23.1 SECTION 23 SPRING SELECTION AND ANALYSIS Proportioning Helical Springs by Minimum Weight 23.1 Determining Safe Torsional Stress for a Helical Spring 23.3 Designing a Spring with a Specific Number of Coils 23.4 Determining Spring Dimensions Before and After Load Applications 23.5 Analysis and Design of Flat Metal Springs 23.6 Sizing Torsional Leaf Springs 23.7 Designing a Spring to Fire a Projectile 23.12 Spring Support of Machinery to Control Vibration Forces 23.13 Spring Selection for a Known Load and Deflection 23.14 Spring Wire Length and Weight 23.15 Helical Compression and Tension Spring Analysis 23.16 Selection of Helical Compression and Tension Springs 23.17 Sizing Helical Springs for Optimum Dimensions and Weight 23.19 Selection of Square- and Rectangular- Wire Helical Springs 23.20 Curved Spring Design Analysis 23.21 Round- and Square-Wire Helical Torsion- Spring Selection 23.23 Torsion-Bar Spring Analysis 23.26 Multirate Helical Spring Analysis 23.27 Belleville Spring Analysis for Smallest Diameter 23.28 Belleville Spring Computations for Disk Deflection, Load, and Number 23.30 Ring-Spring Design Analysis 23.32 Liquid-Spring Selection 23.34 Selection of Air-Snubber Dashpot Dimensions 23.37 Design Analysis of Flat Reinforced- Plastic Springs 23.39 Life of Cyclically Loaded Mechanical Springs 23.42 Shock-Mount Deflection and Spring Rate 23.43 PROPORTIONING HELICAL SPRINGS BY MINIMUM WEIGHT The detent helical spring in Fig. 1 is to be designed so that force P 1 for the extended condition is to be 20 lb (88.9 N). After an additional compression of 0.625 in (1.59 cm) the shearing stress in the spring is to be 75,000 lb/in 2 (517 MPa). The spring index, c, is approximately 8, based on past experience, and the modulus of elasticity in shear is 11,500,000 lb/in 2 (79.2 GPa). Find the diameter of the wire, helix radius, and number of active coils for the smallest amount of material in the spring. Calculation Procedure: 1. Find the diameter of the spring wire A spring which will contain the smallest possible amount of material will have design parameters selected so that the maximum force, stress, and deflection are 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 23.2 DESIGN ENGINEERING FIGURE 1 Spring-loaded de- tent. (Product Engineering.) exactly twice the minimum force, stress, and deflection, respectively. With such parameters, the spring wire diameter, 16Pc 1 d ϭ Ί ␲ (S ) s max where d ϭ spring wire diameter, in (cm); P 1 ϭ force in the extended condition, lb (N); c ϭ spring index, dimensionless; ␲ ϭ 3.1416; (S s ) ϭ shear stress in the max spring wire, lb/in 2 (kPa). Substituting, d ϭ (16 ϫ 20 ϫ 8 / 75,000 ␲ ) ϭ 0.1042 in 0.5 (0.265 cm). Referring to a spring wire table, choose No. 12 wire with d ϭ 0.1055 in (0.268 cm) as the nearest commercially available wire size. 2. Compute the mean radius of the spring helix The helix mean radius, 3 ␲ d (S ) s max R ϭ 32P 1 where R ϭ mean radius of the helix, in (cm); other symbols as defined earlier. Substituting, R ϭ ( ␲ ϫ 0.1055 3 ϫ 75,000)/32 ϫ 20 ϭ 0.4323 in (1.09 cm). 3. Determine the number of active coils in the spring Use the relation, N ϭ number of active coils in the spring, ( ␦ Ϫ ␦ ) dG 21 N ϭ 2 2 ␲ R (S ) s max where, ␦ 2 ϭ maximum compression of spring, in (cm); ␦ 1 ϭ minimum compression of the spring, in (cm); G ϭ modulus of elasticity in shear; other symbols as before. Substituting, N ϭ (0.625 ϫ 0.1055 ϫ 11,500,000) / (2 ␲ ϫ 0.4323 2 ϫ 75,000 ϭ 8.61 active coils. 