CHP-29, Machine Design R.S Khurmi 2010
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//-->Contents1066CHAPTERA Textbook of Machine DesignHelical Gears1.2.3.4.Introduction.Terms used in Helical Gears.Face Width of Helical Gears.For mative or EquivalentNumber of Teeth for HelicalGears.5. Propor tions for HelicalGears.6. Strength of Helical Gears.2929.1 IntroductionA helical gear has teeth in form of helix around thegear. Two such gears may be used to connect two parallelshafts in place of spur gears. The helixes may be righthanded on one gear and left handed on the other. The pitchsurfaces are cylindrical as in spur gearing, but the teethinstead of being parallel to the axis, wind around thecylinders helically like screw threads. The teeth of helicalgears with parallel axis have line contact, as in spur gearing.This provides gradual engagement and continuous contactof the engaging teeth. Hence helical gears give smooth drivewith a high efficiency of transmission.We have already discussed in Art. 28.4 that the helicalgears may be ofsingle helical typeordouble helical type.In case of single helical gears there is some axial thrustbetween the teeth, which is a disadvantage. In order toeliminate this axial thrust, double helical gears (i.e.1066TopContentsHelical Gears1067herringbone gears) are used. It is equivalent to two single helical gears, in which equal and oppositethrusts are provided on each gear and the resulting axial thrust is zero.29.2 Terms used in Helical GearsThe following terms in connection with helical gears, as shown inFig. 29.1, are important from the subject point of view.1.Helix angle.It is a constant angle made by the helices with theaxis of rotation.2.Axial pitch.It is the distance, parallel to the axis, between similarfaces of adjacent teeth. It is the same as circular pitch and is thereforedenoted bypc. The axial pitch may also be defined as the circular pitchin the plane of rotation or the diametral plane.Fig. 29.1.Helical gear3.Normal pitch.It is the distance between similar faces of adjacent(nomenclature).teeth along a helix on the pitch cylinders normal to the teeth. It is denotedbypN. The normal pitch may also be defined as the circular pitch in the normal plane which is a planeperpendicular to the teeth. Mathematically, normal pitch,pN=pccosαNote :If the gears are cut by standard hobs, then the pitch (or module) and the pressure angle of the hob willapply in the normal plane. On the other hand, if the gears are cut by the Fellows gear-shaper method, the pitchand pressure angle of the cutter will apply to the plane of rotation. The relation between the normal pressureangle (φN) in the normal plane and the pressure angle (φ) in the diametral plane (or plane of rotation) is given bytanφN= tanφ× cosα29.3 Face Width of Helical GearsIn order to have more than one pair of teeth in contact, the tooth displacement (i.e. the ad-vancement of one end of tooth over the other end) or overlap should be atleast equal to the axialpitch, such thatOverlap =pc=btanα...(i)The normal tooth load (WN) has two components ; one is tangential component (WT) and theother axial component (WA), as shown in Fig. 29.2. The axial or end thrust is given byWA=WNsinα=WTtanα...(ii)From equation(i),we see that as the helix angle increases, then thetooth overlap increases. But at the same time, the end thrust as given byequation(ii),also increases, which is undesirable. It is usually recom-mended that the overlap should be 15 percent of the circular pitch.∴Overlap =btanα= 1.15pcorwhereb=1.15pc1.15× πm... (Qpc=πm)=tanαtanαb= Minimum face width, andm= Module.Fig. 29.2.Face width ofhelical gear.Notes : 1.The maximum face width may be taken as 12.5mto 20m,wheremisthe module. In terms of pinion diameter (DP), the face width should be 1.5DPto2DP, although 2.5DPmay be used.2.In case of double helical or herringbone gears, the minimum face widthis given by2.3pc2.3× πm=tanαtanαThe maximum face width ranges from 20mto 30m.b=TopContents1068A Textbook of Machine Design3.In single helical gears, the helix angle ranges from 20° to 35°, while for double helical gears, it may bemade upto 45°.29.4 Formative or Equivalent Number of Teeth for Helical GearsThe formative or equivalent number of teeth for a helical gear may be defined as the number ofteeth that can be