SHAFTING

PROBLEM SET #2

1. A shaft is use to transmit 200KW at 300rpm by means of 200mm diameter of

sprocket. Determine the force tangent of the sprocket.

P=2π T N
Solution:

200 = 2πT (300/60)

T = 6.366 KN-m

T=Fxr
6.366 = F (0.20/2)
F = 63.66 KN

2. Find the diameter of a steel shaft which will be use to operate a 110KW motor
rotating at 5 rps if torsional stress is 90Mpa.

Solution:

110 = 2π T
P =2πTN

16T
(300/60) T = 3.501

πd3
S=

90,000 = 16(3.50
1)
πd3
d = 0.05829 m

3. A shaft has an ultimate stress of 350Mpa and has a factor of safety of 5. The torque
developed by shaft is 3 KN-m and the outside diameter is 80mm. Find the inside
diameter of the shaft.

S
Solution:
16 TDo
F. =
π(0.084−
S D4)
350000 16(3)(0.08)
5 π[(0.08)4 − D4]
=
D = 0.69624 m
4. What is the speed of 63.42mm shaft transmitted 75KW if stress is not to exceed
26Mpa.

16
Solution:

T
πD 16T
S=
3

π(0.06342)3
26,000 =

T = 1.302 KN-m
P = 2π TN
75 = 2π (1.302)N
N = 9.16641 rps x 60sec/min
N =549.98 rpm

5. A steel shaft transmit 50Hp.at 1400rpm. If allowable stress is 500psi, find the
diameter.

Solution:
2πTN
33000
P=
2π T (1400)
33000
50 =
T = 187.575 ft-lb x 12 in/ft

16T
T = 2250.905 in-lb

πD 3
16(2250.905)
S=

πD3
500 =
D = 2.841 in
6. The shaft of the motor has a length of 20 times its diameter and has a maximum
twist of 1 degrees when twisted at 2KN-m torque. If the modulus of rigidity of the
shaft is 80Gpa, find the shaft diameter.

Solution:
T
Θ
= L
J (2KN−m)
1 G
0
(20d)
x π
= (80000000)
180 πd4
( )
32
 d = 0.06632 m

7. A hollow shaft developed a torque of 5 KN and shaft stress is 40 Mpa. If outside

diameter of the shaft is 100mm, determine the shaft inner diameter.

16TDo
Solution:

π(D4 − D4)
S=
o i
16(5)(0.1)
π[(0.1)4− D4]
40000 =

D = 0.07764 m

8. A hollow shaft has 100mm outside diameter and 80mm inside diameter is used to
transmit 100 KW at 600 rpm. Determine the shaft stress.

Solution:
P = 2π T N
100 = 2π T
(600/60) T = 1.5915
KN-m
16TDo
S = π[(D4 − D4)]
o i
S = 13,728.73 kPa

9. What is the polar moment of inertia of a solid shaft that has a torque of 1.5 KN and
a stress of 25 Mpa?

16T
Solution:

πD316T
S=

πD3
25000 =
D = 0.06735 m

πD4
32
J=
π(0.06735)4
32
J=

J = 2.02668 x 10-3 m3
10. A force tangent to 1 foot diameter pulley is mounted on a 2 inches shaft. Determine
the torsional deflection if G= 83 x106 kPa.

Solution:
T =F (D/2)
T = 5 (0.3048/2)

πD3 π(0.0508
T = 0.762 KN-m

32 )4
J= =

32

TL
J = 6.53814 x 10-7 m4
Θ=
JG

θ/L =(6.53 x 10−7)(83000000)

0.762

θ/L = 0.0140418 rad/m x

(180/π) θ/L = 0.804530/m

11. What is the minimum diameter of a steel shaft which will be use to operate a 14
KN-m torque and will not twist more than 3o in a length of 6m? (G= 83 x 106 kPa)

TL
Solution:
Θ=
JG
14 x 6
30(π/180)
= J(83 x
106)
πd
J = 1.93286 x 4
10-5 = D = 32
0.11845 m

12. A solid shaft 5m long is stressed 60 Mpa when twisted through 4o. Find the shaft
diameter if G=6.83 Gpa

16T
Solution:

πd3
S=
Solving for T in terms of d;
 πd3(6000
0) = 11780.97 eqn. 1
16
T=
TL
Θ=
JG
T(5)
40 (π/180) πd4
[ ](83 x 106)
32
=
T = 113774.606 d4 eqn.2
Equate 1 & 2
113774.606 d4 = 11780.97 d3
D = 0.10354 m

13. Compute the maximum unit sheared in a 2 inches diameter steel shafting that
transmits 24,000 in-lb of torque at 120rpm.

Solution:
S = 16T
3
πd
16(24,000)
π(2)3
S=

S = 15278.87 psi

14. What is the diameter of line shaft that transmit 150KW at 15rps?

Solution:

D 3N
For line shaft:

53.5
P=

D3(15 x
150/0.746 = 60)
D = 2.28 in 53.5

15. A main shaft has 50mm diameter is running at 300 rpm. What is the power that
could be delivered by the shafted.

50
(3000)
Solution:
3
D N (
25.4
P = 80 =)
80
P =28.60 Hp
16. A 2 inches diameter main shaft running 600 rpm is mounted by a 12 inches
diameter pulley by means of a belt. The belt efficiency is 90%. Find the diameter of
the short shaft.

Solution:

D 3N
For short shaft:

38
P=
(1.5)3N
38
44.4 =

N = 500 rpm x 1/60 = 8.33 rps

17. A 1.5” diameter short shaft is used to transmit 44.4 hp. Determine the shaft speed.

Solution:

P=D N
3

38
1.33N
38
44.4 =

N = 499.911 rpm/60 sec

N = 8.33 rps

18. A 3” diameter solid shaft is desired to replace a hollow shaft having 4” outside
diameter. Consider the strength to be the same, determine inside diameter of
hollow shaft.

Solution:

16T 16T
For solid shaft:

πD3 π(3)3
S= = = 0.1886 T

S = 16TD
For hollow shaft;
o
π(D4 − D4)
o i
16T(4)

π(44−Di4)
0.1886 T =
(4)4- Di4 = 108
Di = 3.48 in
19. A motor is used to drive a centrifugal pump that discharges 3000 li/min at a head of
10m. The pump efficiency is 68%and running at 550 rpm. Find the torsional stress
of the shaft if shaft diameter is 35mm.

Solution:
Solving for the power output of the pump:
Q = 3000li/min
P=wQh
P = 9.81(3/60)(10)
P = 4.905 KW
Solving for the power input of the pump:

P=2π TN
Brake power = 4.905/0.68

7.213 = 2 π T
(550/60) T =
0.1252 KN-m
16(0.1252)
S=
π(0.035)3
S = 1487.63 kpa

20. A circular saw blade has a circular speed of 25 m/sec and 500 mm diameter is
driven by a belt that has a slip of 7%. Find the required speed of the driven shaft if
speed ratio is 3.

V=π DN
Solution:

25 = π (0.5) N
N = 954.93 rpm

21. An 800 mm diameter circular saw blade is driven by 1800rpm motor with speed
gear ratio of 1:8. Find the peripheral speed of the blade.

Solution:
N1/N2 = 1.8
1800/N2 = 1.8
N2 = 1000 rpm
V=π DN
V = π (0.8)(1000/60)
V = 41.88 m/sec x 3.281ft
 V = 137.43 ft/sec

22. A machine shaft is supported on bearings 1 m apart is to transmit 190 KW at 300

rpm while subjected to bending load of 500 kg a the center. If shearing stress 40
Mpa, determine the shaft diameter.

Solution:
For simply supported beam with load at center:
M = PL/4 (500 x 0.00981)(1) / 4
M =1.226 KN-m

P =2 π T N
Solving for the torque developed:

190 = 2π T
(300/60) T =

16
6.04788 KN-m
√M2 + T2
πd 3
S=

40000 = 3 √(1.266)2 − (6.04788)2

16

πd
D = 92.27 mm

23. A 100 mm diameter shaft is subjected to a torque of 6 KN-m and bending moment
of 2.5 KN-m. Find the maximum bending stress developed.

16
Solution:

[M+√M2 + T2 ]
πd3
St =
[2.5+√(2.5)2 + (6)2 ]
16
S=
π(0.01)3
t

St = 45,836.62 kPa

24. A shaft has a length of 10 ft. Find the diameter of the shaft that could safely deliver.

L
Solution:
10
8.9 8.95
D = 2/3 =
5
D = 1.181 in

25. A 12 ft shaft is running at 260 rpm. What that this could safely deliver?

Solution:
 L 12
8.9 8.95
D2/3 = =
5
D = 1.341 in
Hp = (D/4.6)4 N
Hp = (1.341/4.6)4 (200)
Hp = 3.374 hp

26. Two circular shafts, one hollow shaft and one solid, are made of the same materials
and have diameter as follows; hollow shaft inside diameter is one-half of the
external diameter. The external diameter is equal to the solid shaft. What is the
ratio of twisting moment of the hollow shaft to that of the solid shaft?

Solution:

1 1
Do = 1/2Di
4 4
π( −Di )
π( Di)4 Di
2
2
32 = 32
Di = 15/16

27. Determine the thickness of a hollow shaft having an outer side diameter of 100 mm
if it is subjected to a maximum torque of 5403.58 N-m without exceeding a shearing
stress of 60 Mpa or a twist of 0.5 degree per meter length of shaft. G=83,000 Mpa.

TL
Solution:
Θ=
JG
(5403.58)(1000d)
0.5 = π(1004−Di4)
[ ][83000]
32
Di =
85mm
Thickness = Do – Di
= 100 – 85
Thickness = 15mm

28. An engine of a motor vehicle with a wheel diameter of 712 mm develops 50 KW at

2,000 rpm. The combined efficiency of the differential and transmission is 75% with
 an overall speed reduction of 25 is to 1. Determine the torque to be developed by
the clutch in N-m.

Solution:

50 = 2 π T
P = 2π T N

(2000/60) T = 0.239
KN-m
T = 239 N-m

29. An engine of a motor vehicle with a wheel diameter of 712 mm develops 50 KW at

2,000 rpm. The combined efficiency of the differential and transmission is 75% with
an overall speed reduction of 25 is to 1. Determine the draw bar pull developed in
KN.

Solution:
Power on wheels = 50(0.75) = 37.50 kw
Speed of wheels = 80 rpm
37.50 = 2π Tw (80/60)
Tw = 4.476 KN-m
Tw =F x r
4.476 = F x (0.712/2)
F =12.57 KN

30. A 102 mm diameter solid shaft is to be replaced with a hollow shaft equally strong
(torsion) and of the same material. The outside diameter of the hollow shaft is to
be 127 mm. What should be the inside diameter? The allowable sharing stress is
41.4 Mpa.

16
Solution:

T
πd 16T
S=
3

π(102)3
41.4 =

T = 8626422.927 N-m
16(862422.927)(127)
4 4
41.4 = π(127 − Di )
Di =105.815 mm
31. A steel shaft operates at 188 rad/sec and must handle of 2 KW power. The sharing
stress is not to exceed 40 MN/m2 . Calculate the minimum shaft based on pure
torsion.

P =2 π T N
Solution:

2 = 2 π T(188/2
π) T = 0.01064
KN-m
16T
S=
πD3
40 x103 = 16(0.0106
4)
πD3
D = 0.1106 m

32. A round steel shaft transmits 373 watts at 1800 rpm. The torsional deflection is not
to exceed 1 deg in a length equal to 20 diameters. Find the shaft diameter.

P=2π T N
Solution:

0.375 = 2 π T(1800/60)
T =0.001989 KN-m
0.001989(20)
10(π/180) = πD4
32 (80 x106)
D=
0.0066206 m

33. A 25 mm diameter shaft is to be replaced with a hollow shaft of the same material,
weighing half as much but equally strong in torsion. The outside diameter of the
hollow shaft is to be 38 mm. Find the inside diameter.

Solution:

16T 16T
For solid shaft:

πD3 π(25)3
S= =
S = 3.259 x10-4
For hollow shaft:

3.259 x104 T = 16T(38)

Di= 33.64 mm π(38 Di

4 4

)
34. What factor of safety is needed for a 1.998 in diameter shaft with an ultimate
strength of 50,000 psi transmit 40,000 in-lb torque.

16T
Solution:
16(40000
=πD )
3
S=
π(1.998)
3

50000
S = 25541.33831 psi

25541.3383
Fs =
Fs = 1.9576 lb

35. A tubular, shaft having an inner diameter of 30 mm and an outer diameter of 42

mm, is to be used to transmit 90 KW of power. Determine the frequency of rotation
of the shaft so that the shear stress cannot exceed 50 Mpa.

Solution:

36. A solid transmission shaft is 3.5 inches in diameter. It is desired to replace it with a
hollow shaft of the same material and same torsional strength but its weight should
not be half as much as the solid shaft. Find the outside diameter and inside
diameter of the hollow shaft in millimeters.

16T 16TDo
Solution:

πd3 π(Do4Di4)
=
Do4 – Di4 = d3Do
Do4- Di4 = (3.5)3 Do
Do4 – Di4 = 42.87 Do eqn.1
(π 2 − )) L w 1 (π d2)
(Do
Di =
2 4
4 2

Di4 = (Do2 - 6.125)2 eqn.2

Do 4- (Do2 – 6.125)2 = 42.87Do
Do2 – 3.5Do – 3.0635 =0
−(−3.5)±√(−3.5)2−4(1)(−3.0625)
2(1)
Do =

Do =4.225 in = 107 mm

Solving for the inner diameter:

Do2 – Di2 = 6.125
 (4.225)2 – Di2 = 6.125
Di =3.425 in = 87 mm

37. A 76 mm solid steel shaft is to be replaced with a hollow shaft of equal torsional
strength. Find the percentage of weight saved, if the outside of the hollow shaft is
100 mm.

