Kinematics & Dynamics of Machines BMEE207L Dr. J.NAVEEN., M.E., Ph.D (UPM Malaysia) Assistant Professor (Senior Grade) School of Mechanical Engineering, Vellore Institute of Technology, Vellore, India 632014 naveen.j@vit.ac.in, gandhi.naveen66@gmail.com Course Objectives: • 1. To enable students to understand the fundamental concepts of mechanisms. • 2. To facilitate students to understand the functions of cams, gears, and flywheel. • 3. To impart knowledge on design of mechanisms and dynamic loads acting on the mechanism. • 4. To give an insight on the concepts of balancing, vibration and speed governing devices Expected Course Outcome: • At the end of the course, the student will be able to • 1. Examine the kinematic behaviour of various planar mechanisms. • 2. Construct velocity and acceleration diagrams for various planar mechanisms. • 3. Analyse kinematics of cam and gear-train mechanisms. • 4. Investigate the dynamic forces acting on planar mechanisms. • 5. Analyse the balancing of masses and vibrations of mechanical systems. • 6. Assess the characteristics of governors and gyroscopic effects Module:1 Mechanisms and kinematics • Introduction, mechanisms and machines, terminology, planar mechanism - Kinematic diagram and inversion, Mobility, Coincident joints, Grubbler and Grashoff’s law, Four bar, single and double slider mechanisms and their inversions Module:2 Velocity and Accelerations in Mechanisms • Velocity and acceleration in planar mechanisms - Relative velocity method, Coriolis component of acceleration, Kennedy’s Theorem, Instantaneous Centre method Module:3 Kinematic analysis of Cams and Gears • Cams: Types of cams – Types of followers – Definitions – Motions of the followers – Layout of cam profiles. Gear: terminology, fundamental of gearing, involute profile, interference and undercutting, minimum number of teeth, contact ratio - Gear trains: simple, compound and epicyclic. Module:4 Synthesis of Planar mechanisms • Two position and Three position synthesis of planar mechanism - Graphical and analytical methods - Freudenstein equation Module:5 Dynamic Force Analysis • Introduction-D’ Alembert’s principle-static and inertial force analysis of reciprocating engine Equivalent dynamic system. Turning moment diagram-four stroke engine-multicylinder engine-design of flywheel of IC engine-design of flywheel rimdesign of flywheel of punching press. Module:6 Balancing and Vibration • Static and Dynamic Balancing of Rotating Masses, Balancing of Reciprocating Masses. Introduction to vibration - Terminologies - Single degree of freedom- damped and undamped- free and forced vibration – Vibration isolation and Transmissibility. Transverse vibrations of shafts – Whirling of shaft -Torsional vibration of single rotor and two rotors’ systems Module:7 Governors and Gyroscope • Governors: Centrifugal Governors- types and its characteristics - Working principle of electronic governor. Gyroscope – Gyroscopic Effects on the Movement of airplanes and Ships – Gyroscope Stabilization Module:8 Contemporary issues: Module:1 Basics of Mechanisms • Mechanism • If number of bodies are assembled in such a way that the motion of one causes constrained and predictable motion to the others, it is known as a mechanism .A mechanism transmits and modifies a motion • Machine: A machine is a mechanism or collection of mechanisms, which transmit and modifies the mechanical energy into desired work Slider Crank Mechanism • Structure: It is an assemblage of a number of resistant bodies (known as members) having no relative motion between them and meant for carrying loads. A railway bridge, a roof truss, machine frames • Kinematics is the study of motion, without considering the forces which produce that motion. • Dynamics of machines involves the study of forces acting on the machine parts and the motions resulting from these forces. Plane motion: • 1) Translation • 2) rotation and • 3) combination of translation and rotation. Types of Constrained Motion • Completely Constrained motion • Completely constrained motion is when the motion of the pair is limited to one direction, irrespective of the direction of the applied force. • For example, a rectangular bar moving in one direction in a rectangular hole is a completely constrained motion. The bar cannot rotate or move in any other direction. • Incompletely constrained motion • In an incompletely constrained motion, the motion between the pair can take place in more than one direction. A circular shaft moving in a circular hole is an example of incompletely constrained motion • Successfully/Partially constrained motion • In partially constrained motion, when there is no external force applied the motion will be in more than one