Chapter 4 Acceleration Analysis (Text)

Transcription

1 Chapter 4 Acceleration Analysis (Text) Let us begin by describing the acceleration of a point on a body that is pinned to ground (Fig. 4.). Point A on link can be located with respect to the origin A 0 of the fixed complex plane in polar form by the vector R A, made up of angle θ A measured positively counterclockwise from the fixed real x axis and the radius R A : R A = R A e iθa (4.1) The velocity of point A was obtained in Chap. 3 by taking the derivative of the position vector with respect to time. If link is a rigid member, only 0A is changing with time, so the absolute linear velocity of point A on link is ( ω =d θ A /dt) V A =R A ω 1 ie iθa = i ω 1 R A (4.) The magnitude of the velocity vector is R A ω 1 and the direction is perpendicular to the vector R A in the sense of the angular velocity ω 1. From now on, to simplify the notation, in most cases we write ω to mean ω 1 (i.e., angular velocity with respect to the fixed frame of reference). If there is another second index, say ω 3 it means the angular velocity of link with respect to link 3. The acceleration of point A may be found by taking the derivative of the velocity [Eq. (4.)] with respect to time. Again, since all links are assumed to be rigid, the only components that change with time are θ A and If the angular acceleration is defined as the rate of change of the angular velocity with respect to time (α =d ω /dt), the absolute linear acceleration of point A on link with respect to ground is V A =R A ω 1 ie iθa = i ω 1 R A (again) (4.) (4.3) Normal (Centripetal) Tangential Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 1

2 Figure 4. Velocity difference of A with respect to A0. Figure 4.3 Acceleration difference of A B A Figure 4.5 Geometric construction of the acceleration difference between points A and B of link in Fig Figure 4.4 Accelerations of two different points on one rigid link. Notice that in general there are two components of acceleration of a point on a rigid body rotating about a ground pivot (Fig. 4.3). One component has a magnitude of R A α and a direction tangent to the path of A pointed in the sense of the angular acceleration. This component is called the tangential acceleration. Its presence is due solely to the rate of change of the angular velocity. The other component has a magnitude of R A ω and, because of its minus sign, a direction opposite to that of the original position vector RA. This component, which always points toward the center of rotation because R A and ω are always positive, is called the normal or centripetal acceleration. This component is present due to the changing direction of the velocity vector (because point A is moving along a ircu1ar arc). An equivalent form for the magnitude of the normal acceleration is V A /R A. A special case occurs when point A moves in a straight line; then the radius of curvature of the path of A is infinite (R A = ) and the normal acceleration is equal to zero. Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery

3 Let us determine the total absolute acceleration of point, B on the same link (see Fig. 4.4). r A B r iθ B = R ( ω + iα )e = ( ω + iα ) R (4.4) B What is the acceleration difference of B with respect to A (A BA )? You might ask the same question in this way: Suppose that I were sitting at point A (as if on a merry-goround) in a swivel chair that always faces east, in the direction of the inertial x axis. If I were watching point B, what would be the acceleration of point B with respect to my fixedly oriented, orbiting frame of reference? This would be the acceleration difference of point B with respect to point A, or A BA. The total absolute acceleration of point B may be written in terms of the acceleration difference as B A B = A A + A BA (4.5) Each term of Eq. (4.5) has two possible components since link is a rigid member, so that the full expression is A B n + A B t = A A n + A A t + A BA n + A BA t. or A BA n + A BA t = A B n + A B t - A A n - A A t (4.6) where the superscripts n and t refer to the normal and tangential components, respectively, described previously. Initially, a graphical solution of the acceleration problem will be employed. To keep track of which components are known or unknown, an M will be written below a term if the magnitude is known and a D written if the direction is known. Thus from Fig. 4.4, A BA n + A BA t = A B n + A B t - A A n - A A t (4.7) D D D D M M M M A short line segment with an arrow pointing in the direction of the component is also helpful. Note that the respective signed magnitudes on the right side of Eq. (4.7) are This complex equation [Eq. (4.7)] has four real unknowns: the direction and magnitude of A BA n and A BA t. Unless two of these four unknowns can be found some other way, Eq. (4.7) cannot be solved since it represents only two scalar equations. The directions of the normal and tangential components of the relative acceleration of B with respect to A are found by inspection, however. The normal component must be directed toward A (since the reference position is A). The tangential component is perpendicular to the normal acceleration in the sense of the angular acceleration α. Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 3

