Bangalore Institute of Technology

Department of Mechanical Engineering

MODULE-2

Subject: Machine Design (BME602)

CIE Marks: 50 (25T+25A) SEE Marks:100(50)
Course Co-Ordinator : Dr. Nagaraja C Reddy (Dr. NCR)
Email: cnraaja@gmail.com, Mob:9449991551
1 Dr. NCR
 Syllabus
MODULE-2
Design of shafts: Torsion of shafts, solid and hollow shaft design with steady loading based on strength and
rigidity, ASME and BIS codes for power transmission shafting, design of shafts subjected to combined bending,
torsion and axial loading, Discussion on engineering applications.
Design of keys and couplings:
Keys: Types of keys and their applications, design considerations in parallel and tapered sunk keys, and Design
of square and rectangular sunk keys.
Couplings: Rigid and flexible coupling types and applications, design of Flange coupling, and Bush and Pin type
coupling.

2 Dr. NCR
 Design of shaft
A shaft is a rotating machine element, usually of a circular cross-section, which transmits power. Shafts are subjected to
tensile, bending, or torsional stresses or a combination of these stresses.
Classification:
i) Transmission shafts:
• These transmit power between the source and the machines absorbing power.
• These shafts carry machine elements like pulleys, gears, cranks, etc.
ii) Machine shafts:
• These are integral parts of the machine itself
• For example, cam-shaft, crank-shaft, etc.
The design of the transmission shaft consists of determining the correct shaft diameter based on strength, rigidity, and stiffness.
Shaft material: The commonly used material for shafts is low or medium-carbon steel, i.e., C-30, C-35, C-40, C-45, C-50. When
greater strength is needed, suitably alloy steel may be used.
Causes of failure: The main cause of the failure of the shaft is fatigue, which arises due to the following reasons: i. Due to the
presence of cyclic overloads ii. Due to the wrong adjustment of bearings and their insufficient clearances. iii. Due to stress
concentration which arises due to undercuts, keyways, holes, etc. Dr. NCR
3
 Design of shaft
Working Stresses for Shaft:
Case A. A.S.M.E. code of shaft design: The American Society of Mechanical Engineers established a code for the design of
shafts. The code suggests guidelines for deciding the allowable shear stress to be used in the design of the shaft.
According to this code, the permissible shear stress for shafts without keyways can be taken as 0.3y or 0.18u whichever is the
minimum.
Thus, ed= 0.3y or 0.18u whichever is minimum.
If a keyway is present on the shaft then a minimum of these values is further reduced by 25%,
Thus ed= 0.75 x minimum value.
Similarly, According to this code, the permissible stress for shafts without keyways can be taken as 0.6y or 0.36u whichever is
the minimum.
Thus, ed= 0.6y or 0.36u whichever is minimum.
If a keyway is present on the shaft then a minimum of these values is further reduced by 25%,
Thus ed= 0.75 x minimum value
Case B. ASME Code for commercial steel shafting: i. without keyway: ed = 55 Mpa and ed = 110 Mpa.
ii. With keyway: ed = 40 Mpa and ed = 80 Mpa. 4 Dr. NCR
 Design of shaft
Design formulae used:

I) Solidshaft:
é 16 ù1/3
i. The diameter of the shaft 'D' subjected to only torsion is given by D=ê (KM ú ® E.14.9
êëπτed t t )ú
û
é 32 ù1/3
ii.The diameter of the shaft 'D' subjected to only bending is given by D=ê (KbMb)ú ® E.14.10
êëπσed úû

iii. The diameter of the shaft 'D' subjected to combined bending and torsion:
ìï 16 é ù ü
ï
1/3
a. According to maximumnormal stress theory,D=í {2
êKbMb+ (KbMb) +(KM
ïî πσed ë t t)
2 1/2
}
úý ® E.14.11
ûïþ

ìï 16 ü 1/3
ïý ® E.14.12
b. According to maximumshear stress theory,D= í
îï πτed
K{M 2
+ KM 2 1/2
( b b) ( t t) þï }
é584ML ù1/4
iv. The diameter of the shaft 'D' basedon torsional rigidity,D=ê t
ú ® E.14.17
êë Gθ úû Dr. NCR
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 Design of shaft
where, D=diameter of shaft,mm
ed =d =Allowable or design shear stress,Mpa Note:TorqueMt = 9550N×103, N-mm
n
Mt =Torque transmitted,Nmm
Mb =bending moment,Nmm
σed =σd =Allowable normal stress,Mpa
Kb =Cm =Shock and endurance factor for bending moment ® T.14.2
Kt =Ct =Shock and endurance factor for twisting moment ® T.14.2
L=length, mm
G=Modulus of Rigidity,Mpa(if not given assume,G=80-85Gpa)
=angular deflection,degree
II) Hollowshaft:
é 16 ù1/3
i. The outside diameter of the shaft 'D'o subjected to simple torsion, D=êê KM ú ® E.14.34
4 ( t t )ú
( )
êëπτed 1- K ú
û
Where, K=Ratio of inside diameter to outside diameter, Di <1(K=0 for solid shaft)
Do
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 Design of shaft
é 32 ù1/3
ii.The outside diameter of the shaft 'D'o subjected to simple bending, D=êê K M ú ® E.14.35
4 ( b b)ú
( )
êëπσed 1- K úû
iii. The outside diameter of the shaft 'D'o subjected to combined bending and torsion:
ìï é ü
ùï
1/3

a. According to maximumnormal stress theory,D=í 16

( )
ïî πσed 1- K4 { 2
êKbMb+ (KbMb) +(KM
ë t t) }
2 1/2
úý ® E.14.36
ûïþ
ìï ü 1/3
ïý ® E.14.37
b. According to maximumshear stress theory,D= í 16
( )
ïî πτed 1- K4 { K M 2
( b b) +(KM
t t) }
2 1/2
ïþ
é 584ML ù1/4
iv. The outside diameter of the shaft 'D'o based on torsional rigidity,D=êê t ú ® E.14.42
ú
( )
êë 1- K4
Gθú
û

7 Dr. NCR
 Design of shaft
Designprocedurefor shaft:
1. Selectionof suitableequation:
2. Selectionof material, allowableshear stress(τed) anddesignstress(σed):
σy σu
( )
If material is not given, assume C-40 steel, σed= = for σy and σu Ref.T.1.5,ODHB ,\ τed=0.5σed
n n
According to ASME code(commercial steel shaft):τed=55 Mpa σed =110 Mpa ® Without keyway
τed=40 Mpa σed =80 Mpa ® With keyway
>If σu andσy aregiven: find the value of 0.3σy and 0.18σu, then select least value among two.
\ τed =0.75´ least value ® With keyway
­
Keyway factor
find the value of 0.6σy and 0.36σu, then select least value among two.
\ σed =0.75´ least value ® With keyway
­
Keyway factor Dr. NCR
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 Design of shaft
3.Selectionof shock andendurancefactor(Kt andKb):(Ref.T.14.2)
If any condition not given, assume suddenly applied load with minor shocks only.(Note: Kb=Cm=1.75 and Kt =Ct =1.25)
4. Findtorque'M't i.e. beingtransmittedas input torqueis allways equal tooutput torque.: T1

casei. If power and speed is given, WKT, Torque: Mt = 9550N´ 103 ,N-mm T1+T2
n
caseii. If power is transmitted through pulleys, WKT,Torque,Mt = (T1 −T)r2 p , N-mm . T2
2
Where, T1 = Tight side tension,N. T2 =Slacksidetension,N. rp = Radiusof pulley,mm.
Also,ratio of belt tensionis T1 = e →Flat belt and T1 = e/sin →V−Belt.
T2 T2
Where,
μ=Coefficient offriction between pulley and belt. θ=Angle oflap or angle of contact,radians. 2α=Groove angleof V-Pulley.
Note:i. =180o = for equal diameter of pulley.
ii. If the value of ''and'',and power'N' and speed'n'are not given,assume, T1 =3.
T2
caseiii. If power transmitted through gears,then Torque, Mt = Ft r, Nmm.
where, F=Tangentialforceonthegearorpinion,
t N. F=Radialforce,
r N.
9 Dr. NCR
 Design of shaft
Fr = Ft Tan, N. where, = Pressureangleof gear or pinion =141 or 20o,
2
r = Radius of gearorpinion,mm. = d,d = mz = PCD.where, m= module,mm. Z= numberof teeths.
2
5. Drawhorizontal and vertical loadingdiagramandfind bending moment'M': b

Bending moment'M'is
b detremined bycalculating theloads on members (Gears andpulleys)mountedon theshaft.

casei. Calculation of loads on pulley:Ifshaft receives or trasnmit power verticallyfrombelowor above.