16T 16TDo
Solution:

πd3 π(Do4− Di4)

=
16T 16T(100)
π(76)3 π(1004−Di4)
=
(100)4 – Di4 = (76)3(100)
Di =86.55 mm
mS = ( (76)3 ) L w = 4536.36 w L
π
4
mH = [ (D − Di )]L w
π 2 2
4
o
=[ (1002 −
π

4
)] L w = 1970.64 w L
86.55 2
4536.36−1970.64

4536.3
Percentage of weight saved = =56.56%
6

38. A solid steel shaft whose Sy = 300 M pa and Su = 400 Mpa, is used to transmit 300
KW at 800 rpm. Determine its diameter.

Solution

BELTS
PROBLEM SET #6

1. Find the angle of contact on the small pulley for a belt drive a center distance of 72
inches if pulley diameters are 6 in. and 12 in. respectively.

Solution:
Θ= 180 – 2 sin-1 12−6
[ 2(72) ]

Θ =175.220
2. Determine the belt length of an open belt to connect the 6cm and 12 cm diameter
pulley at a center distance of 72 cm.

Solution:
(12−6)
L =1.57(12+ 6) + 2(72) 2

4(72)
+

L =172.385 cm

3. A 12 cm pulley turning at 600 rpm is driving a 20 cm pulley by means of belt. If total

belt slip is 5%, determine the speed of driving gear.

Solution:
D1 N1 = D2N2 (12)
(600) =20 N2
N2 = 360 rpm
Considering the 5% slip:
N2 =360(1- 0.05)
N2 = 342 rpm

4. The torque transmitted in a belt connected to 300 mm diameter pulley is 4 KN-m. Find
the power driving pulley if belt speed is 20 m/sec.

V=π D N
Solution:

20 = π (0.3)
N N = 21.221
rps P =2 π T
N
P = 2 π (4) (21.221)
P =533.34 KW

5. A 3/8 inch flat belt is 12 inches wide and is used on 24 inches diameter pulley rotating at
600rpm. The specific weight of belt is 0.035 lb/in3 . the angle of contact is 150 degrees. If
coefficient of friction is 0.3 and stress is 300 psi, how much power can it deliver?

V=π D N
Solution:

V = π (24/12) (600)
V = 62.83ft/sec
 𝞀v2
e𝐹𝝷−1
F1 - F2 =b t(S − )(
)
2.68 (0.30)(150)(
π
)−1
2
𝑒𝑓𝜃
F1 –F2 = (12)(3/8)(300 0.035(62.83)
)
18

− )
0

(
e π

2.68 (0.30) 180

)
𝑒
(150)(

(F −F )v
F1 – F2 = 608.26 lbs
62.83
(608.26)( )
1 2 60
33000 33000
Hp = =
Hp =69.50

6. A belt connected pulleys has 10 cm diameter and 30 cm diameter. If center distance is

50 cm, determine the angle of contact of smaller pulley.

D2−D1
Θ= 180 ± 2 sin-1[ ]
Solution:

2C
30−10
Θ= 180 – 2 ]
sin [
-1
250)

Θ= 157 deg

7. A compressor is driven by an 18 KW motor by means of V-belt. The service factor is 1.4

and the corrected horsepower capacity of V-belt is 3.5. determine number of belts
needed.
Solution:
P = (18 x 1.4) = 25.2 KW
P =25.5/(3.5 x 0.746) = 9 belts

8. A pulley 600 mm in diameter transmits 50 KW at 600 rpm by means of belt. Determine

the effective belt pull.

50=2 π T
Solution

(600/60)
T=0.79577
T=F x r
0.795775 = F x 0.3
F= 2.65 KN
9. A pulley have an effective belt pull of 3 KN and an angle of belt contact of 160 degrees.
The working stress of belt is 2 Mpa. Determine the thickness of belt to be used if 350
mm and coefficient of friction is 0.32.

F1 F1
Solution:
= =efθ
bt F2
S=

3
= e (0.32) (160) (π/180)
F2
1.2275
(0.35)t
2=

T =7.24 mm

10. A pulley has a belt pull of 2.5 KN. If 20 Hp motor is used to drive the pulley, determine
the belt speed.
Solution:
P = (F1 – F2)v
20 x 0.746 = 2.5 V
V = 5.97 m/sec x 3.281ft
V =19.58 ft/sec
11. A belt connects a 10 cm diameter and 30 cm diameter pulleys at a center distance of 50
cm. Determine the angle of contact at the smaller pulley.

30−10
]
Solution:
Θ= 1800- 2 sin-
1
[ 2(50)

Θ= 156.930

12. An 8 in diameter pulley turning at 600 rpm is belt connected to a 14” diameter pulley. If
there is a 4% slip, find the speed of 14 in pulley.
D 1N 1 = D 2N 2
8(600) = 14N2
N2 = 342.86 (1- 0.04)
N2 = 329 rpm

13. A 3/8” thick flat belt is 12” wide and is used on a 24” diameter pulley rotating at 600
rpm. The specific weight of the belt is 0.035 lb/in3 . the angle of contact is 150o .If
the coefficient of friction is 0.3 and the safe stress is 300 psi, how much power can it
deliver?
 F1 – F2 =b t [s − ][ ]
𝞀v e
Solution:
2 𝐹𝝷−1

e𝐹𝝷
2.68
V = π (24/12)(600/60)
V =62.83 ft/sec

F1 – F2 = (3/8) (12) [300

0.035(62.83)2 e150 x 0.3 x π/180−1
− ] e150 x 0.3 x π/180 ]
[
F1 –F2 = 608.26 lbs 2.68
(608.26)(62.83 x 60)

3300
Hp = =69.50 Hp
0

14. The standard width of a B85 premium quality V-belt = 21/32 inches

15. Determine the width of a 6 ply rubber belt require for a ventilating fan running at 150
rpm driven from a 12 inch pulley on a 70 hp at 800 rpm. The center distance between
pulley is 2 ft and the rated belt tension is 78.0 lb/in width.

Solution:

16. A 25.4 cm pulley is belt-driven with a net torque of 339 N-m. The ratio of tension in the
tight side of the belt is 4:1. What is the maximum tension in the belt?

Solution:
F1 4
F2 =1
F max = Sd (b t)
T = (F1 –F2) r
F1 = 4F2
339 =(4F2 – F2)(0.258/2)
F2 = 889.76 N
F1 = 4(889.76)
F1 = 3559.055 N

17. A 500 rpm shaft is fitted with a 30 inches diameter pulley weighing 250 lb. This pulley
delivers 35 hp to a load. The shaft is also fitted with a 24 in pitch diameter gear weighing
200 lb. This gear delivers 25 hp to a load. Assume that the tension of the tight side of
the belt is twice that on the slack side of the belt, determine the concentrated load
produces on the shaft by the pulley and the gear in lb.
 Solution:

18. An open belt drive connects a 450 mm driving pulley to another driven pulley 1000 mm
in diameter. The belt is 300 mm wide and 10 mm thick. The coefficient of friction of the
belt drive is 0.3 and the mass of the belt is 2.8 kg/m of the belt length. Other data are:
A. Center distance between shaft = 4 meters
B. Maximum allowable tensile stress on the belt = 1,500 kPa
C. Speed of the driving pulley = 900 rpm

Determine the maximum power that can be transmitted in KW.

Solution:

Sw = working stress
14.7 psi
101.325 kpa
= 1500 kpa x =217.52 psi

]
[
bt= 550h e𝐹𝝷
𝐹𝝷
p
2
v(S−12p v /g) e −1
(12ρv /g) = [12(0.034)(69.57)2/ (32.2)] = 60.836 psi
2

][ ]
e0.3(3)
(300/25.4)(10/25.4) =[
550hp

69.573(217.52−60 e0.3(3)−1
836)
Hp =54.7 = 40.8 KW

19. An electric motor running at 1200 rpm drives a punch press shaft at 200 rpm by means
of a 130 mm wide and 8 mm thick belt. When the clutch is engaged the belt slips. To
correct this condition, an idler pulley was installed to increase the angle of contact but
the same belt and pulley were used. The original contact angle on the 200 mm motor
pulley is 160 degrees. The original contact ratio 2.4 and the net tension is 12 N/mm of
the belt width. If an increase in transmission capacity of 20% will prevent slippage
determine the new angle of contact.

Solution:
efθ = f(160)(π/180) = 2.4

F1
f= 0.313

2.4
F1 - = 1.56
F1 = 2.674 KN
T = 1.20 T = 1.20(0.156) = 0.1872 KN-m
(F1- F2) r = T’
(2.674 – F2)(0.100) = 0.1872
 F2 = 0.802 KN
F1 fθ
F2 = e
2.674
0.802
= e0.3135 θ
Θ = 3.841 rad x 180/π = 2200

20. An ammonia compressor is driven by a 20KW. The compressor and the motor RPM are
360 and 1750, respectively. The small sheave has a pitch diameter of 152.4 mm. If the
belt to be use is standard C-120, determine the center distance between sheaves.

SOLUTION:

21. An air compressor is driven by a 7.5 HP electric motor, with a speed of 1750 rpm and a
standard A-60V- belts. The pitch diameter of the small sheave is 110 mm and the larger
sheave is 200 mm. Service factor is 1.2. Determine the arc of contact.

Solution:

200+110
b = 4L – 6.28(D + d)

25.
= 4(60) – 6.28( )
b = 163.35 in 4
√ 200−110 2
2
163.35+ (163.35) − )
25.4
c= 32(

16
c = 20.34 in
200−110
(
θ = 180 – 25.4
)(60)
20.34
θ=
169.550

22. A pulley 610 mm in diameter transmits 40 KW at 500 rpm. The arc of contact between
the belt pulley is 0.35 and the safe working stress of the belt is 2.1 Mpa. Find the
tangential force at the rim of the pulley in Newton.

V=π D N
Solution:

V = π (0.61)(500/60)
V = 15.92 m/sec
40 = (F1 –F2) (15.92)
F1 – F2 = 2.505 KN = 2505 N
 23. A pulley 610 mm in diameter transmits 40 KW at 500 rpm. The arc of contact between
the belt pulley is 144 degrees, the coefficient of friction between belt and pulley is 0.35
and the safe working stress of the belt is 2.1 Mpa. What is the effective belt pull in
Newton.
Solution:
F1 = 2505 N

24. A reciprocating ammonia compressor is driven by a squirrel cage induction motor rated
15HP at 1750 rpm, across the line starting motor pulley 203.2 mm diameter, compressor
pulley 406.4 mm diameter. find the width of belt.

25. A pulley 610 mm in diameter transmits 37 KW at 600 rpm. The arc of contact between
the belt and pulley is 144 degrees, the coefficient of friction between the belt and pulley
is o.35 , and the safe working stress of the belt is 2.1 Mpa . find: A. The tangential force
at the rim of the pulley. B. The effective belt pull. C. The width of the belt used if its
thickness is 6 mm.

26. A nylon- core flat belt has an elastomeric envelop; is 200 mm wide, and transmits 60 KW
at a belt speed of 25 m/s. The belt has a mass of 2 kg/m of the belt length. The belt is
used in crossed configuration to connect a 300 mm diameter driving pulley to a 900 mm
diameter driven pulley at a shaft spacing of 6m. A. Calculate the belt length and the
angles of wrap. B. Compute the belt tensions based on a coefficient of friction of 0.33.
27. The tension ratio for a belt in a pulley with a 200o angle and frictional coefficient of 0.3
is:
Solution:
F1
F2
= e f θ = e (0.3)(200)(π/180) =2.85

BEARINGS:

1. The main bearing of a one cylinder steam are 152 mm diameter by 280 mm long and
support a load of 4400 kg. Find the bearing.

Solution:
 9.81
4400 x
1000
S =(0.152)(0.28)
S =507.10 kPa

2. A bearing 150 m diameter and 300 mm long support s a load of 5000 kg. if coefficient of
friction is 0.18, find the torque to rotate the shaft.
Solution:
F = 5000 x 9.81/ 1000
F = 49 05 KN
T = (49.05)(0.18)(0.150/2)
T = 662 N-m

3. A bearing journal rotates at 460 rpm is use to support a load of 50 KN. It has a diameter
of 20 cm and length of 40 cm. find the friction loss in kw per bearing. Use f = 0.12

Solution:
T = (50)(0.12)(0.20/60)

P = 2 π (0.6)(460/60)
T = 0.6 N-m

P = 14.45 KW

4. A bearing has a per unit load of 550 kpa. The load on bearing is 20 KN and it has a
diameter ratio of 0.0012. if the diametric clearance is 0.120 mm, find the length of
journal.
Solution:
Dc = Cd /D
0.0012 = 0.120/D
D = 100 mm
F
LD
P=
20
0.1L
550 =
L = 0.36363 m

5. A bearing whose shaft rotates at 508 rpm, has a friction loss of 15 kw. The bearing load
is 30 KN and friction of 0.14. Find the bearing diameter.
 15 = 2 π T (500/60)
SOLUTION:

T =0.2865 KN-m
0.2865 = (30)(0.14)d
D =136.42 mm

6. A shaft revolving at 1740 rpm is supported by bearing with a length of 150 mm and
diameter of 64 mm. if the load is high and SAE oil no.20(u = 2.4 x 10-0 reyns) is used and
diametric clearance is 0.136mm, find the power loss due to friction.

