# ME 3031 DESIGN OF MACHINE ELEMENTS SAMPLE QUESTIONS & ANSWERS

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ME 3031
DESIGN OF MACHINE ELEMENTS
SAMPLE QUESTIONS & ANSWERS
B.Tech Year I
(SECOND SEMESTER)
2
ME 3031 Sample Problems and Solutions
Spur Gear
1.
A pair of mating spur gears have 14 ½ ° full depth teeth of 10 module. The pitch
diameter of the smaller gear is 160mm. If the transmission ratio is 3 to 2. Calculate (a)
number of teeth of each gear, (b) addendum, (c) whole depth, (d) clearance, (e) outside
diameters, (f) root diameters, (g) dedendum.
Solution:
(a) Np = Dp/m = 160/10 = 16
Ng = Dg/m = 160 x 1.5/10 = 24
(b) add = m = 10
(c) whole depth = 2.157 x m = 21.57mm
(d) clearance = 0.157 x m = 1.57mm
(e) outside diameters,
Dop = Dp + 2add = 160 + 2 x 10 = 180mm
Dog = Dg + 2add = 240 + 2 x 10 = 260mm
(f) root diameters,
Dip = Do – 2 whole depth = 160 – 2x21.57 = 116.86mm
Dig = Dg – 2 whole depth = 260 – 2x21.57 = 216.86mm
(g) ded = 1.157 x m = 11.57mm
2.
A spur pinion of cast steel (so = 140MN/m2) is to drive a spur gear of cast iron (so =
55MN/m2). The transmission ratio is to be 2-1/3 to 1.The diameter of the pinion is to be
105mm and 20kW will be transmitted at 900rev/min of the pinion. The teeth are to be 20 deg
full depth involute form. Design for the greatest number of teeth. Determine the necessary
module and face width of the gears for strength requirement only.
Solution:
sop = 140MN/m2 , sog = 55MN/m2 , rpmp = 900, V.R = 7/3, Dp = 105mm, 20kW
Dg = 105 x 7/3 = 245mm, rpmg = 900/(7/3) = 385.714
Different material
∴check load carrying capacity.(soy)
assume Np = 15 ⇒ yp = 0.092
Ng = 15 x 7/3 = 35 ⇒ yg = 0.119
(so y)p = 140 x 0.092 = 12.88 MN/m2
(so y)g = 55 x 0.119 = 6.545 MN/m2 < (so y)p.
∴ Gear is weaker. Base design on gear.
3
Known diameter case.
2
⎛ 1 ⎞
⎜
⎟ = skπ
⎜ m2y ⎟
Ft
⎠ all
⎝
πx 245x10 −3 x 385.714
= 4.948m / s < 10m / s
60
3
3
s = so x
= 55x10 6 x
= 20.76 x10 6 N / m 2
3+ V
3 + 4.948
9550xkW
9550x 20
k = 4, Ft =
=
= 4042.328 N
rpmxD / 2 385.714 x 245x10 −3 / 2
V=
6
2
⎛ 1 ⎞
⎜
⎟ = 20.76 x10 x 4 xπ = 202747.513
⎜ m2y ⎟
4042.328
⎝
⎠ all
assume y = 0.1
m = 7mm
std module series ………, 6,7,8, ….
Try m =7
Ng = Dg/m = 245/7 = 35 ⇒yg = 0.119
(1/m2y)ind = 1/(7x10-3)2 0.119 = 171497.17 < (1/m2y)all
∴Design is satisfied. Decrease module.
A module of 6 cannot be used due to the required velocity ratio.
Try m = 5
Ng = Dg/m = 245/5 = 49 ⇒yg = 0.12942
(1/m2y)ind = 1/(5x10-3)2 0.12942 = 309071.24 > (1/m2y)all
∴Design is not satisfied.
∴Take smallest module = 7mm.
Reduce k value.
kred = kmax x (1/m2y)ind/(1/m2y)all = 4 x 171497.17/202747.513 = 3.383
Face width, b
b = kred x π x m = 3.383 x π x 7 = 74.4mm
Use b = 75mm
4
3.
A cast steel pinion (so = 103MN/m2) rotating at 900rev/min is to drive a cast iron gear
(so = 55MN/m2) at 144rev/min. The teeth are to have 20 deg stub involute profiles and the
maximum power to be transmitted is 25kW. Determine the proper module, number of teeth,
and the face width for these gears from the stand point of strength, dynamic load and wear.
Pinion is surface hardened to BHN 250.
Solution:
sop = 103MN/m2 , sog = 55MN/m2 , rpmp = 900, rpmg = 144, 25kW, V.R = 6.25
Different material
∴check load carrying capacity.(soy)
assume Np = 15, Ng ≠ whole number.
Np = 16 ⇒ yp = 0.115
Ng = 100 ⇒ yg = 0.161
(so y)p = 103 x 0.115 = 11.845 MN/m2
(so y)g = 55 x 0.161 = 8.855 MN/m2 < (so y)p.
∴ Gear is weaker. Base design on gear.
Unknown diameter case.
s ind =
2M t
kyπ 2 Nm 3
9550 xkW 9550 x 25
=
= 1657.986 Nm
Mt =
rpm
144
s ind =
2x1657.986
4x 0.161xπ 2 x100xm 3
=
5.217
m3
assume V.F = ½
55x10 6 5.217
=
⇒ m = 5.75mm
2
m3
std module series …..,5, 5.5, 6, 7,…….
Try m = 6
Dg = Ng m = 100 x 6 = 600mm
V = (π x 600 x 10-3 x 144)/60 = 4.524m/s < 10m/s
s all = 55x10 6 x
3
= 21.93x10 6 psi
3 + 4.524
5.217
= 24.153x10 6 psi > s all
s ind =
(6 x10
−3 3
)
∴Design is not satisfied. Increase m.
