Mechanics of Machinery (ME 203) Past Paper Concept Problem PDF
Document Details
National Institute of Technology Karnataka
2024
Dr. Abhilash Singh
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Summary
This document is a course and evaluation plan, along with example problems. It covers concepts and principles of mechanics of machinery, providing problems, diagrams, and analysis concerning mechanical systems and components, for a third-semester undergraduate course, likely in Mechanical Engineering, at the National Institute of Technology Karnataka.
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Course and Evaluation plan (July 2024) Mechanics of Machinery) (ME 203) IIIrd Sem M1, M2 & M3 Capture audience interest using a suitable background picture Event / Meeting Title, Venue, Date Present...
Course and Evaluation plan (July 2024) Mechanics of Machinery) (ME 203) IIIrd Sem M1, M2 & M3 Capture audience interest using a suitable background picture Event / Meeting Title, Venue, Date Presenter Name, Designation Dr. Abhilash Singh, Assistant Professor MED, National Institute of Technology Karnataka Problems Qus.- from figure, a tow truck with an 8 miters boom, which is inclined at a 25° angle. Use trigonometry to determine the horizontal distance that the boom extends from the truck. (All dimensions in miter and angle in Degree) Determine the Horizontal Projection of the Boom horizontal projection cos 25° = 8 horizontal projection = 7.25 m Determine Horizontal Projection of Truck and Boom The horizontal distance from the front end of the truck to the end of the boom is = 6 m + 7.25 m = 13.25 m Determine the Overhang 13.25 m - 11 m = 2.25 m 1 Problems Qus.- Figure shows a front loader. Use trigonometry to determine the required length of the cylinder to orient the arm AB in the configuration shown. Determine Length BC (All dimensions in inch and angle in Degree) By focusing on the triangle created by points A, B, and , it is apparent that this is a Case 3 problem. The third side can be found by using the law of cosines: 1 Problems Qus.- Three forces act on a hook as shown in Figure. Using the analytical component method, determine the net effect of these forces 1 Kinematic Diagram :- A kinematic diagram or skeleton diagram is a simplified drawing that shows the essential components needed for kinematic analysis for any machine/mechanism. Useful to visualize The arrangement of links and their inter-connection. The movement of the parts of the machine. 1 Symbol/Conventions used in Kinematic Diagram- 1 Symbol/Conventions used in Kinematic Diagram- 1 Symbol/Conventions used in Kinematic Diagram- Solve it after the class 1 Steps to draw the kinematic diagram C B 3 2 1 A 4 D Draw the kinematic diagram https://www.youtube.com/watch?app=desktop&v=DYMyoAD9f9c Draw the kinematic diagram Problems Qus.- Figure shows a toggle clamp used to securely hold parts. Analytically determine the displacement of the clamp surface as the handle rotates downward, 15°. Kinematic Diagram Ans = 11.62 mm (displacement) 1 Instruments - Scale Protractor Set square Compass Type of Joints –Kinematic Pairs Lower Pairs–motion is transmitted through an area Higher Pairs–motion is transmitted through a line or a contact, pin and slider joints. point contact; gears, rollers, and spherical joints. Cams and followers to control the opening and closing of valves, optimizing engine performance Mobility or Degree of Freedom (DOF) It refers to the number of independent movements allowed to a mechanism or a component of the mechanism. It quantifies how many parameters are needed to uniquely define the position and orientation of every part of the mechanism. For a planar mechanism, mobility (M) can be calculated using Gruebler's Equation (applicable for 1 DOF): M=3(n−1)−2jp−jh n is the number of links (including the ground frame), jp is the number of lower pairs (single-degree-of-freedom joints, such as revolute or prismatic joints(sliding joints) jh is the number of higher pairs (two-degree-of-freedom joints, such as cam or gear pairs). if DOF = 0 : No movement, Structure, Example - Bridges and Buildings, Application- Where stability require if DOF = 1 : Single Input Control, Predictable Motion , Example - Four-Bar Linkage, Application- Automotive Engines if DOF = 2 : Dual Input Control, Complex Motion , Example - Bicycle Steering Mechanism, Application- Robotic Arms Graphical Problem Figure shows a kinematic diagram of a mechanism that is driven by moving link 2. Graphically reposition the links of the mechanism as link 2 is displaced 30° counterclockwise. Determine the resulting angular displacement of link 4 and the linear displacement of the point E. DOF ( to check planner mechanism) n = 6 jp = (6 pins + 1 sliding) = 7 jh = 0 Kinematic diagram for Problem Continue ……… Velocity Analysis Linear Velocity of a General Point The direction of the linear velocity of a point is the same as the direction of its instantaneous motion. The velocities of points A and B are denoted as VA and VB, respectively. Note that although they are on the same link, both these points can have different linear velocities. Points that are farther from the pivot travel faster. For a point on a rotating link, the linear velocity is always tangent to the circular path of the point. This means the velocity vector will be perpendicular to the radius vector from the axis