最小体积二级圆柱齿轮减速器的最优设计
二级斜圆柱齿轮减速器,高速轴输入功率P1 = 4.5kw,高速轴转速n1 = 1450rpm,总传动比iΣ= 31.5,此轮的齿宽系数ψa= 0.4;齿轮材料和热处理大齿轮45号钢正火HB = 187~207,小齿轮45号钢调质HB = 228~255.总工作时间不少于10年。要求按总中心距aΣ最小来确定总体
方案
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中的各个主要参数。
减速器的总中心距计算公式为
aΣ = a1 + a2 =
式中mn1,mn2——高速级与低速级的齿轮法面模数,mm
i1,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
β——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,β]T = [x1,x2,x3,x4,x5,x6]T
2. 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≤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:
14≤Z1≤22
16≤Z3≤22
5.8≤i1≤7
80≤β≤150
由此建立12个不等式约束条件式
g1(X) = x1 – 2 ≥0
g2(X) = 5 – x1 ≥0
g3(X) = x2 – 3.5≥0
g4(X) = 6 – x2 ≥0
g5(X) = x3 – 14≥0
g6(X) = 22 – x3≥0
g7(X) = x4 – 16≥0
g8(X) = 22 – x4≥0
g9(X) = x5 – 5.8≥0
g10(X) = 7 – x5 ≥0
g11(X) = x6 –0.1396≥0
g12(X) = 0.2618 – x6≥0(已将角度化成弧度)
(2)按齿面接触强度公式
δH =
得到高速级和低速级齿面接触强度条件分别为
② eq \f([δH]2mn23Z33i2ψa,8(925)2K2T2)
– cos3β≥0
式中,[δH]——许用接触应力,Mpa
T1,T2——分别为高速轴I和中间轴II的转矩,N·mm
K1,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
δF1 =
δF2 = δF1
得到高速级和低速级大小齿轮的弯曲强度条件分别为
③ eq \f([δF]1ψa y1,3 K1T1)
(1 + i1) mn13Z12 – cos2β≥0
④ eq \f([δF]2ψa y2,3 K1T1)
(1 + i1) mn13Z12 – cos2β≥0
和 ⑤ eq \f([δF]3ψa y3,3 K2T2)
(1 + i2) mn23Z32 – cos2β≥0
其中[δF]1,[δF]2,[δF]3,[δF]4——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 – de2/2≥0
得 mn2Z3(1 + i2) – 2 cos β(E + mn1) –mn1Z1i1≥0 ⑦
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①至⑦代入有关数据:
[δH] = 518.75 N·mm2
[δF]1= [δF]3= 153.5 N·mm,[δF]2= [δF]4= 141.6 N·mm2
T1 =29638 N·mm,T2 = 28749i1 N·mm
K1 = 1.225,K2 = 1.204
y1=0.248,y2=0.302,y3=0.256,y4=0.302
E = 50mm
得
g13(X) = 4.331×10-7x13x33x5 – cos3x6 ≥0
g14(X) = 1.431×10-5x23x43 – x5cos3x6 ≥0
g15(X) = 1.398×10-4(1 + x5)x13x32 – cos2x6 ≥0
g18(X) = 1.570×10-4(1 + x5)x13x32 – cos2x6 ≥0
g16(X) = 1.514×10-4(31.5 + x5)x23x42 – x52cos2x6 ≥0
g19(X) = 1.647×10-4(31.5 + x5)x23x42 – x52cos2x6 ≥0
g17(X) = x2x4 (31.5 + x5) – 2x5cos x6 (x1+50) –x1x3x52≥0
g18(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。
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 Command1_Click()
N = 6: E = 0.01
Dim X(13, 6), F(13), A(6), B(6)
For j = 1 To 6
A(j) = InputBox(A(j), "输入估计边界的下界")
B(j) = InputBox(B(j), "输入估计边界的上界")
Next j
180: For I = 1 To 12
190: GoSub 730