Solution:

7. A shaft is supported by in 3 diameter and 4 in long bearing that uses oil with viscosity of
2.5x 10-6 reyns. The bearing has a diametric clearance of 0.008 in. At what speed should
the shaft rotates so that the friction power is limited only by 1 hp.
8. A 76 mm bearing using oil with an absolute viscosity of 0.70 poise running at 500rpm
gives satisfactory operation with a bearing pressure of 14 kg/cm2. The bearing clearance
is 0.127 mm. Determine the unit pressure at which the bearing should operate if the
speed is changed to 600 rpm.
Solution:

S = p (Cd )
μn D

S1 = S2
2 2
( ) ( )
0.70 x 500 0.70 x 600 76
76 =
14 0.127 p2 0.127
P2 = 16.8 kg/cm2

9. A 76 mm bearing using oil with an absolute viscosity of 0.70 poise running at 1500 rpm
gives satisfactory operation with a bearing pressure of 14 kg/cm 2. The bearing
clearance is 0.127 mm. if the bearing is given a total clearance of 0.076 mm and speed
to 600 rpm, what change should be made in oil?

Solution:
When the total clearance is change to 0.076:
2 2
( ) ( )
0.70 x 500 μ2 0.70 x 600 76
76 =
14 0.127 16.8 0.076
μ = 0.251 poise
10. A journal bearing 50 mm in diameter and 25 mm long supports a radial load of 1500 kg.
if the coefficient of bearing friction is 0.01 and the journal rotates at 900 rpm, find the
horsepower loss in the bearing.

Solution:
F = 1500 x 9.81/1000
F = 14.715 KN
T = 14.715(0.01)(0.005/2)

P = 2 π (3.679 x 10-4)(900/60)
T = 3.679 x 10-4 KN-m

P = 0.03467 KW

11. A bearing 2.085 inches in diameter and 1.762 in long supports a journal running at 1200
rpm. It operates satisfactorily with a diametric clearance of 0.0028 in and a total load of
1,400 lbs. At 160℉ operating temperature of the oil film, the bearing modulus ZN/P
was found to be 16.48. Determine the bearing stress.

Solution:
1400
Ρ=
(2.085)
= 381.07 psi
(1.762)

12. A bearing 2.085 inches in diameter and 1.762 in long supports a journal running at 1200
rpm. It operates satisfactorily with diametric clearance of 0.0028 in and a total radial
load of 1,400 lbs. At 160℉ operating temperature of the oil film, the bearing modulus
ZN/P was found to be 16.48. Determine the viscosity centipoises

Solution:

vn
BM = 16.48

p
BM =
v(1200)
381.07
16.48 =
v = 5.233

13. The main bearings of an engine are 152 mm diameter and 280 mm long and supports a
load of 4400 kg midway between the bearings. Find the bearing pressure in Kpa.

Solution:
 9.81
4400 x
1000
(0.152)(0.28)
S= = 507.10 kPa

14. A sleeve bearing has an outside diameter of 38.1 mm and length of 50 mm, the wall
thickness is 3/16 inch. The bearing is subjected to a radial load of 450 lb. Determine the
bearing pressure.

Solution:
ro = 1.5/2 = 0.75 in
ri = ro – t = 0.75 – 3/16 = 0.5625 in
Di = 2 ri = 2(0.5625)
Di = 1.125 in
450
P= F/A = 2(1.125 = 200 psi
)

15. A 2.5° diameter by 2 in long journal bearing is to carry a 5500 lb load at 3600 rpm
using SAE 40 lube oil at 200℉ through a single hole at 25 psi. Compute the bearing
pressure.

5500
(2.5)
P=
= 1100 psi
(2)

16. A journal bearing with diameter of 76.2 mm is subjected to a load of 4900N while
running at 200 rpm. If its coefficient of friction is 0.02 and L/D= 2.5, find its projected
area in mm2.
Solution:
L/D= 2.5
L =2.5D
L =2.5(76.2)
L = 190.5 mm
A = LD
A = (190.5) (76.2)
A = 14,516.1 mm
ROLLER CHAINS:

1. A chain and sprocket has 18 teeth with chain pitch of ½ in. Find the pitch diameter
of sprocket.
Solution:
900
2/3
P=( )
N
900 2/3
0.5 = ( )
N
N =76367.5
V=ptN
V = (0.5) (18) (76367.5)

687307.77 = π D
V =687307.77

(76367.5) D =
2.879 in

2. A chain and sprocket has 24 teeth with chain pitch of ½ in. If the sprocket turns at 600
rpm, find the speed of chain.
Solution:
V=ptN
V = (0.5) (24) (600)
V =7200 in/min x ft
12in
V = 601.72 fpm
3. A chain and sprocket has a pitch diameter of 9.56 in and a pitch of ¾ in. How many teeth
are there in sprocket?

p
Solution:
180
sin(
D=
)
T
3/4
180
sin( )
9.56 =
T
T = 40 teeth

4. A chain and sprocket has pitch diameter of 28.654 in and there are 90 teeth available.
Find the pitch of the chain.
Solution:
 p
18
sin(
28.654 =
0
)
90
P =1.00 in

5. A fan requires at least 4.5 hp to deliver 18,000 CFM air running at 320 rpm. For a service
factor of 1.15, find the designed horsepower of the sprocket.
Solution:
P= 4.5hp (1.15)
P =5.175 hp

6. A 20-tooth driving sprocket that rotates at 600 rpm and pitch chain of ½ in drives a
driven sprocket with a speed of 200 rpm. Find the diameter of the driven sprocket.
Solution:
1/2
D= 180
sin( )
20
D = 3.19622 in
N1D1 =N2D2
(3.19622)(600) = Ds(200)
Ds = 9.55 in

7. A chain with speed 800 fpm has driving sprocket turns at 900 rpm. If the pitch of chain is
¾ in, find the number of teeth of driving sprocket.

V=π D N
Solution:

800 = π D (900)
D =0.2829 ft
3/4
(0.2829)(12) 180
= sin )
( T

T =14
teeth

8. Find the standard distance between sprocket having 4 in and 16 in diameter.

Solution:
d
2
C=D+
4
2
C =16 + = 18 in

9. An 18 teeth sprocket driving with 98 teeth sprocket at a center distance of 34 pitches.

Find the chain length in pitches.
 Solution:
(T−t)
L = 2C + T+ 2
t
40C
+
2
98+18 (98−18)2
2 40(34)
L = 2(34) + +

L =131 pitches

10. A sprocket with 8 in diameter and 1 pitch of ¾ in drives another sprocket at standard
center distance of 48 pitches. Find the diameter of larger sprocket.
Solution:
d
(D+ )
2
p
C=
(D+8/2)
3/4
48 =
D =32 in

11. A driving sprocket with 12 teeth drives another sprocket with 48 teeth by means of a
chain having a pitch of ½ in. If chain length is 72 in, find the center distance between
sprockets.

C = (2L − T − t + √(2L − T − 0.8(T − t)2)

p
Solution:

− t)3
√ ( ) 3 2
8
C = 1/2
[2(72) - 48-12 + [2 − 48 − − 0.80(48 − 12)
8
72 12]
C =28.36 in

12. A certain farm equipment which requires 2200 N-m torque at 500 rpm has diesel engine
to operate at 1500 rpm as its prime mover. A No. 60 roller chain with total length of 60
pitches and a small sprocket of 23 teeth are to be used with operating temperature to
remain constant at 45℃. Determine the number of teeth of larger sprocket.
Solution:
RC60
V = p t N = (6/8)(23)(1500)

V=π D N
V = 431.25

431.25 = π D
(500/60) D =
16.473 in
6/8
16.473 180
= sin )
( T

T = 69 teeth
13. A pump that operates ahead of 30 m is use to deliver 80 liters per second of water at an
efficiency of 85%. The pump is driven by a diesel engine by means of chain and sprocket
with a service factor of 1.3. If horsepower per strand is 25, find the approximate number
of strands needed.
14. A 3-strand chain and sprocket turns at 600rpm has a service factor of 1.4 If 14 hp/strand
chain is to be used, find the torque can the chain and sprocket be deliver?
Solution:
2π T
(14)(3) = (600)
33000
T= 367.65 ft-lb (1.4)
T= 262.61 ft-lb
15. A certain farm equipment which requires 2200 Newton meter torque at 500 rpm has a
diesel engine to operate at 1500 rpm as its prime mover. A No. 60 roller chain with a

operating temperature to remain at 45 degrees ℃. Determine the no. of teeth of the

total length of 60 pitches and a small sprocket with 23 teeth are to be used with an

larger sprocket.
Solution:
RC60
V = p t N = (6/8)(23)(1500)

V=π D N
V = 431.25

431.25 = π D
(500/60) D =
16.473 in
6/8
16.473 = 180
sin )
( T

T = 69 teeth

16. A 4 inches diameter shaft is driven at 3600 rpm by a 400HP motor. The shaft drives a 48
inches diameter chin sprocket and having an output efficiency of 85%. The output force
of the driving sprocket and the output of the driven sprocket are:

V= π D N= π (4)(3600)
Solution:

V= 45238.9
P= FV
(400)(33000)=F (45238.9)
F=291.78 lb
Ds Ns = Dc Nc
 4(3600) =48 Nc

V= π (48) (300)
Nc =300 rpm

(291.78)(45238.94)
V=45238.94

33000
P= =340hp
17. In chain of ordinary proportions the term RC 40 means a pitch of .
Solution:

=
RC40= 𝟒" 𝟏
𝟖 𝟐 in
18. The chain speed of an RC 80 chain on a 21-tooth sprocket turning at 600 rpm is
.
Solution:

RC80= 8
8
P 1
= 6.709
=
180
D=
18
0
sin( )
sin( )
600 21
V= π (6.709)( )
T

60
1m
)
39.37
=210.769 in/sec (

V=5.353 m/s
19. What maximum chain number can be used if the small sprocket is to run at 1200

900
rpm Solution:
0 )
P=( )2/3=( 90 2/3 =0.753 in
N 1200
RC60
20. A motor transmits 40 hp to an air conditioning apparatus by means of a roller chain. The
motor runs at 720 rpm. The pitch diameter of the sprocket on the motor should not
exceed 6 ¾ in. If the drive calls for a service factor of 1.2 and a chain pitch of 1 in,
determine the number of teeth of the small sprocket.

P
Solution:
D= 180
sin( )
T

= sin(180)
3 1
64
T=21 teeth
21. A 10-hp engine with a speed of 1200 rpm is used to drive a blower with a velocity ratio
of 3:1. The pitch diameter of the driven sprocket is 85 mm and the center distance
 between the sprocket is 260 mm. Use an RC 40 drive and service factor of 1.2. Find the
number of strands needed.

hp
Solution:

stran = 0.004(T) (N ) (P)

1.06 1 0.9 (3-0o.07p)

= 14
= 0.004(20.95)1.06(1200)0.9(0.5)(3-0.07)(0.5)

𝐡
𝐩
𝐬𝐭𝐫𝐚𝐧𝐝𝐬
22. A silent chain SC 4 is used for a design hp of 12. What chain width is needed if the 21-
tooth driving sprocket runs at 1200 rpm?
23. With a sprocket bore of 2”, the minimum number of sprocket teeth for RC40 is .

Solution:
4d
p
Tmin= +6

RC40
4(2)
= 4 +6
8

Tmin=1200 rpm

24. A 10-hp engine with a speed of 1200 rpm is used to drive a blower with a velocity ratio
of 3:1. The pitch diameter of the driven sprocket is 85 mm and the center distance
between the sprocket is 260 mm. Use an RC 40 drive and a service factor of 1.2 Find the
number of teeth of the driving sprocket:
Solution:
RC40
4 4
8
8 180
P= (0.885) (39.37)=
sin(
)
T
𝐓=𝟐𝟎.𝟗
𝐓=𝟐𝟏 𝐭𝐞𝐞𝐭𝐡
PROBLEM SET # 4

BOLTS AND POWER SCREW

1. Determine the permissible working stress of a UNC that has a stress

area of 0.606 in2 if material used is carbon steel.

Sw = C Ar0.418

For carbon steel, C

=5000 Sw = 5000(0.606

in2)0.418

Sw = 4055.49 psi

2. The stress area of UNC bolt is 0.763 in2, if material used is carbon steel,
determine the applied load in the bolt.

Fa = C Ar1.418

For carbon steel, C

=5000 Fa =

5000(0.763 in2)1.418

Sw = 3407.138 lbs

3. Compute the working strength of 1.5 in. bolt which is screwed up

tightly in packed joint when the allowance working stress is 13,000 psi.

W = St ( 0.55d2 – 0.25d )

W = 13,000 [0.55(1.5)2 –

0.25(1.5)] W = 11,212.5 lbs

4. Determine the diameter of bolt needed to tightly packed the joint of

bolt working strength is 90.76 KN and working stress 82.71 Mpa.

W = St ( 0.55d2 – 0.25d )

20,400.18 = 11,999.378 [0.55(d)2 – 0.25(d)]

d = 2 in.
5. A 12cm x 16cm air compressor have 5 bolts in cylinder head with yield
stress of 440 Mpa. If the bolt stress area is 0.13 in 2, determine the
maximum pressure inside the cylinder.

Sy= 440,000 (14.7/101.325)

Sy= 63,834.196 psi

𝑆𝑦(𝐴)3/2
6
Fe
=
63,834.196
(0.13)3/2 = 498.67 lbs
6
Fe
=

F = 498.67 lbs (5) = 2493.37 lbs

2493.37 𝑙𝑏𝑠
P = F/A = π 12 𝑐𝑚 = 671.97
psi )2
( )(
4 2.54 𝑐𝑚

6. The cylinder head of ammonia compressor has core gasket as of 80 cm2

and flange pressure of 90 kg/cm2. Determine the torque applied on the
bolt if the nominal diameter is 3/5 inch and there are 5 bolts.

