Download Machine Design and shop practice and more Exercises Computer science in PDF only on Docsity!
SHAFTING
PROBLEM SET
- 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 = 2 π T N
200 = 2πT (300/60)
T = 6.366 KN-m
T = F x r
6.366 = F (0.20/2)
F = 63.66 KN
- 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.
S =
16 T
π d
3
π d
3
d = 0.05829 m
- 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
F. S
16 T D
o
π (0.
4
− D
4
π ¿ ¿
D = 0.69624 m
- What is the speed of 63.42mm shaft transmitted 75KW if stress is not to exceed
26Mpa.
Solution:
S =
16 T
π D
3
16 T
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
- A steel shaft transmit 50Hp.at 1400rpm. If allowable stress is 500psi, find the
diameter.
Solution:
P =
2 πTN
2 π T ( 1400 )
T = 187.575 ft-lb x 12 in/ft
T = 2250.905 in-lb
S =
16 T
π D
3
πD
3
D = 2.841 in
- 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:
TL
JG
J =
π (0.06735)
4
J = 2.02668 x 10
-
m
3
- A force tangent to 1 foot diameter pulley is mounted on a 2 inches shaft. Determine
the torsional deflection if G= 83 x
6
kPa.
Solution:
T =F (D/2)
T = 5 (0.3048/2)
T = 0.762 KN-m
J =
πD
3
π (0.0508)
4
J = 6.53814 x 10
m
4
TL
JG
θ/L =
(6.53 x 10 ¿ ¿− 7 )( 83000000 )¿
θ/L = 0.0140418 rad/m x (180/π)
θ/L = 0.
0
/m
- 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 3
o
in a length of 6m? (G= 83 x 10
6
kPa)
Solution:
TL
JG
0
(π/180) =
14 x 6
J ( 83 x 10
6
J = 1.93286 x 10
π d
4
D = 0.11845 m
- 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 =
16 T
πd
3
Solving for T in terms of d;
T =
π d
3
= 11780.97 eqn. 1
TL
JG
0
(π/180) =
T ( 5 )
[
πd
4
]
( 83 x 10
6
T = 113774.606 d
4
eqn.
Equate 1 & 2
113774.606 d
4
= 11780.97 d
3
D = 0.10354 m
- Compute the maximum unit sheared in a 2 inches diameter steel shafting that
transmits 24,000 in-lb of torque at 120rpm.
Solution:
S =
16 T
πd
3
S =
π ( 2 )
3
S = 15278.87 psi
- What is the diameter of line shaft that transmit 150KW at 15rps?
Solution:
For line shaft:
P =
D
3
N
D
3
( 15 x 60 )
D = 2.28 in
- A main shaft has 50mm diameter is running at 300 rpm. What is the power that
could be delivered by the shafted.
0.1886 T =
16 T ( 4 )
π ( 4
4
− Di
4
4
4
Di = 3.48 in
- 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 = w Q h
P = 9.81(3/60)(10)
P = 4.905 KW
Solving for the power input of the pump:
Brake power = 4.905/0.
P = 2 π T N
7.213 = 2 π T (550/60)
T = 0.1252 KN-m
S =
π (0.035)
3
S = 1487.63 kpa
- 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 = π D N
25 = π (0.5) N
N = 954.93 rpm
- 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:
N
1
/N
2
1800/N
2
N
2
= 1000 rpm
V = π D N
V = π (0.8)(1000/60)
V = 41.88 m/sec x 3.281ft
V = 137.43 ft/sec
- 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
S =
πd
3 √
M
2
+ T
2
π d
3
√
2
D = 92.27 mm
- 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:
S
t
πd
3
[M+
√
M
2
+ T
2
]
S
t
π ¿ ¿
[2.5+
√
2
2
]
S
t
= 45,836.62 kPa
- A shaft has a length of 10 ft. Find the diameter of the shaft that could safely deliver.