5
Try m = 7
Dg = Ng m = 100 x 7 = 700mm
V = (π x 700 x 10-3 x 144)/60 = 5.278m/s < 10m/s
3
= 19.93x10 6 psi
3 + 5.278
s all = 55x10 6 x
s ind =
5.217
(7 x10
−3 3
)
= 15.21x10 6 psi > s all
∴Design is satisfied. Take smallest module = 7
Reduce k
kred =kmax x sind / sall = 4 x 15.21/19.93 = 3.052
Face width, b
b = kred x π x m = 3.052 x π x 7 = 67.13mm
Use b = 68mm
Dynamic check
Endurance force, Fo
Fo = so b y π m = 55 x 106 x 68 x 10-3 x 0.161 x π x 7 x 10-3 = 13.242kN
Wear force, Fw
F w = Dp b K Q
Dp = Np m = 16 x 7 = 112
Q=
2N g
Np + Ng
=
2 x100
= 1.724
16 + 100
K = 1310 (Table 1.1)
Fw = 112x10-3x68x10-3x1310x103x1.724 = 17.2 kN
Dynamic force, Fd
Fd = Ft +
21V(bC + Ft )
21V + bC + Ft
Ft = Mt/(D/2) = 1657.986/350x10-3 = 4737.103N
V = 5.278m/s ⇒ ep = 0.07 (Fig 3.5), m =7, we should use carefully cut gear (Fig 1.6)
m = 7, carefully cut ⇒ e = 0.035
steel pinion & C.I gear, φ = 20 deg stub, e = 0.035 ⇒ C = 283.5kN/m
6
Fd = 4737.103 +
21x 5.278(68x 283.5 + 4737.103)
21x 5.278 + 68x 283.5 + 4737.103
= 14.75kN
Fw > Fd and % error = (Fd – Fo)/Fd =10.22% Fo ≈ Fd. The design will probably be satisfied.
For more accuracy, we increase face width up to b = kmax x π x m = 4 x π x 7 = 87.96mm
Use 80mm
Fo = 15.579kN
Fw = 20.235kN
Fd = 15.73kN
Fo ≈ Fd , Fw > Fd ⇒ Design is satisfied.
m = 7, b = 80mm, Np = 16, Ng = 100 ⇐ Ans:
4.
2
A spur steel pinion (so = 200 MN / m2 ) is to drive a spur steel gear ( so = 140 MN / m
) . The diameter of the pinion is to be 100 mm and the center distance 200mm . The pinion is
to transmit 5 kW at 900 rev/min .The teeth are to be 20° full depth. Determine the necessary
module and width of face to give the greatest number of teeth. Design for strength only, using
the Lewis equation. Ans . m = 2 , b = 21.2 mm (use 22 mm)
Solution : This prob: can be solved with the same procedure as Prob:2.
5.
Two spur gear are to be used for a rock crusher drive and are to be of minimum size.
The gear are to be designed for the following requirements: power to be transmitted 18 kW,
speed of pinion 1200 rev/min, angular velocity ratio 3.5 to 1, tooth profiles 20° stub, so value
for pinion 100 MN/m2, so value for gear 70MN/ m2. Determine the necessary face width and
module for strength requirements only, using the Lewis equation.
Ans . m = 5 , b = 57mm
Solution : This prob: can be solved with the same procedure as Prob:3 (strength check only)
6.
A pair of spur gears transmitting power from a motor to a pump impeller shaft is to be
designed with as small a center distance as possible .The forged steel pinion (so = 160 MN/
m2 ) is to transmit 4 kW at 600 rev/min to a cast steel gear(so = 100MN/m2) with a
transmission ratio of 4 ½ to 1, and 20° full depth involute teeth are to be used. Determine
the necessary face width and module for strength only, using the Lewis equation .
Ans . m = 3,b = 30.9 mm(use 31 mm)
Solution: This prob: can be solved with the procedure same as prob:3
7
Helical Gear
1.
A pair of helical gears are to transmit 15 kW. The teeth are 20°stub in diametral plane
and have a helix angle of 45°. the pinion has an 80mm pitch diameter and operates at 10,000
rpm. The gear has a 320mm pitch diameter. If the gears are made of cast steel So =
100MN/m2, determine a suitable module and face width. The pinion is heat treated to a
brinell of 300 and the gear has a brinell hardness of 200.
Solution:
power = 15kW, φ = 20°stub , ψ = 45°, Dp = 80mm, Dg = 320mm, rpm p = 10,000
so = 100MN/m2, BHNp = 300, BHNg = 200
Pinion and gear are made of the same material. So, design is based on pinion.
Known diameter case.
Lewis equation
⎛ 1 ⎞
s k π2
⎜
⎟
=
cos Ψ (1)
⎜ 2 ⎟
Ft
m
y
⎝
⎠ all
⎛ 5.6 ⎞
⎟⎟ , V = π Dp rpmp/60 = 41.887 m/s
s = s o ⎜⎜
⎝ 5.6 + V ⎠
s = 46.388 x 106 MN/m2
k = 6 (maximum)
9550 kW
Ft =
= 358.125N
rpm x D/2
⎛ 1 ⎞
⎜
⎟
= 5.424x10 6 , assume y = 0.15 ⇒ m = 1.11
⎜ 2 ⎟
⎝ m y ⎠ all
Standard module :
preffrred
1, 1.25, 1.5, 2 , ……
second choice 1.125, 1.375, ….
Try m = 1.25
Np = Dp/m, Np = 64, Nfp = 181, φ = 20deg stub⇒ y = 0.166033
⎛ 1 ⎞
⎜
⎟
= 3.855x10 6 <
⎜ 2 ⎟
⎝ m y ⎠ ind
⎛ 1 ⎞
⎜
⎟
, design is satisfied.
