1
2 Power transmission chains have two things in common; side bars or link plates, and pin and bushing joints. The chain articulates at each joint to operate around a toothed sprocket. The pitch of the chain is the distance between the center of the articulating joints Power-transmission chains have several advantages: relatively unrestricted shaft center distances, compactness, ease of assembly, elasticity in tension with no slip or creep, and ability to operate in a relatively high-temperature atmosphere. Basic Types -There are six major types of power- transmission chains, with many modifications and special shapes for specific applications. A seventh type, the bead chain, is often used for light-duty applications. Pitch
3 Chain types
4 Basic Types Detachable Pitch range from.902 4.063in 23 to 103 mm and ultimate strength of 700 to 17,000 psi. Made in steel, does not require lubrication. Used in farm machinery. The ends of the detachable link are referred to as bar end & hook end Pintle Speeds to about 450 ft/min (2.2m/s) and heavier loads. Made up of individual cast links having a full, round barrel end with offset sidebars: inter-coupled with steel pins. Ends of pintle chain links referred as the barrel end and open end. Pitch range from (25 to 150mm) and strength of 3600 to 30,000 psi (25 200MPa)
5 Basic Types Offset-Sidebar Used widely as drive chain on construction machinery. Operate at speeds to 1000 ft/min (5 m/s) and transmit loads to about 250 hp (185 kw). Each link has two offset sidebars, one bushing, one roller, one pin, and if the chain is detachable, a cotter pin. Some offset-sidebar chains are made without rollers
6 Roller: Basic Types Pitch range from.25 3.00in (6 to 75mm) pitch and strength of 925 to 130,000psi (6-900MPa) in single width. Available in multiple widths Small pitch sprocket Can operate at 10,000rpm at 1000 to 1200hp. Assembled with roller links and pin links, if detachable, cotter pins are used Self lubrication is possible
7 Basic Types Double-Pitch Same as roller chains except that t the pitch is twice as long. Have the same diameter pins and rollers, the same width rollers, and the same thickness of link plates. Inverted-Tooth (Silent) High-speed chains, used predominantly for prime-mover, power takeoff drives, such as on power cranes or shovels, machine tools, and pumps. Drives transmitting up to 1200 hp (900 kw) areinuse
8 Sprockets Basic sprocket types used with precision steel roller chains conform to ANSI standards Used for mounting flanges, hubs or other devices, the plate sprocket is a flat, hubless sprocket Small and medium sized hub sprockets are turned from bar stock or forgings or, are made from welding a bar-stock hub to a hot rolled plate For small, light load applications, one hub extension is needed and large sprockets have two hub projections equidistant from the center plane They are made from gray iron castings or cast steel Sprockets made of sintered metal and plastics are economical Plastics offer advantage as they require minimum lubrication and are used when cleanliness is essential
9 Design of Roller Chain Drives The design of a roller chain drive consists: Selection of the chain and sprocket sizes; determination of chain length, center distance, method of lubrication; and arrangement of chain casings and idlers. Belt drives based on linear speeds in fpm or m/s. Chain drives based on rotating speed (rpm) of the smaller sprocket (driver member in most installations). Design of chain drives based on the following factors: Average hp (kilowatts) to be transmitted (Fig. 20-2-5), RPM of the driving and driven members, Shaft diameter. Permissible diameters of sprockets.
10 Design of Roller Chain Drives Load characteristics, whether smooth and steady, pulsating, heavy-starting, ti or subject to peaks. Fig. 20-2-9. 2 Lubrication, whether periodic, occasional, or copious. When chains are exposed to dust, dirt, or injurious foreign matter, chain cases should be used Life expectancy: the amount of service required or the Life expectancy: the amount of service required, or the total life. It is much better to "over-chain" than to skimp on the size of the chain used.
11 Design of Roller Chain Drives The rpm and size of the smaller or faster moving sprocket determines the chain size. Smaller pitch chains in single or multiple l widths are adaptable for increased speed drives and for any speed drives where smooth (quiet) performance is needed Large pitch chains are adaptable for slow and medium speed drives Multiple-width roller chains popular, operate at different speeds with high loads and smooth in action (low noise).