Shaft to be designed Driven +

Driver T2 T1
T1+T2
Receives Power- Driven.
T1 T2 Transmits Power- Driver. Shaft to be designed
T1+T2+Wp Driver +

Fv=T1+T2 - wp
Driven + Fv=T1+T2+wp
Wp
FH=0 FH=0
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 Design of shaft
caseb. If thepulleyis receivingor transmitting power fromor toanother pulleymounted behindit through aninclined belt drive
inclinedat an angleθ to thehorizontal(slopingupward)
T2 (T1+T2 )sin (T1+T2 )cos
T1+T2 T1+T2 T1+T2

 
(T1+T2 )cos (T1+T2 )sin
T1
Fv = (T1+T2)cos -Wp
Fv = (T1+T2)sin -Wp
wp Wp Wp FH = (T1+T2)sin
FH = (T1+T2)cos

Note: i. If (T1+T2)Sin > Wp The vertical loads acts upwards Fr

ii. If (T1+T2)Sin < Wp The vertical loads acts downwards
Driven
casec. Caluculation of loads on gear or pinion:
Fv = Fr = Ftan
t , N (Driver) Ft Ft (Driven)

FH = Ft = Mt , N. where,Torque Mt = Ft r, N.mm Driver

r
d = r = radius of gear or pinion, mm. d= PCD=mz. where m=module, mm. z=number of teeth.
2 Fr
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 Design of shaft
Note: Tilting of views: + +
LSV RSV
i. Left side view(LSV)- -
ii. Right side view(RSV)-
+ - +

- -

6. Draw bending moment diagram(BMD):

i. Finding out the reaction RA and RB. Use + ve for clockwise rotation and – ve for counterclockwise rotation.
ii. Bending moments will be zero at the endpoints
iii. Find out the maximum resultant bending moment. i.e. Mbmax
7. Find the size of shaft ‘D’:
The value of shaft diameter ‘D’ obtained from the above information is rounded off to the nearest next standard size. Ref.T.14.6
NOTE:
i. Effect of weight: any weight due to gravity acts vertically downwards only.
ii. If the direction of rotation is not given, then select any direction of rotation and observation, but in the loading diagram, loads
should be acts always downwards that is essential to find the maximum bending moment ‘Mbmax’
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 Design of shaft
Discussion on engineering applications:

Mechanical engineering is a crucial field that involves the design of shafts, which are essential components in transmitting power
and motion between various parts of a machine. Proper design ensures reliability, efficiency, and safety in mechanical systems,
from everyday machinery to advanced aerospace systems. The design of shafts is crucial in ensuring efficient power transmission,
durability, and safety in mechanical systems, making them a vital component in various engineering applications.
1. Power Transmission Systems:
• Gearboxes: use shafts to transmit torque and rotational motion.
• Automotive driveshafts transmit power from the engine to the wheels.
• Industrial machinery uses shafts in conveyor systems, pumps, and compressors.
2. Rotating Machinery:
• Turbines convert rotational energy into electrical energy or mechanical work.
• Electric motors and generators use rotor shafts to handle high rotational speeds and torque.
• Fans and blowers connect the motor to fan blades, enabling airflow in HVAC systems and industrial cooling applications.

13 Dr. NCR
 Design of shaft
3. Machine Tools:
• Lathes and milling machines use shafts in spindles to hold and rotate cutting tools.
• Drilling machines use shafts to transmit torque to the drill bit.
4. Aerospace Applications:
• Jet engines and turboprops transmit power from the turbine to the compressor and other components.
• Helicopter rotor systems transmit power from the engine to the rotor blades.
5. Marine Engineering:
• Propeller shafts in ships and submarines transmit power from the engine to the propeller.
• Pumps and compressor shafts are used in marine systems for fluid handling and cooling.
6. Robotics and Automation:
• Robotic arms use shafts in joints and actuators to transmit motion and torque.
• CNC machines ensure accurate movement and positioning of tools.
7. Agricultural Machinery:
• Tractors and harvesters use shafts in power take-off systems.
• Irrigation systems use shafts in pumps and motors.
14 Dr. NCR
 Design of shaft
8. Renewable Energy Systems:
• Wind turbines convert wind energy into electrical power.
• Hydroponic systems convert water kinetic energy into electricity.
9. Material Handling Equipment:
• Cranes and hoists use shafts in lifting mechanisms to transmit torque and support heavy loads.
• Conveyor belts drive the rollers, moving materials along the conveyor.
10. Medical Devices:
• Surgical tools use shafts in rotary instruments like drills and saws.

15 Dr. NCR
 Design of shaft
(1) A steel shaft 900 mm long between bearings receives the power of 18 KW at 900 rpm through a 20o involute spur gear of
200 mm diameter located 250 mm to the right of the left-hand bearing by a gear directly below the shaft. This power is
transmitted by a 400 mm diameter pulley to another pulley, which is behind the shaft and above it. The belt drive axis is
inclined at 60o to the horizontal. The pulley is located at a distance of 250 mm to the left of the right-hand bearing. The
angle of the lap of the belt is 180o and the coefficient of friction is 0.3. The shaft rotates in the clockwise direction as seen
from the left side bearing. Design the shaft if the allowable shear stress for the shaft material is 60 Mpa.
Solution: Power N=18 KW, Speed n=900 rpm, ed=60 Mpa, d=?

1. WKT, when the shaft is subjected to combined  = 20o FDI  = 0.3

200 400
bending and torsion, according to maximum shear
Driver
A Driven B
C
stress theory, D

ìï 16 ü 1/3
ïý ® E.14.12
LSV

D= í { K M 2
+ KM 2 1/2

ïî πτed ( b b) ( t t) ïþ } 250 400 250

VLD
2.Any condition is not given, assume sudden applied
695.18 1883.12
RA
load with minor shock, Ref.T.14.2 RB
HLD
Kb=Cm=1.75 and Kt=Ct=1.25 1910 1087.22
RA RB
Torque Mt = 9550N103 =191103 N-mm.
n 16 Dr. NCR
 Numerical on Design of shaft
3.Tofindmaximumbendingmoment 'M': b
+ Fr

i. Consider Gear at C, LSV

Driven
WKT, Torque Mt =Ft ´ r, \ FH= F= M t =191´ 103
=1910 N. + - C
t
r 100 (Driven) Ft Ft(Driver)

v r t tanα = 695.18 N.
F=F=F - Driver
ii. Consider pulleyat D,
3
WKT, Torque Mt =(T-T 1 2) p
r , T+T 191×10 Fr
1 2= =955,
200 T1
WKT, T1 =eμθ, T1 =2.566T, 2 \ T2 =609.72 Nand T1 = 1564.72 N. T1+T2 (T1+T2)Sin60o
T2 T1+T2
T2
\ Fv =(T1 +T2)Sin60 =1883.12 N. and FH =(T1 +T2)Cos60 =1087.22 N.
o o 60o D

▪ Consider Vertical load diagram(VLD):

(T1+T2)Cos60o
Taking moment about point A, - 695.18x250 – 1883.12x 650 – RBx900=0
RB= - 1553.14 N. For equilibrium condition, RA+695.18+1883.12+RB=0, RA= - 1025.16 N.
➢ Find bending moment, MbA=0 and MbB=0, Mbc=RA x 250 = -1025.16 x 250 = - 256290 N-mm.
MbD= RB x 250= - 1553.14 x 250= - 388285 N-mm.
17 Dr. NCR
 Numerical on Design of shaft
• Consider HLD, RA =−1681.45 N. RB =−1315.77 N. MbA = MbB = 0, MbC =−420362.5 N-mm, MbD =−328942.5 N-mm.
FindResultant bendingmoment:
Mbc = (Mbc)V2 +(Mbc)H2 =492330.372 Nmm. MbD= (MbD)V2 +(MbD)H2 =508868.78 Nmm.
\ Maximum bending moment, Mbmax =508868.78 N-mm.
Substitute all values in main equation, we get D= 42.77 mm. Std D= 45 mm.(Ref.T.14.6)
(2) The machine shaft running at 600 rpm is supported on bearings 750 mm apart. 15 KW is supplied to the shaft
through a 450 mm pulley located 250mm to the right of the right bearing. The power is transmitted from a shaft
through a 200 mm spur gear located 250 mm to the right of the left bearing. The belt drive is at an angle of 60o above
the horizontal. The pulley weighs 800N to provide some flywheel effect. The ratio of tensions is 3:1. the gear has a 20o
tooth form and mates with another gear located directly above the shaft. The shaft material has an ultimate strength of
500 MN/m2 and a yield point of 310MN/m2. Determine the necessary diameter of the shaft. Use K b=1.5 and Kt=1.0
Solution: u = 500 Mpa, y =310 Mpa, D=?
=20o Wp=800N
200 450
T1 =3
A C
Driver B T2
LSV D Driven

