外文翻译---行星齿轮变速箱的设计研究

附录 A

英文资料

Research on Design of Planetary Gearbox

Nanjing Artillery College RaoZhenggang

［Abstract］The design and calculations of planetary gearbox are detailedly discussed in this thesis.The calculations of structure

parameters,transmissionratio,torque of each shaft,braketorque,lockingmoment,and transmission power are explained here.Hence,this paper is of much importance to the design of planetary gearbox of caterpillar vehicle,tank,artillery and engineering mechanics.

Key words: Planetary gearbox Structure parameter Torque Power

c)Clutch L locks internal gear rign b and swiveling jib H(see figure 3,c)

When clutch L combines gear ring b and swiveling jib H into one

whole,n b=n H.Then torque T A of input shaft A can transmit through two ways:one is to transmit to output shaft B through central gear a,planet gear g and swiveling jib H;the other is to B by central gear a,planet gear g,internal gear ring b and clutch L.At this moment,T A=T a,T B=-T A and locking moment T L is equal to the torque of internal gear ring T b,T L=T b.According to formula (4-3),the locking moment T L is

T L=T b=pT a(4-11)

From the formula above,when central gear a inputs and swiveling jib H outputs,the locking moment T L of the first scheme is the lowest and the size is smaller.So as to 2K-H(A) planetary gearbox,the first scheme is more reasonable.

(1)Suppose internal gear ring b inputs,swiveling jib H outputs(see figure

4),T A=T b,T B=T H;Let us discuss the locking moment T1 of friction clutch of three different locking pattern.

a)Clutch L locks intermal gear ring b and centra gear a(see figure 4,a),that is

n a=n b=n H;then the planet bar becomes a whole and rotates.So

T B=-T A,T1=T a.According to formula (4-2),T L is:

(4-12)

So,

b)Clutch L locks central gear a and swiveling jib H(see figure 4,b),that is n H=n a and T L=T a.So T L is

So,

(4-13)

c)Clutch L locks internal gear ring b and swiveling jib H(see figure 4,c),that is n H=n b and T L=T b.Since the clutch L links input shaft A and output shaft B,locking moment T L is equal to the torque transmitted by shaft A,namely T L=T b=T A.So the locking moment T L is

T L=T A

From the formula above,when internal gear ring inputs and swiveling jib outputs,locking moment T L of the first scheme is the lowest.So it is the best to adopt the first locking scheme.

After T L is calculated,the necessary control units can be adopted to obtain the friction clutch with smaller size.But,infact,the best locking scheme can't be always realized due to difficulty instructrualdistribution.Under this situation,the locking scheme should be adopted to get more compact structure of clutch and smaller size of the gearbox.

5Calculation of efficiency of planetary gearbox

As to multi-degree planetary geabox,the ratio of output power P B and input power P A is called the transmission efficiency of planetary https://www.360docs.net/doc/553734425.html,ly,

(5-1)

Where,T A、n A—torque and rotational speed of input shaft A；

T B、n B—torque and rotational speed of output shaft B.

Ratio n A/n B=i p is kinematics transmission ratio;ratio-T B/T A=i p～is dynamics

transmission ratio.Since T B has contains the friction loss of planetary

gearbox,puttingip and i p～into formula(5-1),we can get the famous M.A.Kpe-Йюec formula:

(5-2)

So the efficiency of each step of multi-degree planetary g earbox ηp is equal to the ratio of its dynamics transmission ratio i p～and kinematics transmission ratio i p.

As it is said above,its kinematics transmission ratio i p is function of characteristic parameter p of planet bar,that is:

i p=f(p1、p2、…p n)(5-3)

According to the relation of power equation,p is equal to i p multiplying (ηH)x then the function of p is obtained:

(5-4)

Where,ηH—transmission efficiency of planet bar in relative motion.As to 2K-H(A) planet bar,ηH=ηH ag×ηH gb=0.97×0.99=0.96

The equation of index x j is

(5-5)

j=1,2,…，n

The mathematical meaning of sign is symbol.

