This problem involves designing a gear drive system to meet specific power, speed, and ratio requirements. 1. The key specifications are: 15 kW power at 1200 rpm driving a compressor at 300 rpm, with a gear ratio of 4:1. The shafts are 400mm apart. The pinion is forged steel with 210 MPa allowable stress, and the gear is cast steel with 140 MPa stress. 2. A two-stage gear train layout is proposed to achieve a 9:1 ratio from an input of 960 rpm to transmit 2 kW power. The shafts are 200mm apart with coaxial input/output. 3. The solution involves calculating the module, pitch diameter, number

1. SPUR GEAR
Problem and Solution
Dodi Mulyadi, MT

2. Problem and Solution 1
Problem:
A single reduction spur gear are given:
Gear ratio = 10 : 1;
Distance between centres = 660 mm approximately;
Pinion transmits 500 kW at 1800 r.p.m.;
Involute teeth of standard proportions (addendum = m) with
pressure angle of 22.5°;
Permissible normal pressure between teeth = 175 N per mm
of width.
Find :
1. The nearest standard module if no interference is to occur;
2. The number of teeth on each wheel;
3. The necessary width of the pinion; and
4. The load on the bearings of the wheels due to power
transmitted.

3. Problem and Solution 1
Solution:
 Minimum number of teeth on the pinion in order to avoid
interference,
















1sin2
11
1
2
2

GG
G
A
T w
P
 Given :
G = TG / TP = DG / DP = 10
L = 660 mm
P = 500 kW = 500 × 103 W
NP = 1800 r.p.m.
φ = 22.5°
WN = 175 N/mm width

4. Problem and Solution 1
 Minimum number of teeth on the pinion and on gear
















15.22sin2
10
1
10
1
110
12
2
(min)
x
TP
143.13(min) PT
PG GxTT (min)
1401410(min)  xTG

5. Problem and Solution 1
 Standard value of the module
(min)P
P
T
D
m 
22
PG DD
L 
2
10
2
PG DD
L 
 We know that
PDL 5.5
5.5
L
DP 
mmDP 120
5.5
660

57.8
14
120
m
 Since the nearest
standard value of
the module is 8 mm,
therefore we shall
take
m = 8 mm

6. Problem and Solution 1
 Actual number of teeth
m
D
T P
P 
8
120
PT
xGTT PG 
 Number of teeth on the
pinion
 Number of teeth on the
gear
1015xTG 
15PT
150GT

7. Problem and Solution 1
 Width of the pinion
 Tangential load (WT)
xCs
v
P
WT 
1
60
1800)12.0(
10500 3
x
x
x
WT


(Assume Cs = 1)
44209TW N
cos
T
N
W
W 
0
5.22cos
44209
NW
47851NW N

8. Problem and Solution 1
 Since the normal pressure
between teeth is 175 N
p e r m m o f w i d t h ,
therefore necessary width
of the pinion,
175
NW
b 
175
47851
b
b = 273.4 mm
 Load on the bearings of
the wheels
sin.NR WW 
0
5.22sin47851 xWR 
18312RW N

9.  PROBLEM: A gear drive is required to transmit a maximum
power of 22.5 kW. The velocity ratio is 2:1 and r.p.m. of the
pinion is 200. The approximate centre distance between the
shafts may be taken as 600 mm. The teeth has 20° stub
involute profiles. The static stress for the gear material (which
is cast iron) may be taken as 60 MPa and face width as 10 times
the module. Assuming steady load conditions and 8–10 hours
of service per day.
 Find
 module, face width, number of teeth on each gear.
 Check the design for dynamic and wear loads. The deformation
or dynamic factor in the Buckingham equation may be taken as
80 and the material combination factor for the wear as 1.4.

10.  Solution:
Given :
P = 22.5 kW = 22 500 W
V.R. = DG/DP = 2
NP = 200 r.p.m.
L = 600 mm ;
σOP = σOG = 60 MPa = 60 N/mm2
b = 10 m
C = 80
K = 1.4
Cs = 1 (refer table)

11. 1. Calculate design tangential tooth load
xCs
v
P
WT 
60
, PP ND
vwhere


we know length,
22
GP DD
L 
where, 2
P
G
D
D
PG DD 2
2
2
2
PP DD
L 
PD
2
3
600 
400PD mm
60
)200)(4.0(
,

vthen
m/s2.4v
1
2.4
22500
xWT 
5357TW N

12. 2. Apply the Lewis equation:
PVOPT ymbCW ..... 
Check Velocity factor, Cv
Since v is less than 12 m / s,
therefore velocity factor,
v
CV


3
3
2.43
3

VC
417.0VC
Check the pinion tooth factor for
20° stub involute profiles,
P
P
T
y
841.0
175.0 
Check TP ==>
mm
D
T P
P
400

