Tài liệu Handbook of Mechanical Engineering Calculations P19 pptx

P • A • R • T4 DESIGN ENGINEERING Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. DESIGN ENGINEERING 19.3 SECTION 19 SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION STRESSES IN SOLID AND HOLLOW SHAFTS AND THEIR COMPONENTS 19.3 Shaft Torque and Shearing Stress Determination 19.3 Choice of Shaft Diameter to Limit Torsional Deflection 19.4 Shaft Diameter Needed to Transmit Given Load at Stated Stress 19.5 Maximum Stress in a Shaft Produced by Bending and Torsion 19.6 Comparison of Solid and Hollow Shaft Diameters 19.7 Shaft Key Dimensions, Stresses, and Factor of Safety 19.8 Shaft Key Minimal Length for Known Torsional Stress 19.9 Shaft Stress Resulting from Instantaneous Stopping 19.9 SHAFT APPLICATIONS IN POWER TRANSMISSION 19.10 Energy Stored in a Rotating Flywheel 19.10 Shaft Torque, Horsepower, and Driver Efficiency 19.11 Pulley and Gear Loads on Shafts 19.12 Shaft Reactions and Bending Moments 19.13 Solid and Hollow Shafts in Torsion 19.14 Solid Shafts in Bending and Torsion 19.15 Equivalent Bending Moment and Ideal Torque for a Shaft 19.17 Torsional Deflection of Solid and Hollow Shafts 19.18 Deflection of a Shaft Carrying Concentrated and Uniform Loads 19.19 Selection of Keys for Machine Shafts 19.20 Selecting a Leather Belt for Power Transmission 19.21 Selecting a Rubber Belt for Power Transmission 19.23 Selecting a V Belt for Power Transmission 19.26 Selecting Multiple V Belts for Power Transmission 19.29 Selection of a Wire-Rope Drive 19.31 Design Methods for Noncircular Shafts 19.32 Calculating External Inertia, WK 2 , for Rotating and Linear Motion 19.40 Stresses in Solid and Hollow Shafts and Their Components SHAFT TORQUE AND SHEARING STRESS DETERMINATION A hydraulic turbine in a hydropower plant is rated at 12,000 hp (8952 kW). The steel vertical shaft connecting the turbine and generator is 24 in (60.96 cm) in Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 19.4 DESIGN ENGINEERING diameter and rotates at 60 r/min. What is the maximum torque and shearing stress in the shaft at full load? Calculation Procedure: 1. Compute the torque in the shaft at full load Use the relation, hp ϭ 2 ␲ NT/ 33,000, where N ϭ rpm of the shaft; T ϭ torque, lb/ ft (kg/ m). Solving for torque, T ϭ hp(33,000)/ 2 ␲ (N). Or, t ϭ 12,000 ϫ 33,000/ 6.28 ϫ 60 ϭ 1.05 ϫ 10 6 lb ⅐ ft (1423 kN ⅐ m). 2. Find the maximum shearing stress in the shaft Maximum shear stress occurs in the shaft at full load, i.e., 12,000 hp (8952 kW). Use the relation S ϭ RT/I, where S ϭ maximum shear stress, lb/in 2 (kPa); R ϭ shaft radius, in (cm); T ϭ torque in shaft at maximum load, lb ⅐ ft (N ⅐ m); I ϭ polar moment of inertia of the shaft, in 4 (cm 4 ). For a circular section, the polar moment of inertia I ϭ [ ␲ (d) 4 ]/ 32. For this 24- in (60.96-cm) shaft, I ϭ [3.14 ϫ 24 4 ]/32 ϭ 32,556 in 4 (82691 cm 4 ). Substituting in the stress equation, S ϭ (24/ 2)(1,050,000)(12 in /ft) /32556 ϭ 4644 lb/ in 2 (31.99 MPa). Related Calculations. Use the relations here to compute the torque and shear stress in shafts of any material: steel, iron, aluminum, copper, Monel, stainless steel, plastic, etc. Obtain the polar moment of inertia by computation using the standard equations for various shapes available in any mechanical engineering handbook. The shear stress in this shaft is relatively low, a characteristic of hydraulic turbines. This low shear stress partly explains why hydraulic turbines have some of the longest lives of