Index x j=1,its symbol is decided by (5-5)；if the value of sign is more than 0,sign is “+”，and xj=+1;otherwise,xj=-1。

6Sample of the design of planetary gearbox

Suppose the transmission sketch of planetary gearbox of one certain caterpillar truck is shown as figure 5.Its degerrs of freedom is w=3,the number of steps is

n d=4(three forwarding steps and one reverse step).Try to calculate (1) number of control units m,number of planet bars k and number of basic components

n0;(2)transmission ratio i p of each step;(3)torque T,brake moment T2 and locking moment T L of each component;(4)transmission defficiencyηp of each step.

Solution:(1)Calculation of the numbers of control units m,planet bars k and basic components n0.

When degree of freedom w=3,the number of control components m can be calculated through formula (2-6,a),that is

Put n d=4 into the formula above,

m2-m-8=0

I t can be obtained from the equation that:m≈3.4；Because C2m＞m＞W，the number of control units is:：m=W+1=4

Then the number of friction clutches L can be got according to formula(2-8)

L=W-1=3-1=2

The number of brakes Z can be calculated throu-gh(2-9)

Z=m-L=4-2=2

The number of planet bars k is obtained by(2-12)

k=n d-(W-1)=4-(3-1)=2

Finally,the number of basic motive components n0 can be calculated through formula(1-3).

n0=W+k=3+2=5

It is shown on figure 5 that its basic components are A、B、1、2 and 3.

(2)Transmission ratio i p of each step

The module m of the gear in two planet bars of the said planetary gearbox is the same and m=4mm.The characteristic parameters are p1=p2=3.105.Because p1=Z b1/Z a1 and p2=Z b2/Z a2and normally the number of teeth of central gear a Z a＞17.we suppose Z a1=Z a2=19.Then the number of teeth of internal gear ring b is:

Z b1=p1z a1=3.105 19=59

Z b2=p2z a2=3.105 19=59

The number of teeth of planet gear Zg is

The same as above,，Z g2=20

Since the planetary gearbox has three forward steps and one reverse step,each step can be reached by combining two control components.The method to combine the control units is shown in table 6-1.

The transmission ratios of each step of the planetary gearbox is calculated as

followes:

The transmission ratio of the first step(direct step):

The first step of planetary gearbox is obtained by locking two friction clutches L1 and L2.They combine central gear a1,a2 internal gear ring b2 and output shaft B into a whole,then two planet bars X1 and X2 can rotate as one body.There is no relative motion among each unit,that is n A=n H1=n b1=n a1=n a2=n H2=n b2=n B,then the transmission ratio is

i=i AB=i H1b2=1

The transmission ratio of the second step:

The second step of planetary gearbox is obtained by locking brake Z2 and Clutch L2.The twoplanet bars participate in work.After input from input shaft A,the transmission power P distributed through swiveling jib in two ways:one is through gear g1,a1,a2 and clutch L2 to output shaft B;the other is through gear g1,b1,swiveling jib H2,gear g2,a2 and clutch L2 to output shaft B.Now it is known that brake Z2 is locked,that is n b2=0.

According to the kinematics equation of planet bar(3-1),the transmission ratio of the second step can be calculated.It is:

Then through the connecting relations among units,it can be got that：n a1=n a2和n b1=n b2.From the equation above,the transmission ratio of the second step is:

The transmission ratio of the third step:

The third step of the planetary gearbox is obtained by locking brake Z1 and clutch L2.At this moment there is only one planet bar working.Transmission power P is input from shaft A,then passes swiveling jib H,gear g1,a1 and a2,clutch L2 and reaches output shaft B.Now it is known that brake Z1 is locked,that is n b1=0.Then through the connecting relation,it can be got that n a1=n a2.Also through the kinematics equation of planet x1(3-1):

n a1-(1+p1)n H1=0

And the transmission ratio of the third step i3 is:

The transmission ratio of the reverse step：

The reverse step of planetary gearbox is obtained by locking brake Z1 and clutch L1.There are two planet bars participating in work.Transmission power P is input form shaft A,passes through swiveling jib H1,gear g1,a1 and a2,g2,b2,clutch L1and reaches output shaft B.Now it is known that brake Z1 is locked,that is n b1=n H2=0，and planet bar x2 becomes semi-planet gear transmission(fixed-shaft transmission).So the transmission ratio of the reverse step i-1 should be calculated as:

Where,

Then i-1 is:

(3)Calcuations of touqueT,locking moment T L and brake moment T Z of each unit

Now suppose the torque of input shaft T A of the plantary gearbox is known,that is T A can be obtined through formula(4-1).Since the torque T,locking moment T L and brake moment T Z are different at each step,they should be calculated spearately according to their force bearing situation.

