JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

MODULE III

KEYS AND COUPLINGS

DEFINITIONS AND USEFUL INFORMATION

 Key – is a rigid connector between a shaft and the hub of another component such as pulley, cam, or gear.
 Key is a demountable machinery part which, when assembled into keyseats, provides a positive means for transmitting
torque between the shaft and hub.
 The most common function of a key is to prevent relative rotation of a shaft and the member to which it is connected,
such as the hub of a gear, pulley, or crank.
 Its purpose is to prevent relative rotation between the two parts.
 The key fits into mating grooves in the shaft and mating member called the keyway and transmits torque by shear across
the key.
 Keyseat is an axially located rectangular groove in a shaft or hub.
 Keyway is a groove in the shaft and the mating member to which the key fits.
 A key seat must be provided in the shaft and a key-way in the hub of the other part.
 Seat screw is frequently used to seat the key firmly in the keyway and to prevent axial motion of the parts.
 Selection of types of key in any installation depends on several factors, such as power requirements, tightness of fit,
stability of connection, and cost.
 Keys may be classified into constant cross section or variable cross section.
 The width of the square and flat keys is approximately one-fourth the shaft diameter.

Types of Key with Constant Cross Section

1. Square key is the most commonly used in general industrial machinery. The key is sunk half in the shaft and half in
the hub.
2. Flat key is a key used where the weakening of the shaft by the keyway is serious and where added stability of the
connection is desired, as in machine tools.
3. Round key is a key used for fastening cranks, hand wheels, and other parts that do not transmit heavy torques. Some
manufacturers employed this type of key for heavy-duty shafts over 6 inches because the absence of the sharp
corners reduces the stress concentration below that which would exist had a square key and flat key been used.
Keyway for this key may be drilled and reamed after assembly of the mating parts.
4. Barth key is a square in which the two bottom corners are beveled to ensure that the key will fit tightly against the
top of the keyway when the drive is in either direction, and lessen the tendency to twist. It does not require a tight
fit, and the small clearance permits easy assembly and removal.

Types of Keys with Variable Cross Section

1. Kennedy keysv, also called as tangential keys, are tapered square keys that are assembled with the diagonal
dimension virtually in a circumferential direction. This key is used for heavy-duty application.
2. Gib-head keys are tapered square or flat keys with head.
3. Pin key is usually a drive fit key, either straight or tapered (taper of ¼ to 3/16 inch per ft); longitudinally assembled
on the shaft, can be sized about ¼ of the shaft diameter and can transmit heavy power.
4. Woodruff key is a key widely used in the automotive and machine tool industries. It fits into a semi-cylindrical seat
in the shaft.
5. Feather key is a key that allows the hub to move along the shaft but prevents rotation on the shaft. It is used to
permit moving element, say gear, into or out of engagement with its mate, to engage or disengage, say, a jaw clutch.
6. Saddle key is used for light power; it is tapered and either hollow, with a radius of curvature slightly smaller than the
shaft radius, or flat that is assembled flat on the shaft.
7. Spline fitting is composed of a spline shaft formed by milling and mating hub with internal splines formed by
broaching. Splines are actually number of keys integral with the shaft. These splines have the advantages of greater
strength and a self-centering feature.

Figure 3.1. Flat key

D
Key
Hub
b L

T
t
T Shaft

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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

TORQUE TO BE TRANSMITTED THROUGH THE KEY

 Torque analysis for design calculation of key is either on shear or compression basis.

1. Transmitted Torque Based on Shear Stress

 Transmitted torque is due to the load or force, F n, that tends to shear the key on the middle point of its thickness,
in which the shearing area is the key length times half the thickness.

