Machinery''''s Handbook 27th Episode 1 Part 3 pptx

FORCE SYSTEMS 155 Finding the Resultant of Nonparallel Forces Not Meeting at a Common Point: The diagram shows a system of noncoplanar, nonparallel, noncon- current forces F 1 , F 2 , etc. for which the resultant is to be determined. Generally speaking, the resultant will be a noncoplanar force and a couple which may be further com- bined, if desired, into two forces that are skewed. This is the most general force sys- tem that can be devised, so each of the other systems so far described represents a special, simpler case of this general force system. The method of solution described below for a system of three forces applies for any number of forces. 1) Select a set of coordinate x, y, and z axes at any desired point O in the body as shown in the diagram. 2) Determine the x, y, and z coordinates of any convenient point on the line of action of each force as shown for F 2 . Also determine the angles, θ x , θ y , θ z that each force makes with each coordinate axis. These angles are measured counterclockwise from the positive direction of the x, y, and z axes. The data is tabulated, as shown in the table accompanying Step 3, for convenient use in subsequent calculations. 3) Calculate the x, y, and z components of each force using the formulas given in the accompanying table. Add these components algebraically to get ∑F x , ∑F y and ∑F z which are the components of the resultant, R, given by the formula, Force Coordinates of Force F Components of F Fxyz θ x θ y θ z F x F y F z F 1 x 1 y 1 z 1 θ x1 θ y1 θ z1 F 1 cos θ x1 F 1 cos θ y1 F 1 cos θ z1 F 2 x 2 y 2 z 2 θ x2 θ y2 θ z2 F 2 cos θ x2 F 2 cos θ y2 F 2 cos θ z2 F 3 x 3 y 3 z 3 θ x3 θ y3 θ z3 F 3 cos θ x3 F 3 cos θ y3 F 3 cos θ z3 ∑F x ∑F y ∑F z The resultant force R makes angles of θ xR , θ yR , and θ zR with the x, y, and z axes, respectively, and passes through the selected point O. These angles are determined from the formulas, R ΣF x () 2 ΣF y () 2 ΣF z () 2 ++= θcos xR ΣF x R÷= θcos yR ΣF y R÷= θcos zR ΣF z R÷= Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 156 FORCE SYSTEMS General Method of Locating Resultant When Its Components are Known: To determine the position of the resultant force of a system of forces, proceed as follows: From the origin, point O, of a set of coordinate axes x, y, z, lay off on the x axis a length A representing the algebraic sum ∑F x of the x components of all the forces. From the end of line A lay off a line B representing ∑F y , the algebraic sum of the y components; this line B is drawn in a direction parallel to the y axis. From the end of line B lay off a line C represent- ing ∑F z . Finally, draw a line R from O to the end of C; R will be the resultant of the system. 4. Calculate the moments M x , M y , M z about x, y, and z axes, respectively, due to the F x , F y , and F z com- ponents of each force and set them in tabular form. The formulas to use are given in the accompanying table. In interpreting moments about the x, y, and z axes, consider counterclockwise moments a plus ( + ) sign and clockwise moments a minus (− ) sign. In deciding whether a moment is counterclockwise or clockwise, look from the positive side of the axis in question toward the negative side. Force Moments of Components of F (F x , F y , F z ) about x, y, z axes F M x = yF z − zF y M y = zF x − xF z M z = xF y − yF x F 1 M x1 = y 1 F z1 − z 1 F y1 M y1 = z 1 F x1 − x 1 F z1 M z1 = x 1 F y1 − y 1 F x1 F 2 M x2 = y 2 F z2 − z 2 F y2 M y2 = z 2 F x2 − x 2 F z2 M z2 = x 2 F y2 − y 2 F x2 F 3 M x3 = y 3 F z3 − z 3 F y3 M y3 = z 3 F x3 − x 3 F z3 M z3 = x 3 F y3 − y 3 F x3 ∑M x ∑M y ∑M z 5. Add the component moments algebraically to get ∑M x , ∑M y and ∑M z which are the components of the resultant couple, M, given by the formula, The resultant couple M will tend to produce rotation about an axis making angles of β x , β y , and β z with the x, y, z axes, respectively. These angles are determined from the formulas, M ΣM x () 2 ΣM y () 2 ΣM z () 2 ++= β x cos ΣM x M = β y cos ΣM y M = β z cos ΣM z M = Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY FRICTION 157 Friction Properties of Friction.—Friction is the resistance to motion that takes place when one body is moved upon another, and is generally defined as “that force which acts between two bodies at their surface of contact, so as to resist their sliding on each other.” According to the conditions under which sliding occurs, the force of friction, F, bears a certain relation to the force between the two bodies called the normal force N. The relation between force of friction and normal force is given by the coefficient of friction, generally denoted by the Greek letter µ. Thus: Example:A body weighing 28 pounds rests on a horizontal surface. The force required to keep it in motion along the surface is 7 pounds. Find the coefficient of friction. If a body is placed on an inclined plane, the friction between the body and the plane will prevent it from sliding down the inclined surface, provided the angle of the plane with the horizontal is not too great. There will be a certain angle, however, at which the body will just barely be able to remain stationary, the frictional resistance being very nearly over- come by the tendency of the body to slide down. This angle is termed the angle of repose, and the tangent of this angle equals the coefficient of friction. The angle of repose is fre- quently denoted by the Greek letter θ. Thus, µ = tan θ. A greater force is required to start a body moving from a state of rest than to merely keep it in motion, because the friction of rest is greater than the friction of motion. Laws of Friction.—The laws of friction for unlubricated or dry surfaces are summarized in the following statements. 1) For low pressures (normal force per unit area) the friction is directly proportional to the normal force between the two surfaces. As the pressure increases, the friction does not rise proportionally; but when the pressure becomes abnormally high, the friction increases at a rapid rate until seizing takes place. 2) The friction both in its total amount and its coefficient is independent of the areas in contact, so long as the normal force remains the same. This is true for moderate pressures only. For high pressures, this law is modified in the same way as in the first case. 3) At very low velocities the friction is independent of the velocity of rubbing. As the velocities increase, the friction decreases. Lubricated Surfaces: For well lubricated surfaces, the laws of friction are considerably different from those governing dry or poorly lubricated surfaces. 1) The frictional resistance is almost independent of the pressure (normal force per unit area) if the surfaces are flooded with oil. 2) The friction varies directly as the speed, at low pressures; but for high pressures the friction is very great at low velocities, approaching a minimum at about two feet per second linear velocity, and afterwards increasing approximately as the square root of the speed. 