First, make a transmission plan
Second, the choice of motor
Three. Calculate the total transmission ratio and allocate the transmission ratios of all levels.
Four. Calculation of motion parameters and dynamic parameters
Verb (abbreviation of verb) Design and calculation of transmission parts
Design and calculation of intransitive verb axle .......................................................................................................................................................................
Seven. Selection and checking calculation of rolling bearing 19
Eight. Selection and calculation of key connection
Design theme: V-belt-the fourth group single-stage cylindrical reducer
Designer of Dezhou Vocational College of Science and Technology Qingdao Campus: # # #
Lecturer:%%%
20071feb
Calculation process and calculation instructions
First, make a transmission plan
The third group: design a single-stage cylindrical gear reducer and a first-stage belt drive.
(1) working conditions: continuous unidirectional operation, stable load, no-load start, service life 10 year, small batch production, two shifts, and allowable error of conveyor belt speed of 5%.
(2) original data: working tension f =1250n; Tape speed v =1.70m/s;
Drum diameter D = 280mm.
Second, the motor selection
1. Motor type selection: Y series three-phase asynchronous motor.
2, motor power selection:
(1) Total power of transmission equipment:
η Total = η Belt×η2 Bearing×ηGear×ηCoupling×ηDrum
=0.95×0.982×0.97×0.99×0.98×0.96
=0.82
(2) Working power required by the motor:
P work =FV/ 1000η total
= 1250× 1.70/ 1000×0.82
= 2.6 kW
3, determine the motor speed:
Calculate the working speed of the drum:
N tube = 60× 960V /πD
=60×960× 1.70/π×280
= 1 1 1r/min
According to the reasonable range of transmission ratio recommended in Table 2-3 of P7 book, take the transmission ratio range I'a=3~6 of the cylindrical gear-driven primary reducer. If the V-belt transmission ratio I' 1=2~4, the timing range of the total transmission ratio is I'a=6~24. Therefore, the optional range of motor speed is n cylinder = (6 ~ 24) ×11= 666 ~ 2664 rpm.
The synchronization speed in this range is 750, 1000 and 1500 rpm.
According to the capacity and speed, it is found that there are three suitable motor models in the relevant manual: therefore, there are three transmission ratio schemes: considering the size, weight and price of the motor and transmission device and the transmission ratio of belt drive and reducer, it can be seen that the second scheme is more suitable, so n= 1000r/min is selected.
.
4, determine the motor model
According to the selected motor type, required rated power and synchronous speed, the motor model is selected as Y 132S-6.
Its main performance: rated power: 3KW, full load speed of 960r/min, rated torque of 2.0. The mass is 63kg.
Third, calculate the total transmission ratio and allocate large power ratios at all levels.
1, total transmission ratio: I total =n electric /n cylinder = 960/11= 8.6.
2, the huge dynamic ratio of distribution at all levels
(1) According to the instruction, gear I =6 (i=3~6 for single-stage reducer is reasonable).
(2)∫I total =i file ×I band
∴i Belt =i Total /i Gear =8.6/6= 1.4
Four. Calculation of motion parameters and dynamic parameters
1, calculate the speed of each axis (rpm)
NI=n motor = 960 rpm
NII = nickel/iodine band =960/ 1.4=686 (rpm)
NIII=nII/i gear =686/6= 1 14 (rpm)
2. Calculate the power (KW) of each shaft.
PI=P work =2.6KW
PII =π×ηband = = 2.6×0.96 = 2.496 kw
PIII=PII×η bearing×ηgear =2.496×0.98×0.96.
= 2.77 kW
3, calculate the shaft torque (n? mm)
TI = 9.55× 106 pi/nI = 9.55× 106×2.6/960
=25729N？ millimetre
TII=9.55× 106PII/nII
=9.55× 106×2.496/686
=34747.5N？ millimetre
TIII = 9.55× 106 PIII/nIII = 9.55× 106×2.77/ 1 14
=232048N？ millimetre
V. Design and calculation of transmission parts
1, design and calculation of pulley drive
(1) Select the common V-belt profile.