machines, in some cases more than 100 years. CHOICE OF SHAFT DIAMETER TO LIMIT TORSIONAL DEFLECTION A solid cast-iron circular shaft 60 in (152.4 cm) long carries a solid circular head 60 in in diameter at one end, Fig. 1. The bar is subjected to a torsional moment of 60,000 lb in (6780 Nm) which is applied at one end. It is desired to keep the torsional deflection of the circular head below 1 ⁄ 32 in (0.079 cm) when the shaft is transmitting power over its entire length in order to prevent chattering of the as- sembly. What should the diameter of the shaft be if the working stress is taken as 3000 lb /in 2 (20.7 MPa) and the transverse modulus of elasticity is 6 million lb /in 2 (41,340 MPa)? Calculation Procedure: 1. Determine the shaft diameter based on the torque at full load The torque in the shaft ϭ (S w )(polar moment of inertia of the circular shaft, I )/ (working stress, C). For this shaft, torque ϭ 60,000 ϭ 3000 ␲ (d 3 /16). Solving for (d 3 ) ϭ (60,000 ϫ 16)/ (3000 ϫ 3.14) ϭ 4.66 in (11.84 cm). Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION 19.5 SI Values 3 ft (0.91 m) 300 lb (136.2 kg) FIGURE 1 Shaft carrying solid circular head. 2. Find the shaft diameter based on the allowable torsional deflection For torsional stiffness, ␪ ϭ (1/ 32)/ r, where r ϭ radius of the head, in (cm). This equals (1/32)(30) ϭ 1/ 960 radian. Since arc length along the head is ␪ (r), 1/32 ϭ ␪ (r) ϭ ␪ (30). Find the shaft diameter based on the torsional deflection from (d 4 ) ϭ 32(T)(S) / ␲ (E)( ␪ ), where T ϭ given shaft torsion; S ϭ given wheel diameter; E ϭ given transverse modulus of elasticity; ␪ ϭ 1/ 960. Substituting, d 4 ϭ 32(60,000)(60)/ 3.14(6,000,000)(1/ 960) ϭ 5870; then d ϭ 8.75 in (22.2 cm). Since 8.75 in (22.2 cm) is greater than 4.66 in (11.84 cm), the shaft must be designed for torsional stiffness, i.e., its diameter must be increased to at least 8.75 in (22.2 cm). Related Calculations. Use this general approach to size shafts to resist tor- sional deflection beyond a certain desired level. Increasing torsionals stiffness can reduce shaft chatter. With increased emphasis on noise reduction in manufacturing plants by EPA, torsional stiffness of shafts is receiving greater attention today. SHAFT DIAMETER NEEDED TO TRANSMIT GIVEN LOAD AT STATED STRESS What diameter steel shaft is required to transmit 2200 hp (1641 kW) at 2000 r/min with a maximum fiber stress in the shaft of 15,000 lb /in 2 (103.4 MPa)? Calculation Procedure: 1. Determine the torque in the shaft Use the relation, T ϭ hp(33,000)/ 2 ␲ (rpm), or T ϭ 2200(33,000)/ 6.28(2000) ϭ 5780 lb ⅐ ft (7831 N ⅐ m). Note that as the power transmitted rises, torque will in- crease if the rpm is constant, but if the rpm increases along with the power trans- Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION 19.6 DESIGN ENGINEERING mitted, the torque can remain fairly constant, depending on the relative increase of each. 2. Compute the shaft diameter required Use the relation, T ϭ SZ p , where S ϭ stress in shaft in given units; Z p ϭ polar section modulus of the shaft, in 3 (cm 3 ), and Z p ϭ ␲ (d 3 )/ 16. Solving these two relations for d 3 ϭ T(12 in per ft)(16) /S ( ␲ ). Or d 3 ϭ 5780(12)(16)/ 15,000(3.14) ϭ 2356; d ϭ 2.86 in (7.26 cm). A 3-in (7.62-cm) shaft would be chosen, unless space restrictions prevented using a shaft of this diameter. Related Calculations. Use this procedure to determine a suitable diameter for shafts of any material: cast iron, Monel, stainless steel, plastic, etc. MAXIMUM STRESS IN