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. SPRING SELECTION AND ANALYSIS SPRING SELECTION AND ANALYSIS 23.3 4. Find the volume of the spring Volume of a helical spring is given by V ϭ 0.5 ϫ ␲ 2 ϫ d 2 ϫ RN, where the symbols are as given earlier. Substituting, V ϭ 0.5 ϫ 3.14 2 ϫ 0.1055 2 ϫ 0.4323 ϫ 8.61 ϭ 2.044 in 3 (30.0335 dm 3 ). Related Calculations. Because of the many variables involved, the proportion- ing of a helical spring is usually done by trial and error. An unlimited number of springs can be obtained that will satisfy a given set of equations for stress and deflection. However, all springs found this way will contain more material than necessary unless the conditions for the mathematical minimum are fulfilled. The functioning of a helical spring usually occurs when the spring is extended, and is least able to exert a force. For example, in the indexing mechanism shown in Fig. 1, when the ball is in the detent, the holding power of the mechanism depends on the force exerted by the spring. After relative motion between the parts occurs, the ball is out of the detent and the additional force from the increased compression serves no useful purpose and may be a disadvantage. In this condition the shearing stress in the wire is usually the governing factor. Many similar applications arise in which the design is controlled by (a) the spring force in the extended condition and (b) the shearing stress for the most compressed condition. It has been proved for this instance, that the spring will contain the smallest possible amount of material when the design parameters are selected so that the maximum force, stress, and deflection are exactly twice the minimum force, stress, and deflection, respectively. Although not directly specified, the designer can usually estimate a suitable value for the spring index, c. Taking the maximum load as twice the minimum load, the wire-diameter, d, is computed using the equation given in step 1, above. Next the helix radius, R, is computed using the d value from step 1. Then the number of active coils is computed, step 3. Lastly, the minimum volume of the spring is determined, as in step 4. This procedure is the work of M. F. Spotts, Northwestern Technological Institute, as reported in Product Engineering magazine. DETERMINING SAFE TORSIONAL STRESS FOR A HELICAL SPRING The load on a helical spring is 1600 lb (726.4 kg) and the corresponding deflection is to be 4 in (10.2 cm). Rigidity modulus is 11 ϫ 10 6 lb/in 2 (75.8 GPa) and the maximum intensity of safe torsional stress is 60,000 lb/in 2 (413.4 MPa). Design the spring for the total number of turns if the wire is circular in cross section with a diameter of 0.625 in (1.5875 cm) and a centerline radius of 1.5 in (3.81 cm). Calculation Procedure: 1. Find the number of active coils in the spring Use the relation y ϭ N(64)( P)(r 3 )/G(d 4 ), where y ϭ total deflection of the spring, in (cm); N ϭ number of active coils in the spring; P ϭ load on spring, lb (N); r ϭ radius as axis to centerline of wires, in (cm); G ϭ modulus of elasticity of spring material in shear, lb/in 2 (kPa); d ϭ spring wire diameter, in (cm). Solving for the number of active coils, N ϭ y(G)(d 4 )/64(P)(r 3 ). Or N ϭ 4(11 ϫ 11 6 )(0.625 4 )/ 64(1600)(1.5 3 ) ϭ 19.4 active coils. 