thuyết minh đồ án chi tiết máy đề 4 (tiếng anh)

thuyết minh đồ án chi tiết máy đề 4 (full tiếng anh), dùng để kham khảo cho đồ án thiết kế chi tiết máy khi viết bằng tiếng anh. Lưu ý: kiểm tra và chỉnh sửa nếu có ý định dùng làm bản thuyết minh thuyết minh đồ án chi tiết máy đề 4 (full tiếng anh), dùng để kham khảo cho đồ án thiết kế chi tiết máy khi viết bằng tiếng anh. Lưu ý: kiểm tra và chỉnh sửa nếu có ý định dùng làm bản thuyết minh thuyết minh đồ án chi tiết máy đề 4 (full tiếng anh), dùng để kham khảo cho đồ án thiết kế chi tiết máy khi viết bằng tiếng anh. Lưu ý: kiểm tra và chỉnh sửa nếu có ý định dùng làm bản thuyết minh thuyết minh đồ án chi tiết máy đề 4 (full tiếng anh), dùng để kham khảo cho đồ án thiết kế chi tiết máy khi viết bằng tiếng anh. Lưu ý: kiểm tra và chỉnh sửa nếu có ý định dùng làm bản thuyết minh

Index………1

Input data ………2

Part I: Dynamics……… 3

I Choose motor……….3

II Distribute transmission ratio……… 3

Part II: Calculate transmission……….……….……… ……… 4

I Chain drive design……….4

II Gear box transmission ……… 8

A Rapid transmission………… ………

8 B Slow transmission………

14 III Calculate shaft………21

A Shaft I………

21 B Shaft II……….25

C Shaft III………31

D Test shaft……… 35

Part III: Calculate gear box………

37 Appendix………39

Input data

diagram as shown in Figure 02

1- Electric motor.

3- Reducing gear box.

Working 300 days per year, 2 shift a day, 8 hours per shift

Transmision error permit Δu = (2÷3) %.u = (2÷3) %

ηol =0,99 - 1 roller bearing effiency.

ηbr =0,97 - 1 pair of gear effiency

ηx =0,96 - chain drive effiency

The equivalent power: [2.4]

u sb = 8 ÷40, choose u sb =15 [table 2.2]

Preliminary spindle speed:

nsb = usb.nlv = 15*83.3 = 1249.5 (rpm)

With

n sb: Preliminary spindle speed of motor

n lv: Spindle speed of main shaft

According to working principle, power of motor must be greater than needed power

2 Distribute transmission ratio

2.1 System’s transmistion ratio

=> ∆u =|Uht −Upp Uht |*100% =2.7%

⇒ Appropriate with ∆U = (2÷3)%

 Shaft 3: P3 = η 34 Plv = Plv

3.8 0.99∗0.96=4 kW

 Shaft 2: P2 = η 23 P 3 = P 3

4 0.99∗0.97=4.16 kW

 Shaft 1: P1 = η12 P 2= P 2

4.16 0.99∗0.97=4.33 kW

 Motor shaft power: Pdcct= P 1

4.33 0.99∗1=4.37 kW

Statistic table

Shaft

Parameter

U Um1=1 U12=3.2 U23=2.6 Ux=2

T (Nmm) 29286.6 29018.6 44608.13 223391.8 424444.4

PART II: CALCULATE TRANSMISION

I CHAIN DRIVE DESIGN

Chain type:

Small load and light impact => Roller chain

171=1.17 With:

n01 = 200 (page 91)

n3 = 171

Calculate coefficient of use k

Formular [4.8] and [table 4.6]

kc=1,25 Two shift working

kbt=1,3 Dusty woring environment, lubricate quality II

=> k = 1.1,25.1.1,5.1,25.1,3=3,05

Coefficient of unequal load distribute to chain row

Choose 2 row chain: kd = 1,7

∆ = (0,002÷0,004)*a =1,54÷3.048(mm)

