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.
Solution:
P=2πTN
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:
P =2πTN
110 = 2π T (300/60)
T = 3.501
16T
S= 3
πd
16(3.501)
90,000 =
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.
Solution:
S
16 TDo
=
F.S π(0.084 − 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.
Solution:
S=
16T
πD3
16T
26,000 =
π(0.06342)3
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:
P=
2πTN
33000
2π T (1400)
50 =
33000
T = 187.575 ft-lb x 12 in/ft
T = 2250.905 in-lb
16T
S=
πD3
16(2250.905)
500 =
πD3
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:
Θ=
10 x
TL
JG
π
180
=
(2KN−m)(20d)
(80000000)(
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.
Solution:
S=
16TDo
π(D4o − D4i )
40000 =
16(5)(0.1)
π[(0.1)4 − D4 ]
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[(D4o − D4i )]
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?
Solution:
16T
S=
πD3
16T
25000 =
πD3
D = 0.06735 m
J=
J=
πD4
32
π(0.06735)4
32
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)
T = 0.762 KN-m
πD3 π(0.0508)4
J=
=
32
32
J = 6.53814 x 10-7 m4
TL
Θ=
JG
0.762
θ/L = (6.53 x 10−7 )(83000000)
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)
Solution:
Θ=
TL
JG
30(π/180) =
14 x 6
J(83 x 106 )
J = 1.93286 x 10-5 =
πd4
32
D = 0.11845 m
12. A solid shaft 5m long is stressed 60 Mpa when twisted through 4 o. Find the shaft
diameter if G=6.83 Gpa
Solution:
S=
16T
πd3
Solving for T in terms of d;
T=
Θ=
πd3 (60000)
16
= 11780.97
eqn. 1
TL
JG
T(5)
40 (π/180) = πd4
[ 32 ](83 x 106 )
T = 113774.606 d4
Equate 1 & 2
113774.606 d4 = 11780.97 d3
D = 0.10354 m
eqn.2
13. Compute the maximum unit sheared in a 2 inches diameter steel shafting that
transmits 24,000 in-lb of torque at 120rpm.
Solution:
16T
S= 3
πd
16(24,000)
S=
π(2)3
S = 15278.87 psi
14. What is the diameter of line shaft that transmit 150KW at 15rps?
Solution:
For line shaft:
D3 N
P=
53.5
D3 (15 x 60)
150/0.746 =
53.5
D = 2.28 in
15. A main shaft has 50mm diameter is running at 300 rpm. What is the power that
could be delivered by the shafted.
Solution:
50
D3 N (25.4)(3000)
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:
For short shaft:
D3 N
P=
38
(1.5)3 N
44.4 =
38
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=
D3 N
38
1.33 N
44.4 =
38
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:
For solid shaft:
16T
16T
S=
=
= 0.1886 T
3
πD
π(3)3
For hollow shaft;
16TDo
S=
π(D4o − D4i )
16T(4)
0.1886 T =
π(4 4 −Di4 )
(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:
Brake power = 4.905/0.68
P=2πTN
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.
Solution:
V=πDN
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=πDN
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
Solving for the torque developed:
P =2 π T N
190 = 2π T (300/60)
T = 6.04788 KN-m
16
S = 3 √M 2 + T 2
πd
16
40000 = πd3 √(1.266)2 − (6.04788)2
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.
Solution:
St =
16
πd3
[M+√M2 + T2 ]
16
St = π(0.01)3 [2.5+√(2.5)2 + (6)2 ]
St = 45,836.62 kPa
24. A shaft has a length of 10 ft. Find the diameter of the shaft that could safely deliver.
Solution:
D2/3=
L
=
10
8.95 8.95
D = 1.181 in
25. A 12 ft shaft is running at 260 rpm. What that this could safely deliver?
Solution:
D2/3 =
L
=
12
8.95 8.95
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:
Do = 1/2Di
1
2
π( Di)4
=
32
Di = 15/16
1
2
π( Di4 −Di4 )
32
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.
Solution:
TL
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:
P = 2π T N
50 = 2 π T (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.
Solution:
S=
16T
πd3
41.4 =
16T
π(102)3
T = 8626422.927 N-m
16(862422.927)(127)
41.4 =
4
π(127 − Di4 )
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.
Solution:
P =2 π T N
2 = 2 π T(188/2 π)
T = 0.01064 KN-m
16T
S=
πD3
16(0.01064)
40 x103 =
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.
Solution:
P=2πTN
0.375 = 2 π T(1800/60)
T =0.001989 KN-m
0.001989(20)
10(π/180) = πD4
(80 x106 )
32
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:
For solid shaft:
16T
16T
S=
=
3
πD
π(25)3
S = 3.259 x10-4
For hollow shaft:
16T(38)
3.259 x104 T =
π(384 Di4 )
Di= 33.64 mm
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.
Solution:
16T
16(40000)
S=
=
3
πD
π(1.998)3
S = 25541.33831 psi
50000
Fs =
25541.3383
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.
