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DESIGN PROCEDURE OF ROPE DRIVE
STEP-1: Selection of Wire rope(PSG 9.1)
From pg no 9.1 select the wire rope based on application select wire rope designation
6x37, 6x19, 6x19F for cranes and hoisting purpose .6x19 means 6 is no of strands, 19 is no of
wires in each strand
STEP-2: Calculation of design load
Assume large factor of safety (say 15), If not given.
Calculate the DESIGN LOAD:
Design load=factor of safety x load to be lifted.
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STEP-2: Selection of wire rope diameter (PSG. 9.4 to 9.6).
Take the design load as the breaking strength.
Obtain the rope diameter by referring data book (PSG. 9.4 to 9.6)..For corresponding wire rope
designation like 6x37, 6x19, 6x19F
STEP-3: Calculation of drum diameter
By referring PSG.9.1, Take the ratio of drum diameter to the rope diameter.
Obtain the drum diameter D.
Drum diameter should be sufficiently larger to reduce the bending stress.
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STEP-4: Selection of the area of useful cross-section of the rope (A)
Determine the area of useful cross section of the rope using the following table.
TYPE OF CONSTRUCTION
METALLIC AREA OF ROPE (A) IN mm2
6*7
0.38d2
6*19
0.40d2
6*37
0.40d2
STEP-5:Calculation of wire diameter (d w)
Determine the diameter of the wire, using the formuladw=d/1.5√i
Where,
i=number of wires in the rope= no. of strands x no. of wires in each strand.
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STEP-6: Selection of weight of rope (Wr) (Take Approx. Weightkgf/m value)
Weight of the rope is obtained by referring data book (PSG 9.4 to 9.6).
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STEP-7: Calculation of various loads and effective loads:
Calculate the various loads using the relations given below:
Direct load, Wd=W+Wr.
i)
ii)
Bending load, Wb=b*A
iii)
Acceleration load, Wa=((W+Wr)*a/g)
iv)
Starting load, Ws=2(W+Wr)
Effective load on the rope during normal working
i)
Effective load on normal working, Wne=Wd+Wb.
ii)
Effective load on starting, Wse=Ws+Wb.
iii)
Effective load on acceleration, Wae=Wa+Wd+Wb.
STEP-8: Calculation of working factor of safety and check for design
Calculate the working factor of safety,
(Fs)w= Braking load for the selected rope (PSG 9.4-9.6) /Effective load during acceleration.
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This (Fs)wshould not be less than the recommended factor of safety given in PSG(9.1).
If the working factor of safety (Fs)w is not satisfactory, choose some other rope with greater
braking strength (or) no. Of ropes should be increased. (FS)W≥recommended FOS PSG(9.1).
STEP-9: Calculation of number of ropes
No. Of ropes= Recommended factor of safety / Working factor of safety.
DESIGN PROCEDURE OF CHAIN DRIVES
STEP-1: Selection of transmission ratio (i)
Select a preferred transmission ratio (i) from
PSG 7.74.
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STEP-2: Selection of number of teeth on pinion (Z1)
Select the no. of teeth on the pinion’s sprocket.
NOTE:Where the space is a problem, Z1min=7.
STEP-3:Determination of number of teeth on the sprocket (Z2)
Determine the no of teeth on the driven sprocket by using transmission ratio(i) →Z2=i x Z1.
Z2 should not be greater than Z2max.Z2max=100 to 120.
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STEP-4: Selection of standard pitch (p)
PSG 7.74
Determine the range of chain pitch using the formula
𝑎
𝑎
Optimum centre distance, a=(30 to 50)pPmax=
Pmin= always take Pmax value
50
30
From the pitch range obtained, consulting table, obtain max.
Speed of the rotation of the pinion n1max.
PSG 7.74
STEP-5: Selection of chain
Assume any chain based on standard pitch value from PSG 7.71-7.73
N.
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STEP-6: Calculation of total load, PT.
Total load on the driving side of the chain,
PT=Pt+PC+PS.
Where,
Pt=tangential force due to power transmission.
Pt=1020N/v pg no 7.78
N-transmitted power
in kW.
v-chain velocity
in m/s.
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Pc-centrifugal tension=mv /gpg no 7.78
m-mass of chain perm PSG 7.71 to 7.73
v-velocity of chain.
Ps-tension due to sagging.pg no 7.78
Ps=k x W x a
N.
K - co-efficient of saggingpg no 7.78
W-weight per m length if chain.
a-centre distance.
STEP-7:Calculation of service factor, ks.
Ks=k1 x k2 x k3 x k3 x k4 x k5 x k6.
Where,
k1-load factor.
k2-factor for distant regulation.
k3-factor for centre distance of sprocket.
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PSG 7.76 & 7.77.
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k4-factor for position of sprocket.
k5-lubrication factor.
k6-rating factor.
STEP-8:Calculation of design load and checking
Calculate the design load,
Design load= total load on chain (PT.) x service factor.
For the selected chain, the breaking load Q is obtained from PSG7.71 to 7.73.
Actual factor of safety= Q/design load.
This actual F.O.S should not be lesser than the recommended F.O.S (PSG7.77)given in
If it is not satisfactory, one more chain may be added to the existing one or increase the pitch.
