AG31201 Module 2 Mechanical Power Transmission and Power Take-Off Drives PDF

15-10-2024 AG31201 Module 2 Mechanical Power Transmission and Power Take-Off Drives 1  15-10-2024 Bearings & Seals Belt Drives Chain Drives Power take off dev...

15-10-2024 AG31201 Module 2 Mechanical Power Transmission and Power Take-Off Drives 1  15-10-2024 Bearings & Seals Belt Drives Chain Drives Power take off device 2  15-10-2024 Bearings & Seals A mechanical member which supports load providing a relative motion Turbocharger Threshing drum Car wheel S. Sahoo, Ph.D. Engine Indian Institute of Technology Kharagpur, INDIA 3  15-10-2024 Bearings & Seals Bearing Classifications Sliding contact / Plain bearing: sliding between fixed and rotating member, leading to friction, and heating Rolling Contact bearing/ Anti friction bearing: balls/rollers interposed between fixed and rotating member Radial bearing: Cycle pedal Axial/Thrust bearing: Drone fan Source: Mechanical Design Engineering S. Sahoo, Ph.D. Handbook Indian Institute of Technology Kharagpur, INDIA 4  15-10-2024 Bearings & Seals Bearing Classifications Solid bush bearing: Split bush bearing: Source: Mechanical Design Engineering S. Sahoo, Ph.D. Handbook Indian Institute of Technology Kharagpur, INDIA 5  15-10-2024 Bearings & Seals Bearing Classifications Boundary lubricated: No lubricant present (low speed, and heavy load) Mixed film: lubricant is present, working surface partially contact each other at least part of the time (Engine parts in splash lubrication system) Hydrodynamic: working surfaces are completely separated (large propulsion m/c) Hydrostatic: pressurized lubricating oil is supplied to support steady load, without relative motion ( pressurized lubrication system) Magnetic bearing: supports a load using magnetic levitation (compressor, AC dyno) Source: Mechanical Design Engineering S. Sahoo, Ph.D. Handbook Indian Institute of Technology Kharagpur, INDIA 6  15-10-2024 Bearings & Seals Hydrodynamic bearing Source: Machine Design Book S. Sahoo, Ph.D. by Robert L Norton Indian Institute of Technology Kharagpur, INDIA 7  15-10-2024 Bearings & Seals Bearing material properties Tin babbit Lead babbit Lead bronze Copper lead Source: Machine Design Book by R.S. Khurmi S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 8  15-10-2024 Bearings & Seals Lubricants Objectives: Reduce friction Carry away heat released due to friction Protection against corrosion Types: Liquid e.g.: mineral oil, synthetic oils Semi-liquid e.g.: grease ( high viscosity, high pressure, no oil dripping) Solid e.g.: graphite (under extreme temperature & pressure) Properties: Viscosity Oiliness Density Flash/Fire point Pour/ Freeze point S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 9  15-10-2024 Bearings & Seals Design of Plain bearings D = diameter of bearing d = diameter of journal L = length of bearing Z = Absolute viscosity of the lubricant, in kg/m-s, N = Speed of the journal in rpm., p = Bearing pressure on the projected bearing area in N/mm2, (Load on the journal ÷ l × d) Diametral clearance (c) = D – d (0.025mm per cm of journal diameter) Radial clearance (c1) = R – r Diametral clearance ratio = c/d Bearing characteristics number = ZN/p Source: Machine Design Book S. Sahoo, Ph.D. by Robert L Norton Indian Institute of Technology Kharagpur, INDIA 10  15-10-2024 Bearings & Seals Design of Plain bearings McKee relation S. Sahoo, Ph.D. k= 0.002 for l / d ratios of 0.75 to 2.8 Source: Machine Design Book by Robert L Norton Indian Institute of Technology Kharagpur, INDIA The Tribology Handbook 11  15-10-2024 Bearings & Seals Design value for journal bearings Heat