4 Figure 4.5 shows the graphical solution to Eq. (4.7) yielding the normal and tangential acceleration difference components. A convenient center O A ) of the acceleration diagram is located arbitrarily and all the components are drawn to scale. Both AA and AB (combined normal and tangential components) are drawn from A since they are total absolute accelerations. Equation (4.7) tells us that the difference between the total acceleration of B and the total acceleration of A is the unknown ABA (normal plus tangential). Intersections of the line of the direction of the normal acceleration with the line of the direction of the tangential acceleration yield the proper magnitudes of these components. Thus the graphical solution of Eq. (4.7) is complete. Suppose that Eq. (4.7) had been written as Example 4.1 Link J in Fig. 4.6 is in planar motion with respect to ground. Accelerometers located at points A and B measure total absolute linear accelerations as depicted by the vectors A A and A B in Fig (a) Find the angular acceleration of this link. (b) Find the angular velocity of the link. Solution Referring to Eq. (4.7), the components of the acceleration difference A BA n and A BA t may be obtained from the total absolute accelerations A A and A B, as in Fig The resulting magnitudes of the tangential and normal components of the acceleration difference yield the answers to (a) and (b): Figure 4.6 Accelerations of points A and B of link 3 in general planar motion. Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 4

5 The sense of ω j is not known from the information given. The sense of α j is found by observing the directions of A BA t in Fig In digital computation, the algebraic sign of α j. (and thus its sense) is found by using the vector forms of A BA t and i α j R BA as follows: A BA t = i α j R BA where of R BA is the vector locating point B with respect to point A (i.e., pointing from A to B), Therefore, Figures 4.6 and 4.7 show that A BA t and i R BA are collinear. Visualizing A BA t from Fig. 4.7 placed at point B on Fig. 4.6, one can see that this vector imposes a ccw angular acceleration to link J. B Figure 4.7 Geometric construction of the acceleration difference of point B with respect to point A of link 3 in Fig RELATIVE ACCELERATION* To find linear or angular acceleration of an output link of a mechanism when input link acceleration is given, some relative acceleration calculations are usually performed. Suppose that we wish to determine the instantaneous acceleration of the slider of the slider- crank mechanism of Fig. 4.8 (see Sec. 3.5 for the velocity analysis Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 5

6 of this mechanism) with a counterclockwise angular velocity ω and a clockwise angular acceleration of α of link specified. The slider is identified as link 4 while the known velocity and acceleration information is given for link. This is apparently a case 4 analysis. The path of point B on link 4 relative to point A on link is not easily visualized without including link 3. To make the analysis simpler, this example will be solved by superimposing several case and case 3 solutions as follows: Step 1. Determine A A (a case analysis). The acceleration of point A as part of (signified as A A is A A = A A n + A A r A r iθ = R ( ω + iα )e = ( ω + iα ) R (4.11) A A Step. Determine A A3.This is a case 3 analysis, but since point A is a revolute joint joining links and 3, then A A3 = A A (4.1) Step 3. Determine A B3. Recognizing that point B can be thought of as a point on the slider (link 4) as well as a point of link 3, we will first find the total acceleration of B as part of 3 (case ) by using the equation A Figure 4.8 Geometric construction of Figure 4.9 Geometric construction slider velocity. Note that VA is of slider acceleration in the perpendicular to link and VBA is mechanism of Fig Dr. perpendicular Mostafa S. Habib to link 3. MEG 373 Kinematics and Dynamics of Machinery 6

7 The normal and tangential components of A A3 are known from steps 1 and. The direction of A BA n must be toward A and the tangential component is not known as a magnitude but its direction is known as follows, r A BA r iθ 3 = R ( ω + iα )e = ( ω + iα ) R (4.14) BA BA Step 4. Determine A(B4). This is a case 3 analysis, but since point B is a revolute joint, A B4 = A B3 (4.15) which completes our analysis. Example 4. Determine the instantaneous angular acceleration of the coupler and output links of the four- bar linkage shown in Fig. 4.10, given ω = 600 rpm or 6.8 rad/sec cw and α = 000 rad/sec cw. Also find the linear acceleration of points B and C. Results of the velocity analysis, required in the forthcoming acceleration analysis, are shown in the velocity polygon of Fig Solution The total acceleration of point A as part of link is A A = A A n + A A r A = R ( ω + iα )e A A iθ r = ( ω + iα ) R A Figure 4.10 Geometric velocity analysis of four-bar mechanism. Figure 4.11 Geometric construction of accelerations in the four-bar mechanism of Fig Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 7

