Chapter Outline Shigley’s Mechanical Engineering Design

Types of Gears Figs. 13–1 to 13–4 Spur Helical Worm Bevel Shigley’s Mechanical Engineering Design

Nomenclature of Spur-Gear Teeth Fig. 13–5 Shigley’s Mechanical Engineering Design

Tooth Size Shigley’s Mechanical Engineering Design

Tooth Sizes in General Use Table 13–2 Shigley’s Mechanical Engineering Design

Standardized Tooth Systems (Spur Gears) Table 13–1 Shigley’s Mechanical Engineering Design

Standardized Tooth Systems Common pressure angle f : 20º and 25º Old pressure angle: 14 ½º Common face width: Shigley’s Mechanical Engineering Design

Conjugate Action When surfaces roll/slide against each other and produce constant angular velocity ratio, they are said to have conjugate action. Can be accomplished if instant center of velocity between the two bodies remains stationary between the grounded instant centers. Fig. 13–6 Shigley’s Mechanical Engineering Design

Conjugate Action Forces are transmitted on line of action which is normal to the contacting surfaces. Angular velocity ratio is inversely proportional to the radii to point P, the pitch point. Circles drawn through P from each fixed pivot are pitch circles, each with a pitch radius. Fig. 13–6 Shigley’s Mechanical Engineering Design

The most common conjugate profile is the involute profile. Can be generated by unwrapping a string from a cylinder, keeping the string taut and tangent to the cylinder. Circle is called base circle. Fig. 13–8 Shigley’s Mechanical Engineering Design

Involute Profile Producing Conjugate Action Fig. 13–7 Shigley’s Mechanical Engineering Design

Circles of a Gear Layout Fig. 13–9 Shigley’s Mechanical Engineering Design

Sequence of Gear Layout Pitch circles in contact Pressure line at desired pressure angle Base circles tangent to pressure line Involute profile from base circle Cap teeth at addendum circle at 1/P from pitch circle Root of teeth at dedendum circle at 1.25/P from pitch circle Tooth spacing from circular pitch, p = p / P Fig. 13–9 Shigley’s Mechanical Engineering Design

First point of contact at a where flank of pinion touches tip of gear Tooth Action First point of contact at a where flank of pinion touches tip of gear Last point of contact at b where tip of pinion touches flank of gear Line ab is line of action Angle of action is sum of angle of approach and angle of recess Fig. 13–12 Shigley’s Mechanical Engineering Design

A rack is a spur gear with an pitch diameter of infinity. The sides of the teeth are straight lines making an angle to the line of centers equal to the pressure angle. The base pitch and circular pitch, shown in Fig. 13–13, are related by Fig. 13–13 Shigley’s Mechanical Engineering Design

Internal Gear Fig. 13–14 Shigley’s Mechanical Engineering Design

The contact ratio is the average number of pairs of teeth in contact. Arc of action qt is the sum of the arc of approach qa and the arc of recess qr., that is qt = qa + qr The contact ratio mc is the ratio of the arc of action and the circular pitch. The contact ratio is the average number of pairs of teeth in contact. Shigley’s Mechanical Engineering Design

Contact ratio can also be found from the length of the line of action The contact ratio should be at least 1.2 Fig. 13–15 Shigley’s Mechanical Engineering Design

Interference(Girişim) Contact of portions of tooth profiles that are not conjugate is called interference. Occurs when contact occurs below the base circle If teeth were produced by generating process (rather than stamping), then the generating process removes the interfering portion; known as undercutting. Fig. 13–16 Shigley’s Mechanical Engineering Design

Interference For 20º pressure angle, the most useful values from Eqs. (13–11) and (13–12) are calculated and shown in the table below. Minimum NP Max NG Integer Max NG Max Gear Ratio mG= NG/NP 13 16.45 16 1.23 14 26.12 26 1.86 15 45.49 45 3 101.07 101 6.31 17 1309.86 1309 77 Shigley’s Mechanical Engineering Design

