The strategy for analytically determining the acceleration of various points on a mechanism is identical to the method outlined in the previous section. The difference is that vector polygons only need to be roughly sketched.
The magnitude and angles can be solved using the ana-lytical methods introduced in Chapter 3 and incor-porated in Chapter 6 and earlier sections in this chapter.
The most effective manner of presenting the analytical method of acceleration analysis is through an example problem.
EXAMPLE PROBLEM 7.9
The mechanism shown in Figure 7.15 is used to feed cartons to a labeling machine and, at the same time, to prevent the stored cartons from moving down. At full speed, the driveshaft rotates clockwise with an angular velocity of 200 rpm. At the instant shown, determine the acceleration of the ram and the angular acceleration of the connecting rod.
40°
8''
Ram Connecting rod
3''
FIGURE 7.15 Mechanism for Example Problem 7.9.
(a) (b) (c)
3 B
A
3" 8"
C 2 4
1 ω2
β
AC
AC/Bt
ABn = 1314.6
AC/Bn = 307.5 0v
VC/B VB = 62.8
VC
40°
50° 76.1°
76.1°
40°
13.9° FIGURE 7.16 Diagrams for Example Problem 7.9.
SOLUTION: 1. Draw a Kinematic Diagram
The portion of the mechanism that is under consideration includes the drive crank, the pusher ram, and the link that connects the two. Once again, notice that this is the common in-line, slider-crank mechanism. A kinematic diagram is shown in Figure 7.16a.
2. Decide on a Method to Achieve the Desired Acceleration
As in Example Problem 7.8, the acceleration of the ram (link 4) is strictly translational motion and is identical to the motion of point C. The acceleration of point C, which also resides on link 3, can be determined from knowing the acceleration of point B. Point B is positioned on both links 2 and 3. Therefore, the acceleration of point B can be determined from knowing the motion of the input link, link 2.
3. Analyze the Mechanism Geometry
The angle between link 3 and the horizontal sliding surface of link 4,β in Figure 7.16a, can be determined from the law of sines.
b = sin-1arABsin 40°
rBC b = sin-1a(3 in.) sin 40°
(8 in.) b = 13.9°
Acceleration Analysis 189 4. Determine the Velocity of Points B and C
Calculate the magnitude of the velocity of point B using the following equation:
The direction of VBis perpendicular to link 2 and consistent with the direction ofω2, down and to the right. The velocity of point C is parallel to the horizontal sliding surface, and the velocity of C with respect to B is perpendicular to the link that connects points B and C. Calculating this angle,
By understanding the directions of the vectors of interest, a velocity polygon can be assembled (Figure 7.16b). The magnitude of the third angle in the velocity polygon can be determined because the sum of all angles in a triangle is 180°.
The magnitudes of the velocities can be found from the law of sines.
Solve for the unknown velocities with the following:
5. Calculate Acceleration Components
The next step is to construct an acceleration diagram that includes points B and C. Calculate the magnitudes of the known accelerations using the following equations:
AC/Bn = 1VC/B22
rBC = 149.6 in./s22
8.0 in. = 307.5 in./s2 13.9°
constant velocity)
(because the driving link is rotating at ABt = a2rAB = 10 rad/s22 (3.0 in.) = 0
rotation, point A)
(directed toward the center of ABn = 1VB22
rAB = 162.8 in./s22
3.0 in. = 1314.6 in./s2 40°
VC/B= VBa sin 50°
sin 76.1°b = 62.8 in./s a sin 50°
sin 76.1°b = 49.6 in./s 76.1°
VC = VBasin 53.9°
sin 76.1°b = 62.8 in./s asin 53.9°
sin 76.1°b = 52.3 in./s : 180° - (50° + 76.1°) = 53.9°
90°+ (- b) = 90° + (-13.9°) = 76.1°
VB = v2rAB = 120.9 rad/s2(3.0 in.) = 62.8 in./s 50°
v2 = p
30 (200 rpm)= 20.9 rad/s
a R
Q
Note that point C does not have a normal acceleration because the motion is strictly translational.
6. Using Vector Methods, Solve the Relative Acceleration Equation The relative acceleration equation for points B and C can be written as
In forming an acceleration diagram, vector placement arbitrarily starts on the right side of the equation. At the origin of the acceleration diagram, vector , which is completely known, is placed. Because no tangential component of the acceleration of point B exists, is ignored. Vector , which is also completely known, is placed at the end of . At the end of , the vector is placed; however, only the direction of this vector is known. It is directed perpendicular to the normal component, , and thus, perpendicular to the line that connects B and C. The angle has been calculated as
The first term on the left side of the equation can be ignored because there is no normal component of the acceleration of point C. Therefore, the vector representing the tangential acceleration of point C is placed at the
90°+ (- b) = 90° + (-13.4°) = 76.1°
AC/Bn AC/Bt
AC/Bn ABn
AC/Bn ABt
ABn
ACn + 7 ACt = ABn + 7 ABt + 7 AC/Bn + 7 AC/Bt
(directed from C toward B) b
190 CHAPTER SEVEN
origin. However, only the direction of this vector is known: It is parallel to the horizontal surface that link 4 is constrained to slide upon. The vector polygon is illustrated in Figure 7.16c. The unknown vector magnitudes, and , can be determined using the methods presented in Chapter 3. First, each vector can be separated into horizontal and vertical components, as shown in Table 7.2.