If AA = 0 Then GoTo 190
Next I
I0 = 1: I1 = 1: I2 = 1: K = 0
240: For I = 1 To 12
If F(I) > F(I0) Then I1 = I0: I0 = I: GoTo 280
If F(I) > F(I1) Then I1 = I
If F(I) < F(I2) Then I2 = I
280: Next I
K = K + 1: R = 0
For j = 1 To 6
R = R + (X(I0, j) - X(I2, j)) ^ 2
Next j
If Sqr(R) < E Then ZZ = 0: GoTo 670
340: For j = 1 To 6
X(0, j) = 0
For I = 1 To 12
X(0, j) = X(0, j) + X(I, j)
Next I
X(0, j) = (X(0, j) - X(I0, j)) / 11
Next j
I = 0
GoSub 760
If AA = 1 Then GoTo 480
For j = 1 To 6
A(j) = X(I2, j): B(j) = X(0, j): X(1, j) = X(I2, j)
Next j
GoTo 180
480: H = 1.3: I = 13
490: For j = 1 To 6
X(13, j) = X(0, j) + H * (X(0, j) - X(10, j))
Next j
GoSub 760
H = H / 2
If AA = 0 Then GoTo 490
If F(13) >= F(I1) Then GoTo 610
For j = 1 To 6
X(I0, j) = X(13, j)
Next j
F(I0) = F(13)
GoTo 240
610: If H > E Then GoTo 490
730: For j = 1 To 6
X(I, j) = A(j) + Rnd(1) * (B(j) - A(j))
Next j
760: AA = 0
g1 = X(I, 1) - 2
g2 = 5 - X(I, 1)
g3 = X(I, 2) - 3.5
g4 = 6 - X(I, 2)
g5 = X(I, 3) - 14
g6 = 22 - X(I, 3)
g7 = X(I, 4) - 16
g8 = 22 - X(I, 4)
g9 = X(I, 5) - 5.8
g10 = 7 - X(I, 5)
g11 = X(I, 6) - 0.1396
g12 = 0.2618 - X(I, 6)
g13 = 4.331 * 10 ^ -7 * X(I, 1) ^ 3 * X(I, 3) ^ 3 * X(I, 5) - Cos(X(I, 6)) ^ 3
g13 = 1.431 * 10 ^ -5 * X(I, 2) ^ 3 * X(I, 4) ^ 3 - X(I, 5) * Cos(X(I, 6)) ^ 3
g15 = 1.398 * 10 ^ -4 * (1 + X(I, 5)) * X(I, 1) ^ 3 * X(I, 3) ^ 2 - Cos(X(I, 6)) ^ 2
g16 = 1.514 * 10 ^ -4 * (31.5 + X(I, 5)) * X(I, 2) ^ 3 * X(I, 4) ^ 2 - X(I, 5) ^ 2 * Cos(X(I, 6)) ^ 2
g17 = X(I, 2) * X(I, 4) * (31.5 + X(I, 5)) - 2 * X(I, 5) * Cos(X(I, 6)) * (X(I, 1) + 50) - X(I, 1) * X(I, 3) * X(I, 5) ^ 2
If 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 >= 0 And g14 >= 0 And g15 >= 0 And g16 >= 0 And g17 >= 0 Then GoTo 800
Return
800: AA = 1
F(I) = (X(I, 1) * X(I, 3) * (1 + X(I, 5)) + X(I, 2) * X(I, 4) * (1 + 31.5 / X(I, 5))) / (2 * Cos(X(I, 6)))
Return
For j = 1 To 6
C = X(I0, j): X(I0, j) = X(I1, j): X(I1, j) = C
Next j
C = F(I0): F(I0) = F(I1): F(I1) = C: ZZ = 1
670: Print
For j = 1 To 6
Print "x*("; j; ")="; X(I2, j)
Next j
Print K; "F*="; F(I2)
If ZZ = 1 Then GoTo 340
End Sub
运算结果如下:
齿数取整数,模数取标准模数,取mn1=2.5,mn2=5,取Z1=15,Z3=18,i1=6.24,β=0.164215rad,则i2=5.05,Z2=94,Z4=91,螺旋角β=9024' 32''。
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