1、COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling ServicesManual on Design and Manufacture of Torsion Bar Springs and Stabilizer Bars SAE HS-796 Report of the Spring Committee, last revised May 2000 SAE Information Report First edition published 1947
2、Second edition published 1966 Third edition published 1982 Fourth edition published 1991 Fifth edition published 2000 Society of Automotive Engineers, Inc. Warrendale, Pa. COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling ServicesAll technical reports
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5、ity for infringement of patents, trademarks, and copyrights. Copyright O 2000 Society of Automotive Engineers, Inc. 400 Commonwealth Drive Warrendale, PA 15096-0001 U.S.A. Phone: (724) 776-4841 E-mail: publicationssae.org http:llwww.sae.org Fa: (724) 776-5760 ISBN 0-7680-0629-5 All rights reserved.
6、Printed in the United States of America. Permission to photocopy for internal or personal use, or the internal or personal use of specific clients, is granted by SAE for libraries and other users registered with the Copyright Clearance Center (CCC), provided that the base fee of $SO per page is paid
7、 directly to CCC, 222 Rosewood Dr., Danvers, MA O1 923. Special requests should be addressed to the SAE Publications Group. 0-7680-0629-5100 $SO. COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling ServicesSPRING COMMITEE Richard L. Van Eerden (Chairper
8、son), Mentor Suspen- Michael L. Holly, General Motors Corporation Yuyi Lin, University of Missouri-Columbia Gordon D. Millar, Stelco Inc. Roy Lee OmdorE, Jr. Richard F. Pierman, Oxford Suspension Division Kenny E. Siler Fred B. Valdes, Triangle Auto Spring Co. sion Systems Co. Inc. TORSION BAR SPRIN
9、G COMMITTEE Richard L. Van Eerden (Chairperson), Meritor Suspen- Kenneth Campbell Michael L. Holly, General Motors Corporation Gordon D. Millar, Stelco Inc. Kenny E. Siler Steven A. Yollick, Meritor Suspension Systems Co. Inc. Warren J. Young, Freightliner Corporation sion Systems Co. Inc. COPYRIGHT
10、 SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling ServicesTABLE OF CONTENTS CHAPTER 1-INTRODUCTION . .:. 1 CHAPTER 2-DESIGN CALCULATIONS 1. Symbols Used in Formulae . 2. Round Bar in Torsion 3. Rectangular Bar in Torsion A. Straight Bar of Solid Rectangular Cr
11、oss Section . B. Straight Laminated Bar 4. Torsion Bar Spring and Le 4 5. Sample Computation . A. Requirements for Round Bar B. Computation for Round Bar C. Computation for Lami D. Computation for Lam . 9 . 10 6. Operating Stresses . A. Suspension Springs B. Other Torsion Bar Springs . . CHAPTER 3-D
12、ESIGN OF END FASTENING . 15 1. End Configuration. 15 A. Serrated End C B. Hexagonal End Co A. Serrated Anchor . . 16 B. Hexagonal Anchor 2. Anchor Member 3. Transition Section. CHAPTER 4-CONTROL OF ASSEMBLY POSITION 21 1. Methods to Insure Correct Vehicle Height . 21 A. Infinitesimal Adjustment by S
13、crew . C. Equal Serrations at B D. Blocked Serrations at 22 E Torsion Bar Suspension Settling . 22 2. Marking of Windup Direction . .22 B. Adjustment by Vernier Steps . E. Variable Height Systems CHAPTER 5-MATERIAL AN 23 1. General Requirements . . 23 A. Material . 23 B. Value of Shear Modulus . 24
14、C. End Fabrication . 24 D. Heat Treatment . . 25 E. Straightening . 25 E Shot Peening G. Presetting . H. Corrosion Protection I. Testing COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling Services2 . Presetting 26 A . Load Deflection Curve 26 B . Stres
15、s Distribution 27 C . Selection of Preset strain 29 3 . Induction Hardened Torsion Bar 30 CHAPTER positive as shown in Fig. 2.3, rad astatic = Angle a when PshtiC is applied, positive as shown in Fig. 2.1, rad = Angle between reference line and lever at zero load position; positive as shown M Fig. 2
16、.1, rad e = Windup angle when P is applied (= a + ), rad Ostatic = Windup angle when Pst.tic is applied (=.astatic + B) (Fig. 2.1), rad Ojounce = Windup angle when the maximum operating JOUNCE POSITION h, (SEE CHAPTER 2, SECTION 5A) STATIC LOAD TORSI ON REFERENCE LINE - h, (SEE CHAPTER 2, SECTION 5A
17、) v ACTUAL POSITION WHICH WOULD AT ZERO LOAD BE REACHED IF “k“ WERE CONSTANT Fig. 2.1-Relation of various load positions at end of lever 3 COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling Servicesload is applied, with the lever in jounce posi- tion (
18、= ajounee + B) (Fig. 2.1), rad = Diamegr of round bar (outside diameter of tubular bar), mm = Inside diameter of tubular bar, mm = Diameter of tapered bar at large end, mm = Diameter of tapered bar at small end, mm = Thickness of rectangular bar or square bar, = Width of rectangular bar (long side o