=> choose a = 760 (mm)

d Test number of chain impact per 1 sec:

Formular [4.15] and [table 4.10] with p=25.4, [i]=30

15∗x =

23∗171 15∗72 =3.6(¿)<[i]=30

e Test durability of chain

=> satisfy durable condition.

f Test contact reliability

E = 2,1.105 (MPa) elastic module

kr = 0,42 affect of sprocket’s teeth coefficient (z1=23)

A = 306 (mm2) Area of hinged head [table 4.14]

Use heat treated steel 45 with hardness HB170 will archieve contact stress [σH]

= 500 MPa, guarantee durability for drive sprocket

Kx= 1,15 coefficient of weight when chain drive lies horizontal

or with angle below 40ο

Number of chains x 72Drive sprocket split

II Gearbox transmission

A Rapid transmission – V-shaped spur gear

3 Determine allowed stress

[Table 5.2], with heat treated steel C45 archieved hardness HB 180÷350

σ o HLIM =2 HB 1+70 ; S h=1.1 ;σ o FLIM =1.8 HB ;Sf =1.75

Choose hardness of small gear HB1=250, big gear HB2=235

σ o HLIM 1=2 HB 1+70=570 MPa , σ o HLIM 2=2 HB 2+70=540 MPa

c = 1 number of match time.

540∗1 1.1 =490.9MPa

[σ H]=[σ H 1]+[σ H 2]

2 =504.5 MPaFormular [5.10]

423∗1∗1 1.75 =241.7(MPa)Formular [5.12], [5.14]:

4 Determine basic parameters

Prelimitary determine shaft distance aw: [5.15]

√[σ H T 1∗khb]2∗u 12∗ψbaba

5 Determine matching parameters

a Determine module m

Formular [5.18]

m = (0,01÷0,02).aw1 = (0,01÷0,02).80 = (0.8÷1.6) mm

[table 5.7], choose m1=1.25 (mm)

b Determine teeth and incline angle β

Preliminary choose β= 30o and cosβ=0,86

2 aw 1=

1.25∗114 2∗80 =>β 27.1ο

6 Test contact durability

Formular [5.25], tangential stress on working pace of gear:

[table 5.4], Zm = 274 MPa1 /3

[5.27], tgβb = cosαt .tgβ

- at=atw=T g−1(cosβ tgα )=t g−1(cos 27.1tg20 )=22.23 °

=>βb=25.3 (incline angle on base cylindrical gear)

[5.26], Zh=√sin 2 atw 2 cosβb =√ 2cos (25.3)

sin (2∗22.23) =1.6 (coefficient of contact shape)

[5.35], v= π∗dw 1∗n1

60000 =

π∗38.1∗1425

60000 =2.84(m/s)[Table 5.9] choose exact lever 9,

khα = 1.13

kfα = 1.37

[5.37], vh=σh∗go∗v∗√aw 1 u12=0.002∗61∗2.84∗√3.280=1.7

[table 5.11] σh = 0.002; [table 5.12] go = 61

[5.36], khv =1+ vh∗bw 1∗dw 1

1.7+0.5∗80+38.1 29018.6∗1.09∗1.1=1.07[5.34], kh=khβ∗khα∗khv =1.07∗1.13∗1.07=1.29

Replace all value in [5.25], get σh=¿ 456.25 MPa

[5.1],V = 2.84 m/s => Zv = 1, kinetic exact lever 9, choose contact exact lever 8, need process to archieve roughness Ra = 2.5÷1.25 μmm, so Zr = 0.95

da < 700mm, kxh=1

 Satisfy contact condition

7 Test bending durability

To ensure the bending durability for teeth, bending stress forming at teeth feet determine by [5.38] and [5.39]

cos3β=

87 cos(27.1)3≈ 124

B Slow transmission – straight tooth gear

3 Determine allowable stress

[Table 5.2], with heat treated steel C45 archieved hardness HB 180÷350

σ o HLIM =2 HB 1+70 ; S h=1.1 ;σ o FLIM =1.8 HB ;Sf =1.75

Choose hardness of small gear HB1=250, big gear HB2=235

σ o HLIM 1=2 HB 1+70=570 MPa , σ o HLIM 2=2 HB 2+70=540 MPa

540∗1 1.1 =490.9MPa

[σ H]=[σ H 1]+[σ H 2]

