These slides is about power trains found in vehicles and automobiles, Slides of Thermodynamics

Download These slides is about power trains found in vehicles and automobiles and more Slides Thermodynamics in PDF only on Docsity! MODULE 5 MACHINE ELEMENTS Trains DEFINITION OF TRAINS M ACHINE ELEMENTS A train is a series of cylinders or cones, gears, pulleys, or similar devices serving to transmit power from one shaft to another. A pair of rolling cylinders, gears, or pulleys is really a train, but usually the term train is applied only to those combinations in which there are more than one pair. DEFINITION OF TRAINS M ACHINE ELEMENTS Variable belt drive in motorcycle automatic transmission. https://youtu.be/9cKbzUgFdS0 TRAIN DRIVES M ACHINE ELEMENTS Driver and driven – Refer to figure below. Pulley 1 causes pulley 2 to turn, thus 1 is the driver and 2 is the driven. In the same manner, pulley 3 causes pulley 4 to turn, thus 3 is the driver and 4 is the driven. Thus, for this setup two of the wheels are drivers and two are driven wheels. Idler – Refer to figure below. Gear 2 is both a driver and a driven gear. Gear 2 is an idler gear. It is placed between gear 1 and 3 so that they will rotate in the same direction. An idler is also used to reduce the size of gears required to connect two shafts with fixed center distance and a desired velocity ration. An idler does not affect the velocity ratio. 1 2 3 1 23 4 1 23 4 TRAIN DRIVES M ACHINE ELEMENTS The train value maybe defined as the ratio of the absolute angular speed of the last wheel or driven to the angular speed of the first wheel or driver. It is the reciprocal of the speed ratio. Train Value 𝑒 = 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑎𝑠𝑡 𝑤ℎ𝑒𝑒𝑙 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑓𝑖𝑟𝑠𝑡 𝑤ℎ𝑒𝑒𝑙 = 1 𝑆𝑝𝑒𝑒𝑑 𝑅𝑎𝑡𝑖𝑜 The train value is positive when the first and the last shaft/wheel turns in the same direction; when they turn in opposite direction the train value will be negative. Refer to figure. A, B, and C are shafts and 1, 2, 3 and 4 are the pulleys. The speed ratios will be: 𝑁1𝐷1 = 𝑁2𝐷2 → 𝑁2 = 𝑁1(𝐷1/𝐷2) A B C 𝑁3𝐷3 = 𝑁4𝐷4 → 𝑁4 = 𝑁3(𝐷3/𝐷4)and Eq. 1 Eq. 2 Eq. 3 But 3 and 2 are on the same shaft B, thus 𝑁3 = 𝑁2. Therefore 𝑁4 = 𝑁2(𝐷3/𝐷4) Substituting the value of 𝑁2, 𝑁4 = 𝑁1(𝐷1/𝐷2)(𝐷3/𝐷4) 𝑁4/𝑁1 = (𝐷1𝐷3)/(𝐷2𝐷4) Eq. 4 