DEFINITIONS Dynamics of Machinery Analyses the forces and couples on the members of the machine due to external forces (static force analysis), also analyses the forces and couples due to accelerations of machine members ( Dynamic force analysis) Rigid Body Deflections of the machine members are neglected in general by treating machine members as rigid bodies (also called rigid body dynamics). In other words the link must be properly designed to withstand the forces without undue deformation to facilitate proper functioning of the system. In order to design the parts of a machine or mechanism for strength, it is necessary to determine the forces and torques acting on individual links. Each component however small, should be carefully analysed for its role in transmitting force. The forces associated with the principal function of the machine are usually known or assumed.

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. Note that when the word "force" is discussed it will refer to a "generalised force" which will also include moments. a) Joint (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. For example, consider a revolute joint in a planar mechanism. In such a joint there is a rotational freedom and any moment along the axis of the revolute joint will not be transmitted from one link to the other, but there will be a force transmission in any general direction which will be determined by the forces acting on the links. If the revolute joint is in a spatial mechanism, there will be moment reaction components perpendicular to the revolute joint axis and a force reaction in a general direction (e.g. there will be three force components and two moment components). In case of a prismatic joint in a planar mechanism there will be no reaction force component along the axis of the slide but a force perpendicular to the slider axis and a couple along the z-axis will be transmitted between the links joined.

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. In case of springs, the magnitude and the direction of the force acting will depend on the geometry of the mechanism at the instant considered. In case of a string, when the string is tight, the force will be in the direction of the string and its sense must be such that it keeps the string in tension; otherwise the string force will be zero. In machine systems, if the force distribution within the rigid body considered is not our concern, the weight of the rigid body can be considered to be equivalent to a force applied at the center of gravity of the rigid body, in the sense and direction of the gravity field. c) Friction or Resisting Force: In general the resisting forces are those that result due to motion and which resist the motion. Since the rigid body assumption is made, one can neglect the internal friction forces that will exist within the body. In such a case friction forces are at the joints in the direction of the relative motion but in opposite sense or in the members that are specially designed to create the friction force (dampers). Friction forces will be discussed in more detail in coming section . d) Inertial Forces. Are the forces due to the inertia of the rigid bodies involved. These forces will be discussed in the coming sections.

FORCE ANALYSIS Apart from static forces, mechanism also experiences inertia forces when subjected to acceleration, called dynamic forces. Static forces are predominant at lower speeds and dynamic forces are predominant at higher speeds. Here, the analysis is aimed at determining the forces transmitted from one point to another, essentially from input point to output point. This would be the starting point for strength design of a component/ system, basically to decide the dimensions of the components Force analysis is essential to avoid either overestimation or under estimation of forces on machine member. Underestimation: leads to design of insufficient strength and to early failure. Overestimation: machine component would have more strength than required. Over design leads to heavier machines, costlier and becomes not competitive

General Principle of force analysis: A machine / mechanism is a three dimensional object, with forces acting in three dimensions. For a complete force analysis, all the forces are projected on to three mutually perpendicular planes. Then, for each reference plane, it is necessary that, the vector sum of the applied forces in zero and that, the moment of the forces about any axis perpendicular to the reference plane or about any point in the plane is zero for equilibrium.

TWO FORCE MEMBER

Equilibrium of a Two-Force Body Consider a plate subjected to two forces F1 and F2 For static equilibrium, the sum of moments about A must be zero. The moment of F2 must be zero. It follows that the line of action of F2 must pass through A. Similarly, the line of action of F1 must pass through B for the sum of moments about B to be zero. Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.

Equilibrium of a Three-Force Body Consider a rigid body subjected to forces acting at only 3 points. Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero. Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D. The lines of action of the three forces must be concurrent or parallel.

THREE FORCE MEMBER

TWO FORCE and ONE MOMENT (TORQUE) MEMBER F1 = F2 =F and T= F x h

FORCES BETWEEN MEMBERS

The all Gravity forces (mg) were neglected compared to the Joint forces.

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One can reduce the number of equations to be solved if the free-body diagrams are analysed to some detail. One need not write the forces in terms of its x and y components if the direction is known and one can identify the forces that are of equal magnitude before attempting for a solution. The free-body diagrams of the links in the four-bar mechanism are redrawn below.