2 =504.5 MPaFormular [5.10]

423∗1∗1 1.75 =241.7(MPa)Formular [5.12], [5.14]:

4 Determine basic parameters

Prelimitary determine shaft distance aw: [5.15]

5 Determine matching parameters

+ Small sprocket:

170∗2 2∗(2.6+1)≈ 48+ Big sprocket:

2 =175 (mm)

6 Test contact durability

Formular [5.25], tangential stress on working pace of gear:

With:

[table 5.4], Zm = 274 MPa1 /3

[5.26], Zh=√sin 2 atw 2 cosβb =√ 2cos (0)

sin (2∗20) =1.71 (coefficient of contact shape)[5.32], εa=(1.88−2.6(281 +

[5.35], v= π∗dw 2∗n 2

60000 =

π∗97.2∗548

60000 =0.87(m/s)[Table 5.9] choose exact lever 9,

Replace all value in [5.25], get σh=¿ 435.5 MPa

[5.1],V = 1.59 m/s => Zv = 1, kinetic exact lever 9, choose contact exact lever 8, need process to archieve roughness Ra = 2.5÷1.25 μmm, so Zr = 0.95

da < 700mm, kxh=1

 Satisfy contact condition

7 Test bending durability

To ensure the bending durability for teeth, bending stress forming at teeth feet determine by [5.38] and [5.39]

Coefficient of gear match:

yεε= 1

1 1.75=0.57 (coefficient of teeth incline)

Z2 87 125

III Calculate shaft

1 Forces diagram

2 Outside forces applied on shaft

- Force from chain drive: Fx = 2T3/dx = 2395.23 (N)

- Force from joint: Fkn = 2T1/dn = 725.5 (N)

3 Determine shaft length and force acting point

Material: steel C45, σb=600 MPa

Torsional stress [T] = 15÷30 MPa

2 drive gear Z = 27, module m1 = 1.25, incline angle β = 27.1

Torque T = 29018.6 Nmm

Preliminary determine shaft diameter: (choose [T] = 20 MPa)

dsb≥√3 0.2∗T[T]=20 mm

Determine values on shaft

Split ring diameter: dw1=38.1

Tangental force applied on incline teeth gear:

Moment equilibrium equation: (Y-direction)

At B:

Shaft diameter at B: dB ≥3

√0.1∗MtdB[σF]=21 mm

0.1∗[σF]=29.2mm ([table 7.5] choose [σF¿ =63)

Choose flat key:

▪ For join with diameter d = 20 mm:

B = 6, h = 6, t1 = 3.5, t2 = 2.8Test:

Formular [7.34] for calculate durability:

[τcc¿ = 87 MPa: allowable cutting strength

 Key satisfy requirement

3 Choose bearing and test:

Total axial force: Fat =Fa 1−Fa 2=0

 Al support point with bearing only sustain centripetal force, choose single row deep groove ball bearings for point B and E

[appendix table P2.1] choose 6305 with:

D = 62 mm – outer diameter

- Formular [8.3], conventional load is:

Q=( X∗V ∗Fr+Y∗Fa)∗kd∗kt

Inner ring spin, so V = 1

Only withstand centripetal force, so X = 1, Y = 0

[table 8.4], kt=1(t ≤100 ° C); kd=1.1(dyεnamic load)

Static load is calculated by [8.19] and [table 8.7], with Fa = 0:

Qt= Xo∗Fr +Yo∗Fa ; Xo=0.6

 Qt = 909 N

 Satisfed requirement

B Shaft II

1 Calculate shaft diameter

Material: steel C45, σb=600 MPa

Torsional stress [T] = 15÷30 MPa

2 driven gear Z2 = 87, module m1 = 1.25, incline angle β = -27.1

1 drive gear Z3 = 28, module m2 = 2

Torque T2= 44608.13 Nmm

Preliminary determine shaft diameter: (choose [T] = 20 MPa)

dsb≥√3 0.2∗T 2[T]=22.3 mm

Determine values on shaft

Split ring diameter: dw2 = 122.16, dw3 = 56

Tangental force applied on gear:

Torque at gears: ma 3=ma 5= Fa∗dw 2

2 =22715.3 (N)Moment equilibrium equation: (Y-direction)

 REy=395.3 (N)

Forces equilibrium equation:

 RBy = 395.3 (N)

Moment equilibrium equation: (X-direction)

 REx = 587

Forces equilibrium equation:

Calculate equivalent moment at points:

At C:

Shaft diameter at C: dC ≥3

√0.1∗MtdC[σF]=26.16 mm([table 7.5] choose [σF¿ =63)

 Choose dA = dE = 20, dB = dD = 25, dC = 30 mm

2 Choose key and test:

[table 7.15]

Choose flat key:

▪ For 2 gear with diameter d1 = 25 mm:

B = 8, h = 7, t1 = 4, t2 = 2.8

▪ For 1 gear with diameter d2 = 30 mm:

B = 8, h = 7, t1 = 4, t2 = 2.8Test:

Formular [7.34] for calculate durability:

 Key satisfy requirement.

3 Choose bearing:

Total axial force: Fat =Fa 1−Fa 2=0

 Al support point with bearing only sustain centripetal force, choose single row deep groove ball bearings for point A and E

[appendix table P2.1] choose 6304 with:

Choose FrE to test:

- Formular [8.3], conventional load is:

Q=( X∗V ∗Fr+Y∗Fa)∗kd∗kt

Inner ring spin, so V = 1

Only withstand centripetal force, so X = 1, Y = 0

[table 8.4], kt=1(t ≤100 ° C); kd=1.1(dyεnamic load)

Static load is calculated by [8.19] and [table 8.7], with Fa = 0:

Qt= Xo∗Fr +Yo∗Fa ; Xo=0.6

 Qt = 424.2 N < Co

 Satisfed requirement

C Shaft III

1 Calculate shaft diameter:

Material: steel C45, σb=600 MPa

Torsional stress [T] = 15÷30 MPa

1 driven gear Z4 = 73, module m2 = 2

Torque T3 = 223391.8 Nmm

Preliminary determine shaft diameter: (choose [T] = 20 MPa)

√0.2∗T 3[T]=30.3 mm

Determine values on shaft

Split ring diameter: dw4=250

Tangental force applied on incline teeth gear:

Moment equilibrium equation: (Y-direction)

At D:

Shaft diameter at D: dD ≥3

√0.1∗MtdD[σF]=31.3 mm([table 7.5] choose [σF¿ =63)

 Choose dA = dC = 40, dB = 45 mm, dD = 35 mm

2 Choose key and test:

[table 7.15]

Choose flat key:

▪ For 1 gear with diameter d1 = 45 mm:

B = 14, h = 9, t1 = 5.5, t2 = 3.8

▪ For 1 joint with diameter d2 = 30 mm:

B = 10, h = 8, t1 = 5, t2 = 3.3Test:

Formular [7.34] for calculate durability:

[τcc¿ = 87 MPa: allowable cutting strength

 Key satisfy requirement

Choose FrA to test:

- Formular [8.3], conventional load is:

Q=( X∗V ∗Fr+Y∗Fa)∗kd∗kt

Inner ring spin, so V = 1

Only withstand centripetal force, so X = 1, Y = 0

[table 8.4], kt=1(t ≤100 ° C); kd=1.1(dyεnamic load)

 Q = 2445.4 (N)

Dynamic load is calculated by [8.1]:

m = 3 (ball bearing)

Static load is calculated by [8.19] and [table 8.7], with Fa = 0:

Qt= Xo∗Fr +Yo∗Fa ; Xo=0.6

+ Determine safety coefficient at critical section:

Base on the structure and momentum diagram We need to check the endurance of some section:

Shaft I: section at C

Shaft II: section at C

Shaft III: section at B.

