WEEK 1 Dynamics of Machinerykisi.deu.edu.tr/hasan.ozturk/makina dinamigi/Makina... ·...

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1 WEEK 1 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

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WEEK 1Dynamics 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

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Course Policy

Two Midterms (20%) +Homework (10%) + 1 Final (50%)

Prof.Dr. Hasan ÖZTÜRK

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A mechanism is a device which transforms motion to some desirablepattern and typically develops very low forces and transmits little power.

A machine typically contains mechanisms which are designed to providesignificant forces and transmit significant power

Some examples of common mechanisms are a pencilsharpener, a camera shutter, an analog clock, a foldingchair, an adjustable desk lamp, and an umbrella.

Some examples of machines which possess motions similar to the mechanisms listed above are a food blender, a bank vault door, an automobile transmission, a bulldozer, and a robot.

Prof.Dr. Hasan ÖZTÜRK

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Degree of freedom a kinematic pair:

The degree of freedom (DOF) of a rigid body is the number of independent parameters that define its configuration

Degree of freedom of a rigid body:

Six independent parameters are required to define the motion of the ship.

An unrestrained rigid body in space has six degrees of freedom: three translating motions along the x, y and z axes and three rotary motions around the x, y and z axes respectively.

The degrees of freedom (DOF) of a kinemtaic pairs defined as the number of independent movements it has.

Prof.Dr. Hasan ÖZTÜRK

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Kinematics: Kinematic analysis involves determination of position, displacement, rotation, speed, velocity, and acceleration of a mechanism.

Theory of machines is separated into two section

Dynamics is also separated into two section

Kinetics: It is that branch of theory of machines which deals with the inertia forces which arise from the combined effect of the mass and motion of the machine parts.

Kinetics analysis will be used for this lecture.

Dynamics: is that branch of theory of machines which deals with the forces and their effects, while acting upon the machine parts in motion.

Statics: is that branch of theory of machines which deals with the forces and their effects, while the machine parts are rest.

Prof.Dr. Hasan ÖZTÜRK

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Newton's Three Laws of Motion:

Prof.Dr. Hasan ÖZTÜRK

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=F ma

1 2= ⇒ + = ⇒ =∑ netFF ma F F ma am

m

Prof.Dr. Hasan ÖZTÜRK

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Law 3: Reaction is always equal and opposite to action; that is to say, the actions of two bodies upon each other are always equal and directly opposite.

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Rigid Body: is that body whose changes in shape are negligible compared with its overall dimensions or with the changes in position of the body as a whole, such as rigid link, rigid disc…..etc. Links: are rigid bodies each having hinged holes or slot to be connected together by some means to constitute a mechanism which able to transmit motion or forces to some another locations.

Prof.Dr. Hasan ÖZTÜRK

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FORCE AND MOMENT VECTORS

A force is characterized by its magnitude and direction, and thus is a vector. In an (x, y)-plane the force vector, F, can be represented in different forms

The characteristics of a force are its magnitude, its direction, and its point of application. The direction of a force includes the concept of a line along which the force is acting, and a sense. Thus a force may be directed either positively or negatively along its line of action.

Prof.Dr. Hasan ÖZTÜRK

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Two equal and opposite forces along two parallel but noncollinear straight lines in a body cannot be combined to obtain a single resultant force on the body. Any two such forces acting on the body constitute a couple. The arm of the couple is the perpendicular distance between their lines of action, shown as h in the Figure, and the plane of the couple is the plane containing the two lines of action.

= ×

A BAM R F .=Aor M h F

= ×

A BAM R F

( )sin .= =A BA

orM R F h Fθ

Prof.Dr. Hasan ÖZTÜRK

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FORCES IN MACHINE SYSTEMS A machine system is considered to be a system of an arbitrary group of bodies (links), which will be considered rigid. We are involved with different types of forces in such systems. a) Reaction Forces: are commonly called the joint forces in machine systems since the action and reaction between the bodies involved will be through the contacting kinematic elements of the links that form a joint. The joint forces are along the direction for which the degree-of-freedom is restricted. e.g. in constrained motion direction

All lower pairs and their constraint forces:

(a) revolute or turning pair with pair variable θ

Prof.Dr. Hasan ÖZTÜRK

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(b) prismatic or sliding pair with pair variable z

(c) cylindric pair with pair variables θ and z

(d) screw or helical pair with pair variables θ and z

Prof.Dr. Hasan ÖZTÜRK

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(e) planar or flat pair with pair variables x, z, and θ .

(f) spheric pair with pair variables θ, φ and ψ

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Reaction Forces

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b) Physical Forces : As the physical forces acting on a rigid body we shall include external forces applied on the rigid body, the weight of the rigid body, driving force, or forces that are transmitted by bodies that are not rigid such as springs or strings attached to the rigid body.

weight

external forces

Spring forceProf.Dr. Hasan ÖZTÜRK

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c) Friction or Resisting Force: The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it.

32 32( )= = −frictionF R Fµ

d) Inertia Forces are the forces due to the inertia of the rigid bodies involved.