19、f cross = Active length of bar (defined in Chapter 3, = Length of lever (Fig. 2.3), mrn = Deflection at end of lever from reference line when P is applied, measured parallel to di- rection of P (= R sin a); positive as shown in Fig. 2.3, mm = Deflection at end of lever from reference line when Pstat
20、ic is applied, measured parallel to direction of PShuc (= R sin ashtic); positive as shown in Fig. 2.3, mm = Deflection at end of lever from zero load posi- tion to reference line, measured parallel to direction of P (= R sin B), mm fa + fb = Total static deflection at end of lever, from zero load p
21、osition to static load position, measured parallel to direction of Pshtic, mm = Effective static deflection at end of lever, equal to load divided by spring rate prevailing at that load, mm T = Torque applied to bar (= P R cos a), N mm k = Spring rate at end of lever (variable), mea- sured parallel
22、to direction of P, N/mm kT = Torsional spring rate of bar (= T/), N mm/rad Y = Shear strain d di d, dh t W L R f mm section), mm Section 3), mm fa fb 6 G MPa ss = Shear stress, MPa 7JZ 73 Ci = Load factor c2 = Rate factor c3 NL = Shear modulus (see Chapter 5, Section 1), = Saint Venants stress coeff
23、icient (Fig. 2.2) 7 Saint Venants stiffness coefficient (Fig. 2.2) defined in Chapter 2, = Static deflection factor I Section 4 = Number of laminae in laminated bar 2. Round Bar in Torsion Table 2.1 shows the relation which exists for the follow- ing conditions: a) Bar is straight. b) Uniform cross
24、section of solid cylindrical, tubular c) If tubular, inside and outside diameters are concen- d) Torsionally loaded only. e) Taperedbars have constant taper, with d, (at dis- tance x from end with diameter dmin) = dmin + (dmm - cylindrical, or solid tapered circular configuration. tric. also Nu lami
25、nae of f2, 2, T2, q2.2 and 73.2; also ND laminae. . . . : Windup angle: and so on Torsional rate: Stress: q3.18 ti G - q3.2 e t2 G s, = q2.i L q2.2 L and so on Stress rate: and so on Unit radian N * rnrnhad MPa MPa/rad 5 COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by I
26、nformation Handling ServicesIn actual practice, the dimensions for a laminated tor- sion bar can be established by the following procedure, considering the case where the complete bar is to have a rate kT and this cross section: From given data the maximum windup angle (8jounce) will be computed at
27、which all laminae may be subjected to the maximum allowable stress (Ss(jounm), with all lami- nae having the same active length (L), the maximum allowable thickness for each lamina is: Any lamina thicker than tma will be overstressed. All laminae are wound up to the same jounce and would carry the s
28、ame maximum stress if the thickness of all laminae could be held to tmW This would result in the most efficient spring or in a spring of minimum weight. In most cases this is not practical and the thickness of the laminae must be adjusted to give the required rate. It should, however, be kept in min
29、d that the stresses should be held as close as possible to the allowable maxi- mum stress in order to keep the spring weight to a mini- mum. The torsional rate of the complete bar is: In general, this result will not coincide with the desired value calculated from P R cos a kT = a+ See Chapter 2, Se
30、ction 4. If the rate as checked by load-deflection test tuns out too low, another lamina will have to be added. If it turns out too high, thinner laminae must be used, or the widths of the laminae must be reduced. It is important to remember that no laminae should exceed tmm, but all laminae should
31、be kept as close as possible to this thickness. The stress is not affected by a change in width (w) as long as the ratio qdqz = 1; this holds true for most practical applications where Consequently, changing w is a convenient way of rate adjustment if the widths of the laminae are not required to be
32、 held to existing commercial sizes. Changing the active length (I,) of the bar, provided the specific design allows this, is also a means of rate adjustment; however, L also affects the stress, and the whole computation will have to be repeated. (w/t) 3.5. 4. Torsion Bar Spring and Lever Fig. 2.3 sh
33、ows a combination of torsion bar spring and lever which is frequently used in suspensions. In this diagram the deflection fand the angles a and are mea- sured from a reference line which is perpendicular to the applied load and passes through the center of the torsion bar. They are counted positive
34、when their relations to the reference line are as shown in Fig. 2.3. Fig. 2.3-Torsion bar spring and lever The load deflection characteristics of this mechanism are not linear but are given by the following: T p=- R cos a T kT = =a+ T k - T-a+ - sin a f R - The function Ci is plotted in Fig. 2.4 aga
