3.3 NUMERICAL BASED ON GEAR TRAINS
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In an epicyclic gear train with gears A and B on an arm C, the arm rotates at -150 rpm. Gear A is fixed. Using a tabular kinematic analysis method: 1) Arm C is fixed, gear A completes 1 rotation so gear B speed is -TA/TB 2) Gear A completes x rotations, so gear B speed is -x*TA/TB 3) Arm C adds y rotations, so final speeds are: gear A = x+y, gear B = y - x*TA/TB Solving the equations gives: x = 150, y = -150, so gear B speed is -270 rpm.
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3.3 NUMERICAL BASED ON GEAR TRAINS
- 1. Sanjivani Rural Education Society’s Sanjivani College of Engineering, Kopargaon-423603 ( An Autonomous Institute Affiliated to Savitribai Phule Pune University, Pune) NAAC ‘A’ Grade Accredited, ISO 9001:2015 Certified Subject :- Theory of Machines II T.E. Mechanical (302043) Unit 3 3.3 Numericals based Epicyclic Gear Train By Prof. K. N. Wakchaure(Asst Professor) Department of Mechanical Engineering Sanjivani College of Engineering (An Autonomous Institute) Kopargaon, Maharashtra Email: wakchaurekiranmech@Sanjivani.org.in Mobile:- +91-7588025393
- 2. Kinematic Analysis (Tabular Method) STEP OPERATION SPEED OF THE ELEMENTS ARM C GEAR A GEAR B 1 ARM C is fixed and gear A competes +1 rotation 0 +1 - 𝐓 𝐀 𝐓 𝐁 2 ARM C is fixed and gear A competes +x rotation 0 +x - 𝐱 ∗ 𝐓 𝐀 𝐓 𝐁 3 Add +y rotation of ARM C +y +y +y 4 Add Step 2 +Step 3 𝐲 x+y y - 𝐱 ∗ 𝐓 𝐀 𝐓 𝐁 𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝟑𝟔, 𝐓 𝑩 = 𝟒𝟓 𝑵 𝑪 = −𝟏𝟓𝟎 𝒓𝒑𝒎 𝑵 𝑩 = 𝒚 − 𝐓 𝐀 𝐓 𝐁 ∗ 𝒙𝑵 𝑪 = +𝒚 𝑵 𝑨 = 𝒙 + 𝒚 𝑵 𝑪 = −𝟏𝟓𝟎 𝑵 𝑨 = 𝟎 𝒙 + 𝒚 = 𝟎𝑵 𝑪 = 𝒚 = −𝟏𝟓𝟎 𝒙 = −𝒚 𝒙 = 𝟏𝟓𝟎 𝑵 𝑩 = 𝒚 − 𝐓 𝐀 𝐓 𝐁 ∗ 𝒙 𝑵 𝑩 = −𝟏𝟓𝟎 − 𝟑𝟔 𝟒𝟓 ∗ (𝟏𝟓𝟎) 𝑵 𝑩=-270 rpm In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. If the arm rotates at 150 r.p.m. in the anticlockwise direction about the centre of the gear A which is fixed, determine the speed of gear B. Gear A Gear B Arm C 𝑨𝒏𝒕𝒊𝒄𝒍𝒐𝒄𝒌𝒘𝒊𝒔𝒆 C B A
- 3. Kinematic Analysis (Algebraic Method) Gear A Gear B Arm C C B A Two conditions needs to define • element is fixed (Preferably Arm) • other has specified motion. Let the arm C be fixed in an epicyclic gear train as. Therefore speed of the gear A relative to the arm C Speed of the gear B relative to the arm C 𝑵 𝑩 − 𝑵 𝑪 gears A and B are meshing directly, they will revolve in opposite directions 𝑵 𝑩−𝑵 𝑪 𝑵 𝑨−𝑵 𝑪 =- 𝐓 𝐀 𝐓 𝐁 Since the arm C is fixed, therefore its speed, NC = 0. 𝑵 𝑩 𝑵 𝑨 =- 𝐓 𝐀 𝐓 𝐁 If the gear A is fixed, then NA = 0. 𝑵 𝑩−𝑵 𝑪 𝑵 𝑨−𝑵 𝑪 =- 𝐓 𝐀 𝐓 𝐁 𝑵 𝑩−𝑵 𝑪 𝟎−𝑵 𝑪 =- 𝐓 𝐀 𝐓 𝐁 𝑵 𝑩 𝑵 𝑪 =1+ 𝐓 𝐀 𝐓 𝐁 𝑵 𝑨 − 𝑵 𝑪 By rearranging the Terms.