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tan = = frictionFF

φ µ

iFa

( )= −iF ma

= − iF ma

µ: coefficient of friction

m: mass, Kga=acceleration, m/s2

F=force, N

Prof.Dr. Hasan ÖZTÜRK

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Inertia Torque

= − i

GF ma

= − i

GT I α

m: mass, Kga=acceleration, m/s2

F=force, NT=Torque, NmI= Moment of Inertia, kgm2

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We shall be using SI systems of units. A list of units relevant to the course is given below

SI System of units

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• For a body of mass m the resistance to rotation about the axis AA’ is

inertiaofmomentmassdmr

mrmrmrI

==

+∆+∆+∆=

∫ 2

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22

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• The radius of gyration for a concentrated mass with equivalent mass moment of inertia is

mIkmkI == 2

Moment of Inertia of a Mass

• Moment of inertia with respect to the y coordinate axis is

( )∫∫ +== dmxzdmrI y222

• Similarly, for the moment of inertia with respect to the x and z axes,

( )( )∫

∫+=

+=

dmyxI

dmzyI

z

x22

22 • In SI units,( )22 mkg ⋅== ∫ dmrI

Prof.Dr. Hasan ÖZTÜRK

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Moments of Inertia of Common Geometric Shapes

Prof.Dr. Hasan ÖZTÜRK

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A free-body diagram is a sketch or drawing of the body, isolated from the rest of the machine and its surroundings, upon which the forces and moments are shown in action.

A free body diagram shows all forces of all types acting on this body

FREE-BODY DIAGRAMS

Prof.Dr. Hasan ÖZTÜRK

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external force and torque, F4 and T2

Slider –Crank Mechanism

All frictions are neglected except for the friction at joint 14

Prof.Dr. Hasan ÖZTÜRK

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STATIC EQUILIBRIUM:A body is said to be in static equilibrium if under a set of applied forces and torques its translational (linear) and rotational accelerations are zeros (a body could be stationary or in motion with a constant linear velocity).

Planar static equilibrium equations for a single body that is acted upon by forces and torques are expressed as

00

0 == =

∑∑ ∑x

y

FF

F

0= =∑ ∑ zM M

Prof.Dr. Hasan ÖZTÜRK

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Two-force member: If only two forces act on a body that is in static equilibrium, the two forces are along the axis of the link, equal in magnitude, and opposite in direction..

Two force and one moment member: A rigid body acted on by two forces and a moment is in static equilibrium only when the two forces form a couple whose moment is equal in magnitude but in opposite sense to the applied moment

If an element has pins or hinge supports at both ends and carries no load in-between, it is called a two-force member

Prof.Dr. Hasan ÖZTÜRK

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Three-force member: If only three forces act on a body that is in static equilibrium, their axes intersect at a single point.

A special case of the three-force member is whenthree forces meet at a pin joint that is connected between three links. When the system is in static equilibrium, the sum of the three forces must be equal to zero.

For example, if the axes of two of the forces are known, the intersection of those two axes canassist us in determining the axis of the third force.

Prof.Dr. Hasan ÖZTÜRK

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Let the force FA be completely specified. And the line of action of FB and the point of application of FC be known. When the moment equilibrium equation is written for the sum of moments about the point of intersection of the line of action of FA and FB (point O), since MO=0, the moment of FC about O must be zero, or the line of action of the force FC must pass through point O. The magnitudes of the forces can then be determined from the force and moment equilibrium equations.

O

0=∑ FProf.Dr. Hasan ÖZTÜRK

their axes intersect at a single point.

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Example: Find all the pin (joint) forces and the external torque M12 ,that must be applied to link 2 of the mechanism (static). AO2=6 m, AB=18 m, BO4=12 m ve BQ=5 m

M12

=120 N

Prof.Dr. Hasan ÖZTÜRK

GRAPHIC SOLUTION

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F34=33.1 NF14=89 N

M12=183 N.m

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ANALYTIC SOLUTION

we sum moments about point O4. Thus

m

m

m

m

m

m

N

N

N N

N N

N

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from the free-body diagram of link 2

m.N

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Example:For the mechanism shown A0A= a2= 80, AB= a3 =100, B0B= a4=120, A0B0= a1= 140, AC= b3 = 70, BC=80 and B0D= b4=90 mm. When θ12=600, from kinematic analysis θ13=29,980 , θ14 = 96.400. Two forces F13=50 N < 2300 and F14= 100 N < 2000 are acting on links 3 and 4 respectively

Prof.Dr. Hasan ÖZTÜRK

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The free-body diagrams of the moving links are shown (STATIC).

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The three equilibrium equations for link 4 are:

Prof.Dr. Hasan ÖZTÜRK

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There are four unknowns in three equations, therefore the equations obtained from one free-body diagram is not enough to solve for the unknowns. Equations 1 and 2 can be used to solve for G14x and G14y , only when F34xy and F34y are determined. The three equilibrium equations for link 3 must also be written (note that F34 and F43 are of equal magnitude).

Where a= 52.620 (using the cosine theorem for the triangle ABC).

Prof.Dr. Hasan ÖZTÜRK

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Equations 4 and 5 can be used to determine F23x and F23yEquations 3 and 6 must be used simultaneously to solve for F34x and F34y. Substituting the known values into equations 3 and 6 results:

Prof.Dr. Hasan ÖZTÜRK

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