35、inst the ratio f/R or sin a as abscissa, so that the curves represent load-deflection diagrams for the end of the lever. The rate will be a minimum when dk/df is equal to zero, or when -3 sin a 2 sin2a + 1 ci = 6 COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Informati
36、on Handling ServicesFig. 2.4-Load factor versus deflection It will be noted from the chart that Ci become equal to zero when = -a. The vertical rate at the end of the lever is given by k = (dP/df). Using the value of P above and the relation f = R sin a P= dP da dP da -= -= f= tan a + 1 R (a + b) co
37、sa R R sin a df - = Rcosa da - R2 1 + (a + ) tan a cos2 a kT x - 1 + (a + B) tan a cos2 a c2 = Fig. 2.5 shows the function CZ plotted against the ratio 7 COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling Services_- -600-50“-40-30 -20 -10 o IO 20 30 40
38、 500 60 i =SIN d Fig. 2.5-Rate factor versus deflection f/R or sin a as abscissa. It shows that the rate of such a torsion bar and lever is by no means constant, and that the minimum rate occurs at a position of lever center- line below the horizontal. When f/R is zero then for all values of , Cz =
39、1.0. The rate will be a minimum when dk/df is equal to zero, or when Cz = 1/(2 sin2 a + 1) only for negative values of sin a. The static deflection at any point is defined as 6 = P/k. Using the above values this becomes cos a 1 6=R a + + 6 = RCS cos a 1 CJ = a + + tana Fig. 2.6 shows the function C3
40、 plotted against the ratio f/R or sin a as abscissa, so that the curves represent a comparison of static vertical deflection with the vertical deflection from the reference line. By definition, the ab- scissa values in Fig. 2.6 actually represent sin astatic or f,/R. The rate will be a minimum when
41、dk/df is equal to zero, or when C, = -3 sin a. 8 COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information Handling Servicesz I- V W 0 J h. W o i SIN d R Fig. 2.Mtatic deflection factor versus deflection It is to be noted that the curves for minimum rate, maximum stat
42、ic deflection, and = 90 deg all intersect at f/R = -1.0 or C3 = 3.0. Note: All angles in previous equations are expressed in radians. In Figs. 2.4-2.6, the angle is given in degrees, but for calculation of loads and rates from the constants Ci and C2 the torsional rate must be expressed in the dimen
43、sions N mmhadian. Where necessary, angles are converted from degrees to radians by the relation 1 ra- dian = 57.296 deg as shown in the examples. 5. Sample Computation A. Requirements for Round Bar Required: Determination of diameter (d) and active length (L) on a torsion bar spring of round cross s
44、ection for a passenger car suspension for the given values: Static load: Pstatic = K“I N Rate at static load: k = 16 N/mm Length of lever: R=400mm Position of lever at static load: astatic = 7 deg = M.122 rad 9 COPYRIGHT SAE International (Society of Automotive Engineers, Inc)Licensed by Information
45、 Handling ServicesDeflection, static load to jounce position: h, = 100 mm Deflection, static load to rebound position (with rebound stop): hr = 125 mm Stress at static load: Ssstatie I 670 MPa Stress at jounce position: c2 = 1.10 kR2 l6 = 2327000 N 1.10 Therefore, kT = - = c2 mm/rad Rechecking with
46、Fig. 2.4 or formula we obtain 2 327000(0.122 + 0.560) = 4000N - 400 0.993 The lever angle at jounce position is obtained from Ujounee = i-2 1.8 deg = H.38 radian The lever angle at the rebound position is obtained from fa - h, 48.8 - 125 -0.191 400 sin areb = = R areb = -11.0deg = -0.19 radian The t
47、orsional windup angle is + ajounee = 32 deg The maximum torque Trounce = kT ( + ajoUnce) = Then, + 21.8 deg = 53.8 deg = 0.94 radians. 2 327000 (0.56 + 0.38) = 2 187000 N mm 16Tjounce - 16 2 187000 Ss(jounee) 3.1416 900 = 12376 d3 = - d = 23.1 Use d = 23 rnrn 16PR 16 - 4OOO * 400 = 670MPa also it im
48、proves the environmental clearance conditions as its cross section approaches a circle. For this sample computation it will be assumed that a 35% reduction in active length is required (from 900 to 585 mm), that the stress at jounce position is to be limited to 860 MPa, and that w/t will exceed 4.0
49、so that 72 = 73. The requirements for load, angular position, and deflec- tion are to remain unchanged; therefore, the torsional rate is to be maintained at kT = 2 327 000 N mmhad. I Then (which is within 195% of the required rate, thus satisfac- tory) 7.0 7.04 Ss, = 860 - - 855MPa (This solution is satisfactory.) With density of steel = 7850 kg/m3: Mass = 7.85 585 352 = 5.6kg Required Clearance = diagonal of the square 1.414 35 = 49.5 mm 6. Operating Stresses A. Suspension Springs 860 - 585 = 7.04mm Suspension springs are generally loaded
