Robotics 2 E Part 5 pptx

110 Dynamic Analysis of DrivesExercise 3E-3a The given mechanism in Figure 3E-3a consists of a DC electromotor with a char-acteristic T= 4 - 0.1 CD Nm, a speed reducer carrying out a rat

110 Dynamic Analysis of Drives

Exercise 3E-3a)

The given mechanism (in Figure 3E-3a) consists of a DC electromotor with a

char-acteristic T= 4 - 0.1 CD Nm, a speed reducer carrying out a ratio 1:4, and a driven pinion

with a radius jR = 2 cm and moment of inertia together with the attached wheel of/! = 0.01 kg m2 The pinion of radius 2 cm is engaged with a rack of mass ml = 5 kg The

rack lifts a mass ra2 = 5 kg by means of a flexible cable The rotor of the motor, its shaft,

and the wheel fastened to it have a moment of inertia I0 = 0.001 kg m2 Find the speed

of the mass m^ after the motor has been working for 0.1 second; find the distance the mass ml travels in this time At the beginning of the process the motor is at rest.

Answer the same questions for the case in which the drive is performed by an ACelectromotor The characteristic of the latter is (see Expression (3.48)):

where

Exercise 3E-3b)

A screw jack driven by a DC electromotor is shown in Figure 3E-3b) The

charac-teristic of the motor is T = 4 - 0.1 a> Nm The speed a> developed by the motor is reduced

by a transmission with a ratio 1:3 The "screw-nut" device lifts a mass m = 200 kg The

moment of inertia of the rotor and the wheel attached to it IQ = 0.001 kg m2; the moment

of inertia of the screw and its driving wheel I1 = 0.01 kg m2; and the pitch of the screw

is h = 10 mm Find the height the mass will travel during time t= 0.5 sec At the

begin-ning of the process the motor is at rest

Cross-area of the piston = 50 cm2,

Moving mass M= 200 kg, and

Coefficient of hydraulic resistance \// = 150 Nsec2/m2

Calculate the time needed to develop a piston speed V= 5 m/sec; Estimate the time

needed to obtain a displacement s = 0.1 m

Exercise 3E-4a)

A hydraulic drive is shown in Figure 3E-4a The cylinder with an inner diameter

D0 = 0.08 m is used to move a piston rod with mass m l = 100 kg The piston rod (D = 0.02 m) serves as a rack engaged with a gear wheel block with a ratio of radii R/r=2.5 and r= 0.04 m The moment of inertia of the block /= 0.2 kg m2 The blockdrives a mass ra2 = 50 kg The hydraulic pressure on the input of the device

p = 200 N/cm2 Coefficient of hydraulic resistance in the piping y = 120 Nsec2/m2 Find

112 Dynamic Analysis of Drives

the time needed to achieve a speed of the piston V= 2m/sec when the height of the

mass increases; find the distance travelled by the piston At the beginning of the processthe piston is at rest

Exercise 3E-5

Figure 3E-5 shows a pneumatic cylinder serving as an elevator Pressure

P r = 50 N/cm2 to this elevator is supplied from an air receiver 2 located about L = 10 m

away The initial position of the piston 1 = 0.1 m The mass of the elevator handles

m = 400 kg for case a), and m = 550 kg for case b) The stroke smax is about 1.5 m Otherpertinent data are:

Inner diameter of the cylinder D = 0.15 m,

Inner diameter of the pipe d = 0.012 m,

Absolute temperature of the air in the receiver T r = 293° K, and

Coefficient of aerodynamic resistance a = 0.5 sec/m.

Calculate the time needed to lift the mass (for both cases separately) from themoment in time that the valve 3 is actuated until the time the mass reaches point smax.

Exercise 3E-5a)

The pneumatic system in Figure 3E-5a) consists of a volume V c = 0.2 m2, a pipe with

a diameter d = 0.5" and length L = 20 m, and a coefficient of resistance a = 0.5 sec/m The provided pressure P r = 50 N/cm2 The absolute air temperature in the system T r

= 293° K Find the time needed to bring the pressure P c in the volume V c to the value

Pr = 50N/cm2

Exercise 3E-5b)

The pneumatically actuated jig in Figure 3E-5b) is used to support a weight

Q = 5,000 N The designations are clear from the figure The inner diameter of the

cylin-der D = 0.125 m, the initial volume of the cylincylin-der V c = 0.002 m3, the diameter of the

piping d = 0.5", the constant air pressure in the system P r = 60 N/cm2, the air

temper-ature T c = 293° K, and the coefficient of aerodynamic resistance in the piping

a = 0.5 sec/m The distance from the valve to the cylinder L = 20 m Find the time

needed to close the jig from the time the valve is actuated (the real travelling distance

of the piston s = 0).

