Module 6 Acceleration Analysis PDF

BULACAN STATE UNIVERSITY CITY OF MALOLOS, BULACAN COLLEGE OF ENGINEERING MECHANICAL ENGINEERING DEPARTMENT MACHINE ELEMENTS 1 MODULE 6 PREPARED BY: ENGR. ALDRIN C. BERNARDO  Machine...

BULACAN STATE UNIVERSITY CITY OF MALOLOS, BULACAN COLLEGE OF ENGINEERING MECHANICAL ENGINEERING DEPARTMENT MACHINE ELEMENTS 1 MODULE 6 PREPARED BY: ENGR. ALDRIN C. BERNARDO  Machine Elements 1 1 UNIT 6ACCELERATION ANALYSIS LESSONS COVERED 6.1 Linear Acceleration 6.2 Acceleration Of A Link 6.3 Normal And Tangential Acceleration 6.4 Relative Motion 6.5 Relative Acceleration Using Graphical Method 6.6 Coriolis Acceleration Using Graphical Method DURATION: 4 hours PRETEST This is a pre – test. We just want to know if you are equipped before you drive through the lesson. Using graphical method, determine instantaneous velocity of the Whitworth mechanism shown in the figure below. Crank BC rotates in uniform angular speed of 1 rad/s in counter clockwise direction. Dimensions: BC = 40 cm, AD = 25 cm, DQ = 75 cm.  Machine Elements 1 2 INTRODUCTION Acceleration analysis involves determining the manner in which certain points on the links of a mechanism are either “speeding up” or “slowing down.” Acceleration is a critical property because of the inertial forces associated with it. In the study of forces, Sir Isaac Newton discovered that an inertial force is proportional to the acceleration imposed on a body. This phenomenon is witnessed anytime you lunge forward as the brakes are forcefully applied on your car. The determination of accelerations in a linkage is the purpose of this chapter. The primary procedure used in this analysis is the relative acceleration method, which utilizes the results of the relative velocity method introduced in Chapter 5. Consistent with other chapters in this book, both graphical and analytical techniques are utilized. OBJECTIVES/COMPETENCIES Upon completion of this chapter, the student will be able to: 1. Define linear, rotational, normal, tangential, Coriolis, and relative accelerations. 2. Use the relative acceleration method to graphically solve for the acceleration of a point on a link, knowing the acceleration of another point on that link. 3. Use the relative acceleration method to graphically determine the acceleration of a point of interest on a floating link. 4. Understand when the Coriolis acceleration is present, and  include it in the analysis. 5. Use the relative acceleration method to analytically solve for the acceleration of a point. LESSON 6.1: LINEAR ACCELERATION Linear acceleration, A, of a point is the change of linear velocity of that point per unit of time. Chapter 5 was dedicated to velocity analysis. Velocity is a vector quantity, which is defined with both a magnitude and a direction. Therefore, a change in either the magnitude or direction of velocity produces an acceleration. The magnitude of the acceleration vector is designated a = |A|. Linear Acceleration of Rectilinear Points Consider the case of a point having straight line, or recti- linear, motion. Such a point is most commonly found on a link that is attached to the frame Machine Elements 1 3 with a sliding joint. For this case, only the magnitude of the velocity vector can change. The acceleration can be mathematically described as (6.1) However, because then (6.2) For short time periods, or when the acceleration can be assumed to be linear, the following relationship can be used: (6.3) Because velocity is a vector, equation (6.1) states that acceleration is also a vector. The direction of linear acceleration is in the direction of linear movement when the link accelerates. Conversely, when the link decelerates, the direction of linear acceleration is opposite to the direction of linear movement. Linear acceleration is expressed in the units of velocity (length per time) divided by time, or length per squared time. In the U.S. Customary System, the common units used are feet per squared second (ft/s2) or inches per squared second (in./s2). In the International System, the common units used are meters per squared second (m/s2) or millimeters per squared second (mm/s2). For comparison purposes, linear acceleration is often stated relative to the acceleration due to gravity g = 32.17 ft/s2 = 386.4 in./s2 = 9.81 m/s2. Thus, a 10g acceleration is equivalent to 3864 in./s2. Constant Rectilinear Acceleration Rewriting equation (7.3), the