Solution:
16T
16TDo
=
3
πd
π(Do4 Di4 )
Do4 – Di4 = d3Do
Do4- Di4 = (3.5)3 Do
Do4 – Di4 = 42.87 Do eqn.1
π
1 π
(4 (Do2 − Di2 )) L w = 2 (4 d2 )
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)
Do =
2(1)
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.
Solution:
16T
16TDo
=
3
πd
π(Do4 − Di4 )
16T
16T(100)
=
3
π(76)
π(1004 −Di4 )
(100)4 – Di4 = (76)3(100)
Di =86.55 mm
π
mS = ( 4 (76)3 ) L w = 4536.36 w L
π
mH = [ 4 (Do2 − Di2 )]L w
π
=[ 4 (1002 − 86.552 )] L w = 1970.64 w L
Percentage of weight saved =
4536.36−1970.64
4536.36
=56.56%
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 [
Θ =175.220
12−6
2(72)
]
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:
L =1.57(12+ 6) + 2(72) +
(12−6)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.
Solution:
V=πDN
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?
Solution:
V=πDN
V = π (24/12) (600)
V = 62.83ft/sec
F1 - F2 =b t(S −
ρv2
2.68
)(
efθ−1
𝑒 𝑓𝜃
F1 –F2 = (12)(3/8)(300 −
)
0.035(62.83)2
2.68
e
)(
π
)−1
180
π
(0.30)(150)(
)
180
𝑒
(0.30)(150)(
)
F1 – F2 = 608.26 lbs
62.83
(F1 −F2 )v (608.26)( 60 )
Hp =
=
33000
33000
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.
Solution:
Θ= 180 ± 2 sin-1[
D2 −D1
Θ= 180 – 2 sin-1[
]
2C
30−10
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.
Solution
50=2 π T (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.
Solution:
F1 F1
S=
= = e fθ
bt F2
3
= e (0.32) (160) (π/180)
F2
1.2275
2= (0.35)t
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.
Solution:
Θ= 1800- 2 sin-1[
30−10
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.
D1N1 = D2N2
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?
Solution:
ρv2
efθ−1
F1 – F2 =b t [s − 2.68] [ efθ ]
V = π (24/12)(600/60)
V =62.83 ft/sec
F1 – F2 = (3/8) (12) [300 −
0.035(62.83)2
2.68
][
e150 x 0.3 x π/180 −1
e150 x 0.3 x π/180
]
F1 –F2 = 608.26 lbs
Hp =
(608.26)(62.83 x 60)
33000
=69.50 Hp
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
= 1500 kpa x 101.325 kpa =217.52 psi
550hp
efθ
b t = v(S−12p v2/g) [efθ −1]
(12ρv2/g) = [12(0.034)(69.57)2/ (32.2)] = 60.836 psi
550hp
e0.3(3)
(300/25.4)(10/25.4) =[69.573(217.52−60 836)] [e0.3(3)−1]
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
f= 0.313
F
F1 - 2.41 = 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
= efθ
F2
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:
b = 4L – 6.28(D + d)
200+110
= 4(60) – 6.28(
25.4
)
b = 163.35 in
200−110 2
)
25.4
163.35+ √(163.35)2 − 32(
c=
16
c = 20.34 in
θ = 180 –
(
200−110
)(60)
25.4
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.
Solution:
V=πDN
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:
4400 x
S=
9.81
1000
(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)
T = 0.6 N-m
P = 2 π (0.6)(460/60)
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
P = LD
550 =
20
0.1L
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.
SOLUTION:
15 = 2 π T (500/60)
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:
μn
D
S = p (C )
d
S1 = S 2
0.70 x 500
14
P2
76
2
(0.127) =
0.70 x 600
p2
2
76
(0.127)
= 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:
0.70 x 500
14
76
2
(0.127) =
μ = 0.251 poise
μ2 0.70 x 600
16.8
76
(0.076)
2
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)
T = 3.679 x 10-4 KN-m
P = 2 π (3.679 x 10-4)(900/60)
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:
Ρ=
1400
= 381.07 psi
(2.085)(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:
BM = 16.48
vn
BM =
p
v(1200)
16.48 =
381.07
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:
4400 x
S=
9.81
1000
(0.152)(0.28)
= 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.
P=
5500
= 1100 psi
(2.5)(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)
V =687307.77
687307.77 = π D (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?
Solution:
p
D=
180
sin( T )
9.56 =
3/4
180
sin( 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:
28.654 =
p
180
sin( 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:
D=
1/2
180
)
20
sin(
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.
Solution:
V=πDN
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
C=D+2
4
C =16 + 2 = 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)2
T+t
L = 2C + 2 + 40C
98+18
L = 2(34) +
2
(98−18)2
+ 40(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
2
(D+ )
C=
48 =
p
(D+8/2)
3/4
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.
