Gear Design
50T
4
Problem: Pinion 2 is shown in the figure
3
Output
(gear)
runs at 1750 rpm and transmits 2.5 kW to
60T
idler gear 3. The teeth are cut on the 200
20T
full-depth system and have a module of
2
Input
(pinion)
m = 2.5 mm.
a) Draw a free-body diagram of gear 3 and show all the forces and compute the torques on
the output gear 4
dp
T = Wt ∗
Wt : tangential force
2
dp : pitch diameter
H = T *ω
H = Wt *
H: Power
ω : angular rotation
d p ⎛ 2π
⎞
*⎜
*ω2 ⎟
2 ⎝ 60
⎠
ω (rev/min) =
H =
ω
(rev/sec)
60
d pπ * ω
= 2π
*W
2
t
60
(rad/sec)
where,
3
60 * 10
ω
H : (kN)
ω 2 : (rev/min)
dp : (mm)
Wt : (kN)
W
t
=
3
60 * 10
πd
p
* H
*ω
dp= N*m:
2
N: number of teeth
m: module (mm)
W
t
=
60 * 10
3
* H
π (N * m )*ω
2
1
W
t
=
60 * 10
3
* ( 2 .5 )
= 0.546 kN
π ( 20 * 2 . 5 ) * 1750
Thus, the tangential force of gear 2 on gear 3 is 0.546 kN. The radial force is;
W
r
= W t * tan 20
0
= 0 . 546 * tan 20
= 0 . 199 kN
0
The tangential reaction of gear 4 on gear 3 is also equal to Wt
Wt
Gear
3
Fx
Fy
Wr
Wt
Wr
Forces on shaft of gear 3;
FX = - ( - 0.546 + 0.199) = 0.347 kN
FY = - ( 0.199-0.546 ) = 0.347 kN
The output Torque:
ω 4 r2 N 2
= =
ω 2 r4 N 4
N
N
2
= (−
4
N
N
2
3
) * (−
N
N
T2 = Wt * r2
T4 = Wt * r4
2
3
4
) =
20
60
T4 =
r4
N
* T2 = 4 * T2
r2
N2
T4 =
60
50
* T2 = 3 * (0.546 * )
20
2
T2 =13.65 Nm
T4 =40.95 Nm
b) Compute the mean and alternating loads on 1) pinion, 2) gear and, 3) idler.
The repeated loads on pinion and gear:
R=0=
σ min
σ max
(since σ min = 0 )
Wt 0.546
=
= 0.273 kN
2
2
W 0.546
Wtalternating = t =
= 0.273 kN
2
2
Wtmean =
The fully-reversed loads on the idler:
σ min
σ max
= σ max
R = -1 =
σ min
3
Wtalternating = Wt = 0.546 kN
Wtmean = 0 kN
c) Determine a suitable face width and the bending stresses in the idler gear (“3”)
σb =
Wt * Pd K a * K m
*
* KS * KB * KI
F*J
Kv
Recommended face-width range is
δ
Pd
<F<
Take
16
Pd
F=
Pd = diametral pitch = 25.4/m
12
Pd
8m
16m
<F<
25.4
25.4
( SI units )
The relation between Pd (U.S. specification) and the module “m” (S.I. unit) is
25.4
= 2.5 mm/teeth
Pd
Hence, Pd = 10.16 teeth/inches.
m=
The face width (F):
12
F=
= 1.81 inches = 3 cm = 30 mm
10.16
The load distribution factor (Km)
4
Using Table 11 – 16
Km=1.6
(for F=1.81 inches = 30 mm)
The application factor:
(assume uniform loading; table 11 -17)
Ka = 1
The velocity factor:
⎛
A
KV = ⎜
⎜ A + 200 * V
t
⎝
⎞
⎟
⎟
⎠
B
Vt : pitch-line velocity
dp
d
Vt =
* ω = 3 * ω3
2
2
Gear 3:
d3 = m N3 = 2.5*50 = 125 mm
N
ω3 = 2 * ω 2
N3
20
ω 3 = * 1750
50
rev
ω 3 = 700rpm *
min
2π
ω 3 = 700 *
= 73.266 rad/sec
60
Vt = 4.58 m/s = 274.76 m/min
Note that Vt ( ft / min) = Vt (m / sec) *196.85
Vt = 901.5 ft/min
since 1 ft = 12 inches = 30.48 cm
The quality index (Qv)
Using Table 11-7 (page 710)
Vt
Qv
0- 800
6-8
901.5 ft/min ?