ACt
AC/Bt
7.8 ALGEBRAIC SOLUTIONS FOR COMMON
MECHANISMS
For the common slider-crank and four-bar mechanisms, closed-form algebraic solutions have been derived [Ref. 12].
They are given in the following sections.
7.8.1 Slider-Crank Mechanism
A general slider-crank mechanism was illustrated in Figure 4.20 and is uniquely defined with dimensions L1, L2, and L3. With one degree of freedom, the motion of one link must be specified to drive the other links. Most often the crank is driven and θ2,ω2, and α2are specified. To readily address TABLE 7.2 Acceleration Components for Example Problem 7.9
Vector
Reference Angle (θx)
Horizontal Component ah= a cosUx
Vertical Component av= a sinUx
anBAnB 220° 1007.0 845.0
AnC/B 166.1° -298.5 73.9
AC/Bt 76.1° .240 aC/Bt .971 aC/Bt
AC 180° -aC 0
Separate algebraic equations can be written for the horizontal and vertical components.
The vertical component equation can be solved algebraically to give the magnitude
This result can then be substituted into the horizontal equation to give the magnitude
7. Clearly Specify the Desired Results Formally stated, the motion of the ram is
Notice that because the acceleration is in the opposite direction of the ram movement and velocity, the ram is decelerating.
8. Calculate the Angular Acceleration
Finally, the motion of the connecting arm is calculated.
where the direction is consistent with the velocity of C relative to B, counterclockwise. Also
where the direction is consistent with the tangential acceleration of C relative to B, counterclockwise.
a3 = aC/Bt
rCB = 794.1 in./s
8.0 in. = 99.3 rad/s2, counterclockwise v3 = vC/B
rCB = 49.6 in./s
8 in. = 6.2 rad/s, counterclockwise Ac= 1496.1 in./s ;
Vc= 52.3 in./s : aC = 1496.1 in./s2 aC/Bt = 794.1 in./s2 vertical comp.: 0 = (-845.0) + (+73.9) + A+0.971 aC/Bt B
horizontal comp.: + aC = (-1007.0) + (-298.5) + A+0.240atC/BB
AC = ABn + 7 AC/Bn + 7 AC/Bt
Acceleration Analysis 191 the slider-crank mechanism, position, velocity, and
accelera-tion equaaccelera-tions (as a funcaccelera-tion ofθ2,ω2, and α2) are available.
As presented in Chapter 4, the position equations include (4.6) (4.7) As presented in Chapter 6, the velocity equations are
(6.12) (6.13) The acceleration equations are then given as [Ref. 12]
(7.21)
(7.22) Note that an in-line slider-crank is analyzed by substituting zero for L1in equation (4.6).
7.8.2 Four-Bar Mechanism
A general four-bar mechanism was illustrated in Figure 4.22 and is uniquely defined with dimensions L1, L2, L3, and L4. With one degree of freedom, the motion of one link must be specified to drive the other links. Most often the crank is driven and θ2,ω2, and α2are specified. To readily address the four-bar mechanism, position, velocity, and acceleration equations (as a function ofθ2,ω2, and α2) are available. As presented in Chapter 4, the position equations are
(4.9) (4.10)
(4.11)
(4.12) As presented in Chapter 6, the velocity equations are
(6.14)
(6.15) The acceleration equations can be presented as
(7.23)
7.9 ACCELERATION OF A GENERAL POINT ON A FLOATING LINK
Recall that a floating link is not directly connected to the fixed link. Therefore, the motion of a floating link is not limited to only rotation or translation, but a combination of both. In turn, the direction of the motion of points that reside on the floating link is not generally known. Contrast this with the motion of a point on a link that is pinned to the fixed link. The motion of that point must pivot at a fixed distance from the pin connection. Thus, the direction of motion is known.
During the acceleration analyses presented in the preceding sections, the underlying premise of the solution is that the direction of the motion is known. For a general point on a floating link, this is not true. For these cases, two relative acceleration equations must be used and solved simultaneously.
To illustrate the strategy of determining the acceleration of a general point on a floating link, consider the kinematic sketch of the four-bar linkage shown in Figure 7.17.
Link 3 is a floating link because it is not directly attached to link 1, the fixed link. Because points A and B both reside on links attached to the fixed link, the accele-ration of these points can be readily determined. That is, using the methods in the previous two sections, both the direction and magnitude of and can be established.
However, point C does not reside on a link that is directly attached to the fixed link. Therefore, the exact path of motion of point C is not obvious. However, two relative acceleration equations can be written as
(7.25) (7.26)
In equation (7.25), both the magnitude and direction of aCis unknown along with the magnitude of Equation (7.26) introduces an additional unknown, namely the magnitude of Overall, two vector equations can be written, each with the capability of determining two
AC/A.t
FIGURE 7.17 Point on a floating link.