Initial tension = 90(80)

= 7200 kg x 2.205 lbs/kg = 15876

lbs Tension per bolt = 15876 lbs ÷ 5 bolts

= 3172.2 lbs/bolt

Solving for the torque applied per

bolt, T

T = 0.2 FiD

T = (0.2)(3175.2)

(3/5) T = 381.034

in-lb

7. The total power to turn the power screw is 50 N-m. If the linear speed of
the screw 7 ft/min and lead of 8 mm, find the horsepower input of the
power screw.
V =NL
1 1
25.4 12
V = N ( 8m x x )

N = 304.8 rpm

Pi = 2𝜋TN

Pi = 2𝜋(0.050)(304.8/60)
1 𝑘𝑤
Pi = 1.5959 kw x
𝑜.746 𝐻𝑝 = 2.139 Hp

8. A single square thread power screw has lead of 6 mm and mean diameter
of 34 mm. If it is used to lift a load of 26 KN and coefficient of friction of
thread is 0.15, determine the torque required to turn the screw.

Solving for lead angle;

𝐿 6
𝜋𝐷 𝜋(34
Tan x = = 0.05617

𝑚 )
=

T = torque required

𝑊𝐷𝑚 tan 𝑥+𝑓

T= 2 1−𝑓
( tan 𝑥 )

26(0.034) 0.05617+0.16
T= 2 1−0.15
( (0.05617) )

T = 0.0919 KN-m = 91.90 N-m

9. An acme thread power screw that has a mean diameter of 25 mm and

pitch of 5 mm is used to lift a load of 5 kg. If friction on thread is 0.1,
determine the torque required to turn the screw.

Tan x = 𝐿 5
𝜋𝐷𝑚 =𝜋(25 = 0.0636
)

For ACME thread, 𝜃 =

14.5o W = 500 x 0.00981

= 4.905 KN
WDm = 4.905(0.025) = 0.12265 KN-m

0.12
2
[ )+(0.10)
(cos 14.5)
(0.0636
2
]
T=

cos 14.5 − 0.10(0.0636)

T = 0.01030 KN-m = 10.30 N-m

10. A double square thread power screw has a mean radius of 80 mm and
a pitch of 10 mm is used to lift a load of 80 KN. If friction of screw is
0.13 and collar torque is 20% of input torque, determine the input
torque.

Tan x = 𝐿
𝜋𝐷𝑚

L =2p

𝑊𝐷𝑚 tan 𝑥+𝑓

T= 2 1−𝑓
( tan 𝑥 )
Dm = 2r

WDm = 80(0.80 x 2) = 12.8 KN-m

12.
8 0.03979 𝑥 0.13
(1−0.13 (0.03979) )
2
T=

T = 1.092 KN-m

Solving for total

torque, TT TT = T + TC

Tc = 0.20TT

TT = T + 0.20TT

0.8TT = 1.092

TT = 1.365373 KN-

m TT =1365.375

N-m
11. The root diameter of a double thread power screw is 0.55 in. The
screw has a pitch of 0.2 in. Determine the major diameter.

L = 2p = 2(0.2) = 0.4

For square

thread; Do = Di

+ L/2

Do = 0.55 + (0.4/2)

Do = 0.75 in

12. A power screw consumes 6 hp in raising a 2800 lb weight at the rate

of 30 ft/min. Determine the efficiency of the screw.

Hp = power output

𝑊𝑥 30 𝑥
2800
𝑣𝑒𝑙
Po = 2.545 Hp
33,000
33,00
= =

2.545
𝑃𝑜
eff = x 100% =
x 100% = 42.42%

𝑃𝑖 6
13. A square thread power screw has a pitch diameter of 1.5 and a lead
of 1 in. Neglecting collar friction, determine the coefficient of friction
for threads if thread efficiency is 63.62%.

𝐿 1
𝜋𝐷 𝜋(1.5
Tan x = = 0.2122

𝑚 )
=

Using the formula of eff,

tan 𝑥 (1−𝑓 tan 𝑥)

𝑓𝑐𝐷𝑐
tan 𝑥 + 𝑓 + ( )(1−𝑓 𝑡𝑎𝑛𝑥)
eff =

𝐷𝑚

Since collar friction is neglected, 𝑓𝑐

0 Then the quantity, ( ) (1 − 𝑓

𝑓𝑐𝐷𝑐
=
𝑡𝑎𝑛𝑥) = 0
𝐷𝑚

0.2022[ 1−𝑓(0.2122)]
0 2122+𝑓+0
0.6362
=
0.2122 + 𝑓 = 0.333 – 0.07𝑓

𝑓 = 0.133

14. A square thread power screw has an efficiency of 70% when the friction of
thread is
0.10 and collar friction is negligible. Determine the lead angle.
tan 𝑥 (1−𝑓 tan 𝑥)
𝑓𝑐𝐷𝑐
tan 𝑥 + 𝑓 + ( )(1−𝑓 𝑡𝑎𝑛𝑥)
eff =
𝐷𝑚

tan 𝑥[ 1−0.10(tan 𝑥)]

0.70 = tan 𝑥+0.10+0
0.10 tan2x + 0.07 = tan x – 0.10 tan2x

0.10 tan2x – 0.3 tan x +0.07

−(0.3)±√(0.3)2−4(0.10)
x=
(0.07)
Using quadratic formula; tan

2(0.10)

tan x =0.255

x = tan-1

0.255 x =

14.30

15. A 12cm x 16cm air compressor is operating with maximum pressure

of 10 kg/cm2. There are 5 bolts which held the cylinder head to the
compressor. What is the maximum load per bolt in KN?

Load per bolt = ?

𝜋
4
Tension = [ (12)2 ](10 kg/cm2)

Tension = 1130.973 kg x 2.205

lb/kg Tension = 2493.796 lbs

2493.796
𝑙𝑏𝑠
Tension per
= 498.76 lbs/bolt
5 𝑏𝑜𝑙𝑡𝑠
bolt =
1 𝑘𝑔
 2.205
Tension per bolt = 498.76 x 0.00981 KN/kg
𝑙𝑏
lbs x
Tension per bolt = 2.219 KN

16. A double thread screw driven by a motor at400 rpm raises the load
of 900 kg at a speed of 10 m/min. the screw has pitch diameter of 36
mm. determine the lead angle.

V=NL

10 = 400

L = 0.025 m = 25 mm

𝑙𝑒𝑎𝑑
tan x = =25 = 0.221
𝜋 𝐷𝑚 𝜋 (36)

x = 12.46°

17. Find the horsepower lost when a collar is loaded with 1000 lb, rotates
at 25 rpm and has a coefficient of friction of 0.15. The outside diameter
of the collar is 4 in and the inside diameter is 2 in.

𝑓𝑐 𝑊 (𝑟𝑜−𝑟𝑖)
Tc = 2

0.15 (1000 𝑙𝑏)

Tc (2+1) = 18.75 ft-lb
2(12)
=

2𝜋𝑇𝑁 2𝜋(18.75)
(25)
P = = 0.08925 Hp
33,000 33,000
=

18. Compute the working strength of 1” bolt which is screwed up tightly in

a packed joint when the allowable working stress is 13,000 psi.

W = St ( 0.55d2 – 0.25d )

W = 13,000 [0.55(1)2 – 0.25(1)]

W = 3900 lbs
19. What is the working strength of 2” bolt which is screwed up tightly in
a packed joint when the allowable working stress is 12,000 psi.

W = St ( 0.55d2 – 0.25d )

W = 13,000 [0.55(2)2 –

0.25(2)] W = 20,400 lbs

20. What is the frictional HP acting on the collar loaded with 100 kg
weight? The collar has an outside diameter of 100 mm and an internal
diameter of 40 mm. The collar rotates at 1000 rpm and the coefficient
of friction between the collar and the pivot surface is 0.15.
(𝑟𝑜−𝑟𝑖)
= fW[ ]
2
T=
fWrf

[
(0.05−0.02)
]
T = 0.15(1000)
(0.0091) 2

T = 0.00515 KN-

m P = 2𝜋TN
P = 2𝜋(0.000515)(1000/60)

1 𝐻𝑝
0.746
P = 0.5393
𝐾𝑊
𝐾𝑊 x P =

0.7229 Hp

21. If the pitch of a screw is 2/9, find the thread per inch.

Pitch = 2/9

TPI =?

Pitch 1
𝑇𝑃
𝐼
=

2 1
9 =𝑇𝑃𝐼
TPI = 4.5

22. An eyebolt is lifting a block weighing 350 lbs. The eye bolt is of SAE
1040 material with Su=67 ksi and SY= 55 ksi what is the stress area (in
inch square) of the bolt if it is under the unified coarse series thread?

Fc = applied load

3/2
Fc = 𝑆𝑦𝐴𝑠
6

As = Stress area

2/3
As = 6 ]
[
𝐹𝑐
𝑆
𝑦

2
6 𝑥 350 3
A s= [ ] = 0.1134

55,000
2
in

23. Compute how many 3/8 inch diameter of set screws required to
transmit 3 Hp at a shaft speed of 1000 rpm. The shaft diameter is 1
inch.

𝐷𝑁𝑑3/2
50
Hp =

1(1000)𝑑3/2
50
3=

d = 0.4383 in

0.438
no. of set 3 = 1.7
3/8
screws =

therefore no. of set screws = 2

24. For a bolted connection, specifications suggested that a high grade

material 13mm bolt be tightened to an initial tension of 55,000 N. What
is the appropriate tightening torque?
T = 0.2 Fi D

T = 0.2(55,000 N)

(0.013) T = 134 N-m

25. An eye bolt is lifting 500 lbs weight. The Su=70 ksi and SY=58 ksi. What
is the stress area of the bolt?

3/2
Fc = 𝑆𝑦𝐴𝑠
6

As = Stress area

2/3
As = 6 ]
[
𝐹𝑐
𝑆
𝑦

2
6 𝑥 500 3
A = [ ] = 0.13882

58,000
s in2

26. Find the Hp required to drive a power screw lifting a load of 4000 lbs. A
2.5 inch double square thread with two threads per inch is to be used.
The frictional radius of the collar is 2 inches and the coefficients of
friction are 0.10 for the threads and 0.15 for the collar. The velocity of
the nut is 10 ft/min.

Lead = 2p

Lead = 2(1/2) =

1 in For square

thread:

Dm = Do – 0.5p = 2.5 – 0.5(1/2) = 2.25 in

𝑙𝑒𝑎𝑑 1
tan x = =
𝜋 𝐷𝑚 𝜋 (2.28)
= 0.1415

x = tan-1 0.1415 = 8.05o

 [
tan 𝑥+𝑓
𝑊𝐷
]
T =𝑚
2
1−𝑓 tan 𝑥
 [ 0.145 +
(4000)(2.25)
] = 1102.35 in-lb
0.10
T=
2 1−0.10 (0.1415)

Dc = 2 rc = 2 (2) = 4 in

𝑓𝑐 𝑊 (𝑟𝑜−𝑟𝑖)
Tc = 2

0.15 (4000 𝑙𝑏)

Tc (4) = 1200 in-lb
2
=

TT = T + Tc = 1102.35 + 1200 = 2302.35 in-lb =

191.86 ft-lb V =L N

10 x 12 = (1)N

N = 120 rpm
2𝜋𝑇𝑁 2𝜋(191.86)
(120)
P= = 4.38 Hp
33,000 33,000
=

27. The cylinder of a compressor ha core gasket area of 80 cm 2 and flange

pressure of 90 kg/ cm2. Determine the torque on the bolt if nominal
diameter of bolt used is ¾ in and there are 5 bolts.

Total initial tension = 90(80) = 7200 kg x 2.205 lbs =

15,876 lbs Initial tension per bolt = 15,876 lbs/5 bolts

= 3175.2 lbs

Solving for the initial torque applied per bolt:

T = 0.2 Fi D

T = 0.2(3175.2 lbs)(¾ in)

T = 476.28 in-lbs

28. A double thread Acme screw driven by a motor at 400 rmp raises the
attached load of 900 kg at a speed of 10 meters per minute. The screw
has a pitch diameter of 36
mm. the coefficient of friction is 0.15. The friction torque on the thrust
bearing of the motor is taken as 20% of the total torque input. Determine
the motor power required to operate the screw.
V=NL

10 = 400

L = 0.025 m = 25 mm

𝑙𝑒𝑎𝑑
tan x = 25 = 0.221
=
𝜋 𝐷𝑚 𝜋 (36)

x = 12.46°

Torque required to turn the

screw, Ts For ACME thread, 𝜃 =

14.5o

[ 14.5 − 𝑓 tan 𝑥 ]
𝑊𝐷𝑚 (cos 14.5) (tan 𝑥)+𝑓
T = 2 cos
s

(900)
[ )+(0.15)
(cos 14.5)
(0.036)
(0.221
]
Ts
2
=
cos 14.5 − 0.15(0.221)

Ts = 6.31 kg-m

T = Ts + Tc = torque

input T = Ts + 0.20T

T = 6.31 + 0.20(6.31)

T = 7.572 kg-m x (0.00981) = 0.07428 KN-m

Power = 2 𝜋 T N = 2 𝜋 (0.07428 KN-m) (400/60)

Power = 3.11 KW

29. A 12cm x 16cm air compressor has 5 bolts on cylinder head with yield
stress of 440 Mpa. If the bolt stress area is 0.15 in2, determine the
maximum pressure inside the cylinder.
Sy = 440,000 x

(14.7/101.325) Sy =

63834.196 psi
 𝑆𝑦(𝐴)3/
2
6
Fe=

63,834.196 (0.15)3/2
6
Fe
=

Fe = 618.07 lbs

F = 618.07 lbs (5) =

3090.36 lbs P = F/A

3090.36 𝑙𝑏𝑠
12 𝑐𝑚
( 2
P=
π
( ) )
4 2.54 𝑐𝑚

P = 176.29 psi

30. The root diameter of a double square thread is 0.55 inch. The screw has a
pitch of
0.20 inch. Find the outside diameter and the number of threads per inch.