3) For well lubricated surfaces the frictional resistance depends, to a very great extent, on the temperature, partly because of the change in the viscosity of the oil and partly because, for a journal bearing, the diameter of the bearing increases with the rise of temperature more rapidly than the diameter of the shaft, thus relieving the bearing of side pressure. 4) If the bearing surfaces are flooded with oil, the friction is almost independent of the nature of the material of the surfaces in contact. As the lubrication becomes less ample, the coefficient of friction becomes more dependent upon the material of the surfaces. Influence of Friction on the Efficiency of Small Machine Elements.—Friction between machine parts lowers the efficiency of a machine. Average values of the effi- ciency, in per cent, of the most common machine elements when carefully made are ordi- F µ N×= and µ F N = µ F N 7 28 0.25== = Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 158 FRICTION nary bearings, 95 to 98; roller bearings, 98; ball bearings, 99; spur gears with cut teeth, including bearings, 99; bevel gears with cut teeth, including bearings, 98; belting, from 96 to 98; high-class silent power transmission chain, 97 to 99; roller chains, 95 to 97. Coefficients of Friction.—Tables 1 and 2 provide representative values of static friction for various combinations of materials with dry (clean, unlubricated) and lubricated sur- faces. The values for static or breakaway friction shown in these tables will generally be higher than the subsequent or sliding friction. Typically, the steel-on-steel static coeffi- cient of 0.8 unlubricated will drop to 0.4 when sliding has been initiated; with oil lubrica- tion, the value will drop from 0.16 to 0.03. Many factors affect friction, and even slight deviations from normal or test conditions can produce wide variations. Accordingly, when using friction coefficients in design cal- culations, due allowance or factors of safety should be considered, and in critical applica- tions, specific tests conducted to provide specific coefficients for material, geometry, and/or lubricant combinations. Table 1. Coefficients of Static Friction for Steel on Various Materials Tables 1 and 2 used with permission from The Friction and Lubrication of Solids, Vol. 1, by Bowden and Tabor, Clarendon Press, Oxford, 1950. Table 2. Coefficients of Static Friction for Various Materials Combinations Material Coefficient of Friction, µ Material Coefficient of Friction, µ Clean Lubricated Clean Lubricated Steel 0.8 0.16 Hard carbon 0.14 0.11–0.14 Copper-lead alloy 0.22 … Graphite 0.1 0.1 Phosphor-bronze 0.35 … Tungsten carbide 0.4–0.6 0.1–0.2 Aluminum-bronze 0.45 … Plexiglas 0.4–0.5 0.4–0.5 Brass 0.35 0.19 Polystyrene 0.3–0.35 0.3–0.35 Cast iron 0.4 0.21 Polythene 0.2 0.2 Bronze … 0.16 Teflon 0.04 0.04 Sintered bronze … 0.13 Material Combination Coefficient of Friction, µ Material Combination Coefficient of Friction, µ Clean Lubricated Clean Lubricated Aluminum-aluminum 1.35 0.30 Tungsten carbide-tungsten carbide 0.2–0.25 0.12 Cadmium-cadmium 0.5 0.05 Plexiglas-Plexiglas 0.8 0.8 Chromium-chromium 0.41 0.34 Polystyrene-polystyrene 0.5 0.5 Copper-copper 1.0 0.08 Teflon-Teflon 0.04 0.04 Iron-iron 1.0 0.15–0.20 Nylon-nylon 0.15–0.25 … Magnesium-magnesium 