Generated in a Journal Bearing = Heat dissipated in a Journal Bearing = Heat taken away by lubricating oil = S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 12  15-10-2024 Bearings & Seals Design value for journal bearings S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 13  15-10-2024 Bearings & Seals The load on the journal bearing of a centrifugal pump is 20 kN, with a shaft speed of 900 r.p.m.. SAE10 (absolute viscosity at 55°C = 0.017 kg / m-s) is used as the lubricating oil. Design the journal bearing and calculate the mass of oil required for cooling if maximum temperature rise is limited to 10°C. Ambient temperature of oil = 15.5°C ; Heat dissipation coefficient = 1232 W/m2/°C Maximum bearing pressure= 1.5 N / mm2. Answer: W = 20 kN, N = 900 rpm, Toil=55° C, Z = 0.017 kg/m-s, Tchange = 10°C, Tambient = 15.5°C, C = 1232 W/m2/°C, Pmax = 1.5 N/mm2 Assuming, diameter of journal, d = 100 mm, From table, 1< l/d 15 m/s) = = ⅇ𝑢𝜃 𝑇𝑡2 − 𝑇c 𝑇2 𝑤𝑏𝑡𝑣 2 w = weight of belt per 𝑇c = 106 𝑔 unit length in N, Maximum power transmitted by belt = 𝑇1 − 𝑇2 𝑣 𝑏𝑡𝑣 𝑤𝑣 2 𝑒 𝑢𝜃 − 1 𝑃= = 𝜎𝑑 − 6 1000 1000 10 𝑔 ⅇ𝑢𝜃 S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 26  15-10-2024 Belt Drive Pulley Dimension of pulley: 𝝈𝒕 = 𝝆𝒗𝟐 𝜎𝑡 : Pulley rim tensile stress 𝜌 = Density of the rim material, 7200 kg/m3 for cast iron 𝑣 = Velocity of rim. (pi*D*N)/60, D = dia of rim/pulley As per Indian standards, The following are the diameters of pulleys in mm for flat and V-belts (in mm). 20, 22, 25, 28, 32, 36, 40, 45, 50, 56, 63, 71, 80, 90, 100, 112, 125, 140, 160, 180, 200, 224, 250, 280, 315, 355, 400, 450, 500, 560, 630, 710, 800, 900, 1000, 1120, 1250, 1400, 1600, 1800, 2000, 2240, 2500, 2800, 3150, 3550, 4000, 5000, 5400. The first six sizes (20 to 36 mm) are used for V-belts only S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 27  15-10-2024 Belt Drive Pulley Width of pulley (B): 𝑩 = 𝟏 ⋅ 𝟐𝟓𝒃 𝑏: width of the belt Width of flat cast iron and mild steel pulleys in mm : 16, 20, 25, 32, 40, 50, 63, 71, 80, 90, 100, 112, 125, 140, 160, 180, 200, 224, 250, 315, 355, 400, 450, 560, 630. Thickness of pulley rim (tr): 𝑫 + 𝟑 mm for single belt 𝟐𝟎𝟎 𝑫 + 𝟔 mm for double belt 𝟐𝟎𝟎 Hub: Diameter of the hub ( d1 ) = 1.5 d +25 mm (should be < 2d) Length of the hub ( L ) = (pi/2) d (2/3B < L 450 mm, number of arm= 6 1Τ3 𝑏𝐷 𝑡𝑎𝑏 = 2 ⋅ 94 Thickness of the elliptical arm near boss, for single belt, 4𝑛𝑎 1Τ3 𝑏𝐷 𝑡𝑎𝑟 = 2 ⋅ 94 Thickness of the elliptical arm near rim, for single belt, 2𝑛𝑎 S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 29  15-10-2024 Belt Drive S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 30 15-10-2024 31  15-10-2024 Belt Drive V belts Wedging action & high friction For small distances, where flat belt slips and take large space, V belts perform well due to its wedging action High power transmitting capacity, and pulsating loads (ic engine or compressor) Cushions shock load, do not require lubrication, and less critical to mis-alignment Can not be used for large distance due to high weight per unit length S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 32  15-10-2024 Belt Drive V belts S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 33  15-10-2024 Belt Drive V belts S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 34  15-10-2024 Belt Drive V belts Ratio of driving tensions: 𝑇1 − 𝑇c 𝑢 = ⅇ𝑢𝑎𝜃 𝑢𝑎 = 𝑇2 − 𝑇c sin 𝛽 S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 35  15-10-2024 Tutorial Question 1: An open belt drive system transmits 5 kW power using a flat belt of width and thickness as 80 mm and 5 mm, respectively. The driving shaft speed is 1500 rpm and the driven shaft speed