Solution:
Thickness = 15mm
- 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
- 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 T w
Tw = 4.476 KN-m
Tw =F x r
4.476 = F x (0.712/2)
F =12.57 KN
- 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 =
16 T
π d
3
16 T
π ¿ ¿
T = 8626422.927 N-m
π ( 127
4
− Di
4
Di =105.815 mm
- A steel shaft operates at 188 rad/sec and must handle of 2 KW power. The sharing
stress is not to exceed 40 MN/m
2
. 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
S =
16 T
π D
3
40 x
3
π D
3
D = 0.1106 m
- 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 = 2 π T N
0.375 = 2 π T(1800/60)
T =0.001989 KN-m
0
(π/180) =
π D
4
( 80 x 1 0
6
D= 0.0066206 m
- 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:
Di
4
= (Do
2
2
eqn.
Do
4
2
2
= 42.87Do
Do
2
Do =
√
2
Do =4.225 in = 107 mm
Solving for the inner diameter:
Do
2
2
2
2
Di =3.425 in = 87 mm
- 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:
16 T
π d
3
16 TDo
π ( Do
¿ 4 − Di
4
16 T
π ( 76 )
3
16 T ( 100 )
π ( 100
¿ 4 − Di
4
4
4
3
Di =86.55 mm
m S
= ¿ L w = 4536.36 w L
m H
[
π
( Do
2
− Di
2
]
L w
=¿ L w = 1970.64 w L
Percentage of weight saved =
- A solid steel shaft whose S y
= 300 M pa and S u
= 400 Mpa, is used to transmit 300
KW at 800 rpm. Determine its diameter.
Solution
BELTS
PROBLEM SET
- 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
[
]
0
- 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) +
2
L =172.385 cm
- 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:
D
1
N
1
= D
2
N
2
(12)(600) =20 N
2
N
2
= 360 rpm
Considering the 5% slip:
N
2
N
2
= 342 rpm
- 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 = π D N
20 = π (0.3) N
N = 21.221 rps
P =2 π T N
P = 2 π (4) (21.221)
- 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.
T=F x r
0.795775 = F x 0.
F= 2.65 KN
- 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:
S =
F
1
bt
F
1
F
2
= e
f θ
F
2
= e
(0.32) (160) (π/180)
( 0.35) t
T =7.24 mm
- 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 = (F
1
– F
2
)v
20 x 0.746 = 2.5 V
V = 5.97 m/sec x 3.281ft
V =19.58 ft/sec
- 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:
0
[
]
0
- 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
1
N
1
= D
2
N
2
8(600) = 14N
2
N
2
N
2
= 329 rpm
- 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/in
3
. the angle of contact is 150
o
.If the
coefficient of friction is 0.3 and the safe stress is 300 psi, how much power can it
deliver?
Solution:
F
1
– F
2
=b t
[
s −
ρ v
2
] [
e
fθ − 1
e
fθ
]
V = π (24/12)(600/60)
V =62.83 ft/sec
F
1
– F
2
[
e
150 x 0.3 x π / 180
e
150 x 0.3 x π / 180
]
F
1
–F
2
= 608.26 lbs
Hp =
( 608.26) (62.83 x 60 )
=69.50 Hp
- The standard width of a B85 premium quality V-belt = 21/32 inches
- 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:
- 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:
F
1
F
2
F
max
= S
d
(b t)
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:
e
f
θ
f(160)(
π /180)
f= 0.
F
1
F
1
F
1
= 2.674 KN
T = 1.20 T = 1.20(0.156) = 0.1872 KN-m
(F
1
- F
2
) r = T’
(2.674 – F
2
F
2
= 0.802 KN
F 1
F 2
= e
fθ
= e
0.3135 θ
Θ = 3.841 rad x 180/π = 220
0
- 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:
- 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)
b = 163.35 in
c = 163.35+ √
2
c = 20.34 in
θ = 180 –
(
)
θ = 169.
0
- 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 = π D N
V = π (0.61)(500/60)
V = 15.92 m/sec
40 = (F
1
–F
2
F
1
– F
2
= 2.505 KN = 2505 N
- 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.
and the safe working stress of the belt is 2.1 Mpa. What is the effective belt pull in
Newton.
Solution:
F
1
= 2505 N
- 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.
- 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.
- 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
