RCVD第21章翻译

1、MechanicalMechanicalLead,Lead, W W(FerperxiiGular(FerperxiiGular toto R)R)再次翻译RCVD，悬架弹簧，求指教在这章中将谈到近似计算悬架系统弹簧一些特定性质的方法。我们把注意力主要集中在弹簧刚度最大应力，并且还会涉及疲劳度。我们将会涉及扭力弹簧，螺旋弹簧和钢板弹簧。这章涉及的内容对于设计工程计算是不够的。为了达到这个目标，读者必须查看其他有关文献特别是SAE设计手册还有弹簧制造商的手册。21.1扭力弹簧扭力弹簧中，扭力中长而细的杆的塑性被用以产生线性的弹簧刚度，就如螺旋弹簧那样。扭力弹簧的负载或者说，力，通常围绕着杆的中心

2、线，通过一根平衡臂（在一头或者两头， 在两头的话就是anti-roll bar扭力杆）传导扭矩。|有一个图1F itru re 21.1 Basic 牌门龙 etFy of a t orsion spr实际上，负载并不是总是与 R垂直（R的意义见图一）所以在精确的工程设计计算时扭转角校正要考虑进去。（Ref7）WR成比例，并和基本维度d成反扭力杆自身可以有任何数量的不同截面，从最广泛使用的圆杆到椭圆杆或者长方形的杆。我们将主要讨论圆杆和方杆。考虑到应力，弹簧刚度等，平衡杆的配置现在还并不重要。它唯一的用处就是将 W沿着杆的中心线转换成扭矩（ WR）。当轮载荷（1）直接体现在点 弹簧刚度（2）也

3、正好是悬架刚度（3）。最大应力当外扭矩施加在扭力杆上时，铲伤的阻力矩可以表示为横向集合和最大剪应力f的函数.如果把阻力矩称为 M，我们可以在任何工程手册中找到对于不同杆的表达。圆杆和方杆的阻力矩公式为Circular M -Ibin.Square bar: M = 434.8, lb.-in,1圆杆M=因为阻力矩M等于施加的力矩 RW (忽略摩擦的扭矩)我们可以把最大剪应力表示为2Circular bar: f= 16WR/Ed34 psi. (in. and lb. uniSquare bar; f- 4.8 WRAP, psiCircular bar: f= 16WR/Ed34 psi (

4、in. and 1b. units)Square bar; f- 4.8 WRAP, psi在上面两个等式中，我们看到最大应力被表示成直接与 比 它不是杆长 L的函数（L设计到了扭转角和弹簧刚度的确定）为了在方杆和圆杆之间做一个直观的对比，我们可以关注施加每单位扭矩产生的最大应力。使用上述的等式我们得到f?WR = kCircular bar: k = I6/xd* psi/lbi-in.Square bar: k = 4.8/d, psi/lb.4ji.3图21.2是一个k-杆区域的图，显示了对于相等的横截面区域，圆杆有一个更低的K差别在低的相等区域最明显，但是随着这个equal area的

5、增长而减少比如，在1.5英寸平方的相等区域，圆杆有一个28%的K。这个比方杆的要小。我们在检验弹簧刚度之前不能首先认为这个特点能产生显著的优点。扭杆弹簧的弹簧刚度线性的弹簧刚度 S,对于扭杆/平衡杆装置，表示的是在平衡杆尾部（at the end of the lever arm）每单位直线挠度产生的负载（详细的知识见材料力学挠度和负载一章。好像是第五章。弯曲应力）S = WX = W/R9. lb./in.4只有在角度足够小的时候这个表达式才足够准确。对于大的角度（对应一个来自于垂直W和R偏差）需要采用更加准确的方法。扭转角，在工程手册中针对不同的横截面配置已经给出。比如，对于角度，圆杆和方

6、杆的 公式分别是Figure 2L2 Maximum stressptrr unit applied torque, k, vs. bar area.;q一二sd yCircular: Q - 32TLfed4G , rad.Square: 0 = 7ITL/d*G , rad.where T = applied torque, WRG - Sheaf modulus of the material, pmiTo convert rad. to deg+, multiply by 57.3The equations show that for a given G, 0 increases wi

7、th applied torque, T* and with bar length, L, If the 0 expressions are put intc EQ. (2L6) the spring rates become:Circular: S = (O.O98)d4G/R2L; IbJiit(21.9)Stfuare: S-(0.141 )d4GfL , lb冷(21.10)We can compare the two bars by dividing the first cquarion by the second; using sub- Rcnpts c (Gircular bar