Figure 3.2. Load Analysis on the Key

W

Key
Fn
D
2

Pitch Circle of, say, a Gear

Shaft
Hub

 Transmitted Torque of the is given by the following expression

D Fn D D s bLD
T  s s A s    ss  b L   s Eq. 3.1
2 2 2 2

Where, T = transmitted torque, In-lb, kN-m D = shaft diameter, in., m.

ss = shear stress, psi, MPa As = shearing area, in2., m2
L = length of the key, in., m, mm t = thickness of the key, in., m, mm

Figure 3.3. Shearing Area of Key Shearing Area

t/2 Fn
t
Fn
b

2. Transmitted Torque Based on Compression Stress

 Transmitted torque based on compression is due to the bearing force applied on the bearing area, as shown in
figure 3.4 below, either on the key, hub, or shaft.

Figure 3.4. Bearing Area on the Key

L t/2 Fn
t

Fn b

Bearing
Area

Transmitted Torque,

D D D  t  s tLD

T    Fn    sc A b    sc   L  C Eq. 3.2
2 2  2  2 4
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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

Where, T = Transmitted torque, equal to transmitted torque based on shear, In-lb, kN-m
t = thickness of the key, in., m, mm
sc = compression stress, psi, MPa
Fn = bearing force, which is equal to shearing force, lb, kN

DESIGN CONSIDERATION FOR KEYS

 The following are important points to be considered in the design of keys.

1. The shearing stress in eq. 3.1 may be obtained from the shearing yield stress or strength of the key material.
2. The compression stress in eq. 3.2 is taken according to the weakest of the three parts involved, the shaft, the key, or
the hub.
3. Design factor of safety on the yield strength
Smooth load 1.5 Minor shock loading 2 to 2.25
Severe shock loading 2.25 to 4
4. Yield stress on shear to be used (for ductile material)
sys = syc sys = 0.6 sy (Faires) sys = 0.5 sy (MSST) sys = 0.577 sy (OSST)

DESIGN PROCEDURE FOR FLAT AND SQUARE KEYS

 The following is the recommended procedure in designing flat and square keys.
1. To find shaft diameter, change the cross-sectional dimensions b and t of the flat or square key in accordance with
ASA standards, from Table 3.1 below.
2. Solve for L from equation 1 and equation 2, and use the larger value.

TABLE 3.1. Dimensions of Flat and Square Keys (Table AT 19, p.594, Faires, Or Table 9-2, p199, Black & Adams)
Shaft Tolerance
Diameter (Inch) b (Inch) t (Inch) On b (Inch)

½ -9/16 1/8 3/32 - 0.0020

5/8 – 7/8 3/16 1/8 - 0.0020
15/16 – 1 ¼ 1/4 3/16 - 0.0020

1 5/16 – 1 3/8 5/16 ¼ - 0.0020

1 7/16 – 1 ¾ 3/8 ¼ - 0.0020
1 13/16 – 2 ¼ 1/2 3/8 - 0.0025

2 5/16 –2 ¾ 5/8 7/16 - 0.0025

2 7/8 – 3 ¼ ¾ ½ - 0.0025
3 3/8 – 3 3/4 7/8 5/8 - 0.0030

3 7/8 – 4 ¼ 1 ¾ - 0.0030
4¾-5½ 1¼ 7/8 - 0.0030
5¾-6 1½ 1 - 0.0030

OTHER INFORMATION ON KEYS

 Following are additional pointers to be remembered in the design of keys.

 Hub length – between 1.25D and 2.4D, where D is the shaft diameter.

 If the needed key length is greater than about 2D, use two keys 180o apart or Kennedy keys.

 If the load is other than smooth, the key should fit tightly, either by use of taper keys or by clamping the hub onto
the shaft and key.

 If the key and shaft are of the same material, the length of key required to transmit the full power capacity of the
shaft is determined by equating the shear stress of the key to the torsional stress of the shaft,
2T 16 T  1  D
   and b  . Then L = 1.18D or approximately L = 1.2D. The value 0.75 is the
b L D  D 3  0.75  4
estimated weakening effect of the keyway on the torsional strength of the shaft.

 The common material of keys is cold-finished, low-carbon steel (0.2 % C or less) or heat-treated steels.

