جامعة بابل | University of Babylon



Worm Gear setsIt is possible to obtain a large speed reduction (up to 360:1) and a high increase of torque by means of worm sets. Typical applications of worm gear sets include positioning devices that take advantage of their locking property as can be shown in figure(1). Only a few materials are available for worm sets. The worms are highly stressed and usually made of case-hardened alloy steel.Fig.(1) A single-enveloping worm set. Worm Gear Geometry:The terminology used to describe the worm (Figure(2)) and power screws is very similar. In general, the worm is analogous to a screw thread and the worm gear is similar to its nut.Fig.(2) Notation for a worm gear set: (a) double-threaded worm; (b) worm gear.The axial pitch of the worm gear pw is the distance between correspond-ding points on adjacent teeth.The lead L is the axial distance the worm gear (nut) advances during one revolution of the worm. In a multiple-thread worm, the lead is found by multiplying the number of threads (or teeth) by the axial pitch. The pitch diameter of a worm dw is not a function of its number of threads Nw. The speed ratio of a worm set is obtained by the ratio of gear teeth to worm threads:rs=ωgωw=NwNg=LπdgAs in the case of a spur or helical gear, the pitch diameter of a worm gear is related to its circular pitch and number of teeth :dg=NgpπThe center distance between the two shafts equalsc = (dw + dg)/2 as shown in the figure. The lead angle of the worm (which corresponds to the screw lead angle) is the angle between a tangent to the helix angle and the plane of rotation. The lead and the lead angle of the worm have the following relationships:L=pwNwtanλ=Lπdw=VgVwwhereL = the leadPw = the axial pitchNw = the number of threadsλ = the lead angledw = the pitch diameterV = the pitch line velocityFor a 90° shaft angle figure (2), the lead angle of the worm and helix angle of the gear are equal:λ=ψNote that λ and ψ are measured on the pitch surfaces.Normal pressure angles are related to the lead angle and the Lewis form factor Y, as shown in Table (1).Table(1) Various normal pressure angle for worm setThe worm gears usually contain no less than 24 teeth, and the number of gear teeth plus worm threads should be more than 40. The face width ( b) of the gear should not exceed half of the worm outside diameter dwo.AGMA recommends the magnitude of the minimum and maximum values for worm pitch diameter dW as follows:c0.8753≤dw≤c0.8751.6Where, c represents the distance between the centers of the worm and the gear.Ex:A triple-threaded worm has a lead L of 75 mm. The gear has 48 teeth and is cut with a hob of modulus mn = 7 mm perpendicular to the teeth. Calculate : a. The speed ratio rsb. The center distance c between the shafts if they are 90o apartSolutionFor a 90° shaft angle, we have λ = ψThe velocity ratio of the worm gear set is:rs=NwNg=348=116 Using the following equation to calculate the circular pitch pw=LNw=753=25mmSince Pn=1/mnAnd Pnpn=π hence pn=π mnpn=πmn=7π=21.99mmSincepn=pwcosλ (helical gears)thencosλ=pnpw=21.9925=0.8 or λ=28.4odw=Lπtanλ=75πtan28.4=44.15mmdg=Lπrs=75π/16=381.97mmc=12dw+dg=1244.15+381.97=213.1mmWorm Gear Bending and Wear StrengthsLewis Equation:The bending stresses are much higher in the gear than in the worm. The following slightly modified Lewis equation is therefore applied to worm gear:Fb=σobKfYPnThe value of Y can be taken from Table (1). It is to be noted that the normal diametral pitch Pn is defined by the following equation.Pn=PcosψLimit Load for WearThe wear equation by Buckingham, frequently used for rough estimates, has the following form:Fw=dgbKwwhereFw = the allowable wear load dodg = the pitch diameterb = the face width of the gearKw = a material and geometry factor, obtained from Table (2)Table (2) Worm gear factor KwFor a satisfactory worm gear set, the usual requirement is thatFb≥Fd and Fw≥FdFd=1200+V1200Ft for V>4000fpmThermal Capacity of Worm Gear setsThe power capacity of a worm set in continuous operation is often limited by the heat dissipation capacity of the housing or casing. Lubricant temperature commonly should not exceed about 93°C (200°F). The basic relationship between temperature rise and heat dissipation can be expressed as follows:H=CA?twhereH = the time rate of heat dissipation, lb ft/minC = the heat transfer, or cooling rate, coefficient (lb ft per minute per square foot of housing surface area per oF)A = the housing external surface area, ft2Δt = the temperature difference between oil and ambient air, oFThe values of A for a conventional housing, as recommended by AGMA, may be estimated by the following formula:A=0.3c1.7Here, A is in square feet and (c) represents the distance between shafts (in inches). The approximate values of heat transfer rate C can be obtained from Figure (3). Note from the figure that C is greater at high velocities of the worm shaft, which causes a better circulation of the oil within the housing. The manufacturer usually provides the means for cooling, such as external fins on housing and a fan installed on worm shaft, as shown in Figure (3). Fig.