A SHAFT PRODUCED BY BENDING AND TORSION A1 13 ⁄ 16 -in (4.6-cm) diameter steel shaft is supported on bearings 6 ft (1.82 m) apart. A 24-in (60.96-cm) pulley weighing 50 lb (22.7 kg) is attached to the center of the span. The pulley runs at 40 r/min and delivers 15 hp (11.2 kW) to the shaft, which weighs 8.77 lb /ft (13.1 kg/m). A belt exerts a 250-lb (113.5-kg) force in a vertically downward direction on the pulley. Determine the maximum stress in the shaft produced by the combination of bending and torsional stresses. Calculation Procedure: 1. Determine the maximum bending load at the bearing at each end of the shaft Consider the shaft to be a beam with fixed ends, Fig. 1. The maximum bending mo- ment due to loads occurring at the bearings is given by the beam equation BM ϭ [(wl 2 )/12 ϩ PI/ 8], where w ϭ shaft weight, lb /ft (kg/ m); l ϭ distance between bearings, ft (m); P ϭ weight of load, lb (kg). Solving, BM ϭ [8.77(6 2 )(12 in/ ft)/ 12 ϩ (300 ϫ 6 ϫ 12)/8] ϭ 3015.7 lb ⅐ in (340.8 N ⅐ m). 2. Find the torque delivered by the power input to the shaft The torque, T, delivered by the power input to the shaft is given by T ϭ hp ϫ 33,000/ 2 ␲ ϫ rpm. Or, T ϭ 15 ϫ 33,000/ 6.28 ϫ 40 ϭ 1970.5 lb ⅐ in (222.7 N ⅐ m). 3. Compute the maximum shearing stress due to combined loads Use the relation, maximum shearing stress produced by combined loads, S s ϭ [1/ l p ][(BM 2 ϩ T 2 ] 0.5 .Or,S s ϭ [16/ ␲ (1/13/16) 3 ] ϩ [(3015.7) 2 ϩ (1970.5) 2 ] 0.5 ϭ 3602.4 lb/ in 2 (24.86 MPa). 4. Determine the maximum normal stress due to combined loads The maximum normal stress due to combined loads, S n ϭ [1/ I p ][3015.7 ϩ {(3015.7) 2 ϩ (1970.5) 2 }] 0.5 ϭ 5663.8 lb/ in 2 (39 MPa). Related Calculations. Use this procedure to find the maximum shearing stress and maximum normal stress in shafts made of any materials for which stress data are available. Thus, the relations given here are valid for cast iron, stainless steel, Monel, aluminum, plastic, etc. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION 19.7 COMPARISON OF SOLID AND HOLLOW SHAFT DIAMETERS A solid circular shaft is used to transmit 200 hp (149 kW) at 1000 r/min. (a) What diameter shaft is required if the allowable maximum shearing stress is 20,000 lb / in 2 (137.8 MPa)? (b) If a hollow shaft is used having an inside diameter equal to the outside diameter of the solid shaft determined in part (a), what must be the outside diameter of this shaft if the angular twist of the two shafts is to be equal? Calculation Procedure: 1. Determine the torque required to transmit the power Use the relation T ϭ 33,000(hp)/ 2 ␲ N, where T ϭ torque, lb ⅐ in (N ⅐ m); hp ϭ horsepower (kW) transmitted; N ϭ shaft rpm. Substituting, T ϭ 33,000(200)(12 in/ ft)/ 2(3.14)(1000) ϭ 12,611 lb ⅐ in (1425 N ⅐ m). 2. Find the required diameter of the solid Use two relations to find the required shaft diameter, namely: (1) S s ϭ T(d)/2(I p ), where S s ϭ maximum shear stress, lb/ in 2 (N ⅐ m); d ϭ shaft diameter, in (cm); I p ϭ polar moment of inertia of the solid shaft, in 4 (cm 4 ). (2) The polar moment of inertia of the solid shaft, I p ϭ ␲ (d 4 )/ 32, where the symbols are as given earlier. Combining the two equations gives S s ϭ 16T/ ␲ (d 3 ). Substituting d 3 ϭ (16)(12,611)/ ␲ (20,000) ϭ 3.21; d ϭ 1.475 in (3.75 cm). 