0.6 0.08 Solids on rubber 1– 4 … Nickel-nickel 0.7 0.28 Wood on wood (clean) 0.25–0.5 … Platinum-platinum 1.2 0.25 Wood on wood (wet) 0.2 … Silver-silver 1.4 0.55 Wood on metals (clean) 0.2–0.6 … Zinc-zinc 0.6 0.04 Wood on metals (wet) 0.2 … Glass-glass 0.9–1.0 0.1–0.6 Brick on wood 0.6 … Glass-metal 0.5–0.7 0.2–0.3 Leather on wood 0.3–0.4 … Diamond-diamond 0.1 0.05–0.1 Leather on metal (clean) 0.6 … Diamond-metal 0.1–0.15 0.1 Leather on metal (wet) 0.4 … Sapphire-sapphire 0.2 0.2 Leather on metal (greasy) 0.2 … Hard carbon on carbon 0.16 0.12–0.14 Brake material on cast iron 0.4 … Graphite-graphite (in vacuum) 0.5–0.8 … Brake material on cast iron (wet) 0.2 … Graphite-graphite 0.1 0.1 Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY ROLLING FRICTION 159 Rolling Friction.—When a body rolls on a surface, the force resisting the motion is termed rolling friction or rolling resistance. Let W = total weight of rolling body or load on wheel, in pounds; r = radius of wheel, in inches; f = coefficient of rolling resistance, in inches. Then: resistance to rolling, in pounds = (W × f) ÷ r. The coefficient of rolling resistance varies with the conditions. For wood on wood it may be assumed as 0.06 inch; for iron on iron, 0.02 inch; iron on granite, 0.085 inch; iron on asphalt, 0.15 inch; and iron on wood, 0.22 inch. The coefficient of rolling resistance, f, is in inches and is not the same as the sliding or static coefficient of friction given in Tables 1 and 2, which is a dimensionless ratio between frictional resistance and normal load. Various investigators are not in close agreement on the true values for these coefficients and the foregoing values should only be used for the approximate calculation of rolling resistance. Mechanisms Levers The above formulas are valid using metric SI units, with forces expressed in newtons, and lengths in meters. However, it should be noted that the weight of a mass W kilograms is equal to a force of Wg newtons, where g is approximately 9.81 m/s 2 . Thus, supposing that in the first Types of Levers Examples A pull of 80 pounds is exerted at the end of the lever, at W; l = 12 inches and L = 32 inches. Find the value of force F required to balance the lever. If F = 20; W = 180; and l = 3; how long must L be made to secure equilibrium? Total length L of a lever is 25 inches. A weight of 90 pounds is supported at W; l is 10 inches. Find the value of F. If F = 100 pounds, W = 2200 pounds, and a = 5 feet, what should L equal to secure equilibrium? When three or more forces act on lever: Let W = 20, P = 30, and Q = 15 pounds; a = 4, b = 7, and c = 10 inches. If x = 6 inches, find F. Assuming F = 20 pounds in the example above, how long must lever arm x be made? F:Wl:L= FL× Wl×= F Wl× L = W FL× l = L Wa× WF+ Wl× F == l Fa× WF+ FL× W == F 80 12× 32 960 32 30 pounds=== L 180 3× 20 27== F:Wl:L= FL× Wl×= F Wl× L = W FL× l = L Wa× WF– Wl× F == l Fa× WF– FL× W == F 90 10× 25 36 pounds== L 2200 5× 2200 100– 5.24 feet== Fx× Wa× P+ b× Q+ c×= x Wa× P+ b× Q+ c× F = F Wa× P+ b× Q+ c× x = F 20 4× 30+7× 15+10× 6 7 3 1 3 lbs== x 20 4× 30+7× 15+10× 20 22 i n c h es== Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 160 SIMPLE MECHANISMS example l = 0.4 m, L = 1.2 m, and W = 30 kg, then the weight of W is 30g newtons, so that the force F required to balance the lever is This force could be produced by suspending a mass of 10 kg at F. Table of Forces on Inclined Planes Tensions and