


Example (Midterm 1-2015): Find all the pin (joint) forces and the external torque (T5) that must be applied to gear 5 of the mechanism by using the analytical method. Neglecting friction.

Prof..Dr.Hasan ÖZTÜRK

19

A

3

3 0OM = ⇒∑


4 0OM = ⇒∑

21



Fa: axial, Fr: radial Ft: tangential φt: transverse pressure angle ψ: helix angle

Helical Gears


Straight Bevel Gears

24

Dynamic Force Analysis

25

Sometimes it is convenient to arrange these mass moments of inertia and mass products of inertia into a symmetric square array or matrix format called the inertia tensor of the body:

26

Dynamic Force Analysis D'Alembertls principle: The vector sum of all external forces and inertia forces acting upon a system of rigid bodies is zero. The vector sum of all external moments and inertia torques acting upon a system of rigid bodies is also separately zero.

Ti is in opposite sense of the angular acceleration a

inertia force

inertia torque


0 0+ = − =∑ ∑ ∑

iGF F or F ma

Fi has the same line of action of aG but is in opposite direction

0 0+ = − =∑ ∑ ∑

iGM T or M I α

27

external force and torque, F4 and T2

Slider –Crank Mechanism

All frictions are neglected except for the friction at joint 14



We use the four-bar linkage of the below Figure. The required data, based on a complete kinematic analysis, are illustrated in the Figure and in the legend. At the crank angle shown and assuming that gravity and friction effects are negligible, determine all the constraint forces and the driving torque required to produce the acceleration conditions specifıed.

EXAMPLE

30

We start with the following kinematic information.

Next we calculate the inertia forces and inertia torques. Because the solution is analytical, we do not need to calculate offset distances nor do we replace the inertia torques by couples. The six equations are


Considering the free-body diagram of link 4 alone, we formulate the summation of moments about point 04:

Also, considering the free-body diagram of link 3 alone, we formulate the summation of moments about point A:


PRINCIPLE OF SUPERPOSITION

Linear systems are those in which effect is proportional to cause. This means that the response or output of a linear system is directly proportional to the drive or input to the system. An example of a linear system is a spring, where the deflection (output) is directly proportional to the force (input) exerted on the spring.

The principle of superposition may be used to solve problems involving linear systems by considering each of the inputs to the system separately. If the system is linear, the responses to each of these inputs can be summed or superposed on each other to determine the total response of the system. Thus, the principle of superposition states that for a linear system the individual responses to several disturbances, or driving functions, can be superposed on each other to obtain the total response of the system.


PLANAR ROTATION ABOUT A FIXED CENTER

For fixed-axis rotation, it is generally useful to apply a moment equation directly about the rotation axis O.

Application of the parallel-axis theorem for mass moments of inertia

=∑

GF ma


Center of percussion

P

P

P

P


http://www.google.com.tr/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=2ahUKEwi_ksSwlpvZAhUOLewKHWuiDCQQjRx6BAgAEAY&url=http%3A%2F%2Fslideplayer.com%2Fslide%2F4530470%2F&psig=AOvVaw2wMt2C6hj-jdzSmOFdp8Z8&ust=1518345902379152

https://www.google.com.tr/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=2ahUKEwikoZLD547ZAhXHJ-wKHZKUCowQjRx6BAgAEAY&url=https%3A%2F%2Fsciencedemonstrations.fas.harvard.edu%2Fpresentations%2Fcenter-percussion&psig=AOvVaw2gpttC4-okdpcTgiik5b9P&ust=1517921233871553


SHAKING FORCES AND MOMENTS Of special interest to the designer are the forces transmitted to the frame or foundation of a machine owing to the inertia of the moving links. When these forces vary in magnitude or direction, they tend to shake or vibrate the machine (and the frame); consequently, such effects are called shaking forces and shaking moments.

If we consider some machine, say a four-bar linkage for example, with links 2,3, and 4 as the moving members and link I as the frame, then taking the entire group of moving parts as a system, not including the frame, and draw a free-body diagram, we can immediately write

This makes sense because if we consider a free-body diagram ofthe entiremachine including the frame, all other applied and constraint forces have equal and opposite reaction forces and these cancel within the free-body system. Only the inertia forces, having no reactions, are ultimately extemal to the system and remain unbalanced. These are not balanced by reaction forces and produce unbalanced shaking effects between the frame and whatever bench or other surface on which it is mounted. These are the forces that require that the machine be fastened down to prevent it from moving.