⎜ 2 ⎟
m
y
⎝
⎠ all
so, decrease module and try m = 1.125, Np = 71.111≠ whole number
Try m = 1
Np = Dp/m, Np = 80, Nfp = 226.27, φ = 20deg stub⇒ y = 0.1713
8
⎛ 1 ⎞
⎛ 1 ⎞
⎜
⎟
⎟
= 5.838x10 6 > ⎜
, design is not satisfied.
⎜ 2 ⎟
⎜ 2 ⎟
⎝ m y ⎠ ind
⎝ m y ⎠ all
∴Take the smallest module = 1.25
Reduce k value
kred = [ kmax x (1/m2y)ind ]/ (1/m2y)all = 6 x 3.855 x 106 / 5.424 x 106 = 4.264
Face width, b
b = k π m = 16.75 mm, Use 17mm
Dynamic Check
Endurance Force, Fo
Fo = s o byπm cos ψ = 100x10 6 x17 x10 −3 x 0.166033xπx1.25x10 −3 x cos 45 = 783.769 N
Wear Force, Fw
Fw =
Q=
D p bKQ
cos 2 ψ
2D g
Dp + Dg
=
2 x 320
= 250
80 + 320
tan φ n = tan φ cos ψ = tan 20 cos 45 ⇒ φ n = 14.43
BHN avg = 250, s e s = 2.75BHN avg − 70 = 2.75x 250 − 70 = 617.5MN / m 2
⎡ 1
1 ⎤
⎡
⎤
2
s es 2 sin φ n ⎢
+
⎥ (617.5x10 6 ) 2 sin 14.43⎢
⎥
⎢⎣ E p E g ⎥⎦
⎣ 200 x10 9 ⎦ = 678716.96
K=
=
1.4
1.4
Fw =
80 x10 −3 x17 x10 −3 x 678716.96 x1.6
cos 2 45
= 2953.776 N
Dynamic Force, Fd
Fd = Ft +
21v(bC cos 2 ψ + Ft ) cos ψ
21v + bC cos 2 ψ + Ft
v = 41.888m / s ⇒ e p = 0.015 ⇒ precision cut gear is used
m = 1.25, precision cut ⇒ error = 0.01
steel pinion and gear, φ = 20 deg stub, error = 0.01 ⇒ C = 119 kN/m
D
9
Fd = 358.125 +
21x 41.888(17 x10 −3 x119 x10 3 cos 2 45 + 358.125) cos 45
21x 41.888 + 17 x10 −3 x119 x10 3 cos 2 45 + 358.125
∴ Fo < Fd , Fw > Fd ⇒ Design is not satisfied.
= 1287.496 N
∴increase module
if
m = 1.375, Np = 80/1.375 ≠ whole number
m = 1.5, Np = 80/1.5 ≠ whole number
m = 1.75, Np = 80/1.75 ≠ whole number
m = 2, Np = 80/2 = 40, Nfp = 113.137 ⇒ yp = 0.16204
Fo = 1223.872N, ∴ Fo is still weaker. ∴increase b to 19mm and
Fo = 1367.857N, Fw = 2301.279, Fd = 1366.512N
∴Fo, Fw > Fd ⇒ Design is satisfied.
m = 2, b = 19mm⇐ Ans:
2.
A pair of helical gear with a 23 deg helix angle is to transmit 2.5 kW at 10,000
rev/min of the pinion. The velocity ratio is 4 to 1. Both gears are to be made of hardened
steel, with an allowable so = 100MN/m2 for each gear. The gears 20 deg stub and the pinion is
to have 24 teeth. Determine the minimum diameter gears that may be used and the required
BHN.
power = 2.5 kW, rpmp= 10,000, V.R = 4, φ = 20 deg stub, ψ = 23 deg,
Solution:
so= 100MN/m2 , Np = 24
Strength Check
Both gears are made of the same material. ∴Base design on pinion.
Unknown diameter case
2M t
kyπ Nm 3 cosψ
9550 xkW 9550 x 2.5
Mt =
=
= 2.3875 Nm
rpm
10000
24
N p = 24, N fp =
= 30.77 ⇒ y p = 0.13975
cos 3 23
2 x 2.3875
0.0262
∴ sind =
=
2
3
m3
6 x0.13975 xπ x 24 xm cos 23
S ind =
2
10
assume V .F = 1 / 2, Sind becomes So / 2
so 0.0261
=
⇒ m = 0.805
2
m3
Std module series 1, 1.125, 1.25, 1.375, .....
∴
Try m = 1
D = N m = 24, v =
πx24x10-3 x10000
60
= 12.566m / s
5.6
= 61.236 x106 N / m 2
5.6 + v
0.0261
= 26.1x10 6 N / m 2
sind =
−3 3
(1x10 )
∴ sind > sall ⇒∴ Design is satisfied.
sall = so x
Reduce k value
s
26.1x10 6
k red = k max x ind = 6 x
= 2.557
s all
61.236 x10 6
Face width, b
b = k x π x m = 2.557 x π x 1 = 8.034mm, Use b = 9 mm.
Dynamic Check
Endurance Force, Fo
Fo = s o byπm cos ψ = 100 x10 6 x 9x10 −3 x 0.13975xπx1x10 −3 x cos 23 = 363.725 N
Dynamic Force, Fd
Fd = Ft +
21v(bC cos 2 ψ + Ft ) cosψ
21v + bC cos 2 ψ + Ft
v = 12.566m / s ⇒ e p = 0.03(assume precision cut gear is used)
m = 1, precision cut ⇒ error = 0.01
steel pinion and gear, φ = 20 deg stub, error = 0.01 ⇒ C = 119 kN/m
Fd = 198.958 +
21x12.566(9 x10 −3 x119 x10 3 cos 2 23 + 198.958) cos 23
21x12.566 + 9 x10
∴ Fo < Fd ⇒ Design is not satisfied.