250 500 250

18 Dr. NCR
 Numerical on Design of shaft
1. WKT, The shaft subjected to combined bending and torsion,

  →E.14.11
1/3
According to maximumnormal stress theory, D=  16 KbMb + (KbMb )2 +(KM )
2 1/2

πσed  t t

 16
  
1/3
(KbMb )2 +(KM )
1/2
According to maximumshear stress theory,D=   →E.14.12
2
πτ
 ed
t t

2. If given yand u; find 0.3y = 0.3310 =92Mpa and 0.18u = 0.18500 =90Mpa(least)
τed = 0.75least value=0.7590=67.5Mpa.Similarly, 0.6y =186 and 0.36u =180 Mpa(least), σed =0.75×180=135Mpa.

3. Torque, M= 9550N103=955015=238750 N-mm.

t Fr
n 600
4.DrawVLD, HLD and find bending moments M:
b Driven +
LSV
Consider gear at point C, (Driven) Ft Ft (Driver) Ft
+ -
Torque M=F  r,  F = F = Mt = 238750 = 2387.5 N.
t t H t Driver
r 100 Fr -
and Fv= Fr = Ft tan=2387.5 tan20o=868.97 N. Fr
19 Dr. NCR
 Numerical on Design of shaft
Consider pulley at point D, T1 (T1+T2)Sin60o

WKT, Torque, Mt =(T1 −T2 )rp, T1 −T2 = 238750 =1061.11 and T1 =3,T1 =3T2
T1+T2
T2
225 T2 60o

By solving we get T2 =530.56 N. and T1 =1591.66 N. 60o

Fv =(T1 +T2 )Sin60o −800 =1037.89 N. and FH= (T1 +T2 )Cos60o =1061.11 N.
(T1+T2)Cos60o

Wp=800 N

=20o Wp=800N
Consider VLD, 200 T1 =3 450

Find reaction forces, RA and RB by taking moments about point A, A C

Driver B T2
LSV D Driven

868.97250−RB 750−1037.891000 =0,RB =−1094.22 N.

Consider equilibriumcondition,F=0, 250 500 250

RA +RB +1037.5 =868.97,RA =925.31 N. 868.97 N

Find bending moment 'M'b , MbA = 0 and MbD =0

VLD
RB 1037.89 N
RA 2387.5 N 1061.11 N
MbC = RA 250 =925.31250 = 231322.5 Nmm. HLD
RA RB
and MbB =−1037.91250 =−259477.5 Nmm.
20 Dr. NCR
 Numerical on Design of shaft
Consider HLD, Find reaction forces, RA and RB by taking moments about point A,
2387.5250−RB 750+1061.111000 = 0,RB = 2210.63 N.
RA +RB = 2387.5+1061.1, RA =1237.97 N.
Find bending moment 'M'b , MbA = 0 and MbD = 0
MbC = RA 250 =1237.97250 =309492.5 Nmm. and MbD =1061.1250 = 265275 Nmm.

Find resultant bending moment,Mb = (Mbv )2 +(MbH)2 ,

MbC = (231327.5)2 +(309492.5)2 =386386.424 Nmm.

MbD = (259375 )2 +(265275)2 =371081.37 Nmm.

Mbmax =386386.424 Nmm.

Substitute all values in E.14.11 and E.14.12, we get, D=36 mmand D=36 mm. Std, D= 40 mm.(Ref.T.14.6)
21 Dr. NCR
 Numerical on Design of shaft
(3) A horizontal piece of commercial shafting is supported by two bearings 1.5 m apart. A keyed gear, 20o involute and 175
mm diameter is located 400 mm to the left of the right bearing and is driven by a gear directly behind it. A 600 mm
diameter pulley is keyed to the shaft 600 mm to the right of the left bearing and drives another pulley with a horizontal belt
directly behind it. The tension ratio of the belt is 3:1. The drive transmits 45 KW at 330 rpm clockwise direction when
viewed from the right side(i.e., slack side on top). Take Kb=Kt=1.5. calculate the necessary shaft diameter and angular
deflection in degrees. Use allowable shear stress 40 Mpa and G=80 x 103 Mpa.
Solution: Power N=45 KW, Speed n=330 rpm, G=80 Gpa, ed=40 Mpa, D=? , =?

1. The shaft subjected to combined bending and torsion, T1 =3  600

 = 20o FDI
 175
According to maximumshear stress theory, T2
 16
 2 1/2 
1/3 A Driver C D Driven B

(KbMb ) +(KM t t)  →E.14.12

RSV
D=  2
πτ
 ed 
600 400

9550N103=955045=1302272.72 N-mm.
1500

2. Torque, M=t
14883.12 N
n 330 VLD
5417 N
RB
3. DrawVLD, HLD and find bending moments M: b HLD
RA

Consider the pulley drive at point C; RA 8681.82 N RB

WKT,Torque, Mt =(T1 −T2 )rp, 1302272.72=(T1 −T2 ) 600

2 22 Dr. NCR
 Numerical on Design of shaft

T1 −T2 = 4340.91, WKT, T1 =3, T1 =3T,

2 By solving, we get, T2 = 2170.45N. and T1 = 6511.36 N.
T2 T1
+
RSV

FV = 0 and FH = T1 +T2 = 6511.36+2170.45 =8681.82 N. T1+T2 - +

Consider the gear drive at point D; WKT, Torque, Mt = Ft rG, 1302272.72 = Ft 175, T2 -
2
Tangential load, F=14883.12
Ft (Driver)
t N.
Radial load, Fr = Ftan
t =14883.12 tan20o =5417 N. Fr
Driven Driver
Fr

Findbendingmoments'M'; b F (Driven) t

Consider VLD, Find Reactions RA and RB, Taking moments about point A,
14883.121100−RB 1500 =0,RB =10914.28 N. Also RA +RB =14883.12, RA =3968.83 N.
Bendingmoment, MbA = MbB = 0, MbC = RA 600 =3968.83600 = 2381299.2 Nmm.
MbD = RB 400 =10914.28400 =−4365715.2 Nmm.
Consider HLD, Find Reactions RA and RB, Taking moments about point A,
54171100−RB 1500−8681.82600 =0,RB = 499.74 N. Also RA +RB +8681.82 =5417, RA=−3764.56 N.
Bending moment, MbA = MbB = 0, MbC = RA 600 =−3764.56600 =−2258736 Nmm. 23 Dr. NCR
 Numerical on Design of shaft
MbD = RB 400 = 499.74400 =−199896 Nmm.

