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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. 18 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. 22 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. 23