3. Compute the outside diameter of the hollow shaft A hollow shaft having an inside diameter of 1.475 in (3.75 cm), that is the same as the outside diameter of the solid shaft, is desired. The outside diameter of the hollow shaft is to be such that its angular twist shall equal that of the solid shaft. Or, in equation form, ␪ ds ϭ T ds (L ds )/G e )(I ps ), where ␪ ds ϭ angular twist of the solid shaft, degrees; T ϭ torque in solid shaft, lb ⅐ in (N ⅐ m); G e ϭ modulus of elasticity of the shaft material in shear, lb /in 2 (kPa); other symbols as before. For the circular shaft, ␪ dh ϭ T dh (L dh )/G e (I pdh ). Symbols are the same as earlier, except that the sub- script h refers to the hollow shaft. Since the torque on the shaft and the shaft length are identical for both shafts which are made of the same material, by equating the angular twist equations to each other, I ps ϭ I ph ,or ␲ (d 4 )/32 ϭ [ ␲ Ϫ where D ϭ outside 44 (D )(d )]/32, dh dh diameter of the hollow shaft, in (cm). Substituting, and solving for the outside diameter of the hollow shaft gives D ϭ 1.754 in (4.45 cm). The hollow-shaft thick- ness will be (1.754 Ϫ 1.475)/ 2 ϭ 0.139 in (0.35 cm). Related Calculations. Use this general approach to determine the outside di- ameter of a hollow shaft, compared to that of a solid shaft. While the same angular twist was specified for these two shafts, different angular twists can be handled using the same general procedure. Any materials can be analyzed with this pro- cedure: steel, cast iron, plastic, etc. Hollow shafts often find favor today in an environmentally conscious design world. Thus, Richard M. Phelan, Professor of Mechanical Engineering, Cornell University, writes: ‘‘Most shafts are solid. But in situations where weight and reliability are of great importance, hollow shafts with a ratio of d inner / D outer ϭ 0.6 are often used. The (shaft) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION 19.8 DESIGN ENGINEERING weight decreases more rapidly than the strength because the material near the center is not highly stressed and carries only a relatively small part of the total bending and torque loads. The reliability of the material is increased by using hollow shafts. Very large shafts are usually machined from a forged billet, and boring out is specified to remove inclusions, holes, etc., left in the center of the billet, the last region to solidify upon cooling. A hollow shaft also permits more uniform heat treatment and simplifies inspection of the finished part. The shaft may be made hollow by boring, forging, or using cold-drawn seamless tubing. Unless the seamless tubing can be purchased so close to the required final dimensions that very little machining needs to be done, the high material cost may make it less practical than boring a hole in a solid piece.’’ SHAFT KEY DIMENSIONS, STRESSES, AND FACTOR OF SAFETY A solid steel machine shaft with a safe shear stress of 7000 lb/in 2 (48.2 MPa) transmits a torque of 10,500 lb ⅐ in (1186.5 N ⅐ m). (a) Find the shaft diameter for these conditions. (b) A square key is used whose width is equal to one-fourth the shaft diameter and whose length equals 1.5 times the shaft diameter. Find the key dimensions and check the key for its induced shear and compressive stresses. (c) Obtain the factors of safety of the key in shear and in crushing, allowing the ultimate shearing stress of 50,000 lb/in 2 (344.5 MPa) and a compression stress of 60,000 lb/ in 2 (413.4 MPa). Calculation Procedure: 1. Find the shaft diameter for the given conditions Use the relation d 3 ϭ 16T ␲ (S s ), where d ϭ shaft diameter, in (cm); T ϭ torque on shaft, lb ⅐ in (N ⅐ m); S s ϭ shear stress, lb/ in (kPa). Substituting, d 3 ϭ 16(10,500)/ ␲ (7000) ϭ 7.643; d ϭ 1.969 