(TIỂU LUẬN) đồ án môn học CHI TIẾT máy THIẾT kế hệ dẫn ĐỘNG

TRƯỜNG ĐẠI HỌC BÁCH KHOA HÀ NỘI VIỆN CƠ KHÍ BỘ MÔN CƠ SỞ THIẾT KẾ MÁY VÀ RÔ BỐT ĐỒ ÁN MƠN HỌC CHI TIẾT MÁY HỌC KÌ: 20202 Người hướng dẫn Thông tin sinh viên Sinh viên thực Mã số sinh viên Lớp chuyên ngành Lớp tín Ngày kí duyệt đồ án: ……./……./20… Ký tên ĐÁNH GIÁ CỦA THẦY HỎI THI TABLE OF CONTENTS Contents ABSTRACT INTRODUCTION SELECTING THE ENGINE AND TRANSMISSION DESIGNING THE CHAIN DRIVE DESIGNING THE HELICAL GEARBOX DESIGNING THE SHAFT 11 SELECTING AND ANALYZING THE BEARINGS 17 THE OUTSIDE STRUCTURES OF GEARBOX 18 CONCLUSION 18 REFERENCES 19 TABLE OF FIGURES Fig 1: Distance between point of force application………………………………12 Fig Internal torque diagram of the first shaft (I)……………………… 13 Fig Internal torque diagram of the second shaft (II)………………………… 16 ABSTRACT The flexible power transmission elements comprises of belt drives, chain drives and rope drives Since these elements are flexible they are called flexible drives They are commonly used in short distance power transmission Wire ropes can also be used for long distance power transmission Unlike rigid transmission elements(gears), the design of flexible drives are simple and reduces the cost When compared with rigid transmission elements, they are quiet and absorbs shocks which reduces mechanical vibrations Flexible drives not have infinite life, hence its necessary to consider the wear, aging, loss of elastic property and also the environment in which the drive is employed, while designing/selecting the flexible drives INTRODUCTION Transmission systems transfer mechanical power from a source to another machine components For example let us consider a car, in which the power from engine is transmitted to wheels through clutch, gearbox, prop shaft and differential, these components are called transmission elements If we consider a lathe, power from motor is provided to a chuck through belt drive and gearbox, these components are transmission elements A number of elements like gears in a gear box together form a transmission system SELECTING THE ENGINE AND TRANSMISSION In mechanical drive systems, electric motors are very commonly used There are many different types of electric motors, however, due to their many advantages compared to other types of electric motors (simple structure, low cost, easy maintenance, reliable operation ), three – phase induction motor is the most commonly used The process of calculating and selecting electric motors for the drive system is done through the following steps of calculation: - Engine power - Number of preliminary synchronous revolutions of the motor - Requirements on starting torque, overload and installation method 1.1 The engine power - The engine power is calculated by this equation: Pct = P t η +) Pct is the necessary engine power +) Pt is the calculated power in the working process +) η is the drive system efficiency This is The necessary engine power (Pct) equals to The calculated power in the working process (Pt) devided by The drive system efficiency (η) The calculated power in the working process (Pt) is calculated as shown below: Pt =Plv = F.v 1800.2,47 = 10001000 =4,45(kW ) As F is The conveyor belt pulling force, v is The speed of conveyor belt From the Table 2.3[1], p.19, select The drive system efficiency (η) = 0.84 Calculating The necessary engine power (Pct) as followed: Pct = P t η = 4,45 0.84 =5,30 (kW ) 1.2 An overview of motor shaft revolution - The revolution of final drive shaft: nlv = 60000 v z.p = 60000.2,47 12.110 =112(rev / min) As The revolution of final drive shaft (nlv) equals to 6000 times The velocity of large sprocket (v) devided by Number of Large Sprocket Teeth (z) times Sprocket pitch (p) - From the Table 2.4[1], select: +) Speed ratio of chain drive: =uxt=3 +) Gear ratio: ubr=4 - The whole drive transmission (u): u=uxt.ubr=3.4=12 - The revolution of motor shaft: nem=u.nlv=112.12=1344 (rev/min) Searching the Table P1.1[1], for a combination of 1344 rev/min and 5,3 kW provides a 4A112M4Y3 electric motor Results: The 4A112M4Y3 electric motor P = 5,5 kW, n = 1425 rev/min 1.3 Corrected transmission nem 1425 ucorrected = nlv = =12,7 112 The chain drive transmission (uxt) is selected uxt = u 12,7 The gear ratio (ubr) ubr = = =4,2 uxt All the torques (T), power transmissions (P), revolutions (n), drive transmission (u) of each shaft is followed the Table 1.1 below: Table1.1 Torques (T), power transmissions (P), revolutions (n), drive transmission (u) of each shaft Shaft U P (kW) n (v/p) T (Nmm) DESIGNING THE CHAIN DRIVE When selecting roller chains, the following parameters should be taken into account Machine to be used Impact Type Prime Motor Type Power Transmission(kW) Diameter and Rotary Speed of High-Speed Shaft Diameter and Rotary Speed of Low-Speed Shaft Inter-Shaft Distance Based on these conditions, Roller chain is selected 2.1 Sprocket teeth Using the selection guide table(Table 3)or the power transmission efficiency tables, select the chain and the number of small sprocket teeth that satisfy the rotary speed of the high-speed shaft and the corrected power transmission(kW) The chain pitch should be as small as possible, as long as the required power transmission efficiency is achieved This should minimize noise and ensure smooth transmission of power (If a single chain does not provide the required power transmission efficiency, use multiple chains instead If the installation space requires that the inter-shaft distance as well as the outer diameter of sprocket be minimized, use small-pitch multiple chains.) There should be a minimum wrap angle of 120˚ between the small sprocket and the chain Based on these parameters and from Table 5.4[1]: - Number of small procket teeth: z1 = 25 - Number of large procket teeth: z2 = 75 2.2 Main parameters of the chain drive a) Pitch From the Table 5.5[1] with these conditions from Table 1.1 provides the chain: Pitch p = 25,4 mm Roller diameter dc = 7,95 mm Roller width B = 22,61 mm Power transmission efficiency [P] = 19 kW b) Distance between Shaft centers a= p [ x− z + z 2 + √( x− z + z ) −2(z −z ) 2 ] =1121,42 (mm) π 2.3 All parameters of the chain drive After calculating others parameters of the chain drive and looking up from some tables All parameters of the chain drive are show in the table below Table 2.1 All parameters of the chain drive Parameter Type Pitch Number of roller Chain length Distance between Shaft centers Number of small sprocket teeth Number of large sprocket teeth Lực tác dụng lên trục DESIGNING THE HELICAL GEARBOX 3.1 Materials Based on Table 6.1[1], the material of each gear is selected below: - The driving gear +) Steel grade: C45 +) Heat treatment: Hardened after machining +) Brinell hardness: HB = 241 ÷ 285, select HB1 = 265 +) Tensile strength: 850 MPa +) Yield strength: 580 Mpa - The driven gear +) Steel grade: C45 +) Heat treatment: Hardened after machining +) Brinell hardness: HB = 192 ÷ 240, select HB2 = 230 +) Tensile strength: 750 MPa +) Yield strength: 450 Mpa 3.2 Calculating Contact stress [σ H ]and Bending stress [σ F ] Contact stress [σ H ]and Bending stress [σ F ]of the driving gear [ σ ] H1 = [σ F1]= Contact stress [σ H ]and Bending stress [σ F ]of the driven gear [ σ [ σ = ] H2 F2 = ] Initial Contact stress [σ H 1]+[σH2 [σ H ]sb= 3.3 Center distance a =K (u+1) w a √ [σ H ]sb2 uψba So aw = 110 mm is chosen 3.4 Other parameters of gears a) Module m = (0,01 0,02).aW = (0,01 0,02).110= 1,1 2,2 (mm) From the table of standard gear module, provides m = (mm) b) Number of teeth Assume β=10 °=¿ cos β=0,9848 So the number of teeth of driving gear is a cos β w Z1 = = 2.110 0,9848 =20,83 m(u+1) 2.