8、) and s (square bar) we havs for the sameG, and the saniR2JSc/Ss =Q695(d?/iJ(2141)对于一个给定的 G,角度随着施加的扭矩T和杆长L增加。如果角度的公式被放入21.6中，弹簧刚度变成了21.921.10我们可以通过把第一个式子除以秒来比较两种杆。使用sub-scripts c（圆杆）和（方杆）对于同一材料 G，和相同的R平方L如果我们把相等区域标准用在先前的部分我们得至21。11J或者说，对于相同的区域（圆的直径是方形边的 1.127倍）对于相同的G/R平方L圆杆有1 3%更强的弹簧刚度。这个事实，和之前圆杆有更低

9、的K （每施加一单位扭矩产生的应力）的结论一起，说明了除非方杆有特定的配置优势，比如更简便或者更轻的机械固定硬件（比如端部固定件）或者平衡杆硬件，方杆对圆杆没有特别的优势。If we apply the equal area ciiterion used in the previous section we have忒A =吃ajiddg/dj = 4/TC = L273from which处/d? = 1.2732 = L.621Wlrn this result is substituted into Eq. (2L11) we haveSc- 0.695(L621)Sa- M27 Ss(2

10、1,12)Il other words, for the same area (the diameter of lh-e circle is LI27 times the side of the square), and for the same G/R2L the circular bar lus a 13% stiffer spring ratu This fact, together with the previous conclusion that the circular bar has a lower k (stress per unit applied torque),昭 ys

11、that there is no particular advantage of the square bar over the circular bai uni诙5 the square bar has certain installation advantages such as simpler or lighter m時dnie曲 grounding hardware (end fastening or lever arm hiirdwarc,剪切模量和应力水平不同钢材的剪切模量随着材料的不同变化很小。ShearShear ModulusModulus andand StressStre

12、ss LevelsLevelsThe shear modulus of different steels varies little from one to aoother. Ref 19 gives the following data:TYD& of inaterialShear Modulus, psiCold Rolled Steel11.5 x 106Stainless Steel (18-8)10.6x106Carbon heat-treated11011.9 x 10#Titmum5.86.2x 106如下图冷轧钢不锈钢碳热处理在设计和应力计算中的应用，我们推荐11.0*10的六

13、次方这个值当然有些其他的材料性能也在考虑之中，比如烘烤和加工（ machi neability）参见表7 （文中没有）对于扭杆弹簧，典型的最大操作应力水平在表21.1中给出（在参考表 7中是表2.2）表中应力的单位是Mpa乘以145得到psi乘用车和卡车的范围大概是800-900Mpa 这和116，000-130，OOOpsiTableTable 21.121.1 OperatingOperating StressStress LevelsLevelsRoundRound TorsionTorsion BarsBars (from(from Ref.Ref. 7)7)LoadLoad Dire

14、ctionDirectionSAESAE SteolSteol (Typteal)(Typteal)ClassClassNo.No.OperatingOperating ShearShear StressStress MaxMax (MPa)(MPa)SettlingSettling % % ofof MaxMax DeflectionDeflectionSingioSingio OnlyOnlyReverseReverseShotShot PeenedPeenedPresetPresetDiameterDiameter UptoUpto 4040 mmmmDiameterDiameter U

15、pUp toto 6565 mmmmTypicalTypical ApplicationApplication1 1 1250125012501250I I 1010 MaxMaxYitSYitSYesYesYesYesF206HF206H (MOD)(MOD)I I 12001200| | 120012001010 MaxMaxY&iY&iYesYesYesYesI Ip p1150115010Mx10MxYgYgYosYos伽j11001100110011001QMx1QMxYesYesYes伽；H43400H43400 (ESRF(ESRFH43500H43500 ESR? requir

16、edQ1C450Q1C450 totoG10900G10900HatchHatch covers,covers, bds.bds. doors.doors. macWnmacWn componentscomponents IcrWo*IcrWo* twtw IK OUM m6ciorcnotrciorcnotr mrycwmbimrycwmbitoto haroTwOBCthofhnMwwityharoTwOBCthofhnMwwity whrtHMNvwJ*whrtHMNvwJ* *y*yb b 1ESH1ESHCteoiioCteoiio 如)AeowliAeowliPBi-UPaiOOO

17、aKPBi-UPaiOOOaKSampleSample CalculationsCalculationsFor some sample calculations we will use a neutral steer car in which:WT = 1500 lb, (total weight)a 5.6 ft. (CG to front axle)Z = 8 ft. (wheelbase)b = 2.4 ft. (CG to rear axle)The calculations are for round torsion bar springs at e front suspension

18、s.1. Front wheel weight = W = (1(b )(WT) = (1/2X2.4/8.0)(1500) = 225 lb.2. We will select the wheel spring rate on the basis of a desired undamped natural frequency of 2 Hz:3= jK/m = 2兀=12.56 rad./sec.Squaring both sides, we haveK= 12.562(225)/32.2(12) = 91.9 lb./in.3. Using this spring rate and Eq.