Prob. # 3.1] A key is to be designed for a 12.7 cm shaft that will transmit power of 150 kW at 360 rpm. If the allowable shear
stress for the key is 920 kg/cm 2 and the allowable compressive stress is 1200 kg/cm 2, determine a) the length of the key; and
b) the axial force to remove the hub from the shaft if the coefficient of friction is 0.45. Key dimensions: b = 1 ¼ inches =
3.175 mm, and t = 7/8 inches = 2.2225 cm. (Board Prob. Oct. 1985)

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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

Solution:

a) For the length of the key

2T D
Based on shear, L  Key
bLD Hub
30 P 30 150
Where, T    3.979 kN - m
n   360 T
T Shaft
s s  920 kg/cm 2  90 220 720 Pa

2  3979
Then, L   0.021875 m  21.875 mm
0.03175  90 220 720  0.127 

4T
Based on Compression or Crushing, L 
t sc D

Where, sc = 1 200 kg/cm2 = 117 679 200 Pa

4  3 979
Then, L   0.04792 m  47.92 mm
0.022225 117 679 200  0.127 

Therefore, use key length of, say, L = 48 mm ans.

b) For the axial force to remove the hub

F F

2T
Fa = f (2 F) = axial force Where, F 
D
4 f T 4  0.45  3.979
Then, Fa    56.4 kN
D 0.127

Prob. # 3.2] A 76.2 mm diameter shafting of SAE 1040 grade, cold drawn, having a yield point of 50 ksi has a key with a
dimensions of ¾ inch x ¾ inch x 5 inches. Compute the minimum yield point in the key in order to transmit the torque of the
shaft. The factor of safety is 2 and syc = 0.5sy. (Board Prob. Oct. 1997)

Solution:
Solving for the torque to be transmitted through the shaft,

 ss D3 0.5 s y 0.5  50 000 

T Where, s s    12 500 psi
16 N 2

 s s D 3  12 500  3 3
Then, T    66 267.97 in - lb ans
16 16

For the yield strength of the key, based on shear,

2T 2  66 267.97  s ys 0 .5 s y
ss    11 780.97 psi Where, s s  
bLD 0.75  5  3 N N

Nsy 2 11 780.97 

Then, s y    47 123.89 psi  47.12 ksi ans
0.5 0 .5

For the yield strength of the key, based on compression,

4T 4  66 267.97  sy
sc    23 561.94 psi Where, s c 
tLD 0.75  5  3 N

Then, sy = N sc = 2 (23 561.94) = 47 123.89 psi = 47.12 psi ans.

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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

Prob. # 3.3] A rectangular key was used in a pulley connected in a lineshaft with a power of 125 kW at a speed of 900 rpm. If
the shearing stress of the shaft is 40 N/mm2 and the key to be 22 N/mm2, determine the length of a rectangular key with width
equals to one-fourth of the shaft diameter. (Board prob. October 1992)

Solution:
30 P 30 125
For the transmitted Torque, T    1.3263 kN - m
n   900

1 1
 16 1326.3 3
For the shaft diameter, D   16 T 
3

     0.05527 m  55.27 mm  2.2 inches

 ss     40  1000
2
 


2T D
For the length of the key, L  , Where b 
b ss D 4
8T 8 1326.3
Then, L    0.15788 m
22 1000  0.05527
2 2 2
ss D

L = 157.88 mm = 6.22 inches ans.

Prob. # 3.4] A cast iron pulley transmits 65.5 Hp at 1750 rpm. The 1045 as rolled shaft (s y = 59 ksi = 59 000 psi) to which it
is to be keyed is 1.75 inches in diameter; key material is cold drawn 1020 (s y = 66 ksi = 66 000 psi). Compute the length of
flat and square key (b = 3/8 inch, t = ¼ inch) needed.

