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Shaft Design

VIEWS: 35 PAGES: 6

									VI DESIGN CALCULATIONS

6.1 Shaft Design

   The most critical design parameter for the shaft is the maximum allowable deflection. If the
shafts experience a large deflection, then the linkage propulsion system will jam and cease to
function. Through observation of loading, it has been determined that the drive shafts will
experience the greatest loads, and therefore, the greatest deflections, so the analysis must
concentrate on these components. The loading of the drive shaft occurs as is seen in Figure
6.1.1. Here, F1 and F3 are the forces due to the linkage drive system, and F 2 is the lad due to the
tension in our drive belt. All of these forces are acting in the lengthwise direction of the robot.
Points A and B are the reactions by the structure of our robot.
   It should be noted that the directions of the forces F 1 – F3 might be opposite to the directions
depicted in Figure 6.1. There are four different cases that are of concern in this analysis, so the
directions and magnitudes of the forces will be determined on a case by case basis. The four
load cases are presented below in Table 6.1.1.


                                                      Case I
                 F1                 F2                 F3                  RA                 RB
              -5.819 N           8.896 N               0N               -2.133 N           -0.944 N
                                                     Case II
                 F1                 F2                 F3                  RA                  RB
                 0N              8.896 N            -5.819 N            -8.321 N            5.244 N
                                                    Case III
                 F1                 F2                 F3                 RA                  RB
              5.819 N            8.896 N               0N              -14.140 N           -0.575 N
                                                    Case IV
                 F1                 F2                 F3                  RA                 RB
                 0N              8.896 N            5.819 N             -7.952 N            -6.76 N
       Table 6.1.1: Above are the four crit ical load cases of the drive shafts. Negative values indicate that the
       force direction is in the positive Y direct ion.

   The sign convention used for the analysis is that a negative force value indicates that it is in
the positive Y direction. Using this information, the Shear and Moment diagrams for all four
cases were generated. They are shown in Figures 6.1.2 to 6.1.5 below.




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The next step in the shaft design is to create a free body force diagram (Figure 6.1.6), and then
determine the beam deflection equation for this type of loading. The deflection function for this
beam can be found by integrating the moment function.


                EI * M  x    F1 x  R A x  0.476  F2 x  1.757  RB x  15.476

                                                             EI x   M x dx

                                 F1                  RA                        F2                           RB
               EI *  x                            x  0.476                x  1.757                   x  15.476        C1
                                             2                         2                            2                       2
                                    x
                                 2                   2                         2                             2

                                                              EIYx    x dx

                              F1             RA                        F2                           RB
            EI * Y x                       x  0.476                x  1.757                   x  15.476          C1 x  C 2
                                     3                         3                           3                          3
                                 x
                              6              6                         6                            6


We must now determine what C 1 and C2 are using boundary conditions. The boundary
conditions are the same for all four cases.


Condition 1: Y = 0 at the point x = 0.476 cm
Condition 2: Y = 0 at the point x = 15.476 cm

                                                           F1
                                                     0                        C1 0.476  C 2
                                                                           3
                                                              0.476
                                                           6
                                                                       and
                               F                            R                  F2
                          0   1 15.476                    A 15                                  C1 15.476   C 2
                                                       3               3                       3
                                                                                  13.719
                                6                            6                 6
                                                         Thus, we yield:
                                                 C1  41.18F1  37.5RA  28.69F2
                                                               and

                                             C2  19.58F1  17.85RA  13.67F2

   Using these equations, the values from Table IS, and setting the bending angle function equal
to zero, the point of maximum deflection for the beam could be found. From here, it was
desirable that the beam deflect no more than 0.1 mm from any of the four load cases, so using a
Young’s Modulus of 200Gpa for steel, and the deflection function, the minimum shaft radius
was found. These results are all in Table 6.1.2.


                                                                           19
                                              Case I
            C1             C2         Max Deflect @             E                Y     Radius
                                                                       N
          -64.39          30.53           6.84 cm         2.0 *10 7          0.01 cm   1.89 mm
                                                                      cm 2
                                             Case II
            C1             C2         Max Deflect @             E                Y     Radius
                                                                       N
          -56.81          27.05           7.26 cm         2.0 *10 7          0.01 cm   1.89 mm
                                                                      cm 2
                                            Case III
            C1             C2         Max Deflect @             E                Y     Radius
                                                                       N
          -35.40          16.98           6.86 cm         2.0 *10 7          0.01 cm   1.67 mm
                                                                      cm 2
                                            Case IV
            C1             C2         Max Deflect @             E                Y     Radius
                                                                       N
          -42.97          20.46           6.28 cm         2.0 *10 7          0.01 cm   1.69 mm
                                                                      cm 2
                           Table 6.1.2: Results from the beam bending function



Thus, any beam of diameter greater than 3.78 mm will be sufficient for this type of loading.