in (5.00 cm). A 2-in (5.08-cm) diameter shaft would be used. 2. Determine the key dimensions The width of the key is to be one-fourth of the shaft diameter, or 2.0 in/4 ϭ 0.5 in (1.27 cm). Length of the key is to equal 1.5 times the shaft diameter, or 1.5 ϫ 2 ϭ 3.0 in (7.62 cm). Now that we know the key dimensions, we can check it for induced shear and compressive stresses. The tangential force set up at the outside of the P t , lb (kg), is P t ϭ torque, lb ⅐ in/ radius of shaft, in. Or P t ϭ 10,500/ 1 ϭ 10,500 lb (4540 kg). The shear stress of the key is given by S s ϭ (P t )/bL, where b ϭ key width, in (cm); L ϭ key length, in (cm). Substituting, P t ϭ 10,500/ 0.5(3) ϭ 7000 lb/in 2 (48.2 MPa). Find the crushing stress from S c ϭ (2P t )bL, where S c ϭ crushing stress, lb/in (kPa). Substituting, S c ϭ 2(10,500)/ (0.5 ϫ 3) ϭ 14,000 lb/ in 2 (96.5 MPa). 3. Compute the factor of safety for both types of stresses The factor of safety ϭ allowable stress/actual stress. For shear, factor of safety, F s ϭ 50,000/ 7000 ϭ 7.14. For crushing, the factor of safety, F c ϭ 60,000/ 14,000 ϭ 4.29. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION 19.9 Related Calculations. Use this general procedure to size keys for rotating shafts made of any of the popular materials: steel, cast iron, aluminum, plastic, etc. Recommended dimensions of various types of shaft keys can be obtained from handbooks listed in the References section of this book. SHAFT KEY MINIMUM LENGTH FOR KNOWN TORSIONAL STRESS What is the minimum length for a 0.875-in (2.22-cm) wide key for a 3.4375-in (8.73-cm) diameter gear-driving shaft designed to operate at a torsional working stress of 11,350 lb/ in 2 (78.2 MPa)? The allowable shear stress in the key is 12,000 lb/in 2 (1.74 MPa). Calculation Procedure: 1. Determine the torque on the shaft Use the relation, T ϭ (d 3 )S t /5.1, where T ϭ shaft torque, lb ⅐ in (N ⅐ m); d ϭ shaft diameter, in (cm); S t ϭ torsional working stress of the shaft, lb/in 2 (kPa). Substi- tuting, T ϭ (3.4375 3 )(11,350)/ 5.1 ϭ 90,397 lb ⅐ in (10.2 kN ⅐ m). 2. Compute the tangential force on the key Use the relation, P ϭ T/r, where P ϭ tangential force on the key, lb (kg); r ϭ shaft radius, in (cm). Substituting, T ϭ 90,397/ 1.71875 ϭ 52,595 lb (23878 kg). 3. Find the length of the key to satisfy the given conditions Use the relation, L ϭ P/b(S s ), where L ϭ key length, in (cm); P ϭ tangential force computed in step 2; S s ϭ allowable shear stress in key, lb/ in 2 (kPa). Substituting, L ϭ 52,596/ 0.875(12,000) ϭ 5.00 in (12.7 cm). Using the rule that L ϭ 1.5d, then L ϭ 1.5 ϫ 3.4375 ϭ 5.16 in (13.1 cm). Therefore, the minimum computed length of 5.0 in (12.7 cm) because it closely approximates the length based on a ratio to the shaft diameter. The difference is negligible. Related Calculations. Use this general procedure to size keys for any type of shaft having a known torsional stress. SHAFT STRESS RESULTING FROM INSTANTANEOUS STOPPING A flywheel weighing 200 lb (90.8 kg) whose radius of gyration is 15 in (38.1 cm) is secured to one end of a 6-in (15.2-cm) diameter shaft; the other end of the shaft is connected through a chain and sprocket to a motor rotating at 1800 r/min. The motor sprocket is 6 in (15.2 cm) in diameter and the shaft sprocket is 36 in (91.4 cm) in diameter. Total shaft length between flywheel and sprocket is 72 in (182.9 cm). Determine that maximum stress in the shaft resulting from instantaneous stop- ping of the motor drive, assuming that the sprocket and chain