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. SPRING SELECTION AND ANALYSIS 23.4 DESIGN ENGINEERING 2. Check for the safe limit of torsional stress in the spring Use the relation S, ϭ 16(P)(r)/ ␲ (d 3 ). Solving, S ϭ 16(1600)(1.5)/ ␲ (0.625 3 ) ϭ 50,066 lb/in 2 (344.95 MPa). This torsional stress is less than the safe limit of 60,000 lb/in 2 (413.4 MPa) chosen for this spring. Hence, the spring is acceptable. The total number spring turns for two inactive coils (one at each end) is N ϩ 2 ϭ 19.4 ϩ 2 ϭ 21.4, or 22 coils. Related Calculations. Use this general approach for any helical spring you wish to analyze for safe torsional stress. DESIGNING A SPRING WITH A SPECIFIC NUMBER OF COILS A coiled spring with an outside diameter of 1.75 in (4.45 cm) is required to work under a load of 140 lb (622.7 N) and have seven active coils with the ends closed and round. Determine unit deflection, total number of coils, and length of spring when under load when G ϭ 12 ϫ 10 6 lb/in 2 (82.7 GPa) and mean spring diameter (outside diameter Ϫ wire diameter) is to be 0.779 in (1.98 cm). Calculation Procedure: 1. Find the safe shearing stress in the spring Use the relation, S t ϭ 8PD/ ␲ d 3 , where S ϭ fiber stress in shear, lb/in 2 (kPa); P ϭ axial load on spring, lb (N); D ϭ mean diameter of spring as defined above, in (cm); d ϭ diameter of spring wire, in (cm). Substituting, S ϭ 9(140)(0.779)(2)/ ␲ (0.192 3 ) ϭ 78,475 lb /in 2 (540.7 MPa). 2. Determine the deflection of the spring Use the relation, y ϭ 4( ␲ )(N)(r 2 )(S)/Gd, where y ϭ total deflection, in (cm); N ϭ number of active coils; G ϭ rigidity modulus of the spring material in shear, lb / in 2 (kPa); other symbols as defined earlier. Substituting, y ϭ 4( ␲ )(7)(0.779 2 ) (78,475)/12 ϫ 10 6 (0.192) ϭ 1.818 in (4.6 cm), or about 1 13 ⁄ 16 in. 3. Compute the total number of coils in the spring For the ends of a spring to be closed and ground smooth, 1.5 coils should be taken as inactive. In compression springs the number of active coils depends on the style of ends, as follows: open ends, not ground—all coils are active; open ends, ground—0.5 coil inactive; closed ends, not ground—1 coil inactive; closed ends, ground—1.5 coils inactive; squared ends, ground—2 coils inactive. Since this spring is to have ends that are closed and round; i.e., closed and ground, use 1.5 inactive coils. Hence, the total number of coils will be 7 ϩ 1.5 ϭ 8.5, say 9 coils. 4. Find the length of the spring when under load The solid height of the spring when it is entirely compressed is 9 coils ϫ 0.192 ϭ 1.728 in (4.39 cm), say 1 3 ⁄ 4 in. The total deflection under load is 1 13 ⁄ 16 in (4.6 cm). Using a total free space between coils of 1 in (2.54 cm), the free length of the spring will be 1 3 ⁄ 4 ϩ 1 13 ⁄ 16 ϩ 1 ϭ 4 9 ⁄ 16 in (11.59 cm). Hence, the length of the spring when under load ϭ 4 9 ⁄ 16 Ϫ 1 13 ⁄ 16 ϭ 2 3 ⁄ 4 in (6.98 cm). Related Calculations. The procedure given here is valid for sizing any spring when it must have a specified number of coils. 