At the first step,i1=1(direct step).At this moment,clutch L1 and L2 lock the central gear a2,internal gear b2 and output shaft B.Transmission ratio P transmits from input shaft A,thuough clutch L1 and L2,then reaches output shaft B.Because T B=-T A here.the locking moment T L1 and T L2 should be calculated depending on output shaft B.That is locking moment T L1=T b2,T L2=T a2.So T B=T a2,and locking moment T L2 is:

T L2=T a2=T B=-T A

Accordign to formula(4-3),T b2=p2T a2

So,the locking moment T L1 is:

T L1=p2T a2=-p2T A=-3.105T A

Since the brakes Z1 and Z2 are not locked,T Z1=T Z2=0。

At the second step,i2=0.4278,At this time,brake Z2 locks internal gear b2 and

clutch L2 locks central gear a2 and output shaft B.Since T H1=T A,T L2=T B=-i2T A,locking moment T L2 is

T L2=T B=-i2T A=-0.4278T A

Since T a2=T a1，according to formulas(4—2) and (4—3)：

And since brake moment T Z2=T b2，

T Z2=T b2=-0.7564T A

At the third step,the transmission ratio i3=0.2463.Brake Z1 locks internal gear b1 while clutch L2 locks central gear a2 and output shaft B.So T H1=T A，T L2=T a2=T a1=T B。According to formula(4-2)：

T A,So the locking moment is：

Since T Z1=T b1，according to formula(4-2) and (4-3)：

,and the brake moment is:

At reverse step:i-1=-0.7564.At this time,brake Z1 locks internal gear b1 while clutch L1 locks internal gear b2 and output shaft B.So T H1=T A，T L1=T b2 and T Z1=T b1。The same as above:

So the locking moment is

And the brake moment is

That is

(4)Calculation of transmission efficiency of each step

According to the formula(5-2),the transmission efficiency can be calculated.That is the claculation equation is

The transmission efficiency of the first step(direct step)

Since there isn't relative motions among the units,the kinematics transmission ratio is

i p=i1=1

The dynamics transmission ratio is

So the transmission efficiency of the first step is(the friction loss is ignored)：η1=1 The transmission efficiency of the second step：At the second step,the kinematics transmission ratio is

And its dynamics transmission ratio is：

(a) The symbol of the coefficient is decided as:

Put ηH and p1、p2 into equation (a),then

So the transmission efficiency of the second step can be obtained through formula(5-2)：

The transmission efficiency of the third step：

At the third step,the kinematics transmission ratio is

And its dynamics transmission ratio is：

The symbol of x1 is deceded as：

Then the kinematics transmission ratio is:

So the transmission efficiency of the third step is：

The transmission efficiency of the reverse step：

At the reverse step the kinematics transmission ratio is

And its dynamics transmission ratio is：

(c)

The symbol of x1 and x2 is decided as：

＝Sign[-0.7564]=-1

Then the kinematics transmission ratio is:

So the transmission efficiency of the reverse step is:

After the calculations of the structural parameters,transmissionratio,moment and transmission efficiency of each step，we can draw several concluusions:

(a)The characteristic parameters of two planet bars x1 and X2 are

p1=p2=3.105,which are within the reasonable range p=1.6～5.

(b)Since the input unit of this planetary gearbox is swveling jib H1，the transmission ratio of each step is less than 1(excluding the driect step).It means the second,third and the reverse steps of the gearbox all have the effect of acceleration.

(c)The transmission efficiency of each step of the planetary gearbox is

high.Aftercomparison,it shows η1＞η2＞η3＞η-1；And η2=0.9736 is close to

η3=0.9696.The transmission efficiency η-1=0.9307，which is the lowest.However,in caterpillar trunks,the chance to use the reverse step is the least.So all values of transmission efficiency are reasonable.