Solution:
63 000 Hp 63 000  65.5
For the transmitted torque, T    2358 in-lb
n 1750

For the Factor of Safety, Assume minor shack load, N = 2.25

s ys 0.5 s y 0.5  66

For the allowable shearing stress of the key, s s     14.67 ksi
N N 2.25

For the allowable compression or bearing stress,

Since 66 ksi > 59 ksi, base the design on the shaft material.

s yc sy 59
sc     26.22 ksi
N N 2.25

Considering Flat Key

2  2358
2T
L   0.489  0.5 inch
For the Length of key based on shear,  3
ss b D
14670   1.75
8
4T 4  2358
L   0.822 inch
For the length of key based on compression, sc t D 1
26220   1.75
4
Therefore, use L = 0.822 inch ans.

Considering Square key

2  2358
2T
L   0.489  0.5 inch
For the Length of key based on shear,  3
ss b D
14670   1.75
8
4T 4  2358
L   0.548 inch
For the length of key based on compression, sc t D 3
26220   1.75
8
Therefore, use, L = 0.548 inch ans.

Prob. # 3.5] A belt pulley is fastened to a 7.46125 cm shaft, running at 200 rpm, by means of a key 19.05 mm wide by 12.7
cm long. The permissible stresses in the key are 55.1 MPa in shear and 96.5 MPa in compression. a) Determine the power to
be transmitted. b) What depth of key is required?
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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

Solution:
nT
a) For the power to be transmitted, P 
30

b L D ss 0.01905  0.127   0.0746125   55100 

Where, T    4.973 kN  m
2 2

  200  4.973
Therefore, P   104.15 kW ans
30

b) For the thickness or depth of the key

t L D ss
Based on compression, T 
4
2T 4  4.973
Then, t    0.01023 m  10.23 mm
L D sc 0.127  0.0746125  96 500

SHAFT COUPLING

 Coupling is a mechanical device for uniting or connecting parts of a mechanical system.

 Coupling provides for connection of shafts of units that are manufactured separately, such as a motor and a generator,
and to provide for disconnection for repairs or alterations.
 Coupling provides for misalignment of the shafts as to introduce mechanical flexibility.
 Coupling reduces the transmission of shack loads from one shaft to another.
 Coupling introduces protection against overloading.
 Coupling alters the vibration characteristics of rotating units.

Types of Coupling
 There are two main types of coupling, namely: rigid coupling and flexible coupling.
1. Rigid Coupling is a permanent coupling that by virtue of its construction has essentially no degree of angular, axial
or rotational flexibility and it must be used with collinear shafts.
o It has no flexibility and resilience.
o It is necessary for the shafts that are to be connected to be in good alignment, both laterally and angularly, in
order to avoid excessive loads on the coupling, on the shafts, or on the shaft bearings.
2. Flexible Coupling is a coupling that allows angularity for the misalignment of the shafts.

Kinds of Rigid Coupling

 The following are kinds of rigid coupling.
1. Flanged Coupling is a rigid coupling that consists of two halves of flanges and is connected to each other by bolts.
2. Collar coupling – is a rigid coupling that consists of cylindrical collar pressed over the ends of the two collinear
shafts being connected, approximately one-half of the collar contacting each other.

Kinds of Flexible Coupling

 The following are some of the kinds of flexible coupling.
1. Chain Coupling – is a flexible coupling consists of essentially of two chain sprockets connected with short
continuous length of roller or silent chains.
2. Oldham Coupling – is a flexible coupling that can be employed for connecting two parallel shafts with axial
eccentricities from zero to a reasonable amount.
3. Flexible Disk Coupling - is an all metal coupling with the intermediate flexible elements being thin steel disks.
4. Universal Joint – is a flexible coupling used to connect shafts whose axes intersect, that is, whose angular
misalignment is permanent.
5. Hydraulic Coupling – a coupling that employs a fluid to provide angular flexibility between the input and output
shafts.