is 500 rpm. The coefficient of friction between the belt and pulley is 0.20, and the wrap angle of the belt is 168°. If diameter of the smaller pulley is 200 mm, maximum stress induced in the belt will be _________ MPa. Question 2: An open V belt is wrapped around V pulleys having effective diameters of 0.25 m and 0.65 m, and their centers are 1 m apart. Assuming ideal conditions, the wrap angle in degree for the smaller pulley is _______________. [ round off to two decimal places] Question 3: A V belt drive transmits 10 kW power at a belt velocity of 8 ms-1. The angle of contact on the smaller pulley is 170° and the groove angle of the pulley is 38°. The coefficient of friction between the pulley and the belt is 0.28 and the maximum permissible stress is limited to 4 Mpa. Negelcting centrifugal effect of the belt, the minimum cross- sectional area of the V belt in mm2 _________. S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 36  15-10-2024 Chain Drive Chain drive Used for short-distance power transmission (max up to 8 meters) Maximum velocity up to 25 m/s Maximum power up to 110 kW 1. No slip unlike belt drive 2. Maintains constant velocity ratio (8 to 10 in one step) 3. Transmits motion to several shafts at a time by one chain only 4. Can operate at extreme conditions of temp & pressure 5. High cost 6. Maintenance (lubrication, slack adjustment) 7. Velocity fluctuation when improperly stretched S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 37  15-10-2024 Chain Drive Classification of chain drive (based on its use) Hoisting & Hauling Chains Chain with oval Chain with square links links (kinking) Conveyor Chains (Malleable CI, 0.8-3 m/s) Hook type Detachable chains Closed Joint type https://www.youtube.com/watch?v=q6PJF_kHdzI S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 38  15-10-2024 Chain Drive p : Pitch of chain D : Pitch circle dia of chain sprocket d1 :dia of chain roller T : No. of teeth D0: Sprocket outside diameter 𝑁1: Smaller sprocket speed (rpm) 𝑁2 : Larger sprocket speed (rpm) 𝑇1 : No. of teeth on smaller sprocket 𝑇2 : No. of teeth on larger sprocket 𝑵𝟏 𝑻𝟐 𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐫𝐚𝐭𝐢𝐨 = = A, B, C, D are the hinge centres of the 𝑵𝟐 𝑻𝟏 chain 𝟏𝟖𝟎𝟎 𝑫 = 𝒑 𝒄𝒐𝒔𝒆𝒄 𝑻 𝟑𝟔𝟎𝟎 𝜽= 𝑻 𝑫𝟎 = 𝑫 + 𝟎. 𝟖𝒅𝟏 Sprocket outside diameter S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 39  15-10-2024 Tutorial Question 1: A chain drive is used to transmit power from a DC motor to the shaft of a solar energy operated thresher by reducing the speed from 240 rpm to 120 rpm. The number of teeth on the driving sprocket is 20 and the pitch circle diameter of the driven sprocket is 600 mm. The pitch of the chain used, in mm, is _________. (Rounded off to 2 decimal places) S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 40  15-10-2024 Chain Drive Length of the chain (L) = pitch of the chain (p) * number of links (K) 𝑇1 +𝑇2 2𝑥 𝑇2 −𝑇1 2 𝑝 Number of links (K) = + + (approximated to nearest even number) 2 𝑝 2∗𝜋 𝑥 2 2 𝑝 𝑇1 + 𝑇2 𝑇1 + 𝑇2 𝑇2 − 𝑇1 Center to center distance (x) = 𝐾− + 𝐾− −8 4 2 2 2∗𝜋 To accommodate initial sagging of the chain, 2-5 mm reduction in center to center distance is done. Xmin may be taken as (d1+d2)/2+30 to 50 mm (d1 and d2 are pitch circle dia of smaller and larger sprockets) Xmin may be taken 30 to 50 times pitch S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 41  15-10-2024 Chain Drive Factor of safety is the ratio of breaking strength of the chain to the total 𝒘𝑩 load on driving side = 𝒘 𝒘𝑩 for roller chain = 106*p2 newton 𝒘 = 𝑭𝑻 + 𝑭𝑪 + 𝑭𝒔 = tangential driving force + centrifugal tension force + force due to sagging = (P/v) + (mv2) + (k * mg * x) P= power transmitted m = mass of chain in kg per meter length x = center to center distance k = 2 to 6, when the center line of the chain is inclined to the horizontal at an angle less than 40º k = 1 to 1.5, when the center line of the chain is inclined to the horizontal at an angle greater than 40º S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 42  15-10-2024 Chain Drive S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 43  15-10-2024 Chain Drive Tutorial: Design a chain drive to actuate a compressor from 15 kW electric motor running at 1000 r.p.m., the compressor speed being 350 r.p.m. The minimum centre distance is 500 mm. The compressor operates 16 hours per day. The chain tension may be adjusted by shifting the motor on slides. Solution: Rated Power = 15 kW, N1 = 1000 rpm, N2 = 350 rpm, x = 500 mm, Hour of operation per day = 16 Step 1: Calculate velocity ratio, i.e.; N1/N2 = 1000/350 = 2.85 ≈ 3 Step 2: Select the minimum number of teeth on the smaller sprocket or pinion from Table 21.5 Step 3: Find the number of teeth on the larger sprocket, i.e, 3*(1000/350) ≈ 72 teeth Step 4: Determine the design power by using the service factor, such that Design power = Rated power × Service factor S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 44  15-10-2024 Chain Drive Design Power = 15 * (1.5*1*1.25) = 28.125 kW Step 5: Choose the type of chain, number of strands for the design power and r.p.m. of the smaller sprocket from Table 21.4 ( two strand will be used of 15.65kW, 12 B) Step 6: Note down the parameters of the chain, such as pitch, roller diameter, minimum width of roller etc. from Table 21.1. S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 45  15-10-2024 Chain Drive Step 7: Find pitch circle diameters and pitch line velocity of the smaller sprocket d1 = p cosec (180/T) = 152 mm, d2 = 436 mm, v1=(pi*d1*N1/60) = 7.96 m/s Step 8: Determine the load (W) on the chain by using the following relation, i.e. W = Rated power / Pitch line velocity 15 kW/ 7.96 m/s = 1.844 kN, Step 9: Calculate the factor of safety by dividing the breaking load (WB) to the load on the chain ( W ). This value of factor of safety should be greater than the value given in Table 21.2. i.e, 57.8/1.844 ≈ 32 S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 46  15-10-2024 Chain Drive Step 7: Fix the centre distance between the sprockets. Xmin may be taken 30 to 50 times pitch. i.e., 30*19.05 = 572 mm Taking sagging condition, 572 - 4= 568 mm Step 8: Determine the length of the chain Number of links: ≈ 110 Length = 19.05*110 ≈ 2096.6 mm ≈ 2.096 m S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 47  15-10-2024 Assignment Assignment: A chain drive using bush roller chain transmits 5.6 kW of power. The driving shaft on an electric motor runs at 1440 r.p.m. and velocity ratio is 5. The centre distance of the drive is restricted to 550 ± 2% mm and allowable pressure on the pivot joint is not to exceed 10 N/mm2. The drive is required to operate continuously with periodic lubrication and driven machine is such that load can be regarded as fairly constant with jerk and impact. Design the chain drive by calculating leading dimensions, number of teeth on the sprocket and specify the breaking strength of the chain. Assume a factor of safety of 13. S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 48  15-10-2024 Overload Safety Devices Types of safety devices a) device that depends upon the shearing of a replaceable connecting member in the drive line b) device that used the principle of spring force to hold two members c) devices work on friction to hold two members