Resultant bendingmoments, MbR= (MbV)2 +(MbH)2 , MbA = MbB = 0, MbC = 2381299.22 +22587362 =3282144.74 Nmm.
MbD = 4365715.22 +1998962 =4370289.2Nmm. Mbmax = 4370289.2Nmm.
D=95.5 100 mm.
Substitute all values in E.14.12, we get ,
  1004 = 5841302272.72500,
1/4
Findangular deflection'θ', WKT, Torsional rigidity D=584ML →E.14.17

t
 Gθ  80103 
angular deflection θ=0.1425o.
(4) A shaft is supported by two bearings placed 1 m apart. A 600 mm diameter pulley is mounted at a distance of 300
mm to the right of left hand bearing and this drives a pulley directly below it with the help of a belt having a
maximum tension of 2.25 KN. Another pulley 400 mm diameter is placed 200 mm to the left of right hand bearing and
is driven with the help of an electric motor and belt, which is placed horizontally to the right. The angle of contact for
both the pulleys is 180o and  = 0.24. Determine the suitable diameter for a solid shaft, allowing working stress of 63
Mpa in tension and 42 Mpa in shear for the material of the shaft. Assume that the torque on one pulley is equal to that
of the other pulley.
Solution: T1=2.25 KN=2250 N, =  =180o=3.142 rad, = 0.24, ed =42 Mpa, ed =63 Mpa, D=?
24 Dr. NCR
 Numerical on Design of shaft
 600
1. The shaft subjected to combined bending and torsion,  400
C
According to maximumnormal stress theory, LSV
A Driver D Driven B

 16 
 
2 1/2 
1/3
D=  KbMb + (KbMb ) +(KM 2
t t)  →E.14.11
 ed 
πσ  300
1000
200

3308.6 N
According to maximumshear stress theory, VLD
4964.16 N

  
1/3 RA RB

D=  16 (KbMb )2 +(KM
t t)
2 1/2
 →E.14.12
HLD
πτ
 ed  RA RB

2. Assuming, suddenly load with minor shocks,

 Kb =1.75 and Kt =1.25(Ref.T.14.2) C

3. DrawVLD, HLD and find bending moments M: b

T1+T2
Consider pulleyat point C, WKT, Ratio of belt tension, T1 = e, 2250 = e0.243.142,T2 =1058.60 N.
T1 T2

T2 T2 +
Torque Mt =(T1 −T2 )rp =(2250−1058.60) 600 =357420 Nmm.(Torque on both pulleys are same)
LSV

2 + -
Fv = T1 +T2 = 2250+1058.60 =3308.6 N. and FH = 0.
- 25 Dr. NCR
 Numerical on Design of shaft
T3

Consider pulleyat point D, WKT, T3 = e = 2.125, T3 = 2.125T,

4 3 T +T
4
T4 D

Torque Mt =(T3 −T4 )rp,357420 =(2.125T4 −T4 ) 400, 4 T

2
T4 =1588.53 N. and T3 =3375.63 N. Fv = 0, and FH = T3 +T4 =3375.63+1588.53= 4964.16 N.
Tofindmaximumbendingmoment: Consider VLD, Find Reactions RA and RB, Taking moments about point A,
3308.6300−RB 1000 =0, RB =992.58 N and Also, RA + RB =3308.6, RA = 2316.02 N.
Bendingmoment,
MbA = MbB = 0,MbC = RA 300 = 2316.02300 = 694806 Nmm, MbD = RB 200 =992.58200 =198516 Nmm.
Consider HLD, Find Reactions RA and RB, Taking moments about point A,
4964.16800−RB 1000 =0, RB =3971.33 N and Also, RA + RB = 4964.16, RA =992.83 N.
Bendingmoment,
MbA = MbB = 0,MbC = RA 300 =992.83300 = 297849 Nmm, MbD = RB 200 =3971.33200 = 794266 Nmm.
Resultant bendingmoments,
MbR= (MbV)2 +(MbH)2 ,MbA = MbB =0, MbC = 6948062 +297849 2 =755955.55 Nmm.
26 Dr. NCR
 Numerical on Design of shaft
MbD = 1985162 +794266 2 = 818698.41Nmm.Mbmax = 818698.41Nmm.
Substitute all values in E.14.11 and E.14.12, we get , D=62 mm. and D=58 mm.
(5) A hoisting drum of 0.5 m in diameter is keyed to a shaft which is supported in two bearings and driven through a
12:1 reduction ratio by an electric motor. Determine the power of the driving motor, if the maximum load of 8 KN is
hoisted at a speed of 50 m/min and the efficiency of the drive is 80%. Also, determine the torque on the drum shaft and
the speed of the motor in rpm. Determine also the diameter of the shaft made of machinery steel, the working stresses of
which are 115 Mpa in tension and 50 Mpa in shear. The drive gear whose diameter is 450 mm is mounted at the end of
the shaft such that it overhangs the nearest bearing by 150 mm. The combined shock and fatigue factors for bending
and torsion may be taken as 2 and 1.5 respectively.
Solution: Drum diameter D=500 mm, Gear ratio i=12:1, motor power Nm=?, W=8000 N, V=50 m/min=0.834 m/sec,
mech=80%=0.8, Torque on drum MtDrum=?, Speed of motor nm=?, d=?, ed=115 Mpa, ed=50 Mpa, dg=450 mm, overhang l=150
mm, Cm=2, Ct=1.5, assume =20o.
12:1
The shaft subjected to combined bending and torsion,
Driver pinion Motor
Hoisting drum,  500
According to maximumnormal stress theory,
 16 
 
2 1/2 
1/3
D=  KbMb + (KbMb ) +(KM t t)  →E.14.11
2 Driven gear

πσed  
  450
V=50m/min
150
8 KN
27 Dr. NCR
 Numerical on Design of shaft

  
1/3
According to maximumshear stress theory,D=  16 (KbMb )2 +(KM )
2 1/2
 →E.14.12
πτ
 ed
t t

1.Calculate the torque on the drumshaft,
Mtd = W D =8000 500 = 2000103 Nmm.(Torque on both drumand gear are same)
2 2
2.Calculate the speed of the motor, WKT, V= Dnd , 0.834=500nd ,nd =31.86 rpm.
60000 60000
Gear ratio , i =nm , nm =i nd =1231.86 =382.32 rpm.
nd
3.Calculate the power of the driving motor, power at drumis, Nd = 2 n dM td = 2 31.862000 10 3
= 6.67 KW.
60000 60000
Efficiency of the drive is given by,=0.8= Nd = 6.67,Nm =8.34 KW.
Nm Nm d
4.Calculate the diameter of shaft, WKT, tangential tooth load on gear is, Mt = Ft  g , 2000103 =Ft  450,Ft =8.888103 N.
2 2
Fr = Ftan
t =8.888103 tan20o =3234.96 N.
Bending moment, MbH = Ft 150 =8.888103 150 =1333200 Nmm. and MbV = Fr 150 = 485244 Nmm.
Substitute all values in E.14.11 and 14.12, we get, D=75 mm and D=68 mm. 28 Dr. NCR
 Numerical on Design of shaft
(6) A power transmission shaft 1800 mm long, is supported at two points A and B. Whereas A is at a distance of 300 mm
from the left end of the shaft, B is at the right end. A power of 50 KW is received at 500 rpm, through a gear drive
located at the left end of the shaft. The gear mounted on the shaft here has a pitch diameter of 300 mm and weighs 700
N. The driver gear is located exactly behind. 30 KW of this power is given out through a belt drive located at a distance
of 600 mm from the left support. The pulley mounted on the shaft has a diameter of 400 mm and weighs 1000 N. The
belt is directed towards the observer below the horizontal and inclined at 45o to it. The ratio of belt tensions is 3. The
remaining power is given out through a gear drive located at a distance of 400 mm from the right support. The driver
gear has a pitch diameter of 200 mm and weighs 500 N. The driven gear is located exactly above. Selecting C40
steel(y=328 Mpa) and assuming a FOS of 3, determine the diameter of a solid shaft for the purpose. Take Kb=1.75,
Kt=1.5, and pressure angle =20o for the gears.
Solution: speed n=500 rpm, y=328 Mpa, FOS=3, D=?, =20o , Kb=Cm =1.75, , Kt=Ct=1.5.

1. The shaft subjected to combined bending and torsion,

 16
 
2 1/2 
1/3
According to maximumshear stress theory,D=  (KbMb ) +(KM
2
t t)  →E.14.12
πτed 

2. Allowable or design stress, ed = y = 328.6 =109.53 Mpa.ed = ed =54.77 Mpa. (Driver)Ft
n 3 2
C
Fr Fr
3.Maximumbending moment 'Mb '; Driver Driven

Consider gear at point C, Torque Mt = 9550N103 = 955050103 =955000 N mm.

(Driven) Ft
WG=700 N
n 500 29 Dr. NCR
 Numerical on Design of shaft
T1 =3W =1000 N
p Wp=500 N

WKT tangential force, Ft = Mt = 955000 = 6366.67 N.