8 Figure 4.11 shows the acceleration polygon, which represents a graphical solution to Eq. (4.16). The intersection of lines of A t B and A t BA marks the junction of these vectors and determines their magnitudes. Also, The total absolute acceleration of C can be determined from either of the following acceleration difference equations. Thus A C = 11,100 cm/sec. Figure 4.1 shows an enlarged drawing of the acceleration image triangle of the coupler ( a'b'c' in Fig. 4.11) to help illustrate the solution of these equations. Triangle a b c of Fig is similar to triangle ABC of the coupler link and also similar to triangle abc of the velocity polygon (Fig. 4.10). Therefore, a more rapid method of finding the total absolute acceleration of a point on a link when the total absolute accelerations of at least two other points are known is to utilize the acceleration image triangle technique. For example, if the absolute acceleration of the center of mass of the coupler (say point G of Fig. 4.10) is required, and the acceleration polygon has proceeded as far as solving Eq. (4.16), then image triangle a b g can be drawn similar to ABG. One must be careful to make sure that the triangles are not flipped on the wrong side of a b. This will be avoided if the velocity image and the acceleration image maintain the same directionally cyclic order as the coupler triangle. Here ABG, abg, and a b g are all directionally counterclockwise. The similarity of the acceleration image to the original coupler can be proved by noting from Eq. (4.3) that Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 8

9 Figure 4.1 Acceleration image &i b c of the coupler triangle AABC of the mechanism in Fig The expressions for the angular accelerations for the coupler link (a3) and the output link (a4) of this same four-bar mechanism (Fig. 3.8) are similarly derived (see Prob. 4.47): Note that these equations are in terms of a and known angular velocity and position data. Equations (4.1) and (4.) are easily programmed for automatic computation, yielding angular acceleration information of the entire four-bar mechanism for any number of positions 0 during the cycle of motion. 4.4 CORIOLIS ACCELERATiON Thus far, acceleration analysis has been restricted to examples in which acceleration differences and relative accelerations have been between two points that have a fixed distance between them. In many mechanisms, however, lengths between points on different members do not remain constant (although the members themselves are rigid). Figure 4.1 3a shows the path of point P as it moves with respect to a reference coordinate system (x,iy). To derive the acceleration of point F, this path* must be known (this is a case 4 analysis described in Table 3.1). If the path of P were traced by a coupler point of the four-bar mechanism shown in Fig b, we are already prepared to find the acceleration of point P (as was done in Example 4.) by case Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 9

10 and case 3 analyses. Let us now consider the case where a similar path might be generated by point P on the inverted slider kits Figure 4.13 Point P moving with respect to fixed frame x,iy; (a) as a free point; (b) as a coupler point of a four-bar mechanism. crank mechanism of Fig with link as the input. Figure 4.15 shows this linkage without link 4. Let us determine the absolute acceleration of point P on the slider. The position P is defined by the vector z: z=ze8 (4.3) The absolute velocity, acceleration, and further derivatives of the changing position vector of point P may be obtained by taking successive time derivatives of Eq. (4.3). In this example, both the angle 8 and the length z change as slider 3 moves along link. If the following terms are defined, then the terms of Fig are derived [144]. Figure 4.14 Inverted slider-crank mechanism. Figure 4.15 Vector representation of the position of point P on the slider of Fig Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 10

11 Figure 4.16 Time derivatives of the position vector of point Pin Fig There are two components of velocity (Fig. 4.17a): (1) the tangential velocity with magnitude zω, and direction ±ie iθ, the ± depending on the sense of ω (recall that ccw = +); and () the sliding or radial velocity with magnitude of v and a direction along link in the sense of v positive outward (z increasing) and negative inward (z decreasing). The derivatives of the velocity terms yield several acceleration terms (see Figs and 4.17b). Two familiar terms are the normal (or centripetal) acceleration (magnitude zω and direction -e iθ ) and the tangential acceleration (magnitude zα and the direction perpendicular to e iθ in the sense of α). The sliding or radial acceleration is one new term that has magnitude a and is directed along the link, outward if v is increasing (positive acceleration), and vice versa. The other two components are equal and are combined to form the Coriolis acceleration: vωie iθ. The magnitude of this component is vω and its direction is perpendicular to e iθ directed one way or the other depending on the directions of v and ω. Figure 4.17 Vector representation of (a) the velocity and (b) the acceleration components of pomt P of Figs and Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 11

12 Figure 4.18 Determining the direction and sense of the Coriolis acceleration. In all cases, A c = vωie iθ where v is positive if it points in the outward direction from the center of rotation and ω is positive ccw. is perpendicular to e directed one way or the other depending on the directions of v and to. Figure 4.18 can be used as a guide for determining which way the Coriolis component is directed. For example, if the velocity vector is directed outward (with respect to the center of rotation), which means that it is positive, and the angular velocity counterclockwise (which is also positive), the Coriolis component has the unit vector ilk, where lk e 0, and therefore points to the left in Fig Another popular method is to rotate the velocity v 90 in the direction of to; then the resulting direction is that of the Coriolis acceleration. Coriolis acceleration* (*Conolis acceleration affects our daily lives through its influence on both weather and ocean currents. For example, as air attempts to move from a high-pressure area to a low, it is deflected by Coriohs acceleration so as to move in a circular path, counterclockwise in the northern hemisphere and clockwise in the southern hemisphere.) is usually believed to be conceptually difficult, but a rule of thumb will help: There is a Coriolis component of acceleration when the vector that locates the point is both rotating and changing length with respect to the fixed inertial frame of reference. This is true of the example of Figs and 4.15, since the vector z is changing length as well as rotating as the mechanism moves. Mechanism designs have been known to fail due to the mistaken omission of the Coriolis component in an acceleration analysis, leading to incorrect inertia forces. The following examples contain analyses in which the Coriolis component is present. It is noteworthy that the complex-number treatment provides the magnitude and direction of the Coriolis term automatically, without resorting to a rule of thumb for determining its direction and whether or not it is present. The graphical approach requires determination of direction, as illustrated in Fig Example 4.3 Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 1