velocity change that occurs during a period of constant acceleration is expressed as (6.4) Additionally, the corresponding displacement that occurs during a period of constant acceleration can be written as (6.5) Equations (6.4) and (6.5) can be combined to give Machine Elements 1 4 (6.6) Since rectilinear motion is along a straight line, the direction of the displacement, velocity, and acceleration (r, v, a) can be specified with an algebraic sign along a coordinate axis. Thus, equations (6.4), (6.5), and (6.6) can be written in terms of the vector magnitudes (r, v, a). EXAMPLE PROBLEM 6.1 An express elevator used in tall buildings can reach a full speed of 15 mph in 3s. Assuming that the elevator experiences constant acceleration, determine the acceleration and the displacement during the 3s. SOLUTION: 1. Calculate Acceleration Assuming that the acceleration is constant, equation (7.3) can be accurately used. Because the elevator starts at rest, the velocity change is calculated as Then, the acceleration is calculated as 2. Normalize the Acceleration with Respect to Gravity When people accelerate in an elevator, the acceleration is often “normalized” relative to the acceleration due to gravity. The standard acceleration due to gravity (g) on earth is 32.17 ft/s2 or 9.81 m/s2. Therefore, the acceleration of the elevator can be expressed as 3. Calculate the Displacement during the 3-Second Interval The displacement can be determined from equation (6.5). Acceleration and the Velocity Profile As stated in equation (6.1), the instantaneous acceleration is the first derivative of the instantaneous velocity with respect to time. Occasionally, a closed-form equation for the instantaneous velocity of a point is available. In Machine Elements 1 5 these cases, the derivative of the equation, evaluated at the specified time, will yield the instantaneous acceleration. More often, especially in programmable actuators used in automation, velocity profiles are specified as introduced in Chapter 6. Recall that the displacement for a certain time interval is the area under the v-t curve for that time interval. Conversely, the acceleration at a certain time is the slope of the v-t curve. EXAMPLE PROBLEM 7.2 An automated assembly operation requires linear motion from a servo actuator. The total displacement must be 10 in. For design reasons, the maximum velocity must be limited to 2 in./s, and the maximum acceleration or deceleration should not exceed 4 in./s2. Plot the velocity profile for this application. SOLUTION: 1. Determine the Motion Parameters during Speed-Up For the standard velocity profile for a servomotor, the speed-up portion of the motion is constant acceleration. Rewriting and substituting the magnitudes of the velocity v and acceleration a into equation (6.3) gives the time consumed during speed-up. Equation (6.5) is used to calculate the magnitude of the displacement during speed-up. 2. Determine the Motion Parameters during Slow-Down For a standard velocity profile, the slow-down portion of the motion is constant acceleration. The time consumed during slow-down is The magnitude of the displacement during slow-down is 3. Determine the Motion Parameters during Steady-State Because 0.5 in. of displacement occurs during speed-up and another 0.5 in. during shut-down, the remaining 9.0 in. of displacement is during constant Machine Elements 1 6 velocity motion. Equation (5.2) is used to calculate the time consumed during constant velocity. 4. Determine the Motion Parameters during Steady-State Using the velocity and time information for this sequence, the velocity profile shown in Figure 6.1 is generated. FIGURE 6.1 Velocity profile LESSON 6.2: ACCELERATION OF A LINK Recall from Section 5.3 that any motion, however complex, can be viewed as a combination of a straight line movement and a rotational movement. Fully describing the motion of a link can consist of specifying the linear motion of one point and the rotational motion of the link about that point. As with velocity, several points on a link can have different accelerations, yet the entire link has the same rotational acceleration. Angular Acceleration Angular acceleration, α, of a link is the angular velocity of that link per unit of time. Mathematically, angular acceleration of a link is described as (6.7) However, because then (6.8) Machine Elements 1 7 For short time periods, or when the angular