1 / 66

ENT142/3 ENGINEERING DYNAMICS Programme: B.Eng.(Hons) (Mechanical)

ENT142/3 ENGINEERING DYNAMICS Programme: B.Eng.(Hons) (Mechanical) School of Mechatronic Engineering Universiti Malaysia Perlis UniMAP. Course Description. Course : ENT142/3 Engineering Dynamics. Lecturer: Dr. Mohd Afendi Bin Rojan. Contact Hours: 3 hours or 4 hours/week.

kcallender
Download Presentation

ENT142/3 ENGINEERING DYNAMICS Programme: B.Eng.(Hons) (Mechanical)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ENT142/3 ENGINEERING DYNAMICS Programme: B.Eng.(Hons) (Mechanical) School of Mechatronic Engineering Universiti Malaysia Perlis UniMAP

  2. Course Description Course : ENT142/3 Engineering Dynamics. Lecturer: Dr. Mohd Afendi Bin Rojan. Contact Hours: 3 hours or 4 hours/week. Tuition Pattern: 3 hours lecture or 2 hours lecture + 2 hours tutorial. Credits: 3 credits Pre-Requisites : None Website: www.afendirojan.wordpress.com

  3. Time table

  4. Course Outcomes CO1: Ability to analyze problems related to rectilinear kinematics, law of motions, and also concepts of mechanics and vector mechanics. CO2: Ability to evaluate problems related to kinematics of particle, involving force and acceleration, work and energy, and also impulse and momentum. CO3: Ability to evaluate problems related to planar kinetics or a rigid body, involving force and acceleration, work and energy, and also impulse and momentum.

  5. 2011/2012 results FAIL 0% FAIL 43.6% FAIL 26%

  6. 2012/2013 results FAIL 0% FAIL 44.1% FAIL 9.1%

  7. Syllabus Chapter 1. Kinematics of a Particle 1.1 Introduction. 1.2 Rectilinear Kinematics: Continuous Motion. 1.3 Rectilinear Kinematics: Erratic Motion 1.4 General Curvilinear Motion 1.5 Curvilinear Motion: Rectangular Component 1.6 Motion of a Projectile 1.7 Curvilinear Motion: Normal and Tangential Components Chapter 2. Kinetics of a Particle: Force and Acceleration 2.1 Newton’s Law of Motion 2.2 The Equation of Motion 2.3 Equations of Motion for a System of Particles 2.4 Equations of Motion: Rectangular Coordinates 2.5 Equation of Motion: Normal and Tangential Coordinates

  8. Chapter 3. Kinetics of a Particle: Work and Energy 3.1 The Work of a Force 3.2 Principle of Work and Energy 3.3 Principle of Work and Energy for a System of Particles 3.4 Power and Efficiency 3.5 Conservative Forces and Potential Energy 3.6 Conservation of Energy. Chapter 4. Kinetics of a Particle: Impulse and Momentum 4.1 Principle of Linear Impulse and Momentum 4.2 Principle of Linear and Momentum for a System of Particles 4.3 Conservation of Linear Momentum for a System of Particles 4.4 Impact

  9. Chapter 5. Planar Kinematics of a Rigid Body 5.1 Rigid-Body Motion 5.2 Translation 5.3 Rotation About a Fixed Axis 5.4 Relative-Motion Analysis: Velocity 5.5 Relative-Motion Analysis: Acceleration Chapter 6. Planer Kinetics of a Rigid Body: Force and Acceleration 6.1 Moment of Inertia 6.2 Planar Kinetic Equations of Motion 6.3 Equation of Motion: Translation

  10. Chapter 7. Planar Kinetics of a Rigid Body: Work and Energy 7.1 Kinetic Energy 7.2 The Work of a Force 7.3 The Work of Couple 7.4 Principle of Work and Energy 7.5 Conservation of Energy Chapter 8. Planar Kinetics of a Rigid Body: Impulse and Momentum 8.1 Linear and Angular Momentum 8.2 Principle of Impulse and Momentum 8.3 Conservation of Momentum

  11. Required Textbook : • R. C. Hibbler, Engineering Mechanics: Dynamics, Latest Ed., Prentice Hall, 2010. Recommended Books: • J. L. Meriam and L. Glenn Kraige, Engineering Mechanics: Dynamics, 2001, John Willey & Sons, Inc. • F. P. Beer, E. R. Johnston and W. E. Clausen, Vector Mechanics for Engineers: Dynamics, 2004, Mc Graw Hill.