Pulleys 1 and 3 are drivers, and pulleys 2 and 4 are driven. REVERTED GEAR TRAIN M ACHINE ELEMENTS Reverted gear trains are gear trains where both the input and output shaft axes are either collinear or coincident. They are used in back gear of lathe and drilling machine, and also in automobile gear box. Four speed gear box is a common application of reverted gear train. Engine power flows from driver shaft, via a parallel shaft called Lay shaft to the collinear output shaft called the Main shaft. Simple setup of a reverted gear train. Main Shaft Lay Shaft 𝑟𝐴 + 𝑟𝐵 = 𝑟𝐶 + 𝑟𝐷 𝑇𝐴 + 𝑇𝐵 = 𝑇𝐶 + 𝑇𝐷 Eq. 7 Eq. 8 Since the distance between centers of the shafts of gears 1 and 2 as well as gears 3 and 4 is the same, therefore: Also, the circular pitch or module of all gears is assumed to be the same, therefore the number of teeth on each gear is directly proportional to its circumference or radius. https://youtu.be/0k2kcvde180 REVERTED GEAR TRAIN M ACHINE ELEMENTS Gears C, E, and G can slide on splined main shaft, to engage with different gears to produce different gear trains & ratios as follows. Four speed gear box Gear 4 via A–C engaged with a dog clutch Gear 3 via A–B–D–C Gear 2 via A–B–F–E Gear 1 via A–B–H–G Dog Clutch Spline REVERTED GEAR TRAIN M ACHINE ELEMENTS The power flows from the engine to the input shaft, then transmitted directly to the output shaft by means of engagement of dog clutch between gears A and C. In this configuration, the lay shaft is still turning since the gears A and B are always engaged. Fourth Gear Gear 4 1:1 via A–C engaged with a dog clutch This configuration produces the fastest speed. Train Value 𝐺 = 1 Gear Ratio 𝑒 = 1 REVERTED GEAR TRAIN M ACHINE ELEMENTS First Gear Train Value 𝐺 = 1 𝑒 = 4 Gear Ratio 𝑒 = 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑑𝑟𝑖𝑣𝑒𝑟𝑠 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡𝑒𝑒𝑡ℎ 𝑜𝑓 𝑑𝑟𝑖𝑣𝑒𝑛𝑠 = + 15 𝑥 15 30 𝑥 30 = +0.25 Gear 1 1:4 via A–B–H–G The power flows from the engine to the input shaft, then takes a detour to the lay shaft through the engagement of gears A and B. From the lay shaft power is then transmitted to the output shaft by the engagement of gears H and G. This configuration outputs the highest torque at the cost of speed. REVERTED GEAR TRAIN M ACHINE ELEMENTS EXAMPLE 4 The speed ratio of the reverted gear train as