+ Choose type of fit:

Bearings are fitted on the shaft with type k6, sprocket, shaft connector and gear use fit k6 with keys

From the table 7.6

All shaft is processed on milling machine At critical section, we choose the

roughness Ra = 2,5…0,63μmm Follow the table 7.8, we choose Kx = 1,06

Shaft is processed to increase surface durability by high current prequency

Tension fit

Keyway

Tensionfit

6

1.96

2.56

10.14

2.5

From the result in the table, all the critical section satisfied the endurance condition

PART III: Calculate gear box

Choose casted box: GX 15-32

Mounting suface go throught shaft center

Center distance: a = aw1 + aw2= 80 + 175 = 255 mm

Box body thickness: δ=0.03∗aw+3=8.3 mm choose δ = 13

Box lid thickness: δ 1=0.9∗δ=7.47 mm choose δ1 = 11

Reinforce tendon:

Thickness: e=(0.8+1)δ=15 mm

Height: h < 58 mm

Slope: ≈ 2 °

Base bolt: d 1>0.04∗a+10=20 choose d1 = 24

Bearing side bolt: d2 = (0,7 ÷0,8) * d1 = 16.8÷19.8 choose d2 = 20

Lid and body attach bolt: d3 = (0,8÷0,9) * d2 = 14.4÷16.2 choose d3 = 16

Bearing lid secure bolt: d4 = (0,6÷0,7) * d2 = 10.8÷12.6 choose d4 = 10 (8 for small lid)

Checkout lid bolt: d5 = (0,5÷0,6) * d2 = 9 * 10.8 choose d5 = 10

Flange witdth of body: S3 = (1.4÷1.8) * d3 = 22.4÷28.8 choose S3 = 26

Flange witdth of lid: S4 = (0,9÷ 1) * S3 = 23.4÷26 choose S4 = 26

Bolt mounting surface beside bearing:

Flange of body lid: K3 = K2 – (3÷5) = 52÷50 choose K3 = 50

Bottom surface thickness of box without protrusion:

S1 = (1.3÷1.5) * d1 choose S1 = 38

Bottom surface width: K1 = 3 * d1 = 72

Gap between parts:

+ Gear and box: ∆ = (1,1÷1,2) * δ = 9.1÷11 choose ∆ = 10

+ Top of biggest gear and box: ∆1 ≥ (3÷5) * δ choose ∆1 = 30

+ Between gears: ∆ > δ choose ∆ = 10

Number of base bolts:

Z = (500 + 400) / (200÷300)

Choose Z = 4

With:

L = 500 Prelimitary box lengh

B = 300 Prelimitary box width

Bearing attach for inner ring spin and withstand cycle load, outer ring is static and withstand static load

1 Thiết kế đồ án chi tiết máy – TS Văn Hữu Thịnh – TS Nguyễn Minh Kỳ

2 Tính toán thiết kế hệ dẫn động cơ khí - Trịnh Chất – Lê Văn Uyển, nhà xuất bản Giáo dục Việt Nam tập 1

3 Tính toán thiết kế hệ dẫn động cơ khí - Trịnh Chất – Lê Văn Uyển, nhà xuất bản Giáo dục Việt Nam tập 2

4 Dung sai kỹ thuật đo - Trần Quốc Hùng, nhà xuất bản đại học Sư phạm Kỹ thuật Tp.HCM