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STEP-9:Check for bearing stress
Check for bearing stress in the roller using formula,
𝑑𝑒𝑠𝑖𝑔𝑛 𝑙𝑜𝑎𝑑(𝑃𝑡𝑋𝐾𝑠)
=Pt x kS/A=
𝑎𝑟𝑒𝑎 (𝐴)
Where, A-bearing area taken from :
This  is compared with permissible value given in
PSG 7.71 to 7.73.
PSG7.77.
STEP-10:Calculation of length of chain PSG 7.75.
Calculate the length of the chain in terms of no. Of links (correct to even number).
lp=2ap+(Z1Z2/2)+((Z2-Z1)/2𝜋)2/ap.
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STEP-11:Calculation of centre distance (a)
Also calculate the final centre distance corrected to an even no. of links using the equation.
a=((+√(2-8m)) x p/4.
PSG 7.75.
2.
m=((Z2-Z1)/2𝜋)
=lp-((Z1-Z2)/2)
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STEP-12:Calculate diameter of sprocket
Dia. of small sprocket, d1=p/(sin (180/Z1))
Dia. of largest sprocket, d2=p/(sin (180/Z2)).
Sprocket outside diameter, do=d+0.8dr.
dr-diameter of roller.
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STEP-13: Specification of chain drive
Write all the finding value of chain drive
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PSG 7.78
PSG 7.71-7.73
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UNIT – 2
DESIGN PROCEDURE OF SPUR GEAR
STEP 1:Calculation of Gear Ratio
Use i=N1/N2=Z2/Z1
Assume Z1 =18 if not given
STEP 2:Selection of Material
If the material is not given select a suitable material from
PSG.8.5
Mostly assume steel 40Ni Cr1 Mo 28 case hardened surface hardness case(600)
STEP 3:Calculation of Gear Life
If the gear life is not given assume the gear life as 20,000 hrs.
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STEP 4:Calculation of Initial Design Torque
[Mt] = Mt х K х Kd
Kd х K = 1.3 (initially assume the symmetric scheme from PSG 8.15
Ko = 1.5 (assuming medium shock or medium load from table)
𝑃𝑋60000
Mt= 2⊓𝑁1 Nmm
STEP 5:Calculation Of Design Bending Stress [σb] and Design Contact Stress [σc]
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To find [σb]: Refer the formula
PSG.8.18
𝟏.𝟒 𝑲𝒃𝒍
[σb]= 𝒏 𝑲𝛔 σ-1𝑲𝒃𝒍 − 𝑡𝑎𝑏𝑙𝑒 𝑛𝑜 22 𝑝𝑔 𝑛𝑜 8.20 𝑲𝛔 − table 21 𝐩𝐠 𝐧𝐨 𝟖. 𝟏𝟗assume 0≤ X≤ 0.1
To finding the 𝐾σ value
n-table no 20 pg no 8.19 To finding the σ-1=0.35 σu +120where σu=600 N/mm2for alloy
steelσu=350 N/mm2 for cast iron
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To find [σc] : Refer the formula
PSG .8.16
2
2
[σc]=CR HRC Kclkgf/cm or CB HB Kclkgf/cm depending upon the material corresponding
HRC or HB value take anyone of the [σc] formula To find the CR HRC and CB HB value
from pg no 8.16 table no 16
To find Kcl-table 17 pg no 8.20 take steel >350 surface hardness if material not given
STEP 6:Calculation of Centre Distance (a)Refer the formula PSG.8.13 table no 8
3
0.74 2 E[Mt]
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a≥(i+1) √( [σc] )
i¥
take E-table 9 pg no 8.14
here take [σc] - from step 5 take i-from step 1
take [Mt]-from step 4 take ¥=0.3
STEP 7:Selection of Number of Teeth (Z1 & Z2)
If not given assume Z1 = 20
Z2 = i х Z1
STEP 8:Calculation of Module (m)Refer the formula
PSG.8.22
𝟐𝒂
m=
take a-from step 6 take Z1, Z2-from step 7, here find the module
𝒁𝟏+𝒁𝟐
value and then standard that finding value from pg no 8.2 table no 1 from preferred(1) row.
STEP 9:Recalculation of Centre Distance (a) Refer the formula
By using the step 8 module formulas recalculate the centre distance
𝑍1+𝑍2
a= m( 2 ) take m-from step 8 take Z1, Z2-from step 7
STEP 10:Calculation of b,d1, v and ψp
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PSG8.22
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b=ψхa
d1 = m х Z1
v = ∏ хd1 х N1 / 60
ψp= b / d1
for ψ refer PSG 8.14 table 10
STEP 11:Selection of Suitable Quality of Gear
Refer
PSG.8.16 table no 15
For the corresponding velocity find the quality of gear like IS QUALITY 5,6,8,10
STEP 12:Recalculation of Design Torque [Mt]
[Mt] = Mt х K х Kd
K = Refer the value from
Kd = Refer the value from
PSG.8.15 table no 14
PSG.8.16 table no 15
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STEP 13:Check for Bending Stress Refer PSG. 8.13A for spur gears
𝒊+𝟏
σb=𝒂𝒎𝒃𝒚 [𝐌𝐭] ≤[σb] take a-from step 9 , m-from step 8 , b-from step 10
Take [Mt]-from step 12 take y-from pg no 8.18 corresponding Z1 and addendum
modification coefficient X=0 by use these value find σband compare this stress to step 5 [σb]
if σb≤[σb] design safe
STEP 14:Check for Contact Stress Refer PSG.8.13 table no 8 for spur gears
σc=0.74
𝒊+𝟏
𝒂
𝒊+𝟏
√
𝒊𝒃
𝑬[𝐌𝐭]≤ [σc] take i-from step 1 a-from step 9 b-from step 10
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E-from step 6 [Mt]-from step 12 by use these value find σcand compare this stress
to step 5 [σc] if σc≤[σc] design safe
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STEP 15:Calculation of Basic Dimensions of the Gear Pair Refer the formulas for spur gear.