12 Design of Roller Chain Drives Size of Sprockets - Use a minimum size sprocket of 17 teeth to obtain smooth operation at high speeds. 19 or 21-tooth sprockets used for greater life expectancy and smoother operation (lesser tooth impact slower speed). On slow-speed and special-purpose installations or where space limitations are involved, sprockets smaller than 17-tooth can be used. The normal maximum number of teeth is 120. Ordinary practice indicates: Ratio of driver to driven sprockets should be no more than 6 to 1. Recommended chain wrap on the driver is 120.
13 Design of Roller Chain Drives Chain Tension Chains should never run with both sides tight. Provide adjustable centers to permit proper initial slack and to allow for periodic adjustment due to natural chain wear. The chain sag should be equivalent to approximately 2% of the center distance (Fig. 20-2-7). Idler sprockets used as a means of taking up chain slack when it is not possible to provide adjustable centers.
14 Design of Roller Chain Drives Center Distances Best results are obtained by using a center distance of 30 to 50 times the pitch of the chain used. 80 times the pitch is considered maximum. Chain Length Function of the number of teeth in both sprockets and of the center distance. Chain must consist of an integral number of pitches, with an even number preferable to avoid the use of an offset link.
15 Design of Roller Chain Drives
16 Design of Roller Chain Drives Chain Length Formula - Compute the chain length in terms of chain pitches and then to multiply pythe result by the chain pitch to obtain the length in inches (millimeters). (Fig. 20-2-8). 1. Divide the center distance in inches (millimeters) by the pitch of the chain, obtaining C. 2. Add the number of teeth in the small sprocket to the number of teeth in the large sprocket, obtaining M. 3. Subtract the number of teeth in the small sprocket from the number of teeth in the large sprocket, obtaining value F to obtain the corresponding value of S. M S 4. Chain length in pitches equals: 2 C + + 2 C Round-off the length to the next higher integer, preferable even number. The center distance must then be corrected. 5. Multiply the number of pitches by the chain pitch used in order to get the chain length in inches (millimeters).
17 Design of Roller Chain Drives
18 Design of Roller Chain Drives Drive Selection The horsepower (kilowatt) ratings relate to the speed of the smaller sprocket,, and drive selections are made on this basis (speed- reducing or speed-increasing). Dependent on loads imposed on the chain by the type of input power and the type of equipment to be driven. Service factors are used to compensate for these loads.
19 Design of Roller Chain Drives Required horsepower (kw) rating of the chain(fig. 20-2-9): Required hp (kw) rating = hp (kw) to be transmitted X service factor Multiple strand factor Figures 20-2-10 to 20-2-13 (pp. 754-757) show the hp and kw ratings for few of the many roller chains available. For more info, refer to catalogs. The hp and kw rating charts (Figs. 20-2-14 and 20-2-15) provide a quick means of finding the probable chain requirements (See pp. 758 and 759.)
20 Chain Drive Design Example 1 Select aroller chain di driven by anelectric motor to transmit 5 hp from a countershaft to the main shaft of a wiredrawing machine. The countershaft is 1.5 in. in diameter and operates at 1200 rpm. The main shaft is also 1.5 in. and must operate between 378 and 382 rpm. Shaft centers, once established, are fixed, and by initial calculations must be approximately 22.5 in. The load on the main shaft is uneven and presents peaks that places it in the heavy-shock load category. Solution Step 1: Service Factor - The corresponding service factor from Fig. 20-2-9 for heavy-shock load and electric motor is 1.5
21 Chain Drive Design Solution: Step 1: Service Factor Corresponding service factor from Fig. 20-2-9 for heavy-shock load and electric motor = 1.5 Step 2: Design Horsepower Design horsepower = 5 X 1.5 = 7.5 hp
22 Chain Drive Design
23 Chain Drive Design Step 3: Tentative Chain Selection Given the sprocket speed (driver) = 1200rpm, and design horse power = 7.5hp., using the horsepower rating chart (Fig. 20-2-14) yields the chain number to be 40. Given chain number = 40, the chain pitch =.50in from Fig. 20-2-12. Multiple-strand chain selected but using single strand data, required horsepower rating per strand is: Required horsepower rating = Multiple strand factor from Fig.20-2-9. Design hp multiple-strand factor Or refer to the right-handhand columns shown in Fig. 20-2-14 2 or 15.