Wg=700 N  400  = 20o FDI
 300 T2  200
rg 150 C
Driven Driver D Driver
A E B
LSV
Fr = Ftan
t = 6366.67tan20 = 2317.28 N.
o

vertical component Fv = Ft + WG = 6366.67 +700 = 706667

. N. 300 600 400
1800
Horizontal component FH = Fr = 2317.28 N. 7066.67 5051.72 1890.37
T 1
Consider Pulleyat point D, VLD
R
D 45 o 2317.28 R 4051.72
A 3820 B

Torque Mt = 9550N103 T +T 1 HLD 2

n W =1000 N T
R A R B

= 955030103 =573000 Nmm.

p 2

500
Also Mt =(T1 −T2 )rp,573000 =(T1 −T2 )200, T1 −T2 = 2865, and T1 =3, by solving we get T2 =1432.5 N. and T1 = 4297.5 N.
T2
+
Thus, Fv =(T1 +T2 )Sin45o +1000 =5051.72 N. FH =(T1 +T2 )Cos45o = 4051.72N. LSV
+ -
Consider geraat point E, Power available at C, NE=NC-ND=50-30=20 KW.
-
30 Dr. NCR
 Numerical on Design of shaft
Torque M= 9550N×103=9550×20×103=382000 Nmm. F = Mt = 382000 =3820 N. Fr
t t
n 500 rg 100
Fr = Ftan
t =1390.37 N. Fv = Fr + Wp =1390.37+500 =1890.37 N. and FH = Ft = 3820 N. Driven

(Driven) Ft Ft(Driver)
Tofindmaximumbendingmoment:
E
Consider VLD, Find Reactions RA and RB, Taking moments about point A, Driver

−(7066.67300) +(5151.72600) +(1890.371100) −(RB 1500) = 0, RB =1993.63, F r+ WG

Also RA +RB = 7066.67+5151.72+1890.37, RA =12015.13 N.

BendingMoment, MbC = MbB = 0, MbA =−7066.67300 =−2120001 Nmm.
MbD =−(7066.67900) +(12015.13600) =849075 Nmm. MbE =1993.63400 = 797452 Nmm.
Consider HLD, Find Reactions RA and RB, Taking moments about point A, RB =3958.57 N,RA =6230.43 N.
BendingMoment, MbC = MbB = 0, MbA =−2317.28300 =−695184 Nmm.
MbD =−(2317.28900) +(6230.43600) =1652706 Nmm. MbE =3958.57400 =1583428 Nmm.

31 Dr. NCR
 Numerical on Design of shaft

Resultant BendingMoment: MbR= (MbV)2 +(MbH)2 , MbC = MbB = 0,

MbA = (−2120001)2 +(−695184)2 = 2231072.62 Nmm. MbD = (849075)2 +(1652706)2 =1858054.22 Nmm and

MbE = 7974522 +15834282 =1772900 Nmm. Thus maximumbending moment occurs at A, i.e. Mbmax = 2231072.62
and maximumtwisting moment occurs at C,Mt =955000 Nmm.
substitute all values in E.14.12, we get, D=72.86 80 mm.

32 Dr. NCR
 Numerical on Design of shaft
(7)A shaft is driven using a motor, placed vertically below it as shown in the figure. The diameter of the pulley is 1.5 m
and has belt tensions of 5.4 KN and 1.8 KN on tight and slack sides of the belt respectively. Find the diameter of the
shaft, assuming a maximum allowable shear stress of 42 Mpa.
Solution: Diameter of pulley d=1500 mm, T1=5400 N, T2=1800 N, D=?, ed=42 Mpa, beam length L=400 mm.

1.The shaft subjected to combined bending and torsion,

According to maximumshear stress theory,
 16
  
1/3
(KbMb )2 +(KM t t)
1/2
D=  2
 →E.14.12
πτed  400 T1 T2

2.Assuming sudenly applied load with minor shocks Kb=1.75 and Kt =1.25 →Ref.t.14.2

t (T1 −T2 )rp =(5400 −1800)750=2.7×10 Nmm.

6
3.Torque: WKT M=
4.Bending moment: Mb =(T1 +T2 )L =(5400+1800)400 = 2.88106 Nmm. Substitute all values in above equation we get

D=86 mm. standard size of shaft, D=90 mm.

33 Dr. NCR
 Numerical on Design of shaft
(8HW)Design a shaft to transmit power from an electric motor to a lathe head stock through a pulley by means of a belt
drive. The pulley weighs 200n and is located at 300 mm from center of bearing. The diameter of the pulley is 200 mm
and the maximum power transmitted is 1 KW at 120 rpm. The angle of lap of the belt is 180o and coefficient of friction
between the belt and the pulley is 0.3. The shock and fatigue factors for bending and twisting are 1.5 and 2.0. The
allowable shear stress in the shaft may be taken as 35 Mpa.
Solution: Wp=200N, L=300 mm, dia. of pulley d=200 mm, N=1 KW, n=120 rpm, =180o=3.142 rad. =0.3, Kb=1.5, Kt=2,
ed=35 Mpa. D=?
ANS, D=51.12 mm and std D=56 mm.

300 T1 T2
Wp=200 N

(9HW)A power transmission shaft is supported in bearings 2 m apart and carries a pulley weighing 1 KN at its
midpoint and it receives power by a belt drive. The shaft transmits power to another machine by means of a flexible
coupling just outside the right bearing. The power transmitted is 20 KW at 120 rpm. The ratio of belt tensions is 3:1.
Estimate the size of the shaft if the permissible stress in shear is 54 Mpa. Also calculate the twist in the shaft if G=0.8 x
105. take Cm and Ct as 1.5 and the pulley diameter is 200 mm.
Solution: Wp=200 mm, L=2000 mm, N=20 KW, n=120 rpm, ed=54 Mpa, G=0.8 x 105Mpa, Kb=Kt=1.5, D=?, =?.

34 Dr. NCR
 Numerical on Design of shaft
Hint: Mbv = WL
v and M = WHL
bH
T1 =3W =1000 N
p
 200
4 4 T2
ANS: D=131.2=140 mm, =0.0302o. C
A B

1000
2000
1000
VLD
31800 RB
RA
HLD
RA RB

(10)A hollow C15 steel shaft transmits 15 KW at 250 rpm. It is supported on two bearings 750 mm apart. A 500 mm
diameter pulley whose weight is 1000 N is keyed to the shaft at a distance of 100 mm to the left of the left-hand bearing.
A pinion having 75 teeth and a 4 mm module is mounted at 150 mm to the left of the right-hand bearing. The pulley is
driven by a belt downward at an angle of 60o to the horizontal and towards the observer. The pinion drives a gear
placed directly over it. The ratio of belt tension is 3, diameter ratio is 0.5. Use ASME code. Assume suddenly applied
load with minor shock. Consider the keyway effect. Assume the shaft rotates in the clockwise direction when viewed
from the right bearing. Determine the size of the shaft. Assume the pressure angle is 20o.
Solution: Material-C15 steel, m=4 mm, Z=75, N=15 KW, n=250 rpm, K=0.5, Do=?, Di=?. 35 Dr. NCR
 Numerical on Design of shaft
1.Theshaft subjectedtocombined bendingandtorsion, T1 =3
T2
According tomaximumshear stress theory, Wp=1000 N
 500
 16
  
1/3

D=  ( ) ( )
1/2  →E.14.37 Driver

2 2

 ed ( ) K M + KM C Driven
πτ 1− K4 b b t t

A D B RSV

According to maximumnormal stress theory,

 16
  
1/3 100 750 150

D= K M + (K M )2 +(KM )2  →E.14.36

1/2

 ed ( )
4  b b 
πσ 1− K  b b t t

4969.86 1390.37
VLD
2. Material- C-15, σu =420Mpa and σy =235Mpa. 2292 RA
3820
RB

HLD
→0.3σy =70.5Mpa and 0.18σu =75.6Mpa.Take theleast valueamong the two, we haveR A RB

τed =0.75×70.5=52.88Mpa.
→0.6σy =141Mpa and 0.36σu =151.2Mpa.Take theleast valueamong the two, we have
σed =0.75×141=105.75Mpa.
3.Condition, Assumesuddenlyappliedload with minor shocks, Kb =Cm =1.75and Kt =Ct =1.25.
36 Dr. NCR
 Numerical on Design of shaft

4. Torque, Mt = 9550N×103 = 9550×15×103 =573000 N-mm.

n 250
5. Drawhorizontal and vertical loadingdiagramandfind bending moment'M': b Consider Pulley at point C,
Torque on pulley Mt =(T1 −T2 )rp =(3T2 −T2 )250, T1 =3, T2 =1146 N and T1 =3438 N.
T2
Thus, Fv =(T1 +T2 )Sin+ Wp =(3438+1146)Sin60 +1000 = 4969.86 N.
+
o T 2 60 o C
RSV