Dm = Di + ½(p) = 0.55 + ½(0.2) = 0.65 in

Do = Dm + (Dm – Di) = 0.65 + (0.65 + 0.55)

= 0.76 in TPI = 1/p = 1/0.2 = 5

31. A single square thread power screw is to raised load of 70 KN. The
screw has a major diameter of 36 mm and a pitch of 6 mm. The
coefficient of thread friction and collar friction are 0.13 and 0.10
respectively. If the collar mean diameter is 90 mm and the screw turns
at 60 rpm, find the combined efficiency of screw and collar.

lead = p = 6 mm
𝑙𝑒𝑎𝑑
6
tan x = = = 0.053

𝜋 𝐷𝑚 𝜋 (36)

tan 𝑥 (1−𝑓 tan 𝑥)

𝑓𝑐𝐷𝑐
tan 𝑥 + 𝑓 + ( )(1−𝑓 𝑡𝑎𝑛𝑥)
eff =
𝐷𝑚

=
eff
0.0 53 [1−0.13(0.053)]
0.10(90)
0.053 +0.13 + [ ]+ [1−0.13(0.053)]
36
eff =

32. A single thread trapezoidal metric thread has a pitch of 4 mm, and a
mean diameter of 18 mm. It is used as a translation screw in conjunction
with collar having an outside diameter of 37 mm and an inside diameter
of 27 mm. Find the required torque in N-m to raise a load of 400 kg if the
coefficient of friction is 0.3 for both thread and collar.

solving for the collar torque of the screw:

[
𝑊𝐷𝑚 tan 𝑥+𝑓
Tc =

2 ]
1−𝑓 tan 𝑥 37 27
+
0.3(400
)
[
(0.3)(400)
2 ]
( )
2
2 1−0.10 (0.02274)
Tc =

Tc = 1920 kg-mm

For ACME of trapezoidal thread, 𝜃 = 14.5o

𝑙𝑒𝑎𝑑
4
tan x = = = 0.0707

𝜋 𝐷𝑚 𝜋 (18)

[ 14.5 − 𝑓 tan 𝑥 ]
𝑊𝐷𝑚 (cos 14.5) (tan 𝑥)+𝑓
T = 2 cos
s

(400) (cos 14.5)

(0.0707)+(0.30)
Ts (18) [ ]
=
2 cos 14.5 − 0.30(0.0707)

Ts = 1401 kg-mm

Solving the total

torque:

T = Ts + Tc = torque input

T = 1401 kg-mm + 1920 kg-mm

T = 3321 kg-mm x 9.81 N/kg ÷ 1000

mm/m T = 32.58 N-m

33. The collar of the capstan has an outside diameter of 8 in and an inside
diameter of 6 in. The total load is 2000 lbs. If the coefficient of friction is
0.10 and lead is 0.5 in, what is the torque required to turn the screw?

L = p = 0.5
𝐷𝑖+ 𝐷𝑜 6+8

2 2
Dm = = = 7 in

Solving for the lead angle of the screw:

𝑙𝑒𝑎𝑑
0.5 = 0.02274
𝜋 𝐷𝑚 𝜋 (7)
tan x = =

solving for the torque required to turn the screw:

[
tan 𝑥+𝑓
𝑊𝐷
]
=𝑚
2
T

1−𝑓 tan 𝑥

(2000)
(7)
]
0.02274 + 0.10
1−0.10 (0.02274)
[
T=

T = 861.138 in-lb

34. Calculate the bolt area in mm2 of each of 20-bolts used to fasten the
hemispherical joints of a 580 mm pressure vessel with an internal
pressure of 10 Mpa and a material strength of 80 Mpa and a bolt
strength of 150 Mpa.

35. A double-thread Acme-form power screw of 50 mm major diameter is

used. The nut makes one turn per cm of axial travel. A force of 60 kg is
applied at the end of750 mm wrench used on the nut. The mean
diameter of the collar is 90 mm. If the coefficient of friction at the thread
and the collar are 0.15 and 0.13, respectively. Determine the weight
being lifted.

W =?

1 turn/cm
=
TPI
 1 𝑖𝑛
1
(1𝑐𝑚 𝑥 )
0.3937 𝑐𝑚

1 1
𝑇𝑃 2.5
Pithc = = 0.3937
= 𝐼 4
L = 2p = 2(0.3937) = 0.7874 in
1 1
Dm = Do - p = 1.968 in - (0.3937)
2 2

Dm = 1.77115 in
𝐿 0.7874
𝜋𝐷 𝜋(1.7711
Tan x =

𝑚 5)
=

x = 8.0545
tan 𝑥 (𝑐𝑜𝑠𝜃−𝑓 𝑠𝑖𝑛𝑥)
𝑓𝑐𝐷𝑐
𝑡𝑎𝑛𝑥 cosθ + 𝑓 𝑐𝑜𝑠𝑥+[ ](𝑐𝑜𝑠𝜃−𝑓 𝑠𝑖𝑛𝑥)
eff =
𝐷𝑚

(tan 8.0545)[cos 14.4−0.15(sin 8.0545)

0.13(3,54)
(tan 8.0545)(𝑐𝑜𝑠14.4) + 0.15(𝑐𝑜𝑠8.0545)+[ ]
eff
[(𝑐𝑜𝑠14.4)−0.15(𝑠𝑖𝑛8.0545)]
=
1.77115

eff = 0.276766

𝑊 𝐿𝑒𝑎𝑑
𝐹(2𝜋𝑟)
eff
=

𝑊 (0.7874)
𝑙𝑏𝑠
(60 𝑘𝑔 𝑥 2.205 )[(2𝜋)
0.276766
(29.527)]
=
𝑘𝑔

x 0.00981𝐾𝑁
W = 8627.3265 x
1 𝑘𝑔
𝑘𝑔
2.205 𝑙𝑏𝑠

W = 38.3827 KN

36. Determine the stress area in in2 of a 1-8UNC bolt.

As = ?

1-8 UNC bolt

𝜋
As = [D- 0.9743(P)]2
4
𝜋
As = [1- 0.9743(0.125)]2
4

As = 0.6057 in2
37. Determine the stress in psi of a 1-4UNC bolt if applied force is 4000 lb.
ST = ?

1 – 4 UNC

D = 1, TPI = 4

1 1
𝑇 4
Pitch = = 0.25

𝑃𝐼
=
𝜋
As = [D- 0.9743(P)]2
4
𝜋
As = [1- 0.9743(0.25)]2 = 0.4494 in2
4

A =𝜋
24
(Dm)
𝜋
0.4494 in2 = (Dm)2
4

Dm = 0.7564 in

W = St [ 0.55d2 – 0.25d ]

4,000 = St [ 0.55(0.7564)2 –

0.25(0.7564) ] St = 31,852.8 psi

38. The outside diameter of a collar is 4 in and the inside diameter is 2 in.
Find the horsepower lost when a collar is loaded with 1000 lbs rotates
at 25 rpm, and has a coefficient of friction of 0.15

𝑓𝑐 𝑊 (𝑟𝑜−𝑟𝑖)
Tc = 2

0.15 (1000 𝑙𝑏)

Tc (2+1) = 18.75 ft-lb
2(12)
=

2𝜋𝑇𝑁 2𝜋(18.75)
(25)
P = = 0.08925 Hp
33,000 33,000
=
PROBLEM SET # 7

BRAKES
1. A brake has a difference in band tension of 4 KN. The drum diameter is 1 meter
and rotating at 300 rpm. Determine the power needed to drive the drum.

T = (F1 - F2)r = (4)0.5 =2 KN-m

P = 2𝜋TN

P = 2 𝜋 (2)

(300/60) P =

62.832 KW

2. In a brake, the tension on tight side is thrice the slack side. If coefficient of
friction is 0.25, find the angle of contact of the band.

F1/F2 = ℯ𝑓∅
𝜋
0.25(∅ 𝑥 )
3/1 = ℯ 180

∅ =251.78 deg.

3. A simple band brake has a 76 cm drum and fitted with a still band 2/5 cm thick
lined with a brake lining having a coefficient of friction of 0.25. The arc of contact
is 245 degrees. The drum is attached to a 60 cm hoisting drum, that sustains a
rope load of 820 kg. Find the tension at the slack side.

Torque = f x r

Torque = 820(60/2) = 24,600 kg-cm = 2,413.26 N-m

𝜋
𝑓(245 𝑥
F1/F2 = 𝑓 )
ℯ ∅ ℯ 180
= 2.912
=

F1 = 2.912

F2 T = (F1 -
F2)r
 F1 - F2 = 6350.7 N

2.912F2 - F2 =

6350.70 F2 =

3321.50 N

4. A steel band has a maximum tensile stress of 55 Mpa and thick of 4 mm, If the
tension in tight side is 6 KN, what width of band should be used?

𝐹1
S =𝑤𝑡

6
𝑤(0.004
55,000

)
=

w = 0.027273

m w = 27.273

mm

5. A band brake has a straight brake arm of 1.5 m and is placed perpendicular to
the diameter bisecting the angle of contact of 270 degrees which is 200 mm
from the end of slack side.

F2 = tension ay slack side

∑ 𝑀o = 0

F1 (L) = F2 (a)

200(1.5) =

F2(0.2) F2 =

1500 N

6. The band of a band brake has 210 degrees of contact with its owndrum.It is
found that the pull on the tight side is 800 lbs and the pull on the slack side is
285 lbs. Determine the coefficient of friction.
F1/F2 = ℯ𝑓∅
 800/285 = 𝑓(210 𝑥 𝜋
)
ℯ 180

2.807 = ℯ3.665𝑓

Taken in both side:

ln2.807 = ln ℯ3.665𝑓

ln2.807 = 3.665𝑓 (ln ℯ)

𝑓 = 0.281

7. The ratio of band tension in a band brake is 4.If the angle of contact is 260
degrees, determine the coefficient of friction.

F1/F2 = ℯ𝑓∅
𝜋
𝑓(260 𝑥 )
180
ℯ
4=

f = 0.305

8. A band brake has an angle of contact of 280 degrees and is to sustain a torque
of 10,000in- lb. The band bears against a cast iron drum of 14 in diameter. The
coefficient of friction is
0.30. Find the difference in band tension.

T = (F1 - F2)r

1,000 = (F1 - F2)(14 in./

2) F1 - F2 = 1428 lbs

9. A band brake is installed on a drum rotating at 250 rpm and a diameter of 900
mm. The angle of contact is 1.5л rad and one end of band brake is fastened to a
fix pin while the other end to the brake arm 150 mm from the fixed pin. The
coefficient of friction is 0.25 and the straight brake arm is 1000 mm long and is
placed perpendicular to the diameter bisecting the angle of contact. Determine
the minimum force in Newtons applied at the end of the brake arm necessary to
stop the drum if 60 KW is being absorbed.
P = 2𝜋TN

60 = 2 𝜋 (T)

(250/60) T =

2.2918 KN-m

F1/F2 = ℯ𝑓∅
𝜋
0.25(270 𝑥 )
180
ℯ
F1/F2 =

F1/F2 =

3.248 F1 =

3.248F2 T

= (F1 - F2)r

2.2918 = (F1 - F2)

(0.900/2) F1 - F2 =

5.0929

3.248F2 - F2 =

5.0929 F2 = 1.2465

KN

F1 = 3.248F2 =

3.248(1.2465) F1

=4.04849 KN

Summation of moments about pivot point = 0

F (L) = F1(a)

F = 4.04849(0.15 m)

F = 0.60727 KN = 607,27 N
10.A band brake use to fastened to an 900 mm brake drum with tight side is 3
times the slack side of force. The torque transmitted is 2 KN-m and a steel band
with a maximum tensile stress of 60 Mpa and 3 mm thick will be used. What
should be its width in mm?

F1 = 3F2

T = (F1 - F2)r

2 = (F1 - F2)

(800/2) F1 - F2 =

5 KN

3F2 - F2 = 5

F2 = 2.5 KN

F1 = 3F2 = 3(2.5) = 7.5 KN

S = F/A =

F1/bt 60,000

7.5
𝑏(0.003)
=

b = 0.0416666 m

b = 41.67 mm

11.A simple band brake has a 76 cm drum with coefficient of friction of 0.25 and arc
of contact of 245˚. The drum sustains a load of 820 kg and rotates at 260 rpm.
Find the power absorbed by the band.

F1/F2 = ℯ𝑓∅
𝜋
0.25(245 𝑥 )
18
ℯ 0
F1/F2 = = 2.912

F1 = 2.915F2
T=Fxr=

8.0442(0.75/2) T =

3.057 KN-m
 P = 2𝜋TN

P = 2 𝜋 (3.057)

(260/60) P =

83.2278 KW

12.Determine the internal load expressed in lb-ft 2 of a brake that requires 900 in-lb
of torque to stop a shaft operating at 840 rpm in a period of 3.5 seconds.

(𝑤𝑟2) 𝑁
308 (𝑡)
T=

T = 900/12 = 75 ft-lb

(𝑤𝑟2) (840)
308 (3.5)
75
=

w r2 = 96.25 ft2-lb

CLUTCH

1. A cone clutch has an angle of 12˚ and coefficient of friction of 0.42. Find the
axial force required if the capacity of the clutch is 8KW at 500 rpm. The mean
diameter of the active sections is 300 mm. Use uniform wear method.