∴increase module
−3
3
2
x119 x10 cos 23 + 198.958
= 1103.436 N
11
Try m = 2
D = Nm = 48, v =
πx 48x10 −3 x10000
= 25.133m / s
60
2.3875
Ft =
= 99.479 N
48x10 −3 / 2
b max = k max xπxm = 6 xπx 2 = 37.699mm ⇒ Use b = 36mm
Fo = 2909.784 N
Fd = 3176.42 N
∴ Fo < Fd ⇒∴ Design is not satisfied
Try m = 2.5
D = Nm = 54, v =
Ft =
πx 54 x10 −3 x10000
= 28.274m / s
60
2.3875
= 88.426 N
54 x10 −3 / 2
b max = k max xπxm = 6 xπx 2.5 = 42.4mm ⇒ Use b = 30mm
Fo = 2727.918 N
Fd = 2708.13N
∴ Fo > Fd ⇒∴ Design is satisfied for the strength and dynamic effect.
For satisfactory wear effect
Fw ≥ Fd
D p bKQ
2
cos ψ
≥ Fd ; Q =
2x 216
= 1.6; tan φ n = tan φ cos ψ = tan 20 cos 23 ⇒ φ n = 18.523
54 + 216
54x10 −3 x30x10 −3 xKx1.6
cos 2 23
K ≥ 885511.906
≥ 2708.13
⎡ 1
1 ⎤
s es 2 sin φ n ⎢
+
⎥
⎣⎢ E p E g ⎦⎥
≥ 2708.13
1.4
s es ≥ 624.687MN / m 2
2.75BHN − 70 ≥ 624.687 ⇒ BHN ≥ 252.61
module = 2.5, face width, b = 30mm and required BHN = 252.61 ⇐ Ans:
12
3.
Two parallel shafts are connected by a pair of steel helical gears.The pinion transmits
10kW at 4000 rev/min of the pinion. Both gears are made of the same materials, hardened
steel with an allowable so = 100MN/m2. If the velocity ratio is 4 ½ to 1, determine the
smallest diameter gears that may be used having sufficient strength. No less than 30 teeth are
to be used on either gear, the teeth are of 20˚ stub in diametral plane, and the helix angle is
45˚. Use the Lewis equation.
Ans : m =1.5 , b= 26mm
Also determine the required average brinell hardness of the two gears to provide a
satisfactory design from the standpoint of wear .Assume a precision cut gear .C = 119 kN/m2
Ans : K =660kN/m2 , Fd =2530N, average BHN =214
Solution: This prob can be solved with the procedure same as prob:2
13
Bevel Gears
1.
Two cast iron bevel gears having pitch diameters of 80 and 100mm respectively are to
transmit 2
3
1
kW at 1100rpm of the pinion. The tooth profiles are of 14 deg composite form.
4
2
Take So = 55MN/m2.
(a) Determine the face width b and the required module from the stand point of strength
using the Lewis equation.
(b) Check design from the stand point of dynamic load and wear. C = 110kN/m and K =
1330kN/m2.
Solution:
(a) Strength Check
Dp = 80mm, Dg = 100mm, power = 2.75kW, rpmp = 1100, φ = 14.5 deg comp,
So = 55MN/m2
Same material. ∴Base design on pinion.
Known diameter case.
⎛ 1 ⎞
sbπ ⎛ L - b ⎞
⎜⎜
⎟⎟ =
⎜
⎟
⎝ my ⎠ all Ft ⎝ L ⎠
L=
L
1
D 2p + D g2 = 64.031mm, = 21.34mm ⇒ Use b = 21mm
3
2
Ft =
9550 kW
9550x 2.75
=
= 596.875N
rpm x D / 2 1100x80x10 −3 / 2
V=
πDrpm πx80x10 − 3 x1100
=
= 4.607m / s
60
60
s = s o x V.F = 55x10 6 x
6
= 31.11MN / m 2
6 + 4.607
⎛ 1 ⎞
31.11x10 6 x 21x10 − 3 xπ ⎛ 64.031 − 21 ⎞
⎟⎟ =
⎜⎜
⎜
⎟ = 2310.845
596.875
⎝ 64.031 ⎠
⎝ my ⎠ all
assume y = 0.1 ⇒ m = 4.327
Std module series
….,4,4.5,5,5.5,6,7,…
14
Try m = 5
16
D 80
=
= 16, N fp =
= 20.489 ⇒ y p = 0.091
100 / 2 x 64.031
m 5
⎛ 1 ⎞
⎛ 1 ⎞
1
⎟⎟
⎟⎟
⎜⎜
=
= 2197.802 < ⎜⎜
⎝ my ⎠ all
⎝ my ⎠ ind 5x10 − 3 x 0.091
N=
∴Design is satisfied.
Decrease module. Try m = 4
20
D 80
=
= 20, N fp =
= 25.612 ⇒ y p = 0.098
100 / 2 x 64.031
m 4
⎛ 1 ⎞
⎛ 1 ⎞
1
⎟⎟
⎟⎟
⎜⎜
=
= 2551.02 > ⎜⎜
⎝ my ⎠ all
⎝ my ⎠ ind 4 x10 − 3 x 0.098
N=
∴Design is not satisfied.
