Chuong 3 truyen dong banh rang

1 10 Advanced Gear Analysis • Epicyclic Gearing • Tooth Strength Analysis Epicyclic Gears 174 Epicyclic Gearset An epicyclic gear set has some gear or gears whose center revolves about some point Here is a gearset with a stationary ring gear and three planet gears on a rotating carrier The input is at the Sun, and the output is at the planet carrier INPUT The action is epicyclic, because the centers of the planet gears revolve about the sun gear while the planet gears turn Sun Finding the gear ratio is somewhat complicated because the planet gears revolve while they rotate Planet CARRIER Ring Epicyclic Gears 175 Epicyclic Gearset Let’s rearrange things to make it simpler: 1) Redraw the planet carrier to show arms rotating about the center 2) Remove two of the arms to show only one of the planet arms OUTPUT INPUT INPUT OUTPUT Epicyclic Gears 176 The Tabular (Superposition) Method To account for the combined rotation and revolution of the planet gear, we use a two step process Planet OUTPUT First, we lock the whole assembly and rotate it one turn CounterArm Clockwise (Even the Ring, which we know is fixed) INPUT We enter this motion into a table using the convention: CCW = Positive Sun CW = Negative Ring Rotate Whole Assembly CCW Sun Planet Ring Arm +1 +1 +1 +1 Epicyclic Gears 177 The Tabular (Superposition) Method Next, we hold the arm fixed, and rotate whichever gear is fixed during INPUT Turn operation one turn clockwise Planet Here, we will turn the Ring clockwise one turn (-1), holding the arm fixed The Planet will turn NRing / NPlanet turns clockwise • Since the Planet drives the Sun, the Sun will turn (NRing / NPlanet) x (- Ar m • NPlanet / NSun) = - NRing / NSun turns (counter-clockwise) • The arm doesn’t move Sun We enter these motions into the second row of the table Ring Rotate Whole Assembly CCW Hold Arm, Rotate Ring CW Sun Planet Ring Arm +1 +1 +1 +1 NRing / NSun - NRing / NPlanet -1 Epicyclic Gears 178 The Tabular (Superposition) Method Finally, we sum the motions in the first and second rows of the table Sun Planet Ring Arm Rotate Whole Assembly CCW Hold Arm, Rotate Ring CW +1 +1 +1 +1 NRing / NSun - NRing / NPlanet -1 Total Motion + NRing / NSun - NRing / NPlanet +1 Now, we can write the nSun =1 + relationship: n Arm = N Ring N Sun 1+ N Ring × n Arm , or × nSun N Sun If the Sun has 53 teeth and the Ring 122 teeth, the output to input speed ratio is +1 / 3.3 , with the arm moving the same direction as the Sun Epicyclic Gears 179 Planetary Gearset OUTPUT The configuration shown here, with the input at the Sun and the output at the Ring, is not epicyclic It is simply a Sun driving an internal Ring gear through a set of three idlers INPUT Sun Planet The gear ratio is: nring nsun − N sun N planet − N sun = × = N planet N ring N ring Ring n = speed; N = # Teeth Where the minus sign comes from the change in direction between the two external gears If the Sun has 53 teeth and the Ring 122 teeth, the ratio is -1 / 2.3 Epicyclic Gears 180 Another Epicyclic Example Here are three different representations of the same gearset: ➀ ➃ ➃ ➁ ➁ ➂ ➂ ➂ ➀ ➀ ➄ ➄ ➄ ➁ ➃ Given: Ring = 100 Teeth (Input) Gear = 40 Teeth Gear = 20 Teeth Ring = 78 Teeth (Fixed) Always draw this view to get direction of rotation correct Arm = Output See the Course Materials folder for the solution Epicyclic Gears 181 Epicyclic Tips • If you encounter a gear assembly with two inputs, use superposition Calculate the output due to each input with the other input held fixed, and then sum the results • Typically, when an input arm is held fixed, the other output to input relationship will not be epicyclic, but be a simple product of tooth ratios OUT IN IN • Use the ± sign with tooth ratios to carry the direction information Epicyclic Gears 182 Gear Loading • Once you determine the rotational speeds of the gears in a train, the torque and therefore the tooth loading can be determined by assuming a constant power flow through the train • Power = Torque x RPM, so Torque = Power / RPM • If there are n multiples of a component (such as the idlers in the planetary gearset example) , each component will see 1/n times the torque based on the RPM of a single component • From the torque, T, compute the tangential force on the teeth as W t = T/r = 2T/D , where D is the pitch diameter Epicyclic Gears 183 The Tabular Method - Example First, we lock the assembly and rotate everything one turn CCW Rotate Whole Assembly CCW Hold Arm, Rotate Fixed Ring CW Ring Gear Gear Ring Arm +1 +1 +1 +1 +1 Total Motion    Next, we hold the arm fixed, and turn the fixed component (Ring 4) one   turn CW, to fill in row two Epicyclic Gears 184 The Tabular Method - Example First, we lock the assembly and rotate everything one turn CCW Ring Gear Gear Ring Arm +1 +1 +1 +1 +1 Rotate Whole Assembly CCW Hold Arm, Rotate Fixed Ring CW Total Motion   Next, we hold the arm fixed, and turn the fixed component    (Ring 4) one turn CW, to fill in row two Epicyclic Gears 185 The Tabular Method - Example We turn Ring one turn CW (-1)   • Ring drives Gear 3, which turns + N4/N3 x (-1) = - N4/N3 rotations • Gear is on the same shaft as Gear 3, so it also turns - N4/N3 rotations • Gear drives Ring 1, which turns    + N2/N1 x n2 = + N2/N1 x (- N4/N3) = - N2N4/N1N3 rotations • The arm was fixed, so it does not turn We enter these motions into row two of the table: Rotate Whole Assembly CCW Hold Arm, Rotate Ring CW Ring Gear Gear Ring Arm +1 +1 +1 +1 +1 - N2N4/N1N3 - N4/N3 - N4/N3 -1 Epicyclic Gears 186 The Tabular Method - Example Then we add the two rows to get the total motion Rotate Whole Assembly CCW Hold Arm, Rotate Ring CW Total Motion And we can write the relationship: Ring Gear Gear Ring Arm +1 +1 +1 +1 +1 - N2N4/N1N3 - N4/N3 - N4/N3 -1 0 +1 - N2N4/N1N3 - N4/N3 - N4/N3 N2 N4 nRing1 =1 − × n Arm , or N1 N n Arm = 1− N2N4 × nRing1      N1 N Epicyclic Gears 187 The Tabular Method - Example For our example We compute: n Arm = Ring 1: N1 = 100 Teeth Gear 2: N2 = 40 Teeth Gear 3: N3= 20 Teeth Ring 4: N4 = 78 Teeth n Arm =   1− × nRing1 N2N4 N1 N 1− 40 × 78 × nRing1 100 × 20 −1 n Arm = × nRing1 = × nRing1 −1.56 0.56 n Arm = − 1.786 × nRing1    The output arm rotates almost twice as fast as the input ring, and in the opposite direction • Output direction is dependent on the numbers of teeth on the gears! Epicyclic Gears 188 [...]...11 12 13 14 15 16 17 18 19 20 21 22 23 24 26 27 28 29 30 31 32 33 34 : Hệ số hình dạng bề mặt tiếp xúc 35 36