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. SPRING SELECTION AND ANALYSIS SPRING SELECTION AND ANALYSIS 23.5 DETERMINING SPRING DIMENSIONS BEFORE AND AFTER LOAD APPLICATION A cylindrical helical spring of circular cross-section wire is to be designed to safely carry an axial compressive load of 1200 lb (5338 N) at a maximum stress of 110,000 lb/in 2 (757.9 MPa). The spring is to have a deflection scale of approxi- mately 150 lb/in (262 N/cm). Spring proportions are to be: Mean diameter of coil/diameter of wire ϭ 6 to 8; spring length when closed/mean diameter of coil ϭ 1.7 to 2.3. Determine (a) mean diameter of coil; (b) diameter of wire; (c) length of coil when closed, and (d ) length of coil before load application. Use G ϭ rigidity modulus of steel in shear ϭ 11.5 ϫ 10 6 (79,235 MPa). Calculation Procedure: 1. Find the trial wire size In this trial design we can assume the ratio of D / d to be 6:8, as given. Then d ϭ D/7, using the midpoint ratio, where D ϭ mean diameter of spring coil/diameter of spring wire. Use the relation, D ϭ S( ␲ )(d 3 )/8(P), where S ϭ spring stress, lb/ in 2 (kPa); D ϭ mean diameter of the spring ϭ outside diameter Ϫ wire diameter, in (cm); d ϭ diameter of spring wire, in (cm). Substituting, D ϭ (110,000)(3.1416)[(D/7) 3 ]/8(1200); D ϭ 4.45 in (2.02 cm). The wire diameter is d ϭ 4.45/7 ϭ 0.635 in (1.6 cm). The nearest wire standard wire size is 0.6666 in (1.69 cm). Solving for D / d ϭ 4.45 / 0.6666 ϭ 6.68. This lies within the limits of 6 to 8 set for this spring for D /d. 2. Determine the spring scale Use the relation P/y ϭ G(d 4 )/N(64)(r 3 ), where y ϭ total deflection of the spring, in (cm); N ϭ number of coils; other symbols as before. For this spring, P / y ϭ 150. Substituting, 150 ϭ (11.5 ϫ 10 6 )(0.625 4 )/N(64)(2.38 3 ). Solving, N ϭ 13.6 active coils. Total turns for this spring, assuming closed ends ground ϭ 13.6 ϩ 1.5 ϭ 15.1. 3. Evaluate the length closed/mean diameter ratio This ratio must be between 1.7 and 2.3 if the spring is to meet its design require- ments. Using the relation, (d)(total turns) / D, we have (15.5)(0.6666)/4.45 ϭ 2.26, which is within the required limits. Then, the length of the closed coil ϭ 15.1 ϫ 0.6666 ϭ 10.065 in (25.56 cm). 4. Find the coil length before load application To find the length of the spring coil before the load is applied we add the closed length of the coil to the total deflection of the spring. To find the total deflection, use the relation, y ϭ N(64)(P)(2.38 3 )/11.5 ϫ 10 6 (0.6666 4 ); or y ϭ 13.6(64)(1200)(2.38 3 )/11.5 ϫ 10 6 (0.6666 4 ) ϭ 6.2 in(15.75 cm). Then, the length of the coil before load application ϭ 10 ϩ 6.2 ϭ 16.2 in (41.1 cm). Related Calculations. The steps in this procedure are useful when designing a spring to meet certain preset requirements. Using the given steps you can develop a spring meeting any of several dimensional or stress parameters. 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. SPRING SELECTION AND ANALYSIS 23.6 DESIGN ENGINEERING ANALYSIS AND DESIGN OF FLAT METAL SPRINGS Determine the width and thickness of the leaves of a six-leaf steel cantilever spring 13-in (33-cm) long to carry a load of 375 lb (1668 N) with a deflection of 1.25 in (3.2 cm). The maximum stress allowed in this spring is 50,000 lb / in 2 (344.5 MPa); the modulus of elasticity of the steel spring material is 30 ϫ 10 6 lb/in 2 (206,700 MPa). 