pressures in pounds. The table below makes it possible to find the force required for moving a body on an inclined plane. The friction on the plane is not taken into account. The column headed “Tension P in Cable per Ton of 2000 Pounds” gives the pull in pounds required for moving one ton along the inclined surface. The fourth column gives the perpendicular or normal pressure. If the coefficient of friction is known, the added pull required to overcome friction is thus easily determined: Per Cent of Grade. Rise, Ft. per 100 Ft. Angle α Tension P in Cable per Ton of 2000 Lbs. Perpendicular Pressure Q on Plane per Ton of 2000 Lbs. Per Cent of Grade. Rise, Ft. per 100 Ft. Angle α Tension P in Cable per Ton of 2000 Lbs. Perpendicular Pressure Q on Plane per Ton of 2000 Lbs. 1 0.57 20.00 1999.90 51 27.02 976.35 1745.49 2 1.15 40.00 1999.60 52 27.47 993.76 1735.64 3 1.72 59.99 1999.10 53 27.92 1011.07 1725.61 4 2.29 79.98 1998.40 54 28.37 1028.27 1715.42 5 2.86 99.96 1997.50 55 28.81 1045.37 1705.05 6 3.43 119.93 1996.40 56 29.25 1062.37 1694.51 7 4.00 139.89 1995.10 57 29.68 1079.26 1683.80 8 4.57 159.83 1993.60 58 30.11 1096.05 1672.93 9 5.14 179.76 1991.91 59 30.54 1112.72 1661.88 10 5.71 199.67 1990.01 60 30.96 1129.28 1650.67 11 6.28 219.56 1987.91 61 31.38 1145.73 1639.30 12 6.84 239.42 1985.62 62 31.80 1162.07 1627.76 13 7.41 259.27 1983.12 63 32.21 1178.29 1616.06 14 7.97 279.09 1980.43 64 32.62 1194.39 1604.19 15 8.53 298.88 1977.54 65 33.02 1210.37 1592.17 16 9.09 318.64 1974.45 66 33.42 1226.23 1579.98 17 9.65 338.36 1971.17 67 33.82 1241.97 1567.64 18 10.20 358.06 1967.69 68 34.22 1257.59 1555.15 19 10.76 377.72 1964.01 69 34.61 1273.07 1542.49 20 11.31 397.34 1960.13 70 34.99 1288.44 1529.68 21 11.86 416.92 1956.06 71 35.37 1303.67 1516.72 22 12.41 436.46 1951.79 72 35.75 1318.77 1503.61 23 12.95 455.96 1947.33 73 36.13 1333.74 1490.35 24 13.50 475.41 1942.68 74 36.50 1348.58 1476.94 25 14.04 494.81 1937.82 75 36.87 1363.28 1463.38 26 14.57 514.16 1932.78 76 37.23 1377.84 1449.67 27 15.11 533.46 1927.54 77 37.60 1392.27 1435.82 28 15.64 552.71 1922.11 78 37.95 1406.56 1421.83 29 16.17 571.90 1916.49 79 38.31 1420.71 1407.69 30 16.70 591.04 1910.67 80 38.66 1434.71 1393.41 31 17.22 610.12 1904.67 81 39.01 1448.57 1379.00 32 17.74 629.13 1898.47 82 39.35 1462.29 1364.44 33 18.26 648.09 1892.08 83 39.69 1475.86 1349.75 34 18.78 666.97 1885.51 84 40.03 1489.29 1334.93 35 19.29 685.80 1878.75 85 40.36 1502.56 1319.97 36 19.80 704.55 1871.79 86 40.70 1515.69 1304.87 37 20.30 723.23 1864.65 87 41.02 1528.66 1289.65 38 20.81 741.84 1857.33 88 41.35 1541.48 1274.30 39 21.31 760.38 1849.82 89 41.67 1554.14 1258.82 40 21.80 778.84 1842.12 90 41.99 1566.65 1243.22 41 22.29 797.22 1834.24 91 42.30 1579.01 1227.49 42 22.78 815.52 1826.18 92 42.61 1591.20 1211.64 43 23.27 833.74 1817.93 93 42.92 1603.24 1195.67 44 23.75 851.88 1809.50 94 43.23 1615.12 1179.58 45 24.23 869.93 1800.89 95 43.53 1626.83 1163.37 46 24.70 887.90 1792.10 96 43.83 1638.38 1147.04 47 25.17 905.77 1783.14 97 44.13 1649.77 1130.60 48 25.64 923.56 1773.99 98 44.42 1660.99 1114.05 49 26.10 941.25 1764.67 99 44.71 1672.05 1097.38 50 26.57 958.85 1755.17 100 45.00 1682.94 1080.60 F 30g 0.4× 1.2 10g 98.1 newtons.=== Q coefficient of friction× additional pull required.= Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY SIMPLE MECHANISMS 161 Inclined Plane—Wedge W = weight of body Neglecting friction: If friction is taken into account, then Force P to pull body up is: Force P 1 to pull body down is: Force P 2 to hold body stationary: in which µ is the coefficient of friction. W = weight of body W = weight of body Neglecting friction: With friction: Neglecting friction: With friction: Neglecting friction: With friction: Coefficient of friction = µ. Neglecting friction: With friction: Coefficient of friction = µ = tan φ. Force Moving Body on Horizontal Plane.—F tends to move B along line CD; Q is the component which actually moves B; P is the pressure, due to F, of the body on CD. PW h l × W αsin×== WP l h × P αsin P cosecα×=== QW b l × W αcos×== PWµαcos αsin+()= P 1 W µαcos αsin–()= P 2 W αsin µαcos–()= PW αsin βcos ×= WP βcos αsin ×= QW αβ+()cos βcos ×= Coefficient of friction µ= φtan= PW αφ+()sin βφ–()cos ×= PW h b × W αtan×== WP b h × P αcot×== Q W αcos W αsec×== Coefficient of friction µ= φtan= PW αφ+()tan= P 2Q b l × 2Q αsin×== QP l 2b × 1 2 P cosecα×== P 2Q µαcos αsin+()= P 2Q b h × 2Q αtan×== QP h 2b × 1 2 P αcot×== P 2Q αφ+()tan= QF αcos×= PF 2 Q 2 –= Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY 162 SIMPLE MECHANISMS Wheels and Pulleys Note: The above formulas are valid using metric SI units, with forces expressed in newtons, and lengths in meters or millimeters. (See note on page 159 concerning weight and mass.) The radius of a drum on which is wound the lifting rope of a windlass is 2 inches. What force will be exerted at the periphery of a gear of 24 inches diameter, mounted on the same shaft as the drum and transmitting power to it, if one ton (2000 pounds) is to be lifted? Here W = 2000; R = 12; r = 2. A, B, C and D are the pitch circles of gears. Let the pitch diameters of gears A, B, C and D be 30, 28, 12 and 10 inches, respectively. Then R 2 = 15; R 1 = 14; r 1 = 6; and r = 5. Let R = 12, and r 2 = 4. Then the force F required to lift a weight W of 2000 pounds, friction being neglected, is: The velocity with which weight W will be raised equals one-half the veloc- ity of the force applied at F. n = number of strands or parts of rope (n 1 , n 2 , etc.). The velocity with which W will be raised equals of the velocity of the force applied at F. In the illustration is shown a combination of a double and triple block. The pulleys each turn freely on a pin as axis, and are drawn with differ- ent diameters, to show the parts of the rope more clearly. There are 5 parts of rope. Therefore, if 200 pounds is to be lifted, the force F required at the end of the rope is: F:Wr:R= FR× Wr×= F Wr× R = W FR× r = R Wr× F = r FR× W = F 2000 2× 12 333 pounds== F Wr× r 1 × r 2 × RR 1 × R 2 × = W FR× R 1 × R 2 × rr 1 × r 2 × = F 2000 5× 6× 4× 12 14× 15× 9 5 pounds== F 1 ⁄ 2 W= F:W αsec :2= F W αsec× 2 = W 2F αcos×= F 1 n W×= 1 n F 1 ⁄ 5 200× 40 pounds== Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY SIMPLE MECHANISMS 163 Differential Pulley Screw Geneva Wheel In the differential pulley a chain must be used, engaging sprockets, so as to prevent the chain from slipping over the pulley faces. Chinese Windlass The Chinese windlass is of the differential motion principle, in that the resultant motion is the difference between two original motions. The hoisting rope is arranged to unwind from one part of a drum or pulley onto another part differing somewhat in diameter. The distance that the load or hook moves for one revolution of the com- pound hoisting drum is equal to half the differ- ence between the circumferences of the two drum