Increasing the pressure angle to 25º allows smaller numbers of teeth 9 Interference Increasing the pressure angle to 25º allows smaller numbers of teeth Minimum NP Max NG Integer Max NG Max Gear Ratio mG= NG/NP 9 13.33 13 1.44 10 32.39 32 3.2 11 249.23 249 22.64 Shigley’s Mechanical Engineering Design

Interference can be eliminated by using more teeth on the pinion. However, if tooth size (that is diametral pitch P) is to be maintained, then an increase in teeth means an increase in diameter, since P = N/d. Interference can also be eliminated by using a larger pressure angle. This results in a smaller base circle, so more of the tooth profile is involute. This is the primary reason for larger pressure angle. Note that the disadvantage of a larger pressure angle is an increase in radial force for the same amount of transmitted force. Shigley’s Mechanical Engineering Design

Common ways of forming gear teeth Sand casting Shell molding Forming of Gear Teeth Common ways of forming gear teeth Sand casting Shell molding Investment casting Permanent-mold casting Die casting Centrifugal casting Powder-metallurgy Extrusion Injection molding (for thermoplastics) Cold forming Shigley’s Mechanical Engineering Design

Common ways of cutting gear teeth Milling Shaping Hobbing Cutting of Gear Teeth Common ways of cutting gear teeth Milling Shaping Hobbing Shigley’s Mechanical Engineering Design

Shaping with Pinion Cutter Fig. 13–17 Shigley’s Mechanical Engineering Design

Shaping with a Rack Fig. 13–18 Shigley’s Mechanical Engineering Design

Hobbing a Worm Gear Fig. 13–19 Shigley’s Mechanical Engineering Design

Calculation Torque and Revolution D1 İS pitch diameter D2 is pitch diameter RESULT:Torque of gear in a gear pair works together with changes in proportion to the number of teeth Fig. 13–19 Shigley’s Mechanical Engineering Design

Calculation Torque and Revolution D1 is pitch diameter RESULT:Torque of gear in a gear pair works together with changes in proportion to the number of teeth D2 İs pitch diameter RESULT: Cycles of gears in a pair of gear which works together changes with inversely proportional Fig. 13–19 Shigley’s Mechanical Engineering Design

Example P=10 Kw N=3000 rpm T=? N=? Fig. 13–19 Shigley’s Mechanical Engineering Design

Example Calculating Torque Fig. 13–19 1) P= T1*N1/9550 First find the enter Torque T1=9550*P/N1  => T1=9550*10/3000 => T1=31.8 N-m 2) T1/T2=Z1/Z2 => T2=T1*Z2/Z1 => T2=31.8*40/10 => T2=127.2 N-m 3) Due to Z2 and Z3 Gears are on same shaft T2=T3=127.2N-m 4) T3/T4=Z3/Z4 => T4=T3*Z4/Z3 => T4=127.2*50/16 => T4=397.5 N-m Out Shaft Tork T=397.5N-m Fig. 13–19 Shigley’s Mechanical Engineering Design

Calculating Revolution Example Calculating Revolution 1) N1/N2=Z2/Z1 => N2=N1*Z1/Z2 => N2=3000*10/40 => N2=750rpm 2) Due to Z2 ve Z3 Gears are on same shaft N2=N3=750rpm 3) N3/N4=Z4/Z3 => N4=N3*Z3/Z4 => N4=750*16/50 => N4=240rpm Out Shaft Revolution N=240 rpm Ratio of Reduction i=Input Revolution/OutputRevolution i=3000/240  => i=12.5 Fig. 13–19 Shigley’s Mechanical Engineering Design

Example Pratic Solving Fig. 13–19 i=i1*i2 i=(Z2/Z1)*(Z4/Z3) Noutput=Ninput/ i NÇIKIŞ=3000/12.5=240 rpm Toutput=Tinput*i P= Tinput*Ninput/9550  => Tinput=9550*P/Ninput => Tinput=9550*10/3000 => Tinput=31.8 N-m Toutput=31.8*12.5 Toutput=397.5 N-m Fig. 13–19 Shigley’s Mechanical Engineering Design