24 Chain Drive Design
25 Chain Drive Design Step 4: Final Selection of Chain and Small Sprocket Given chain No. = 40, small sprocket speed =1200 rpm, the computed design = 7.5 hp. Using Fig. 20-2-12. Go down the column headed by the small sprocket speed of 1200 rpm and find the nearest value to the design horsepower 7.69hp. Follow this line horizontally to the left to find the number of teeth for the small sprocket = 20. For intermediate speeds or sprocket sizes not tabulated, interpolate between the appropriate columns or lines.
26 Chain Drive Design 20-2-12
27 Chain Drive Design Check the maximum bore for the selected sprocket : Given chain pitch =.50in, number of small sprocket teeth = 20. Using Fig. 20-2-16. 2 Go down the.50in chain pitch column and go along the 20 teeth row, the intersection = maximum small sprocket bore diameter = 1.78in. Greater that the given 1.50in shaft diameter (Countershaft) Hence sprocket can accommodate the 1.50in shaft. If the selected sprocket will not accommodate the shaft, use a larger sprocket or make a new sprocket and chain selection from the rating table for the next-larger chain number.
28 Chain Drive Design.
29 Chain Drive Design Step 5: Selection of the Large Sprocket Driver speed = 1200 rpm ; Driven speed at a min of 378 rpm; Speed ratio = 1200/378 = 3.17:1 1 min. Large sprocket = 20 X 3.17 = 63.4 teeth Standard sprocket sizes near this number of teeth are either 60 or 70 teeth (Fig. 20-2-17). More economical and time-saving to try to use a combination of standard sprockets.
30 Chain Drive Design Rechecking smaller sprocket, 19-tooth sprocket also OK. (Drop safety margin) Large sprocket of 19 X 3.17 teeth = 60.2 teeth (use 60 teeth). 19- and 60-tooth sprockets are acceptable and standard. More economical to use this combination.
31 Chain Drive Design Step 6: Chain Length in Pitches 19- and 60-tooth sprockets are to be placed on 22.5-in. centers. Chain length in pitches = M S 2C + + where 2 C C = center distance pitch = 22.5.5 = 45 M = total number of teeth on both sprockets = 19 + 60 = 79 F = 60-19 = 41 S = 42.58 (value obtained from table 20-2-8) Substituting values for C, M, and S, we get Chain length in pitches = 79 42.58 2 45 + + = 130.44 2 45 The chain is to couple to an even number of pitches, use 130 pitches because the leeway on the 22.5 in. centers is not critical
32 Chain Drive Design Step 7: Chain Length in Inches (Millimeters) Length of chain = number of pitches X pitch = 130 X.5 = 65 in. For metric chain problems the same procedures are followed except that t Figs. 20-2-10, 20-2-11, and 20-2-15 replace Figs. 20-2-12, 2 20-2-13, 2 and 20-2-142
33 Chain Drive Design
34 Metric units Table
35 Chain Drives Metric units Table
36 Chain Drives Metric units Table
37 Metric units Figure
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39 Gear Drives Gear Drives transmit motion (rotary or reciprocating) from one part to another Reduce or increase revolutions of shaft Gears cylinders or cones having teeth (for positive motion) Most durable (used in automotive and heavy duty Based on shaft position, gears are classified as Spur, Bevel, Worm and Rack & Pinion
40 Gear Drives Spur Gears Spur gear proportions and the shape of gear teeth are standardized, and the definitions, symbols, and formulas are given in Figs. 20-3-3 to 20-3-6 Gears used to transmit motion at constant angular velocity The form of the gear that produces this constant velocity is involute
41 Gear Drives Spur Gears Involute curve traced by a point on a string unwinding from a circle (base circle) Every involute gear has one base circle from which all involutes are generated (not a physical part of the gear) Line of Action tangent to and crosses the base circles at contact Pressure angle - (also referred to as tooth shape ) is the angle at which the pressure from the tooth of one gear is passed on to the tooth of another gear. Spur gears come in two PA s: 14.5º and 20º http://en.wikipedia.org/wiki/involute_gear