in the drive line Shear devices Shear key between shaft and hub Flange-mounted shear pin Power = Torque * w = ( ShearForce * radial _ dis tan ce) * w  = ( S s *( d12 ) * rs ) * w 4 S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA Diametral shear pin through hub & shaft 49  15-10-2024 Tutorial Tutorial 1: Specify the location of a 2.38mm flange-mounted steel shear pin (Ss=310 MPa) which will fail at 4.5kN when the speed is 650 rpm. Tutorial 2: What size diametral shear pin would be required if the shaft diameter is 25 mm Tutorial 3: A mild steel flange-mounted single shear pin (Ultimate shear strength= 42 MPa) of 10 mm diameter is used in a flange. The perpendicular distance between the axis of the driving shaft and the shear pin axis is 100 mm. If the speed of the driving shaft is 300 rpm, the maximum power of the shaft could transmit in kW before the failure of the pin is ___________. S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 50  15-10-2024 Overload Safety Devices Universal Joints Cardan or Hooke Joint 1 = angle of rotation of the driving shaft from the initial position tan 2 2 = angle of rotation of the driven shaft = cos  tan 1  = joint angle depends on the type of drive, the torsional flexibility of the shafts, the inertia of the rotating parts, the speed, and the desired life. cos  2 = 1 1 = angular velocity of the input (driving) shaft 1 − sin 2  sin 2 1  = angular velocity of the output (driven) shaft 2 Assignment: Determine ∅𝟏 for maximum & minimum value S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA of (𝝎𝟐 / 𝝎𝟏 ). Also Maximum and Minimum (𝝎𝟐 / 𝝎𝟏 ) 51  15-10-2024 Overload Safety Devices S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 52  15-10-2024 Overload Safety Devices Universal Multijointed Combinations Single Universal Joint Double Universal Joint W Configuration Z Configuration Angular alignment Parallel alignment Angular displacement and velocity fluctuations will cancel if the joint angles are equal Telescopic Driveshaft Angular, Parallel, Axial alignment S. Sahoo, Ph.D. Front Wheel Drive https://www.youtube.com/watch?v=LSbmrUp3Aas Indian Institute of Technology Kharagpur, INDIA Constant Velocity Joints https://www.youtube.com/watch?v=auQ65qno2Eo 53  15-10-2024 Overload Safety Devices Power take off (PTO) drives Transfers power from the tractor or other source of power to an implement Rated speed: 540 rpm (6 spline) 1000 rpm (21 spline) Internal Yoke, External Yoke, Universal Joint, Safety Chain, Safety Shield Transmission driven PTO runs only when the engine tractor is in motion (transmission & PTO controlled by a single clutch) Continuous running Internal Operates when the master clutch is engaged through a single clutch pedal (possible Yoke to stop the tractor) Independent PTO drives Separate clutch for Transmission & PTO S. Sahoo, Ph.D. External Indian Institute of Technology Kharagpur, INDIA Safety Shield Yoke 54  15-10-2024 Overload Safety Devices Power take off (PTO) drives standardization S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 55  15-10-2024 Overload Safety Devices Friction Clutch During normal operations (when there is no torque overload), the limiter uses a friction disc to transmit torque from the driving shaft to the driven member, such as pulleys and sprockets. When the torque limit is exceeded, the friction force is no longer strong enough to transmit the torque, and the driven member slips between the friction discs. After the torque level goes down to the recommended rates, the friction plates resume transmitting rotating motion automatically. S. Sahoo, Ph.D. Indian Institute of Technology Kharagpur, INDIA 56