You must estimate some values.These are Pinion thooth number(Zp) Calculations You must estimate some values.These are Pinion thooth number(Zp) Main Gear thooth number(Zg) Presure Angle(f) Module is main paremater which is variable in equation, try to find best value in calculations Shigley’s Mechanical Engineering Design

Slay Makers Equation(interferance control) Calculations Slay Makers Equation(interferance control) k =1 for full depth teeth. k = 0.8 for stub teeth Slay Makers Equation for Full Sized Gears   Gear Width b=10*M Shigley’s Mechanical Engineering Design

Calculation of Force P= FS*V(power) V= п*D*N/60 P=FS* п*D*N/60 Module in terms of: M=D/Z => D=M*Z  Shigley’s Mechanical Engineering Design

Calculation of Force Fy =FS *CS SERVICE FACTOR Cs GEAR WORKING TIME LOADING CASE 3 Hours/Day 8-10 Hour/Day 24 Hours/Day Continous working 0.8 1 1.25 to light shock 1.5 to moderate shock 1.8  to severe shock 2.0 Maximum Force From Load (FY) Fy =FS *CS Lewis Equation(Dişin dayanabileceği teğetsel kuvvet)   Ft =σs *b*Y*M Ft: Maximum Strenght of Teeth(N) σs : Strenght of the Tooth(N/mm2) b: Lenght of the Teeth(mm) Y: Lewis Foctor of Form(Look Table) M:Module(mm) Shigley’s Mechanical Engineering Design

Calculation of Force Lewis Equation(Dişin dayanabileceği teğetsel kuvvet)   Ft =σs *b*Y*M Number of theeth Ft: Maximum Strenght of Teeth(N) σs : Strenght of the Tooth(N/mm2) b: Lenght of the Teeth(mm) Y: Lewis Foctor of Form(Look Table) M:Module(mm) Shigley’s Mechanical Engineering Design

Calculation of Force Lewis Equation Ft =σs *b*Y*M Maximum Force From Load (FY) Lewis Equation   Ft =σs *b*Y*M Number of Teeth Barth Equation (Coming from Lewis) You can add Fatique to Lewis Equation Ft max =Ft * Cv          =>Ft max =σs *b*Y*M*Cv  Cv  : : Speed Coefficient Shigley’s Mechanical Engineering Design

Speed Coefficient (CV) Determination Shigley’s Mechanical Engineering Design

We must find module under this condition           Iteration Must be We must find module under this condition Shigley’s Mechanical Engineering Design

Affect of Dynamic Forces From Tooth Buckinghm Equation Under the condution of low speed just load forces exist But If the tangential speed more than limits, You must consider dynamic forces Buckinghm equtions is used to find dynamic forces. FY: maximum tangential force generated by the load Fi: dynamic forces generated by the speed C: Some mistakes coming from manufacturing k=0.107*e => Ф=14.50 Full teeth k=0.111*e => Ф=20 Full teeth k=0.115*e => Ф=20 Basık (Stub) Shigley’s Mechanical Engineering Design

Affect of Dynamic Forces From Tooth Buckingham Equation If the pinion and main gear is to be made of the same material of which the elastic module (E) is equal to each other. Thus, from the same steel material and steel gears Ф = 20 full-length "c" coefficients occurs as follows. FY: maximum tangential force generated by the load Fi: dynamic forces generated by the speed Shigley’s Mechanical Engineering Design

Affect of Dynamic Forces From Tooth Buckingham Equation k=0.107*e => Ф=14.50 Full teeth k=0.111*e => Ф=20 Full teeth k=0.115*e => Ф=20 Basık (Stub) Shigley’s Mechanical Engineering Design

Affect of Dynamic Forces From Tooth Buckingham Equation Acording to e(mm), manifacturing precision must be select from table which is given below The maximum amount of error that may occur in the processing (mm) Module M Commercial Gears Elaborated Gears Precision machining Grinding Gears 4 0.05 0.025 0.0125 5 0.056 6 0.064 0.030 0.0150 7 0.072 0.035 0.0170 8 0.080 0.038 0.0190 9 0.085 0.041 0.0205 10 0.090 0.044 0.0220 Shigley’s Mechanical Engineering Design