direction, but when an external force is applied the motion is restricted to a single direction. • Ex; foot step bearing, • Piston in a cylinder of an IC engine is made to have only reciprocating motion and no rotary motion due to constrain of the piston pin • Rigid Bodies-does not suffer any distortion • Resistant Bodies—semi rigid bodies-Flexible Link • • • • A link is assumed to be completely rigid, or under the action of forces it does not suffer any deformation, signifying that the distance between any two points on it remains constant. No relative motion between the joints Binary link: Link which is connected to other links at two points. Ternary link: Link which is connected to other links at three points. • Quaternary link: Link which is connected to other links at four points. Kinematic Pair • The two links or elements of a machine, when in contact with each other, are said to form a pair. • Have relative motion • Classification of kinematic pair ❖ Based on nature of contact between elements a.Lower pair If the joint by which two members are connected has surface contact (b) Higher pair. at a point or along a line …between two gear teeth , cam and follower, wheel rolling on a surface Based on relative motion between pairing elements: (a) Siding pair. (b) Turning pair (revolute pair). (c) Rolling Pair .Ball Bearing • screw pair.(helical pair) • (links have both turning and sliding motion) Spherical pair.(when one link in the form of sphere turns inside the fixed link it is a spherical pair) Based on the nature of mechanical constraint. • Closed pair. • When the elements of a pair are held together mechanically .it is known as closed pair. • One of the element is solid and full and the other is hollow or open • Unclosed pair • When two links of a pair are in contact either due to spring action or force of gravity.. The links are not held together mechanically. Joint • Binary Joint(B)(If two links are joined at the same connection…) • Ternary joint (T)(if three links are joined at a connection…It is considered equivalent to 2 binary joints, since fixing of anyone link constitutes two binary joints with each of the other two links ) • Quaternary Joint (Q) (if four links are joined at a connection…..) 1 T= 2 B 1Q= 3 B Problem Degrees of Freedom • A free body in space can have six degrees of freedom. I.e., linear positions along x, y and z axes and rotational/angular positions with respect to x, y and z axes. Unconstrained rigid body in space describes 6 DOF. They are 3Translational and 3 rotational. • Unconstrained rigid body in a plane has 3 DOF. They are 2Translational (about x,y-axis). and one – Rotational . • • • • • N- Total number of links Nb, Nt F- Degrees of Freedom P1 - Lower pair P2- Higher Pair DOF of a spatial Mechanisms :• • • • • • In a Mechanism one link is fixed Therefore number of Movable link= N-1 Number of Degree of freedom= 6(N-1) DOF of a spatial Mechanisms :DOF=6(N-1)-5P1-4P2-3P3-2P4-P5 N=Total Number of links in a mechanism. P1=Number of Pairs having one degree of freedom P2=Number of Pairs having two degree of freedom and so on . DOF of Planar Mechanism • If there are N links in a mechanism, Number of movable links are only (N-1) [because one link is fixed in mechanism] • DOF for (N-1) links in a plane=3(N-1) • In Planar Mechanisms Lower Pairs will have 1DOF and Higher pairs will have 2 DOF • Kutzback’s Criterion DOF=3(N-1)-2L-H REDUNDANT LINK: • Sometimes a mechanism may have one or more links which do not introduce any extra constraints. Such links are known as redundant links and should not be counted to find the DOF. • The Mechanisms has ‘5’ links, but the function of the mechanism is not affected even if any one of the links 2, 4 or 5 are removed. Thus effective number of links in this case is ‘4’`with 4 turning pairs and thus has one degree of freedom Redundant Degree Of Freedom. • Sometime one or more links of a mechanism can be moved without causing any motion to the rest of the links of the mechanisms. Such a link is said to have redundant degree of freedom. • For Mechanisms possessing redundant degree of freedom, the effective degree of freedom is given by • F=3(N-1)-2L-H-Fr • Fr- Number of Redundant degree of freedom Problem 1 • Calculate the following 1. Number of Binary Links Nb 2. Nb=4 2. Number of Ternary Links (Nt) Nt= 4 3. Number of Quaternary or other links No=0 4. Total no of links N=8 5. Number of Loops L-4 7. Degrees of freedom F= 3(N-1)- 2P1- P2 N= 8 P1= number of lower pair freedom or no of pairs which is having 1 degree of freedom P1= 11 P2=0 F=-1 Problem 2 Problem 2 • Calculate the following Number of Binary Links Nb=4 Number of Ternary Links (Nt) 4,3,2,(3 locations :1 ter) 4 Number of Quaternary or other