Take smallest module = 5 and b = 21 mm
(b) Wear Force, Fw
Fw =
Q=
0.75D p bKQ
cos α p
2 N fg
N fp + N fg
, N fg =
20
= 32.016
80 / 2 x 64.031
2x 32.016
= 1.22, K = 1330 (given)
20.489 + 32.016
0.75x80x10 − 3 x1330x10 3 x1.22
Fw =
= 2617.767 N
0.781
Q=
Dynamic Force, Fd
Fd = Ft +
21V(bC + Ft )
21V + bC + Ft
Fd = 596.875 +
21x 4.607(21x110 + 596.875)
21x 4.607 + 21x110 + 596.875
= 2463.508N
∴Fw >Fd ⇒∴Design is satisfied from the stand point of wear and dynamic effect.
15
2.
A pair of bevel gears is to be used to transmit 9kW. Determine the required module
and gear diameters for the following specifications.
Pinion
Gear
No. of teeth
21
60
Material
steel
cast iron
2
So
85mN/m
55MN/m2
Brinell hardness
200
160
Speed
1200 rpm
420 rpm
Tooth profile
1
14 deg comp
2
same
Solution:
Strength check
Different material. ∴Check load carrying capacity (So y)
N fp =
N fp =
Np
Ng
Ng
Np
N 2p + N g2 = 22.249 ⇒ y p = 0.093
N 2p + N g2 = 181.625 ⇒ y g = 0.11964
(soy)p = 85 x 0.093 = 7.905
(soy)g = 55 x 0.11964 = 6.5802 < (Soy)p
Since the gear is weaker, base design on gear
Unknown diameter case.
s ind =
3
⎛L − b⎞ L
= (assume)
⎜
⎟;
m byπN ⎝ L ⎠ L − b 2
2M t
2
mN p
2
21m
L
⎛ 60 ⎞
1 + ⎜ ⎟ = 31.784m, b = = 10.595m
2
2
3
⎝ 21 ⎠
9550 kW 9550 x 9
Mt =
=
= 204.64 Nm
rpm
420
L=
s ind =
1 + (V.R ) 2 =
2 x 204.64
3
m x10.595x 0.11964xπx 60
x
3 2.596
=
2
m3
assume V.F = 1/2 and s ind becomes 55x10 6 / 2
55x10 6 2.569
=
⇒ m = 4.537 , Std module series … ,4,4.5,5, ….
2
m3
16
Try m = 4.5
D g = Nxm = 60 x 4.5 = 270
πx 270x10 −3 x 420
= 5.938m / s
60
6
s all = 55x10 6 x
= 27.644 x10 6 N / m 2 (assume cut teeth)
6 + 5.938
2.569
s ind =
= 28.192 x10 6 N / m 2
−3 3
(4.5x10 )
V=
∴sind >sall ⇒ Design is not satisfied.
Try m = 5
D g = Nxm = 60 x 5 = 300
πx 300 x10 −3 x 420
= 6.598m / s
60
6
s all = 55x10 6 x
= 26.195x10 6 N / m 2 (assume cut teeth)
6 + 6.598
2.569
= 20.552 x10 6 N / m 2 < S all
s ind =
−3 3
(5x10 )
V=
∴Design is satisfied.
Take smallest module = 5
b = 10.595 x 3 = 52.975mm⇒ Use b = 52 mm
L = 31.784 x 5 = 158.92mm
After determining the module and face width from the strength check, it is necessary to make
dynamic check.
Dynamic Check
Endurance force, Fo
⎛L − b⎞
Fo = s o byπm⎜
⎟
⎝ L ⎠
⎛ 158.92 − 52 ⎞
Fo = 55x10 6 x 52x10 −3 x 0.11964 xπx 5x10 −3 ⎜
⎟ = 3616.119 N
⎝ 158.92 ⎠
The limiting wear load, Fw
Fw =
0.75D p bKQ
cos α p
17
Q=
2 N fg
=
N fp + N fg
2 x181.625
= 1.782, cosα p = N p / N fp = 0.944
22.249 + 181.625
K = 600(200BHN steel, C.I, 14.5 deg comp)
Fw =
0.75x105x10 −3 x 52x10 −3 x 600x10 3 x1.782
= 4638.108N
0.944
The dynamic load, Fd
Fd = Ft +
21V(bC + Ft )
21V + bC + Ft
V = 6.598 m/s⇒permissible error ep = 0.06 (carefully cut and precision cut can be used)
Try with precision cut
Steel pinion and cast iron gear, φ =14.5 deg comp,e = 0.01⇒ C = 76kN/m
Ft =
204.64
150x10 −3
= 1364.267 N
Fd = 1364.267 +
21x 6.598(52 x 76 + 1364.267)
21x 6.598 + 52x 76 + 1364.267
= 4847.545 N
∴Fo < Fd , Fw > Fd ⇒ Design is not satisfied.
To satisfy the dynamic effect, Fo will have to be increase to equal Fd . This may be
accomplished by using a larger module or better material. For a given material, we consider
to increase module.
Try m = 5.5
D g = 60 x 5.5 = 330, V = 7.257m / s, L = 174.814mm, b = 58mm
Ft = 204.64 / 165x10 −3 = 1240.242 N
Fo = 4406.546 N
Fw = 5691.443N
Fd = 5023N
From these value we can see that Fo is still weaker. So, increase module.
Try m = 6
D g = 60 x 6 = 360, V = 7.917m / s, L = 190.704mm, b = 63mm
Ft = 204.64 / 180 x10 −3 = 1136.889 N
Fo = 5232.695 N
Fw = 6743.096 N
Fd = 5186.772 N
Fo , Fw > Fd , Design is now satisfied.
18
Module = 6, face width = 63 mm ⇐Ans:
3.
Check Example (2) by means of the procedure recommended by AGMA using m =5.
Take Cm = 0.3.
Solution:
(a) The power rating from the standpoint of strength for the values calculated in Example
(2) may be determined by AGMA equation.