According to the textbook, kA= 1.2.
Pd=KAP= 1.2×3=3.9KW
From the textbook: choose a-type v-type belt
(2) Determine the reference diameter of the pulley and check the belt speed.
According to the textbook, the recommended reference diameter of the small pulley is
75 ~ 100 mm
Then take DD1=100 mm.
dd2=n 1/n2？ DD1= (960/686) ×100 =139mm
Take dd2= 140mm from table 5-4 of teaching material P74.
Actual driven wheel speed N2' = n1DD1/dd2 = 960×100/140.
= 685.7 rpm
The speed error is N2-N2'/N2 = 686-685.7/686.
=0.00040.05 (allowed)
Tape speed v: v = π DD1n1/60×1000.
=π× 100×960/60× 1000
= 5.03 m/s
In the range of 5~25m/s, the belt speed is appropriate.
(3) Determine the belt length and central moment.
According to the textbook
0.7(DD 1+dd2)≤A0≤2(DD 1+dd2)
0.7( 100+ 140)≤A0≤2×( 100+ 140)
So there are: 168mm≤a0≤480mm.
Excerpted from the textbook P84 (5- 15):
l0 = 2 A0+ 1.57(DD 1+dd2)+(dd2-DD 1)2/4a 0
=2×400+ 1.57( 100+ 140)+( 140- 100)2/4×400
= 1024 mm
According to table 7-3 of the textbook, Ld= 1 120mm.
According to the textbook P84 (5- 16):
a≈A0+Ld-L0/2 = 400+( 1 120- 1024/2)
=400+48
= 448 mm
(4) Check the wrap angle of the small pulley
α 1 = 1800-dd2-DD 1/a×600
= 1800- 140- 100/448×600
= 1800-5.350
= 174.650 & gt； 1200 (applicable)
(5) Determine the number of bands.
According to the textbook (7-5), P0 = 0.74kw.
According to the textbook (7-6) △P0=0. 1 1KW
According to the textbook (7-7), kα = 0.99.
According to the textbook (7-23), KL = 0.91.
Formula from textbook (7-23)
Z= Pd/(P0+△P0)KαKL
=3.9/(0.74+0. 1 1) ×0.99×0.9 1
=5
(6) Calculate the pressure on the shaft
According to the textbook, q=0. 1kg/m, and the initial tension of a single v-belt is given by the formula (5- 18):
F0 = 500 PD/ZV(2.5/kα- 1)+qV2
=[500×3.9/5×5.03×(2.5/0.99- 1)+0. 1×5.032]N
= 160 cattle
Then the pressure FQ acts on the bearing,
FQ = 2zf 0 sinα 1/2 = 2×5× 158.0 1 sin 167.6/2
= 1250N
2. Design and calculation of gear transmission
(1) Select gear material and accuracy grade.
Considering the low transmission power of the reducer, the gear adopts soft tooth surface. The pinion is quenched and tempered with 40Cr, and the tooth surface hardness is 240~260HBS. The big gear is made of 45 steel, quenched and tempered, and the tooth surface hardness is 220HBS；; Choose level 7 accuracy according to the teaching material. Tooth surface roughness ra ≤1.6 ~ 3.2 μ m.
(2) Design according to contact fatigue strength of tooth surface.
By d1≥ 76.43 (kt1(u+1)/φ du [σ h] 2)1/3.
Determine the relevant parameters as follows: transmission ratio I tooth =6.
Take the number of pinion teeth Z 1=20. Number of teeth of big gear:
Z2=iZ 1=6×20= 120
Actual transmission ratio I0= 120/2=60.
Transmission ratio error: I-i0/I = 6-6/6 = 0%
Tooth ratio: u=i0=6
Take φd=0.9 from the textbook.