Solution:
p
C = 8 (2L − T − t + √(2L − T − t)3 − 0.8(T − t)2 )
1/2
C = 8 [2(72) - 48-12 +√[2(72) − 48 − 12]3 − 0.80(48 − 12)2
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 = 431.25
V=πDN
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 (600)
(14)(3) =
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
total length of 60 pitches and a small sprocket with 23 teeth are to be used with an
operating temperature to remain at 45 degrees ℃. Determine the no. of teeth of the
larger sprocket.
Solution:
RC60
V = p t N = (6/8)(23)(1500)
V = 431.25
V=πDN
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:
Solution:
V= π D N= π (4)(3600)
V= 45238.9
P= FV
(400)(33000)=F (45238.9)
F=291.78 lb
Ds Ns = Dc Nc
4(3600) =48 Nc
Nc =300 rpm
V= π (48) (300)
V=45238.94
(291.78)(45238.94)
P=
=340hp
33000
17. In chain of ordinary proportions the term RC 40 means a pitch of _______.
Solution:
𝟒"
𝟏
RC40=
= in
𝟖
𝟐
18. The chain speed of an RC 80 chain on a 21-tooth sprocket turning at 600 rpm is
_______.
Solution:
8
RC80=
8
P
1
D=
180 =
180 = 6.709
sin( T )
sin( 21 )
600
V= π (6.709)(
)
60
1m
=210.769 in/sec (
)
39.37
V=5.353 m/s
19. What maximum chain number can be used if the small sprocket is to run at 1200 rpm
Solution:
900
900
P=( N )2/3=(1200)2/3=0.753 in
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.
Solution:
P
D=
180
sin( )
T
3
1
6 =
4
sin(180)
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.
Solution:
hp
= 0.004(T)1.06(N1)0.9(P)(3-0o.07p)
strand
= 0.004(20.95)1.06(1200)0.9(0.5)(3-0.07)(0.5)
𝐡𝐩
= 14
𝐬𝐭𝐫𝐚𝐧𝐝𝐬
22. A silent chain SC 4 is used for a design hp of 12. What chain width is needed if the 21tooth driving sprocket runs at 1200 rpm?
23. With a sprocket bore of 2”, the minimum number of sprocket teeth for RC40 is _______.
Solution:
4d
Tmin= +6
p
RC40
=
4(2)
4
8
+6
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
P= (0.885) (39.37)=
180
8
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 in2, determine the maximum pressure inside the
cylinder.
Sy= 440,000 (14.7/101.325)
Sy= 63,834.196 psi
Fe=
Fe=
𝑆𝑦 (𝐴)3/2
6
63,834.196 (0.13)3/2
6
= 498.67 lbs
F = 498.67 lbs (5) = 2493.37 lbs
P = F/A =
2493.37 𝑙𝑏𝑠
π
12 𝑐𝑚
( 4 )(2.54 𝑐𝑚)2
= 671.97 psi
6. The cylinder head of ammonia compressor has core gasket as of 80 cm 2 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
V = N ( 8m x 25.4 x 12 )
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;
Tan x =
𝐿
𝜋𝐷𝑚
=
6
𝜋(34)
= 0.05617
T = torque required
T=
T=
𝑊𝐷𝑚
2
tan 𝑥+𝑓
(1−𝑓 tan 𝑥 )
26(0.034)
2
0.05617+0.16
(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.
𝐿
5
Tan x = 𝜋𝐷 = 𝜋(25) = 0.0636
𝑚
For ACME thread, 𝜃 = 14.5o
W = 500 x 0.00981 = 4.905 KN
WDm = 4.905(0.025) = 0.12265 KN-m
T=
0.122 (cos 14.5) (0.0636)+(0.10)
2
[
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.
𝐿
Tan x = 𝜋𝐷
𝑚
L =2p
T=
𝑊𝐷𝑚
2
(
tan 𝑥+𝑓
1−𝑓 tan 𝑥
)
Dm = 2r
WDm = 80(0.80 x 2) = 12.8 KN-m
T=
12.8
2
0.03979 𝑥 0.13
(1−0.13 (0.03979))
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
Po =
𝑊 𝑥 𝑣𝑒𝑙
33,000
=
𝑃
30 𝑥 2800
33,000
eff = 𝑃𝑜 x 100% =
𝑖
2.545
= 2.545 Hp
x 100% = 42.42%
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%.
