Machine elements

VILNIUS GEDIMINAS TECHNICAL UNIVERSITY DEPARTMENT OF PRINTING MACHINES COURSE PROJECT OF MACHINE ELEMENTS Vilnius 2006 Toothed-belt and worm gear transmission Task: T=200 Nm ω3=ωout=6 rad/s M-TB-WG-RC Kinematic scheme Introduction Worm-gear drives have wide application in mechanics because they have noisless operation, they can have rather high transmission ratios and not very big dimensions. Their work is even and uniform. Among shortcomings we have to note not big efficiency, they produce big amount of heat, they are rather expensive, because their production should be very accurate. Designed worm-gear will be used in the factory for moving crane base. Moving crane is carrying a load, so the movement of crane should be uniform, even. As worm-gear work is noisless, it increases comfortability for workers. Working medium is not “active”. Selection of an electric motor Effciency: η= η1·η2·η3; η1=0.99 – efficiency of bearings; η2=0.95; η3=0.95(1-u/200), where u=30 for worm-gear, then η3=0.95(1-30/200)=0.81; η=0.992·0.95·0.81=0.76. Rotational frequency of output shaft: Power of output shaft: P3= ω3·T; P3=6·200=1200W=1.2kW. Needful power of electrical motor: ; Electric motor 4A80132 is selected. It’s characteristics are: Power P1=2.2kW; Rotational frequency nsin=3000 r.p.m.; Sliding s=4.3%. Real rotational frequency: ; Determination of transmission ratio and torque Transmission ratios Common transmission ratios of the drive: . Selected transmission ration for the belt drive: u1=2 then transmission ratio of worm-gear will be . The number of teeth for the worm-gear: z2=z1·u2=2·25.05=50.10; Assume z2=50. Real transmission ration of worm-gear: . Real rotational frequency of output shaft: Real rotational frequency differs from the needed one by: . Error is allowable. Rotational frequencies: n1=2871r.p.m.; ; Angular velocities: ; ; . Torques: ; ; η2 – efficiency of belt drive; ; η1 – efficiency of bearing; η3 – efficiency of worm-gear. Real torque T3 differs from required one by: ; Error is allowable. Calculation of a reducer Transmission ratio of reducer: u2=25; Number of starts of warm z1=2; Number of teeth of worm-wheel z2=50. Material selected: For worm – steel 45, hardness > HRC45; For worm-wheel – tin free bronze BrAZ9-4P (casted in sand form); Assume sliding velocity vsl=4m/s. From tables we obtain: admissible contact stresses σadmH=161MPa. Admissible tension stresses: σadm=σ0e·kre; kre – regime coefficient calculating bending stresses; kre=0.543; σ0e=98.1MPa σadm=98.1·0.543=53.3MPa Distance between axes T3=199.1Nm; Assuming that number of modulus q=8; Θ=57 – coefficient of deformation of the worm; x=0.6 – coefficient, evaluating the load deviation (deviations of load are not big). Coefficient of load concentration: ; . Assuming 7th degree of accuracy for the worm-gear. Dynamic coefficient: Kd=1.1. Coefficient of loading: K=Kcon·Kd=1.27·1.1=1.4. Torque for calculations: Tcal3=T3·K=199.1·1.4=278.7Nm. Distances between axis calculated with respect to contact stresses: ; Main parameters of worm-gear Engage modulus: . According to VST2144-76 it is assumed that: ms=5 and q=10. Real distance between axis: . Main parameters of the worm pitch diameter: d1=ms·q=5·10=50mm. Diameter of addendum circle: da1=ms·(q+2)=5·(10+2)=60mm. Diameter of dedendum circle: df1=ms·(q-2.4)=5·(10-2.4)=38mm. Length of cut part: L≥(11+0.06·z2)·ms=(11+0.06·50)·5=70mm. This length should be increased by 25 mm, so