800-2000
8-10
Take Qv = 8
(12 − 8) 2 / 3
= 0.63
4
A = 50 +56(1-B) =7 0.72
B=
5
⎞
⎛
70.72
KV = ⎜⎜
⎟⎟
⎝ 70.72 + 200 * 4.58 ⎠
0.63
= 0.8
The size factor KS=1 for all gears
The rim thickness factor KB= 1
(assume no rim)
The idler factor KI = 1 for gears “2” and “4”, KI = 1.42 for the gear “3”.
The bending geometry factor J for 200 presume angle (full depth teeth with HPSTC
loading)
pinion : 20 T (200)
gear : 60 T (200)
idler : 50 T (200)
First, calculate the Jpinion, and Jgear based on pinion (gear “2” in the figure) and idler (gear “3”
in the figure) interactions:
idler
(gear)
50T
20T
pinion
Using Table 11- 9 (p.716)
Jpinion ≅ 0.34
Jgear(idler) ≅ 0.39
σb =
Wt * Pd K a * K m
*
* KS * KB * KI
F*J
Kv
since F=
σb =
12
Pd
Wt * Pd2 K a * K m
*
* KS * KB * KI
12 J
Kv
also,
Pd =
25.4
m
6
2
⎛ 25.4 ⎞
Wt * ⎜
⎟
⎝ m ⎠ * Ka * Km * K * K * K
σb =
S
B
I
12 J
Kv
σb =
Wt * (53.76) K a * K m
*
* KS * KB * KI
Kv
m2 * J
m = 2.5 mm
then
σ b is in MPa
Wt = 0.546 kN
σb
pinion
σb
σb
gear
0.546 * (53.76) 1 * 1.6
*1 *1 *1
*
0. 8
( 2.5) 2 * (0.34)
= 27.63
pinion
pinion
σb
=
=
MPa
0.546 * (53.76) 1 * 1.6
*
* 1 * 1 * 1.42
( 2.5) 2 * (0.39)
0 .8
= 34.2
MPa
Second, calculate the Jpinion, and Jgear based on idler (“3”) and gear (“4”) interactions:
60T
Using Table 11- 9 (p.716)
50T
Jpinion(idler) ≅ 0.42
Jgear ≅ 0.43
gear
7
idler
Note: The idler has a slightly different J factor when considered to be the “gear” in mesh with
the smaller pinion (0.42) than when considered to be the “pinion” in mesh with the larger gear
(0.43).
σb
=
idler
0.546 * (53.76) 1 * 1.6
*
* 1 * 1 * 1.42
2
0.8
(2.5) * (0.42)
= 31.75 MPa
σb
gear
=
0.546 * (53.76) 1 * 1.6
*
*1 *1 *1
2
0 .8
(2.5) * (0.43)
= 21. 84 MPa
d) All gears are made of steel (AGMA grade 2 with Brinell Hardness number of 250 HB).
The service life required is 5 years of one-shift operation. Operating temperature is 2000 F.
Calculate the safety factor for bending stress (Assume 99 % reliability).
Sf b=
KL
* S ′f b
KT * K R
S ′f b = 280 MPa
(from the Figure 11-25, pp.733 for HB = 250)
ω piniom =1750 rpm
ω idler =700 rpm
ω gear =583.3 rpm
60 min
2080hr
*
* 5 yr
hr
shift − year
=1.092*109 cycles
N pinion = 1750rpm
N pinion
Nidler = 0.4368*109 cycles
8
Ngear = 0.364*109 cycles
The life factor:
KL(pinion)= 1.3558*N-0.0178
(Fig. 11-24, N>108 commercial app)
= 0.936
KL(idler) = 0.951
KL(gear) = 0.955
The temperature factor:
KT=1
(T<2500)
The reliability factor:
KR=1
(%99 reliability)
262
= 9.48
27.63
S f b pinion = 262 MPa
N b pinion =
S f b idler = 266 MPa
N b idler =
266
= 7.77
34.2
S f b gear = 267.4 MPa
N b gear =
267.4
= 12.24
21.84
Hence, idler is the most critical (it is exposed to fully reversed stresses).
e) Determine the surface stresses for
1) pinion- idler mesh
2) idler-gear mesh
and the safety factor for surface stresses.