Here find the all the nomenclature value from pg no 8.22
Module ( m )
Number of teeth : Z1, Z2
Pitch circle diameter : d1,d2
Face width (b)
Height factor :fo
Center distance (a )
Bottom clearance (c)
Tool depth (h)
Tip diameters da1, da2
Root diameter (df1,df2)
DESIGN PROCEDURE OF HELICAL GEAR
STEP 1:Calculation of Gear Ratio
Use i=N1/N2=Z2/Z1
STEP 2:Selection of Material
If the material is not given select a suitable material from
PSG.8.5
Mostly assume steel 40Ni Cr1 Mo 28 case hardened surface hardness case(600)
STEP 3:Calculation of Gear Life
If the gear life is not given assume the gear life as 20,000 hrs and
Find the gear life as per the given value
STEP 4:
Calculation of Initial Design Torque
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[Mt] = Mt х K х Kd
Kd х K = 1.3 (initially assume the symmetric scheme from PSG 8.15
Ko = 1.5 (assuming medium shock or medium load from table)
𝑃𝑋60000
Mt= 2⊓𝑁1 Nmm
STEP 5: Calculation Of Design Bending Stress [σb] and Design Contact Stress [σc]
To find [σb]: Refer the formula
PSG.8.18
𝟏.𝟒 𝑲𝒃𝒍
[σb]=
σ-1𝑲𝒃𝒍 − 𝑡𝑎𝑏𝑙𝑒 𝑛𝑜 22 𝑝𝑔 𝑛𝑜 8.20 𝑲𝛔 − table 21 𝐩𝐠 𝐧𝐨 𝟖. 𝟏𝟗
𝒏 𝑲𝛔
assume 0≤ X≤ 0.1 To find the 𝐾σ valuein-table no 20
pg no 8.19
To finding the σ-1=0.35 σu +120where σu=600 N/mm2 for alloy steelσu=350 N/mm2 for
cast ion
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To find [σc] : Refer the formula
PSG .8.16
[σc]=CR HRC Kclkgf/cm2 or CB HB Kclkgf/cm2 depending upon the material corresponding
HRC or HB value take anyone of the [σ c] formula
To find Kcl-table 17 pg no 8.20 take
steel >350 surface hardness if material not given
To find the CR HRC and CB HB value from pg no 8.16 table no 16
STEP 6:Calculation of Centre Distance (a) Refer the formula PSG.8.13 table no 8
𝟑
𝟎.𝟕 𝟐 𝐄[𝐌𝐭]
a≥(i+1) √([𝛔𝐜])
𝐢¥
here take [σc] - from step 5 take i-from step 1
STEP 7:
take [Mt]-from step 4 take ¥=0.3
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take E-table 9 page no 8.14
Selection of Number of Teeth (Z1 & Z2)
Z2 = i х Z1
If not given assume Z1 = 20
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STEP 8: Calculation of Module (m) Refer the formula PSG.8.22
𝟐𝒂 𝒄𝒐𝒔ß
mn= 𝒁𝟏+𝒁𝟐 take a-from step 6 assume ß = 150 take Z1, Z2-from step 7 here find the module
value and then standard that finding value from pg no 8.2 table no 1 frompreferred(1) row
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STEP 9: Recalculation of Centre Distance (a) Refer the formula PSG8.22for helical gear
By using the step 8 module formula recalculate the centre distance
𝑴𝒏 𝒁𝟏+𝒁𝟐
a= 𝒄𝒐𝒔ß 𝟐 take mn-from step 8 take Z1, Z2-from step 7 assume ß = 150
STEP 10:Calculation of b, d1, v and ψp
b=ψхa
a-from step 6 for ψ refer PSG 8.14 table 10
d1 = mn х Z1
mn-from step 8 Z1- from step 7
v = ∏ х d1 х N1 / 60
N1-given value
ψp = b / d1
STEP 11: Selection of Suitable Quality of Gear
Refer
PSG.8.16 table no 15
For the corresponding velocity find the quality of gear like IS QUALITY 5,6,8,10
STEP 12: Recalculation of Design Torque [Mt]
[Mt] = Mt х K х Kd
K = Refer the value from
PSG.8.15 table no 14
Kd = Refer the value from PSG.8.16 table no 15
𝑷𝑿𝟔𝟎𝟎𝟎𝟎
Mt= 𝟐⊓𝑵𝟏 Nmm-from step 4
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STEP 13: Check for Bending Stress PSG. 8.13A for helical gear
𝒊+𝟏
σb=𝟎. 𝟕 𝒂𝒎𝒃𝒀𝒗 [𝐌𝐭] ≤ [σb]
take a-from step 9 , mn-from step 8 , b-from step 1.
