LECTURE – 4: WORKED OUT PROBLEMS ON INVOLUTE SPUR GEARS
CONTENTS OF EARLIER LECTURES
1. Interference 2. Methods of eliminating interference 3. Standard tooth systems for spur gears 4. Profile shifted gears 5. Involutometry 6. Design of gear blanks
GEAR CALCULATION – QUESTION 1
In a drive, a velocity ratio of 2.5 with a centre distance of 70 mm is desired. (a) Determine the pitch diameter of the gears with 20o full depth involute teeth, (b) Is there any interference in the system? If so, how will you avoid it? (c) Determine the contact ratio, (d) Find the dedendum, addendum and root diameters, and the tip clearance, (e) If the centre distance is increased by 1.5%, what will be the new pressure angle? o
• • •
φ = 20 ; Data: i = 2.5; C = 70 mm; involute full depth tooth system ; Q: m = ?; Z1 = ? ; Z2 = ?; d1 = ? and d2 = ?; Interference exists or not ?; CR = ?
•
Solution : C = (r 1 +r 2) = 0.5 m ( Z 1 + Z2) = 0.5 m (Z1 + i Z1) = 70 mm i.e., m x Z1 = 40 Possible solutions for standard modules as can be seen from the table are : Solution I: m = 2mm, Z1 = 20, Z2 = i. Z1 = 2.5 x 20 = 50 d1 = m Z1 = 2 x 20 = 40 mm d2 =i d1 = 2.5 x 40 = 100 mm
• • • • • •
STANDARD MODULES IN mm 0.3 1.25 3.5 7 14 26 45 •
0.4 0.5 0.6 0.7 0.8 1.0 1.5 1.75 2.0 2.25 2.5 3.0 4.0 4.5 5.0 5.5 6.0 6.5 8 9 10 11 12 13 15 16 18 20 22 24 28 30 33 36 39 42 50 Further increase is in terms of 5 mm
Minimum number of teeth that can engage with the gear of 50 teeth without interference is given by 4k( 4k(z2 + k) z12 + 2 z1 z2 = sin 2 φ
•
For full depth gears, k = 1. Substituting the values Z2 = 50, φ = 20oin the above equation,
z12
2 z1 x50
z 2 + 100 z 1 1
4 x 1( 50 1 ) sin2 20 o
1744 0
•
i.e., Z1min = 15 is the minimum permissible without interference.
•
Since Z1 = 20 > Z1min (15) no interference will occur.
Contact ratio, CR : Referring to the figure the path of contact : La = ua + ur L a
(r
L a
( 20
1
9.798
a )2
mm
2 2)
r 2 cos2
(r
1
20
2
cos 2
2
20
o
a )2
( 50
r 2 cos2 2
2)
2
50
2
(r
r ) sin
1
cos 2
20
2
o
( 20
50 ) sin 20
o
CR
L a p cos
L a m cos
9.798
x 2 x cos 20 o
9.798
1.660
5.901
For full depth tooth addendum: a = 1m = 1x2 = 2mm Dedendum : b = 1.25m = 1.25x2 = 2.5 mm Clearance: c = 0.25 m = 0.25x2 = 0.5 mm Pinion root circle diameter: dr1 = d1- 2b = 40 – 2x2.5 = 35 mm Pinion addendum diameter: da1 = d1+ 2a = 40 + 2x2 = 44 mm Gear root circle diameter: dr2 = d2- 2b = 100 – 2x2.5 = 95 mm Gear addendum diameter: da2 = d2+ 2a = 100 + 2x2 = 104 mm • • • • • • •
•
Solution II: C = (r 1 +r 2) = 0.5 m ( Z 1 + Z2)
= 0.5 m (Z1 + i Z1) = 70 mm Simplifying, m x Z1 = 40 another possible solution for std module m = 2.5 mm, Z1 = 16, Z2 = i x Z1 = 40, d1 = m Z1 = 2.5x16=40 mm, d2 = m Z2 = 2.5 x 40 =100 mm Minimum number of teeth that can engage the gear of 40 teeth without with interference is given by 4k(z2 + k) z12 + 2 z1 z2 = sin 2 φ o
For full depth gears k = 1. Substituting Z 2 = 40, φ = 20 , in the above equation, z 2 + 1
2 z
1
x 40 =
z 2 + 80 z 1 1
•
4(40 + 1 ) sin2 20o
1402 0
i.e., Z1min = 15 is the minimum permissible without interference.
Since Z1 = 16 > Z1min (15) no interference will occur. Contact ratio, CR : Referring to the figure the path of contact: La = ua + ur
a )2
L a
(r
L a
( 20
1
r 2 cos2 1
2.5 )2
(r
2
a )2
20 2 cos 2 20 o
r 2 cos2
( 50
(r
2
1
2.5 )2
r ) sin 2
50 2 cos 2 20 o
( 20 50 ) sin 20 o
11.854 mm
CR
L a p cos
L a m cos
11.854
x 2.5 x cos 20o
11.854 5.901
1.607
For full depth tooth Addendum : a =1m = 1x2.5 =2.5mm Dedendum : b =1.25m =1.25x2.5 = 3.125 mm Clearance: c = 0.25 m = 0.25x2.5 = 0.625 mm Pinion root circle diameter: dr1= d1- 2b = 40 – 2x3.125 = 33.75 mm Pinion addendum diameter: da1= d1+ 2a = 40+2x2.5 = 45 mm Gear root circle diameter: dr2= d2- 2b = 100 – 2x3.125 = 93.75 mm Pinion addendum diameter: da2= d2+ 2a = 100 + 2x2.5 = 105 mm New Pressure angle when C is increased by 1.5% is : -1
o
Ø = cos ( r 1 cos20 / r 1 1.015) o = 22.21
GEAR CALCULATIONS- QUESTION 2 o
A 21 teeth gear has 20 full depth involute teeth with a module of 12mm . (a) Calculate the radii of pitch circle, base circle and addendum circle (b) Determine the tooth thickness at base circle, pitch circle and addendum circle. Comment on the top land thickness.
r b
r
cos
inv
t
Data: o
Pressure angle : Ø =20 = 0.349 rad. Z1 = 21, m= 12mm. Solution:
r 1 = 0.5.m.Z1 = 0.5 x 12 x 21=126 mm o r b = r 1.CosØ = 126 x cos 20 = 118.4 mm a = 1m = 1 x 12 = 12 mm
tan
t 2r( 1 2 r 1
inv
inv
)
r a = r 1 + a = 126 + 12 = 138 mm p1 = π.m = π. 12 = 37.14 mm t1 = 0.5 p1 = 0.5 x 37.14 = 18.57 mm inv Ø = tan Ø - Ø = tan 0.349 – 0.349 = 0.015 rad At the base circle Ø b= 0, Hence tooth thickness tb at the base circle is: t
b
t 2r ( 1 b 2 r 1
2 x118.4 x(
inv
inv
18.57 2x126
b
)
0.015 0 )
21.0 mm
At the addendum circle the pressure angle : -1 -1 Øa=cos (r b/r a) =cos ( 118.4 /138) =0.54 rad inv Øa = tan 0.54 – 0.54 = 0.059 Hence tooth thickness ta is given by: t
a
2r a
t 1 2 r 1
inv
inv a
Hence tooth thickness ta is:
t
a
2 x 138
x
18.57 2 x 126
0.015
0 .059
8. 20 mm
Minimum recommended top land thickness: tamin = 0.25 m = 0.25 x 12 = 3 mm Since ta (8.2) > tamin (3 ), the tip thickness is permissible.