acceleration is assumed to be linear, the following relationship can be used: (6.9) Similarly to the discussion in Section 7.2, the direction of angular acceleration is in the direction of motion when the angular velocity increases or the link accelerates. Conversely, the angular acceleration is in the opposite direction of motion when the angular velocity decreases, or the link is decreasing. In planar analyses, the direction should be described as either clockwise or counterclockwise. Angular acceleration is expressed in the units of angular velocity (angle per time) divided by time, or angle per squared time. In both the U.S. Customary System and the International System, the common units used are degrees per squared second (deg/s2), revolutions per squared second (rev/s2), or the preferred unit of radians per squared second (rad/s2). Constant Angular Acceleration Rewriting equation (6.7), the angular velocity change that occurs during a period of constant angular acceleration is expressed as (6.10)  Additionally, the corresponding angular displacement that occurs during a period of constant angular acceleration can be written as Equations (6.10) and (6.11) can be combined to give EXAMPLE PROBLEM 6.3 An electric motor drives the grinding wheel clockwise, as shown in Figure 6.3. It will speed up to 1800 rpm in 2 s when the power is turned on. Assuming that this speed-up is at a constant rate, determine the angular acceleration of the grinding wheel. Also determine the number of revolutions that the wheel spins before it is at full speed. Machine Elements 1 8 FIGURE 6.3 Grinding wheel SOLUTION: 1. Calculate the Acceleration Since acceleration is typically specified in rad/s2, convert the speed of the grinding wheel to rad/s with the following: With constant acceleration, equation (6.9) can be used, giving The direction of the acceleration is clockwise, which is in the direction of motion because the grinding wheel is speeding up. 2. Calculate the Displacement during the 2-Second Interval The number of revolutions during this speed-up period can be determined through equation (6.11). LESSON 6.3: NORMAL AND TANGENTIAL ACCELERATION As presented in Section 6.2., the velocity of a point moving in a general path can change in two independent ways. The magnitude or the direction of the velocity vector can change over time. Of course, acceleration is the time rate of velocity change. Thus, acceleration is commonly separated into two elements: normal and tangential components. The normal component is created as a result of a change in the direction of the velocity vector. The Machine Elements 1 9 tangential component is formed as a result of a change in the magnitude of the velocity vector. Tangential Acceleration For a point on a rotating link, little effort is required to determine the direction of these acceleration components. Recall that the instantaneous velocity of a point on a rotating link is perpendicular to a line that connects that point to the center of rotation. Any change in the magnitude of this velocity creates tangential acceleration, which is also perpendicular to the line that connects the point with the center of rotation. The magnitude of the tangential acceleration of point A on a rotating link 2 can be expressed as (6.13) It is extremely important to remember that the angular acceleration, α, in equation (6.13) must be expressed as units of radians per squared time. Radians per squared second is the most common unit. Similarly to the discussion in Section 6.2, tangential acceleration acts in the direction of motion when the velocity increases or the point accelerates. Conversely, tangential acceleration acts in the opposite direction of motion when the velocity decreases or the point decelerates. Normal Acceleration Any change in velocity direction creates normal acceleration, which is always directed toward the center of rotation. Figure 6.4a illustrates a link rotating at constant speed. The velocity of point A is shown slightly before and after the configuration under consideration, separated by a small angle dθ2. Because the link is rotating at constant speed, the magnitudes of VA’ = VA” are equal. Thus, VA’ = VA”. FIGURE 6.4 Normal acceleration Figure 6.4b shows a velocity polygon, vectorally solving for the change in velocity, dv. Notice that the change of the velocity vector, dv, is directed toward the center of link rotation. In