P = 2𝜋TN

8 = 2 𝜋 (T)

(500/60) T =

0.15279 KN-m

𝐹𝑎 𝑓 𝑟𝑓
sin 12
T=

0.3
𝐹𝑎(0.45)( )
2
sin 12
0.15279 =
 Fa = 0.50423

KN Fa = 504.23

N-m

2. How much torque can a cone clutch transmitted if the angle of conical elements
is 10 degrees. The mean diameter of conical section is 200 mm and an axial
force of 600 N is applied. Consider a coefficient of friction of 0.45.

𝐹𝑎 𝑓 𝑟𝑓
sin 12
T=

0.2
600(0.45)( )
2
T= sin 10
T = 155.487 N-m

3. A clutch has an outside diameter of 8 in and inside diameter of 4 in. An axial

force of 500 lb is used to hold the two parts together.

T = Fa f rf

1 𝐷3− 𝑑3 1 (8)3− (4)3

[ ] [
3
𝐷− (8) − (4)2] = 3.111111
rf =
2
=
2
2 𝑑 3

T = 500 lbs x 0.4 x 3.111111 in

T = 622.22 in-lb

4. A disc clutch has 6 pairs of contacting friction surfaces. The friction radius is 2 in
and the coefficient of friction is 0.30. An axial force of 100 lb acts on the clutch.
The shaft speed is 400 rpm. What is the power transmitted by the clutch?

T = n Fa f rf

T = 6 x 0.3 x 100 lb x 2 in
 T = 360 in-lb ÷ 12 in/ft = 30 ft-lb

2𝜋𝑇 2 𝜋 (30 𝑓𝑡−𝑙𝑏)

𝑁 (400)
33,00 33,000
P= =
0

P = 2.285 Hp

5. A cone clutch has cone elements at an angle of 12˚. The clutch transmits 25 Hp
at a speed of 1200 rpm. The mean diameter of the conical friction section is 16
in and the coefficient of friction is 0.35. Find the axial force needed to engage
the clutch.

2𝜋𝑇𝑁
33,000
P=

2 𝜋 (𝑇)
25 (1200)
33,000
=

T = 109.42 ft-lb x 12 in/ft = 1,313.03 in-lb

𝐹𝑎 𝑓 𝑟𝑓
sin 12
T=

0.16 𝑖𝑛
𝐹𝑎(0.35)( )
2
sin 12
1313.03 =

𝐹𝑎 = 97.5 lbs

6. A band clutch has an angle of contact of 270˚ on a 15 in diameter drum. The

rotational speed of the drum is 250 rpm and the clutch transmits 8 Hp. The band
is 1/16 in thick and has a design stress of 5000 psi. How wide should the band
be? Use coefficient of friction to be 0.40.

2𝜋𝑇𝑁
33,000
P=

2 𝜋 (𝑇)(250)
33,000
8=
T = 168.02 ft-lb
 F1/F2 = ℯ𝑓∅
𝜋
0.4(270 𝑥 )
18
ℯ 0
F1/F2 = = 5.342

F1 = 5.342F2

T = (F1 - F2) r

168.02 = 1(F - F )( )(15 in/2)

15 𝑖𝑛
12 𝑖𝑛/𝑓𝑡
2

F1 - F2 = 268.832 lb

5.342F2 - F2 =

268.832 F2 = 61.914

lb

F1 = 5.342F2 =

5.342(61.94) F1 = 330.75

lb

S = F/A = F1/bt

330.75
5,000 1

𝑏( )
=
16

b = 1.058 psi

7. The angle of contact of a band clutch is 250 degrees. The cross section of the
band is 1/16 in x 1.5 in. The design stress for the band material is 8000 psi. If the
drum is 16 inch in diameter and rotates at 350 rpm, what is the power capacity
of the clutch if f = 0.40.
S = F/A = F1/bt

𝐹
8,000 (1.5)( 1
= )
16
 F = 750 lbs

T=Fxr

T = 750(16in/2)
6,000 𝑖𝑛−𝑙𝑏
12 𝑖𝑛/𝑓𝑡
T= = 500 ft-lb

2𝜋𝑇𝑁
33,000
P=

2 𝜋 (500 𝑓𝑡−𝑙𝑏)(350)
33,000
P=

P =33.3199 Hp

8. A simple disc clutch has an outside diameter of 200 mm and inside diameter of
100 mm. The friction is 0.40 and applied load is 1,500 N. Find the torque
transmitted using uniform wear.

T = Fa f rf

1 𝐷3− 𝑑3 1 (0.200)3− (0.100)3

] [ ]
3 [ (0.200) −
𝐷−
rf =
2 2
= = 0.0778 m
2 𝑑 3 2 (0.100)

T = 1500 N x 0.40 x

0.0778 m T = 46.667 N-m

9. Find the power capacity of a cone clutch with mean diameter of 250 mm if the
conical elements inclined at 8 degrees and axial force of 450 N. The rotational
force of drive is 200 rpm and the friction is 0.20.

𝐹𝑎 𝑓 𝑟𝑓
sin ∅
T=

0.25
450 𝑁(0.20)( )
2
T= sin 8
 T = 80.834 N-m = 0.080834 KN-m

P = 2𝜋TN

P = 2 𝜋 (0.080834 KN-m)

9200/60) P = 1.69 KW

10.Find the cone angle of a clutch having a major diameter of 300 mm and minor
diameter of 250 mm and length of contact of 250 mm.

(𝐷−𝑑)/2
sin ∅
w=

(300−250)/
250 2
sin ∅
=

∅ = 5.74 degree

11.Find the power capacity of uniform wear of a cone clutch having major diameter
of 250 mm, minor diameter of 200 mm, length of contact 125 mm, f = 0.30
speed of 870 rpm and has an operating force of 500 N.

(𝐷−𝑑)/2
sin ∅
w=

(250−200)/
250 2
sin ∅
=

∅ = 0.112957 deg.

1 𝐷3− 𝑑3 1 (0.250)3− (0.200)3

[ ] [ ] = 0.11296
3
𝐷− (0.250
rf =
2
− (0.200)2
=
2 𝑑 3 )2

𝐹𝑎 𝑓 𝑟𝑓
sin ∅
T=
 0.5 𝐾𝑁(0.30)
(0.11296) = 0.0847 KN-m
sin 8
T=

P = 2𝜋TN

P = 2 𝜋 (0.0847 KN-m)

(870/60) P = 7.719 KW

12.Using a uniform wear, find the power capacity of a single disc clutch with an
outside diameter of 200 mm and inside diameter of 100 mm, rotating at 1160
rpm (f = 0.35). The axial operating force is 800 N.

1 𝐷3− 𝑑3 1 (0.200)3− (0.100)3

] [ ]
3 [ (0.200) −
𝐷−
rf =
2 2
= = 0.0778 m
2 𝑑 3 2 (0.100)

T = Fa f rf = 0.8 KN x 0.35 x 0.0779 m = 0.021784 KN-m

P = 2𝜋TN = 2 𝜋 (0.021784 KN-m) (1160/60)

P = 2.646 KW

13.In a band clutch, the ratio of the pull on the right side of the band to that of the
slack side is 4:1. The band contacts the drum for 250 degrees. What is the
coeffi=cient of friction?
F1/F2 = 4/1

F1/F2 = ℯ𝑓∅
𝜋
𝑓(250 𝑥 )
ℯ
4/1 =
180

ln 4 = ln ℯ4.3622𝑓
f = 0.318
14.Assuming uniform wear, find the power capacity of a single disc clutch with an
outside diameter of 200 mm and 100 mm respectively, a rotational of 1160 rpm,
a coefficient of friction of 0.35 and an axial operating speed of 800 Newtons.

Using uniform

wear: T =

𝑟𝑜+ 𝑟𝑖
fF[ ]
2

]
0.10 +
(800) [
T = (0.35)
0.05 = 21 N-m = 0.021 KN-m
2

P = 2𝜋TN = 2 𝜋 (0.021 KN-m) (1160/60) = 2.55 kw

15. Find the cone angle of the clutch having a major diameter of 280 mm and minor
diameter of 200 mm and length of contact of 250.

(𝐷−𝑑)/2
sin ∅
w=

(280−200)/
250 2
sin ∅
=

∅ = 9.21 degree

16.The large diameter and face of the disk of multiple disk clutch are 255 mm and
25 mm respectively. The load applied to the spring of the clutch axially is 1688
N. Assuming that there are 10 pairs of friction 0.15 and turning Hp at 1000 rpm.
Find the power transmitted by the clutch using uniform pressure method.

1 𝐷3− 𝑑3 1 (0.255)3− (0.025)3

] [ ]
3 [ (0.255) −
𝐷−
rf =
2 2
= = 0.08574
2 𝑑 3 2 (0.025)

T = n Fa f rf = 9 x 0.15 x 1.688 KN x 0.08574 = 0.1954 KN-m

 P = 2𝜋TN = 2 𝜋 (0.1954 KN-m) (1000/60)

P = 20.46 KW

17.An engine of a motor vehicle develops 50 Kw at 2000 rpm. Determine the axial
force on the clutch in KN. The outside and inside diameter of the clutch are 300
mm and 240 mm respectively. There are 4 pairs of mating surfaces with f =
0.30.

1 𝐷3− 𝑑3 1 (0.300)3− (0.240)3

] [ ]
3 [ (0.300) −
𝐷−
rf =
2 2
= = 0.1355
2 𝑑 3 2 (0.240)

P = 2𝜋TN

50 = 2 𝜋 (T)

(2000/60) T =

0.2387 KN-m

T = n Fa f rf

0.2387 KN-m = 4 x 0.3 x Fa x

0.135 Fa = 1.468 KN

PROBLEM SET # 8

MACHINE SHOP, PRESSURE VESSEL & SPRING

1. A horizontal cantilever beam, 20 ft long is subjected to a load of 5000 lb

located to its center. The dimension of the beam is 8 x 12 inches respectively. W
= 20 lb/ft, find its flexural stress.
F = total load at the center
F = 5000 + 120(120) = 7400 lbs

M = maximum moment

M = 7400 (20/2) = 74,000 ft-lb

2
9
S = Mc/l = 96.35 psi

2. A 10 m simply supported beam has a uniform load of 3 KN/m extended from

left end to 4 m and has a concentrated load of 10 KN, 2 m from the right end.
Find the maximum moment at the 10 KN concentrated load of: (El = 9,000 KN-
m2)

R1 (10) = (4 x 2)(8) + 10(2)

R1 = 8.4 KN

R2 (10) = (4 x 2)(2) + 10(8)

R2 = 9.6 KN

3. A 12 m simply supported beam with uniform load of 2.5 KN/m from the right
end to left end has a maximum deflection of: ( El = 11,000 KN-m 2 )

Y = maximum deflection

5𝑤𝐿4
384 𝐸𝑙
Y=

5(2.5)(12)4
Y = 384 (11,000)

Y = 0.0613636 m

Y = 61.3636 mm
4. An 9 m simply supported beam has a uniform load of 3 KN/m from the right
end to left end and concentrated load of 15 KN at a center has a maximum
deflection of: ( El = 7,000 KN-m2
)

𝑃𝐿=3 (15)(9)3
48
Y
= 48(7,000 = 0.0325446 m
𝐸𝑙 )
1

Considering the effect of uniform distributed load of 2 kN/m.

5𝑤𝐿
= 4 5(3)(9)4 = 0.036613 m
384
Y
= 384
𝐸𝑙 (7,000)
2

Y = Y1 +Y2

Y = 0.0325446 m + 0.036613 m

Y = 0.069157 m = 69.157 mm

5. A 14 m cantilever beam has a concentrated load of 25 KN at the med

span. Find the maximum slope of the beam. ( El = 10,000 KN-m 2 )

𝑃𝐿4
∅=
8 𝐸𝑙

25(14)4
∅=
8
= 0.6125 rad
(10,000)

Maximum slope = 61.25 mm

6. A 10 m cantilever beam has a uniform load of 4 KN/m fro left to right

end. Find the maximum slope of the beam. ( El = 12,000 KN-m 2 )

𝑤𝐿3
∅=
6 𝐸𝑙
 4(10)3
∅=
6 (12,000)

∅ = 0.0556 rad

7. A 11 m cantilever beam has a uniform load of 2.3 KN/m fro left to right end
and with concentrated load of 11 KN at the center. Find the maximum slope of
the beam. ( El = 12,000 KN-m2 )

𝑤𝐿
∅1 3
6
𝐸𝑙
=

2.3(11)3
∅1
6
= 0.0425 rad
(12,000)
=

𝑃𝐿
∅2 2
8
𝐸𝑙
=

11(11)2
∅2
8
= 0.01386 rad
(12,000)
=

∅ = ∅1 + ∅2

∅ = 0.0425 rad + 0.01386 rad

∅ = 0.0564 rad

8. A vertical load of 100 N acts at the end of a horizontal rectangular cantilever

beam 3 m long and 30 mm wide. If the allowable bending stress is 100 Mpa, find
the depth of the beam.
For cantilever beam with load act at the free end:
P = 1000 N = 1 kN

M = PL

M = (1)(3) = 3 kN-m

6𝑀
𝑆=
𝑏ℎ2
6 (3)

(0.03)(ℎ)2
100,000 =

h = 0.07745 m = 77.45 mm

9. A simply supported beam is 50 mm by 150 mm in cross section and 4 m long.

If the flexural stress is not to exceed 8.3 Mpa, find the maximum mid-span
concentrated load.