Power =
ms(rpm) p D p byπ(L − 0.5b) ⎛ 5.6 ⎞
⎜⎜
⎟⎟
19100L
⎝ 5.6 + V ⎠
5x10 −3 x 55x10 6 x1200 x105x10 −3 x 52 x10 −3 x 0.11964 xπ(158.92 − 0.5x 52)
19100 x158.92
=
⎛
⎞
5.6
⎜⎜
⎟⎟
⎝ 5.6 + 6.598 ⎠
=20.33kW
(b) For durability the AGMA power rating is
Power = 0.8CmCBb ; Cm = 0.3
D1p.5 (rpm) p ⎛ 5.6 ⎞
⎜⎜
⎟⎟
CB =
0.032
⎝ 5. 6 + V ⎠
=
⎞
(105x10 −3 )1.5 1200 ⎛
5.6
⎜⎜
⎟⎟ = 874.68
0.032
⎝ 5.6 + 6.598 ⎠
Power = 0.8x0.3x874.68x52x10-3 = 10.91kW
The above equations indicate that m of 5 and a face width of 53mm is more than safe for
9kW according to AGMA.
4.
Two steel bevel gears with a 14 ½ deg full depth tooth profile have been designed for
strength to transmit 20kW at 1250rpm of the pinion. For the pinion so = 175MN/m2 and N =
23; for the gear so = 140MN/m2 and N = 32. The module is 6 and the face width is 40mm.
Find their required hardness to satisfy the AGMA durability recommendations.
Solution: 20kW, rpmp = 1250, sop = 175MN/m2, N = 23, sog = 140MN/m2, N = 32,
m = 6, b = 40mm
AGMA equation; Power = 0.8CmCBb
Dp = Nm = 23x6=138mm, V = π x 138 x 10-3 x 1250/60 = 9.033m/s
19
CB =
D1p.5 rpm p
0.032
(138x10 −3 )1.5 1250
5.6
=
= 1203.129
x
x
0.032
5.6 + V
5.6 + 9.033
5.6
20 = 0.8 x Cm x 1203.129 x 40 x 10-3
Cm =0.5
This material indicates a required brinell hardness of 285 for the gear and 335 for the
pinion. These values correspond to a Cm = 0.5 from the material factor table (Table 3.1).
5.
The 14 ½ deg composite tooth, steel bevel gears are to transmit 15kW with a
transmission ratio of 3. The speed of the pinion is 1800rpm. If the gears are to be hardened to
give a material factor of 0.4. Determine the probable minimum pinion diameter starting with
the AGMA durability equation.
Solution: 15kW, V.R = 3, rpmp = 1800, Cm = 0.4
Power = 0.8CmCBb
b=
Dp
L Dp
=
1 + V.R 2 =
x 1 + 3 2 = 0.527 D p
3
6
6
CB =
D1p.5 rpm p
0.032
xV.F
assume V.F = ½
∴15 = 0.8x 0.4 x
D1p.5 x1800
0.032
x
1
x 0.527 D p
2
Dp ≈ 100mm ⇒ V = π x 100 x 10-3 x 1800/60 =9.425m/s ⇒
V.F =
6.
5.6
5.6 + 9.425
= 0.646 instead of assumed 0.5 which is on safe side.
A right angle speed reducer uses hardened alloy steel precision cut bevel gears. The
transmission ratio is 5 to 1 and the pinion rotates at 900 rev/min while transmitting 40kW. If
the teeth are of 20˚ full depth form and the pinion has a diameter of 115mm, what must be the
module and width of the face of the gears, using Lewis equation? so = 200MN/m2 for both
gears. What brinell hardness is required for satisfactory wear?
Ans : m= 2.5 ,b = 100 mm, average BHN = 253 using Fw= Fd = 15.9 kN
7.
Two shafts at right angles are to be connected with a pair of bevel gears having
20˚full depth teeth. The velocity ratio is to be 4 ½ to 1. The pinion is to be made of steel (so =
20
100 MN /m2) and the gear is to be made of semi- steel (so = 85 MN/m2). The pinion is to
transmit 4 kW at 900 rev/min Determine the minimum diameters, module and face width that
should be used based on strength only. Then using the AGMA durability equation estimate
the minimum diameter that would be required for wear ,assuming a material factor of 0.4.
Ans : For strength : m=3, Dp= 48mm, Dg= 216mm, b=36.9 mm(use 35mm). For wear
Dp= 654mm
Solution: This prob: can be solved with the procedure same as prob: 5.
8.
A pair of straight tooth bevel gears at right angles is to transmit 1 ½ kW at 1200
rev/min of the pinion. The diameter of the pinion is 75mm and the velocity ratio is 3 ½ to 1.
The tooth form is 14 ½˚ composite type. Both pinion and gear are cast iron (55 MN/m2).
Determine the module and face width from the standpoint of strength only , using the Lewis
equation .
Ans : m = 1 , b = 45.5 mm ( use 45 mm)
Solution: This prob: can be solved with the procedure same as prob:1
21
Worm Gear
1.
A worm gear speed reducer unit has a gear center distance of 250mm and a
transmission ratio of 14. Estimate the power input without the gear overheating, assuming
that the strength and wear capacity are not exceeded.
Solution:
From standpoint of heat dissipation,
Permissible power input =
2.
3650C1.7 3650(0.25)1.7
=
= 18.198kW
V.R + 5
14 + 5
A hardened steel worm and phosphor bronze gear reducer unit has a transmission
ratio of 40 to 1 and a center distance of 300mm. The worm speed is a 1500rev/min and its
diameter is 125mm. What is the safe power input based on AGMA from the standpoint of
wear?