𝑍1
Take [Mt]-from step 12 take yv-from pg no 8.18 corresponding Zv1=𝐶𝑂𝑆 3ß(pg no 8.22) value
and addendum modification coefficient X=0 by use these value find σband compare this stress
to step 5 [σb] if σb≤[σb] design safe
STEP 14: Check for Contact Stress Refer PSG.8.13 table no 8 for helical gear
σc=0.7
𝒊+𝟏
𝒂
𝒊+𝟏
√
𝒊𝒃
𝑬[𝑴𝒕]≤ [σc]
Take i-from step 1 a-from step 9 b-from step 10. E-from step 6 [Mt]-from step 12 by use
these value find σc and compare this stress to step 5 [σc] if σc≤[σc] design safe
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STEP 15: Calculation of Basic Dimensions of the Gear Pair Refer the formulas for helical
gear
Here find the all the nomenclature value from pg no 8.22like
Normal module: mn
Number of teeth: Z1, Z2
Pitch circle diameter: d1,d2
Center distance (a)
Bottom clearance (c)
Tool depth (h)
Tip diameters da1, da2
Root diameter (df1,df2)
Height factor: fo
Virtual number of teeth: Zv₁, and Zv₂
UNIT – 3
DESIGN PROCEDURE FOR BEVEL GEARS
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STEP 1: Calculation of Gear Ratio and Pitch angles or (Reference angle) (pg no 8.39)
Use i=N1/N2=Z2/Z1Pitch angle or (Reference angle)=>tan δ 2 = I and δ1 = 90°- δ2
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STEP 2: Selection of Material
If the material is not given select a suitable material from
PSG.8.5
Mostly assume steel 40Ni Cr1 Mo 28 case hardened surface hardness case(600)
STEP 3: Calculation of Gear Life
If the gear life is not given assume the gear life as 20,000 hours & Find the gear life as
per the given value
STEP 4: Calculation of Initial Design Torque
[Mt] = Mt х K х Kd
Kd х K = 1.3 (initially assume the symmetric scheme from PSG 8.15
Ko = 1.5 (assuming medium shock or medium load from table)
𝑃𝑋60000
Mt= 2⊓𝑁1 Nmm
STEP 5: Calculation Of Design Bending Stress [σb] and Design Contact Stress [σc]
To find [σb]: Refer the formula
PSG.8.18
𝟏.𝟒 𝑲𝒃𝒍
[σb]= 𝒏 𝑲𝛔 σ-1𝑲𝒃𝒍 − 𝑡𝑎𝑏𝑙𝑒 𝑛𝑜 22 𝑝𝑔 𝑛𝑜 8.20 𝑲𝛔 − table 21 𝐩𝐠 𝐧𝐨 𝟖. 𝟏𝟗
Assume 0≤ X≤ 0.1 to finding the 𝐾σ valuen-table no 20
pg no 8.19
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To finding the σ-1=0.35 σu +120where σu=600 N/mm2 for alloy steel σu=350 N/mm2 for cast
iron
To find [σc] : Refer the formula
PSG .8.16
[σc]=CR HRC Kclkgf/cm2 or CB HB Kclkgf/cm2 depending upon the material corresponding
HRC or HB value take anyone of the [σc] formula To find Kcl-table 17 pg no 8.20 take steel
>350 surface hardness if material not given
To find the CR HRC and CB HB value from pg no 8.16 table no 16
STEP 6: Calculation of Cone Distance (R) for straight bevel and spiral bevel gear
Refer the formula PSG.8.13 table no 8
𝟑
𝟎.𝟕𝟐
𝟐 𝑬[𝑴𝒕]
R≥¥y√(𝒊𝟐 + 𝟏) √((¥𝒚−𝟎.𝟓)[𝛔𝐜])
𝑹
𝒊
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Assume ¥y =𝒃 =3 ,
take i-from step 1, take [σc]-from step 5 , take E- Equivalent
young’s modulus from table 9 pg no 8.14,take [Mt]-from step 4
Selection of Number of Teeth (Z1& Z2) and Virtual no of teeth (ZV1& ZV2)
Z2 = i х Z1If not given assume Z1 = 20
For the right angled bevel gears,
𝐙𝟏
𝐙𝟐
Zv1 =𝐂𝐨𝐬 𝛅𝟏
Zv2 =𝐂𝐨𝐬 𝛅𝟐pg no 8.39 Take δ1, δ2 value from step 1
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STEP 7:
Step: 8Calculation of transverse module(mt) pg no 8.38
𝑹
mt =
this formula derive from cone distance formula
𝟐
𝟎.𝟓√𝒁𝟏 +𝒁𝟐^𝟐
take R- value from step 6 ,Z1 , Z2 value from step 7find the transverse module
value select std value from pg no 8.2 table no from preferred(1) row
pg no 8.38
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Step: 9 Revision of cone distance (R)
R=0.5mt√𝑍12 + 𝑍2^2
Mt- from step 8 Z1 , Z2 value from step 7
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STEP 10: Calculation of b, mav,d1av ,v and ψp
𝑹
b =¥𝐲 R-value from step 9 and assume ¥y=3
𝒃 𝒔𝒊𝒏 𝛅𝟏
mav= mtpg no 8.38 this formula derive freom transverse module formula take mt𝒁𝟏
from step 8
d1av = mav х Z1
mav-from step 10 Z1- from step 7
v = ∏ х d1av х N1 / 60 N1-given value put d1av should be in m to finding the velocity