(3) Torque T 1
t 1 = 9550×P/n 1 = 9550×2.6/960
=25.n？ m
(4) load coefficient k
Take k= 1 from the textbook.
(5) allowable contact stress [σH]
[σH]= σHlimZNT/SH in the textbook:
σhlim 1 = 625 MPaσhlim 2 = 470 MPa
Contact fatigue life coefficient found in textbooks;
ZNT 1=0.92 ZNT2=0.98
For general-purpose gears and general industrial gears, the safety factor SH= 1.0 is selected according to the general reliability requirements.
[σH] 1 =σhlim 1 znt 1/SH = 625×0.92/ 1.0 MPa
=575
[σH]2 =σhlim 2 znt 2/SH = 470×0.98/ 1.0 MPa
=460
Therefore:
d 1≥766(kt 1(u+ 1)/φdu[σH]2) 1/3
= 766 [/kloc-0 /× 25.9× (6+1)/0.9× 6× 4602]1/3mm
= 38.3 mm
Modulus: m = d1/z1= 38.3/20 =1.915mm.
According to Table 9- 1 of the textbook, take the standard modulus: m=2mm.
(6) Check the bending fatigue strength of the tooth root.
According to the style of textbooks
σF =(2kt 1/bm2z 1)YFaYSa ≤[σH]
Determine relevant parameters and coefficients
Diameter of dividing circle: d1= mz1= 2× 20mm = 40mm.
D2 = mz2 = 2×120mm = 240mm.
Tooth width: b = φ DD1= 0.9× 38.3mm = 34.47mm.
Take b=35mm b 1=40mm.
(7) Tooth profile coefficient YFa and stress correction coefficient YSa
According to the number of teeth Z 1=20, we can get Z2= 120 from the table.
yfa 1 = 2.80 ysa 1 = 1.55
YFa2=2. 14 YSa2= 1.83
(8) Allowable bending stress [σF]
According to the textbook P 136(6-53):
[σF]= σFlim YSTYNT/SF
According to the textbook:
σflim 1 = 288 MPaσflim 2 = 19 1 MPa
According to figure 6-36, YNT 1=0.88 YNT2=0.9.
The stress correction coefficient of the test gear YST=2
According to the general reliability, the safety factor SF= 1.25 is selected.
Calculate the allowable bending stress of two wheels
[σF] 1 =σflim 1 ystynt 1/SF = 288×2×0.88/ 1.25 MPa
=4 10Mpa
[σF]2 =σflim 2 ystynt 2/SF = 19 1×2×0.9/ 1.25 MPa
= 204 MPa
Substitute the obtained parameters into equation (6-49).
σf 1 =(2kt 1/bm2z 1)yfa 1 ysa 1
= (2×/kloc-0 /× 2586.583/35× 22× 20) × 2.80×1.55 MPa.
= 8Mpa & lt[σF] 1
σF2 =(2kt 1/bm2z 2)yfa 1 ysa 1
= (2×1× 2586.583/35× 22×120) × 2.14×1.83 MPa.
= 1.2 MPa & lt； [σF]2
Therefore, the bending fatigue strength of gear tooth root is sufficient.
(9) Calculate the central moment a of gear transmission.
a = m/2(z 1+Z2)= 2/2(20+ 120)= 140mm
(10) Calculate the peripheral speed v of the gear.
v =πd 1n 1/60× 1000 = 3. 14×40×960/60× 1000
= 2.0096 m/s
Six, the design and calculation of the shaft
Design and calculation of input shaft
1. Calculate the shaft diameter according to the torque.
Choose 45# tempering with hardness of 2 17~255HBS.
Look up the table according to the textbook and take c= 1 15.
d≥ 1 15(2.304/458.2) 1/3mm = 19.7mm
Consider the keyway and increase the diameter by 5%, then
d = 19.7×( 1+5%)mm = 20.69
∴ Select d=22mm.