have no ability to Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION 19.10 DESIGN ENGINEERING absorb impact loading. Assume a shear modulus of 12,000,000 lb /in 2 (82,680 MPa). Neglect the effect of shaft kinetic energy. Calculation Procedure: 1. Determine the torsional impact caused by the sudden stop When the flywheel or any other rotating mass is stopped short, the stored kinetic energy in the rotating mass is converted to torsional impact. The magnitude of this energy is given by E ϭ W( ␳ 2 )( ⍀ 2 )/2g, where W ϭ flywheel weight, lb (kg); ␳ ϭ radius of gyration of the flywheel, ft (cm); ⍀ ϭ angular velocity, radians per second; g ϭ acceleration of gravity, 32.2 ft /s 2 (9.8 m/ s 2 ). Substituting, E ϭ 200 (1.25 2 )(2 ␲ 300/ 60(1/ 12) ϭ 57,394 lb ⅐ in (6486 N ⅐ m). 2. Find the modulus of resilience for the shaft The shaft offers resilience to torsional twist, as detailed in Marks’ ‘‘Mechanical Engineers’ Handbook.’’ Resilience, U in lb ⅐ in (N ⅐ m) is the potential energy stored in the deformed body, the shaft. The amount of resilience equals the work required to deform the shaft from zero stress to stress S. So the modulus of resilience, U p , in lb ⅐ in/in 3 (N ⅐ m/cm 3 ), or unit resilience, is the elastic energy stored in an in 3 (cm 3 ) of the shaft material at the elastic limit. The unit of resilience for a solid shaft is U p ϭ (S s ) 2 /4G, where S s ϭ maximum shear stress developed on instanta- neous stopping, lb /in 2 (kPa); G ϭ modulus of elasticity of the shaft material, lb / in 2 (kPa). Find the full volume of the shaft from V ϭ 0.785 ϫ 6 2 (72) ϭ 2035 in 3 (33,348 m 3 ). Then, substituting in the unit resilience equation for the entire shaft, U p total ϭ (2035)(1/ 12,000,000) ϭ 57,394 lb/in. Solving for S t ϭ (4 ϫ 12,000,000 2 0.25(S ) t ϫ 57394/ 2035) 0.5 ϭ 36,794 lb/ in 2 (4158 N ⅐ m). Thus, the maximum stress in the shaft at instantaneous stopping will be 36,794 lb /in 2 (4159 N ⅐ m). Related Calculations. Sudden stopping of a rotating member can cause ex- cessive stress that may lead to failure. Therefore, it is important that the stress caused by sudden stopping be analyzed for every design where the possibility of such stopping exists. The time needed to compute the stress that might occur is small compared to the damage that might result if sudden stopping does occur. Further, having the calculations on file proves that he engineer took time to look ahead to see what might happen in the event of sudden stopping. Shaft Applications in Power Transmission ENERGY STORED IN A ROTATING FLYWHEEL A 48-in (121.9-cm) diameter spoked steel flywheel having a 12-in wide ϫ 10-in (30.5-cm ϫ 25.4-cm) deep rim rotates at 200 r /min. How long a cut can be stamped in a 1-in (2.5-cm) thick aluminum plate if the stamping energy is obtained from this flywheel? The ultimate shearing strength of the aluminum is 40,000 lb /in 2 (275,789.9 kPa). Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION [...]... POWER TRANSMISSION 19.11 Calculation Procedure: 1 Determine the kinetic energy of the flywheel In routine design calculations, the weight of a spoked or disk flywheel is assumed to be concentrated in the rim of the flywheel The weight of the spokes or disk is neglected In computing the kinetic energy of the flywheel the weight of a rectangular, square or circular rim is assumed to be concentrated at the... the weight is concentrated at a radius of 48 / 2 Ϫ 10 / 2 ϭ 19 in (48.3 cm) from the centerline of the shaft to which the flywheel is attached Then the