links N0=0 Total no of links N=8 Number of Joints/Pairs P1= P1=10 • • • • • Degrees of freedom F= 3(N-1)-2P1-P2 N=8 P1=10 F=1 Problem 3 Problem 3 Calculate the following 1. Number of Binary Links Nb= 7 2. Number of Ternary Links (Nt)=2 3. Number of Quaternary or other links=2 4. Total no of links=N=11 5. Number of Loops=5 6. Number of Joints/Pairs= 15 (11B+2 Ter) Problem 4 calculate the DOF • • • • • F= 3(N-1)-2P1-P2 N=7 P1 or lower pair or binary joints= 8 P2 =0 DOF= 2 (2 separate input motions are required to produce constrained motion) Calculate the DOF- Problem 5 Calculate the DOF- Problem 5 • DOF= 3(N-1)-2J-H • N=6 ; J=7 (5B+1T) • ; H=0; DOF= 1 Problem 6 • DOF= 3(N-1)-2J-H • N=3 ; J=2 ; H=1; DOF= 1 Problem 7 • DOF= 3(N-1)-2J-H • N=8 ; J=8B+1 T=10 • ; H=0; DOF= 1 Problem 8 • DOF= 3(N-1)-2J-H • N=8 ; J=8B+2B=10 ; H=0; DOF= 1 Problem 9 N=7; J= 8; H=1; DOF=1 Problem 10 • • • • N=4 J=4 H=0 Redundant degrees of freedom =1 (The rod can slide without causing any motion to other link .so it has redundant DOF) • Dof= 3(n-1)-2J-H-Fr • Dof=0 A kinematic chain may be defined as a combination of kinematic pairs, joined in such a way that each link forms a part of two pairs and the relative motion between the links or elements is completely or successfully constrained. Mechanism and Inversion ❖When one link is fixed in a kinematic chain it forms a mechanism ❖different mechanisms by fixing different links in a kinematic chain, is known as inversion of the mechanism Types of Kinematic Chains 1. Four bar chain or quadric cyclic chain, 2. Single slider crank chain, and 3. Double slider crank chain. Four Bar Chain or Quadric Cycle Chain ❖ Grashof ’s law for a four bar mechanism, the sum of the shortest and longest link lengths should not be greater than the sum of the remaining two link lengths if there is to be continuous relative motion between the two links. ❖ A very important consideration in designing a mechanism is to ensure that the input crank makes a complete revolution relative to the other links. ❖ AD (link 4 ) is a crank. ❖ The link BC (link 2) which makes a partial rotation or oscillates is known as lever or rocker or follower ❖ The link CD (link 3) which connects the crank and lever is called connecting rod or coupler. ❖ The fixed link AB (link 1) is known as frame of the mechanism. ❖ When the crank (link 4) is the driver, the mechanism is transforming rotary motion into oscillating motion. Grashof’s Law Find the inversion of this chain Inversions of Four Bar Chain1.Beam engine (crank and lever mechanism). 2.Coupling rod of a locomotive. ❖ The links AD and BC (having equal length) act as cranks and are connected to the respective wheels. ❖ The link CD acts as a coupling rod . ❖ The link AB is fixed in order to maintain a constant centre to centre distance between them. ❖ This mechanism is meant for transmitting rotary motion from one wheel to the other wheel 3. Watt’s indicator mechanism (Double lever mechanism). ✓ ✓ ✓ ✓ ✓ Fixed link at A link AC, link CE and link BFD. It may be noted that BF and FD form one link because these two parts have no relative motion between them. ✓ The links CE and BFD act as levers. ✓ The displacement of the link BFD is directly proportional to the pressure of gas or steam which acts on the indicator plunger. Single Slider Crank Chain Single Slider Crank Chain • A single slider crank chain is a modification of the basic four bar chain. It consist of one sliding pair and three turning pairs. • It is, usually, found in reciprocating steam engine mechanism. • This type of mechanism converts rotary motion into reciprocating motion and vice versa Single Slider Crank Chain • the links 1 and 2, links 2 and 3, and links 3 and 4 form three turning pairs while the links 4 and 1 form a sliding pair • The link 1 corresponds to the frame of the engine, which is fixed. The link 2 corresponds to the crank ; link 3 corresponds to the connecting rod and link 4 corresponds to cross-head. • As the crank rotates, the cross-head reciprocates in the guides and thus the piston reciprocates in the cylinder. Inversions of Single Slider Crank Chain 1. Pendulum pump or Bull engine. ❖The inversion is obtained by fixing the cylinder or link 4 (i.e. sliding pair). ❖when the crank (link 2) rotates, the connecting rod (link 3) oscillates about a pin pivoted to the fixed link 4 at A and the piston attached to the piston rod (link 1) reciprocates. 2. Oscillating cylinder engine. • In this mechanism, the link 3 forming the turning pair is fixed. • The link 3 corresponds to the connecting rod of a reciprocating steam engine mechanism. • When the crank (link 2) rotates, the piston attached to piston rod (link 1) reciprocates and the cylinder (link 4) oscillates. 