L≥70+25=95mm; Assume L=95mm. Slope angle of winding of the worm: λ=11˚18’36”. Main parameters of worm gear-wheel Pitch diameter: d2=z2·ms=50·5=250mm. Diameter of addendum circle: da2=ms·(z2+2)=5·(50+2)=260mm. Diameter of dedendum circle: df2=ms·(z2-2.4)=5·(50-2.4)=238mm. External diameter: ; Assume Dex=267mm. Width of the crown of worm-wheel: b2=da1·0.75=60·0.75=45mm. Circle velocity of the worm: v2=ω2·0.5·dd1=150.25·0.5·0.05=3.76m/s. Velocity of sliding: . When we have such velocity, admissible contact stress σadmc=161MPa; Coefficient f=0.023·1.4=0.03; Angle of friction ρ=1˚41’. Efficiency of reducer: . Verifying calculations for a worm-gear When z1=2 and q=10, coefficient of deformation for worm θ=86, x=0.6, z2=50. Coefficient of load concentration: . Assume 7th degree of accuracy. vsl=3.83m/s; Dynamic coefficient Kd=1.1. Load coefficient: K=Kcon·Kd=1.079·1.1=1.187. Torque for calculations: Tcal3=T3·K=199.1·1.187=236.3Nm. Forces acting in the engagement Peripheral velocity of worm-wheel is equal to axial force on worm: . Axial force of worm-wheel is equal to peripheral velocity of the worm: . Radial forces of worm-wheel: Fr1=Fr2=F2·tanαs=1593·tan20˚=579.8N. Peripheral force of worm-wheel for calculations: F2cal=1593·1.187=1891N. Contact stresses in worm gear: ; . Reducer number of worm-wheel teeth: . Working stresses from bending in worm-wheel teeth: ; y=0.45 coefficient of teeth form; . Orientational calculation of shafts Diameter of the output end of worm-wheel shaft (torque and bending are acting): Equivalent moment: ; -bending moment which is equal to torque transmitted by this shaft; ; . Material of the shaft is steel 45 which has strength limit: σu=590MPa. Admissible stress: σadm=54MPa. Diameter of the output end of worm-wheel shaft: . Diameter of the output end of worm-wheel shaft assume d3=40mm; Diameter of shaft under bearings d3b=45mm; Diameter under the worm-wheel 50mm; Worm and worm shaft are manufactured as one component. Diameter of the input end of worm shaft: Assume diameter of input end of worm shaft d2=20mm; Diameter of shaft under worm d2z=38mm; Diameter of transition part dtr2=34mm; Diameter under the bearings d2b=25mm. Design of dimensions of worm-wheel Crown is pressed on worm-wheel; d=50 – diameter of shaft under worm wheel. Diameter of the hub: dh=(1.6…1.8)d=(1.6…1.8)50=80…90mm; Assume dh=80mm. Length of the hub: Lh=(1.2…1.8)d=(1.2…1.8)50=60…90mm; Assume Lh=60mm. Width of the disk: c=0.3b2; c=0.3·45=13.5mm; Assume c=14mm. Diameter of the bolt: db=(1.2…1.5)m=(1.2…1.5)5=6.0…7.5mm; Assume db=6.0mm. Length of the bolt: lb=(0.3…0.4)B=(0.3…0.4)45=13.5…18mm; Assume lb=14mm. The calculations of acting forces, torques and bending moments Worm Shaft Reactions in vertical plane: ΣMA=0; Fr1·a-Fa1·d1/2-RBy·(a+b)=0; 579.8·0.1185-1593·0.05/2-RBy·(0.1185+0.1185)=0; RBy=122N. ΣMB=0; RAy·(a+b)-Fr1·b-Fa1·d1/2=0; RAy·(0.1185+0.1185)-579.8·0.1185-1593·0.05/2=0; RAy=458N. Bending moments in vertical plane: MA=0; Mc’=-54.3Nm; Mc”=14.5Nm; MB=0. Reactions in horizontal plane: ΣMA=0; -F1·a+RBx·(a+b)=0; -400·0.1185+RBx·(0.1185+0.1185)=0; RBx=200N. ΣMB=0; -RAx·(a+b)+F1·b=0; -RAx·(0.1185+0.1185)-400·0.1185=0; RAx=200N. Bending moments in horizontal plane: MA=0; MC=23.7Nm; MB=0. Summary reactions: ; ; Axial components of radial reactions: s=0.83·e·Fr; s1=0.83·0.3·500=125N; s2=0.83·0.3·234=58N; s1≥s2, therefore axial force Fa1=125N; Fa2= s1+Fa1=125+1593=1718N;

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