shown is to be 12. The module pitch of gears A and B is 3.125 mm and of gears C and D is 2.5 mm. If the speed ratio between gears A and B and between gears C and D are to be the same, calculate the suitable numbers of teeth for the gears. No gear is to be less than 24 teeth. Solution 𝑁𝐴/𝑁𝐷 = 12 𝑚𝐴 = 𝑚𝐵 = 3.125 𝑚𝑚 𝑚𝐶 = 𝑚𝐷 = 2.5 𝑚𝑚 We know that circular pitch, 𝑃𝐶 = 2𝜋𝑟 𝑇 = 𝜋𝑚 or 𝑟 = 𝑚𝑇 2 where 𝑚 is the module. 𝑟𝐴 = 𝑚𝐴. 𝑇𝐴 2 ; 𝑟𝐵 = 𝑚𝐵. 𝑇𝐵 2 ; 𝑟𝐶 = 𝑚𝐶 . 𝑇𝐶 2 ; 𝑟𝐷 = 𝑚𝐷 . 𝑇𝐷 2 Since the sum of the radius of each pair must be equal: 𝑚.𝑇𝐴 2 + 𝑚.𝑇𝐵 2 = 𝑚. 𝑇𝐶 2 + 𝑚.𝑇𝐷 2 𝑇𝐴 + 𝑇𝐵 = 𝑇𝐶 + 𝑇𝐷 REVERTED GEAR TRAIN M ACHINE ELEMENTS Since the speed ratios to be the same: Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth, therefore 12 = 𝑁𝐴 𝑁𝐷 = 𝑁𝐴 𝑁𝐵 𝑥 𝑁𝐶 𝑁𝐷 𝑁𝐴 𝑁𝐵 = 𝑁𝐶 𝑁𝐷 and 12 = 𝑁𝐴 𝑁𝐵 𝑥 𝑁𝐴 𝑁𝐵 = 𝑁𝐴 𝑁𝐵 2 ∴ 𝑁𝐴 𝑁𝐵 = 12 𝑇𝐵 𝑇𝐴 = 𝑇𝐷 𝑇𝐶 = 12 Eq. i We know that the distance between shafts, 𝑟𝐴 + 𝑟𝐵 = 𝑟𝐶 + 𝑟𝐷 = 200 𝑚𝑚 or 𝑚𝐴. 𝑇𝐴 2 + 𝑚𝐵. 𝑇𝐵 2 = 𝑚𝐶 . 𝑇𝐶 2 + 𝑚𝐷 . 𝑇𝐷 2 = 200 𝑚𝐴. 𝑇𝐴 +𝑚𝐵. 𝑇𝐵 = 𝑚𝐶 . 𝑇𝐶 +𝑚𝐷. 𝑇𝐷 = 400 3.125 𝑇𝐴 + 𝑇𝐵 = 2.5 𝑇𝐶 + 𝑇𝐷 = 400 𝑇𝐴 + 𝑇𝐵 = 400/3.125 = 128 𝑇𝐶 + 𝑇𝐷 = 400/2.5 = 160 Eq. ii Eq. iii From Eq. i, 𝑇𝐵 = 12 𝑇𝐴, substituting this to Eq. ii. REVERTED GEAR TRAIN M ACHINE ELEMENTS Substituting Eq. i to Eq. ii, 𝐷𝐵 2 (240 − 𝐷𝐵)2 = 2.5 → 𝑫𝑩= 𝟏𝟒𝟕 𝒎𝒎; 𝑫𝑨≅ 𝟗𝟑𝒎𝒎 Solving for 𝑇𝐴 and 𝑇𝐵, 𝑇𝐴 = 𝐷𝐴 𝑚 = 93 𝑚𝑚 2.5 𝑚𝑚/𝑡𝑒𝑒𝑡ℎ = 37.2 𝑡𝑒𝑒𝑡ℎ ≈ 𝟑𝟕 𝒕𝒆𝒆𝒕𝒉 𝑇𝐵 = 𝐷𝐵 𝑚 = 147 𝑚𝑚 2.5 𝑚𝑚/𝑡𝑒𝑒𝑡ℎ = 58.8 𝑡𝑒𝑒𝑡ℎ ≈ 𝟓𝟗 𝒕𝒆𝒆𝒕𝒉 For Second Gear: Gear 2 1:1.85 via A–B–F–E 𝐷𝐹 + 𝐷𝐸 2 = 120 → 𝐷𝐹 + 𝐷𝐸 = 240 → 𝐷𝐹 = 240 − 𝐷𝐸 1.85 = 𝑁𝐴 𝑁𝐵 𝑥 𝑁𝐹 𝑁𝐸 → 𝐷𝐵 𝐷𝐴 𝑥 𝐷𝐸 𝐷𝐹 (in terms of diameter.) Eq. iii Eq. iv 𝐷𝐵 𝐷𝐴 = 147 93 = 1.58,Since 1.85 = 1.58 𝑥 𝐷𝐸 𝐷𝐹 REVERTED GEAR TRAIN M ACHINE ELEMENTS Substituting Eq. iii to Eq. iv, 1.85 = 1.58 𝑥 𝐷𝐸 240 − 𝐷𝐸 → 𝑫𝑬 = 𝟏𝟐𝟗. 𝟒𝟒𝟔 𝒎𝒎; 𝑫𝑭= 𝟏𝟏𝟎. 𝟓𝟓𝟒 𝒎𝒎 Solving for 𝑇𝐴 and 𝑇𝐵, 𝑇𝐸 = 𝐷𝐸 𝑚 = 129.446 𝑚𝑚 2.5 𝑚𝑚/𝑡𝑒𝑒𝑡ℎ = 51.78 𝑡𝑒𝑒𝑡ℎ ≈ 𝟓𝟐 𝒕𝒆𝒆𝒕𝒉 𝑇𝐹 = 𝐷𝐹 𝑚 = 110.554 𝑚𝑚 2.5 𝑚𝑚/𝑡𝑒𝑒𝑡ℎ = 44.22 𝑡𝑒𝑒𝑡ℎ ≈ 𝟒𝟒 𝒕𝒆𝒆𝒕𝒉 For Third Gear: Gear 3 1:1.36 via A–B–D–C 𝐷𝐷 + 𝐷𝐶 2 = 120 → 𝐷𝐷 + 𝐷𝐶 = 240 → 𝐷𝐷 = 240 − 𝐷𝐶 1.36 = 𝑁𝐴 𝑁𝐵 𝑥 𝑁𝐷 𝑁𝐶 → 𝐷𝐵 𝐷𝐴 𝑥 𝐷𝐶 𝐷𝐷 (in terms of diameter.) Eq. v Eq. vi 𝐷𝐵 𝐷𝐴 = 147 93 = 1.58,Since 1.36 = 1.58 𝑥 𝐷𝐶 𝐷𝐷 REVERTED GEAR TRAIN M ACHINE ELEMENTS Substituting Eq. v to Eq. vi, 1.36 = 1.58 𝑥 𝐷𝐶 240 − 𝐷𝐶 → 𝑫𝑪 = 𝟏𝟏𝟏𝒎𝒎; 𝑫𝑫= 𝟏𝟐𝟗𝒎𝒎 Solving for 𝑇𝐴 and 𝑇𝐵, 𝑇𝐶 = 𝐷𝐶 𝑚 = 111 𝑚𝑚 2.5 𝑚𝑚/𝑡𝑒𝑒𝑡ℎ = 44.4 𝑡𝑒𝑒𝑡ℎ ≈ 𝟒𝟒 𝒕𝒆𝒆𝒕𝒉 𝑇𝐷 = 𝐷𝐷 𝑚 = 129 𝑚𝑚 2.5 𝑚𝑚/𝑡𝑒𝑒𝑡ℎ = 51. 