机械优化设计作业

1、4-2已知某汽车行驶速度x与每公里耗油量的函数关系为f(x)=x + 20/x,试用0.618法确定速度x在每分钟0.21公里时的经济速度 x*。精度£ =0.01.Private Sub Comma nd1_Click() a = 0.2 b = 110: a1 = b - 0.618 * (b - a) a2 = a + 0.618 * (b - a) f1 = a1 + 20 / a1 f2 = a2 + 20 / a2If f1 < =f2 The n b = a2: a2 = a1: f2 = f1: a1 = b - 0.618 * (b - a): f1 = a1

2、 + 20 / a1If f1 > f2 The n a = a1: a1 = a2: f1 = f2: a2 = a + 0.618 * (b - a): f2 = a2 + 20 / a2 If Abs(b - a) >= 0.01 The n GoTo 10Prin t "a*=" (a + b) / 2t = (a + b) / 2Prin t "f*=" t + 20 / tEnd Sub运行结果:. 996749540627092 f*= 21.D6igT0T2B9O15最终结果:a*=0.996749540627092f*=21

3、.061970726901515-1试用变量轮换法求目标函数f(X) = 4 +29x12242、-4X2 + X1 + 2x2 - 2X1X2 + X1 - 2x1 X2 的最优解。初始2 -点 X(0) = -2,2.2 T 精度 & = 0.000001。Private Sub Comma nd1_Click()n = 2: E = 0.000001Dim X(2), X0(2), P(2, 2)Forj = 1 To 2X0(j) = InputBox(X0(j)," 输入 X0 的 x、y 坐标") X(j) = X0(j)Next jGoSub 630K

4、 = 0Fori = 1 To 2Forj = 1 To 2P(i, j) = 0If i = j The n P(i, j) = 1Next j, i260: F0 = FFori = 1 To 2GoSub 470Next iR = 0: K = K + 1: Print FForj = 1 To 2R = R + (X(j) - X0(j) A 2Next jR = Sqr(R)If R < E The n GoTo 420i = nGoSub 470Forj = 1 To 2X0(j) = X(j)Next jGoTo 260470: H = 1: FX = 0 480: GoS

5、ub 600F1 = F: FX = FX + 1If F1 > F0 And FX > 1 The n GoTo 540If F1 > F0 And FX = 1 The n H = -HF0 = F1: H = 2 * HGoTo 480540: H = -0.5 * HIf 10000 * Abs(H) < E The n Return560: GoSub 600F0 = F1: F1 = FIf F1 < F0 The n GoTo 560GoTo 540600: For j = 1 To 2 X(j) = X(j) + H * P(i, j) Next

6、j 630: F = 4 + 2 / 9 * X(1) - 4 * X(2) + X(1) A 2 + 2 * X(2) A 2 - 2 * X(1) * X(2) + X(1) A 4 - 2 * X(1) A 2 * X(2)Return420: For j = 1 To 2Prin t "x*(" j; ")=" X(j)Next jPrint K; "f*=" FEnd Sub运行结果如下：-5 146848195091.21 E5T2T72755033-7. 44934968051013-7.8431&4622B51

7、43-6 0323078783148-8.12261795B5S9S5-& 16448497469126-8.1S3ST24039145-8 192802730968-8. 1K901T9229461-6 1987760776023-8 19963709547631-0.20002906424000-B.20D20T9597DTSB-6. 2COZ3956T54O42-8 20032678315644-6.2QO343750TTE&S-B 200351485B7215 2C03550U561&-8 20D356&1361345-6.20035735104019-

8、6 £0035763437559-0.2OO35703TO22T7-8 £0035790634650-6.20035793T961&1-8 2C035796236501-8.20035795892702-8 2003579619191-6.2OO3579632T0&3-8 2003579639062-8.2C03579641B4充-6.20035756431165-8.20035796437049-8 20035796439656-8.20035796440939-8.2003579644145-8.20035796441712x*( 1 )= 2.3141

