Question 16.1: Cam-Driven Fourbar Slider With Motion Functions Applied to t...

1 The equations for the position of the crank-slider linkage were developed in Section 4.6, its velocity in Section 6.7, and acceleration in Section 7.3 and are repeated here for your convenience.

\begin{gathered}\theta_{3_{1}}=\arcsin \left(\frac{a \sin \theta_{2}-c}{b}\right)      (4.16a) \\d=a \cos \theta_{2}-b \cos \theta_{3}        (4.16b)\\\omega_{3}=\frac{a}{b} \frac{\cos \theta_{2}}{\cos \theta_{3}} \omega_{2}          (6.22a)\\\dot{d}=-a \omega_{2} \sin \theta_{2}+b \omega_{3} \sin \theta_{3}              (6.22b)\\\alpha_{3}=\frac{a \alpha_{2} \cos \theta_{2}-a \omega_{2}^{2} \sin \theta_{2}+b \omega_{3}^{2} \sin \theta_{3}}{b \cos \theta_{3}}                (7.16d)\\\ddot{d}=-a \alpha_{2} \sin \theta_{2}-a \omega_{2}^{2} \cos \theta_{2}+b \alpha_{3} \sin \theta_{3}+b \omega_{3}^{2} \cos \theta_{3}                  (7.16e)\end{gathered}

The terminology used in the equations is shown in Figure 16-3.

2 The cam program has two segments: 220° of constant velocity at 49.066 05°/s for 30° of follower arm (crank) motion, and a fifth-degree polynomial return motion over 140°. Total angular displacement of the crank is 33.2° in segment 2.

3 The boundary conditions for the polynomial in segment 1 are shown in Table 16-3.
The boundary conditions for the polynomial in segment 2 are shown in Table 16-4.
The normalized displacement equations for the cam functions as derived with program Dynacam with x = θ/β are as follows:†

Segment 1:

\begin{aligned}&S=30.00 x \operatorname{deg} \\&V=\frac{\omega}{\beta}(30)=\frac{2 \pi}{3.8397}(30)=49.066 deg / sec \\&A=\frac{\omega^{2}}{\beta^{2}}(0) deg / sec ^{2}\end{aligned}                            (16.1)

Segment 2:

\begin{aligned}&S=-294.546 x^{5}+736.364 x^{4}-490.909 x^{3}+19.091 x+30.00 deg \\&V=\frac{\omega}{\beta}\left(-1472.73 x^{4}+2495.46 x^{3}-1472.73 x^{2}+19.091\right) deg / sec \\&A=\frac{\omega^{2}}{\beta^{2}}\left(-5890.91 x^{3}+7486.38 x^{2}-2945.46 x\right) deg / sec ^{2}\end{aligned}                              (16.2)

4 These S, V, A equations are the driving functions for the input crank of the linkage where w = 2p rad/sec for a 1-Hz cycle, and b is the angular duration of the motion in radians—3.8397 rad for segment 1 and 2.4435 rad for segment 2.

5 The equations for the kinematic behavior of the linkage from step 1 must be solved at a set of discrete points (e.g., every degree of the cam cycle) to determine the behavior of the slider d. Equation 16.1 applies to the first 220° (constant velocity) and equation 16.2 to the last 140°.

6 This is obviously a job for a computer and one was used to solve these equations. A good approach is to use an equation solver such as Matlab, Mathcad, or TKSolver. Plots of the polynomial motions from program Dynacam are shown in Figure 16-4.

7 Figure 16-5a shows the displacement of the follower arm and the slider superposed and normalized to percent for comparison. The nonlinear behavior of the linkage distorts the entire function. Figure 16-5b shows the normalized velocities of follower arm and slider superposed. Note that the follower arm has constant angular velocity during the first segment but the slider does not have constant linear velocity.

8 Figure 16-6 shows the percent error between the crank arm and slider in both displacement and velocity. The differences in this example are not extreme, mainly because the crank only rotates through 30° during the constant velocity motion. The larger the angle of crank rotation, the larger the distortion in the output functions.

9 If the magnitude of the deviation from constant velocity is small enough to still allow proper function, then this solution will be adequate as is. If the error is not tolerable, then the solution is to apply the desired motion function to the slider and calculate the linkage’s deviation in reverse to redefine the cam function to correct it. The next example will do this.