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43 Gear Drives Drawing Gear Teeth Exact form is time consuming, approximate methods are used and the two most common methods are shown in the figure below First layout root, pitch and outside circles. Mark circular thickness on the pitch circle. With the line tangent to pitch circle draw pressure line (14.5 or 20 )
44 Gear Drives Drawing Gear Teeth Draw the base circle tangent to this pressure line With radius = 1/8 th the pitch dia and center on base circle draw arcs through the circular thickness, from base circle to top of the teeth Below the base circle, the teeth is drawn with a radial line ending in a fillet to the root diameter
45 Gear Drives Drawing Gear Teeth More closer approximation use grants representation R and r for inch gears value divided by the diametral pitch; for metric, the value times the module
46 Gear Drives Working Drawing of Spur Gears Sectional view is enough. If web or arm details are needed, the front view is drawn Teeth need not be shown in the front view, phantom lines for outside and root circles and center line for pitch circle to be used (ANSI) In section view, root and outside circle are shown as solid lines Gear blank is dimensioned in the drawing and teeth information in table as they are made by 2 different processes
47 Gear Drives Working Drawing of Spur Gears
48 Gear Drives Working Drawing of Spur Gears The only differences in terminology between inch and metric-size gear drawings are the terms diametrical pitch (DP) and module (MDL) For inch gears, DP is used. DP is a ratio of the number of teeth to a unit length of pitch diameter. DP = N/PD For metric gears, MDL is used. It is the length of pitch diameter per tooth measured in millimeters. MDL = PD/N From these definitions it can be seen that the MDL is equal to the reciprocal of the DP and not its metric dimensional equivalent. If the DP is known, the MDL can be obtained. MDL = 25.4 DP Gears here are designed in the inch and have a standard DP instead of a preferred standard MDL. Therefore, DP is referenced beneath the MDL when gears designed with standard inch pitches are used
49 Gear Drives Working Drawing of Spur Gears For gears designed with standard MDL, the DP need not be referenced on the gear drawing. The standard modules for metric gears are 0.8, I, 1.25, 1.5, 2.25, 3, 4, 6, 7, 8, 9, 10, 12, and 16 For metric gear calculations, the DP is soft-converted to MDL in Fig. 20-3-2 for comparison purposes and for use in assignments Spur Gear Calculation Center Distance - The distance between the two shaft centers is sum of pitch diameter of the two gears divided by 2 Example 1 - A 12-DP, 36-tooth pinion mates with a 90-tooth gear Find the center distance
50 Gear Drives Spur Gear Calculation Pitch diameter (PD) = number of teeth (N) DP For gear 1 (PD) = 36 12 = 3.00 in. (pinion- smaller gear) For gear 2 (PD) = 90 12 = 7.50 in. (gear) Sum of the two PDs = 3.00 in. + 7.50 in. = 10.50 in. Center distance = 10.50/2 = 5.2 Ratio of the gear is the relationship between any of the following: RPM of the Gears Number of teeth on the gears Pitch diameter of the gears The ratio is obtained by dividing idi the larger value of these 3 parameters by the corresponding smaller value
51 Gear Drives Spur Gear Calculation Determining Pitch and Outside diameter Pitch diameter (PD) can be found if number of teeth (N) and diametral pitch (DP) or module (MDL) is known The outside diameter (OD) is equal to the pitch diameter plus 2 addendums The addendum for a gear tooth is 1/DP or the MDL
52 Power Transmission Capacity Spur Gears Gear drives are required to operate under such a wide variety of conditions that it is very difficult and expensive to determine the best gear set for a particular application. The most economical procedure is to select standard stock gears with an adequate load rating for the application Approximate hp (kw) ratings for spur gears of various sizes (numbers of teeth), at several operating speeds (rpm) are given in catalogs with the spur gear listings. Ratings for gear sizes and/ or speeds not listed may be estimated from the values shown on page 768 in Fig. 20-4-1