So....
400
841.0
175.0
m
yP 
myP 0021.0175.0 

13. 2. Apply the Lewis equation:
PVOPT ymbCW ..... 
5357 = 60 x 0.417 x 10m x π x m x (0.175 - 0.0021m)
5357 = 137.5m2 - 1.65m3
Solving this equation by hit and trial method, we find that m = 6.5
 Take the nearest recommended module, m = 8 mm
 Face width, b = 10m = 10 × 8 = 80 mm
 Number of teeth on the pinion, TP = DP / m = 400 / 8 = 50
 Number of teeth on the gear, TG = DG / m = 800 / 8 = 100

14. 3. Calculate the dynamic load (WD):
T
T
TD
WCbv
WCbv
WW



.21
).(21
535780802.421
)53578080(2.421
5357



xx
xx
WD
10631DW N

15. 4. Find the static tooth load
WS = σe.b.π m.yP
Flexural endurance limit (σe) for cast iron is 84 MPa
or 84 N/mm2
WS = 84 x 80 x π x 8 x (0.175- 0.0021 x 8)
WS = 26705 N
WS > WD ==> OK!

16. 5. Find the wear tooth load
WW = DP.b.Q.K
Check the ratio factor
1
2


VR
xVR
Q
12
22


x
Q 33.1Q==> ==>
WW = 400 x 80 x 1.33 x 1.4
WW = 59584 N
Since both WS and WW are greater than WD, therefore the design is
safe.
WW > WD ==> OK!

17.  PROBLEM:
A bronze spur pinion rotating at 600 r.p.m. drives a cast iron
spur gear at a transmission ratio of 4 : 1. The allowable static
stresses for the bronze pinion and cast iron gear are 84 MPa
and 105 MPa respectively.
The pinion has 16 standard 20° full depth involute teeth of
module 8 mm. The face width of both the gears is 90 mm.
 Find the power that can be transmitted from the standpoint
of strength.

18.  SOLUTION:
Given:
NP = 600 r.p.m.
V.R. = TG / TP = 4
σOP = 84 MPa = 84 N / mm2
σOG = 105 MPa = 105 N/mm2
TP = 16
m = 8 mm
b = 90 mm
Cs = 1 (assume)

19. 1. Calculate design tangential tooth load
xCs
v
P
WT 
60
,Check PP ND
v


we know,
DP = m x TP
DP = 8 x 16
DP = 128 mm
60
600821.0
,Then
xx
v


20.4v m/s
S
T
C
vW
P
.
so, 

20. 2. Apply the Lewis equation:
ymbCW VOT ..... 
Check Velocity factor, Cv
Since v is less than 12.5 m / s,
therefore velocity factor,
v
CV


3
3
02.43
3

VC
724.0VC
Check the pinion tooth factor for
20° full depth involute teeth,
P
P
T
y
291.0
154.0 
16
291.0
154.0 Py
097.0Py
Tangential load on the pinion
PVOPPT ymbCW .....)( 
097.0890724.084)( xxxxxW PT 
7870)( PTW N

21. 2. Apply the Lewis equation: ymbCW VOT ..... 
Check the gear tooth factor for
20° full depth involute teeth,
G
G
T
y
291.0
154.0 
P
G
TVR
y
.
291.0
154.0 
164
291.0
154.0
x
yG 
14.0Gy
Tangential load on the gear:
GVOGGT ymbCW .....)( 
14.0890724.0105)( xxxxxW GT 
14198)( GTW N
Since WT(P) is less than WT(G)
therefore the pinion is weaker.

22. 3. Power that can be transmitted
S
T
C
vW
P
.

1
20.47870 x
P 
31637P W

23. 1. A 15 kW and 1200 r.p.m. motor drives a compressor at 300
r.p.m. through a pair of spur gears having 20° stub teeth. The
centre to centre distance between the shafts is 400 mm. The
motor pinion is made of forged steel having an allowable
static stress as 210 MPa, while the gear is made of cast steel
having allowable static stress as 140 MPa. Assuming that the
drive operates 8 to 10 hours per day under light shock
conditions.
Find from the standpoint of strength: module; face width and
number of teeth and pitch circle diameter of each gear.
Check the gears thus designed from the consideration of wear.
The surface endurance limit may be taken as 700 MPa.

24. 2. A two stage reduction drive is to be designed to transmit 2
kW; the input speed being 960 r.p.m. and overall reduction
ratio being 9. The drive consists of straight tooth spur gears
only, the shafts being spaced 200 mm apart, the input and
output shafts being co-axial.
a) Draw a layout of a suitable system to meet the above
specifications, indicating the speeds of all rotating
components.
b) Calculate the module, pitch diameter, number of teeth,
blank diameter and face width of the gears for medium
heavy duty conditions, the gears being of medium grades
of accuracy.
c) Draw to scale one of the gears and specify on the drawing
the calculated dimensions and other data complete in
every respect for manufacturing purposes.

25. Thank you