of rotation to the point. Several points on a link can have different linear velocities, being a rigid body, the entire link has the same angular velocity. The angular velocity (ω) of a link is the angular displacement of that link per unit of time. Velocity Analysis Velocity of link In order to determine the relative motion of the ends of a link, one of the ends is assumed to be moving relative to the other end. Let a link AB rotate about A in anti-clockwise direction such that the end B rotates relative to A. Then the direction of relative motion of B with respect to A is perpendicular to AB. Therefore, the direction of relative velocity of a point w.r.t any other point on a link is always along perpendicular to the straight line joining the two points. Let the velocity of B w.r.t A be represented by ab. Then ab is drawn perpendicular to AB to any convenient scale as shown in the velocity diagram. If the ω = angular velocity of link AB about its end A, then velocity of B w.r.t A = vba = AB.ω. Similarly, velocity of any intermediate point C w.r.t A = = vca = AC.ω. It is represented by the vector ac in the velocity diagram Hence, the point c divides the vector ab in the same ratio as the point C divides the link AB. Velocity Analysis Relationship between linear and angular velocities v = rω v = I V I= magnitude of the linear velocity of the point of consideration r = distance from the center of rotation to the point of consideration ω = angular velocity of the rotating link that contains the point of consideration (radians per time) Linear velocity is always perpendicular to a line that connects the center of the link rotation to the point of consideration. The linear velocity of a point on a link in pure rotation is often called the tangential velocity. The figure illustrates a cam mechanism used to drive the exhaust port of an internal combustion engine. Point B is a point of interest on the rocker plate. At this instant, the cam forces point B upward at 30 mm/s. Determine the angular velocity of the rocker plate and the velocity of point C. Velocity Analysis Solution of the Problem: - ω2 = 1.5 rad/s, cw vC = 22.5 mm/s Kinematic Diagram Mechanism Crank-rocker mechanism: In a four bar linkage, if the shorter side link revolves and the other one rocks (i.e., oscillates), it is called a crank-rocker mechanism. Slider-crank mechanism- An arrangement of mechanical parts designed to convert straight-line motion to rotary motion, as in a reciprocating piston engine, or to convert rotary motion to straight-line motion. Velocity Analysis Problem In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and AD are of equal length. Find the angular velocity of link CD when angle BAD = 60. Velocity Analysis Relative Velocity The difference between the motion of two points is termed relative motion. Relative velocity is a term used when the velocity of one object is related to that of another reference object, which can also be moving. VA = absolute velocity of point A VB = absolute velocity of point B VB/A = VB - VA Figure shows a cargo lift mechanism for a delivery truck. At this instant, point A has a velocity of 12 in./s in the direction shown, and point B has a velocity of 10.4 in./s, also in the direction shown. Determine the angular velocity of the lower link and the relative velocity of point B relative to point A. Velocity Analysis Relative Velocity Figure shows a cargo lift mechanism for a delivery truck. At this instant, point A has a velocity of 12 in./s in the direction shown and point B has a velocity of 10.4 in./s, also in the direction shown. Determine the angular velocity of the lower link and th relative velocity of point B relative to point A. Velocity Analysis Velocity analysis method - Kinematic diagram Velocity diagram Continue ……… Velocity Analysis Problem The crank and connecting rod of a theoretical steam engine are 0.5 m and 2 m long respectively. The crank makes 180 r.p.m. in the clockwise direction. When it has turned 45° from the inner dead center position, determine : 1. Velocity and acceleration of the piston 2. Angular velocity and angular acceleration of the connecting rod 3. Velocity and acceleration of point E on the connecting rod 1.5 m from the gudgeon pin 4. Position and linear velocity of any point G on the connecting rod which has the least velocity relative to crank shaft Instantaneous center of rotation Instantaneous centre : It is a point in on a space about which whole body is assumed to be in pure rotation. It is also called as virtual center. Velocity analysis by instantaneous center of rotation Velocity analysis by instantaneous center of rotation Number of Instantaneous Centres in a Mechanism In a pin jointed four bar mechanism, as shown in the fig. below AB 300mm, BC = CD = 360mm, and AD =600mm. The angle BAD = 60. The crank AB rotates uniformly at 100 r.p.m. Locate all the instantaneous centres and find the angular velocity of the link BC and velocity of C. 600mm the angular velocity of the link BC = 6.04 RAD/S and velocity of C =2.35 M/S Q: -Locate all the instantaneous centres of the slider crank mechanism. The lengths of crank OB and connecting rod AB are 100 mm and 400 mm respectively. If the crank rotates clockwise with an angular velocity of 10 rad/s, and making an angle 45 from IDC find: 1. Velocity of the slider A, 2. Angular velocity of the connecting rod AB