53 Power Transmission Capacity Spur Gears Pitch-line velocities exceeding 1000 ft/min (5 m/s) for 14.5 PA (pressure angle), or 1200 ft/min (6 m/s) for 20 PA, are not recommended for metallic spur gears. Ratings are listed for speeds below these limits The ratings given (or calculated) should be satisfactory for gears used under normal operating conditions, i.e., when they are properly mounted and lubricated, carrying a smooth load (without shock) for 8 to 10 hours a day The charts shown in Fig. 20-4-1 indicate the approximate hp (kw) ratings of 16- and 20-tooth steel spur gears of several tooth sizes operating at various speeds
54 Power Transmission Capacity Spur Gears They may be used to determine the approximate DP or MDL of a 16- or 20-tooth steel pinion that will carry the hp (kw) required at the desired speed. The intersection of the lines representing values of rpm and hp (kw) indicates the approximate gear DP (MDL) required The number of teeth th normally should not be less than 16 to 20 in a 14.50 pinion, or less than 13 in a 20 pinion Ratings shown for spur gears in catalogs normally are for class 1 service. For other classes of service, the service factors in Fig. 20-4-2 (p. 769) should be used
55 Power Transmission Capacity Selecting the Spur Gear Drive
56 Power Transmission Capacity Selecting the Spur Gear Drive 1. Determine the class of service 2. Multiply the horsepower (kilowatts) required for the application by the service factor 3. Select spur gear pinion with catalog rating hp (kw) determined in step 2 4. Select driven spur gear with a catalog rating hp (kw) determined in step 2
57 Power Transmission Capacity Selecting the Spur Gear Drive Example 1 - Select a pair of 20 spur gears that will drive a machine at 150rpm. Size of driving motor = 25hp, 600rpm. Service factor = 1 Solution - Since service factor = 1, we need not increase or decrease the design hp. Refer charts in Fig. 20-4-1A, which h show design data for 20 spur gears having 16 and 20 teeth. Selecting a pinion having 16 teeth and reading vertically on the column showing 600 rpm to hp rating of 25, we find that required DP = 4. (Select the DP equal to or greater than hp) Pinion: N = 16, DP = 4, PD = 16 4 = 4.000 in. Ratio = 600/150 = 4:1 Gear: N = 16 X 4 = 64, DP = 4, PD = 64 4 = 16.000 in required
58 Power Transmission Capacity Selecting the Spur Gear Drive
59 Power Transmission Capacity Selecting the Spur Gear Drive
60 Power Transmission Capacity Selecting the Spur Gear Drive
61 Power Transmission Capacity Example 3 A 7.5-kW; 900-r/min motor is attached by means of 14.5 spur gears to a punch that operates 24 hours a day. Gear ratio: 4 to 1. Assuming the punch is being operated at motor capacity. Solution: Machine operating conditions = Service class 3 = Service factor of 1.3. Therefore: Kilowatts required for design purposes = 7.5 1.3 = 9.75 kw. Pinion mounted on motor and runs at 900 r/min. Refer to Fig. 20-4-2B. Select either a pinion having a module of 5.08 and N of 20 or a module of 6.35 and N of 16. First, module of 5.08mm: Pinion: N = 20, MDL = 5.08, PD = 101.6 mm. Gear travels at 225 r/min, or one-fourth of pinion revolutions per minute. Gear: N = 4 20=80 80, MDL=508 5.08, PD=80 5 5.08 = 406.44 mm
62 Power Transmission Capacity Selecting the Spur Gear Drive Example 3 Second, using a module of 6.35mm: Pinion: N = 16, MDL = 6.35, PD = 101.6 mm. Gear: N = 64, MDL = 6.35, PD = 406.4 mm. Both sets of gears are of the same diameter, the overall size is not a factor. Extra strength of gear set with module of 6.35 not required. Considerable savings by selecting the gear and pinion having the module of 5.08. Hence gear and pinion with module of 5.08 selected.