fact, the normal acceleration will always be directed toward the center of link rotation. This is because, as the point rotates around a fixed pivot, the velocity vector will change along the Machine Elements 1 10 curvature of motion. Thus, the normal vector to this curvature will always be directed toward the fixed pivot. In Figure 7.4a, because Δθ relationship can be stated: Because acceleration is defined as the time rate of velocity change, both sides should be divided by time: Using equation (5.6), the relationships between the magnitude of the linear velocity and angular velocity, the following equations for the magnitude of the normal acceleration of a point can be derived: (6.14)  (6.15) Total Acceleration As previously mentioned, acceleration analysis is important because inertial forces result from accelerations. These loads must be determined to ensure that the machine is adequately designed to handle these dynamic loads. Inertial forces are proportional to the total acceleration of a body. The total acceleration, A, is the vector resultant of the tangential and normal components. Mathematically, it is expressed as (6.16) EXAMPLE PROBLEM 6.4 The mechanism shown in Figure 6.5 is used in a distribution center to push boxes along a platform and to a loading area. The input link is driven by an electric motor, which, at the instant shown, has a velocity of 25 rad/s and accelerates at a rate of 500 rad/s2. Knowing that the input link has a length of 250 mm, determine the instantaneous acceleration of the end of the input link in the position shown. FIGURE 6.5 Transfer mechanism SOLUTION: Machine Elements 1 11 1. Draw a Kinematic Diagram and Calculate Degrees of Freedom The kinematic diagram for the transfer mechanism is shown as Figure 6.6a. Notice that it is the familiar four-bar mechanism.  FIGURE 6.6 Diagrams 2. Determine the Tangential Acceleration of Point A Because the input link (link 2) is in pure rotation, the acceleration components of the end of the link can be readily obtained. Equation (6.13) can be used to determine the magnitude of the tangential acceleration. Because the link is accelerating, the direction of the vector is in the direction of the motion at the end of the link, which is perpendicular to the link itself. Thus, the tangential acceleration is 3. Determine the Normal Acceleration of Point A Equation (6.14) can be used to determine the magnitude of the normal acceleration Normal acceleration always occurs toward the center of rotation. Thus, normal acceleration is calculated as The components of the acceleration are shown in Figure 6.6b. 4. Determine the Total Acceleration of Point A The total acceleration can be found from analytical methods presented in Chapter 3. A sketch of the vector addition is shown in Figure6.6c. Because the normal and tangential components are orthogonal, the magnitude of the total acceleration is computed as Machine Elements 1 12 The angle of the total acceleration vector from the normal component can be calculated as The direction of the total acceleration vector from the horizontal axis is Formally, the total acceleration can then be written as LESSON 6.4: RELATIVE MOTION As discussed in detail in Chapter 6, the difference between the motion of two points is termed relative motion. Relative velocity was defined as the velocity of one object as observed from another reference object that is also moving. Likewise, relative acceleration is the acceleration of one object as observed from another reference object that is also moving. Relative Acceleration As with velocity, the following notation is used to distinguish between absolute and relative accelerations: AA = absolute acceleration (total) of point A AB = absolute acceleration (total) of point B AB/A = relative acceleration (total) of point B with respect to A = acceleration (total) of point B “as observed” from point A From equation (5.10), the relationship between absolute velocity and relative velocity can be written as Taking the time derivative of the relative velocity equation yields the relative acceleration equation. This can be written mathematically as (6.17) Typically, it is more convenient to separate the total accelerations in equation (6.17) into normal and tangential components. Thus, each acceleration is separated into its two components, yielding the following: (6.18)  Machine Elements 1 13 Note that equations (6.17) and (6.18) are vector equations and the techniques discussed in Chapter 3 must be used in dealing with these equations. EXAMPLE PROBLEM 6.5 Figure 6.7 shows a power hacksaw. At this instant, the electric motor rotates counterclockwise and drives the free end of the motor crank (point B) at a velocity of 12 in./s. Additionally, the crank is accelerating at a rate of 37 rad/s2. The top portion of the hacksaw is moving toward the left with a velocity of 9.8 in./s and is accelerating at a rate of 82 in./s2. Determine the relative acceleration of point C with respect to point B. SOLUTION: 1. Draw a Kinematic Diagram and Identify the Degrees of Freedom Figure 6.8a shows the kinematic diagram of the power hacksaw. Notice that it is the familiar slider-crank mechanism with one degree of freedom. FIGURE 6.7 Power saw Machine Elements 1 14  FIGURE 6.8 Kinematic diagram 2. Determine the Tangential Acceleration of Point B From the kinematic diagram, it should be apparent that point B travels up and to the left as link 2 rotates counterclockwise. Because the motor crank (link 2) is in pure rotation, the components of the acceleration at the end of the link can be readily obtained. Equation (6.13) can be used to determine the magnitude of the tangential acceleration. Because the link accelerates, the direction of the vector is in the direction of the motion at the end of the link. Thus, the tangential acceleration is calculated as 3. Determine the Normal Acceleration of Point B Equation (6.15) can be used to determine the magnitude of the normal acceleration. Normal acceleration is always directed toward the center of rotation. Thus, normal acceleration is Link 2 is isolated and the components of this acceleration are shown in Figure 6.8b. 4. Specify the Acceleration of Point C Point C is constrained to linear motion. Therefore, point C does not experience a normal acceleration. The total acceleration is given in the problem statement as 5. Construct the Velocity Polygon for the Acceleration of C Relative to B To determine the relative acceleration, equation (6.16) can be written in terms of points B and C and rearranged as Because two acceleration components of point B exist, the equation is written as  Machine Elements 1 15 A vector polygon is formed from this equation (Figure 6.8c). The unknown vector can be determined using the methods presented in Chapter 3. Either a graphical or analytical solution can be used to determine vector AC/B. 6. Solve for the Unknown Vector Magnitudes Arbitrarily using an analytical method, the acceleration AC/B can be found by separating the vectors into horizontal and vertical components. See Table 6.1. Separate algebraic equations can be written for the horizontal and vertical components as follows: The magnitude of the acceleration can be found by The direction of the vector can be determined by Finally, the relative acceleration of C with respect to B is LESSON 6.5: RELATIVE ACCELERATION USING GRAPHICAL METHOD EXAMPLE PROBLEM 6.6 Using graphical method, determine the instantaneous acceleration of C for the crank and rocker mechanism shown in Figure 6.9. Link AB rotates at uniform angular speed of 1 rad/s in counter clockwise direction. Dimensions: AB = 25 cm, BC = 60 cm, CD = 45 cm, AD = 50 cm, 0 = 120°  Machine Elements 1 16 Figure 6.9 SOLUTION: Relative acceleration equation: Step 1. Tangential acceleration at is zero because crank AB is rotating at constant velocity. Normal and tangential acceleration at C/B: atC/B = ? (Known direction, perpendicular to BC, but not in magnitude Normal and tangential acceleration at C atC = ? (Known direction, perpendicular to CD, but not in magnitude Step 2: Acceleration scale (Drawn in 8.5" x 11" Letter Sized Paper): 3 mm = 1 cm/s2 Choose convenient location Q and sketch acceleration anB with magnitude Qb and direction parallel to link AB Step 3: From the tip of vector Qb, sketch normal acceleration a'C/which is parallel to link BC and with magnitude bc. Step 4: Machine Elements 1 17 Sketch the direction of tangential acceleration acle which is perpendicular to link BC, the magnitude will be determined later. Step 5: From Q, plot the normal acceleration a'c knowing its magnitude and direction. The magnitude is Qd and the direction is parallel with link CD. Step 6: From the tip of vector Qd, sketch normal acceleration a'c which is perpendicular to link CD. Intersection e determines the magnitude of