Tan x =
𝐿
𝜋𝐷𝑚
=
1
𝜋(1.5)
= 0.2122
Using the formula of eff,
eff =
tan 𝑥 (1−𝑓 tan 𝑥)
𝑓 𝐷
𝐷𝑚
tan 𝑥 + 𝑓 + ( 𝑐 𝑐 )(1−𝑓 𝑡𝑎𝑛𝑥)
Since collar friction is neglected, 𝑓𝑐 = 0
𝑓𝐷
Then the quantity, ( 𝐷𝑐 𝑐) (1 − 𝑓 𝑡𝑎𝑛𝑥) = 0
𝑚
0.6362 =
0.2022[ 1−𝑓 (0.2122)]
0 2122+𝑓+0
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.
eff =
tan 𝑥 (1−𝑓 tan 𝑥)
𝑓 𝐷
tan 𝑥 + 𝑓 + ( 𝐷𝑐 𝑐 )(1−𝑓 𝑡𝑎𝑛𝑥)
𝑚
0.70 =
tan 𝑥[ 1−0.10(tan 𝑥)]
tan 𝑥+0.10+0
0.10 tan2x + 0.07 = tan x – 0.10 tan2x
0.10 tan2x – 0.3 tan x +0.07
Using quadratic formula; tan x =
−(0.3)±√(0.3)2 −4(0.10)(0.07)
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/cm 2.
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
Tension = [ 4 (12)2 ](10 kg/cm2)
Tension = 1130.973 kg x 2.205 lb/kg
Tension = 2493.796 lbs
Tension per bolt =
2493.796 𝑙𝑏𝑠
5 𝑏𝑜𝑙𝑡𝑠
Tension per bolt = 498.76 lbs x
= 498.76 lbs/bolt
1 𝑘𝑔
2.205 𝑙𝑏𝑠
x 0.00981 KN/kg
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
L = 0.025 m = 25 mm
tan x =
𝑙𝑒𝑎𝑑
𝜋 𝐷𝑚
=
25
𝜋 (36)
= 0.221
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.
Tc =
Tc =
P =
𝑓𝑐 𝑊 (𝑟𝑜 −𝑟𝑖 )
2
0.15 (1000 𝑙𝑏)(2+1)
2(12)
2𝜋𝑇𝑁
33,000
=
= 18.75 ft-lb
2𝜋(18.75)(25)
33,000
= 0.08925 Hp
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.
T = fWrf = fW[
(𝑟𝑜 −𝑟𝑖 )
2
]
T = 0.15(1000)(0.0091) [
(0.05−0.02)
2
]
T = 0.00515 KN-m
P = 2𝜋TN
P = 2𝜋(0.000515)(1000/60)
P = 0.5393 𝐾𝑊 x
1 𝐻𝑝
0.746 𝐾𝑊
P = 0.7229 Hp
21. If the pitch of a screw is 2/9, find the thread per inch.
Pitch = 2/9
TPI =?
Pitch =
2
9
=
1
𝑇𝑃𝐼
1
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
Fc =
𝑆𝑦 𝐴𝑠 3/2
6
As = Stress area
As = [
6𝐹𝑐
𝑆𝑦
2/3
]
2
As = [
6 𝑥 350 3
55,000
] = 0.1134 in2
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.
Hp =
3=
𝐷𝑁𝑑3/2
50
1(1000)𝑑3/2
50
d = 0.4383 in
no. of set screws =
0.4383
3/8
= 1.7
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?
Fc =
𝑆𝑦 𝐴𝑠 3/2
6
As = Stress area
As = [
6𝐹𝑐
𝑆𝑦
2/3
]
2
As = [
6 𝑥 500 3
58,000
] = 0.13882 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
tan x =
𝑙𝑒𝑎𝑑
𝜋 𝐷𝑚
=
1
𝜋 (2.28)
= 0.1415
x = tan-1 0.1415 = 8.05o
T=
𝑊𝐷𝑚
2
[
tan 𝑥+𝑓
1−𝑓 tan 𝑥
]
T=
(4000)(2.25)
2
[
0.145 + 0.10
] = 1102.35 in-lb
1−0.10 (0.1415)
Dc = 2 rc = 2 (2) = 4 in
Tc =
Tc =
𝑓𝑐 𝑊 (𝑟𝑜 −𝑟𝑖 )
2
0.15 (4000 𝑙𝑏)(4)
2
= 1200 in-lb
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
P=
2𝜋𝑇𝑁
33,000
=
2𝜋(191.86)(120)
33,000
= 4.38 Hp
27. The cylinder of a compressor ha core gasket area of 80 cm2 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
L = 0.025 m = 25 mm
tan x =
𝑙𝑒𝑎𝑑
𝜋 𝐷𝑚
=
25
= 0.221
𝜋 (36)
x = 12.46°