M = maximum moment

M = PL/4

𝑃 (4)
M= 4

M=P
6𝑀
𝑆=
𝑏ℎ2
6 (𝑃)

(0.05)(0.20)2
8,300 =

P = 2.77 KN

10.An occupant moves toward the center of the center to a merry go around at
10 m/s. If the merry go around rotates at 8 rpm. Compute the acceleration
component of the occupant normal to the radius.

V=2𝜋 RN

10 = 2 𝜋 R (8/60)
R = 11.94

a = acceleration

𝑉
a =2

(10)2
= 8.377 m/s2
𝑅 11.9
=
4

11.A car travels with an initial velocity of 36 km/hr. If it is decelerating at the

rate of 2 m/s2, how far in meters does it travel before stopping?

Vf2 = Vo2 – 2aS

Vo = 36 km/hr = 10 m/s

(0)2 = (10)2 – 2 (2) S

S = 25

12.A block weighing 60 lbs rest on horizontal surface. The force needed to move
along the surface is 15 lbs. Determine the coefficient of friction.

F=fN=fW

15 = f (60)

f = 0.25
13.A baseball is thrown straight upward with a velocity of 25 m/s. Compute the
time elapse for the baseball to return. Assume for a zero drag.

Vf = Vo – g t

0 = 25 – 9.81 t

t = 2.54

14.A wheel accelerates from rest with a = 4 rad/sec.sec. Compute how many
revolutions are made in 5 seconds.

1 rev = 2 𝜋 rad

∅ = w1t + ½ d t2

w1 = 0(from rest)

∅ = 0 + ½ (4/2 𝜋) (6)

∅ = 11.459 rev

15.What minimum distance can a truck slide on a horizontal asphalt road if it is

traveling at 20 m/s. The coefficient of sliding between asphalt and rubber is at
0.5. The weight of the truck is 8500 kg.

∑Fh = 0

Fr = FR
fw=𝑤 a
𝑔

a = f g = 0.5(9.81) = 4.905 m/s

after the slide it will stop so that V2 = 0

V2 = V12 – 2aS

(0)2 = (20)2 – 2 (4.905) S

S = 40.7747 m

16.A liquid full is to be rotated in the vertical plane. What minimum angular
velocity in rpm is needed to keep the liquid not spilling if the rotating arm is 1.5
meters?

Fc = Fg

𝑚𝑉2
= mg
𝑅

V2 = gR

V2 = 9.81(1.5)

V = 3.836 m/s

Solving for N:

V=2𝜋 RN

3.836 = 2 𝜋 (1.5) (N)

N = 0.407 rev/s x 2 𝜋
N = 2.557 rad/s

17.Compute the speed a satellite to orbit the earth at an elevation of 150 km.
Earth’s radius is at 6500 km. Assume no change of gravity with the elevation.

Fc = Fg

𝑚𝑉2
= mg
𝑅

V2 = gR

V2 = 9.81(6,500,000 + 150,000)

V =8072.79 m/s

18.Compute the cutting speed in fpm of a workplace with 5 inches and running at
120 rpm.

V=2𝜋 DN

V = 2 𝜋 (5/12) (120)

V = 157.08 ft/min

19.Determine the time in seconds to saw a rectangular bar 7 in wide and 1.5 in
thick, if the length of cut is 7 in the power hacksaw does 120 strokes/min and
the feed/stroke is 0.154 mm.

Time = length to be cut / cutting rate

Time per stroke = 0.154 mm = 0.006063 in

7
𝑠𝑡𝑟𝑜𝑘𝑒𝑠 𝑖
(12 ) )
Time = = 9.6211 min = 577.27 sec
𝑛
0 (0.006063
mi𝚗 𝑠𝑡𝑟𝑜
𝑘𝑒

20.Compute the drill penetration in in/min when a drill turns at 1000 rpm and
the feed of 0.004 in per revolution.

Drill penetration = (feed rate) N

Drill penetration = (0.004 in/rev)(1000 rev/min)

Drill penetration = 4 in/min

21.In an Oxygen-acetylene manual welding method, to weld a 5 ft. long seam in

a 0.375 in thick steel plate at a consumption rate of 10 ft3/ft for oxygen and 8
ft3/ft for acetylene. Compute the total combined gas consumption in ft 3.

V = total gas consumption

V = (Vo + Va)L

V = (10 +8)(5) = 90 ft3

22. Compute the manual cutting in time in minutes of a steel plate 4 ft x 8 ft by

3/4 in thick with a hand cutting speed of 3 to 4 mm/sec, cutting crosswise.

Cutting speed = (3 + 4) / 2 = 3.5 mm/sec

Length = 8 ft(12)(25.4) = 2438.40 mm

Cutting time = length / cutting speed = 2,438.40 / 3.5

696.69
𝑠𝑒𝑐
Cutting time =

60
= 11.61 min

𝑠𝑒𝑐/𝑚𝑖𝑛

23. How long will it take to mill a 1/2 by 3” ling keyway in a 4 diameter shafting
with a 24 tooth cutter turning at 90 rpm and 0.006 in feed per tooth.

Time = length to be cut / cutting rate

3 𝑖𝑛
(24 𝑡𝑒𝑒𝑡ℎ/𝑟𝑒𝑣)(90 𝑟𝑒𝑣/𝑚𝑖𝑛)(0.006 𝑖𝑛/𝑡𝑜𝑜𝑡ℎ)
Time =

Time = 0.231 min x 60 sec/min = 13.89 sec

24.A 6 ft by 12 ft steel plate is to be divided into three parts equally by cutting

crosswise. If the cutting is 0.5 in per in, how long will it take to do the job?

L = total length to be cut

L = 6 + 6 = 12 ft

Time = length to be cut/cutting time

12 𝑓𝑡 𝑥 12 𝑖𝑛/𝑓𝑡
Time = 0.5 𝑖𝑛/𝑚𝑖𝑛

Time = 288 min

25.Determine the internal pressure of cylindrical tank 500 mm internal diameter,
20 mm thick and 3 m length if stresses limited to 140 Mpa.

𝑃 𝐷𝑖
S = 2𝑡𝜂

𝑃 (0.50)

2 (0.02)
140,000 =

P = 11,200 kpa = 11.20 Mpa

26.Determine the bursting steam pressure of as steel shell with dimeter of 20

inches and made of 5/16 steel plate. The joint efficiency is at 80% and a tensile
strength is 63,000 psi.

𝑃 𝐷𝑖
S = 4𝑡𝜂
𝑃 (20)

4 (0.3125)(0.80)
63,000 =

P = 3,150 psi

27.A water tank 10 m x 12 m is completely filled with water. Determine the

minimum thickness of the plate if stress is limited to 50 Mpa.

P =wh = 9.81(12) = 117.71 kPa

𝑃 𝐷𝑖
S = 2𝑡
 117.72(10)
50,000 = 2𝑡

t = 0.011772 m = 11.77 mm

28.Determine the safe wall thickness of a 50 inches steel tank with internal
pressure of 8 Mpa. The ultimate stress is 296 Mpa. The factor of safety to use is
3.

𝑆
𝑃 𝐷𝑖
𝐹𝑆 = 2𝑡

296 8 (50)
3 = 2𝑡

t = 2.02 in

29.The internal pressure of a 400 mm inside diameter cylindrical tank is 10 Mpa

and thickness is 25 mm. Determine the stress developed if joint efficiency is
95%.

𝑃 𝐷𝑖
S =2𝑡𝜂

10 𝑀𝑝𝑎 (400)
S=
2(25)(0.95)

S = 84.21 Mpa

30.A water tank 12 m x 15 m is completely filled with water. Determine the

minimum thickness of the plate if stress is limited to 40 Mpa.

P =wh = 9.81(15) = 147.15 kPa

 𝑃 𝐷𝑖
S = 2𝑡

147.15 kPa(12)
40,000 = 2𝑡

t = 0.0220725 m = 22.0725 mm

31.A spherical tank 15 mm thick has an internal pressure of 5 Mpa. The joint
efficiency is 96% and stress is limited to 46875 Kpa. Find the inner diameter of
the tank.

𝑃 𝐷𝑖
S = 4𝑡

5000 (𝐷𝑖)
46,875 =
4 (0.015)(0.96)

Di = 0.540 m = 540 mm

32.A spherical tank has a diameter of 10 m and 80 cm thickness is 25.4 cm. If

tangential stress of tank is 16 Mpa, find the maximum internal pressure of the
tank can carry if the wall thickness is 25.4 cm.

𝑃 𝐷𝑖
S = 4𝑡

𝑃 (10)
16,000
4
=
(0.254)

P = 1,625.60 kpa
33.A cylindrical tank has an inside diameter of 5 in and is subjected to internal
pressure of 599 psi. If maximum stress is 1200 psi, determine the required
thickness.

𝑃 𝐷𝑖
S = 4𝑡

500
1,200 (5)
=
2𝑡

t = 1.04166 in

t/Di = 1.04166 in / 5 = 0.2087

Therefore, use thick-wall formula:

𝐷
t= √( 𝑆𝑡 + 𝑃𝑖 )
[ ( 𝑆 𝑡 − 𝑃𝑖 ) − 1]
2

5
t = √(1200 + 500)
[ (1200 − 500) − 1]
2

t = 1.395 in

34.A thickness of cylindrical tank is 50 mm. The internal diameter is 300 mm

and has an internal pressure of 30 Mpa. Determine the maximum internal
pressure.

𝐷
t= ( 𝑆 𝑡 + 𝑃𝑖 )
[ ( 𝑆 𝑡 − 𝑃𝑖 ) − 1]
2
.050 = √(30,000 + 𝑃𝑖 )
0.30
[ (30,000 − 𝑃𝑖 ) − 1]
2
0.333 =
[ √(30,000 + 𝑃𝑖 )
− 1]
(30,000 − 𝑃𝑖 )

1.333 = (30,000 +
[ √
𝑃𝑖 )
− 1]
(30,000 − 𝑃𝑖 )

Squaring both sides:

3000 + 𝑃𝑖
1.333 =3000 − 𝑃𝑖

30,000 + Pi = 53,333.33 – 1.77777Pi

Pi = 8400 Kpa = 8.4 Mpa

35.A round vertical steel tank has an inside diameter of 3 cm and is 25 Mpa, find
the minimum thick required.

P = wh = (750 x 9.81 / 1000)(6) = 44.145 kpa

𝑃 𝐷𝑖
S = 2𝑡

44.145 (3)
25,000
2𝑡
=

t = 2.648 x 10-3 m = 2.648 mm

36.A cylinder having an internal diameter of 30 inches is subjected to an internal

pressure of 8,000 psi. If the hoop stress at the inner is 13,000 psi, find the
external pressure.

ri = 18/2 =

9 ro = 30/2
= 15
 2 2
2 𝑃𝑖 𝑟 − 𝑃𝑜(𝑟2+ 𝑟 )
Sto 𝑖 2
𝑟 −𝑟
𝑜
2 𝑖

=
𝑜 𝑖

2(13,000)(9)2−800(152+
Sto
92)
=
(152+ 92)

Sto = 6,625 psi

37.A cylinder having an internal diameter of 16 in and wall thickness of 6 inches

is subjected to an internal pressure of 65 Mpa and external pressure of 13 Mpa.
Determine the hoop stress at the outer.

ri = 16/2 = 8
ro = (16/2)+6 = 14

2 2
2 𝑃𝑖 𝑟 − 𝑃𝑜(𝑟2+ 𝑟 )
Sto 𝑖 2
𝑟 − 𝑟2
𝑜 𝑖

=
𝑜 𝑖

2(65)(8)2− 13(142+
Sto
82)
=
(142+ 82)

Sto = 37.424 Mpa

38.The pressure inside the cylindrical tank varies from 800 kpa to 3200 kpa
continuously. The diameter of shell is 1.6 m and 2.3 m high. Determine the
minimum thickness of the tank plate.

P = wh

P = 9.81(1.6)
P = 24.525 Kpa
 𝑃 𝐷𝑖
S = 2𝑡

24.525 (2.5)
3,200
2𝑡
=

t = 6.13125 x 10-3 m = 6.13125 mm

PROBLEM SET # 9

SPUR GEAR AND SPRING

1. Find the distance between centers of a pair of gears, one of which has 12 teeth and
the other is 37 teeth. The diametral pitch is 7.

D1 = T1/DP = 12/7 =

1.7143 D2 = T2/PD =

37/7 = 5.2857

𝐷1+𝐷2 1.7143 + 5.2857

C= 2 = 2 = 3.50 in

2. Determine the pitch diameter of a gear with 28 teeth, 4 diametral pitch.

D = T/P = 28/4 = 7 in.

3. Two parallel shaft have an angular velocity ratio of 3 to 1 are connected by gears
the largest by which has 36 teeth. Find the number of teeth of smaller gear.

T1 N1 = T2

N2 3 (N1) =

1(36)

N1 = 12

4. Two parallel shafts have a center distance of 15 in. One of the shaft carries a 40-
tooth 2 diametral pitch gear which drives a gear on the other shaft at a speed of
150 rpm. How fast is the 40-tooth gear turning?

𝑇𝑔+𝑇𝑝
2𝐷𝑃
C=

𝑇𝑔+40
2(2)
15
=

Tg = 20

Tg Ng = Tp Np

20(Ng) =

40(150)

Ng = 300 rpm

5. A 14.5 degrees full-depth involute gear has an outside diameter of 8.5 and
diametral pitch of 4. Find the number of teeth.