ace and ac. Acceleration polygon is shown in Figure 6.10 Figure 6.10 Step 7: The magnitude and direction of the total acceleration ac is determined by laying out a line from Q to e,  Machine Elements 1 18 LESSON 6.6: CORIOLIS ACCELERATION USING GRAPHICAL METHOD When a point on a link slides along a rotating link, such as in crank and Slotted fever mechanism and Whitworth quick return mechanism, then the coriolis component of acceleration must be considered. Figure 6.11 The Coriolis Effect is the deflection of moving objects when viewed from a rotating reference point. Figure 6.11 shows the movement of the slider along another rotating link, point C is known as the coincident point. The expression for Coriolis acceleration is: EXAMPLE PROBLEM 6.7 Using graphical method, determine the instantaneous acceleration of slider C for the crank and slotted lever mechanism shown in Figure 6.7. Crank BC rotates in uniform angular speed of 1 rad/s in counter-clockwise direction, BC = 20 cm, AD = 80 cm, DQ = 40 cm, θ= 45°  Machine Elements 1 19 Figure 6.12 Solution: Step 1: Write the acceleration equation that connects C and C': Tangential acceleration at C is zero because crank AB is rotating at constant velocity. Normal and tangential acceleration of point-driven C': atc = ? (Known direction, perpendicular to AD, but not in  magnitude) Normal and tangential acceleration at C/C’: atc/c' = 0 (Since the path of the slider is straight line along CA) anc/c' = ? (Known direction, along to AD, but not in magnitude) Solving for Coriolis Acceleration: Machine Elements 1 20 Hence, we arrive with the working equation for acceleration: Step 2: Acceleration scale (Drawn in 8.5" x 11" Letter Sized Paper): 4 mm = 1 cm/s2. Choose convenient location P and sketch acceleration ac with magnitude Pc and direction parallel to link BC towards pivot B. Step 3: Again, from P, sketch acceleration anc, with magnitude Pcn and direction parallel to link AD. Step 4: From the tip of vector Pcn, sketch normal acceleration atc, which is perpendicular to link AD and with unknown magnitude. Step 5: Since the magnitude of atc and anc/c’ are unknown (but we known directions), it is convenient to plot acor at the tip vector Pc and then sketch directions of atc and anc/c’ intersection of the vectors determines the magnitude of each. The Coriolis acceleration, acor is perpendicular with link AD, is presented as cc”.  Figure 6.13 Machine Elements 1 21 Step 6: From c” draw acceleration anc/c' parallel with link AD. The mersection c' determines the magnitude of acceleration ac. Step 7: The magnitude and direction of total acceleration of D can determined by proportion. This is presented by line c'd. Step 8: From here onward, use the following relative acceleration: Normal and tangential acceleration of Q/D: atQ/D = ? (Known direction, perpendicular to link QD, but not in magnitude) Step 9: From the tip of d, sketch acceleration anQ/D parallel to link this is presented by line de. Step 10: From the tip of e, lay out acceleration and perpendicular to link QD with unknown magnitude. Step 11: Finally, instantaneous acceleration of the slider, aQ is determined by drawing a line Pq parallel along the line of stroke until it intersects line eg. Acceleration polygon is shown in Figure 6.13. Machine Elements 1 22 POST-TEST 1. Using graphical method, determine the instantaneous acceleration of slider C for the offset slider and crank mechanism shown in Figure 6.4. Link AB rotates at uniform angular speed of 1 rad/s in counter clockwise direction. Dimensions: AB = 20 cm, BC = 65 cm, e = 30 cm 2. Using graphical method, determine instantaneous acceleration of the Whitworth mechanism shown in Figure 6.14. Crank BC rotates in uniform angular speed of 1 rad/s in counter clockwise direction. BC = 40 cm, AD = 25 cm, DQ = 75 cm. REFERENCES Myszka, D. H. (2012). Machines And Mechanisms Applied Kinematic Analysis Bautista, A. J. C. (2015). Kinematics of Machine Elements 1. Graphical Approach. Booklore Publishing Corp. James, V. D. (1954). Elements of Mechanism. New York and London: Wiley & Sons, Inc. Machine Elements 1 23 Myszka, D. H. (2012). Machines and Mechanisms, Fourth Edition. New Jersey: Prentice Hall. Norton, R. L. (1999). Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, Second Edition. McGraw Hill Inc. Reinholtz, H. M. (1987). Mechanism and Dynamics of Machinery. New York: Wiley & Sons, Inc. Machine Elements 1 24