6 𝑡𝑒𝑒𝑡ℎ ≈ 𝟓𝟐 𝒕𝒆𝒆𝒕𝒉 The third gears actually doesn’t need to be solved since gear C = F, and gear D = E. The center distance, which is 120 mm, should be maintained for each mating pair when we solve for the actual diameters from the number of teeth. 𝐷𝐴, 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐷𝐻 = 𝑚.𝑇𝐴 = 2.5 37 = 92.5 𝑚𝑚 𝐷𝐵, 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐷𝐺 = 𝑚.𝑇𝐵 = 2.5 59 = 147.5 𝑚𝑚 𝐷𝐴 + 𝐷𝐵 2 = 𝐷𝐻 + 𝐷𝐺 2 = 92.5 + 147.5 2 = 𝟏𝟐𝟎𝒎𝒎✓ EPICYCLIC GEAR TRAIN M ACHINE ELEMENTS Simple setup of planetary gear Epicyclic gear trains are useful for transmitting high velocity ratios with gears of moderate size in a comparatively lesser space. The epicyclic gear trains are used in back lathes, differential gears of the automobiles, hoist, pulley block, etc. A simple setup of an epicyclic gear train is shown, where gear A and arm C have common axis at 𝑂1 about which they can rotate. If the arm is fixed, the gear train is simple and gear A can drive gear B or vice-versa, but if gear A is fixed and the arm is rotated about the axis of gear A, then the gear B is forced to rotate upon and around gear A. Simple setup of an epicyclic gear train. EPICYCLIC GEAR TRAIN M ACHINE ELEMENTS The velocity ratio of an epicyclic gear train can be determined by two methods: 1. Tabular method 1. Tabular method. Consider the epicyclic gear shown on the right. Let 2. Algebraic method 𝑇𝐴 = number of teeth on gear A 𝑇𝐵 = number of teeth on gear B Step 1. Suppose the arm is fixed, therefore the axis of both the gears are also fixed relative to each other. When gear A makes one revolution counterclockwise (+1) , the gear B will make −𝑇𝐴 / 𝑇𝐵 revolutions clockwise (by convention counterclockwise is positive (+) while clockwise is negative (-)). Enter this on the first row of the table. Step 2. If the gear A makes +𝑥 revolutions, then the gear B will make −𝑥(𝑇𝐴 / 𝑇𝐵) revolutions. Enter this on the second row of the table. In other words, multiply each motion entered in the first row by 𝑥. Step 3. Each element is given +y revolutions and entered in the third row. Step 4. Add the motion of each element of the gear train and enter this on the fourth row. When the two conditions about the motion of rotation of any two elements are known, then the unknown speed of the third element may be obtained