Torque required to turn the screw, Ts
For ACME thread, 𝜃 = 14.5o
Ts =
Ts =
𝑊𝐷𝑚 (cos 14.5) (tan 𝑥)+𝑓
2
[
cos 14.5 − 𝑓 tan 𝑥
]
(900)(0.036) (cos 14.5) (0.221)+(0.15)
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
Fe=
6
Fe =
63,834.196 (0.15)3/2
6
Fe = 618.07 lbs
F = 618.07 lbs (5) = 3090.36 lbs
P = F/A
P=
3090.36 𝑙𝑏𝑠
π
12 𝑐𝑚
( 4 )(2.54 𝑐𝑚)2
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
tan x =
eff =
𝑙𝑒𝑎𝑑
𝜋 𝐷𝑚
=
6
𝜋 (36)
= 0.053
tan 𝑥 (1−𝑓 tan 𝑥)
𝑓 𝐷
tan 𝑥 + 𝑓 + ( 𝐷𝑐 𝑐 )(1−𝑓 𝑡𝑎𝑛𝑥)
𝑚
eff =
0.053 [1−0.13(0.053)]
0.10(90)
]+ [1−0.13(0.053)]
36
0.053 +0.13 + [
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:
Tc =
𝑊𝐷𝑚
2
[
tan 𝑥+𝑓
1−𝑓 tan 𝑥
]
37
27
(0.3)(400) 0.3(400)( 2 + 2 )
Tc =
[
]
2
1−0.10 (0.02274)
Tc = 1920 kg-mm
For ACME of trapezoidal thread, 𝜃 = 14.5o
tan x =
Ts =
Ts =
𝑙𝑒𝑎𝑑
𝜋 𝐷𝑚
=
4
𝜋 (18)
= 0.0707
𝑊𝐷𝑚 (cos 14.5) (tan 𝑥)+𝑓
[
2
cos 14.5 − 𝑓 tan 𝑥
]
(400)(18) (cos 14.5) (0.0707)+(0.30)
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
Dm =
𝐷𝑖 + 𝐷𝑜
2
=
6+8
2
= 7 in
Solving for the lead angle of the screw:
tan x =
𝑙𝑒𝑎𝑑
𝜋 𝐷𝑚
=
0.5
𝜋 (7)
= 0.02274
solving for the torque required to turn the screw:
T=
T=
𝑊𝐷𝑚
2
[
tan 𝑥+𝑓
1−𝑓 tan 𝑥
(2000)(7)
2
[
]
0.02274 + 0.10
]
1−0.10 (0.02274)
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
1 𝑖𝑛
(1𝑐𝑚 𝑥 0.3937 𝑐𝑚)
Pithc =
1
𝑇𝑃𝐼
=
1
2.54
= 0.3937
L = 2p = 2(0.3937) = 0.7874 in
1
1
Dm = Do - 2p = 1.968 in - 2(0.3937)
Dm = 1.77115 in
Tan x =
𝐿
𝜋𝐷𝑚
=
0.7874
𝜋(1.77115)
x = 8.0545
eff =
tan 𝑥 (𝑐𝑜𝑠𝜃−𝑓 𝑠𝑖𝑛𝑥)
𝑓 𝐷
𝑡𝑎𝑛𝑥 cosθ + 𝑓 𝑐𝑜𝑠𝑥+[ 𝐷𝑐 𝑐 ](𝑐𝑜𝑠𝜃−𝑓 𝑠𝑖𝑛𝑥)
𝑚
eff =
(tan 8.0545)[cos 14.4−0.15(sin 8.0545)
0.13(3,54)
(tan 8.0545)(𝑐𝑜𝑠14.4) + 0.15(𝑐𝑜𝑠8.0545)+[ 1.77115 ] [(𝑐𝑜𝑠14.4)−0.15(𝑠𝑖𝑛8.0545)]
eff = 0.276766
eff =
𝑊 𝐿𝑒𝑎𝑑
𝐹(2𝜋𝑟)
0.276766 =
𝑊 (0.7874)
𝑙𝑏𝑠
(60 𝑘𝑔 𝑥 2.205 𝑘𝑔 )[(2𝜋)(29.527)]
1 𝑘𝑔
𝐾𝑁
W = 8627.3265 x 2.205 𝑙𝑏𝑠 x 0.00981 𝑘𝑔
W = 38.3827 KN
36. Determine the stress area in in2 of a 1-8UNC bolt.
As = ?
1-8 UNC bolt
𝜋
As = 4 [D- 0.9743(P)]2
𝜋
As = 4 [1- 0.9743(0.125)]2
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
Pitch =
1
𝑇𝑃𝐼
=
1
4
= 0.25
𝜋
As = 4 [D- 0.9743(P)]2
𝜋
As = 4 [1- 0.9743(0.25)]2 = 0.4494 in2
𝜋
A = 4 (Dm)2
𝜋
0.4494 in2 = 4 (Dm)2
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
Tc =
Tc =
P =
𝑓𝑐 𝑊 (𝑟𝑜 −𝑟𝑖 )
2
0.15 (1000 𝑙𝑏)(2+1)
2(12)
2𝜋𝑇𝑁
33,000
=
= 18.75 ft-lb
2𝜋(18.75)(25)
33,000
= 0.08925 Hp
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 =
3/1 =
ℯ
ℯ 𝑓∅
0.25(∅ 𝑥
𝜋
)
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
F1/F2 =
ℯ 𝑓∅ = ℯ 𝑓(245 𝑥 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?