FH =(T1 +T2 )Cos=(3438+1146)Cos60o = 2292 N. - +

T +T 1 2

Wp=1000 N
Consider Gear at point D, -
T
F 1
r

Torque Mt = Ft rG, FH=Ft = Mt = 573000 =3820 N.( PCD, d= mz =475=300 mm)

rG 150 Driven

Fv = Fr = Ftan
t =3820tan20o =1390.37 N. F t
Ft
(Driver) (Driven)
D
Consider VLD, Driver
Taking moments about point Aand equating to zero, we have, −4969.86100+1390.37600−RB 750 = 0.
Fr
RB = 449.65 N. also RA +RB = 4969.86+1390.37, RA =5910.58 N.
Moments:MbC = MbB = 0, MbA = 4969.86100 = 496986 N-mm. MbD = 449.65 150 = 67447.5 N-mm.
37 Dr. NCR
 Numerical on Design of shaft
Consider HLD, Taking moments about point Aand equating to zero, we have, −2292100−RB 750+3820600 = 0
RB = 2750.4 N. also, RA +RB = 2292+3820, RA =3362 N.
Moments: MbC = MbB = 0,MbA = 2292100 = 229200 N-mm. MbD = 2750.4150 = 412560 N-mm.
Resultant Moments: MbC = MbB = 0,MbA = 547291N-mm, MbD = 418037 N-mm.
Substitute all values in main equations, we get Do=49.71 mmand Do = 48.04 mm. Std, Do =50 mm.

(11) A 100 mm diameter solid circular shaft can carry a torque T without exceeding a certain maximum shear stress.
What portion of this torque T can a hollow shaft having a wall thickness of 10 mm and the same outer diameter carry?
Both the shafts should have the same maximum shear stress.
Solution:
 
1/3
Solidshaft, subjected to torsion only,D= 16 (KM )
t t  →E.14.9, Mt =196349.54ed ( neglect Kt )
πτed 
 16 
1/3

4 ( t t)
Hollowshaft, subjected to torsion only,D= KM  →E.14.34 Mt =115924.76ed,
πτed (1−K ) 
MtHollow =115924.76ed = 0.59, Since ed(solid) =ed(Hollow) M
MtSolid 196349.54ed tHollow =0.59×MtSolid
38 Dr. NCR
 Numerical on Design of shaft
(12) A hollow shaft of diameter ratio 3/8 is required to transmit 500 KW at 110 rpm, the maximum torque being 20%
greater than the mean. The shear stress is not to exceed 60 Mpa and angle of twist in a length of 3 m is not to exceed
1.4o. Calculate the shaft diameter if G=84 Gpa.
Solution: K=3/8=0.375, N=500 KW, n=110 rpm, Mtmax=1.2Mt, L=3000 mm, =1.4o,G=84x103 Mpa, d=?

, t = 9550N 103 = 9550500103 =43.41103 N-mm,

WKT, TorqueM
n 110
But,Mt max =1.2Mt =1.243.41103 =52.10103 N-mm,
Shaft diameter:
 16   
1/3 1/3

4 (
a) Based on Torsion: Diameter, D= 4 ( t t)
KM  →E.14.34, D= 16 1.2552.1010 ) =178mm.
πτed (1−K )  60(1−0.375 )
6
 

 584ML   
1/4 1/4

b)Based on rigidity: Diameter, D=  →   58452.1010 3000  =168 mm.

6

(1−K4 )Gθ (1−0.3754 )84103 1.4

t E.14.42, D=

Among two values, select maximumdiameter, i.e. D=178 mm.

 Std. diameter D=180 mm, but, K= Di 0.375 = Di Di = 67.5 mm.

Do 180 39 Dr. NCR
 Numerical on Design of shaft
(13) A hollow shaft transmits 200 KW at 150 rpm. The total angle of twist in a length of 5 m is 3o. Find the inner and
outer diameters of the shaft, if the shear stress is limited to 60 Mpa. Take G=80 Gpa.
Solution: N=200 KW, n=150 rpm, L=5000 mm, =3o,G=80x103 Mpa, ed =60 Mpa, Do=Di=?
Shaft diameter:
 16 
1/3

4 ( t t)
a) Based on Torsion: Diameter, Do= KM  →E.14.34,
πτed (1−K ) 

Where, Torque, Mt = 9550N 103 =12.73106 N-mm, Do3 (1−k4 ) =1.35106 →(1)
n
 584ML 
1/4

b) Based on rigidity: Diameter, Do=  →E.14.42,  D (1−k ) =  →(2)

6
154.88 10
(1−K )Gθ
t 3 4
4 o
Do

Equating Eqs.(1)&(2), we have, 1.3510 =

6 154.88106
, Do =114.72 mm, Std, Do =120 mm.
Do

Eq(1)yields.... Do3 (1−k4 ) =1.35106,  K=0.68, Also, K= Di 0.68 = Di ,Di =81.6 82 mm.
Do 120

40 Dr. NCR
 Numerical on Design of shaft
(14) A mild steel shaft has to transmit 75 KW at 200 rpm. The allowable stress in the shaft material is limited to 40 Mpa
and the angle of twist is not to exceed 1o in a length of 20 diameters. Calculate the suitable diameter of the shaft.
Solution: power N=75 KW, n=200 rpm, =40 Mpa, =1o, L=20D, D=?

, t = 9550N 103 = 955075103 =3.58106 N-mm,

WKT, TorqueM
n 200
 16 
(1.253.58106 ) =83 mm.
1/3
 
1/3
a. BasedonTorsion:Shaft diameter, D= (KM t t ) →E.14.9, D=
16
πτed   40 

   (20D)
1/4
 
1/4 6
584ML 584 3.58 10
b. Basedonrigidity:Shaft diameter, D= t
 →E.14.17 D=  D=81.06mm.
 Gθ   78.5103
1 

Select maximumdiameter, D=81.06 mm, and std. diameter D=90 mm.

41 Dr. NCR
 Numerical on Design of shaft
(15) A shaft is mounted on bearings 400 mm apart and carries at its middle a gear of 200 mm pitch diameter. The gear
causes 8000 N radial load and 3000 N tangential load as shown in fig. The shaft is rotating at 500 rpm. If the allowable
shear stress for the shaft is 60 Mpa, determine the diameter of the shaft.
Solution: L=400 mm, d=200 mm, Fr=8000 N, Ft=3000 N, speed n=500 rpm, ed=60 MPA, D=?

1. The shaft subjected to combined bending and torsion, Fr

 16
 
2 1/2 
1/3
According to maximumshear stress theory,D=  (KbMb ) +(KM
2
t t)  →E.14.12 Ft
C
πτed  A B

Assume, Kt=1.25 and Kb=1.75,

Tangential tooth load acting perpendicular to the plane at C
400

Bending moment due to F, M = Fr 400 = 8000400 =8105 N-mm

r bV
4 4
Bending moment due to F,t MbH = Ft 400 = 3000400 =3105 N-mm
4 4
Torque Mt = Ft rg =3000100 =3105 N-mm, Resultant bending moment, Mb = (MbV )2 +(MbH )2 =854.4103 N-mm.
Substitute all values in above equation, we get D=50 mm. 42 Dr. NCR
 Design of Keys
The key is a mechanical element and temporary fastener usually made of cold rolled mild steel bars which are inserted between
the shaft and hub of the pulley or gear to prevent relative motion between them. The key transmits the torque from the shaft to
the hub or from the hub to the shaft. It is always inserted parallel to the axis of the shaft in the provided keyway. To fit the key-
way(key slot), key-ways are cut on the shaft and hub. The key fails by shear or crushing while transmitting torque from the
shaft. Hence it acts as a safety device so that it fails earlier than the other members such as the shaft or pulley. It is cheaper and
can be replaced easily.

43 Dr. NCR
 Design of Keys
Types of keys: The commonly adopted form of keys may be classified into four types.
I . Saddle key: For fitting a saddle key, a keyway is not needed on the shaft. The keyway is machined in the hub and the key is
just hammered between the shaft and hub. The key is likely to slip around the shaft under load. These keys are not suitable for
the transmission of power. Therefore, it is used for comparatively light loads such as temporary fastening in fixing and setting
eccentrics, cams, etc.

Saddle keys are of two types:

a) Hollow saddle key: The bottom of the key is shaped to fit the curved surface of the
shaft.
b) Flat saddle key: It is a taper key that fits in a keyway in the hub and is flat on the
shaft. It has a better grip as compared to a hollow saddle key but still can be used for
comparatively light loads only.