DESIGN CALCULATIONS FOR FLANGE COUPLING

 As shown in the Figure 3.5 below, let

D = shaft diameter, inches, mm Dh = hub diameter, inches, mm
Db = bolts circle diameter, inches, mm d = bolt diameter, inch, mm
w = flange thickness, inch, mm b = width of the key, inch, mm
t = thickness of the key, inch, mm Lh = Hub length, inches, mm
L = length of key, inches, mm Do = outside diameter of the coupling, inches, mm
F = shearing or compressive force on bolts, lb, kN T = coupling transmitted torque, in-lb, kN-m
P = power transmitted, kW Hp = transmitted Hp
N = Factor of safety on shaft Nks = Factor of safety on key based on shear
Nkc = Factor of safety on key based on compression Nbs = Factor of safety on bolts based on shear
Nbc = Factor of safety on bolts based on compression Nh = Factor of safety on the hub
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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

Figure 3.5. Flange Coupling

Bolts Circle
Flange Wedge Bolt
d
Key
Key
Db Dh D

Shaft
Flange

w
Hub
DESIGN EQUATIONS

 The following equations are equations used for the design computations for flanged coupling.

2T
1. Total transmitted force or load on bolt: F 
Db

F 2T
2. Shearing on each bolt: Fb  
nb nbDb

Fb 4 Fb 8T
3. Shearing Stress on Bolts: s sb   2

Ab d  n bd 2Db

Fb F 2T
4. Compressing or bearing stress on bolts or flange: s cb   b 
Ac d w n bd w Db

Fch 2T 2T
5. Shearing Stress on flange with the hub: s sh   
A sh   D h w  D h  w D 2h

Fk 2T 2T
6. Shearing stress on key: s s   
A sk  b L  D b L D

Fk 2T 4T
sc   
7. Compressing or bearing stress on either key, hub, or shaft: A ck  t  btD
 b D
2 

Prob. # 3.6] A flange coupling is to connect two 57 mm shafts. The hubs of the coupling are each 111 mm in diameter and 92
mm hub length. Six 16-mm bolts in a 165-mm diameter bolts circle connect the flanges. The key way is 6 mm shorter than
the hub length and the key is 14 mm x 14 mm. Coupling is to transmit 45 kW at 160 rpm. For all parts, yield point value in
shear is one-half the yield point value in tension or compression that is 448 MPa. Find a) the shearing stress of key and its
factor of safety; b) the bearing stress of key and the factor of safety based on yield point; c) the shearing stress in bolts and
factor of safety based on yield point. (Board problem, November 1983)

Solution:
30P 30  45 
Solving for the transmitted torque of the coupling, T    2.686 kN  m
n   160 
a) Solving for the shearing stress of the key and its factor of safety

Fs 94.25
ss    78 280.73 kN / m2 ans.
A s 0.001204
2T 2  2.686 
Where, Fs = the shearing force of the key, Fs    94.25 kN
Ds 0.057
As = shearing area As = b (L) = 0.014(0.086) = 0.001204 m2

sy 448000
For the Factor of safety, N    2.86 ans.
2ss 2  78 280.73 

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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

14 mm x 14 mm x 86 mm
d = 16 mm

Dh = 111 mm D = 57 mm Db = 165 mm

w = 50 mm
92 mm

b) Solving for the bearing stress and the factor of safety of the key

Fc 94.25
sc    78 280.73 kN / m2 ans.
A c 0.001204

t  0.014 
Where, Ac = compression area on the key, A c  L    0.086    0.000602 m
2

 2  2 
sc = compression stress of the key
sy 448000
For the Factor of Safety, N    2.86 ans.
sc 156 561.5

c) Solving for the shearing stress of bolts and the factor of safety
Fb 32.558
Ssb    26 525.84 kN/ m2 ans.
A sb 0.00120671

 2 
db nb     0.016   6   0.001206371 m2
2
Where, Asb = shearing area of the bolts, A sb 
4 4
Fb = shearing force of the bolts
2T 2  2.686 
Then, Fb    32.558 kN
H 0.165
sy 448 000
For the Factor of Safety, N    8.44 ans.
2ssb 2  26 525.84 

PRACTICE PROBLEMS (Module III)

INSTRUCTION: Solve the following problems and select the correct answer for each question.