2. Structural design of the shaft
Positioning, Fixing and Assembly of Parts on (1) Axis
In a single-stage reducer, the gears can be arranged in the center of the box and symmetrically distributed with respect to the two bearings. The left side of the gear is positioned by the shoulder and the right side is axially fixed by the sleeve. The connection is fixed by Ping Jian as a transition fit, and the two bearings are positioned by the shoulder and the big cylinder respectively, so the transition fit is adopted.
(2) Determine the diameter and length of each section of the shaft.
Section: d 1=22mm, length: L 1=50mm.
∫ h = 2cc =1.5mm.
Section 2: D2 = d1+2h = 22+2× 2×1.5 = 28mm.
∴d2=28mm
At first, 7206c angular contact ball bearing with an inner diameter of 30 mm was selected.
The width is 16mm.
Considering the gear end face and the inner wall of the box, there should be a certain distance between the bearing end face and the inner wall of the box. The sleeve length is 20mm, and the shaft length passing through the sealing cover should be determined according to the width of the sealing cover and considering that the coupling and the outer wall of the box body should have a certain torque. So the length of this section is 55mm, and the length of the gear section should be 2mm smaller than the width of the hub, so the length of the second section is:
L2 = (2+20+16+55) = 93mm.
The diameter of the third section d3=35mm
L3 = l1-l = 50-2 = 48mm.
Ⅳ segment diameter d4 = 45mm.
According to the instruction: c =1.5h = 2c = 2×1.5 = 3mm.
d4=d3+2h=35+2×3=4 1mm
The length is the same as the sleeve on the right, that is, L4 = 20 mm.
However, considering the locating shoulder of the left rolling bearing in this section, the bearing should be easy to disassemble, and the installation size h=3 meets the standard. The diameter of this section should be (30+3× 2) = 36mm.
Therefore, the Ⅳ section is designed as a step, and the diameter of the left section is 36 mm.
ⅴ segment d5=30mm in diameter and L5= 19mm in length.
According to the length of each section of the shaft, the span L= 100mm of the shaft support can be calculated.
(3) according to the bending moment combination strength.
① Find the diameter of the cyclotron: d 1=40mm is known.
② Torque calculation: T2=34747.5N is known? millimetre
③ Find the circumferential force: feet
According to the style of textbooks
ft = 2 T2/D2 = 69495/40 = 1737.375n
④ Find the radial force Fr.
According to the style of textbooks
Fr=Ft？ tanα= 1737.375×tan 200 = 632n
⑤ Because the two bearings of the shaft are symmetrical, La = LB = 50 mm.
(1) Draw the axial stress diagram (as shown in Figure A).
(2) Draw the vertical bending moment diagram (as shown in Figure B)
Bearing reaction force:
Fay =FBY=Fr/2=3 16N.
FAZ = FBZ = feet /2 = 868 newtons.
Symmetry on both sides shows that the bending moment of section C is also symmetrical. The bending moment of section c in the vertical plane is
MC 1 = FAyL/2 = 235.3×50 = 1 1.765n？ m
(3) Draw a horizontal bending moment diagram (as shown in Figure C). The bending moment of section c on the horizontal plane is:
MC2 = FAZL/2 = 63 1.6 1455×50 = 3 1.58n？ m
(4) Draw a bending moment diagram (as shown in Figure D)
MC =(MC 12+MC22) 1/2 =( 1 1.7652+3 1.582) 1/2 = 43.345n？ m
(5) Draw a torque diagram (as shown in Figure E)
Torque: T=9.55×(P2/n2)× 106=35N? m
(6) Draw the equivalent bending moment diagram (as shown in Figure F)
The torsional shear force generated by the torque changes according to the pulse period, and α= 1 is taken as the equivalent bending moment at section c:
Mec=[MC2+(αT)2] 1/2
=[43.3452+( 1×35)2] 1/2 = 55.5n？ m
(7) Check the strength of dangerous section C.
