TUTORIAL 1
To Study about Spur Gear
(1)
(2)
(3)
(4)
Discuss causes and remedies of gear tooth failure.
Explain standard system of gear tooth and advantage and
disadvantage of 14.50 and 200 involutes system.
Explain the gear material and heat treatment.
A spur pinion having 20 teeth is to mesh with a gear having 43 teeth.
The pinion and gear are to be made of plain carbon steels having
ultimate tensile strength of 600 N/mm2 and 400 N/mm2 respectively.
The pinion is to be driven by a three phase induction motor having a
speed of 1440 r.p.m and 10 kW rating. The starting torque of the
motor is twice the working torque. The tooth system is 200 full-depth
involute. If the surface hardness of the gear pair is to be 400 BHN,
design a gear pair with a factor of safety of 1.5. Assume velocity factor
Kv =
3
(3+𝑉)
accounts for the dynamic load.
Use the following relations:
Y = 0.484 (5)
2.87
𝑍
,
K = 0.16 [
𝐵𝐻𝑁 2
]
100
A spur pinion having 21 teeth to be made of plain carbon steel 55C8
(Sut = 720 N/mm2) is to mesh with a gear to be made of plain carbon
steel 40C8 (Sut = 580 N/mm2). The gear pair is required to transmit 22
kw power from an I.C. engine running at 1000 r.p.m. to a machine
running at 300 r.p.m. The starting torque required is 200 % of the rated
torque, while the load distribution factor is 1.5. The factor of safety
required is 1.5. The face width is ten times the module and the tooth
system is 200 full-depth involute. The gears are to be machined to
meet the specifications of grade 6. The gear and pinion are to be case
hardened to 400 BHN and 450 BHN respectively. The deformation
factor ‘c’ for gear pair is 11500 e, N/mm. design the gear pair by using
the dynamic factor Kv =
6
6+𝑉
and Buckingham’s equation for dynamic
load.
Use the following relations:
Y = 0.484 (6)
2.87
𝑍
,
K = 0.16 [
𝐵𝐻𝑁 2
]
100
For grade 6, e = 8.0 +0.63 (m + 0.25√𝑑) 𝜇𝑚
A spur gear pair made of plain carbon steel 55C8 (σut = 720 N/mm2 and
E = 210 GPa ) is required to transmit power from an electric motor
running at 1440 rpm to a machine running at 370 rpm. The tooth
system is 200 full depth involute and no. of teeth on pinion are as
minimum as possible. The service factor and load concentration factor
are 1.25 and 1.2 respectively. The factor of safety required is 1.25 to
1.5. The face width is twelve times the module. The gears are to be
machined to meet the specifications of grade 7. Design the gear pair
by using the velocity factor Kv =
3
(3+𝑉)
and Buckingham’s equation for
dynamic load. Suggest the case hardness for gear pair. Use the
following relations.
Fd =
(7)
21𝑣 (𝑏𝑐+𝐹𝑚𝑎𝑥)
21𝑣+ √𝑏𝑐+𝐹𝑚𝑎𝑥
C = 0.111e [
𝐸𝑝.𝐸𝑔
𝐸𝑝+𝐸𝑔
],
For grade 7 e = 11 + 0.9(m + 0.25√𝑑)
A spur gear having 22 teeth to be made of plain carbon steel 40C8 (Sut
= 580 N/mm2) is to be mesh with a gear having 88 teeth to be made of
grey cast iron FG260 (Sut = 260 N/mm2). The pinion shaft is connected
to 12kw, 1440 rpm electric motor. The starting torque of the motor is
approximately twice the rated torque. The tooth system is 200 full
depth involute. The face width is 10 times module for which the load
distribution factor is 1.4. The gear are to be machined to meet the
specifications of grade 7 for which deformation factor is 240 N/mm.
(i)If factor of safety require against bending failure 1.0, design the gear
pair by using velocity factor =
dynamic load.
6
(6+𝑉)
and Buckingham’s equation for
(ii)If the factor of safety required against pitting failure is 1.5, specify
surface hardness.
Y = 0.484 -
2.87
𝑍
Buckingham’s equation, Fd = Ft +
K = 0.18
(8)
𝐵𝐻𝑁 2
[
]
100
21𝑣+ √𝑏𝑐+𝐹𝑚𝑎𝑥
For steel pinion and cast iron
Standard module are 4,5,6,8,10,12,16
Service factor = 2, Load Concentration factor = 1.4
Design a spur gear pair from the following given data :
Power to be transmitted = 22.5 kW, Pinion speed = 1450 rpm
Speed reduction = 2.5, No. of teeth on pinion = 20
Service factor = 1.5, Face width b = 10m
Pitch line velocity = 5 m/sec (for initial calculation of module)
Maximum permissible error in gear tooth failure = 0.025mm
K = A factor depending upon the form of teeth = 0.111
Velocity factor =
(9)
21𝑣 (𝑏𝑐+𝐹𝑚𝑎𝑥)
3
(3+𝑉)
, where V is the pitch line velocity in m/s
Take endurance surface hardness = 600 Mpa
Lewis form factor = 0.154 – 0.912/No. of teeth for 200 pressure angle
involute tooth system
The material and stresses are as under:
Material
[ 𝜎b ]
Elasticity
Hardness
modulus
Pinion (Fe 410) 135 N/mm2
2.1
× 105 260 BHN
N/mm2
Gear (FG 200) 65 N/mm2
1.1
× 105 250 BHN
N/mm2
Design a spur gear pair to transmit 15 kw power from the following
specifications.
Tooth system = 200 pressure angle full depth involute
Number of teeth on pinion = 25, Speed reduction ratio = 3: 1
Service factor = 1.25, Material of pinion and gear = FG 200
Design bending stress of material = 60 MPa
Surface hardness of pinion and gear = 200 BHN
Endurance strength of the material = 84 MPa
Dynamic load factor = 178 N/mm
Modulus of elasticity = 1.1 × 105 MPa
Assume pitch line velocity as 7.5 m/sec for module calculation.
Velocity factor, Cv =
3
(3+𝑉)
,
Wear load, Fw = dp.f.kw.Q
Lewis form factor, Y = 0.154 -
0.912
𝑍
for 200 pressure angle full depth
involute tooth system
Dynamic load equation, Fd = Ft +
21𝑣 (𝑏𝑐+𝐹𝑚𝑎𝑥)
21𝑣+ √𝑏𝑐+𝐹𝑚𝑎𝑥
(10) A pair of mating carefully cut spur gears has 200 full depth of 4 mm
module. The number of teeth on pinion and gears are 38 and 115
respectively. The face width is 40 mm. if the pinion and gear are made
of steel with fb static = 233 MPa and surface hardness of 300 BHN.
Calculate the safe power that can be transmitted when the pinion is
run at 1200 rpm.