  12. CHAPTER 1 KINEMATICS OF A PARTICLE

  13. 1.1 Introduction & 1.2 Rectilinear Kinetics: Continuous Motion Today’s Objectives: Students will be able to find the kinematic quantities (position, displacement, velocity, and acceleration) of a particle traveling along a straight path. In-Class Activities: • Applications •Relations between s(t), v(t), and a(t) for general rectilinear motion •Relations between s(t), v(t), and a(t) when acceleration is constant

  14. APPLICATIONS The motion of large objects, such as rockets, airplanes, or cars, can often be analyzed as if they were particles. Why? If we measure the altitude of this rocket as a function of time, how can we determine its velocity and acceleration?

  15. APPLICATIONS(continued) A train travels along a straight length of track. Can we treat the train as a particle? If the train accelerates at a constant rate, how can we determine its position and velocity at some instant?

  16. Mechanics: the study of how bodies react to forces acting on them Statics: the study of bodies in equilibrium Dynamics: 1. Kinematics – concerned with the geometric aspects of motion 2. Kinetics - concerned with the forces causing the motion An Overview of Mechanics

  17. The displacement of the particle is defined as its change in position. Vector form: r = r’ - r Scalar form: s = s’ - s POSITION AND DISPLACEMENT A particle travels along a straight-line path defined by the coordinate axis s. The position of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft). The totaldistancetraveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.

  18. The average velocity of a particle during a time interval t is vavg = r/t The instantaneous velocity is the time-derivative of position. v = dr/dt Speed is the magnitude of velocity: v = ds/dt Average speed is the total distance traveled divided by elapsed time: (vsp)avg = sT/  t VELOCITY Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s.

  19. The instantaneous acceleration is the time derivative of velocity. Vector form: a= dv/dt Scalar form: a = dv/dt = d2s/dt2 Acceleration can be positive (speed increasing) or negative (speed decreasing). ACCELERATION Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s2 or ft/s2. As the book indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

  20. Position: Velocity: v t v s s t ò ò ò ò ò ò = = = dv a dt or v dv a ds ds v dt v o v s s o o o o o SUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION •Differentiate position to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds •Integrate acceleration for velocity and position. • Note that so and vo represent the initial position and velocity of the particle at t = 0.

  21. v t ò ò = = + dv a dt v v a t yields c o c v o o s t ò ò = = + + ds v dt s s v t (1/2)a t yields o o c 2 s o o v s ò ò = = + v dv a ds v (v ) 2a (s - s ) yields 2 2 c o c o v s o o CONSTANT ACCELERATION The three kinematic equations can be integrated for the special case when accelerationis constant (a = ac) to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2 downward. These equations are:

  22. EXAMPLE Given: A motorcyclist travels along a straight road at a speed of 27 m/s. When the brakes are applied, the motorcycle decelerates at a rate of -6t m/s2. Find: The distance the motorcycle travels before it stops. Plan: Establish the positive coordinate s in the direction the motorcycle is traveling. Since the acceleration is given as a function of time, integrateit once to calculate the velocity and again to calculate the position.