( 4,2+1) Z1 = 20 (teeth) is chosen Z2 = u.Z1 = 4,2.20 = 84 Z2 = 84 (teeth) is chosen c) Helix angle cos β= => β=arccos (cos β )=arccos(0,9636 )=19 ° d) Check for factor of safety All the calculated strengths is less than allowable strengths, then the design is safe e) Table of parameters of gears Table 3.1 Parameters of gears Parameter Distance center Number of teeth Module Helix angle Gear width Pitch circle diameter Tip diameter Root diameter Contact force Tangential force Radial force Axial frce 10 DESIGNING THE SHAFT 4.1 Initial caculations a) Materials - Symbol: C45 - Tensile strength: 600 Mpa - Allowed torque strength: [τ] = 12 ÷ 30 Mpa b) Distance between each point of force application 11 Fig 1: Distance between point of force application lm 13=(1,2 ÷1,5 )d1=(1,2 ÷ 1,5)25=30 ÷37,5 (mm) => lm13 = 37 (mm) lm 23=(1,2 ÷1,5 )d2=(1,2 ÷ 1,5)30=36 ÷ 45( mm) => lm23 = 40 (mm) lm 12=(1,4 ÷ 2,5) d1=(1,4 ÷ 2,5) 25=35 ÷ 62,5(mm) => lm12 = 45(mm) lm 22=(1,2÷ 1,5) d2=(1,2 ÷ 1,5)30=36 ÷ 45( mm) => lm22= 45 (mm) l cki=0,5 (lmki +b0 )+ k3 +hn ¿ - Second shaft: l 22=lc 22=70(mm) l 23=0,5(l¿¿ m 23+b02 )+ k1 +k2 ¿ = 0,5(40 + 19) + 15 + 10 = 54,5 Chọnl23=55( mm) l 21=2 l23 =110(mm) -First shaft: l 12=lc 12=69(mm) l 13=l23=55(mm) 12 l 11=2.l13=110 (mm) 4.2 Diameter of the first shaft (I) caculation and selection a) Internal torque diagram of the first shaft (I) Fk14 = 153,1 (N) Ft13 = 1628,7 (N) Fr13 = 662,59 (N) Fa13 = 560,81 (N) F x 10=910,39 (N) F y10 =223,62( N) F x 11=565,21( N ) F y11=438,97( N) Fig Internal torque diagram of the first shaft (I) b) Diameter of shaft at each point of force application M j 11= M tđ 11 √ M2x 11+ M2y11=√10563,92 +02=10563,9( Nmm) =√M j 11 +0,75 T211 =√10563,92+ 34447,022 =36030,45( Nmm) ⇒d11= 13 √M 2x 13+ M 2y13=√50071,452 +24143,352=55588,23 (Nmm) Mtđ 13=√ M2J 13 +0,75 T213=√55588,232 +0,75.34447,022=63087,24 ( Nmm) M J 13= ⇒d13= M tđ 13 √0,1[σ] √M 2x14 + M 2y 14=0 ( Nmm Mtđ 14=√M2j14 +0,75 T 142=√0+0,75 M j 14= ⇒d14 = d 10=d11 =18,8(mm) Based on those calculations above, corrected diameter of shaft (I) at each point of force application: d 10=d11=20( mm) d 13=22( mm) d 14=18 (mm) Key selection for the first shaft (I) From Table 9.1a[1] : - At d13=22 mm, the parameters of key are: c) b=6 mm h=6 mm t 1=3,5 mm l=36 mm - At d14=18 mm, the parameters of key are: b=6 mm h=6 mm t 1=3,5 mm l=36 mm Check for factor of safety All t he calculated actual factor of safety is greater than minimum factor of safety, so the design is safe d) 14 4.3 Diameter of the second shaft (II) caculation and selection Because of limited time, only the first shaft is calculated Ft23 = 1628,7 (N) Fa23 = 560,8 (N) Fr23 = 626,97 (N) F x20=2728,53( N) F y 20=2109,04( N ) F x21=310,86( N ) F y 21=−731,16( N) Fig Internal torque diagram of the second shaft (II) d 20=d21=30 (mm ) d 23=32( mm) d 24=28(mm) - Key selection for the second shaft (II) At d13=32 mm, the parameters of key are: b=10 mm h=8 mm t 1=5 mm l=36 mm SELECTING AND ANALYZING THE BEARINGS 5.1 Bearing selection At position 10: F r 10=Fr 0=√ F x 10 + F2y10 =√910,392+ 223,622 =937,45( N ) At position 11: √ F r 11=Fr 1= F x 11 + F2y11=√565,212+ 438,972=715,65( N ) Axial force: 16 F a=Fa 13=560,81( N) Because the axial force is huge, so Tapered roller bearing is chosen 5.2 Bearing parameters for the first shaft (I) From Table P2.11[1], with d = 20 mm: Symbol: 7204 Inner diameter: d= 20 mm Outer diameter: D= 47 mm Dynamic load rating: C= 19,1kN Static load rating: C0= 13,30kN Cone witdth: B= 14mm Cone angle α = 13,500 5.3 Analyzing the bearings a) Dynamic load C d=Q b) m √L=2,3361.10 /√3 855=17,7 ( kN )