S=
𝐹1
𝑤𝑡
55,000 =
6
𝑤(0.004)
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 =
ℯ 𝑓∅
800/285 = ℯ
𝜋
𝑓(210 𝑥 180 )
2.807 = ℯ 3.665𝑓
Taken in both side:
ln2.807 = ln ℯ 3.665𝑓
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 =
4=
ℯ
ℯ 𝑓∅
𝑓(260 𝑥
𝜋
)
180
f = 0.305
8. A band brake has an angle of contact of 280 degrees and is to sustain a torque of 10,000inlb. 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 =
ℯ 𝑓∅
F1/F2 = ℯ
0.25(270 𝑥
𝜋
)
180
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
7.5
60,000 = 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 =
ℯ 𝑓∅
F1/F2 = ℯ
𝜋
0.25(245 𝑥 180 )
F1 = 2.915F2
T = F x r = 8.0442(0.75/2)
T = 3.057 KN-m
= 2.912
P = 2𝜋TN
P = 2 𝜋 (3.057)(260/60)
P = 83.2278 KW
12. Determine the internal load expressed in lb-ft2 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.
T=
(𝑤𝑟 2 ) 𝑁
308 (𝑡)
T = 900/12 = 75 ft-lb
75 =
(𝑤𝑟 2 ) (840)
308 (3.5)
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
T=
𝐹𝑎 𝑓 𝑟𝑓
sin 12
0.3
0.15279 =
𝐹𝑎 (0.45)( 2 )
sin 12
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.
T=
𝐹𝑎 𝑓 𝑟𝑓
sin 12
0.2
T=
600(0.45)( 2 )
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
rf =
1 𝐷3 − 𝑑 3
1 (8)3 − (4)3
3 𝐷 −𝑑
3 (8)2 − (4)2
[ 2
]= [
2
] = 3.111111
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
2𝜋𝑇𝑁
P=
=
33,000
2 𝜋 (30 𝑓𝑡−𝑙𝑏)(400)
33,000
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
P=
33,000
25 =
2 𝜋 (𝑇)(1200)
33,000
T = 109.42 ft-lb x 12 in/ft = 1,313.03 in-lb
T=
𝐹𝑎 𝑓 𝑟𝑓
sin 12
1313.03 =
𝐹𝑎 (0.35)(
0.16 𝑖𝑛
)
2
sin 12
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.
P=
8=
2𝜋𝑇𝑁
33,000
2 𝜋 (𝑇)(250)
33,000
T = 168.02 ft-lb
F1/F2 =
ℯ 𝑓∅
F1/F2 = ℯ
𝜋
0.4(270 𝑥 180 )
= 5.342
F1 = 5.342F2
T = (F1 - F2) r
168.02 = (F1 - F2)(
15 𝑖𝑛
12 𝑖𝑛/𝑓𝑡
2
)(15 in/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
5,000 =
330.75
1
16
𝑏( )
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
8,000 =
𝐹
1
16
(1.5)( )
F = 750 lbs
T=Fxr
T = 750(16in/2)
T=
6,000 𝑖𝑛−𝑙𝑏
= 500 ft-lb
12 𝑖𝑛/𝑓𝑡
P=
P=
2𝜋𝑇𝑁
33,000
2 𝜋 (500 𝑓𝑡−𝑙𝑏)(350)
33,000
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
rf =
1 𝐷3 − 𝑑 3
1 (0.200)3 − (0.100)3
3 𝐷 −𝑑
3 (0.200)2 − (0.100)2
[ 2
]= [
2
] = 0.0778 m
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.
T=
𝐹𝑎 𝑓 𝑟𝑓
sin ∅
450 𝑁(0.20)(
T=
sin 8
0.25
)
2
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.
w=
(𝐷−𝑑)/2
sin ∅
250 =
(300−250)/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.
w=
(𝐷−𝑑)/2
sin ∅
250 =
(250−200)/2
sin ∅
∅ = 0.112957 deg.
rf =
T=
1 𝐷3 − 𝑑 3
1 (0.250)3 − (0.200)3
3 𝐷 −𝑑
3 (0.250)2 − (0.200)2
[ 2
𝐹𝑎 𝑓 𝑟𝑓
sin ∅
]= [
2
] = 0.11296
T=
0.5 𝐾𝑁(0.30)(0.11296)
sin 8
= 0.0847 KN-m
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.
rf =
1 𝐷3 − 𝑑 3
1 (0.200)3 − (0.100)3
3 𝐷 −𝑑
3 (0.200)2 − (0.100)2
[ 2
]= [
2
] = 0.0778 m
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 =
ℯ 𝑓∅
4/1 = ℯ
𝑓(250 𝑥 180 )
ln 4 = ln ℯ
f = 0.318
𝜋
4.3622𝑓
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 = fF[
𝑟𝑜 + 𝑟𝑖
2
]
T = (0.35)(800) [
0.10 + 0.05
2
] = 21 N-m = 0.021 KN-m
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.