9、631630715 x*( 2 )= 4. 33475T197E112T3720035796441712最优解：X1*= 2.314163X2*=4.834757F* = -8.200358最小体积二级圆柱齿轮减速器的最优设计如图所示的二级斜圆柱齿轮减速器，咼速轴输入功率Pi = 4.5kw，高速轴转速ni = 1450rpm ,总传动比i 2= 31.5，此轮的齿宽系数 “ a= 0.4 ;齿轮材料和热处理大齿轮45号钢正火HB = 187207 ,小齿轮45号钢调质HB = 228255.总工作时间不少于 10年。要求按总中心距 a三最小来确定总体 方案中的各个主要参数。减速器的总中心距计

10、算公式为=ai + a2 =12cos 3mnlZi(1 + ii) + mn2Z3(1 + i2)式中mn1, mn2 高速级与低速级的齿轮法面模数，mmi1, i2高速级与低速级传动比Z1 , Z3高速级与低速级小齿轮齿数3齿轮的螺旋角1. 选取设计变量计算总中心距涉及的独立参数有，故取X = mn1, mn2, Z1 , Z3, i1,3 = X1,X2,X3,X4,X5,X6】T2. 建立目标函数f(X) = X1X3 (1+X5) + X2X4 (1 + 31.5/ X5)/(2COS X6)3. 确定约束条件(1)确定约束条件的上下界限 从传递功率与转速可估计2< mn1&l

11、t; 5标准值(2, 2.5, 3, 4, 5)2< mn2 6标准值(3.5, 4, 5, 6)综合考虑传动平稳、轴向力不可太大，能满足短期过载，高速级与低速级大齿轮浸油深度大 致相近，轴齿轮的分度圆尺寸不能太小等因素，取：14W Z1< 2216W Z3< 225 -5.8< il< 78°w B 三 15°6 -由此建立12个不等式约束条件式# -gi（X） = xi -2 > 0g2（X） = 5 -xi > 0g3（ X）= X2 3.50g4（ X） = 6 -X2 > 0g5（X） = X3 -14> 0g

12、6（X） = 22 -X3> 0g7（X） = X4 T60g8（X） = 22 -X4> 0g9（ X ）= X5 -5.80gi0（X） = 7 -X50 gii（X） = X6 -0.13960gi2（X） = 0.2618 -冷0 （已将角度化成弧度）# -# -(2)按齿面接触强度公式# -# -925& = T3心皿 <SH , N/mm2bi得到高速级和低速级齿面接触强度条件分别为233.和mni Zi ii-cos3 f 028(925) K订i%2mn23Z33i2 da8(925) 2K2T2式中，S许用接触应力，MpaTi,T2分别为高速轴I和中

13、间轴II的转矩，N mmKi,K2分别为高速级和低速级载荷系数.(3)按轮齿弯曲强度计算公式-cos301.5 KiTibdi mniyi2< Sf i, N mm# -yi2Si W S 2, N mmy2得到高速级和低速级大小齿轮的弯曲强度条件分别为7 -# -S 1 帕 yi3 KiTiS 2 % y23 KiTi卡3幅y33 K2T232（1 + ii） mni Zi32（1 + ii） mni Zi32（1 + i2）mn2 Z32-cos f02-cos f02-cos0S 4 ©a y43 K2T2其中Si, S2, S3, S4分别为齿轮 1, 2, 3, 4的

14、许用弯曲应力， N/mm2;分别为齿轮1, 2, 3, 4的齿形系数.32（1 + i2）mn2 Z32-cos B02,yi , y2, y3, y4# -(4)按高速级大齿轮与低速轴不干涉相碰的条件a2 -E -de/2 >0得mn2Z3(1 + i2)-2 COS gE + mni) TlnlZiii0式中E低速轴轴线与高速级大齿轮齿顶圆之间的距离，mm;de2高速级大齿轮齿的齿顶圆直径，mm.对式至代入有关数据：雷=518.75 N mm22& 1=刘 3= 153.5 N mm ,沖2= Sf 4= 141.6 N mmTi =29638 N mm, T2 = 2874

15、9i1 N mmK1 = 1.225 , K2 = 1.204y1=0.248, y2=0.302 , y3=0.256, y4=0.302E = 50mm得g13(X) = 4.331 10X13X33X5 -COS3X6 > 0g14(X) = 1.431 10X23X43 -X5COS3X6 > 0g15(X) = 1.3981农(1+X5)X13X32-cos2X6> 0g18(X) = 1.57010(1+X5)X13X32-COS2X6> 0-43 222g16(X) = 1.514 10X(31.5 + X5)X2 X4 -X5 COS X6 > 0