Affect of Dynamic Forces From Tooth Buckingham Equation If the pinion and main gear is to be made of the same material of which the elastic module (E) is equal to each other. Thus, from the same steel material and for steel gears Ф = 20 full-length "c" coefficients occurs as follows. Shigley’s Mechanical Engineering Design

In Tooth Corrosion Strength FW: Abrasion force (N) DP: pinion gear section circle diameter (mm) b: Gear width (mm) Q: Ratio factor K: load stress factor (N / mm2) σ A: Surface strength resistance (N / mm2) Ф: Gear pressure angle E: Material elastic modulus (N / mm2) The surface abrasion resistance is directly related to the surface hardness σ and BHN is 2.6487 times the surface hardness. Shigley’s Mechanical Engineering Design

System works every day between 8 or 10 hour Example Coupling Main gear Die rolls N=310 rpm Electrical motor P=20KW Pinion gear System works every day between 8 or 10 hour Shigley’s Mechanical Engineering Design

Gear reduction ratio:i= NP/NG = 1000/310 => i=3.226:1 Main gear teeth count:ZG =i*ZP = 3.226*31 => ZG =100 ZG /ZP =100/31 =3.226 Slay maker’s equation: 3453.6<7167 No İnterferance SERVICE FACTOR Cs GEAR WORKING TIME LOADING CASE 3 Hours/Day 8-10 Hour/Day 24 Hours/Day Continous working 0.8 1 1.25 to light shock 1.5 to moderate shock 1.8  to severe shock 2.0 Force of Turning the Gears (FY) b=10*M Shigley’s Mechanical Engineering Design

Tangential Force in Lewis Equation (Ft) Ft  =σs *b*Y*M  σs  : 138.3 N/mm2           (Table 1) b= 10*M STRENGTH (σs) AND HARDNESS OF GEAR MATERIAL WHICH WILL USED IN LEWIS FORMULA MATERIAL STRENGTH OF GEAR σ(N/mm^2) HARDNESS (BHN) 20 Grade Iron casting 47.1 200 25 Grade Iron casting 56.4 220 35 Grade Iron casting 225 35 Grade Iron casting (heat treated) 78.5 300 0.2% Carboniferous steel 138.3 100 0.2% Carboniferous steel (heat treated) 193.2 250 Bronze 68.7 80 Phosphorous Bronze 82.4 Manganese Bronze Aluminous Bronze 152.0 180 0.3% Carboniferous wrought steel 172.6 150 0.3% Carboniferous wrought steel (heat treated) 220.0 C30 Steel (heat treated) 220.6 C40 Steel 207.0 C45 Steel 233.4 Alloy Steel-surface hardeness 345.2 650 Cr-Ni alloy 0.45% carboniferous steel 462.0 400 Cr-Va alloy 0.45% carboniferous steel 516.8 450 Plastic 58.8 ZP =31, Ф=200 To get a full gear; Y= 0.358  Table 3 Ft =138.3*10*M*0.36*M         Ft =498*M2 Shigley’s Mechanical Engineering Design

Tangent velocity of pinion and gear are same Tangential velocity (Vt) Tangent velocity of pinion and gear are same Np=1000 rpm ZP=31 gear Cv=Hız constant Shigley’s Mechanical Engineering Design

Fatique Calculation Barth Equation Ft max =Ft*Cv Barth Equation For iteration:   M=6  Shigley’s Mechanical Engineering Design

Pinion section circle diameter: (DP) Dp=M*ZP = 6*31 => Dp=186 mm. Gear width (b): b= 10*M=10*6=> b=60mm. Pinion section circle diameter: (DP)  Dp=M*ZP = 6*31 => Dp=186 mm. The main gear section circle diameter: (DG) DG=M*ZG = 6*100 => Dp=600 mm. Gear distance between axes: C(mm) C(mm)= (Dp +DG )/2 C=(186+600)/2=393 mm. Shigley’s Mechanical Engineering Design