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. SPRING SELECTION AND ANALYSIS SPRING SELECTION AND ANALYSIS 23.7 Calculation Procedure: 1. Find the leaf thickness for this steel spring See the list of equations below. Knowing the deflection, we can use the equation, F ϭ S(l 2 )/E(t), where F ϭ deflection, in (cm); S ϭ safe tensile stress in the spring metal, lb / in 2 (kPa); l ϭ spring length, in (cm); E ϭ modulus of elasticity as given above; t ϭ spring leaf thickness in (cm). substituting, 1.25 ϭ 50,000(13 2 )/30 ϫ 10 6 (t); t ϭ 0.225 in (0.57 cm). W ϭ save load or pull, lb (N) F ϭ deflection at point of application, in (cm) S ϭ safe tensile stress of material, lb /in 2 (kPa) E ϭ modulus of elasticity, 30 ϫ 10 6 for steel (kPa) Other symbols as shown above. 2. Determine the width of each leaf in the spring Use the relation, W ϭ S(N )(b)(t 2 )/6(l ), where the symbols are as defined above and W ϭ safe load or pull, lb (N). Substituting, 375 ϭ 50,000(6)(b)(.225 2 )/6(13); b ϭ 1.93 in (4.9 cm). Related Calculations. Use the spring layout and equations above to design any of the three types of metal springs shown. Flat metal springs find wide use in a variety of mechanical applications. The three types shown above—flat parallel, flat triangular, and flat leaf—are the most popular today. Using the data given here, engineers can design a multitude of flat springs for many different uses. SIZING TORSIONAL LEAF SPRINGS Size a torsional leaf spring, Fig. 2, for a 38-in ϫ 38-in (96.5-cm ϫ 96.5-cm) work platform weighing 145 lb (65.8 kg) which requires an assist spring to reduce the lifting force that must be exerted by a human being. A summation of the moments about a line through both platform pivots, Fig. 3, shows that a lifting force of only 23 lb (102.3 N) is needed if a torsional spring provides an assist moment of M t ϭ 1880 lb ⅐ in (212.4 N ⅐ m). An assist moment is required to raise the platform from either the stowed or the operating position. As the platform is rotated from one position to the other, the assist spring undergoes a load reversal. The required spring wind-up angle, ␪ ϭ ␲ /2 radians; the spring length ϭ 36 in (91.4 cm). AISI 4130 alloy steel heat-treated to 31 Rc minimum with an allowable shear stress of 68,000 lb/in 2 (468.5 MPa) and a shear modulus of G ϭ 11.5 ϫ 10 6 lb/in 2 (79,235 MPa) is to be used for this spring. Size the spring, determining the required dimensions for blade thickness, number of blades, aspect ratio, and blade width. Calculation Procedure: 1. Estimate the blade thickness Use the equation b ␶ ␤ ϭϭ , l 2G ␪ 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. SPRING SELECTION AND ANALYSIS 23.8 DESIGN ENGINEERING FIGURE 2 Torsional leaf springs normally are twisted to a specific wind-up angle. This two-blade spring is wound at right angles. (Machine Design.) FIGURE 3 Work platform requires an assist spring to reduce required lifting force. (Machine Design.) 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. SPRING SELECTION AND ANALYSIS SPRING SELECTION AND ANALYSIS 23.9 where the symbols are as given below. Solving for the blade thickness using the given data, b ϭ (68.000)(36) /2(11.5 ϫ 10 6 )( ␲ /2) ϭ 0.068 in (0.172 cm). Note that this is the blade half-thickness. To keep manufacturing costs low, a stock plate thickness would be chosen. The nearest stock plate thickness, 2b ƒ , is 0.125 in (0.3175 cm). So b ƒ ϭ 0.125/2 ϭ 0.0625 in (0.159 cm), which is less than b ϭ 0.068 in (0.172 cm) the computed required blade half-thickness. Nomenclature a ϭ blade half-width, in (cm) a ƒ ϭ selected blade half-width, in (cm) a r ϭ required blade half-width, in (cm) b