ψp = b / d1av
Step: 11 Selection of Suitable Quality of Gear
Refer PSG.8.16 table no 15
For the corresponding velocity select the quality of gear like IS QUALITY 5,6,8,10
STEP 12: Recalculation of Design Torque [Mt]
[Mt] = Mt х K х Kd
K = Refer the value from
Kd = Refer the value from
𝑷𝑿𝟔𝟎𝟎𝟎𝟎
Mt= 𝟐⊓𝑵𝟏 Nmm-from step 4
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PSG.8.15 table no 14
PSG.8.16 table no 15
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STEP 13: Check for Bending Stress PSG. 8.13A for bevel gear
For straight bevel gear
For spiral bevel gear
σb=
𝑹√𝒊𝟐 +𝟏[𝑴𝒕]
𝟏
((𝑹−𝟎.𝟓𝒃)^𝟐)𝒃𝑴𝒏𝒀𝒗 𝒄𝒐𝒔𝜶
≤ [σb]
σb=
𝟎.𝟕𝑹√𝒊𝟐 +𝟏[𝑴𝒕]
((𝑹−𝟎.𝟓𝒃)^𝟐)𝒃𝑴𝒏𝒀𝒗
≤ [σb]
Taking values to find σb
R-from step 9 i-from step 1 [Mt]-from step 12 b-from step 10 mn=mt-from step 8
𝑍1
take yv-from pg no 8.18 corresponding Zv1=Cos δ1 value take α=150
always take spiral
bevel gear if type doesn’t mention in problem. ifσb ≤ [σb] then design safe. if design not
safe increase the transverse module and then redo from step 9
STEP 14: Check for Contact Stress Refer PSG.8.13 table no 8 for bevel gear
𝟎.𝟕𝟐
𝒊𝒃
) 𝑬[𝑴𝒕]
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σc=(𝑹−𝟎.𝟓𝒃) √(
√((𝒊𝟐+𝟏)𝟐
taking values to find σb
R-from step 9 i-from step 1 [Mt]-from step 12 b-from step 10 E-from step 6
compare this stress to step 5 [σc] if σc≤[σc] design safe
find σcand
STEP 15: Calculation of Basic Dimensions of the Gear Pair Refer the formulas
for bevelgear
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Refer pg no 8.38 find all the dimensions or nomenclature of bevel gear by using the above
finding value calculate all the dimensions of bevel gear from pg no 8.38
Transverse module :mt
Number of teeth : Z1, Z2
Pitch circle diameter : d1= mt x Z1 d2 = Mt x Z2
Cone distance R
Face width b
Tip diameters da1, da2
Pitch angle δ₁ , δ₂
Height factor :fo
Clearance : c
Virtual number of teeth :Zv₁, and Zv₂
Addendum angle
Dedendum angle, Tip angle, Root angle
find all these values
DESIGN PROCEDURE OF WORM GEARS
STEP-1: Selection of Material
PSG. 8.45
If material is not given in the problem assume Worm- steel and Worm wheel – bronze
STEP-2: Calculation of Initial Design Torque PSG.8.44
[Mt]=Mt x K x Kd.
Mt=
Initially, Assume K x Kd = 1.
Nmm by using this formula find Mt
𝑃𝑋60000
2⊓𝑁1
STEP-3: Selection of Z1&Z2. PSG 8.46 table 37
From page no 8.46 Select Z1 for given efficiencies. if efficiencies not given in
problem assume η=0.85 and Z1=3. in your data book Z1&Z2 are in the form of symbol as
Z & z so here consider Z1= Z &Z2= z
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𝑁1
𝑧
𝑍2
Z2 = i x Z1. Here i=𝑁2 = 𝑍 = 𝑍1
pg no 8.46
STEP-4:Selection of design bending stress [σb]& design surface stress[σc]PSG.8.45
To find [σb]
From pg no 8.45 table no 33 for bronze wheel, assume σu<390 N/mm2 assume method
of casting of wheel sand and assume rotation in one direction only condition take
[σb]=500kgf/cm2 this value converted as [σb]=50N/mm2. If worm wheel material is cast iron
take grade 25 σu=250 N/mm2 rotation in one direction only condition take [σb]=30N/mm2
To find [σc]
From pg no 8.45 table no 32 if worm-steel and wheel-cast iron assume sliding
velocity Vs=0.5m/s take [σc] =1200kgf/cm2=120N/mm2
If worm – steel and wheel bronze assumeVs=3m/s take [σc]=1590kgf/cm2=159N/mm2
STEP-5:Calculation of Centre DistancePSG.8.44
𝟐
𝟑
𝟓𝟒𝟎
a =( + 𝟏) √( 𝒛 [𝛔𝐜]) [𝐌𝐭]
𝒒
𝒒
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𝒛
Here taking values fromz-from step 3, assume q=11, [σc]-from step 4, [Mt]-from step 2 .By
using this value find centre distance a
STEP-6:Calculation of Axial Module
PSG.8.43
𝟐𝒂
mx=
(𝒒+𝒛)
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here taking values from a-from step 5 q=11 z-from step 3 after finding thisaxial module std
this value from pg no 8.2
STEP-7:Calculation of Revised Centre Distance (a)
PSG.8.43
Refer from pg no 8.43 takea=0.5mx(q+z) formula find a value here taking values mx-from
step 6, q=11, z-from step 3.
