1、由此建立12个不等式约束条件式g1(X) = x1 3.5 0g2(X) = 8 x1 0g3(X) = x2 3.50g4(X) = 10 x2 0g5(X) = x3 140g6(X) = 22 x30g7(X) = x4 160g8(X) = 22 x40g9(X) = x5 5.80g10(X) = 7 x5 0g11(X) = x6 80 g12(X) = 15 x60 (2)按齿面接触强度公式H = H,N/mm2得到高速级和低速级齿面接触强度条件分别为 cos30 cos30 式中,H许用接触应力,Mpa T1,T2分别为高速轴和中间轴的转矩,Nmm K1,K2分别为高速级和低速
2、级载荷系数.(3)按轮齿弯曲强度计算公式F1 = F1,Nmm2F2 = F1 F2,N得到高速级和低速级大小齿轮的弯曲强度条件分别为(1 + i1) mn13Z12 cos20 和 (1 + i2) mn23Z32 cos20 (1 + i2) mn23Z32 cos20 其中F1,F2,F3,F4分别为齿轮1,2,3,4的许用弯曲应力,N/mm2; y1,y2,y3,y4分别为齿轮1,2,3,4的齿形系数.(4)按高速级大齿轮与低速轴不干涉相碰的条件a2 E de2/20得 mn2Z3(1 + i2) 2 cos (E + mn1) mn1Z1i10 式中E低速轴轴线与高速级大齿轮齿顶圆之
3、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:H = 836 NF1= F3=444Nmm,F2= F4= 410.3NT1 =144700Nmm,T2 = 146789i1 NK1 = 1.225,K2 = 1.204y1=0.248,y2=0.302,y3=0.256,y4=0.302E = 50mm得g13(X) = 5.310-6x13x33x5 cos3x6 0g14(X) = 2.31710-5x23x43 x5cos3x6 0g15(X) = 3.11710-4(1 + x5)x13x32 cos2x6 0g18(X) = 3.42210-5(1
4、+ x5)x13x32 cos2x6 0g16(X) = 3.4510-6(31.5 + x5)x23x42 x52cos2x6 0g19(X) = 3.3210-5(31.5 + x5)x23x42 x52cos2x6 0g17(X) = x2x4 (31.5 + x5) 2x5cos x6 (x1+50) x1x3x520g18(X)、g19(X)和g15(X)、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。 至此已形成了完整的数学模型。2初始搜索区间的确定1) 将x0=3+;4;17;20;6;8;代入目标函数得2) 确定其搜索区间a,b。设初始
5、点,初始步长h=1。 :function a,b= jiana0=0;h=1;a1=a0+h;f0=60.1*a0+1/(1+a0)+1/(5-a0) ;f1=60.1*a1+1/(1+a1)+1/(5-a1) ;if f1=f0 while true h=2*h; a2=a1+h; f2=60.1*a2+1/(1+a2)+1/(5-a2) ; if f1f0 a2=a0-h; %大的a1赋给a2作为边界,a1则缩短再比较,若此时f1f0了,则结束 f2=60.1*a2+1/(1+a2)+1/(5-a2); if f0=f2 a=a2; b=a1; a1=a0; a0=a2; f1=f0; f
6、0=f2;ab得到运行结果:a= -2.80,b=-1.12所以初始区间为-2.80,-1.12。3然后用黄金分割法求其最优解以下是基于c-free4.0上运行的C程序:#include math.hvoid main() float a1,a2,f1,f2,f3,a,b,am; a=-2.8089,b=-1.1236; while(b-a0.01) a1=b-0.618*(b-a), f1=60.1*a1+1/(1+a1)+1/(5-a1), a2=a+0.618*(b-a), f2=60.1*a2+1/(1+a2)+1/(5-a2) , if(f1= 维数n */#define pi 3.
7、1416 float f(float x);void mjtf(int n,float x0,float h,float s,float a,float b);void mhjfgf(int n,float a,float b,float flag,float x);void mpowell(int n,float x0,float h,float flag1,float flag2,float a,float b,float x);float f(float x) float result; result=(x0*x2*(1+x4)+x1*x3*(1+31.5/x4)/2/cos(x5*pi
8、/180) +1/(x0-3.5)+1/(8-x0)+1/(x1-3.5)+1/(10-x1)+1/(x2-14)+1/(22-x2) +1/(x3-16)+1/(22-x3)+1/(x4-5.8)+1/(7-x4)+1/(x5-8)+1/(15-x5) +1/(x0*x2-68.3722)+1/(x1*x3-66.7325)+ 1/(0.0003099*(1+x4)*x0*x0*x0*x2*x2-cos(x5*pi/180)*cos(x5*pi/180) +1/(0.00000354*(31.5+x4)*x1*x1*x1*x3*x3-x4*x4*cos(x5*pi/180)*cos(x5*p
9、i/180) +1/(x1*x3*(x4+31.5)-x4*(2*(x0+50)*cos(x5*pi/180)+x0*x2*x4); return result;/*搜索区间子程序*/void mjtf(int n,float x0,float h,float s,float a,float b) a0=2,b0=10; a1=2,b1=13; a2=14,b2=22; a3=16,b3=22; a4=5,b4=7; a5=8,b5=15;/*多维黄金分割法子程序*/void mhjfgf(int n,float a,float b,float flag,float x) int i; float x1m,x2m,f1,f2,sum; for(i=0;iflag); xi=(float)0.5*(bi+ai); /*已达到,输出极小点*/*鲍威尔法子程序*/void mpowell(int n,float x0,float h,float flag1,
copyright@ 2008-2022 冰豆网网站版权所有
经营许可证编号:鄂ICP备2022015515号-1