sections. F = force at end of handle or wrench; R = lever-arm of F; r = pitch radius of screw; p = lead of thread; Q = load. Then, neglecting friction: If µ is the coefficient of friction, then: For motion in direction of load Q which assists it: For motion opposite load Q which resists it: Geneva wheels are frequently used on machine tools for indexing or rotating some part of the machine through a fractional part of a revolution. The driven wheel shown in the illustration has four radial slots located 90 degrees apart, and the driver carries a roller k which engages one of these slots each time it makes a revolution, thus turning the driven wheel one-quarter revolution. The concentric surface b engages the concave surface c between each pair of slots before the driving roller is dis- engaged from the driven wheel, which prevents the latter from rotating while the roller is moving around to engage the next successive slot. The circular boss b on the driver is cut away at d to provide a clearance space for the projecting arms of the driven wheel. In designing gearing of the general type illustrated, it is advisable to so proportion the driv- ing and driven members that the angle a will be approximately 90 degrees. The radial slots in the driven part will then be tangent to the circular path of the driving roller at the time the roller enters and leaves the slot. When the gearing is designed in this way, the driven wheel is started gradually from a state of rest and the motion is also grad- ually checked. PR× 1 ⁄ 2 WR r–()= P WR r–() 2R = W 2PR Rr– = FQ p 6.2832R ×= QF 6.2832R p ×= FQ 6.2832µrp– 6.2832r µp+ × r R ×= FQ p 6.2832µr+ 6.2832r µp– × r R ×= Machinery's Handbook 27th Edition Copyright 2004, Industrial Press, Inc., New York, NY [...]... 10 .05 11 .10 12 .15 13 .19 14 .24 15 .29 16 .34 17 .38 18 . 43 19 .48 20.52 21. 57 22.62 23. 67 24. 71 0. 73 1. 78 2. 83 3.87 4.92 5.97 7.02 8.06 9 .11 10 .16 11 .20 12 .25 13 .30 14 .35 15 .39 16 .44 17 .49 18 . 53 19 .58 20. 63 21. 68 22.72 23. 77 24.82 0.84 1. 88 2. 93 3.98 5. 03 6.07 7 .12 8 .17 9. 21 10.26 11 . 31 12 .36 13 .40 14 .45 15 .50 16 .55 17 .59 18 .64 19 .69 20. 73 21. 78 22. 83 23. 88 24.92 0.94 1. 99 3. 04 4.08 5 . 13 6 .18 7. 23 8.27 9 .32 10 .37 ... 19 10 5 41 4 93 450 31 8 12 73 406 37 0 33 8 239 955 32 5 296 270 19 1 764 2 71 247 225 15 9 637 232 211 19 3 13 6 546 2 03 18 5 16 9 11 9 478 18 0 16 4 15 0 10 6 424 16 2 14 8 13 5 96 38 2 14 8 13 5 12 3 87 34 7 13 5 12 3 11 3 80 31 8 12 5 11 4 10 4 73 294 11 6 10 6 97 68 2 73 10 8 99 90 64 255 9549 4775 31 8 3 238 7 19 10 15 92 13 64 11 94 10 61 955 868 796 735 682 637 Safe speeds of rotation are based on safe rim speeds shown in Table 1 Safe Speed... 5.24 6.28 7 .33 8 .38 9.42 10 .47 11 .52 12 .57 13 . 61 14.66 15 . 71 16.75 17 .80 18 .85 19 .90 20.94 21. 99 23. 04 24.09 0 .10 1. 15 2.20 3. 25 4.29 5 .34 6 .39 7. 43 8.48 9. 53 10 .58 11 .62 12 .67 13 .72 14 .76 15 . 81 16.86 17 . 91 18.95 20.00 21. 05 22 .10 23 .14 24 .19 0. 21 1.26 2 .30 3. 35 4.40 5.44 6.49 7.54 8.59 9. 63 10 .68 11 . 73 12 .78 13 .82 14 .87 15 .92 16 .96 18 . 01 19.06 20 .11 21. 15 22.20 23. 25 24.29 0. 31 1 .36 2. 41 3. 46 4.50 5.55... 9.74 10 .79 11 . 83 12 .88 13 . 93 14 .97 16 .02 17 .07 18 .12 19 .16 20. 21 21. 26 22 .30 23. 35 24.40 0.42 1. 47 2. 51 3. 56 4. 61 5.65 6.70 7.75 8.80 9.84 10 .89 11 .94 12 .98 14 . 03 15 .08 16 . 13 17 .17 18 .22 19 .27 20 .32 21. 36 22. 41 23. 46 24.50 0.52 1. 57 2.62 3. 66 4. 71 5.76 6. 