Solution:
P=
rpm w KQm
V.R
Where rpmw = 1500, V.R = 40, K = 1.71 (by interpolation in table)
Q = V.R/(V.R+2.5) = 0.94
Vw = π Dw rpmw/60 = π x 0.125 x 1500/60 = 9.817m/s
3.
m=
2.3
2.3
=
= 0.179 ,
2.3 + Vw + 3Vw / V.R 2.3 + 9.817 + 3x 9.817 / 40
P=
rpm w KQm 1500 x1.71x 0.94 x 0.179
=
= 10.8kW ⇐ Ans:
V.R
40
Complete the design and determine the input power capacity of a worm gear speed
reducer unit composed of a hardened steel worm and a phosphor bronze gear (so = 55MN/m2)
having 20° stub involute teeth. The center distance C is to be 200mm, the transmission ratio
R is to be 10, and the worm speed is to be 1750rev/min.
Solution:
(a) Determine Dw, Dg, ma, Nw, Ng, α, b, and L.
Dw ≈ C0.875/ 3.48 = 0.0703m. Axial module ma = Dw/3π = 7.45mm; use ma = 8mm.
Dg (approx)= 2C– Dw (approx) = 2×200 – 70.3 = 329.7mm
22
Now Ng/Nw = 10 = Dg/maNw. Then for various values of Nw, the exact value of Dg can be
determined.
Nw
1
2
3
4
5
Dg(mm)
80
160
240
320
400
Thus the diameter of the gear (Dg (exact))will be taken as 320mm (closest to 330mm) and
the worm diameter (Dw (exact)) will be 2C – Dg (exact) = 80mm, which are close to desire
proportions. The corresponding Nw = 4.
Since the worm will have a quadruple thread,
tan α =
m a N w 8x 4
=
= 0.4 ⇒ α = 21.8 ο
Dw
80
The face width b = 0.73Dw = (0.73) (80) = 58.4 mm; use 60mm
(b) We can estimate the capacity of the gear as follows:
F = sbyπ mn
Vg = π x Dg x rpmg/60 = π x 320 x 10-3 x 175/60 =2.93m/s
⎛ 6
s = s o x⎜
⎜ 6 + Vg
⎝
⎞
6
⎟ = 55x
= 36.954MN / m 2
⎟
+
6
2
.
93
⎠
φ= 20º stub and Ng = 10Nw = 40 giving y = 0.146 and
m n = m a cos α = 8 cos 21.8 = 7.428mm
∴F = 36.954x106x60x10-3x0.146xπx7.428x10-3 = 7554.198N
For a transmitted load F = 7554.198N the dynamic load will be approximately
⎛ 6 + Vg
Fd = ⎜⎜
⎝ 6
⎞
⎟F = 7554.198x ⎛⎜ 6 + 2.93 ⎞⎟ = 11.243kN
⎟
6
⎠
⎝
⎠
The allowable wear load, Fw = DgbB = 320x10-3x60x10-3x687.5x103 =13.2kN
Where B = (1.25)(550) kN/m2 for hardened steel on phosphor bronze and a lead angle
between 10° and 25°. (Table 4.1)
Since the dynamic load does not exceed the allowable wear load, the safe transmitted
load calculated above governs. Hence the allowable power, from the standpoint of gear tooth
strength and wear, is
FVg = (7554.198) (2.93) = 22.133 kW
( c ) According to the AGMA formula for wear, the input capacity of the unit is
23
rpm w KQm
V.R
P=
rpmw =1750, V.R = 10, K = 0.485 (by interpolation in table)
Q = V.R/(V.R+2.5) = 0.8
Vw = π Dw rpmw/60 = π x 0.08 x 1750/60 = 7.33m/s
m=
2.3
2.3
=
= 0.194 ,
2.3 + Vw + 3Vw / V.R 2.3 + 7.33 + 3x 7.33 / 10
P=
rpm w KQm 1750x 0.485x 0.8x 0.194
=
= 13.2kW
V.R
10
(d) The input power from the standpoint of heat dissipation may be estimated:
3650C1.7 3650(0.2)1.7
P=
=
= 17.77 kW
V.R + 5
10 + 5
The above analysis indicates the input power should be limited to 13.2kW in
accordance with AGMA wear recommendations.
4.
A speed reducer unit is to be designed for an input of ¾kW with a transmission ratio
of 27. The speed of the hardened steel worm is 1750rev/min. The worm wheel is to be made
of phosphor bronze. The tooth form is to be 14½º involute. Determine the salient dimensions
from the point of view of strength and wear. Check the design using the AGMA wear and
heat dissipation criteria.
Solution
(a) It is necessary to choose a center distance for trial. Assume that from previous experience
the center distance for this size unit should be about 100mm.
Dw ≈ C0.875/3.48 = 38.3mm
ma = Dw/3π, = 4.064mm, say 4mm
Dg (approx)= 2C – Dw (approx) = 162mm. These are desired proportions.
V.R = Ng/Nw = 27 = Dg/maNw, Dg = 108Nw
For Nw = 1, Dg = 108mm which is significantly different from the required valve and
gives a disproportionately large Dw.
For Nw = 2, Dg = 216mm which is too large since (Dg + Dw) must be less than 2C.
Try ma = 3mm. Dg = 81Nw . Now for Nw = 2⇒ Dg = 162mm, Dw = 38mm which is
satisfactory. Ng = 54.
Face width b = 0.73Dw = 27.74, say 30mm.