By formula (6-3)
σe = Mec/0. 1d 33 = 55.5/0. 1×353
= 12.9 MPa & lt； [σ- 1]b=60MPa
This shaft is strong enough.
Design and calculation of output shaft
1. Calculate the shaft diameter according to the torque.
45# quenched and tempered steel with hardness of (2 17~255HBS) is selected.
Take c= 1 15 according to the textbook.
d≥c(P3/n3) 1/3 = 1 15(2.77/ 1 14) 1/3 = 34.5mm
Take d=35mm.
2. Structural design of the shaft
Part positioning, fixing and assembly of (1) shaft
In the single-stage reducer, the gear can be set in the center of the box, symmetrically distributed relative to the two bearings, and the left side of the gear is positioned by the shoulder.
On the right, the sleeve is used for axial positioning, and the key and transition fit are used for circumferential positioning. The two bearings are positioned by the bearing shoulder and the sleeve respectively, and the transition is used for circumferential positioning.
Fit or interference fit, the shaft is stepped, the left bearing is loaded from the left side, and the gear sleeve, the right bearing and the pulley are loaded from the right side in turn.
(2) Determine the diameter and length of each section of the shaft.
Choose 7207c angular contact ball bearing with an inner diameter of 35mm and a width of 17mm. Considering the end face of the gear and the inner wall of the box, the bearing end
There must be a certain moment between the surface and the inner wall of the box, so if the sleeve length is 20mm, the section length is 4 1mm, and the gear section length is 2mm for the hub width.
(3) According to the combined strength of bending and torsion.
① Find the diameter of the dividing circle: d2=300mm is known.
② Torque calculation: T3 = 271nIs it known? m
③ Find the circumferential force Ft: According to the textbook formula.
ft = 2T3/D2 = 2×27 1× 103/300 = 1806.7n
④ Find the radial force formula.
Fr=Ft？ tanα= 1806.7×0.36379 = 657.2n
⑤∵ The two bearings are symmetrical.
∴LA=LB=49mm
(1) Find the counterforce fax of FBZ FAZ FBY.
Fax = fby = fr/2 = 657.2/2 = 328.6n.
FAZ = FBZ = ft /2 = 1806.7/2 = 903.35 Newton.
(2) When the two sides are symmetrical, the bending moment of the book section C is also symmetrical.
The bending moment of section c in the vertical plane is
MC 1 = FAYL/2 = 328.6×49 = 16. 1N？ m
(3) The bending moment of section C on the horizontal plane is
MC2=FAZL/2=903.35×49=44.26N？ m
(4) calculating the composite bending moment
MC=(MC 12+MC22) 1/2
=( 16. 12+44.262) 1/2
=47. 1N？ m
(5) Calculation of equivalent bending moment: According to the textbook, α= 1.
mec =[MC2+(αT)2] 1/2 =[47. 12+( 1×27 1)2] 1/2
=275.06N？ m
(6) Check the strength of dangerous section C.
Equation (10-3)
σe = Mec/(0. 1d)= 275.06/(0. 1×453)
= 1.36 MPa & lt； [σ- 1]b=60Mpa
This shaft is strong enough.
Seven. Selection and checking calculation of rolling bearing
Depending on the conditions, the expected life of the bearing
16×365× 10=58400 hours
1, calculate the input orientation.
(1) It is known that n Ⅱ = 686 r/min.
Radial reaction force of two bearings: fr1= fr2 = 500.2n.
The first two bearings are 7206AC angular contact ball bearings.
According to the textbook, the axial force inside the bearing
FS=0.63FR, then fs1= fs2 = 0.63fr1= 315.1n.
(2)∫fs 1+Fa = fs2fa = 0
So take any end as the compression end, and now take 1 end as the compression end.
fa 1 = fs 1 = 3 15. 1N FA2 = FS2 = 3 15. 1N
(3) find the coefficients x and y.
fa 1/fr 1 = 3 15. 1N/500.2n = 0.63
FA2/FR2 = 3 15. 1N/500.2n = 0.63
According to the textbook, e=0.68
fa 1/fr 1 & lt； e x 1 = 1 FA2/FR2 & lt； e x2= 1
y 1=0 y2=0
(4) Calculate the equivalent load P 1 and P2.
