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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. Course Policy.
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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
Course Policy Two Midterms (20%) +Homework (10%) + 1 Final (50%) Prof.Dr. Hasan ÖZTÜRK
A mechanism is a device which transforms motion to some desirable pattern and typicallydevelops very low forces and transmits little power. A machinetypically containsmechanisms which are designed to provide significant forces and transmit significantpower Some examples of common mechanisms are a pencil sharpener, a camera shutter,an analog clock, a folding chair, an adjustable desk lamp, and an umbrella. Someexamples of machineswhich possess motions similar to the mechanisms listed above area food blender, a bank vault door, an automobile transmission, a bulldozer, and a robot. Prof.Dr. Hasan ÖZTÜRK
Degree of freedom of a rigid body: The degree of freedom (DOF) of a rigid body is the number of independent parameters that define its configuration 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. Degree of freedom a kinematic pair: The degrees of freedom (DOF) of a kinematic pairs defined as the number of independent movements it has. Prof.Dr. Hasan ÖZTÜRK
Theory of machines is separated into two section Dynamics is also separated into two section Statics: is that branch of theory of machines which deals with the forces and their effects, while the machine parts are rest. Dynamics: is that branch of theory of machines which deals with the forces and their effects, while acting upon the machine parts in motion. Kinematics: Kinematic analysis involves determination of position, displacement, rotation, speed, velocity, and acceleration of a mechanism. 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. Prof.Dr. Hasan ÖZTÜRK
Newton's Three Lawsof Motion: Prof.Dr. Hasan ÖZTÜRK
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. Prof.Dr. Hasan ÖZTÜRK
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
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
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. Prof.Dr. Hasan ÖZTÜRK
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 Prof.Dr. Hasan ÖZTÜRK
(b) prismatic or sliding pair with pair variable z (c) cylindric pair with pair variables and z (d) screw or helical pair with pair variables and z Prof.Dr. Hasan ÖZTÜRK
(e) planar or flat pair with pair variables x, z, and . (f) spheric pair with pair variables, and Prof.Dr. Hasan ÖZTÜRK
Reaction Forces Prof.Dr. Hasan ÖZTÜRK
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 force Prof.Dr. Hasan ÖZTÜRK
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. : coefficient of friction d) Inertia Forcesare the forces due to the inertia of the rigid bodies involved. m: mass, Kg a=acceleration, m/s2 F=force, N Prof.Dr. Hasan ÖZTÜRK
Inertia Torque m: mass, Kg a=acceleration, m/s2 F=force, N T=Torque, Nm I= Moment of Inertia, kgm2 Prof.Dr. Hasan ÖZTÜRK
We shall be using SI systems of units. A list of units relevant to the course is given below SI System of units Prof.Dr. Hasan ÖZTÜRK
Moment of inertia with respect to the y coordinate axis is • Similarly, for the moment of inertia with respect to the x and z axes, • For a body of mass m the resistance to rotation about the axis AA’ is • In SI units, • The radius of gyration for a concentrated mass with equivalent mass moment of inertia is Moment of Inertia of a Mass Prof.Dr. Hasan ÖZTÜRK 20
Moments of Inertia of Common Geometric Shapes Prof.Dr. Hasan ÖZTÜRK
FREE-BODY DIAGRAMS 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 Prof.Dr. Hasan ÖZTÜRK
Slider –Crank Mechanism external force and torque, F4 and T2 All frictions are neglected except for the friction at joint 14 Prof.Dr. Hasan ÖZTÜRK
STATIC EQUILIBRIUM: A body is said to be in static equilibrium if under a set of applied forces and torques itstranslational (linear) and rotational accelerations are zeros (a body could be stationary or inmotion with a constant linear velocity). Planar static equilibrium equations for a single body that is actedupon by forces and torques are expressed as Prof.Dr. Hasan ÖZTÜRK
Two-force member: If only two forces act on a body that is in staticequilibrium, the two forces are along the axis of the link,equal in magnitude, and opposite in direction.. If an element has pins or hinge supports at both ends and carries no load in-between, it is called a two-force member 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 Prof.Dr. Hasan ÖZTÜRK
Three-force member:If only three forces act on a body that is in staticequilibrium, their axes intersect at a single point. A special case of the three-force member is when three forces meet at a pin joint that is connected betweenthree links. When the system is in static equilibrium, thesum of the three forces must be equal to zero. For example, if the axes of two of theforces are known, the intersection of those two axes canassist us in determining the axis of the third force. Prof.Dr. Hasan ÖZTÜRK
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. their axes intersect at a single point. O Prof.Dr. Hasan ÖZTÜRK
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 GRAPHIC SOLUTION Prof.Dr. Hasan ÖZTÜRK
F34=33.1 N F14=89 N M12=183 N.m Prof.Dr. Hasan ÖZTÜRK
ANALYTIC SOLUTION we sum moments about point O4. Thus m m N N m m m m N N N N N Prof.Dr. Hasan ÖZTÜRK
from the free-body diagram of link 2 m.N Prof.Dr. Hasan ÖZTÜRK
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 12=600, from kinematic analysis 13=29,980 , 14 = 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
The free-body diagrams of the moving links are shown (STATIC). Prof.Dr. Hasan ÖZTÜRK
The three equilibrium equations for link 4 are: Prof.Dr. Hasan ÖZTÜRK
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
Equations 4 and 5 can be used to determine F23x and F23y Equations 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