w=
(𝐷−𝑑)/2
sin ∅
250 =
(280−200)/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.
rf =
1 𝐷3 − 𝑑 3
1 (0.255)3 − (0.025)3
3 𝐷 −𝑑
3 (0.255)2 − (0.025)2
[ 2
]= [
2
] = 0.08574
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.
rf =
1 𝐷3 − 𝑑 3
1 (0.300)3 − (0.240)3
3 𝐷 −𝑑
3 (0.300)2 − (0.240)2
[ 2
]= [
2
] = 0.1355
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
S = Mc/l =
2
9
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-m2 )
Y = maximum deflection
Y=
Y=
5𝑤𝐿4
384 𝐸𝑙
5(2.5)(12)4
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
)
Y1 =
𝑃𝐿3
48 𝐸𝑙
=
(15)(9)3
48(7,000)
= 0.0325446 m
Considering the effect of uniform distributed load of 2 kN/m.
Y2 =
5𝑤𝐿4
384 𝐸𝑙
=
5(3)(9)4
384 (7,000)
= 0.036613 m
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-m2 )
∅=
∅=
𝑃𝐿4
8 𝐸𝑙
25(14)4
8 (10,000)
= 0.6125 rad
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-m2 )
∅=
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 )
∅1 =
∅1 =
∅2 =
∅2 =
𝑤𝐿3
6 𝐸𝑙
2.3(11)3
6 (12,000)
= 0.0425 rad
𝑃𝐿2
8 𝐸𝑙
11(11)2
8 (12,000)
= 0.01386 rad
∅ = ∅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
𝑆=
6𝑀
𝑏ℎ2
100,000 =
6 (3)
(0.03)(ℎ)2
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
M=
𝑃 (4)
4
M=P
𝑆=
6𝑀
𝑏ℎ2
8,300 =
6 (𝑃)
(0.05)(0.20)2
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=2𝜋RN
10 = 2 𝜋 R (8/60)
R = 11.94
a = acceleration
a=
𝑉2
𝑅
=
(10)2
11.94
= 8.377 m/s2
11. A car travels with an initial velocity of 36 km/hr. If it is decelerating at the rate of 2 m/s 2,
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.
V f = 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
𝑤
fw=𝑔 a
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
𝑚𝑉 2
= mg
𝑅
V2 = gR
V2 = 9.81(1.5)
V = 3.836 m/s
Solving for N:
V=2𝜋RN
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
𝑚𝑉 2
= 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=2𝜋DN
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
Time =
7
= 9.6211 min = 577.27 sec
𝑠𝑡𝑟𝑜𝑘𝑒𝑠
𝑖𝑛
(120
)(0.006063
)
min
𝑠𝑡𝑟𝑜𝑘𝑒
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 ft3.
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
Cutting time =
696.69 𝑠𝑒𝑐
60 𝑠𝑒𝑐/𝑚𝑖𝑛
= 11.61 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
Time = (24 𝑡𝑒𝑒𝑡ℎ/𝑟𝑒𝑣)(90 𝑟𝑒𝑣/𝑚𝑖𝑛)(0.006 𝑖𝑛/𝑡𝑜𝑜𝑡ℎ)
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
Time =
12 𝑓𝑡 𝑥 12 𝑖𝑛/𝑓𝑡
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.
S=
𝑃 𝐷𝑖
2𝑡𝜂
140,000 =
𝑃 (0.50)
2 (0.02)
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.
S=
𝑃 𝐷𝑖
4𝑡𝜂
63,000 =
𝑃 (20)
4 (0.3125)(0.80)
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
S=
𝑃 𝐷𝑖
2𝑡
50,000 =
117.72(10)
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
𝐹𝑆
=
296
3
𝑃 𝐷𝑖
=
2𝑡
8 (50)
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%.
S=
S=
𝑃 𝐷𝑖
2𝑡𝜂
10 𝑀𝑝𝑎 (400)
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
S=
𝑃 𝐷𝑖
2𝑡
40,000 =
147.15 kPa(12)
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.
S=
𝑃 𝐷𝑖
4𝑡
46,875 =
5000 (𝐷𝑖 )
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.
S=
𝑃 𝐷𝑖
4𝑡
16,000 =
𝑃 (10)
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.