According to the textbook, take f P= 1.5.
According to the style of textbooks
p 1 = fP(x 1fr 1+y 1fa 1)= 1.5×( 1×500.2+0)= 750.3n
P2 = FP(x2fr 1+y2fa 2)= 1.5×( 1×500.2+0)= 750.3n
(5) Bearing life calculation
∫p 1 = P2, so p = 750.3n n.
∵ Angular contact ball bearing ε=3
According to the manual, Cr=23000N of 7206AC model is obtained.
According to the style of textbooks
LH= 16670/n(ftCr/P)ε
= 16670/458.2×( 1×23000/750.3)3
= 1047500h & gt； 58,400 hours
Life expectancy is sufficient.
2. Calculate the output orientation
(1) It is known that n Ⅲ =114r/min.
Fa = 0 FR = FAZ = 903.35 Newton.
Trial production of 7207AC angular contact ball bearing
According to the textbook FS=0.063FR, then
fs 1 = FS2 = 0.63 fr = 0.63×903.35 = 569. 1N
(2) Calculate the axial loads FA 1 and FA2.
∫fs 1+Fa = fs2fa = 0
∴ Either end is the compression end, 1 is the compression end, and 2 is the relaxation end.
Axial load of two bearings: fa1= fa2 = fs1= 569 438+0n.
(3) find the coefficients x and y.
fa 1/fr 1 = 569. 1/903.35 = 0.63
FA2/FR2 = 569. 1/930.35 = 0.63
According to the textbook: e=0.68.
∵fa 1/fr 1 & lt； ∴x 1= 1 East
y 1=0
∫FA2/FR2 & lt； ∴x2= 1 East
y2=0
(4) Calculate the equivalent dynamic load P 1 and P2.
Take fP= 1.5.
p 1 = fP(x 1fr 1+y 1fa 1)= 1.5×( 1×903.35)= 1355n
P2 = fP(x2fr 2+y2fa 2)= 1.5×( 1×903.35)= 1355n
(5) Calculate the bearing life LH
∫P 1 = P2, so P= 1355 ε=3.
According to manual 7207AC bearing Cr=30500N.
According to the textbook: ft= 1
According to the style of textbooks
Lh= 16670/n(ftCr/P) ε
= 16670/76.4×( 1×30500/ 1355)3
= 2488378.6h & gt58400 hours
This bearing is qualified.
Eight, key connection selection and check calculation
Axis diameter d 1=22mm, L 1=50mm.
Check the manual, select C Ping Jian, and get:
Key A8× 7gb1096-79l = l1-b = 50-8 = 42mm.
T2=48N？ M h = 7 mm
According to the textbook P243( 10-5)
σp=4T2/dhl=4×48000/22×7×42
= 29.68Mpa & lt[σR]( 1 10Mpa)
2. The input shaft and gear are connected by Ping Jian.
Axis diameter d3=35mm L3=48mm T=27 1N? m
Look up the manual P5 1 and choose Ping Jian type A.
Key 10×8 GB 1096-79
L = L3-b = 48-10 = 38mm h = 8mm.
σp = 4T/DHL = 4×27 1000/35×8×38
= 10 1.87 MPa & lt； [σp]( 1 10Mpa)
3. The output shaft and gear 2 are connected by Ping Jian.
Axis diameter D2 = 51ml2 = 50mt = 61.5nm.
Consult the manual, and choose Ping Jian type A..
Key16×10 GB1096-79
L = L2-b = 50- 16 = 34 mm h = 10 mm.
According to the textbook
σp = 4T/DHL = 4×6 100/5 1× 10×34 = 60.3 MPa & lt； [σp]