ϭ blade half-thickness, in (cm) b ƒ ϭ selected blade half-thickness, in (cm) F ϭ torsional stress factor, 1 / in 3 (1/cm 3 ) ƒ ϭ normalized torsional stress factor G ϭ shear modulus, lb / in 2 (kPa) R 0 ϭ estimated aspect ratio K ϭ torsional stiffness factor, in 4 (cm 4 ) R r ϭ required aspect ratio k ϭ normalized torsional stiffness s ϭ normalized shear stress factor ␣ ϭ normalized half-width l ϭ spring length, in (cm) ␤ ϭ normalized half-thickness M t ϭ applied moment, lb ⅐ in (N ⅐ m) ␪ ϭ spring wind-up angle, rad m t ϭ normalized applied moment ␶ ϭ torsional stress, lb / in 2 (kPa) n ϭ number of blades ␶ a ϭ allowable torsional stress, lb / in 2 (kPa) n f ϭ selected number of blades R ϭ aspect ratio 2. Select the number of blades for this torsion leaf spring Use the equation, 3l(M / ␪ ) t nR ϭ 4 16Gb ƒ where the symbols are as given above. Substituting, nR ϭ 3(36)[1880 / ( ␲ /2)]/ 16(11.5 ϫ 10 6 )(0.0626 4 ) ϭ 46. Next, use the equation, nR 4 Ͻ R ϭϽ15, 0 n f letting n f ϭ 5; then R 0 ϭ 46/5 ϭ 9.2. 3. Choose the aspect ratio The required aspect ratio, R r , is established by employing the computed value for R 0 in step 2 as an initial estimate in the equation lM t ͩͪͩͪ G ␪ R ϭ i ϩ 1 16 3.36 1 4 nb Ϫ 1 Ϫ ͫͩͪͬ ƒƒ 4 3 R 12R ii where i ϭ 0,1,2, 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. SPRING SELECTION AND ANALYSIS 23.10 DESIGN ENGINEERING Usually, after three or so iterations, computed values R converge to a single value i ϩ 1 which is the required aspect ratio, R r . Substituting, using R 0 ϭ 9.2, as computed in step 2, for R 1 , 36 1880 ͩͪͩͪ 6 11.5 ϫ 10 ␲ /2 R ϭ 1 16 3.36 1 4 5(0.0625 ) Ϫ 1 Ϫ ͫͩ ͪͬ 4 3 9.2 12(9.2 ) 49.108 ϭ 16 3.36 1 Ϫ 1 Ϫ ͩͪ 4 3 9.2 12(9.2 ) ϭ 9.88 Using the R 1 value of 9.88 found above, substitute it in the equation again, 49.108 R ϭ 2 16 3.36 1 Ϫ 1 Ϫ ͩͪ 4 3 9.88 12(9.88 ) ϭ 9.83 ϭ R r Further iterations show that the aspect ratio converges to a value of 9.83. Therefore, let R r ϭ 9.83. 4. Determine the blade width Given R r ϭ 9.83 and 2b ƒ ϭ 0.125 in (0.3175 cm), the required blade width, 2a,is found from 2a r ϭ R r (2b ƒ ). Or, 2a r ϭ (9.83)(0.125) ϭ 0.123 in (0.31 cm). For manufacturing simplicity, the blades will be cut to a 1.25-in (3.18-cm) width from stock 0.125-in (3.18-cm) thick plates. Therefore, let 2a ƒ ϭ 1.25 in (3.18 cm). 5. Check the assist moment and stress The assist spring has been sized to have five 36-in (91.4-cm)-long blades with a 0.125-in (3.18-cm) wide by 0.125-in (3.18-cm) thick cross section. The assist mo- ment provided by such a spring stack is determined from KG ␪ M ϭ t l 4 nG b b 16 ƒƒƒ 3 ϭ ab Ϫ 3.36 1 Ϫ ␪ ͫͩͪͬ ƒƒ 4 l 3 a 12a ƒƒ Ϸ M t required Substituting, 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. SPRING SELECTION AND ANALYSIS [...]... ANALYSIS Determine the major dimensions of a ring spring made of material having an allowable stress of 175,000 lb / in2 (1.21 ϫ 109 Pa), E ϭ 29 ϫ 106 lb / in2 (199.9 ϫ 109 Pa), a coefficient of friction of 0.12, an inside diameter of 7.0 in (17.8 cm), an outside diameter of 9.0 in (22.9 cm) or less, a taper angle of 14Њ, an axial load of 56 tons (50.8 t), and a deflection of not more than 8.0 in (20.3 cm)... positions and dimensions (Product Engineering. ) FIGURE 13 Ring-spring compression constant in terms of the taper angle for various values of the coefficient of friction (Product Engineering. ) tangent of