   The analysis so far, has only taken consideration of forces in the 1 direction of the robot (see
figure 6.1.7), but deflection of the shaft in the 2 (due to the weight of the device) direction is also
important. In this case, the loading diagram of the critical case is the same as in Figure 6.1.1,
however F3 in this case is zero, and F2 . Thus, the same bending equations apply for this as
applied for cases 1 through four. The results for this case are summarized in Table 6.3.




                                                   20
                                   Deflection in the 2 Direction

                   F1                                     -26.7 N

                   RA                                    21.31 N

                   RB                                    -0.901 N

                   C1                                     -45.12

                   C2                                      20.78

                   Max Deflection @                      2.88 cm
                                                                    N
                   E                                   2.0 *10 7
                                                                   cm 2
                   Y                                     0.01 cm

                   Radius                                2.09 mm
                        Table 6.1.3: Summary of the results in the 2 direction

Thus, the critical diameter is 4.18mm.
Table 6.1.4 summarizes this shaft analysis.

6.2 Motor and Gear Selection

   The ultimate governing conditions for the selection of a motor and gearbox for this design
were speed, torque, and power. The goal for Team 1 was to be capable of crossing the 3.5m
beam in 30seconds. Using Analytix computational software, it was determined that the length of
the stride of the robot was 5.88cm/step, and the vertical displacement of the foot was 1.26cm.
Also, it should be noted that there are two steps per complete revolution of the drive link, making
the total length of the robot traveled per revolution 11.76 cm. Thus, the required rotational
velocity of the drive length is calculated by
                                                   R 60s
                                           V *    *
                                                   d 1min

where  = angular velocity of the drive link in revolutions per minute, V is the vehicle velocity,
R = 1 revolution, d = the distance traveled by the device in 1 revolution.

                                             59.52rpm


                                                 21
   The next requirement was that the drive system produce enough torque. Using Analytix, the
maximum torque required by each drive link was determined to be .2 Nm (28 oz- in). Thus,
because two drive links are engaged at all times, the drive system must be capable of producing
0.4 Nm (56 oz- in) of torque.

   The final requirement was that the motor produce sufficient power. The required power was
determined by the equation below.

                                                   P  F *V

Where F is the weight of the vehicle (53.4N) and V is the velocity at which the robot will travel
vertically
                                           1.26cm      1m     1m
                             V  60rpm *             *     *
                                          revolution 60s 100cm

                                V = .0126 m/s and P = 0.673w
Table 6.2 summarizes the motor requirements.

6.3 Linkage Analysis

   Linkage analysis for this application, centered primarily on stress analysis. The first step in
performing this analysis was to determine the maximum forces that each link in the system
would receive for assumed 26.7 N load. This 26.7 N load was selected because it is was
determined to be 2 times the anticipated static loading on each leg. For ease of discussion,
Figure 6.3.1 below shows a rough sketch of the system, with each link being numbered. In
reality, links 1, 2, and 3 will be one triangular piece, but separating them into three distinct parts
makes the analysis much simpler. Assuming a worst case scenario, the maximum force that any
one link experiences is 26.7 N. Using this force and knowing that the links are made of Acrylic,
which has a yield strength of 5.17 x 107 Pascals, the following relationship can be arrived at
                                                      F
                                               yield 
                                                      A
                                          A  2.164 E  7m 2

Thus, the minimal cross sectional area that any link can have is .516 mm2 . The minimal cross
section achieved by any link is18 mm2 , therefore, no link will fail. Table 6.3 summarizes the
linkage analysis.




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6.4 Pseudo code
    When the machine is turned on it will run the MIVDS machine program. The program will
continue to run until the machine is turned off. The program starts by turning on the motor,
which drives the locomotion and continues driving the motor until it detects a source. If it detects
a light source it will drop 2 balls, if it detects a wind source it will determine if the wind source is
hot or cold and then drop 1 or 3 balls respectively. After the balls have been dropped the motor
will be turned back on and repeat this loop until the robot is turned off.


MIVDS Machine () {
While (Machine is on) {
    Turn on Motor;
    If (Reading from light sensor is high) { //detect light
        Turn off motor;
        Drop Ball (2); }
        If (Reading from wind sensor is high) { //detect wind or heat
        Turn off motor;
        If (Reading from temperature sensor is >= 80 degrees Fahrenheit) //detect temperature
            {Drop Ball (1);}
            //else thee is wind
                Else {Drop Ball (3);}}
    Turn off motor;}
Drop Ball (int counter) { //drop specified amount of ping-pong balls
    for (int I=0; I++; I<counter) {
    Turn on solenoid #2; Turn on solenoid #1;
    Wait 2 seconds;
    Turn off solenoid #1; Turn off solenoid #2;
    Wait 2 seconds;}}


6.5 Electronic Control System
    The full electronic schematic can be found in figure 6.5. This figure does not include exact
Op Amp, Voltage divider, resistor specs, etc do to lack of funds for testing materials.



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