(2) Heat transfer coefficient C for worm gear housing. Fig.(3) Worm gear speed reducerThe heat-dissipation capacity H of the housing, as determined by the above equation, in terms of horsepower is given in the following form:(hp)d=CA?t33000This loss of horsepower equals the difference between the input horsepower hpi and output horsepower hpo. Inasmuch as e = hpo/hpi, we have hpd = hpi – e(hpi). The input horsepower capacity is thereforehpi=hpd1-eThe quantity (e) represents the efficiency.Worm Gear EfficiencyThe expression of the efficiency e for a worm gear reduction is the same as that used for a power screw and nut. In the notation of this chapter, it is written as follows:e=cosΦn-ftanλcosΦn+fcotλwheref is the coefficient of frictionφn represents the normal pressure angleThe value of (f ) depends on the velocity of sliding Vs between the teeth:Vs=VwcosλThe quantity Vs is the pitch-line velocity of the worm. Table (3) furnishes the values of the coefficient of friction.Table(3) Worm gear coefficient of friction (f) for different sliding speedsEx: Design Analysis of a Worm Gear Speed ReducerA worm gearset and its associated geometric quantities are schematically shown in Figure (4). Estimatea. The heat-dissipation capacity.b. The efficiency.c. The input and output horsepower.The gear set is designed for continuous operation based on a limiting100o F temperature rise of the housing without fan.Fig.(4)SolutionThe speed ratio of the worm gear set isrs=NwNg=460=115A=0.3c1.7=0.3(8)1.7=10.29ft2From figure (2) C=42ft/(min ft2.F)hpd=CA?t33000=42(10.29)(100)33000=1.31The pitch-line velocity of the worm isVw=πdwnw12=π(3)(1000)12=785.4fpmVs=Vwcosλ=785.4cos15=813fpmFrom table (3) f = 0.0238. Introducing the numerical values into the efficiency Equation to gete=cosΦn-ftanλcosΦn+fcotλOre=cos25-0.0238tan15cos25+0.0238cot15=0.904or 90.4%To calculate the input horse power Use the following Equation :hpi=hpd1-e=1.311-0.904=13.65The output horse power is thenhpo=13.65-1.31=12.3Comments: Because of the sliding friction inherent in the tooth action, usually worm gear sets have significantly lower efficiencies than those of spur gear drives. The latter can have efficiencies as high as 98%.Ex:A worm gear set is schematically illustrated in Figure (5). Finda. The gear ratio, gear diameter, and lead of wormb. The helix angle and center distance of gearsNw = 2, Ng = 40, P = 8, The worm diameter will be dw = 3.5p, where p is the circular pitch of the gear.Fig.(5)Solutionrs=NwNg=240=120P=Ndg=40dgdg=408=5inrs=Lπdg=Lπdg120=Lπ×5L=0.785inSince Pp=π Hence p=πP=π8=0.39indw=3.5p=3.5×0.39=1.37intanλ=Lπdw=0.785π×1.37=0.14λ=8oc=12dg+dw=125+1.37=3.18inEx:A worm gear set consists of gear with 65 teeth and a triple-threaded worm. The center distance 210mm and the face width 25mm. The module is 6mm. The worm rotates at speed 1200rpm. The worm diameter will be dw= c 0.875/2 .Gear and worm are made of bronze and cast iron, respectively. Compute a. The worm diameter dw, gear pitch diameter dg, and lead angle λ. b. Limit load for wear.Solution(a)dw= c 0.875/2=(210)0.872/2=53.816mmc= (dg+dw)/2210=(dg+53.81)/2dg=366.19mmNwNg=Lπdg365=Lπ×366.19L=53mmtanλ=Lπ×dw=53π×53.81=0.31λ=17.4o(b) Fw=dgbKwKw can be obtained from table (2)For cast iron worm and Bronze gear the value of Kw=185psiNote:1N/m 2=0.000145psi or 1psi=6894.77N/m2∴185psi=185×6894.77=1275532.45Nm2=1.275N/mm2Fw=366.19×25×1.275=11672.3NEx:A 10 hp, 1000 rpm electric motor drives a 50 rpm machine through a worm gear reducer with a center distance of 7 in. as shown in figure (5)Determinea. The value of the helix angleb. The transmitted loadc. The power delivered to the driven machined. The worm gear reducer of Figure (6) is to be designed for continuous operation and a limiting 100 oF temperature rise of the housing with a fan. Estimate the heat dissipation capacity. Will overheating be a problem?Fig.(6)Solution:rs=ωgωw=NwNg=Lπdgrs=501000=120=NwNg=3NgOrNg=60teethc=12dw+dgOr7=123+dgdg=11inrs=Lπ×dg120=Lπ×11Or L=1.72intanλ=Lπ×dw=1.72π×3=0.18 or λ=10.34oλ=ψ for shafts with 90oapart(b) Ft=PowerVV=πdwnw12=π×3×100012=785.39fpmFt=10×33000785.35=420lb(c) The power delivered to the driven machine can be evaluated as:hpo=e×hpiThe worm gear set efficiency can be evaluated as:e=cosΦn-ftanλcosΦn+fcotλThe coefficient of friction can be taken from table (3) after calculating the sliding speed.Vs=Vwcosλ=785.4cos10.34=798.3fpm≈800fpmFrom table(3) f=0.024Hence the efficiency can be evaluated as:e=cos20-0.024tan10.34cos20+0.024cot10.34=0.876Hence the power delivered to the machine can be evaluated as:hpo=0.876×10=8.7hpd=CA?t33000A=0.3C1.7=0.3(7)1.7=8.199ft2C can be taken from the chart of figure(2) at nw=1000rpm C=57ft.lb/(min.ft2.F)hpd=57×8.199×10033000=1.41 ................
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