  23. 1) Integrate acceleration to determine the velocity. a = dv / dt => dv = a dt => => v – vo = -3t2 => v = -3t2 + vo v t ò ò = - dv ( 6 t ) dt v o o 3) Now calculate the distance traveled in 3s by integrating the velocity using so = 0: v = ds / dt => ds = v dt => => s – so = -t3 + vot => s – 0 = (-3)3 + (27)(3) => s = 54 m s t ò ò = - + ds ( 3 t v ) dt 2 o s o o EXAMPLE (continued) Solution: 2) We can now determine the amount of time required for the motorcycle to stop (v = 0). Use vo = 27 m/s. 0 = -3t2 + 27 => t = 3 s

  24. 1.3 Rectilinear Kinematics: Erratic Motion Today’s Objectives: Students will be able to determine position, velocity, and acceleration of a particle using graphs. In-Class Activities: • Applications • s-t, v-t, a-t, v-s, and a-s diagrams

  25. APPLICATION In many experiments, a velocity versus position (v-s) profile is obtained. If we have a v-s graph for the rocket sled, can we determine its acceleration at position s = 300 meters ? How?

  26. GRAPHING Graphing provides a good way to handle complex motions that would be difficult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of differentiation and integration as used in dynamics. The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve.

  27. S-T GRAPH Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds/dt). Therefore, the v-t graph can be constructed by finding the slope at various points along the s-t graph.

  28. V-T GRAPH Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or a = dv/dt). Therefore, the a-t graph can be constructed by finding the slope at various points along the v-t graph. Also, the distance moved (displacement) of the particle is the area under the v-t graph during time t.

  29. A-T GRAPH Given the a-t curve, the change in velocity (v) during a time period is the area under the a-t curve. So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle.

  30. s2 ½ (v1²– vo²) = = area under the ò a ds s1 a-s graph A-S GRAPH A more complex case is presented by the a-s graph. The area under the acceleration versus position curve represents the change in velocity (recall òa ds = ò v dv ). This equation can be solved for v1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance.

  31. V-S GRAPH Another complex case is presented by the v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. a = v (dv/ds) Thus, we can obtain a plot of a vs. s from the v-s curve.

  32. EXAMPLE Given: v-t graph for a train moving between two stations Find: a-t graph and s-t graph over this time interval Think about your plan of attack for the problem!

  33. a(ft/s2) 4 3 t(s) -4 3 EXAMPLE (continued) Solution: For the first 30 seconds the slope is constant and is equal to: a0-30 = dv/dt = 40/30 = 4/3 ft/s2 Similarly, a30-90 = 0 and a90-120 = -4/3 ft/s2

  34. s(ft) 3600 The area under the v-t graph represents displacement. Ds0-30 = ½ (40)(30) = 600 ft Ds30-90 = (60)(40) = 2400 ft Ds90-120 = ½ (40)(30) = 600 ft 3000 600 t(s) 30 90 120 EXAMPLE (continued)

  35. 1.4 General Curvilinear Motion & 1.5 Curvilinear Motion: Rectangular Components Today’s Objectives: Students will be able to: a) Describe the motion of a particle traveling along a curved path. b) Relate kinematic quantities in terms of the rectangular components of the vectors. In-Class Activities: • Applications • General curvilinear motion • Rectangular components of kinematic vectors

  36. APPLICATIONS The path of motion of each plane in this formation can be tracked with radar and their x, y, and z coordinates (relative to a point on earth) recorded as a function of time. How can we determine the velocity or acceleration of each plane at any instant? Should they be the same for each aircraft?

  37. How can we determine its position or acceleration at any instant? If you are designing the track, why is it important to be able to predict the acceleration of the car? APPLICATIONS (continued) A roller coaster car travels down a fixed, helical path at a constant speed.

  38. POSITION AND DISPLACEMENT A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectors are used to describe the motion. A particle moves along a curve defined by the path function, s. The position of the particle at any instant is designated by the vector r= r(t). Both the magnitude and direction of r may vary with time. If the particle moves a distance Ds along the curve during time interval Dt, the displacement is determined by vectorsubtraction: D r = r’ - r

  39. VELOCITY Velocity represents the rate of change in the position of a particle. The averagevelocity of the particle during the time increment Dt is vavg = Dr/Dt . The instantaneous velocity is the time-derivative of position v = dr/dt . The velocity vector, v, is always tangent to the path of motion. The magnitude of v is called the speed. Since the arc length Ds approaches the magnitude of Dr as t→0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector!