CONTROL OF CORROSION STRENGTH SERVICE FACTOR Cs GEAR WORKING TIME LOADING CASE 3 Hours/Day 8-10 Hour/Day 24 Hours/Day Continous working 0.8 1 1.25 to light shock 1.5 to moderate shock 1.8  to severe shock 2.0 Shigley’s Mechanical Engineering Design

Ф=200 For full-sized gears: k=0.111*e c=11655*e  e: for error factor Graphics1.   V=9.7 m/ sn => e=0.04mm. => c=11655*0.04=466 N/mm Shigley’s Mechanical Engineering Design

Affect of Dynamic Forces From Tooth Buckingham Equation Shigley’s Mechanical Engineering Design

In Tooth Corrosion Strength FW: Abrasion force (N) DP: pinion gear section circle diameter (mm) b: Gear width (mm) Q: Ratio factor K: load stress factor (N / mm2) σ A: Surface strength resistance (N / mm2) Ф: Gear pressure angle E: Material elastic modulus (N / mm2) Shigley’s Mechanical Engineering Design

In Tooth Corrosion Strength İf Ep=EG Same Material STRENGTH (σs) AND HARDNESS OF GEAR MATERIAL WHICH WILL USED IN LEWIS FORMULA MATERIAL STRENGTH OF GEAR σ(N/mm^2) HARDNESS (BHN) 20 Grade Iron casting 47.1 200 25 Grade Iron casting 56.4 220 35 Grade Iron casting 225 35 Grade Iron casting (heat treated) 78.5 300 0.2% Carboniferous steel 138.3 100 0.2% Carboniferous steel (heat treated) 193.2 250 Bronze 68.7 80 Phosphorous Bronze 82.4 Manganese Bronze Aluminous Bronze 152.0 180 0.3% Carboniferous wrought steel 172.6 150 0.3% Carboniferous wrought steel (heat treated) 220.0 C30 Steel (heat treated) 220.6 C40 Steel 207.0 C45 Steel 233.4 Alloy Steel-surface hardeness 345.2 650 Cr-Ni alloy 0.45% carboniferous steel 462.0 400 Cr-Va alloy 0.45% carboniferous steel 516.8 450 Plastic 58.8 The surface abrasion resistance is directly related to the surface hardness σ and BHN is 2.6487 times the surface hardness. Selected material is 0,2 C Steel Shigley’s Mechanical Engineering Design

In Tooth Corrosion Strength FW=DP*b*Q*K          Dp=186 mm. B=60 mm. Q=1.52 K=0.5  FW=186*60*1.52*0.5  FW=8,481 N Must be required to corrosion: FD =19,719 N > FW=8,481 N   Corrosion Strength is not enough We must change the metarial Shigley’s Mechanical Engineering Design

We must select a harder material For 300BHN STRENGTH (σs) AND HARDNESS OF GEAR MATERIAL WHICH WILL USED IN LEWIS FORMULA MATERIAL STRENGTH OF GEAR σ(N/mm^2) HARDNESS (BHN) 20 Grade Iron casting 47.1 200 25 Grade Iron casting 56.4 220 35 Grade Iron casting 225 35 Grade Iron casting (heat treated) 78.5 300 0.2% Carboniferous steel 138.3 100 0.2% Carboniferous steel (heat treated) 193.2 250 Bronze 68.7 80 Phosphorous Bronze 82.4 Manganese Bronze Aluminous Bronze 152.0 180 0.3% Carboniferous wrought steel 172.6 150 0.3% Carboniferous wrought steel (heat treated) 220.0 C30 Steel (heat treated) 220.6 C40 Steel 207.0 C45 Steel 233.4 Alloy Steel-surface hardeness 345.2 650 Cr-Ni alloy 0.45% carboniferous steel 462.0 400 Cr-Va alloy 0.45% carboniferous steel 516.8 450 Plastic 58.8 We select new material We must select a harder material For 300BHN FW=DP*b*Q*K     → FW=186*60*1.52*1.4     FW=23,816 N      FD =19,719 N < FW=23,816 N Corrosion Strengt is Suitable   Shigley’s Mechanical Engineering Design