Numerical problems on spur gear (type ii), Based on Expression for interference

1. Department of Mechanical Engineering
JSS Academy of Technical Education, Bangalore-560060
Kinematics of Machines
(Course Code:18ME44)

2. Kinematics of Machines
Module 5: Numerical problems on spur gears
(type-II)

3. 1. Two mating gears have 20 and 40 involute teeth of module 10 mm and 20° pressure angle. The addendum on
each wheel is to be made of such a length that the line of contact on each side of the pitch point has half the
maximum possible length. Determine the addendum height for each gear wheel, length of the path of contact, arc
of contact and contact ratio.
Solution
Given:
• No. of teeth on pinion = z1= 20 teeth
• No. of teeth on gear = z2 = 40 teeth
• Pressure angle = = 20°
• Module = m = 10 mm
• Path of contact =? Arc of contact =?
• Contact ratio =?
• Path of approach & recess = ½ max. possible length
• Addendum for each gear = a=?
w.k.t, pitch circle radius of pinion, r1 =
𝑚𝑧1
2
=
10 𝑋 20
2
= 100 mm
pitch circle radius of gear r2 =
𝑚𝑧2
2
=
10 𝑋 40
2
= 200 mm
Given, Path of approach = ½ max. possible length
Path of approach = = ½ x r1 sin 
ra2 = 206.5 mm
Addendum height for larger gear wheel.
wkt, radius of addendum circle of gear, ra2 = r2 + a
a = ra2 - r2 = 206.5 – 200 = 6.5 mm
Given, Path of recess = ½ max. possible length
Path of recess = = ½ x r2 sin 
ra1 = 116.2 mm
Addendum height for smaller gear.
wkt, radius of addendum circle of smaller gear, ra1 = r1 + a
a = ra1- r1 = 116.2 – 100 = 6.2 mm

4. wkt, Length of path of contact = ½ max. possible length on either side of the pitch point.
path of contact = ½ max. possible length + ½ max possible length
= ½ x r1 sin  + ½ x r2 sin 
= ½ x 100 sin 20 + ½ x 200 sin 20
 Path of contact = 51.3 mm
Length of the path of contact
Length of the arc of contact
Contact ratio = Arc of contact / p
=
54.6
31.42
= 1.74
P=  m =  × 10 = 31.42 mm

5. 2. Two 20° pressure angle involute gears in mesh have a module of 10 mm. The addendum is one module. The large gear has
50 teeth and the pinion 13 teeth.
i) Does interference occur ?
ii) If it occur, to what value should the pressure angle be changed to eliminate the interference
Given:
• No. of teeth on pinion = z1= 13 teeth
• No. of teeth on gear = z2 = 50 teeth
• Pressure angle = = 20°
• Module = m = 10 mm
• Addendum = a = 1m = 1x10 = 10mm
• Find new pressure angle  =? To avoid interference.
Solution
w.k.t, pitch circle radius of pinion, r1 =
𝑚𝑧1
2
=
10 𝑋 13
2
= 65 mm
pitch circle radius of gear r2 =
𝑚𝑧2
2
=
10 𝑋 50
2
= 250 mm
radius of addendum circle of gear, ra2 = r2 + a = 250+10=260 mm
radius of addendum circle of pinion, ra1 = r 1+ a = 65+10=75 mm
Maximum addendum circle radius of gear to avoid interference
(ra2 )max = 258.44 mm and actual ra2 = 260 mm
actual radius ra2 exceeds (ra2 )max  interference exists
To calculate the new value of  to avoid interference
260 = √(250^2 + 65^2 sin^2  + 2 𝑋 65 𝑋 250 sin ^2)

6. 3. Two gear wheels mesh externally and are to give a velocity ratio of 3 to 1. The teeth are of involute form ; module = 6
mm, addendum = 1 m, pressure angle = 20; the pinion rotates at 90 rpm.
Find : i) No. of teeth on pinion to avoid interference on it and the corresponding number on the wheel.
ii) The length of path and arc of contact.
iii) The number of pairs of teeth in contact.
iv) The maximum velocity of sliding.
Given:
• Velocity ratio = i = 3
• Module = m = 6 mm
• Addendum = a = 1m = 1x6 = 6 mm
• Pressure angle = = 20°
• Speed of pinion = 1 = 90 rpm
• Addendum coefficient = ag =1
• No. of teeth on pinion = z1= ?
• No. of teeth on gear = z2 = ?
• Path of contact, arc of contact, Contact ratio =?
• Max. velocity of sliding = (Vs ) Max.?
Solution
Minimum number of teeth to avoid interference
Z2 = = 44.9  45 teeth
Gear ratio = i =
𝑧2
𝑧1
; 3 =
45
𝑧1
 z1 = 15 teeth

7. Path of contact = path of approach + path of recess
Path of approach =
Path of recess =
w.k.t, pitch circle radius of pinion, r1 =
𝑚𝑧1
2
=
6 𝑋 15
2
= 45 mm
pitch circle radius of gear r2 =
𝑚𝑧2
2
=
6 𝑋 45
2
= 135 mm
Radius of addendum circle of pinion, ra1 = r1 + a = 45+6 = 61 mm
Radius of addendum circle of gear, ra2 = r2 + a = 135+6 = 141 mm
Path of approach = = 15.37 mm
Path of recess = = 13.12 mm
Path of contact = path of approach + path of recess
Path of contact = 15.37 + 13.12 = 28.49 mm
Max. Velocity of sliding
Since the path of approach is greater than path of recess
Velocity of sliding
vS = (ω1 + ω2) x Path of approach
Gear ratio = i =
1
2
; 3 =
9.42
2
 2 = 3.14 rad/s
vS = (ω1 + ω2) x Path of approach
= (9.42 + 3.14 ) x 15.37 = 193.18 mm/s

8. End of Module