kinetic energy K ϭ Wv2 / (2g), where K ϭ kinetic energy of the rotating shaft, ft ⅐ lb; W ϭ flywheel weight of flywheel rim, lb; v ϭ velocity of flywheel at the horizontal centerline of the rim, ft / s The velocity of a rotating rim is v ϭ 2␲RD / 60, where... Compute the dimensions of the hole that can be stamped A stamping operation is a shearing process The area sheared is the product of the plate thickness and the length of the cut Each square inch of the sheared area offers a resistance equal to the ultimate shearing strength of the material punched During stamping, the force exerted by the stamp varies from a maximum F lb at the point of contact to 0 lb... cm) Use a 6.0-in (15.2-cm) diameter shaft Related Calculations Use this procedure for any solid shaft of uniform cross section made of metal—steel, aluminum, bronze, brass, etc The equation used in step 4 to determine the location of zero shear is based on a strength -of- materials Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill... the length of each rope would be equal to the lift height Hence, with a rope weight of 0.51 lb / ft (0.76 kg / m), the total weight of the rope ϭ (0.51)(500) / 2000 ϭ 0.127 ton (0.115 t) Acceleration of the car from the stopped condition places an extra load on the rope The rate of acceleration of the car is found from a ϭ v2 / (2d), where a ϭ car acceleration, ft / s2; v ϭ final velocity of the car,... Compute the required diameter of the solid shaft For a solid metal shaft, d ϭ (584Tl / G␣)1/3, where l ϭ shaft length expressed as a number of shaft diameters, in; G ϭ modulus of rigidity, lb / in2; ␣ ϭ angle of torsion deflection, degree Usual specifications for noncritical applications of shafts require that the torsional deflection not exceed 1Њ in a shaft having a length of equal to 20 diameters Using... Related Calculations Use this procedure to determine the steady-load torsional deflection of any shaft of uniform cross section made of any metal—steel, bronze, brass, aluminum, Monel, etc The assumed torsional deflection of 1Њ for a shaft that is 20 times as long as the shaft diameter is typical for routine applications Special shafts may be designed for considerably less torsional deflection DEFLECTION OF. .. due to shaft weight For a shaft of uniform weight, ⌬ ϭ 5wl3 / 384EI, where w ϭ total distributed load ϭ weight of shaft, lb Thus, ⌬ ϭ 5(60)(72)3 / [384(30 ϫ 106)(0.7854)] ϭ 0.0129 in (0.328 mm) The deflection per foot of shaft length is ⌬f ϭ 0.0129 / 6 ϭ 0.00214 in / ft (0.178 mm / m) 3 Determine the total deflection of the shaft The total deflection of the shaft is the sum of the deflections caused by the... type of key to use When a key is designed so that its allowable shear stress is approximately one-half its allowable compressive stress, a square key (i.e., a key having its height equal to its width) is generally chosen For other values of the stress ratio, a flat key is generally used Determine the dimensions of the key from Baumeister and Marks—Standard Handbook for Mechanical Engineers This handbook. .. Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website SHAFTS, FLYWHEELS, PULLEYS, AND BELTS FOR POWER TRANSMISSION 19.22 DESIGN ENGINEERING Calculation Procedure: 1 Determine the belt speed The speed of a belt S is found from S ϭ ␲RD, where R ϭ rpm of driving . subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS Downloaded from Digital Engineering Library. is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 19.4 DESIGN ENGINEERING diameter and