S=
𝑃 𝐷𝑖
4𝑡
1,200 =
500 (5)
2𝑡
t = 1.04166 in
t/Di = 1.04166 in / 5 = 0.2087
Therefore, use thick-wall formula:
𝐷
(𝑆 + 𝑃 )
𝑖
t = [√ 𝑡
− 1]
2
(𝑆 −𝑃 )
𝑡
5
𝑖
(1200 + 500)
t = 2 [√(1200 − 500) − 1]
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
(𝑆 + 𝑃 )
t = 2 [√ ( 𝑆𝑡 − 𝑃 𝑖 ) − 1]
𝑡
.050 =
𝑖
0.30
(30,000 + 𝑃 )
[√ (30,000 − 𝑃 𝑖 ) − 1]
2
𝑖
(30,000 + 𝑃 )
0.333 = [√ (30,000 − 𝑃 𝑖 ) − 1]
𝑖
(30,000 + 𝑃𝑖 )
− 1]
(30,000 − 𝑃𝑖 )
1.333 = [√
Squaring both sides:
1.333 =
3000 + 𝑃𝑖
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
S=
𝑃 𝐷𝑖
2𝑡
25,000 =
44.145 (3)
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
Sto =
Sto =
2 𝑃𝑖 𝑟𝑖2 − 𝑃𝑜 (𝑟𝑜2 + 𝑟𝑖2 )
𝑟𝑜2 − 𝑟𝑖2
2(13,000)(9)2 −800(152 + 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
Sto =
2 𝑃𝑖 𝑟𝑖2 − 𝑃𝑜 (𝑟𝑜2 + 𝑟𝑖2 )
Sto =
𝑟𝑜2 − 𝑟𝑖2
2(65)(8)2 − 13(142 + 82 )
(14 2 + 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
S=
𝑃 𝐷𝑖
2𝑡
3,200 =
24.525 (2.5)
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
C=
𝐷1 +𝐷2
1.7143 + 5.2857
=
= 3.50 in
2
2
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?
C=
𝑇𝑔 +𝑇𝑝
15 =
2𝐷𝑃
𝑇𝑔 +40
2(2)
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.
𝑁
3
Speed ratio = 𝑁1 = 1
2
D2 = (N1/N2)D1
D2 = (6/1)D1
D2 = 6D1
C=
𝐷1 +𝐷2
2
2.6 =
6𝐷1 +𝐷2
2
𝑇
1
D1 = 𝐷𝑃
𝑇
10
0.743 = 1
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.
DP =
25.4
𝑀
or M =
25.4
𝐷𝑃
25.4
DP = 6.5 = 3.90769
𝑁+2
Do = 𝐷𝑃 , N=no. of teeth
𝑁+2
195/25.4 = 3.90769
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
1.5708 = 40
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/T
2 = D / 40
D = 80 mm
V=𝜋DN
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 = 2/P
𝜋
Pc = 𝐷𝑃
𝜋
0.7854 = 𝐷𝑃
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.
C=
𝐷−𝑑
2
20 =
50 − 𝑑
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
0.2122 = Ft [
10 𝑖𝑛
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/m2.
y=
8𝐶 3 𝑛
𝐺𝑑
C = Dm/d = 70/10 = 7
𝑦
𝐹
=
1
30
8𝐶 3 𝑛
=
𝐺𝑑
8(7)3 𝑛
(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
Whaal factor =
4𝐶 −1
4𝐶−4
+
0.615
𝐶
=
4(13.86) −1
4(13.86)−4
+
0.615
13.86
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
b.) Pc =
Pc =
𝜋𝐷
𝑇
=
𝜋𝐷
12.7
= 0.2474
2𝜋𝐶
𝑇𝑔 + 𝑇𝑝
0.2474 =
2𝜋𝐶
18 + 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.
K=
4𝐶 −1
4𝐶−4
K= 1.184
+
0.615
𝐶
=
4(8) −1
4(8)−4
+
0.615
8
C =dm/d
8 =Dm/d
Dm = 8d
Using stress formula:
K=
8𝐾𝐹𝐷𝑚
𝜋𝐷3
90,000 =
8(1.184)(50)(8𝑑)
𝜋𝑑 3
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
F1 / y1 = F2 / 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.
F=
C=
K=
K=
200
3/12
𝐷𝑚
𝑑
= 800 lbs
=
7𝑑
4𝐶 −1
4𝐶−4
𝑑
+
8𝐾𝐹𝐷𝑚
𝜋𝐷3
=7
0.615
𝐶
=
4(7) −1
4(7)−4
+
0.615
7
= 1.2128
100,000 =
8(1.2128)(800)(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.
y=
8𝐶 3 𝑛
𝐺𝑑
0.075 =
8(10)(𝐶)3 (12)
(80 𝑥 106 )(0.025)
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
K=
4𝐶 −1
4𝐶−4
+
0.615
𝐶
=
4(8) −1
4(8)−4
+
0.615
8
= 1.184
Ss = stress at solid length
Ss =
Ss =
8𝐾𝐹𝐷𝑚
𝜋𝐷3
8(1.184)(15)(0.10)
𝜋(0.0125)3
= 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.
A.) Ss =
8𝐾𝐹𝐷𝑚
𝜋𝐷3
Dm = Di + d = 2.5 + d
63,000 =
8(1.25)(825)(2.5+𝑑)
𝜋𝑑 3
Try: d = 0.5 in.
63,000 =
8(1.25)(825)(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
y=
8𝐶 3 𝑛
𝐺𝑑
2.0625 =
8(825)(6)3 (𝑛)
(10,800,000)(0.5)
n = 7.8125 coils