附录B

中文翻译

行星齿轮变速箱的设计研究

饶振纲

［摘要］本文较详细地讨论了行星齿轮变速箱的设计计算。文中阐述了行星齿轮变速箱的结构参数计算，各档的传动比计算，各构件的转矩、制动转矩和闭锁转矩计算，以及各档的传动效率计算。同时，还附有具体的设计计算示例。因此，本文对于履带车辆、坦克、自行火炮和工程机械等的行星齿轮变速箱的设计计算均具有较重要的指导意义。

关键词：行星齿轮变速箱结构参数转矩功率

c) 离合器L将内齿圈b与转臂H闭锁(见图3，c)

当离合器L使内齿圈b与转臂H结合成一体时，即有n

b =n

H

。此时，输入轴A

的转矩T

A

可从两条路径传递：一路是经过中心轮a、行星轮g和转臂H传到输出轴B；另一路是经过中心轮a、行星轮g、内齿圈b和离合器L传到输出轴B。此

时，转矩T

A =T

a

,T

B

=-T

A

。而离合器的闭锁力矩T

L

等于内齿圈b的转矩T

b

,即T

L

=T

b

。

据公式(4-3)可得闭锁力矩T

L

的计算公式为

T L =T

b

=pT

a

(4-11)

由上式可见，当中心轮a输入，转臂H输出时，第一闭锁方案(图3，a)的闭锁力矩T

L

为最小，且使其具有较小的结构尺寸。所以，对于2K-H(A)型行星排选取第一闭锁方案较合理。

(2)若内齿圈b输入，转臂H输出(见图4)，即T

A =T

b

、T

B

=T

H

；讨论在下列三

种不同的2闭锁方式下，其摩擦离合器的闭锁力矩T

L

的计算公式。

a) 离合器L将内齿圈b与中心轮a闭锁(见图4，a)，即有n

a =n

b

=n

H

;故该行

星排变成为一个整体旋转，则可得：输出轴转矩T

B =-T

A

。而闭锁力矩T

L

=T

a

，据公

式(4-2)可得T

L

的计算公式为：

所以

(4-12)

b) 离合器L将中心轮a与转臂H闭锁(见图4，b),即有n

H =n

a

和闭锁力矩T

L

=T

a

。

仿上，按公式(4-3)可得闭锁力矩T

L

的计算公式为

所以，

(4-13)

c) 离合器L将内齿圈b与转臂H闭锁(见图4，c),即有n

H =n

b

和T

L

=T

b

。由于

离合器L将输入轴A和输出轴B相连接，所以，闭锁力矩T

L

应等于输入轴A所

传递的转矩T

A ,即有T

L

=T

b

=T

A

。则得得闭锁力矩T

L

的计算公式为

T

L

=T

A

(4-14)

由上式可见，当内齿圈b输入，转臂H输出时，第一闭锁方案的闭锁力矩

T

L

为最小值，故应选取第一闭锁方案最为有利。

为了获得较小尺寸的摩擦离合器，利用以上各式求得所承受的闭锁力矩T

L 后，便能够合理地选取所需的控制元件。但是，实际上由于结构布置上的困难，欲使最有利的行星排构件闭锁，并不是都能实现的。在此情况下，则应当从离合器的结构紧凑和行星齿轮变速箱外形尺寸较小的观点出发选取最合适的闭锁方案。

5 行星齿轮变速箱的效率计算

对于多级行星齿轮变速箱，其输出功率P

B 与输入功率P

A

的比值，则称为该

行星变速箱的传动效率；即得：

(5-1)

式中，T

A 、n

A

—输入轴A的转矩和转速；

T B 、n

B

—输出轴B的转矩和转速。

其中，比值n

A /n

B

=i

p

为其运动学传动比；比值-T

B

/T

A

=

p

为其动力学传动比。因为

转矩T

B 考虑了行星齿轮变速箱的摩擦损失。将i

p

和

p

代入式(5-1)，则得到著名

的克列依涅斯(M.A.Kpe-Йюec)公式：

(5-2)

由此可知，多级行星齿轮变速箱各档的传动效率η

p

等于其动力学传动比与其运

动学传动比i

p

之比值。

如前所述，其运动学传动比i

p

都是行星排特性参数p的函数，即

i p =f(p

1

、p

2

、…p

n

) (5-3)

根据相对运动的功率方程式的关系，行星变速箱的动力学传动比

p

应等于运

动学传动比i

p

关系式中的每个特性参数p值上乘以(ηH)X，则可得其动力学传动

比

p

的函数式为：

(5-4)

