Question 16.3: Slider-Crank Linkage Driven by a Servomotor to Perform a For...
Slider-Crank Linkage Driven by a Servomotor to Perform a Forming Operation
Given: A slider-crank linkage mechanism (Figure 16-11) has been designed to provide the geometry for a suitable motion of a die in a forming press. The required motion begins from a dwell and moves the slider through a 2-in stroke ending at TDC with crank angle \theta_{2}=0 . The die needs to be brought to zero velocity for an instant at 1.588-in stroke then advanced into the material. It is then pulled back to dwell at 1.400 in before returning to zero. The slider motion timing diagram is shown in Figure 16-11. Rather than design a custom cam to provide the motion and dwells to the linkage crank, it is desired to use a servomotor directly driving the input link of the crank-slider mechanism through a gearbox to obtain the necessary motions and dwells. Machine speed is 120 cycles per min or 2 Hz.
Find: The motion function needed to program a servomotor to drive a crank-slider linkage that provides the required linkage output motion program as defined at the slider.
Assume: The position and velocity accuracy required at intermediate points in the stroke requires that the motion function be applied to the slider and mathematically transformed to a function at the servo axis that accounts for linkage geometry.
1 The functions chosen to provide the motions defined in Figure 16-12 are:
Segment 1: Polynomial over 60° with the boundary conditions of Table 16-7.
Segment 2: Dwell at 2 in for 20°.
Segment 3: Cycloidal fall to 1.4 in over 20°.
Segment 4: Dwell at 1.4 in for 20°.
Segment 5: Cycloidal fall to zero over 60°.
Segment 6: Dwell at zero in for 180°.
The S, V, A functions for the slider motion as calculated with program Dynacam are shown in Figure 16-13.
2 The geometry of the slider-crank linkage to be driven by the servomotor was chosen based on packaging requirements and is shown in Figure 16-13. The maximum displacement at TDC when \theta_{2}=0 is d = a + b = 9.8 in. A 2-in stroke requires that the starting displacement d_{\text {start }} = 9.8 – 2 = 7.8 in. Equations 16.4 are used to find the starting crank angle at d_{\text {start }} .
\begin{aligned}\omega_{2} &=\frac{d \cos \theta_{3}}{a\left(\cos \theta_{2} \sin \theta_{3}-\sin \theta_{2} \cos \theta_{3}\right)} \\\omega_{3} &=\frac{a \omega_{2} \cos \theta_{2}}{b \cos \theta_{3}}\end{aligned} (16.4)
\begin{aligned}K_{1} &=a^{2}-b^{2}+c^{2}+d^{2}=2.4^{2}-7.4^{2}+0+7.8^{2}=11.840 \\K_{2} &=-2 a c=-2(2.4)(0)=0 \\K_{3} &=-2 a d=-2(2.4)(7.8)=-37.440 \\A &=K_{1}-K_{3}=11.840-(-37.440)=49.280 \\B &=2 K_{2}=2(0)=0 \\C &=K_{1}+K_{3}=11.840+(-37.440)=-25.600\end{aligned} (a)
\begin{aligned}\theta_{\text {start }_{1,2}} &=2 \arctan \left(\frac{-B \pm \sqrt{B^{2}-4 A C}}{2 A}\right) \\&=2 \arctan \left(\frac{0 \pm \sqrt{0-4(49.280)(-25.600)}}{2(49.280)}\right) \\\theta_{\text {start }_{1}} &=71.564^{\circ} \\& \theta_{\text {start }_{2}}=-71.564^{\circ}\end{aligned} (b)
Because of symmetry, either value could be used depending on the desired direction of crank rotation. We will choose clockwise crank rotation with \theta_{\text {start }}=71.564^{\circ}, \theta_{\text {end }}=0^{\circ} .
3 It is only necessary to convert the displacement function with the linkage geometry in order to create a usable servo control function because the typical servomotor controller needs only that function for control.* Note that the effects of the chosen velocity, acceleration, and jerk functions are all contained within the displacement function because it comes from the integration of those functions.† Most servo controllers are capable of importing a table of displacement data, which in this case will be generated by the displacement function of Figure 16-12 as modified by the linkage geometry. This data table of crank angles can be generated at any resolution desired from every degree to any fraction of a degree.
4 The values of d_{\text {start }} + d as defined by the linear displacement function of Figure 16-12 for each increment of machine angle over the cycle from 0 to 360° are used as input to equation 16.4 to calculate their corresponding angular displacements \theta_{2} t to be applied to the servo axis. Do not confuse the linkage crank angles \theta_{2} t with the timing angles shown as the independent variable in Figures 16-12 and 16-14. Those angles represent fractions of one machine cycle, which is always one revolution of the main timing shaft of the machine. In a fully servomotor-controlled machine with no physical timing shaft, this is often an “imaginary” or virtual “master” axis that serves as a timing reference for all “slave” servomotors to follow.
5 Figure 16-14 shows the displacement functions for both the slider linear displacement and the linkage crankshaft angular displacement over the first five segments. The 180° dwell of segment 6 is omitted since the crank is then stationary at the start position. Each function is normalized to a range of 0 to 1 to allow their superposition. Note the large differences between the two functions as compared to those of the previous two examples. In those cases, the total rotation of the crank was less and avoided the regions near BDC or TDC where geometric distortion is greatest. This example takes the crank to TDC, and the differences between the functions is seen to increase as it approaches TDC.
6 While this design does what the client requested, a close inspection of the crank arm displacement function of Figure 16-14 at 60 and 80 degrees of timing angle shows this to be a dynamically poor solution. There is such a rapid change of slope at those points that it will have very large acceleration. It comes close to violating the fundamental law described earlier in this chapter. This results from the large amount of distortion introduced by the nonlinear slider-crank function. An improvement will be sought in the next example.
| TABLE 16-7 Boundary Conditions |
| Example 16-3, Segment 1 |
| When θ = 0°: s = 0, v = 0, a = 0 When θ = 35°: s = 1.588, v = 0 When θ = 60° s = 2, v = 0, a = 0 |