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STEP-8: Calculation of d, v,γ,Vs.Pg no 8.43
Pitch diameter:
d1= q x mx
d2=z x mx
Pitch line velocity:
v1= πd1N1 / 60 v2= πd2N2 / 60 put diameter should be inm
Lead angle( γ)
:
γ= tan-1 {Z/q}
Sliding velocity Vs:
Vs= v1/cos γ
Taking values from
q=11, mx – from step 6 , z-from step 3 N 1,N2 – Given value
STEP-9: Recalculation of Design Contact Stress [σc] UsingVs. PSG.8.45
For the new sliding velocity from step 8 value you have to calculate the design contact stress
for that step 8 sliding velocity contact value.referpg no 8.45 for the corresponding worm and
wheel material and step 8 slidibg velocity select the new design surface stress [σ c]
STEP-10:Revision of [Mt]PSG.8.44
The calculated step 8 value of v2 value will be less than 3 therefore take kd=1 and k=1 put
these values once again calculate the Mt value by using [Mt]=Mt x K x Kd.
STEP-11:Check for Bending stress PSG.8.4
𝟏.𝟗[𝑴𝒕]
[σb] = 𝑴𝒙𝟑 𝒒 𝒛 𝒀𝒗≤[σb]
Taking values from
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[Mt]-From step 10, mx-from step 6, q=11, z-from step 3, take yv-from pg no 8.18
𝑍
corresponding Zv1=𝐶𝑂𝑆 3 γ value and addendum modification coefficient X=0 by use these
value find σb and compare this stress to step 5 [σb] if σb≤[σb] design safe
STEP-12:Check for Wear σc PSG.8.44
Refer the formula from pg no 8.44 find σc
𝟑
𝒛
σc=
𝟓𝟒𝟎
𝒛
𝒒
( )
√(
((𝒒)+𝟏)
𝒂
) [𝑴𝒕]
AP
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here taking values fromz-from step 3, q=11, a-from step 7, [Mt]-from step 10 by use these
value find σband compare this stress to step 4 value [σc] if σc≤[σc] design safe.
STEP-13:Check for Efficiency
η=0.95 x tan γ/tan (γ+ρ)
ρ=TAN-1(μ)
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STEP-14:Calculation of Cooling Area Required
pg no 8.52
(1-η)x Input power =hcr x Ax(to-ta) by using these these equation find thecooling area. Here
taking values from η-from given or assume value input power- from given power. hcr=Heat
transfer co efficient –from given if hcr not given in the problem don’t do this step (to-ta)-from
given or assume 450
STEP-15: Calculation of Basic Dimensions PSG.8.43
From pg no 8.43 find all the dimension of worm gear like
Axial module(mx),Number of starts(z),Number of teeth on worm wheel(Z),length
of worm (L),centre distance (a),face width(b),Height factor(fo),bottom clearance(c),pitch
diameter(d1,d2),tip diameter (da1,da2),Root diameter(df1,df2). Find all these values from pg no
8.43 formulas
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UNIT - 4
DESIGN OF GEAR BOX
Procedure for typical gear box design.
Step 1:
Calculate the progression ratio (φ)
𝑁
Progression ratio = 𝑁𝑚𝑎𝑥 = φ Z-1(where Z= number of speed)
𝑚𝑖𝑛
Step 2:
Write the structural formulae
AP
P
Z1 = p1 (x2) x p2(x2) xp3(x3) x p4(x4)
X1 = 1, X2=p1, X3= p1p2, X4= p1p2 p3
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Preferred Structural Formulas
➢ 6 speeds: 2 x 3 or 3 x 2
➢ 8 speeds: 2 x 4 or 4 x 2 or 2 x 2 x 2
➢ 9 speeds: 3 x 3
➢ 12 speeds: 3 x 2 x 2 or 2 x 2 x 3 or 2 x 3 x 2
➢ 16 speeds: 4 x 2 x 2 or 2 x 4 x 2 or 2 x 2 x 4
Step 3:
Draw the ray diagram or speed diagram
Step 4:
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Draw the kinematic diagram.
Step 5:
ST
Calculate the number of teeth.
REQUIREMENT TO OBTAIN THE OPTIMIUM DESIGN:
To reduce the large diameter of gear wheels and also limit the pitch line velocity of the
gear drives the following principle to be followed.
1.
2.
3.
4.
No. of gear on the last stage should be minimum.
No .of gear on the shaft should not be more than 3.(but some cases it may be 4)
It is necessary to have Nmax ≥Ninput ≥Nmin ( in all strages except in the 1 st stage)
The transmission ratio between the drive and driven shaft should be maximum
𝑁𝑚𝑎𝑥
𝑁𝑖𝑛𝑝𝑢𝑡
𝑁
≤ 2 and 𝑁 𝑚𝑖𝑛 ≥ .25( in all strages except in the 1st stage)
𝑖𝑛𝑝𝑢𝑡
Problem:Design a 12 speed gear box speed range 100 rpm to 355 rpm
To calculate
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•
•
•
•
Draw ray diagram
Draw kinematic diagram
No. of teeth on each gear
Also calculate the percentage deviation of the obtainable speeds from the calculated
ones.