II. Sunk keys: The sunk keys are placed in such a way that half the portion is in the keyway of the shaft and the other half is in
the keyway of the hub of the rotating element. These keys can transmit a high amount of power as there is no possibility of slip.
Sunk keys can be of various types.
a) Square keys: This is the most commonly used key for power transmission having equal keyway depths in the shaft and the
hub. This is used for light-duty applications and is most common in industrial machinery 44 Dr. NCR
 Design of Keys
b) Rectangular key: This is a modification of a square key and is used where stability of the connection is desired.
c) Gib-head key: It is similar to a square or rectangular key but has a head projecting from the large end of the key. The
extended head provides the means of extracting the key from the same end at which it was installed.
d) Woodruff key: A semi-circular disc of uniform thickness can be used as a key that aligns itself in the shaft. This key is
called a woodruff key. This is extensively used in the automotive and machine tool industries. A woodruff key is easy to
assemble. The key is inserted in the keyway first and then the hub is pressed on it.

III. Tangent key: The tangent keys are fitted to withstand torsion in one
direction only. These keys are used in heavy-duty shafts.

45 Dr. NCR
 Design of Keys
Design of square or rectangular key: When a key is inserted into a keyway which is partly in the shaft and partly in the hub,
then the key is said to be a sunk key(Square or rectangular key).
The keys are subjected to shear stress and crushing or bearing stress while transmitting torque.
Let, b=width of key, h= thickness or height of key, l=length of key, d= diameter of shaft.
Considering the shearing of the key: b
Reaction of hub on key(F)

Stress stress, d == F = F , Where, F=Mt = Mt = 2Mt , Force of shaft on key(F) F h

Area of shear b.l r  d d
 2 Shaft

 d == 2Mt ,Also b= 2Mt →E.19.50, Also Mt = bτdld N-mm.

r +
b.l.d τdld 2 Mt

Fig. Force on key

>Consideringthecrushingof key:
F
Crushing stress or bearing stress, b' =c = = F = 2Mt = 4Mt h
Area of crushing  h l d h l h.l.d
 2  2
b

Also Thickness of key, h= ' →E.19.51, Also Mt = ld N-mm.

'
4M t hσ b
l

σb ld 4 46 Dr. NCR
 Design of Keys
Designprocedure:
Type-I:Lengthof keyor hubis not available.

Step1:Find torque, Mt = 9550N103 N-mm.

n
Step2:Diameter of shaft, d=3 16Mt →E.19.49, Keyway factor, = 0.75.
d

Step3:Based on the shaft diameter, fromT.17.4, Select standard key cross section, b and h.

Step4: Find length of key from, a) Width of key, b=2Mt , then find l=?, d = allowable shear stress for key.
dld

b) Thickness of key, h=4Mt , then find l=?, finally recommend the bigger value as length.
cld
c =b ' = allowable crushing stress or bearing stress for key.
47 Dr. NCR
 Design of Keys
Type-II:Lengthof keyor hubis given.
Step1:Find torque, Mt = 9550N103 N-mm.
n

Step2:Diameter of shaft, d=3 16Mt →E.19.49, Keyway factor, = 0.75.

d
Step3:Based on the shaft diameter, fromT.17.4, Select standard key cross section, b and h.
Step4: Find length of key from, a) Width of key, b=2Mt , then find b=?, d = allowable shear stress for key.
dld
b) Thickness of key, h=4Mt , then find h=?, finally recommend the bigger values as width and height.
lc d
c =b ' = allowable crushing stress or bearing stress for key.

48 Dr. NCR
 Design of Keys
Note:
Designation of key,
A key is designated by specifying the width, thickness, and length of the key.
For example; A parallel key having a width of 20 mm, thickness of 12 mm, and length of 120 mm used with a shaft of diameter of
70 mm is designated as a parallel key of 20 x 12 x 120
i.e. least dimension is –Thickness(h),
Maximum dimension is- length(l),
The remaining dimension is -width(b).
σy
Allowablestress,σ= =σb'=σc, τd =τ=0.5σ
FOS

49 Dr. NCR
 Numerical - Design of Keys
(1) Select a rectangular parallel key to transmit 9 KW at 300 rpm. The yield stress for the steel used is 310 Mpa. Take
FOS as 2.5.
Solution: power N=9 KW, Speed n=300 rpm, y=310 Mpa, FOS =2.5.
Type-I:Lengthof keyor hubis not available.

Allowable stress, b ' = y = 310 =124 Mpa. Also assume =d =0.5b ' = 62 Mpa
FOS 2.5
Step1:Find torque, Mt = 9550N103 =95509103 = 286.5103 N-mm.
n 300

Step2:Diameter of shaft, d=3 16M t =3 16 286.5 103

=32 mm.
d 0.7562
Step3:Based on the shaft diameter, fromT.17.4, Select standard key cross section, b and h.
d=32 mm, Select, b=10 mm, h=8 mm.
2M t
Step4: Find length of key from, a) Width of key, b= , 10= 2 286.5103
, then find l=28.88  29 mm.
dld 62l 32
4M
b) Thickness of key, h= t , 8= 4 286.5 103
, then find l=36.137 mm. Recommend length, l=37 mm.
cld 124l 32
50 Dr. NCR
 Numerical - Design of Keys
(2) Determine the required length of a square key for a shaft diameter of 35 mm. The key is 8 x 8 mm. Allowable shear
and crushing stresses for key material are 55 Mpa and 12 Mpa respectively. The shaft transmits 639 N-m of torque.
Solution: d=35 mm, b=h=8 mm(square key), =d=55 Mpa,  c=b’=120 Mpa, Mt=639 x 103 N-mm, l=?.

2M t
Find length of key from, a) Width of key, b= , 8= 2 639 10 3
, then find l=82.99 83 mm.
dld 55l 35
4M t
b) Thickness of key, h= , 8= 4 639 10 3
, then find l=76.07  77 mm.
cld 120l 35
Recommend length, l=83 mm.
(3) Rectangular cross-section 8 x 7 x 36 is used to transmit 6 KW at 1200 rpm. The shaft diameter is 30 mm. if the
allowable shear and crushing stress for key material are 60 Mpa and 135 Mpa respectively. Find whether key safe or
not.
Solution: N=6 Kw, n=1200 rpm, d=30 mm, Key cross-section l=36 mm, h=7 mm, b=8 mm. d all =60 Mpa, b’=135 Mpa.

Find torque, Mt = 9550N103 =95506103 = 47.75103 N-mm.

n 1200
2M t
Width of key, b= , 8= 2 47.75103
, then find d =11.05 Mpa. d =dind d all(given), Hence key is safe.
dld d 3630
51 Dr. NCR
 Numerical - Design of Keys
2M 2 47.75103
, then find d =11.05 Mpa.
t
Width of key, b= , 8=
dld d 3630 d =dind d all(given), Hence key is safe.

(4)A shaft is required to transmit 16 KW at 500 rpm. Select a suitable key of rectangular cross-section, if the hub length
is 60 MM. take allowable shear and crushing stresses for material used as 72 Mpa and 140 Mpa respectively.
Solution: N=16 KW, n=500 rpm, d=72 Mpa, b’=140 Mpa. l= 60 mm.
Type-II:Lengthof keyor hubis given.
Step1: Find Torque, Mt = 9550N103 =955016103 =305.6103 N-mm.
n 500
Step2:Diameter of shaft, d=3 16M t =3 16 305.5  10 3
= 28 mm.
d 0.7572
Step3:Based on the shaft diameter, fromT.17.4, Select standard key cross section, b and h.
d=28 mm, Select, b=8 mm, h=7 mm.
2M
Step4: Find length of key from, a) Width of key, b= = t 2305.510 3
=5.05 mm.
dld 726028
b) Thickness of key, h= 4M t = 4 305.5 10 3
=5.19 mm. Recommend b=8 mm, h= 7 mm.
cld 1406028
52 Dr. NCR
 Numerical - Design of Keys
(5) Select a rectangular key for transmitting a power of 50 KW at 500 rpm, to mount a hub of length 60 mm on a solid
circular shaft of diameter 50 mm.
Solution: N=50 KW, n=500 rpm, d=50 mm, l=60 mm.
Type-II:Lengthof keyor hubis given.
Step1: Find Torque, Mt = 9550N103 =955050103 =955103 N-mm.
n 500
Step2:Based on the shaft diameter, fromT.17.4, Select standard key cross section, b and h.
d=50 mm, Select, b=14 mm, h=9 mm.
2M
Step3: Find length of key from, a) Width of key, b= = t 2 955  10 3
=11.58 mm.(Assume ed =55 Mpa and b ' =110 Mpa)
dld 556050

b) Thickness of key, h= 4M t = 4955103

=11.56 mm.
cld 1106050
Recommend width of key b=14 mm, thickness of key h=12 mm.(2012)
(6) Find the length of a square key of size d/4 such that the shaft and the key are made up of the same material.
Solution: l=? Key-Square, width b=d/4.
Let d= diameter of the shaft, b= width of the key, h= thickness of the key, d=shear stress for key material, and s=shear stress
for shaft material.
53 Dr. NCR
 Numerical - Design of Keys