σ c = cp
ωt
Ca * Cm * Cs * C f
F *I *d
Cv
Ca = Ka = 1
Cm = Km = 1.6
9
Cv = Kv = 0.8
F = 30 mm
Notes:
• d (pitch diameter) is the smaller of mating gears
• “I” is the smaller of mating gears
• Be careful with the +/- signs in equations 11.22a and 11.22b.
Eq. 11.22a (+ : external, -: internal)
Eq. 11.22b (- : external, +: internal)
Cs= 1 (size factor)
Cf = 1 (surface finish factor)
The elastic coefficient:
Cp =
1
⎡ 1− v
π ⎢(
2
p
⎢⎣ E p
1 − v g2 ⎤
)+(
)⎥
E g ⎥⎦
v =0.3
All gears are made of steel
(Table 11-18 assumes that v = 0.3)
Ep = 30*106 psi = 2*105 MPa
Instead of calculating the Cp using the equation; you can directly get the value from the table if
the material is available.
UNIT
pinion
steel
gear
Cp=2300
psi
Cp=191
MPa
steel
Surface Geometry Factor (I)
I=
cos φ
⎛ 1
1 ⎞⎟
⎜
±
*d
⎜ρ
⎟ p
⎝ p ρg ⎠
ρ p = (rp +
1+ xp
pd
use “+” sign for external gears and “-“ for internal gears
) 2 − (rp * cos φ ) 2 −
ρ g = c sin φ m ρ p
π
pd
* cos φ
-: external gears
10
+ : internal gears
pinion-idler mesh
rp=
dp
2
ridler=
=
m * N pinion
2
=
2.5 * 20
= 25 mm
2
d idler 2.5 * 50 125
=
=
mm
2
2
2
φ = 20 0
m=
25.4
pd
Pd =
25.4
= 10.16
m
The pinion addendum coefficient xp= 0
(xp= 0 for full-depth teeth and xp= 0.25 for 25 % long-addendum teeth)
ρ p = (25 +
1
10.16
) 2 − (25 * cos 20 0 ) 2 −
π
10.16
ρ p = 8.54 mm
ρ gear = c sin φ − ρ1 = (25 +
ρ gear =21.39 mm
I=
cos φ
⎛ 1
1 ⎞⎟
⎜
*d
+
⎟ p
⎜ρ
⎝ p ρg ⎠
125
) * sin 20 0 − 8.54
2
= 0.115
Idler –gear mesh
idler
125
rp=
m
2
150
rp=
mm
2
φ = 20 0
11
* cos 20 0
ρ p = (62.5 +
1
10.16
) 2 − (62.5 * cos 20 0 ) 2 −
π
10.16
* cos 20 0
ρ p =21.37 mm
ρ g = c sin φ m ρ p = (62.5 + 75) * sin 20 0 − 21.37
ρ g = 25.7 mm
Iidler(pinion) =
cos φ
= 0.175
1 ⎞
⎛ 1
+
⎜
⎟ * 62.5
⎝ 21.37 25.7 ⎠
The surface stress for pinion-idler mesh:
σ cp = c p
ωt
Ca * Cm * C s * C f
F * I pinion * d p
Cv
0.546 *10 3 N 1 * 1.6 * 1 *1
30 * 0.115 * 50
0.8
σ cp = 480.56 MPa
= 191 *
The surface stress for idler-gear mesh:
σ cidler = 191 *
0.546 * 10 3 1 *1.6 *1 *1
30 * 0.175 * 62.5
0.8
σ cidler = 348.4 MPa
Sfc=
CL * CH
* S'f c
CT * C R
S ' f c ≅ 820MPa
(for HB= 250 and Grade II ) Table 11-27, page 737
CT=1
(T< 2500 F)
CR=1
(%99 reliability)
CH=1
(Hardness Ratio factor)
12
C L pinion = 1.4488 N −0.023 = 1.4488(1.092 *10 9 ) −0.023
(Fig. 11-26, pp.734)
=0.898
C L idler = 0.917
C L gear = 0.92
= 820 * 0.898 = 736MPa
Sfc
pinion
Sfc
idler
Sfc
gear
=752 MPa
= 754 MPa
2
N c pinion −idler
⎡ 736 ⎤
=⎢
= 2.4
⎣ 480.56 ⎥⎦
2
N c idler − gear
⎡ 752 ⎤
=⎢
= 4.7
⎣ 348.4 ⎥⎦
Hence; pinion-idler surface is more critical.
Note: we take the square of the safety factor because the surface stress is related to the
square root of the load.
13