19、 (21.9) wc haveS = 0.098G/L = 91.9 lb./in.Using a G of 11 x 106 psi this becomesR2L/d4 = 0.098(1 小106)/ 91.9 = 11,730(a)4. Using Eq. (21.3) and setting the maximum stress at 100,000 psi we have4Using Eq. (21.3) and setting the maximum stress at 100,000 psi we have100.000= 16WR/Tid3= 16(225)Rz7td3oro

20、rR/TUP = 27.78(b)Eqs. (a) and (b) can now be used to determine the required d, R, and L In this example we will arbitrarily select an R = 5 in. Then from (b).d3 = 5 加(27.78) = 0.0573 andd = 0.386 in. diameterPutting this d and R (5 in J into (a),L= 11730(0386尸/25 = 10.41 in. lengthSampleSample Calcu

21、lationsCalculationsFor some sample calculations we will use a neutral steer car in which:WT = 1500 lb, (total weight)a 5.6 ft. (CG to front axle)Z = 8 ft. (wheelbase)b = 2.4 ft. (CG to rear axle)The calculations are for round torsion bar springs at e front suspensions.1. Front wheel weight = W = (1(

22、b )(WT) = (1/2X2.4/8.0)(1500) = 225 lb.2. We will select the wheel spring rate on the basis of a desired undamped natural frequency of 2 Hz:3 = jK/m = 2兀=12.56 rad./sec.Squaring both sides, we haveK= 12.562(225)/32.2(12) = 91.9 lb./in.3. Using this spring rate and Eq. (21.9) wc haveS = 0.098G/L = 91

23、.9 lb./in.Using a G of 11 x 106 psi this becomesR2L/d4 = 0.098(1 小106)/ 91.9 = 11,730(a)4. Using Eq. (21.3) and setting the maximum stress at 100,000 psi we have4 Using Eq (21.3 and setting th maximum stress at 100.000 psi we have100.000 = 16WR/nd3 = 16(225)R/WorR/TUP = 27.78(b)Eqs. (a) and (b) can

24、now be used to determine the TCquireci d. R. and L. In this example we will arbitrarily select an = 5 in* Then from (b),d3 = 5(2758) = 0.0573andd = 0 x386 im diameterPutting this d and R (5 in J into (a),L= 11730(0.386)4/25 匸 10.41 tn. length算例21.2螺旋弹簧螺旋弹簧利用线的扭转产生直线弹簧刚度系数。它在汽车的独立悬挂中得到广泛的使用并且也在非独立悬架中

25、使用( Solid axle suspension).最常见的形状就是螺旋的直径是常数 的螺旋状，锥形弹簧圈的直径随线长改变。螺旋弹簧可以被设计成压缩或者伸长的形式。在本书中，我们将注意力放在直径不变，压缩的螺旋状弹簧上(就是不是锥形的螺旋弹簧)下图d - Wire diameierFigure 213, taken frotn Ref. 6* gives the nomjenclature for the helica ceil spring and typical entfe that are used for compression springs.(b) TYPICAL EM DS O

26、F HELI CAL COM PR ESS IO M SPRINGS图21.3 (a)中显示的重要参数是无负载弹簧的自由长度L (base length)d:弹簧的线圈直径(粗细)。D :线圈中心直径(介于外径和内径之间) 有效螺旋数目 N图21.3 ( b)显示了决定有效螺旋圈数目的端部处理。非Jnslde diameterOutside diameterL - Free Length+Space between coilsN f No. of active coilsGround section - CON diameter SPRING DETAILS AND TERMENOLOGY (10 Total coils shown)Closed endsClosed and ground ndeFigure 2L3 Helical coil compresshyn sipyingy (Ref 6).被安装在特殊形状的弹簧座上。有平末端固定(with plain ends ground)被认为是在每一个末端有1/2的闲置螺旋。这样参数N在弹簧刚度的确定中占有了重要的地位，但是从以上可以看出在给N赋值时也有很多的