3. Rotary internal combustion engine or Gnome engine ❖It consists of seven cylinders in one plane and all revolves. ❖while the crank (link 2) is fixed. ❖The connecting rod (link 4) rotates, the piston (link 3) reciprocates inside the cylinders forming link 1. • • • • • • • 4. Crank and slotted lever quick return motion mechanism. This mechanism is mostly used in shaping machines, slotting machines and in rotary internal combustion engines. link AC (i.e. link 3)-fixed. The driving crank CB revolves with uniform angular speed about the fixed centre C. A sliding block attached to the crank pin at B slides along the slotted bar AP and thus causes AP to oscillate about the pivoted point A. A short link PR transmits the motion from AP to the ram which carries the tool and reciprocates along the line of stroke R1R2. The line of stroke of the ram (i.e. R1R2) is perpendicular to AC produced. • In the extreme positions, AP1 and AP2 are tangential to the circle and the cutting tool is at the end of the stroke. • The forward or cutting stroke occurs when the crank rotates from the position CB1 to CB2 (or through an angle β) in the clockwise direction. • The return stroke occurs when the crank rotates from the position CB2 to CB1 (or through angle α) in the clockwise direction. • Since the crank has uniform angular speed, therefore, • Since the tool travels a distance of R1 R2during cutting and return stroke, therefore travel of the tool or length of stroke.= R1R2 = P1P2 = 2P1Q--------[1] From ΔP1AQ Sin (90 – α / 2) = P1Q / P1A P1Q = Sin (90 – α / 2) P1A Apply Equ. 1 Length of stroke=2P1Q =2Sin (90 – α / 2) P1A Sin (90 – α / 2) = Cos α / 2 Length of the stroke = 2 AP1COS α / 2 AP1= AP CB1= CB Problem: In a crank and slotted lever quick return motion mechanism, the distance between the fixed centres is 240 mm and the length of the driving crank is 120 mm. Find the inclination of the slotted bar with the vertical in the extreme position and the time ratio of cutting stroke to the return stroke. If the length of the slotted bar is 450 mm, find the length of the stroke if the line of stroke passes through the extreme positions of the free end of the lever. 5. Whitworth quick return motion mechanism. The link CD (link 2) -is fixed, • The driving crank CA (link 3) rotates at a uniform angular speed. • The slider (link 4) attached to the crank pin at A slides along the slotted bar PA (link 1) which oscillates at a pivoted point D. • The connecting rod PR carries the ram at R to which a cutting tool is fixed. • When the driving crank CA moves from the position CA1 to CA2 (or DP from the position DP1 to DP2) through an angle α in the clockwise direction, the tool moves from the left hand end of its stroke to the right hand end through a distance 2 PD. • Now when the driving crank moves from the position CA2 to CA1 (or the DP from DP2 to DP1 ) through an angle β in the clockwise direction, the tool moves back from right hand end of its stroke to the left hand end. • The ratio between the time taken during the cutting and return strokes is given by Double Slider Crank Chain • A kinematic chain which consists of two turning pairs and two sliding pairs is known as double slider crank chain 1. Elliptical trammels ❖It is an instrument used for drawing ellipses. ❖This inversion is obtained by fixing the slotted plate (link 4). ❖The fixed plate or link 4 has two straight ❖grooves cut in it, at right angles to each other. ❖The link 1 and link 3, are known as sliders and form sliding pairs with link 4. ❖The link AB (link 2) is a bar which forms turning pair with links 1 and 3. ❖When the links 1 and 3 slide along their respective grooves, any point on the link 2 such as ❖P traces out an ellipse on the surface of link 4, as shown in Fig.. A little consideration will ❖show that AP and BP are the semi-major axis and semi-minor axis of the ellipse respectively 2. Scotch yoke mechanism. ❖This mechanism is used for converting rotary motion into a reciprocating motion. 3. Oldham’s coupling ❖An oldham's coupling is used for connecting two parallel shafts whose axes are at a small distance apart. The shafts are coupled in such a way that if one shaft rotates, the other shaft also rotates at the same speed. This inversion is obtained by fixing the link 2, as shown in Fig. ❖The shafts to be connected have two flanges (link 1 and link 3) rigidly fastened at their ends by forging. ❖When the driving shaft A is rotated, the flange C (link 1) causes the intermediate piece (link 4) to rotate at the same angle through which the flange has rotated, and it further rotates the flange D (link 3) at the same angle and thus the shaft B rotates. Hence links 1, 3 and 4 have the same angular velocity at every instant.