54

Force Analysis using the method of Virtual Work

1. If a rigid body is in equilibrium under the action of external forces, the total work done by these forces is zero for a small displacement of the body.

2. Work:

1. With F, x, T, q, vectors and W a scalar. 2. To indicate that we are dealing with infinitesimal displacements

(virtual displacements), use the notation:

3. Now apply the virtual work definition:

∫∫ ⋅=⋅= θTxF dWdW ,

θTxF δδδδ ⋅=⋅= WW

0=⋅+⋅= ∑∑j

jji

iiW θTxF δδδ


55

Virtual Work (cont.)

4. If we divide the virtual work by a small time step, we get:

5. These are all external torques and forces on the body, and include inertial forces and gravity. Rewrite, to clearly show this as:

0i i j j

i j

F v T

i iexternal,friction,weight external,friction G 0⋅ + ⋅ + ⋅ + ⋅ =∑ ∑ ∑ ∑F v T ω F v T ω


56

Example (2015-Midterm 1). Neglect the gravity force for the mechanism shown in the Figure. Links 2 and 3 are uniform. E and D points are the centroids of the triangles. All frictions are neglected. a) Find all the pin (joint) forces and the external torque that must be applied to link 2 of the mechanism by using the analytical method. b) Find the external torque that must be applied to link 2 of the mechanism by using the Virtual Work method.


57

ANSWER- a):


( )43 34F F= −

59


61

ANSWER- b) Virtual Work method :


66

Example (2010-Midterm 1): For the mechanism shown in the figure, Links 2 and 3 are uniform. All frictions and mass of link 2 are neglected. Link 4 can rotate about its axis. Find all the pin (joint) forces and the external torque that must be applied to link 2 of the mechanism by using the grapical method.

α3=545 r/s2

g


Free Body Diagram


68

1 cm: 5 N

43 23 0+ + =

F F F

F23

F43

13.37=F N

F23

F43

13.37=F N

43 4.2 5 21= × =F cm N

23 3.28 5 16.4= × =F cm N

F32

F12

A

O

T2

F32

F12

A

O

T2

0.1 cm 23 12=F F

2 32

2

0.10 0100

0.1 16.4 0.0164100

= ⇒ + × =

= − × = −

∑ OM T F

T N Nm


69

Example (2011-Final): For the mechanism shown in the figure, Links 2 and 3 are uniform. All frictions (except for 3-4) and mass of link 2 are neglected. Link 4 can rotate about its axis. Find all the pin (joint) forces and the external torque that must be applied to link 2 of the mechanism by using the grapical method.

α3=545 r/s2

µ=0.5

g

V34


70

0tan 26.56= ⇒ =φ µ φ

m3g=5 N

F43

F23

3 3 12.4iGF m a N= = 2

3 24.8 /Ga m s=

3α

A

B

C

G3

3 3 3 0.227iGT I Nmα= =

Ffric

FR43

F23

13.37 N

A

B

C

G3

00.227 13.74 cos(53.2 )0.0275 27.5

Nm dd m d mm

= × ×= ⇒ =

d

220

58.80

26.560


71

13.37=F N

FR43

F23

13.37=F N

FR43

F23

FR43

F23

13.37 N

A

B

C

G3

d

220

58.80

26.560

FR43

Ffric

A

B

C

G3

26.560

F43

V34

Ffric


72

FR43

F23

13.37 N

A

B

C

G3

d

220

58.8026.560

13.37=F N

FR43

F23

1 cm: 5 N

43 23 0+ + =

RF F F43 4.3 5 21.5= × =RF cm N

13.37=F N

FR43

F23

23 4.7 5 23.5= × =F cm N

F32

F12

A

O

T2

F32

F12

A

O

T2

1.5 cm

2 32

2

1.50 0100

1.5 23.5 0.3525100

= ⇒ − × =

= × =

∑ OM T F

T N Nm


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