式中，ηH—在相对运动中，行星排的传动效率，对于2K-H(A)型行星排，且有：ηH=ηH ag×ηH gb=0.97×0.99=0.96

指数X

j

的计算公式为：

(5-5)

j=1,2,…，n

sign的数学含义是表示符号。

指数X

j

=1,其符号取决于公式(5-5)；若sign的值大于零，则“sign”为正

号“+”，则得X

j =+1;否则，即得X

j

=-1。

6 行星齿轮变速箱的设计计算示例

已知某军用履带车辆传动装置的行星齿轮变速箱的传动简图如图5所示。其自由度数W=3，档位数n

d

=4(三个前进档和一个倒档)。试计算：(1)该行星变速

箱所需的控制元件数m、行星排数k和基本构件数n

0；(2)各个档位的传动比i

p

；

(3)各构件的转矩T、制动力矩T

Z 和闭锁力矩T

L

；(4)各个档位的传动效率η

p

。

图5 履带车辆传动装置

Fig.5 The caterpilar vehicle transmission

的计算。

解：(1) 控制元件数m、行星排数k和基本构件数n

当自由度数w=3时，其控制元件数m可按公式(2-6,a)计算，即

因n

=4代入上式，则得：

d

m2-m-8=0

由上述方程可求解得：m≈3.4；又因C2

＞m＞W，则可得控制元件数为：m=W+1=4

m

按式(2-8)可求得其摩擦离合器数L为

L=W-1=3-1=2

再按式(2-9)可得其制动器数Z为

Z=m-L=4-2=2

按公式(2-12)可得其行星排数k为

k=n

-(W-1)=4-(3-1)=2

d

为

最后，可按其结构公式(1-3)求得其运动基本构件数n

=W+k=3+2=5

n

由图5中可见，其运动基本构件为A、B、1、2和3共五个基本构件。

计算

(2)各档位的传动比i

p

上述行星齿轮变速箱的两个行星排中的各齿轮的模数m相同，且知，m=4mm，

两个行星排的特性参数为p

1=p

2

=3.105。因p

1

=Z

b1

/Z

a1

和p

2

=Z

b2

/Z

a2

；一般，中心轮

a的齿数Z

a ＞17。现选取：Z

a1

=Z

a2

=19,则得内齿圈b的齿数为：

Z

b1

=p

1

z

a1

=3.105×19=59

Z

b2

=p

2

z

a2

=3.105×19=59

其行星轮的齿数Z

g

为

仿上，Z

g2

=20

由于该行星齿轮变速箱具有三个前进档和一个倒档，每结合两个控制元件便可获得一个档位。其各档位结合控制元件的组合方法如表6-1所示。

表6-1 某行星变速箱各档位结合控制元件的组合方法

Table 6-1 Combining method of control wnitsof one planetary gearbox

现计算上述行星齿轮变速器各档的传动比如下：

一档(直接档)的传动比：

该行星齿轮变速箱的一档是将两个摩擦离合器L

1、L

2

都闭锁起来而得到的。

它们将中心轮a

1、a

2

和内齿圈b

2

，以及输出轴B结合成为一体，从而使得两个行

星排X

1、X

2

连成为一个整体旋转；其各构件之间均没有相对运动，即转速

n A =n

H1

=n

b1

=n

a1

=n

a2

=n

H2

=n

b2

=n

B

。则得其传动比为：

i=i

AB

=i

H1b2

=1

二档的传动比：

该行星齿轮变速箱的二档是将制动器Z

2

制动和离合器L

2

闭锁而得到的；其

两个行星排均参与工作。传递功率P从输入轴A输入后，经转臂H

1

通过两路分

流：一路经齿轮g

1、a

1

和a

2

，离合器L

2

传到输出轴B；另一路经齿轮g

1

、b

1

，转

臂H

2，齿轮g

2

、a

2

和离合器L

2

传到输出轴B。现已知：制动器Z

2

被制动，即转速

n

b2

=0。

据行星排的运动学方程式(3-1)可求得二档的传动比i

2

值。即得：

再由构件之间的连接关系可求得：n a1=n a2和n b1=n H2。由上式经整理后可求得其二档的传动比为：

三档的传动比：

该行星齿轮变速箱的三档是将特制动器Z 1制动和离合器L 2闭锁而得到的；此时只有第一个行星排X 1参与工作。传递功率P 从输入轴A ，经转臂H 1、齿轮g 1、a 1和a 2，离合器L 2传到输出轴B 。现已知：制动器Z 1被制动，即转速n b1=0。再由连接关系得：n a1=n a2。据行星排x 1的运动学方程式(3-1)得：