For key,shear stress Mt = bτdld →E.19.50, d = 2Mt

2 bld

For shaft, shear stress Mt = s →E.19.2, s =16M3 t (Neglect effect of keyway effect)
d 3

16 d
Since the shaft and the key are made up of same material, we have
τd =τs
2Mt =16Mt
bdl d3
1= 8
lb d2

(
1 = 8 b =d 4
l (d 4) d2
)
1= 2
l d
l =1.57d
Note:Ingeneral thelengthof thekeyshouldat least beequal totheshaft diameter for satisfactoryproportions.
54 Dr. NCR
 Numerical - Design of Keys
(7)Prove that a square key is equally strong in shear and compression.
Solution:
we knowthat shear stress of key is Mt = bτdld hσ
→E.19.50 and Compressive stress, Mt = ld →E.19.51
b
'

2 4
Since the key is equally strong in shear and compression, equating the above equations, we have
bτdld = hσb'ld
2 4
bτd = 1hσb' →(1) But for square key, b=h
2
2τd =σb', →(2)
This means crushing strength is twice that of shear strength or the shear stress is half of crushing strength.
Substituting (2) into (1) yields
bτd = 1h2τd
2
b=h Thus, squarekey is equallystronginshear andcompression
55 Dr. NCR
 Numerical - Design of Keys
(8 It is required to design a square key for fixing a gear on a shaft, of diameter 25 mm, and 15 KW of power at 720 rpm
is to be transmitted from the shaft to the gear. The key is made of steel 50C4 (yt=460Mpa) and the FOS is 3. The yield
strength in compression can be assumed to be equal to the yield strength in tension. Determine the dimensions of the
key.
Solution: d=25 mm, N=15 KW, n=720 rpm, yt=460Mpa, FOS=3, yt= yc, Key dimension=?

Type-I:Lengthof keyor hubis not available.

y 460
Allowable stress, b ' = = =153.33 Mpa. Also assume =d =0.5b ' = 76.67 Mpa
FOS 3
Step1:Find torque, Mt = 9550N103 =955015103 =198.96103 N-mm.
n 720
Step2:Based on the shaft diameter, fromT.17.4, Select standard key cross section, b and h. d=25 mm, Select, b=8 mm, h=8 mm.
2M t
Step3: Find length of key from, a) Width of key, b= , 8= 2 198.96  103
, then find l=25.59  26 mm.
dld 76.67l 25
4M
b) Thickness of key, h= , 8=t 4 198.96  10 3
, then find l=25.95  26 mm.
cld 153.33l 25
Recommended length of key is l=26 mm.
Thus the key dimensions are 8x8x26 56 Dr. NCR
 Design of Coupling
A Coupling is a mechanical element used to connect two coaxial shafts at their ends to transmit power. The shafts are coaxial with
or without a small angular or linear misalignment. Generally, it is used to connect the prime mover with any machine. Couplings
do not normally allow disconnection of shafts during operation. However, torque limiter couplings can slip or disconnect when the
preset torque limit is exceeded. The industry uses These torque limiters as a safety device to protect the equipment from accidental
overloading. It also reduces the transmission of vibrations and shocks between two connected shafts.

Uses of Shaft Coupling:

Classification:

1. Rigid Couplings:

It is used to connect two shafts which are perfectly aligned.

Types of rigid couplings are;
➢ Muff or Sleeve Coupling
➢ Split muff or clamp or compression coupling
➢ Flange coupling(unprotected or Protected type flange coupling) 57 Dr. NCR
 Design of Coupling
2. Flexible coupling: It is used to connect two shafts having both lateral and angular misalignment.
Types of flexible couplings are;
➢ Bushed-pin type coupling
➢ Universal coupling(Hooke’s joint)
➢ Oldham coupling

58 Dr. NCR
 Design of Coupling
Flange coupling(CI or unprotected or Protected type flange coupling):

1. Unprotected Type Flange Coupling:

It consists of two similar cast iron flanges. Each flange is mounted on the shaft
end and keyed to it. The two flanges are connected using bolts and nuts as
shown in the figure.

59 Dr. NCR
 Design of Coupling
2. Protected Type Flange Coupling:

It consists of two separate cast iron flanges. Each flange is mounted on the shaft end and keyed to it. One of the flanges has a
projected portion, and the other flange has a corresponding recess.

60 Dr. NCR
 Design of Coupling
Design Procedure: (Unprotective coupling)

The components to be designed in this coupling are i) Shaft ii) Flanges iii) Bolts and iv) Keys.

1. Design of Shaft(d): (all units in N and mm)

a) Torque Transmitted by the coupling: Torque Mt = 9550N103,N−mm. →E19.3C. Find Mt

n
b) Torque Transmitted by the shaft: Torque Mt =  d3
 →E19.2,
16 s
Where, =keyway factor=0.75. d=dia of shaft, mm. s =k =b = allowablestress for shaft.
Find 'd', in mm. Standardise the dia. of shaft.
2. Design of Flange and Hub(D1, D2, D, l ,t):

a) Bolt circle diameter, D1 = 2d+50 → E19.12B b) Hub diameter, D2 = 1.5d+25 → E19.13B

c) Length of Hub, l = 1.25d+18.75 → E19.14D d) Outer diameter of flange, D = 2.5d+75 → E19.14B

e) Thickness of flange, t = 0.5d → E19.17

NOTE: Width of the protective flange t1=0.25d mm. Ref. Fig- F19.1
61 Dr. NCR
 Design of Coupling
g) Check for stresses:

Torque capacity based on shear of flange: M t max = t ( D2 ) f → E19.6, where  f =  f ind = induced shear stress inthe flange.

Find  f and  f ind   f all (allowable stress ).Designis safe. otherwiswe , increase ' t ' and check again.
y
Where, for  f all → ref .T 1.4,  f all = and  y = 0.5 y
FOS

3. Design of Bolt( i, d1,b ):

i) Number of bolts: i = 0.02d+3→ E.19.1B Minimum number of bolts = 4 (should be even number always)

0.5
ii) Diameter of bolt(d1), d1 = → E19.8 Select standard diameter d1 from T18.7
i
iii) Check for shear stress induced in bolt:
  d12   D1 
Torque transmitted through bolts, Mt = i   b   → E19.4 Where, D1= bolt diameter, mm b=induce shear stress, MPa
 4   2 
Find b ind < b all (given), Design is safe NOTE: Core area also select based on d1, from T18.7
62 Dr. NCR
 Design of Coupling
4. Design of key( b, h, l, k ): ➢ Based on the diameter of shaft ‘d’ from T17.4, find width ‘b’ and thickness of the key ‘h’
➢ From T17.5, select the standard length of the key ( l ) or use E19.14D
➢ NOTE: The length of the key should not be less than the length of the hub
➢ Check induce shear stress in the key(k), k=d2
2M t
Width of the key, b = → E19.50 Find induce shear stress k <  all , design is safe.
 k ind ld

Also check induce crushing stress in the key(b):

Thickness of the key, h = 4Mt →E19.51(Note: 'b =k bearingstress)

 'b ind ld
Find induce stress ’b ind < ’b all allowable,  design is safe.Otherwise increase the key dimensions and check again.

NOTE:
i) The design procedure for the protected type flange coupling is the same as that of the unprotected type flange coupling. The
width of the protective flange ‘ t1 ’ is equal to the major diameter of the bolt or t1=0.25d
ii) Length of the KEY = Length of HUB
63 Dr. NCR
Design of Coupling

64 Dr. NCR
65 Dr. NCR