- 4. Kinematic Analysis (Tabular Method) STEP OPERATION SPEED OF THE ELEMENTS ARM C GEAR A GEAR B GEAR D 1 ARM C is fixed and gear A competes +1 rotation 0 +1 - 𝐓 𝐀 𝐓 𝐁 - 𝐓 𝐀 𝐓 𝑫 𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝟑𝟔, 𝐓 𝑩 = 𝟒𝟓 𝑵 𝑪 = −𝟏𝟓𝟎 𝒓𝒑𝒎 In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. Gear B gives the motion to Annular Gear B. If the arm rotates at 150 r.p.m. in the anticlockwise direction about the centre of the gear A which is fixed, determine the speed of gear B and Anular Gear D. C B A Gear A Gear B Arm C Gear D BAD BAD BAD TTT TmTmTm T d mas rrr *2 **4**2**2 _ *2 D 12645*236 DT Arm C is fixed Gear A Completes +1 Rotation Speed of Gear B 𝑵 𝑩 𝑵 𝑨 = 𝐓 𝐀 𝐓 𝐁 Speed of Gear A and Gear B are having External Meshing hence both will rotate in Opposite Direction Speed of Gear B 𝑵 𝑩= − 𝐓 𝐀 𝐓 𝐁
- 5. Kinematic Analysis (Tabular Method) STEP OPERATION SPEED OF THE ELEMENTS ARM C GEAR A GEAR B GEAR D 1 ARM C is fixed and gear A competes +1 rotation 0 +1 - 𝐓 𝐀 𝐓 𝐁 - 𝐓 𝐀 𝐓 𝑫 𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝟑𝟔, 𝐓 𝑩 = 𝟒𝟓 𝑵 𝑪 = −𝟏𝟓𝟎 𝒓𝒑𝒎 In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. Gear B gives the motion to Annular Gear B. If the arm rotates at 150 r.p.m. in the anticlockwise direction about the centre of the gear A which is fixed, determine the speed of gear B and Anular Gear D. C B A Gear A Gear B Arm C Gear D D 12645*236 DT Arm C is fixed Gear A Completes +1 Rotation Gear B and Gear D are having Internal Meshing Speed of Gear D 𝑵 𝑫 𝑵 𝐵 = 𝐓 𝑩 𝐓 𝑫 Speed of Gear D 𝑵 𝑫 = 𝐓 𝑩 𝐓 𝑫 ∗ 𝑵 𝑩 Speed of Gear D 𝑵 𝑫 = 𝐓 𝑩 𝐓 𝑫 ∗ 𝐓 𝐀 𝐓 𝐁 = 𝐓 𝐀 𝐓 𝑫 Speed of Gear D 𝑵 𝑫 = − 𝐓 𝐀 𝐓 𝑫 𝑫𝒐𝒏𝒕 𝑪𝒐𝒏𝒔𝒊𝒅𝒆𝒓 𝒓𝒐𝒕𝒂𝒕𝒊𝒐𝒏 𝑫𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒊𝒔 𝒔𝒕𝒂𝒈𝒆 𝑵 𝑩= − 𝐓 𝐀 𝐓 𝐁