1. A key 24 mm wide, 16 mm deep, and 30.5 cm long is to be used on a 200 Hp, 1160 rpm, squirrel cage induction motion.
The shaft diameter is 98.5 mm. The maximum running torque is 200 % of the full-load torque.
A. What is the full-load torque?
a) 1.23 kN-m b) 2.13 kN-m c) 1.32 kN-m d) 2.31 kN-m

B. What is the maximum running torque?

a) 2.64 kN-m b) 2.46 kN-m c) 4.26 kN-m d) 4.62 kN-m

C. Determine the maximum allowable shear on the key, in MPa.

a) 6.82 MPa b) 6.28 MPa c) 8.26 MPa d) 8.62 MPa

D. Determine the maximum compressive stress, in MPa, on the key.

a) 24.04 MPa b) 20.44 MPa c) 23.03 MPa d) 20.33 MPa

E. What is the maximum shear stress on the shaft?

a) 13.09 MPa b) 10.39 MPa c) 13.90 MPa d) 10.93 MPa

2. A rectangular key was used in a pulley connected to a line shaft with a power of 125 kW at a speed of 900 rpm. If the
shearing stress of the shaft is 40 MPa and the key is 220 MPa.
A. What is the diameter of the shaft?
a) 51.6 mm b) 56.1 mm c) 61.5 mm d) 65.1 mm

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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

B. The length of the rectangular key if the width is one-fourth that of the shaft diameter.
a) 18.13 mm b) 13.81 mm c) 31.18 mm d) 38.13 mm

3. The equivalent of 1 Hp.

a) 33 000 ft-lb/min b) 0.746 Kw c) 2544.4 Btu/hr d. All of the above

4. A number that is divided into criterion of strength in order to obtain a design criterion.
a) Stress concentration factor b) Size factor c) Factor of safety d) None of the above

5. A small countershaft is 1.5 inches in diameter and has an allowable shear stress of 8500 psi. Find the Hp delivered by the
shaft at a speed of 15.7 rad/s.
a) 12.3 Hp b) 13.2 Hp c) 23.1 Hp d) 21.3 Hp

6. Determine the torque received by the motor shaft running at 4250 rpm, transmitting 8.5 kW, through a 25.4 cm diameter,
20o involute gear. The shaft is supported by ball bearings at both ends and the gear is fixed at the middle of 20.32 cm
shaft length.
a) 19.1 N-m b) 0.0191 kN-m c) 1.91 kN-cm d) All of the above

7. The force of a point of a shaper when cutting is 1 500 N. If the length of the stroke is 120 mm, how much work is done
inn one cutting stroke?
a) 180 J b) 1.8 kJ c) 195 J d) 100 J

8. An electric motor is rated when revolving at 1400 rpm. Find the torque available at this rating.
a) 1125 in-lb b) 1215 in-lb c) 2511 in-lb d) 2511 in-lb

9. A shearing machine is used to crop off lengths of a round bar 20 mm diameter. If the ultimate shear strength of the
material is 160 Mpa, calculate the force needed to crop the bar.
a) 52 kW b) 51.3 kN c) 250 kN d) None of the above

10. The shear strength of a plate is 300 Mpa. Calculate the force required to punch a hole 40 mm diameter in a plate 7 mm
thick.
a) 200 kN b) 230 KN c) 250 kN d) None of the above

11. The shear strength of a shaft is not to exceed 30 Mpa. The shaft is 100 mm in diameter. What is the greatest torque that
can be placed on it?
a) 5 900 N-m b) 6 100 N-m c) 6 200 kN-m d) None of the above

12. The maximum bending moment of a beam simply-supported on both ends and subject to a total load of W uniformly
distributed over a length L is:
WL W L2 WL WL
a) b) c) d)
8 8 2 4