  40. ACCELERATION Acceleration represents the rate of change in the velocity of a particle. If a particle’s velocity changes from v to v’ over a time increment Dt, the average acceleration during that increment is: aavg = Dv/Dt = (v - v’)/Dt The instantaneous acceleration is the time-derivative of velocity: a = dv/dt = d2r/dt2 A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function.

  41. RECTANGULAR COMPONENTS: POSITION It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangularcomponents, relative to a fixed frame of reference. The position of the particle can be defined at any instant by the position vector r = x i + y j + z k . The x, y, z components may all be functions of time, i.e., x = x(t), y = y(t), and z = z(t) . The magnitude of the position vector is: r = (x2 + y2 + z2)0.5 The direction of r is defined by the unit vector: ur = (1/r)r

  42. The velocity vector is the time derivative of the position vector: v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt Since the unit vectorsi, j, k are constant in magnitude and direction, this equation reduces to v = vxi + vyj + vzk where vx = = dx/dt, vy = = dy/dt, vz = = dz/dt • • • x y z RECTANGULAR COMPONENTS: VELOCITY The magnitude of the velocity vector is v = [(vx)2 + (vy)2 + (vz)2]0.5 The direction of v is tangent to the path of motion.

  43. The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector): a = dv/dt = d2r/dt2 = axi + ayj + azk where ax = = = dvx /dt, ay = = = dvy /dt, az = = = dvz /dt • • •• •• vx vy x y • •• vz z RECTANGULAR COMPONENTS: ACCELERATION The magnitude of the acceleration vector is a = [(ax)2 + (ay)2 + (az)2 ]0.5 The direction of a is usually not tangent to the path of the particle.

  44. EXAMPLE Given: The motion of two particles (A and B) is described by the position vectors rA = [3t i + 9t(2 – t) j] m rB = [3(t2 –2t +2) i + 3(t – 2) j] m Find: The point at which the particles collide and their speeds just before the collision. Plan: 1) The particles will collide when their position vectors are equal, or rA = rB . 2) Their speeds can be determined by differentiating the position vectors.

  45. EXAMPLE (continued) Solution: 1) The point of collision requires that rA = rB, so xA = xB and yA = yB . x-components: 3t = 3(t2 – 2t + 2) Simplifying: t2 – 3t + 2 = 0 Solving: t = {3  [32 – 4(1)(2)]0.5}/2(1) => t = 2 or 1 s y-components: 9t(2 – t) = 3(t – 2) Simplifying: 3t2 – 5t – 2 = 0 Solving: t = {5  [52 – 4(3)(–2)]0.5}/2(3) => t = 2 or – 1/3 s So, the particles collide when t = 2 s. Substituting this value into rA or rB yields xA = xB = 6 m and yA = yB = 0

  46. EXAMPLE (continued) 2) Differentiate rA and rB to get the velocity vectors. . . vA = drA/dt = = [3i + (18 – 18t)j] m/s At t = 2 s: vA = [3i – 18j] m/s xA . i + yA j • • vB = drB/dt = xBi + yBj = [(6t – 6)i + 3j] m/s At t = 2 s: vB = [6i + 3j] m/s Speed is the magnitude of the velocity vector. vA = (32 + 182) 0.5 = 18.2 m/s vB = (62 + 32) 0.5 = 6.71 m/s

  47. 1.6 MOTION OF A PROJECTILE Today’s Objectives: Students will be able to analyze the free-flight motion of a projectile. In-Class Activities: • Check homework, if any • Applications •Kinematic equations for projectile motion

  48. APPLICATIONS A kicker should know at what angle, q, and initial velocity, vo, he must kick the ball to make a field goal. For a given kick “strength”, at what angle should the ball be kicked to get the maximum distance?

  49. APPLICATIONS (continued) A fireman wishes to know the maximum height on the wall he can project water from the hose. At what angle, q, should he hold the hose?

  50. For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant. CONCEPT OF PROJECTILE MOTION Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., gravity).

More Related