81 7.85 8.90 9.95 11 .00 12 .04 13 .09 14 .14 15 .18 16 . 23 17 .28 18 .33 19 .37 20.42 21. 47 22. 51 23. 56 24. 61 0. 63 1. 67 2.72 3. 77 4.82 5.86 6. 91 7.96 9. 01 10.05... 2864.79 14 32 .39 954. 93 716 .20 572.96 286.48 19 0.99 14 3. 24 11 4.59 95.49 81. 85 71. 62 63. 66 57.29 52.09 47.74 44.07 40.92 38 .20 35 . 81 33 .70 31 . 83 30 .15 Angle ° 1. 00 1. 10 1. 20 1. 30 1. 40 1. 50 1. 60 1. 70 1. 80 1. 90 2.00 2.25 2.50 2.75 3. 00 3. 25 3. 50 3. 75 4.00 4.25 4.50 4.75 5.00 Coefficient 28.64 26.04 23. 87 22. 03 20.46 19 .09 17 .90 16 .85 15 . 91 15.07 14 .32 12 . 73 11 .45 10 . 41 9.54 8. 81 8 .17 7. 63 7 .15 6. 73 6 .35 6.02... foot-pounds = E1 Copyright 2004, Industrial Press, Inc., New York, NY Machinery's Handbook 27th Edition 18 6 FLYWHEELS Dimensions of Flywheels for Punches and Shears A B C D E F G H Max R.P.M J 24 3 31 2 6 11 ⁄4 13 ⁄8 23 4 31 4 31 2 955 30 31 2 4 7 13 ⁄8 11 ⁄2 3 33 4 4 796 36 4 41 2 8 11 ⁄2 13 ⁄4 31 4 41 4 41 2 637 42 41 4 43 4 9 13 ⁄4 2 31 2 41 2 5 557 48 41 2 5 10 13 ⁄4 2 33 ⁄4 43 4 51 2 478 54 43 4 51 2 11 2 21 4 4... 5.25 5.44 23 1. 18 5.50 5 .19 24 1. 12 5.75 4.97 25 1. 07 6.00 4.76 26 1. 03 6.50 4 .39 27 0.98 7.00 4.07 28 0.94 7.50 3. 80 29 0.90 8.00 3. 56 30 0.87 8.50 3. 35 31 0. 83 9.00 3 .16 32 0.80 10 .00 2.84 33 0.77 11 .00 2.57 34 0.74 12 .00 2 .35 35 0. 71 13. 00 2 .17 36 0.69 14 .00 2. 01 37 0.66 15 .00 1. 87 38 0.64 16 .00 1. 74 39 0.62 17 .00 1. 64 40 0.60 18 .00 1. 54 41 0.58 19 .00 1. 45 42 0.56 20.00 1. 37 43 0.54 21. 00 1. 30 44 0.52... 43 4 9 13 ⁄4 2 31 2 41 2 5 557 48 41 2 5 10 13 ⁄4 2 33 ⁄4 43 4 51 2 478 54 43 4 51 2 11 2 21 4 4 5 6 430 60 5 6 12 21 4 21 2 41 2 51 2 61 2 38 2 72 51 2 7 13 21 2 23 4 5 61 2 7 31 8 84 6 8 14 3 31 2 51 2 71 2 8 2 73 96 7 9 15 31 2 4 6 9 9 239 10 8 8 10 16 1⁄2 33 ⁄4 41 2 61 2 10 1⁄2 10 212 12 0 9 11 18 4 5 71 2 12 12 19 1 The maximum number of revolutions per minute given in this table should never be exceeded for... Handbook 27th Edition FLYWHEELS 19 1 Table 2 Safe Speeds of Rotation for Flywheels Outside Diameter of Rim (feet) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 6,600 Safe Rim Speed in Feet per Minute (from Table 1) 5 ,10 0 4,650 4,240 3, 000 12 ,000 30 ,000 210 0 10 50 700 525 420 35 0 30 0 2 63 233 210 19 1 17 5 16 2 15 0 14 0 Safe Speed of Rotation in Revolutions per Minute 16 23 14 80 13 50 955 38 20 812 740 676 478 19 10 5 41. .. E1 W = a = 1. 22W b = ay 2 2 12 Dy CD R For cast-iron flywheels, with a maximum stress of 10 00 pounds per square inch: W = C1 E1 R = 19 40 ÷ D Values of C and C1 in the Previous Formulas Per Cent Reduction C C1 Per Cent Reduction C C1 21 2 0.00000 2 13 0 .12 50 10 0.00000 810 0. 032 8 5 0.00000426 0.0625 15 0.000 011 80 0.0225 71 2 0.00000 617 0.0 432 20 0.000 015 35 0. 017 3 Example 1: A hot slab shear . 11 .00 11 .10 11 .20 11 . 31 11. 41 110 11 .52 11 .62 11 . 73 11 . 83 11 .94 12 .04 12 .15 12 .25 12 .36 12 .46 12 0 12 .57 12 .67 12 .78 12 .88 12 .98 13 .09 13 .19 13 .30 13 .40 13 . 51 130 13 . 61 13. 72 13 .82 13 . 93 14 . 03. 5 71. 90 19 16.49 79 38 . 31 1420. 71 1407.69 30 16 .70 5 91. 04 19 10.67 80 38 .66 14 34 . 71 139 3. 41 31 17.22 610 .12 19 04.67 81 39 . 01 1448.57 13 79.00 32 17 .74 629 . 13 18 98.47 82 39 .35 14 62.29 13 64.44 33 18 .26. 416 .92 19 56.06 71 35 .37 13 03. 67 15 16.72 22 12 . 41 436 .46 19 51. 79 72 35 .75 13 18.77 15 03. 61 23 12 .95 455.96 19 47 .33 73 36 . 13 13 33. 74 14 90 .35 24 13 .50 475. 41 1942.68 74 36 .50 13 48.58 14 76.94 25 14 .04