D = T/DP

8.5 = T/4

T = 34
6. A pear of meshing gears has a diametral pitch of 10, a center distance of 2,6 inch
and velocity ratio of 1:6. Determine the number of teeth of smaller gear.
 3
Speed ratio = 𝑁1
=
𝑁2 1

D2 =

(N1/N2)D1 D2

= (6/1)D1 D2

= 6D1

𝐷1+𝐷2
C= 2

6𝐷1+𝐷2
2.6 = 2
𝑇1
D1 =
𝐷𝑃

𝑇1
10
0.743 =

T1 = 7.43 = 8

7. A spur gear 20 degrees full-depth involute teeth has an outside diameter of 196
mm and a module of 6.5. Determine the number of teeth.

25.4
𝑀 25.4
DP =
or M =𝐷𝑃

25.4
6.5
DP = = 3.90769

𝑁+2
𝐷𝑃
Do =
, N=no. of teeth
𝑁+2
3.90769
195/25.4 =

N + 2 = 30

N = 28 teeth

8. What is the pitch diameter of a 40 teeth spur gear having a circular pitch of 1.5708.
 𝜋𝐷
Pc =
𝑇

𝜋𝐷
40
1.5708 =

D = 20 in

9. How many revolutions per minute is a spur gear turning at, if it has 28 teeth, a
circular pitch of 0.7854 in and a pitch line velocity of 12 ft/sec?

Pc = лD / T

0.7854 = лD /

28

D = 7 in

V=лDN

12 = л(7/12) N

N = 6.548 rps x

60 N = 392.88

rpm

10. How many revolutions per minute is a spur gear turning if it has a module of 2
mm, 40 teeth, and pitch line velocity of 2500 mm/sec?

M=D/

T2=

D / 40

D = 80

mm V = 𝜋

DN

2500 = 𝜋 (80) N

N = 9.947 rev/sec x
60 N = 596.83 rpm
11. A standard 20 degrees full-depth spur gear has 24teeth and circular pitch of
0.7854 in. Determine the working depth.

Working depth =
𝜋
𝐷𝑃
2/P Pc =

𝜋
𝐷𝑃
0.7854 =

DP = 4

Working depth = 2/4 = 0.5 in

12. What is the equivalent diametral pitch of a gear that has a module of 2.5?

M = 25.4 / DP

2.5 = 25.4 /DP

DP = 10.16

13. A 20 degrees full-depth gear has tooth thickness of 0.25 in. find the addendum
distance.

Tooth thickness = 0.25 in

Tooth thickness = 1.

5708 / P

0.25 = 1.5708 P

P = 6.2832

Addendum = 1/P = 1 /

6.2832 Addendum =

0.15915

14. A 14.6 degrees full-depth gear has tooth thickness of 0.25 in. find the addendum
distance.
Tooth thickness = 0.25 = 1.5708 / P
 P = 6.2832

Addendum = 1 / P = 1 / 6.2832

Addendum = 0.15915

15. A gear has a pitch diameter of 10 inches and diametral pitch of 5, Determine
the outside diameter.

D = 10 in , PD = 5

Pitch diameter + outside diameter = 10

16. A 14.5 degrees full-depth has a dedendum of 0.2 inch, determine the tooth
thickness.

Dedendum = 0.2 in = 1.157 / P

0.2 = 1.157 / P

P = 5.785

Tooth thickness = 1.5708 / P = 1.5708 /

5.785 Tooth thickness = 0.2715

17. An internally meshing gear has a center distance of 20 in. If larger gear has a
diameter of 50 inches, find the diameter of internal gear.

𝐷−𝑑
C= 2

50 − 𝑑
20 = 2

D = 10

18. A 10 inches diameter gear is use to transmit 20 KW at 900 rpm. Determine the
tangential force acting on each gear.
 P = 2𝜋TN

20 =2 𝜋(T)

(900/60) T =

0.2122 KN-m

T = Ft x r

10 𝑖𝑛
0.2122 =
Ft [ 2(39.37)]

Ft = 1.671 KN

19. A 200 mm diameter 14.5 degrees involute gear is use to transmit 40 KW at 600
rpm. Determine the total force transmitted on gear.

P = 2𝜋TN

40 =2 𝜋(T)

(600/60) T =

0.63662 KN-m

T = Ft x r

0.63662 = F /

(0.2/2) F = 6.366

KN

20. A square and ground ends spring has a pitch of 20 mm, wire diameter of 12.5 mm.
If there are 12 actual number of coils, find the deflection when spring is
compressed to its solid length.

Actual no. of coils = n

+ 2 12 = n + 2

n = 10
free length = np + 2d

free length = 10(20) +

2(12.5) free length = 225

mm
 Solid length = (n+2)d

Solid length = (10+2)(12.5) = 150 mm

Ys = FL – SL = 225 – 150

Ys = 75 mm

21. A spring with plain ends has 15 active coils diameter of 6 mm and pitch of 10 mm.
If spring rate is 100 KN/m, determine the solid force.

Solid length = (n+1)d = (15+1)(6) =

96 mm Free length = np + d = 15(10)

+ 6 = 156 mm

Ys = FL – FS = 156 – 96 =

60 mm Fs = kys =

100(0.060) = 6 KN

22. A spring rate has a spring rate of 30 KN/m. If wire diameter is 10 mm with a mean
diameter of 70 mm, determine the number of active coils. G = 80 GN/m 2.

8𝐶3𝑛
𝐺𝑑
y=

C = Dm/d = 70/10 = 7

𝑦 8𝐶3𝑛
𝐹 = 𝐺𝑑

1 8(7)3𝑛
30 =(80 𝑥 106)(0.010)
n = 9.72 coils
23. A high alloy spring having square and ground ends and has a total of 16 coils and
modulus of elasticity in shear of 85 Gpa. Compute the whaal factor. The spring
outside diameter is 9.66 cm, wire diameter is 0.65cm.

Solving for spring index:

Dm = Do – d = 9.66 – 0.65 =

9.01 cm C = Dm/d =

9.01/0.65 = 13.86

4𝐶 0.615 4(13.86) 0.615

Whaal factor −1 −1 13.86
+ +

4𝐶− 𝐶 4(13.86)−
= =
4 4

Whaal fator = 1.1023

24. A spring with plain ends has 15 active coils,diameter of 6 mm and pitch of 10
mm.If spring rate is 100 KN/m, determine the solid force.

For plain end type of spring:

Solid length = (n +

1)d Solid length =

(15 +1)6 Solid

length = 96 mm

ys = FL – SL = 156 – 96 =

60 mm Fs = k ys =

100(0.060) = 6 KN

25. A pair of meshing spur gears has a module of 2. The pinion has 18 teeth and the
gear has 27 teeth. Find the ff:
a. Pitch diameter of each gear.
b. Center distance of each gears.
a.) DP = 25.4/M = 25.4 / 2 =

12.7 DP = T/D
 12.7 = 18/Dpinion

Dpinion = 1.417 in = 36 mm

12.7 = 27/Dgear

Dgear = 2.126 in = 54 mm

𝜋 𝜋
b.) Pc 𝐷 𝐷 = 0.2474
𝑇 12.
=
7
=

2𝜋𝐶
𝑇𝑔+
Pc =

𝑇𝑝

2𝜋𝐶
18 +
0.2474

27
=

C = 1.7717 in

C = 45 mm

26. An extension coil spring is to be elongate 5 in. under a load of 50 lb. What is the
spring rate?

Spring rate = F/Y

Spring rate = 50/5 = 10 lb/min.

27. The spring has a load of 50 lb with a spring index of 8. If stress induced is 90,000
psi, determine the wire diameter.

4𝐶 0.615 4(8) 0.615

−1
−1 8
+ +

4(8)−
K
4𝐶− 𝐶
= =
4 4
K= 1.184
 C

=dm/d

=Dm/d

Dm =

8d

Using stress
formula:

8𝐾𝐹𝐷𝑚
𝜋𝐷3
K=

8(1.184)(50)(8𝑑)
𝜋𝑑3
90,000 =

d = 0.1157
in

28. It is found that a load of 50 lb an extension coil spring deflects 8.5. What load
deflect the spring at 2.5 in?

K1 = K2

F 1 / y1 = F 2 / y2

50 / 8.5 = F2 /

2.5 F2 = 14.70

lbs

29. A spring sustain 200 ft-lb of energy with deflection of 3 in. assume that the main
coil diameter is 7 times the wire diameter and allowable stress of 100,000 psi,
determine the wire diameter.

200
F
3/1
= 800 lbs
=
2
 𝐷𝑚 7𝑑

𝑑 𝑑
C= = =7

4𝐶 0.615 4(7) 0.61

−1 5 = 1.2128
−1
+

4(7)− 7
K
4𝐶− 𝐶
= = +

4 4

8𝐾𝐹𝐷𝑚
K = 𝜋𝐷3
 8(1.2128)(800)
100,000 (7𝑑)
𝜋𝑑3
=

d = 0.416 in

30. A weight of 100 lbs strikes a coil spring for a height of 18 inches and deflects the
spring of 6 inches. Find the average force acting on the spring.

W(h+y) = (F/2)y

100(18+6) = (F/2) (6)

F = 800 lbs

31. If it is determined experimentally that a load of 20 kg applied to an extension coil

spring will provide a deflection of 200 m. What load will deflect the spring 60 mm.

F1/Y1 = F2/Y2

20 kg / 200 mm = F2 / 60

mm F2 = 6 kg

32. Three extension coil spring are hooked in series and supports a weight of 70 kg.
One spring has a spring rate of 0.209 kg/mm and the other two have spring rate of
0.643 kg/mm. Find the deflection.

y = total

deflection y =

y1 + y2 +y3

For series connection of

spring: F1 + F2 + F3 = 70 kg

K=

F/y y

= F/K
y = F1/K1 + F2/K2 + F3/K3
 y = 70/0.209 + 70/0.643 +

70/0.643 y = 552.65 mm

33. Four compression coil spring in parallel support a load of 360 kg. Each spring has
a gradient of 0.717 kg/mm.

y1 = y2 = y3 = y4

F1 = F2 = F3 = F4 = 360/4 = 90 kg

y1 = F1/K1 = 90/0.717

y1 = 125.52 = y2 = y3 = y4 = y

34. A spring has a diameter of 25 mm and 12 active coils. If a load of 10 KN is applied it

deflects 75
mm. Determine the mean diameter of the spring. G = 80 GN/m2.

8𝐶3
𝑛
y
𝐺𝑑
=

8(10)(𝐶)3(12)
(80 𝑥 106)(0.025)
0.075
=

C = 5.386

C = Dm/d

5.386 = Dm/25

Dm = 134.65

mm
35. A concentric helical spring is use to support load of 90 KN. The inner spring has
spring rate of
495.8 KN/m and outer spring is 126.5 KN/m. If initially the inner spring is 25 mm
than the other spring, find the percent load carried by inner spring.
 yL = ys + 0.025

K = F/y

Fs = ys Ks = 495.8

ys FL = 126.5 yL

Fs + FL = 90

495.8ys + 126.5 yL = 90

495.8ys + 126.5 (yS + 0.025) = 90

495.8ys + 126.5 yS + 3.1625

= 90 ys = 0.1395 m

Fs = 495.8 (0.1395) = 69.185 KN

% Load Carried = 69.185 / 90 = 76.87%

36. How long a wire is needed to make a helical spring having a main coil diameter of
1 inch if there are 8 active coils.

L = wire length

L = circumference x No. of coils

L = 𝜋 D (n) = 𝜋(1)(8) = 25.13 in

37. A 0.08” diameter spring has a length of 20 in. If density of spring is 0.282 lb/in,
determine the mass of spring.

V = volume of spring
 V = (𝜋/4 d2) L = [𝜋/4 (0.08)2](20) = 0.10053 in3

Solving for mass:

w = m/V

m = V w = (0.10053)(0.282) = 0.0283 lb

38. A square and ground end spring has a free length of 250 mm. There are 10 active
coils with wire diameter of 12.5 mm. If the spring rate is 150 KN/m and the mean
diameter is 100 , determine the solid stress.

For square and ground ends:

Solid length = (n + 2)d = (10 + 2) = 150 mm

Ys = solid length deflection

Ys = free length – solid length = 250 - 150 =

100 mm Fs = force at solid length = k Ys

Fs = 150(0.100) = 15

KN C = Dm/d =

100/12.5 = 8

4𝐶 0.615 4(8) 0.61

−1 5 = 1.184
−1
+

4(8)− 8
K
4𝐶− 𝐶
= = +

4 4

Ss = stress at solid length

8𝐾𝐹𝐷𝑚
Ss = 𝜋𝐷3

8(1.184)(15)(0.10)

𝜋(0.0125)3
Ss
= 2,315,544 Kpa = 2,315.54 Mpa
=
39. A helical compression spring has a scale of 400 lbs/inch, an inside diameter of 2.5
in, a free length of 8 in. and with square and ground ends and the material has a
working strength of 63,000 psi and G = 10,800,000 psi. for a load of 825 lbs, and
for average service, Whaal factor, k = 1.25, determine:
A. The standard size wire diameter.

B. The spring index.

C. Number of active coil.

8𝐾𝐹𝐷
A.) Ss = 𝜋𝐷3 𝑚

Dm = Di + d = 2.5 + d

8(1.25)(825)
63,000 (2.5+𝑑)
𝜋𝑑3
=

Try: d = 0.5 in.

8(1.25)(825)
63,000 (2.5+𝑑)
𝜋(0.5)3
=

63,000 = 63,000

Therefore: d = 0.5 in.

B.) Dm = 2.5 + 0.5 =

3 in C = Dm/d =

3/0.5 = 6

C.) k = F/y

400 =

825/y y =

2.0625 in

8𝐶3𝑛
𝐺𝑑
y=
2.0625
=

8(825)(6)3(𝑛)
(10,800,000)
(0.5)
n = 7.8125 coils