taper angle; hi ϭ height of inner ring, in Thus, wi ϭ [0.537 Ϫ (1.352 tan 14Њ) / 4] / 1.35 ϭ 0.314 (7.98 mm) Use a trial-and-error process to determine the dimensions of the outer ring Do this by assuming... small amount of room to spare As a trial, assume that the width of the spring wire ϭ 0.5 in (1.3 cm) ϭ a Since the spring is square, the height of the spring wire ϭ 0.5 in (1.3 cm) ϭ b With a 4-in (10.2-cm) outside diameter and a spring wire width of 0.5 in (1.3 cm), the mean radius of the spring coil rm ϭ 1.75 in (4.4 cm) This is the radius from the center of the spring to the center of the spring... subject to the Terms of Use as given at the website SPRING SELECTION AND ANALYSIS 23.26 DESIGN ENGINEERING Related Calculations Use the same procedure for square-wire helical torsion springs, but substitute the length of the side of the square for d in each equation where d appears TORSION-BAR SPRING ANALYSIS What must the diameter of a torsion bar be if it is to have a spring rate of 2400 lb ⅐ in /... (3.4 N ⅐ cm / cm3) Related Calculations Use this procedure for torsion-bar springs made of any metal The energy-storage capacity of various springs in terms of the spring weight is as follows: 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... proportional to the number of disks Friction between the surfaces of the disks will reduce the load capacity The total deflection of two or three Belleville springs arranged in parallel equals the deflection of any individual spring in the group In a series and parallel arrangement combined, the deflection range of single springs, the load capacity of the assembly, and the amount of friction damping can... The natural frequency of free vibration of this system is given by ƒϭ 1 2␲ ΊW kg where k ϭ spring modulus, lb / in (kg / cm); g ϭ gravitational constant, in / s2 (cm / s2); W ϭ weight of each spring, lb (kg) The amplitude of the forced vibration of this machine, A, is given by A ϭ xs 1 1 Ϫ ( ƒ1 / ƒ)2 where xs ϭ displacement of the spring, in (cm), caused by steady-state vibration of the machine; ƒ1 ϭ... frequency of the exciting force, cps; ƒ ϭ natural frequency of the system, cps 2 Set up the magnification factor The ratio of ƒ1 / ƒ is called the frequency ratio, and the ratio of 1 / (1 Ϫ [ ƒ1/ƒ]2) may be interpreted as the magnification factor Now, because of proportionality factors, we can say 20 ϭ 1200 1 (2000 / 60)2 1Ϫ ƒ2 ͫ ͬ Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)... 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 FIGURE 9 Recommended bending stresses for torsion springs (Product Engineering. ) SPRING SELECTION AND ANALYSIS 23.25 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)... The number of coils n assumed for this spring is based on past experience with similar springs However, where past experience does not exist, several trial values of n can be used until a spring of suitable deflection and length is obtained Related Calculations Use this general procedure to analyze helical coil compression or tension springs As a general guide, the outside diameter of a spring of this . use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 23.2 DESIGN ENGINEERING FIGURE 1 Spring-loaded. closed/mean diameter of coil ϭ 1.7 to 2.3. Determine (a) mean diameter of coil; (b) diameter of wire; (c) length of coil when closed, and (d ) length of coil before