13 Let us determine the angular acceleration of link 4 of the inverted slider-crank mechanism of Fig. 4.0a, given the peripheral speed of the crank pin, V = 40 cm/sec (constant angular velocity) as shown. Solution Three methods ( The student may want to concentrate on only one method in this and other examples of Coriolis acceleration to avoid possible undue confusion in the early stages of the learning process.) for solving this example will be described; the first two are graphical (but in different directions around the loop), and the third is analytical. Both Table 3.1 and Fig will be useful as guides in setting up the correct equations. Figure 4.1 shows how we might visualize the relative and difference motion for this example. Consider point P. the center of the crank pin, in the position shown in Fig At that instant, three points coincide at P :P on link, P3 on link 3, and P4 on link 4 (Fig. 4.1). As the mechanism moves, P and P3 remain coincident, but P4 moves apart from these, with relative velocity and relative acceleration The velocity and acceleration equations can be written two ways: Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 13

14 Figure 4.0 In (b) and in subsequent velocity and acceleration vector diagrams and equations, it is useful to enter (D, M) for a vector known both in magnitude and direction, and (D) where only the direction is known prior to the construction of the diagram. Point P3 could also be used here in place of F, because the slider, link 3, is pin connected directly to link at this point. However, this approach would be no different from using P. Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 14

15 Notice that when using Eqs. (4.9) and (4.30), the motion of P with respect to O is equal to the motion of P4 with respect to O4 plus the motion of P with respect of P4. The first and second motions are difference motions (case in Table 3.1) Figure 4.1 Point P on link 4 is separated from P on links and 3 in the course of mo04 tion. and, since O and O4 are fixed, they are absolute motions. The third motion, that of P with respect to P4, is relative (case 3). The path generated in difference motion is a circular arc, but relative motion can be more complex. Figure 4.0b shows the graphical solutions of Eq. (4.9), where we observe that Solution of Eq. (4.30) will take more thought. Expanding the total acceleration terms in Eq. (4.30) may yield up to four components (derived in Fig. 4.16): the normal, tangential, sliding, and Coriolis accelerations for each term. Since Eq. (4.30) is made up of terms that represent only case 1,, and 3 motions, none of the terms will contain all four acceleration components. (The complex-number analysis that yielded the equations in Fig involved case 4 motion.) The motion of point P as part of is constrained to a fixed radius at constant peripheral speed, so that only the normal term is present in A (Fig. 4.0c). As for 4, since (O 4 P 4 ) is constant, the sliding and Coriolis terms are zero (Fig. 4.0d). In the tangential term, the direction is known, but the magnitude is not, because α 4 is as yet unknown. The normal term is known. Solving of A4, however, is more complicated. In Fig. 4., P4 and P coincide initially. To visualize the motion of P with respect to P4, we apply a small virtual Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 15

16 displacement represented by the rotation L04, after which P and P are separated by z p4. Focusing on the 04+ sw4 position, we have Figure 4. Visualization of how the Coriolis acceleration originates in the motion off as a point of linlc with respect topas apointoflink4offig. 41. Now, to refer to these results back to position P4, P, we let the virtual displacement txo4 * 0. Then In the limit A84 * 0, (*Becae a =0. ) the last two terms also go to zero, because rp4 * 0 (i.e., there is zero distance between P and P4 after the virtual displacement is set to zero). Equation (4.30) can now be expanded by inserting the results of our analysis thus far (entering on the right side only the nonzero components. Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 16

17 Equation (4.33a) is the expanded form of Eq. (4.30), which has only two unknowns: the directed magnitudes of A4 and Ai4 (i.e., A4 and Ajp4 each is either positive or negative). Figure 4.3 shows the vector solution to Eq. (4.33a). From this figure the angular acceleration of link 4 is A5,4 = (461 cm/sec)(_ie194 ics4z4. For the analytical version of this solution, see Prob. 4.44(a). Figure 4.3 Vector diagram of the solution of Eq (4.33a) (see Figs ). Dr. Mostafa S. Habib MEG 373 Kinematics and Dynamics of Machinery 17

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