13. A keyed gear delivers a torque of 912.4 N-m through its shaft of 63.5 mm outside diameter. If the key has thickness of
15.875 mm and width of 11.1125 mm, find the length of the key. Assume the permissible stress values of 61.2 Mpa for
shear and tension at 99.8 Mpa.
a) 42.22 Mpa b) 47.42 Mpa c) 39.72 Mpa d) None of the above

14. The power to be transmitted by a shaft with a torque of 12 600 in-lb and rotating at 200 rpm.
a) 40 kW b) 40 Hp c) 30 Hp d) None of the above

15. The torsional deflection of 2-inch diameter shaft that transmits 20 Hp and rotates at 200 rpm. The length of the shaft is
30 inches and the shear modulus of elasticity is 12 000 000 psi.
a) 0.010 radian b) 0.10 degrees c) 0.10 radian d) 0.02 radian

16. What is the torque of a solid shaft with a capacity of 2.5 kW and rotating at 1200 rpm?
a) 756 in-lb b) 20 N-m c) 19.89 N-m d) 2.50 kN-m

17. Determine the diameter of a solid shaft, subject to a pure torsion, that rotates 100 rpm and transmits 1.2 Hp, and the
design stress for shear is to be 6 000 psi.
a) 0.863 inch b) 0.756 inch c) 1.45 inches d) 1.032 inches

18. Von Mises theory is the other term used for

a) Maximum principal stress theory b) Octahedral shear-stress theory
c) Maximum shear-stress theory d) None of the above

19. A type of key in which width and thickness are equal is called as:
a) Flat key b) Square key c) Pin key d) Barth key

20. In the design of key, the typical hub lengths are in accordance with the following relation where D is shaft diameter.
a) 1.25D to 2.4D b) 0.5D to 1.25D c) 2.4D to 3.5D d) None of the above

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JCSF MECHANICAL ENGINEERING REVIEWER MACHINE DESIGN: KEYS & COUPLING

21. A line shaft runs at 360 rpm. A 46 cm pulley on the same shaft is belt connected to a 31cm pulley on the countershaft.
From a 15 inches pulley on the countershaft, motion is transmitted to the machine. Compute the rpm of the countershaft.
a) 345.2 rpm b) 542.3 rpm c) 534.2 rpm d) 453.2 rpm

22. Considering the shafting system in question # 21, calculate the required diameter of the pulley on the machine to give a
spindle speed of 660 rpm.
a) 12.14 inches b) 14.12 inches c) 21.14 inches d) 24.12 inches

23. A coupling that allows axial flexibility/movement in the operation. Made of alternate bolting of steel, leather, fabric
and/or plastic material into two flanges.
a) Flexible disk coupling b) Flexible toroidal spring coupling
c) Flexible Oldham coupling d) Elastic material bonded coupling

24. A flanged shaft coupling has ten, 25.4 mm diameter, steel bolts evenly tighten around a 415 mm bolt circle. Determine
the torque capacity of the connection if the allowable shearing stress in the bolt is 50 Mpa.
a) 52.57 kN-m b) 57.52 kN-m c) 72.25 kN-m d) 75.25 kN-m

25. The endurance stress in shear for the maximum shear-stress theory.
a) 0.5sn b) 0.577sn c) 0.6sn d) None of the above

26. Compute the maximum allowable shear stress in a 76.2-mm diameter steel shafting that transmits 2 712 N-m of torque at
99 rpm.
a) 31.22 MPa b) 32.12 MPa c) 22.31 MPa d) 13.22 Mpa

27. The power transmitted by the shaft in question # 26.

a) 21.18 kW b) 28.12 kW c) 18.12 kW d) 81.21 kW

ANSWER

1. A.a, B.b, C.a, D.b, E.a

2. A.a, B.a
3. d
4. c
5. b

6. d
7. a
8. a
9. d
10. d

11. d
12. d
13. d
14. b
15. a

16. c
17. a
18. b
19. b
20. a

21. c
22. a
23. c
24. a
25. a

26. a
27. b

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