n a1-(1+p 1)n H1=0

则可求得其三档的传动比i 3为：

倒档的传动比：

该行星齿轮变速箱的倒档是将制动器Z 1制动和离合器L 1闭锁而得到的；其两个行星排均参与工作。传递功率P 从输入轴A ，经转臂H 1，齿轮g 1、齿轮a 1和a 2、g 2、b 2，再经离合器L 1传到输出轴B 。现已知：制动器Z 1被制动，即转速n b1=n H2=0，行星排x 2变成为准行星齿轮传动(定轴齿轮传动)。因此，其倒档的传动比i -1应按下列关系式求得：

式中， 则可得其倒档的传动比i -1为：

(3) 各构件的转矩T、闭锁力矩T

L 和制动力矩T

Z

的计算

现假设该行星齿轮变速箱输入轴的转矩T

A 为已知，即T

A

可由电动机的额定

功率P和转速n

1按公式(4-1)求得。由于各构件的转矩T、闭锁力矩T

L

和制动力

矩T

Z

在各个档位都是不一样的。所以，它们应按各个档位的受力情况分别计算。

在第一档位时，其传动比i

1=1(直接档)，此时离合器L

1

和L

2

将中心轮a

2

、

内齿轮b

2与输出轴B闭锁。输入功率P从输入轴A，经离合器L

1

和L

2

传到输出

轴B。因为，在此考虑到T

B =-T

A

，故离合器的闭锁力矩T

L1

和T

L2

应从输出轴B方

面来确定。即有闭锁力矩T

L1=T

b2

,T

L2

=T

a2

。因转矩T

B

=T

a2

,所以，闭锁力矩T

L2

为：

T

L2

=T

a2

=T

B

=-T

A

按公式(4-3)可得转矩T

b2=p

2

T

a2

所以，闭锁力矩T

L1

为

T L1=p

2

T

a2

=-p

2

T

A

=-3.105T

A

因制动器Z

1和Z

2

均未制动，故其制动力矩为T

Z1

=T

Z2

=0。

在第二档位时，其传动比i

2=0.4278。此时，制动器Z

2

将内齿轮b

2

制动，离

合器L

2将中心轮a

2

与输出轴B闭锁。因为，T

H1

=T

A

，T

L2

=T

B

=-i

2

T

A

。所以，闭锁力

矩T

L2

为

T L2=T

B

=-i

2

T

A

=-0.4278T

A

因T

a2-T

a1

，按公式(4—2)和(4—3)可得：

因制动力矩T

Z2=T

b2

，所以，

T

Z2

=T

b2

=-0.7564T

A

第三档位时，其传动比i

3=0.2436。此时，制动器Z

1

将内齿轮b

1

制动，离合

器L

2将中心轮a

2

与输出轴B闭锁。因此可得：T

H1

=T

A

，T

L2

=T

a2

=T

a1

=T

B

。由公式(4—

2)得：。所以，其闭锁力矩为：

因制动力矩T

Z1=T

b1

，由公式(4-2)和(4-3)可得：

,所以，其制动力矩为：

在倒档传动时，其传动比i

-1=-0.7564。此时，制动器Z

1

将内齿圈b

1

制动，

离合器L

1将内齿轮b

2

与输出轴B闭锁。因此可得：T

H1

=T

A

，T

L1

=T

b2

和T

Z1

=T

b1

。仿上，

可得：

所以，其闭锁力矩为：

仿上，可得其制动力矩为

即得：

(4) 各档位的传动效率计算

按照前苏联学者M.A.克列依涅斯提出的公式(5-2)可计算各档的传动效率。即传动效率的计算公式为：

一档(直接档)的传动效率：

在一档时，由于各构件间均没有相对运动，故其运动学传动比为：

i p =i

1

=1

其动力学传动比为：

所以，其一档的传动效率(忽略轴承的摩擦功率损失)为：η

1

=1 二档的传动效率：在二档时，其运动学传动比为：

其动力学传动比为：

(a)

先确定系数x

1、x

2

的符号：