In this case to simplify the calculations we note that Fij = -Fji for the joint forces. Furthermore, since link 3 is a two-force member, F23 and F43 are equal, opposite and their line of action is along AB. Hence F23 =F23 <q13, and q13 is known from the kinematic analysis. Also link 2 is a two-force plus a moment member. Therefore: F13 = - G12 Hence: F 43= -F32 =-F34= -G12= -F23 Now, one can solve for the unknown forces if we write the 3 equilibrium equations for link 4 and one moment equilibrium equation for link 2, which are There are four equations with four unknowns (F34, T12, G14,F14 or F34, T12, G14x, G14y). If the magnitudes come out negative, the direction of the force or torque is opposite to that indicated on the free-body diagram.

VECTORAL CALCULATION OF MOMENT y x Where; F is the magnitude of F and v is a unit vector in the direction of the force F. r is the distance from point C to a point on the line of action of F and u is a unit vector in the direction of r. r must be directed from the point that moment will be taken according to this point,toward a point at which force is applied.

Example Figure shows a slider crank mechanism in which the resultant gas pressure 8 x 104 N/m2 acts on the piston of cross sectional area 0.1 m2 . The system is kept in equilibrium as a result of the couple applied to the crank 2, through the shaft at O2. Determine forces acting on all the links (including the pins) and the couple on 2.

Example

Example Since link 3 is acted upon by only two forces, F43 and F23 are collinear, equal in magnitude and opposite in direction

Example

Example Determine the torque T2 required to keep the given mechanism in equilibrium.

THE FREE BODY DIAGRAMS F34x F23x F34y F23y 3 FC=1000N A B F23y F12y 390 FC=1000N A B F23y F12y F12x T2 O2 A 450 2 B F14y F41y F34y 4

For the Connecting Rod (The member 3) F23y F12y F12x T2 O2 A 450 F23x 2 F34x F23x F34y F23y 3 390 FC=1000N A B B F14y F41y F34y 4 F34x For the Connecting Rod (The member 3) For the piston (The member 4) From the eqn.(2) For the Crank (The member 2)

F23y FC=1000N F23x A 3 F34y B A A FC=1000N F34y C 3 B B The another way (Vectoral method) to calculate the moment for the Connecting Rod (The member 3) F23x F34y F23y 3 390 FC=1000N A B To show the best, how to calculate the moment by vectorial method; the forces, position vectors ,and their unit vectors are drawn separately as shown below 3 390 FC=1000N A B C F34y A B

The Second way for solution Three force member Two force and one moment member

Example

Example Free Body Diagrams

For the member 4 We have 4 unknowns and 3 equations, we can not find them still.

F14x F34x F14y F34y 4 P=200N B O4 C B O4 The Calculation of the moment by vector method It gave the same result

The Calculation of the moment by vector method For the member 3 The Calculation of the moment by vector method It gave the same result The number of the unknowns are 6: The number of the equations are 6 Since the member 2 is two force and one Moment member

The Calculation of the moment by vector method For the member 3 F34y S=500N F34x B 3 F23x A F23y The Calculation of the moment by vector method It gave the same result The number of the unknowns are 6: The number of the equations are 6 Since the member 2 is two force and one Moment member

From the eqn (8) From the eqn (2) From the eqn (3) From the eqn (1) From the eqn (4)

For the Crank (The member 2) F23y F12y F12x T2 O2 A 450 F23x 2 For the Crank (The member 2)

Example A four link mechanism is acted upon by forces as shown in the figure. Determine the torque T2 to be applied on link 2 to keep the mechanism in equilibrium. AD=50mm, AB=40mm, BC=100mm, Dc=75mm, DE= 35mm,

Example Free Body Diagrams

Example Free Body Diagrams

Example Free Body Diagrams

Example Free Body Diagrams

Example Free Body Diagrams

Example: Figure shows a two-dof articulated robot having the same link dimensions. The robot is interacting with the environment surface in a horizontal plane. Obtain the joint torques 1,2 for pushing the surface with an endpoint force of F=(Fx,Fy). Assume no friction. It had been taken from http://www.deu.edu.tr/userweb/zeki.kiral/MEC5004/hafta3_2012(1).ppt

Link 0 Link 1 Link 2 x y θ1 L1 L2 θ2 Fx Fy

2 1 2 1 FREE BODY DIAGRAM Fy Fx Link 2 L2 A B A Link 1 L1 B A' B' The sign of the torques is reverse since they are reaction torques.