Solution:
Step 1: Progression ratio
𝑁
Progression ratio = 𝑁𝑚𝑎𝑥 =φz-1
𝑚𝑖𝑛
Z=number of speed=12
100
=φ12-1 = φ11
AP
P
355
φ11 =3.55
φ
= 1.127
(PSG ddb pg no 7.20)
The progression ratio φ =1.122 coincide with R 20 series therefore select R 20 series speeds
N1 = 100 rpm N2 = 112 rpm N3 = 125 rpm N4 = 140 rpm
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N5 = 160 rpm N6 = 180 rpm N7 = 200 rpm N8 = 224 rpm
N9 = 250 rpm N10 = 280 rpm N11= 315 rpm N12 = 355 rpm
Step 2: Structural formulae
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Z1 = p1 (x2) p2(x2) p3(x3) p4(x4)
X1 = 1, X2= p1, X3= p1p2, X4= p1p2 p3
ST
No of speed =12 speed
Preferred structural formulae is =3 x 2 x 2 or 2 x 3 x 2 or 2 x 2 x 3
Let us select 3 x 2 x 2
Where p1 =3, p1 =2, p3 =2, X1 = 1, X2 = 3, X3 = 6
Z= 3(1) .2(3). 2(6)
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Step 3: Ray diagram
2(3)
2(6)
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3(1)
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Procedure for the construction of ray diagram
3rd stage: 2 speed, 6 spaces
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Locate point (A) minimum speed as at 100rpm leave 6 interval and mark point (B)
maximum speed as at 200rpm.
Locate point(C) input speed as at 160rpm and check the conditions
(
𝑁𝑚𝑎𝑥
𝑁𝑖𝑛𝑝𝑢𝑡
𝑁
≤2)
200
(𝑁 𝑚𝑖𝑛 ≥ .25 )
𝑖𝑛𝑝𝑢𝑡
=1.25 ≤ 2
160
100
160
= .625 ≥ .25
Selected input speed is satisfactory
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2rd stage: 2 speed , 3 spaces
Locate point (C) minimum speed as at 160rpm leave 3 interval and mark point (D)
maximum speed as at 224rpm.
Locate point (E) input speed as at 200rpm and check the conditions
𝑁
(𝑁 𝑚𝑎𝑥 ≤ 2 )
𝑖𝑛𝑝𝑢𝑡
(
𝑁𝑚𝑖𝑛
𝑁𝑖𝑛𝑝𝑢𝑡
224
=1.24 ≤ 2
180
160
≥0 .25 )
180
= .88≥ .25
Selected input speed is satisfactory
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1st stage: There are 3 speeds with 1 space interval.
Therefore mark (F) and (G)Assume input speed as the engine speed and mark point (H).
(For first stage it is not necessary to satisfy the condition)
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Step 4: Structural Diagram
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12 possible speeds are:
Z5 Z6
Z5 Z6
Z7 Z8
Z7 Z8
Z5Z6
Z5 Z6
Z7Z8
Z7Z8
Z5Z6
Z5Z6
Z7Z8
Z7Z8
ST
1. Z9 Z10
2. Z9 Z10
3. Z9 Z10
4. Z9 Z10
5. Z11Z12
6. Z11Z12
7. Z11 Z12
8. Z11 Z12
9. Z13 Z14
10. Z13 Z14
11. Z13 Z14
12. Z13 Z14
Z1Z2
Z3Z4
Z1Z2
Z3Z4
Z1Z2
Z3Z4
Z1Z2
Z3Z4
Z1Z2
Z3Z4
Z1Z2
Z3Z4
Step 5: Calculation of no. of teeth on each gear.
Third stage:
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1st pair :( reducing gear)
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑛 𝑔𝑒𝑎𝑟
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑟 𝑔𝑒𝑎𝑟
100
160
20
=Z1, Z1 = 32 teeth
=
𝑁1
𝑁2
𝑍
=𝑍2
1
LET US ASSUME Z2 = 20
2nd pair :( increasing speed gear)
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑛 𝑔𝑒𝑎𝑟
=
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑟 𝑔𝑒𝑎𝑟
𝑁3
𝑁4
𝑍
=𝑍4
3
Z3 +Z4 = Z1 +Z2
Z3 +Z4 = 32
160
Z4
=52−Z4 = 1.25(52 –Z4 ) = Z4
AP
P
200
Z4 =29, Z3 =23
Second stage:
1st pair :( reducing gear)
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑛 𝑔𝑒𝑎𝑟
160
180
20
=Z5, Z5= 23 teeth
𝑁5
=
𝑁6
𝑍
=𝑍6
5
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𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑟 𝑔𝑒𝑎𝑟
LET US ASSUME Z6 = 20
2nd pair :( increasing speed gear)
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑛 𝑔𝑒𝑎𝑟
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑟 𝑔𝑒𝑎𝑟
Z7 +Z8 = 43
Z8
𝑁8
𝑍
=𝑍8
7
ST
224
𝑁7
U
Z6 +Z5 = Z7 +Z8
=
=
= 1.244(43 –Z8 ) = Z8
180 43−Z8
Z8 =24, Z7 =19
First stage:
In this stage there are 3 speeds and 3 pairs of gears are required. To avoid interference of gear
of one shaft with the gear of the other shaft while shifting following condition has to be satisfied
to avoid interference.