generated on the surface of a cylinder having a radius equal to the radius of curvatureat a point at the tip of the minor axis of an ellipse obtained by taking a section of the gear in the normalplane. Mathematically, formative or equivalent number of teeth on a helical gear,TE=T/ cos3αwhereT= Actual number of teeth on a helical gear, andα= Helix angle.29.5 Proportions for Helical GearsThough the proportions for helical gears are not standardised, yet the following are recommendedby American Gear Manufacturer's Association (AGMA).Pressure angle in the plane of rotation,φ= 15° to 25°Helix angle,α= 20° to 45°Addendum= 0.8m(Maximum)Dedendum= 1m(Minimum)Minimum total depth= 1.8mMinimum clearance= 0.2mThickness of tooth= 1.5708mIn helical gears, the teeth are inclined to the axis of the gear.TopContentsHelical Gears106929.6 Strength of Helical GearsIn helical gears, the contact between mating teeth is gradual, starting at one end and movingalong the teeth so that at any instant the line of contact runs diagonally across the teeth. Therefore inorder to find the strength of helical gears, a modified Lewis equation is used. It is given byWT= (σo×Cv)b.π m.y'whereWT= Tangential tooth load,σo= Allowable static stress,Cv= Velocity factor,b= Face width,m= Module, andy'= Tooth form factor or Lewis factor corresponding to the formativeor virtual or equivalent number of teeth.Notes : 1.The value of velocity factor (Cv) may be taken as follows :Cv====6,for peripheral velocities from 5 m / s to 10 m / s.6+v15,for peripheral velocities from 10 m / s to 20 m / s.15+v0.75,for peripheral velocities greater than 20 m / s.0.75+v0.75+0.25,for non-metallic gears.1+v2.The dynamic tooth load on the helical gears is given by21v+b.Ccos2α +WTwherev, bandChave usual meanings as discussed in spur gears.3.The static tooth load or endurance strength of the tooth is given byWS=σe.b.πm.y'4.The maximum or limiting wear tooth load for helical gears is given byDP.b.Q.Kcos2αwhereDP,b, QandKhave usual meanings as discussed in spur gears.Ww=In this case,where(σes)2sinφN⎡11⎤K=⎢E+E⎥1.4G⎦⎣PφN= Normal pressure angle.WD=WT+21v(b.Ccos2α +WT) cosαExample 29.1.A pair of helical gears are to transmit 15 kW. The teeth are 20° stub in diametralplane and have a helix angle of 45°. The pinion runs at 10 000 r.p.m. and has 80 mm pitch diameter.The gear has 320 mm pitch diameter. If the gears are made of cast steel having allowable staticstrength of 100 MPa; determine a suitable module and face width from static strength considerationsand check the gears for wear, givenσes= 618 MPa.Solution.Given :P= 15 kW = 15 × 103W;φ= 20° ;α= 45° ;NP= 10 000 r.p.m. ;DP= 80 mm= 0.08 m ;DG= 320 mm = 0.32 m ;σOP=σOG= 100 MPa = 100 N/mm2;σes= 618 MPa = 618 N/mm2Module and face widthLetm= Module in mm, andb= Face width in mm.TopContents1070A Textbook of Machine DesignSince both the pinion and gear are made of the same material (i.e. cast steel), therefore thepinion is weaker. Thus the design will be based upon the pinion.We know that the torque transmitted by the pinion,P×60 15×103×60==14.32 N-mT=2πNP2π ×10000∴*Tangentialtooth load on the pinion,14.32=358 NDP/ 2 0.08 / 2We know that number of teeth on the pinion,TP=DP/m= 80 /mand formative or equivalent number of teeth for the pinion,TP80 /m80 /m226.4===TE=mcos3αcos345°(0.707)3∴Tooth form factor for the pinion for 20° stub teeth,0.8410.841=0.175−=0.175−0.0037my'P=0.175−TE226.4 /mWe know that peripheral velocity,πDP.NPπ ×0.08×10000==42 m / sv=6060∴Velocity factor,WT=T==0.104...(Qvis greater than 20 m/s)0.75+v0.75+42Since the maximum face width (b) for helical gears may be taken as 12.5mto 20m,wheremisthe module, therefore let us takeb= 12.5mWe know that the tangential tooth load (WT),358 = (σOP.Cv)b.π m.y'P= (100 × 0.104) 12.5m×πm(0.175 – 0.0037m)= 409m2(0.175 – 0.0037m)= 72m2– 1.5m3Solving this expression by hit and trial method, we find thatm= 2.3 say 2.5 mmAns.and face width,b= 12.5m= 12.5 × 2.5 = 31.25 say 32 mmAns.Checking the gears for wearWe know that velocity ratio,DG320==4V.R.=DP80∴Ratio factor,2×V.R. 2×4==1.6Q=V.R.+1 4+1We know that tanφN= tanφcosα= tan 20° × cos 45° = 0.2573∴φN= 14.4°Cv=0.75=0.75*The tangential tooth load on the pinion may also be obtained by using the relation,WT=PπDP.NP, wherev=(in m / s)v60Top
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