FIGURE 3E-5a)

114 Dynamic Analysis of Drives

Exercise 3E-5c)

The machine shown in Figure 3E-5c) must be stopped by a brake The initial speed

of the drum with a moment of inertia ^ = 0.1 kg m2 is n0 = 1,500 rpm The ratio of the speed reducer connecting the drum to the brakes is zjz2 = 3.16 The moment of inertia

of the brake's drum is I2 = 0.01 kg m2 Find the time required for the machine to stop

completely when the braking torque T= 5 + 4 0 Nm (where 0 is the rotation angle); find the time needed for the machine to stop when the braking torque T= 5 + 4 a> Nm (the final speed a>f is about 0.5% of that in the beginning).

Exercise 3E-6

Figure 3E-6 shows a mechanism consisting of a rotating column with a moment of

inertia I0 , to which a lever is connected by means of a hinge The hinge moves so that the angle 6 changes according to the law 6 = at, where a = constant A concentrated mass m is fastened to the end of the lever, whose length is r Write the equation of motion of this system when a constant torque T is applied to the column Solve the

equation for the following data:

FIGURE 3E-5c)

Exercise 3E-6a)

A mass m is moving along the diameter of a disc-like rotating body with a moment

of inertia /„ (see Figure 3E-6a)) The law of motion r(t) of this mass relative to the center

of the disc is r = R 0 cos at where R 0 and co are constant values The mechanism is driven

by a DC electric motor with a characteristic T=T l -T Q co Write the equation of motion

for the disc in this case At the beginning of the process the motor is at rest

Exercise 3E-7

Consider the electromagnet shown in Figure 3E-7 How will its response timechange if:

The number W of winds on its coil is doubled? The voltage £7 is doubled? The mass

m of the armature is doubled? (In each of the above-mentioned cases the rest of the

parameters stay unchanged.)

FIGURE 3E-6a)

FIGURE 3E-7

Kinematics and Control of

Automatic Machines

4.1 Position Function

We begin our discussion with the study of case 4, described in Chapter 1 and shown

in Figure 1.5 The drive, whatever its nature, imparts the required movement to thetools through a mechanical system that controls the sequence and regularity of thedisplacements Every mechanism has a driving link and a driven link The first ques-tion in kinematics is that of the relationship between the input (driving motion) andthe output (driven motion) Let us denote:

x = the input motion, which can be linear or angular,

5 = the output motion, which also can be linear or angular

Thus, we can express the relationship between these two values as:

We call Il(jt) the position function

From Equation (4.1), it follows that

and

The importance of Equation (4.2) is that it expresses the interplay of the forces: bymultiplying both sides of Equation (4.2) by the force (or torque, when the motion is

116

angular), we obtain an equation for the power on the driving and driven sides of themechanism (at this stage frictional losses of power can be neglected) Hence,

From Equation (4.2),

then

Obviously, H'(x) is the ratio between the driving and driven links In the particular case where the input motion can be considered uniform (i.e., x = constant and x = 0), it

follows, from Expression (4.3), that

The designer often has to deal with a chain of n mechanisms, for which

To illustrate this, let us take the Geneva mechanism as an example for calculation

of a n function The diagram shown in Figure 4.1 will aid us in this task It is obviousthat this mechanism can be analyzed only in motion, that is, when the driving link isengaged with the driven one For the four-slot Geneva cross shown on the right side

of the figure, this occurs only for 90° of the rotation of the driving link; during theremainder of the rotation angle (270°) the driving link is idle To avoid impact betweenthe links at the moment of engagement, the mechanism is usually designed so that,

at that very moment, there is a right angle between OtA and 0;A

FIGURE 4.1 Layout of a four-slot Geneva mechanism

118 Kinematics and Control of Automatic Machines

A scheme of this mechanism is presented on the left side of the figure Here 0:A = r for the driving link at the very moment of engagement, and OjA' = r is constant at all intermediate times The number of slots n in the cross determines the angle y/ 0 , i.e.,

and, obviously,

From the triangle O^A we obtain

Applying Equation (4.8) to the triangle 0^^' we can express

where the value of if/ is unknown, and the length of 0^' = h.

From the sine law, we obtain

And thus from Equations (4.9) and (4.10):

Denoting A = r// and simplifying Equation (4.11), we obtain

or

From Equation (4.13) we obtain the following expression for the velocity of the driven

link ij/:

or

When cfy/dt = a> 0 = constant, we obtain

For acceleration of the driven link we obtain

When 0 = a> Q we can simplify the expression to the form

and

Graphical interpretations of Expressions (4.15) and (4.17) are shown in Figure 4.2.This mechanism is very convenient whenever interrupted rotation of a tool is nec-essary Naturally, various modifications of these mechanisms are possible For instance,two driving pins can drive the cross, as in Figure 4.3 The resting time and the time ofrotation for this mechanism are equal More than four slots (the minimum number ofslots is three) can be used Figure 4.4 shows a Geneva mechanism with eight slots Onedriver can actuate four (or some other number of) mechanisms, as in Figure 4.5 Thedurations of the resting times can be made unequal by mounting the driving pins atangle A (see Figure 4.6) One stop will then correspond to an angle (A - 90°), the other to

an angle (270° - A) Another modification of such a mechanism is shown in Figure 4.7

FIGURE 4.2 Speed and acceleration of the drivenlink of the Geneva mechanism shown in Figure 4.1

120 Kinematics and Control of Automatic Machines

FIGURE 4.3 Geneva mechanism with twodriving pins

FIGURE 4.4 Eight-slot Geneva mechanism

FIGURE 4.5 Geneva mechanism with multipledriven links

FIGURE 4.6 Asymmetrical Genevamechanism with two driving pins

FIGURE 4.7 Modified Geneva mechanism.