Z12 – Z10 ≥ 4
Z12 – Z14 ≥ 4
1st pair:(reducing speed)
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𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑛 𝑔𝑒𝑎𝑟
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑟 𝑔𝑒𝑎𝑟
180
355
=
𝑁9
𝑁10
𝑍
= 𝑍10
9
20
=Z9, Z9= 40 teeth ,LET US ASSUME 𝑍10 =20 teeth
2nd pair:(intermediate speed)
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑛 𝑔𝑒𝑎𝑟
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑟 𝑔𝑒𝑎𝑟
=
𝑁11
𝑁12
𝑍
=𝑍12
11
Z9 +Z10 =Z11 +Z12
Z11 +Z12 = 60
200
355
=
Z12
60−Z12
= 0.563(60 – Z12) = Z12
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Z11 =38, Z12 =22
3rd pair:(high speed)
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑛 𝑔𝑒𝑎𝑟
𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑟𝑖𝑣𝑒𝑟 𝑔𝑒𝑎𝑟
=
𝑁13
𝑁14
=
𝑍14
𝑍13
Z9 +Z10 = Z14 +Z13
224
355
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Z14 +Z13 = 60
Z14
=60−Z14 = 0.613(60 –Z14 ) = Z14
Z14 =23, Z13 =27
U
RESULT:
ST
1st stage:
Z14 =23 teeth
Z13 =27 teeth
Z11 =38 teeth
Z12 =22 teeth
Z9 = 40 teeth
𝑍10 =20 teeth
2ndstage:
Z8 =24 teeth
Z7 =19 teeth
Z5 = 23 teeth
Z6 = 20 teeth
3rd stage :
Z4 =29 teeth
Z3 =23 teeth
Z1 = 32 teeth
Z2 = 20 teeth
Step 6: Calculate the percentage of deviation
Obtainable speed calculation.
Os1 = IS x
𝑍10
𝑍9
𝑍
x
𝑍6
𝑍5
𝑍
x
𝑍2
𝑍1
𝑍
= 355 x
20
20
40
23
x
20
Os2 = IS x 𝑍10x 𝑍6 x𝑍4 = 355 x 40x
9
𝑍
5
𝑍
3
𝑍
Os3 = IS x 𝑍10x 𝑍8 x 𝑍2
9
7
1
20
20
23
= 355 x 40x
x
32
= 96.46
29
x 23 = 194.61
24
19
20
x 32 = 140.13
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𝑍
𝑍
𝑍
20
24
29
Os4 = IS x 𝑍10x 𝑍8 x𝑍4 = 355 x 40x 19 x 23 = 282.7
9
7
𝑍
3
𝑍
𝑍
22
Os5 = IS x𝑍12x 𝑍6 x 𝑍2
11
5
𝑍
1
𝑍
𝑍
22
5
𝑍
𝑍
22
Os7 = IS x𝑍12x 𝑍8 x 𝑍2
11
7
𝑍
23
3
𝑍
𝑍
22
20
x 32 = 111.69
29
x 23 = 225.3
24
= 355 x 38x
1
𝑍
23
20
Os6 = IS x𝑍12x 𝑍6 x𝑍4 = 355 x 38x
11
20
= 355 x 38x
20
x
= 162.25
19 32
24
29
Os8 = IS x𝑍12x 𝑍8 x𝑍4 = 355 x 38x 19 x 23 = 327.33
7
𝑍
3
𝑍
𝑍
23
Os9 = IS x𝑍14x 𝑍6 x 𝑍2
13
5
1
𝑍
𝑍
𝑍
23
Os10 = IS x𝑍14x 𝑍6 x𝑍4 = 355 x 27x
13
𝑍
5
𝑍
3
𝑍
Os11 = IS x𝑍14x 𝑍8 x 𝑍2
13
𝑍
7
𝑍
20
= 355 x 27x
1
𝑍
23
23
20
23
= 355 x 27x
23
24
20
x 32 = 119.9
29
x 23 = 241.19
24
19
29
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11
20
x 32 = 174.2
Os12 = IS x𝑍14x 𝑍8 x𝑍4 = 35527x 19 x 23 = 351.63
7
3
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Deviation
% deviation
nobt - nC
Deviation x 100 / nC
100
96.46
-3.54
-3.54
2.
112
111.69
-0.31
-0.27
3.
125
119.90
-5.1
-4.08
4.
140
140.13
0.13
0.09
5.
160
162.25
2.25
1.40
6.
180
174.20
-5.8
-3.22
7.
200
194.61
-5.39
-2.69
8.
224
225.30
1.3
0.58
9.
250
241.19
-8.81
-3.52
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1.
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Sl.No
Calculated
Obtainable speed
speed (nC) rpm
(nobt) rpm
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10.
280
282.70
2.7
0.96
11.
315
327.33
12.33
3.91
12.
355
351.63
-3.37
-0.94
The permissible deviation = ± 10(Φ-1)%
= ± 10(1.127-1)%
= ±1.27
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From the above table we find, the deviations are well within the permissible limits.
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Unit – 5:CAM
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