16、-43222g19(X) = 1.647 1CK(31.5 + X5)X2X4 -X5 COS X6 > 02g17( X ) =X2X4 (31.5 + X5) -2X5COSX6(X1+5O) -C1X3X50g18(X)、g19(X)和g15(X卜g16(X)相比为明显的消极约束，可省略。共取g1(X)至g17(X)的17个约束条件。4. 选用合适的算法求解这一约束问题采用复合形法求解。在进行优化的过程中，6个变量都是作为连续变量处理的，因为齿轮的齿数应为整数，模数应取标准模数，所以最后对结果进行适当调整。Private Sub Comma nd1_Click()N = 6: E

17、= 0.01Dim X(13, 6), F(13), A(6), B(6)Forj = 1 To 6A(j) = In putBox(A(j),"输入估计边界的下界”)B(j) = InputB0X(B(j),"输入估计边界的上界 ”)Next j180: For I = 1 To 12190: GoSub 730If AA = 0 The n GoTo 190NeXt II0 = 1: I1 = 1: I2 = 1: K = 0240: For I = 1 To 12If F(I) > F(I0) The n I1 = I0: I0 = I: GoTo 280If

18、F(I) > F(I1) The n I1 = IIf F(I) < F(I2) The n I2 = I280: Next IK = K + 1: R = 0Forj = 1 To 6R = R + (X(IO, j) - X(I2, j)人 2Next jIf Sqr(R) < E The n ZZ = 0: GoTo 670340: For j = 1 To 6X(0, j) = 0For I = 1 To 12X(0, j) = X(0, j) + X(I, j)Next IX(0, j) = (X(0, j) - X(I0, j) / 11Next jI = 0Go

19、Sub 760If AA = 1 The n GoTo 480Forj = 1 To 6A(j) = X(I2, j): B(j) = X(0, j): X(1, j) = X(I2, j)Next jGoTo180480: H = 1.3: I = 13490: For j = 1 To 6X(13, j) = X(0, j) + H * (X(0, j) - X(10, j)Next jGoSub 760H = H / 2If AA = 0 The n GoTo 490If F(13) >= F(I1) The n GoTo 610Forj = 1 To 6X(I0, j) = X(

20、13, j)Next jF(I0) = F(13)GoTo 240610: If H > E The n GoTo 490730: For j = 1 To 6X(I, j) = A(j) + Rn d(1) * (B(j) - A(j)Next j760: AA = 0g1 = X(I, 1) - 2g2 = 5 - X(I, 1)g3 = X(I, 2) - 3.5g4 = 6 - X(I, 2)g5 = X(I, 3) - 14g6 = 22 - X(I, 3)g7 = X(I, 4) - 16g8 = 22 - X(I, 4)g9 = X(l, 5) - 5.8g10 = 7 -

21、 X(l, 5)g11 = X(l, 6) - 0.1396g12 = 0.2618 - X(l, 6)g13 = 4.331 *10 A-7 * X(l, 1) A 3 *X(l, 3) A 3 * X(l, 5) - Cos(X(l, 6) A 3g13 = 1.431 *10 a-5 * X(l, 2) a 3 *X(l, 4)卜 3 - X(l, 5) * Cos(X(l, 6)卜 3g15 = 1.398 * 10 a -4 * (1 + X(l, 5) * X(l, 1) a 3 * X(l, 3) a 2 - Cos(X(l, 6) a 2g16 = 1.514 *10 a-4

22、* (31.5 + X(l,5) * X(l, 2) a 3 * X(l, 4) a 2 - X(l, 5) a 2*Cos(X(l,6) a 2g17 = X(l, 2) *X(l,4) * (31.5 + X(l,5) - 2 * X(l, 5) * Cos(X(l, 6) * (X(l, 1)+ 50) -X(l,1) * X(l,3) * X(l, 5) a 2lf g1 >= 0 And g2 >= 0 And g3 >= 0 And g4 >= 0 And g5 >= 0 And g6 >= 0 And g7 >= 0 And g8 >= 0 And g9 >= 0 And g10 >= 0 And g11 >= 0 And g12 >= 0 And g13 &