Here, the slots have different lengths and the driving pins are located on different radii.These changes also produce different stop durations Figure 4.8 shows a Geneva mech-

anism with internal engagement The driven body a rotates for an angle iff = 2n /4 while the driving link b passes through an angle <f> = n + i]/ We will return to these mecha-

nisms later

Another example of the derivation of the position function is shown for a shaft mechanism (Figure 4.9) Omitting the intermediate strokes, we obtain for the

crank-coordinates y and x of point M on the connecting rod the following expressions:

where 0 = cot, and CD = angular speed of the driving crank.

Differentiating y and x, with respect to time, we obtain the components of the speed

in the corresponding directions The components of the acceleration can be obtained

FIGURE 4.8 Internal engagement Geneva mechanism

122 Kinematics and Control of Automatic Machines

FIGURE 4.9 Crankshaft mechanism

in the same manner Thus, the position function is expressed separately for each dinate Obviously, this analytical approach can be useful for analyzing other specifickinds of trajectories Figure 4.10 shows the different trajectories of several points belong-ing to the same link: the connecting rod of a four-bar mechanism

coor-[Note: The reader may, of course, use the modern vector loop method to make the

kinematic analyses of designed or chosen mechanisms This approach is especiallyuseful for computerized calculations or animation of the mechanism on the com-puter's screen However, the author's opinion remains that the choice of mathemati-cal approach is a private affair depending on personal taste, predilection, etc In theauthor's opinion the offered approach gives a better physical understanding of thekinematic events His duty was to show that such-and-such things at this-and-thisdesign stage must be calculated.]

In all the examples we have discussed so far, neither the position function nor thekinematic properties (except speeds) can be modified after the dimensions and shapesare established However, this lack of flexibility can be overcome by altering the design.Take, for instance, the mechanism for contour grinding in Figure 4.11 Grinding tool(grinding wheel) 4, with its motor, is mounted on the connecting rod of the crankshaftmechanism The mechanism is adjusted by moving joint 3 and guide 1 and securingthem in the new position by means of set screws 2 The radius of crank 5 can also bechanged by moving it along the slot in rotating table 6

FIGURE 4.10 Trajectories of different points

of a connecting rod of a four-link mechanism

FIGURE 4.11 Mechanism for contour grinding Views of contours

produced when settings of parts 1 and 3 are changed

4.2 Camshafts

It is not always possible to satisfy the desired position function by means of themechanisms discussed in the previous section The requirements dictated by the timingdiagram (see Chapter 2) vary, but they can often be met by using cam mechanisms.The idea underlying such mechanisms is clear from Figure 4.12a), in which a disc cam

is presented schematically A linearly moving follower has a roller to improve frictionand contact stresses at the cam profile-follower contact joint It is easy to see that, byrotating the cam from positions 0 to 11, the follower will be forced to move vertically

in accordance with the radii of the profile Graphical interpretation of the positionfunction has the form shown in Figure 4.12b) During cam rotation through the angle0! (positions 0-1-2-3-4-5) the follower climbs to the highest point; during the angle 02(6-7-8-9-10-11) it goes down, and during the angle 03 the follower dwells (because thisangle corresponds to that part of the profile where the radius is constant) Changingthe profile radii and angles yields various position functions, which in turn producedifferent speeds and acceleration laws for the follower movements Figure 4.13 illus-trates the cosine acceleration law of follower movement The analytical description ofthis law is given by the following formulas:

124 Kinematics and Control of Automatic Machines

FIGURE 4.12 a) Disc cam mechanism; b) The follower motion law

To provide the desired sequence and timing of actions, it is convenient to mountall cams needed for the machine being designed on one shaft, thus creating a camshaft,for example, as shown in Figure 4.14 One rotation of this shaft corresponds to one

cycle of the machine, and thus one revolution lasts one period or T seconds As can

be seen from Figure 4.14, a camshaft can drive some mechanisms by means of cams(mechanisms A, B, C, and D), some by cranks (mechanism E), and some by gears (mech-anism F) Sometimes a single straight shaft is not optimum for a given task Then thesolution shown in Figure 4.15 can be useful Here, motor 1, by means of belt drive 2,drives camshaft 3, which is supported by bearings 4 A pair of bevel gears 5 drive shaft

6 The ratio of transmission of the bevel gears is 1:1; thus, both shafts complete onerevolution in the same time and all cams and cranks (7 and 8) complete their tasks atthe same time

(Figure 4.14a) shows a photograph of a specific camshaft controlling an tive assembly machine serving the process of dripping irrigation devices production)

automa-FIGURE 4.13 Cosine acceleration law carried out by link

driven by cam mechanism