by substituting the given data in the third column of the fourth row. Revolutions of elements Step No. Conditions of motion Arm C Gear A Gear B Ty L. Arm fixed-gear 4 rotates through + 1 0 aa a revolution i.e. 1 rev. anticlockwise B Ty 2. Arm fixed-gear 4 rotates through + x 0 +x aes Ts revolutions B 3. Add + y revolutions to all elements ty ty ty xx 7A 4. Total motion +y x+y y Ts ACHINE ELEMENTS EPICYCLIC GEAR TRAIN M ACHINE ELEMENTS EXAMPLE 1 ∴ Speed of gear B, 𝑁𝐵 = 𝑦 − 𝑥 𝑇𝐴 𝑇𝐵 = 150 + 150 36 45 = +270 𝑟𝑝𝑚 = 𝟐𝟕𝟎 𝒓𝒑𝒎 (𝐚𝐧𝐭𝐢𝐜𝐥𝐨𝐜𝐤𝐰𝐢𝐬𝐞) Take note that by convention anticlockwise is positive (+) and clockwise is negative (−). Speed of gear B when gear A make − 300 rpm: Since the gear A makes 300 rpm clockwise, therefore from the fourth row of the table, 𝑥 + 𝑦 = −300 𝑜𝑟 𝑥 = −300 − 150 = −450 rpm ∴ Speed of gear B, 𝑁𝐵 = 𝑦 − 𝑥 𝑇𝐴 𝑇𝐵 = 150 + 450 36 45 = +510 𝑟𝑝𝑚 = 𝟓𝟏𝟎 𝒓𝒑𝒎 (𝐚𝐧𝐭𝐢𝐜𝐥𝐨𝐜𝐤𝐰𝐢𝐬𝐞) EPICYCLIC GEAR TRAIN M ACHINE ELEMENTS EXAMPLE 1 Algebraic Method 𝑁𝐴 = speed of gear A.Let 𝑁𝐵 = speed of gear B. 𝑁𝐶 = speed of gear C. Assuming arm C to be fixed: = 𝑁𝐴 − 𝑁𝐶Speed of gear A relative to arm C: = 𝑁𝐵 − 𝑁𝐶Speed of gear B relative to arm C: Since gears A and B revolve in opposite directions, ∴ 𝑁𝐵 −𝑁𝐶 𝑁𝐴 −𝑁𝐶 = − 𝑇𝐴 𝑇𝐵 Speed of gear B when gear A is fixed: When gear A is fixed, the arm rotates at 150 rpm in the anticlockwise direction, i.e. 𝑁𝐴 = 0, and 𝑁𝐶 = +150 𝑟𝑝𝑚 Eq. i From Eq. 1 ∴ 𝑁𝐵 − 150 0 − 150 = − 36 45 = −0.8 𝑵𝑩 = −150 −0.8 + 150 = 120 + 150 = 𝟐𝟕𝟎 𝒓𝒑𝒎 Speed of gear B when gear A makes − 300 rpm clockwise: When gear A is fixed, the arm rotates at 150 rpm in the anticlockwise direction, i.e. 𝑁𝐴 = −300 𝑟𝑝𝑚 ∴ 𝑁𝐵 − 150 0 − 150 = − 36 45 = −0.8 ; 𝑵𝑩 = 𝟓𝟏𝟎 𝒓𝒑𝒎 EPICYCLIC GEAR TRAIN In a reverted epicyclic gear train, the arm A carries two gears B and C and a compound gear D-E. The gear B meshes with gear E and the gear C meshes with gear D. The number of teeth on gears B, C and D are 75, 30, and 90 respectively. Find the speed and direction of gear C when B is fixed and the arm A makes 100 rpm clockwise. EXAMPLE 2 Solution 𝑇𝐵 = 75 𝑇𝐶 = 30 𝑁𝐴 = 100 rpm clockwise𝑇𝐷 = 90 Solve first for the number of teeth on gear E. 𝑇𝐵 + 𝑇𝐸 = 𝑇𝐶 + 𝑇𝐷 ∴ 𝑇𝐸 = 𝑇𝐶 + 𝑇𝐷 − 𝑇𝐵 = 30 + 90 − 75 = 45 The table of motions is drawn as follows: Since gear B is fixed, therefore from the 4th row of the table, 𝑦 − 𝑥 𝑇𝐸 𝑇𝐵 = 0 𝑜𝑟 𝑦 − 𝑥 45 75 = 0