24
(b) Check design for gear tooth strength and wear. Permissible transmitted load F is
F = sby πmn
Vg = π x Dg x rpmg/60 = π x 162 x 10-3 x 64.815/60 = 0.55m/s
⎛ 6
s = s o x⎜
⎜ 6 + Vg
⎝
⎞
6
⎟ = 55x
= 50.382MN / m 2
⎟
6 + 0.55
⎠
φ= 14 ½ º stub and Ng = 54 giving y = 0.1112 and
tan α =
m a N w 3x 2
=
= 0.158 ⇒ α = 8.972 ο
Dw
38
m n = m a cos α = 3 cos 8.972 = 2.963mm
∴F = 50.382x106x30x10-3x0.1112xπx2.963x10-3 = 1564.517N
Required transmitted load, F = Transmitted power/ Vg = 750/0.55 = 1363.636N
Estimated dynamic load = Fx
6 + Vg
6
= 1363.636x
6 + 0.55
= 1488.638 N
6
Allowable wear load, Fw = DgbB = (0.162)(0.03)(550,000) = 2673N
Since the allowable wear load is greater than the estimated dynamic load, and since the
allowable transmitted load is greater than the required transmitted load, the design is
satisfactory from the standpoint of strength and wear of the gear teeth. In fact, the
permissible power is
FVg = 1564.517 × 0.55 = 0.86kW
(c) Check AGMA rating for wear.
P=
rpm w KQm
V.R
rpmw =1750, V.R = 27, K = 0.0661 (Table 4.2)
Q = V.R/(V.R+2.5) = 0.915
Vw = π Dw rpmw/60 = π x 0.038 x 1750/60 = 3.482m/s
m=
2.3
2.3
=
= 0.373 ,
2.3 + Vw + 3Vw / V.R 2.3 + 3.482 + 3x3.482 / 27
P=
rpm w KQm 1750 x 0.0661x 0.915x 0.373
=
= 1.462kW
V.R
27
(d) Check for heat dissipation. Permissible input power,
P=
3650C1.7 3650(0.1)1.7
=
= 2.276kW
V.R + 5
27 + 5
25
(e)
In summary
Safe power
Based on
0.86
Gear tooth strength and wear
1.462
AGMA wear rating
2.276
Heat dissipation capacity
The unit as designed is good for 0.86kW according to our estimates. It could be
redesigned for a slightly smaller center distance.
5.
A hardened steel worm rotating at 1250 rev/min transmits power to a phosphor bronze
gear with a transmission ratio of 15 to 1. The center distance is 225mm. Determine the
remaining design and give estimated power input ratings from the standpoint of strength,
endurance, and heat dissipation. The teeth are of 14½ ˚ involute, full depth, form .
Solution: This Prob: can be solved with the procedure same as prob: 3
26
Belt Drives
1. Example 1 Page 61 ME3031lecture notes Part2.
2. Example 2. Page 62 ME 3031 lecture notes Part 2
3. Example 3. Page 63 ME 3031 lecture notes Part2.
4. A leather belt is to transmit 22kW from a 250mm fibre-covered driver pulley
running at 1200 rev/min to a 600 mm steel pulley .The coefficient of friction
between the steel pulley and belt is 0.3 and that between the driver and belt is 0.4. The
distance between shafts is 1.8mm. Assume that the maximum allowable working stress
in the belt is 1.7Mpa and that the belt mass density = 970 kg/m3. Which of the
following belts would be more suitable, a belt 9 mm thick or a belt 12.5 mm thick?
Calculate the required width of belt.
Ans : A 9mm belt would be preferable for a greater belt life ( d/t = 250/9 =28 )
than a 12.5 mm belt (d/t = 250/12.5 = 20 ).The belt width for a 9mm belt is 169mm (
use 170 mm wide ).
5. A V- belt operates on two sheaves having pitch diameters of 250mm and 800mm. The
groove angle of the sheaves is 36˚ and the contact angle of of the small sheave is 140˚.
The maximum allowable belt load is 900N and the V- belt mass is 0.523 kg/m. The
smaller sheave rotates at 1150 rev/min, and 26 kW is to be transmitted. For a
coefficient of friction of 0.2, how many V-belts should be used, assuming each one
takes its proportional part of the load .
Ans : 2.78 belts : use 3 belts
6. An electric motor drives a compressor through V-belts. The following data are
known:
Motor Pulley
Compressor Pulley
Pitch diameter
225 mm
1200 mm
Angle of contact
Coefficient of friction
0.3
0.3
Groove angle
34˚
flat pulley
Power transmitted
17.5kW
Speed
1800 rev/min
Each belt has a mass of 0.3 kg/m and the maximum permissible force is 450 N per
belt .Determine (a) fα/( sin½θ ) for motor pulley and for compressor pulley ,(b) mv2
27
for one belt,(c) the tension on the slack side of the belt ,(d) the power per belt ,(e) the
number of belts required.
Ans : (a) 2.06 for motor pulley , 1.284 for compressor pulley :compressor pulley
governs design . (b) 135N ,(c) 380N ,(d) 1.48 kW/ belt ,(e) 11.8; use 12.
7. An electric motor drives a compressor by means of a flat leather belt .The following
data are known:
Motor Pulley
Compressor Pulley
Diameter
250 mm
1500mm
Angle of wrap
Coefficient of friction
0.25
0.3
Speed
1200 rev/min
( Note that an idler pulley is used.) The belt is 6 mm by 200mm cross section and has
density 970 kg/m3.If the maximum allowable belt tension is 2.7 kN determine the power
capacity .Ans : 24 kW
8. A 1.35m diameter steel flywheel is to be connected to a 0.4m diameter rubber faced
motor pulley by means of a double ply leather belt which has a thickness of 8mm. The
center distance is 3m. The coefficient of friction for leather on steel is 0.2 and for
leather on rubber is 0.4. The leather has an allowable stress of 2.75Mpa, and the joint
efficiency is 80%. Density of leather is 970kg/m3. If 45kW is transmitted with a belt
speed of 24.5m/s, determine:
(a) the maximum permissible working stress,
(b) efα for the pulley which governs the design,
(c) the necessary belt width.
Ans: (1) 2.21 Mpa, (b) 2.0, (c) 282 use 300mm
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