- 6. Kinematic Analysis (Tabular Method) STEP OPERATION SPEED OF THE ELEMENTS ARM C GEAR A GEAR B GEAR D 1 ARM C is fixed and gear A competes +1 rotation 0 +1 - 𝐓 𝐀 𝐓 𝐁 - 𝐓 𝐀 𝐓 𝑫 2 ARM C is fixed and gear A competes +x rotation 0 +x -x* 𝐓 𝐀 𝐓 𝐁 -x* 𝐓 𝐀 𝐓 𝑫 3 Add +y rotation of ARM C +y +y +y +y 4 Add Step 2 +Step 3 0 x+y y-x* 𝐓 𝐀 𝐓 𝐁 y-x* 𝐓 𝐀 𝐓 𝑫 𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝟑𝟔, 𝐓 𝑩 = 𝟒𝟓 𝐓 𝑫 = 𝟏𝟐𝟔 𝑵 𝑪 = −𝟏𝟓𝟎 𝒓𝒑𝒎 C B A Gear A Gear B Arm C Gear D D 𝑵 𝑩 = 𝒚 − 𝐓 𝐀 𝐓 𝐁 ∗ 𝒙𝑵 𝑪 = +𝒚 𝑵 𝑨 = 𝒙 + 𝒚 𝑵 𝑫 = 𝒚 − 𝐓 𝐀 𝐓 𝑫 ∗ 𝒙 𝑵 𝑪 = −𝟏𝟓𝟎 𝑵 𝑨 = 𝟎 𝒙 + 𝒚 = 𝟎𝑵 𝑪 = 𝒚 = −𝟏𝟓𝟎 𝒙 = −𝒚 𝒙 = 𝟏𝟓𝟎 𝑵 𝑫 = 𝒚 − 𝐓 𝐀 𝐓 𝑫 ∗ 𝒙 𝑵 𝑩 = −𝟏𝟓𝟎 − 𝟑𝟔 𝟏𝟐𝟔 ∗ (𝟏𝟓𝟎) 𝑵 𝑩=-192.87 rpm 𝑨𝒏𝒕𝒊𝒄𝒍𝒐𝒄𝒌𝒘𝒊𝒔𝒆
- 7. Kinematic Analysis (Tabular Method) (Simple Trick) STEP OPERATION SPEED OF THE ELEMENTS ARM C GEAR A GEAR B GEAR D 1 ARM C is fixed and gear A competes +1 rotation 0 +1 - 𝐓 𝐀 𝐓 𝐁 - 𝐓 𝐀 𝐓 𝑫 2 ARM C is fixed and gear A competes +x rotation 0 +x -x* 𝐓 𝐀 𝐓 𝐁 -x* 𝐓 𝐀 𝐓 𝑫 3 Add +y rotation of ARM C +y +y +y +y 4 Add Step 2 +Step 3 0 x+y y-x* 𝐓 𝐀 𝐓 𝐁 y-x* 𝐓 𝐀 𝐓 𝑫 C B A Gear A Gear B Arm C Gear D D 𝑵 𝑩 = 𝐓 𝐀 𝐓 𝐁 𝑵 𝑫 = 𝐓 𝐀 𝐓 𝑫 𝑵 𝑨 = 𝟏
- 8. Kinematic Analysis (Tabular Method) (Simple Trick) STEP OPERATION SPEED OF THE ELEMENTS 𝑵 𝑪 𝑵 𝑨 𝑵 𝑩 = 𝑵 𝑬 𝑵 𝑫 𝑵 𝑭 1 ARM C is fixed and gear A competes +1 rotation 0 +1 - 𝐓 𝐀 𝐓 𝐁 - 𝐓 𝐀 𝐓 𝑫 - 𝐓 𝐀 𝐓 𝑩 * 𝐓 𝑬 𝐓 𝑭 2 ARM C is fixed and gear A competes +x rotation 0 +x -x* 𝐓 𝐀 𝐓 𝐁 -x* 𝐓 𝐀 𝐓 𝑫 - 𝐓 𝐀 𝐓 𝑩 * 𝐓 𝑬 𝐓 𝑭 3 Add +y rotation of ARM C +y +y +y +y +y 4 Add Step 2 +Step 3 0 x+y y-x* 𝐓 𝐀 𝐓 𝐁 y-x* 𝐓 𝐀 𝐓 𝑫 y- 𝐓 𝐀 𝐓 𝑩 * 𝐓 𝑬 𝐓 𝑭 C B A Gear A Gear B Arm C Gear D D 𝑵 𝑩 = 𝐓 𝐀 𝐓 𝐁 𝑵 𝑫 = 𝐓 𝐀 𝐓 𝑫 𝑵 𝑨 = 𝟏 E F E Gear F 𝑵 𝑩= 𝑵 𝑬 𝑵 𝑭 = 𝐓 𝐀 𝐓 𝑩 * 𝐓 𝑬 𝐓 𝑭
- 9. Kinematic Analysis (Tabular Method) STEP OPERATION SPEED OF THE ELEMENTS ARM C GEAR A GEAR B 1 ARM C is fixed and gear A competes +1 rotation 0 +1 - 𝐓 𝐀 𝐓 𝐁 2 ARM C is fixed and gear A competes +x rotation 0 +x - 𝐱 ∗ 𝐓 𝐀 𝐓 𝐁 3 Add +y rotation of ARM C +y +y +y 4 Add Step 2 +Step 3 𝐲 x+y y - 𝐱 ∗ 𝐓 𝐀 𝐓 𝐁 𝑮𝒊𝒗𝒆𝒏 𝐓 𝐀 = 𝟑𝟔, 𝐓 𝑩 = 𝟒𝟓 𝑵 𝑪 = −𝟏𝟓𝟎 𝒓𝒑𝒎 𝑵 𝑩 = 𝒚 − 𝐓 𝐀 𝐓 𝐁 ∗ 𝒙𝑵 𝑪 = +𝒚 𝑵 𝑨 = 𝒙 + 𝒚 𝑵 𝑪 = −𝟏𝟓𝟎 𝑵 𝑨 = 𝟎 𝒙 + 𝒚 = 𝟎𝑵 𝑪 = 𝒚 = −𝟏𝟓𝟎 𝒙 = −𝒚 𝒙 = 𝟏𝟓𝟎 𝑵 𝑩 = 𝒚 − 𝐓 𝐀 𝐓 𝐁 ∗ 𝒙 𝑵 𝑩 = −𝟏𝟓𝟎 − 𝟑𝟔 𝟒𝟓 ∗ (𝟏𝟓𝟎) 𝑵 𝑩=-270 rpm 𝑨𝒏𝒕𝒊𝒄𝒍𝒐𝒄𝒌𝒘𝒊𝒔𝒆 C B A Gear A Gear B Arm C Gear D bad bad bad TTT TmTmTm T d mas rrr *2 **4**2**2 _ *2
- 10. Torques in Epicyclic Gear Trains Gear A Gear B Arm C Gear D