SHAFTINGS, KEYS &
SPLINES, COUPLINGS
M A C H I N E D E S I G N & S H O P P R A C T I CE
“Satisfaction lies in the effort, not the attainment.
Full effort is full victory.”
- Mahatma Gandhi
SHAFTINGS
Shaft
• a rotating machine element which is used to transmit
power from one place to another.
Axle
• a stationary machine element and is used for the
transmission of bending moment only.
• It simply acts as a support for some rotating body such
as hoisting drum, a car wheel or a rope sheave.
Spindle
• short shaft that imparts motion either to a cutting tool
(e.g. drill press spindles) or to a work piece (e.g. lathe
spindles).
SHAFTINGS
Types of Shafts:
1. Transmission shafts
• These shafts transmit power between the source
and the machines absorbing power.
• Example: counter shafts, line shafts, over head
shafts and all factory shafts
2. Machine shafts
• These shafts form an integral part of the machine
itself.
• Example: crank shaft
SHAFTINGS
Analysis for shaftings:
• Ductile materials
Based on strength is controlled by the maximum
shear theory.
• Brittle materials
Designed on the basis of the maximum normal stress
theory.
STRESSES ON SHAFTS
Shear Stress (torsional load)
Tc
Ss =
I
𝟏𝟔𝐓
𝐒𝐬 =
𝛑𝐃𝟑
𝐒𝐬 =
𝟏𝟔𝐓𝐃𝐨
𝛑 𝐃𝟒𝐨 − 𝐃𝟒𝐢
Bending Stress (tension or compression)
Mc
Sb =
I
𝟑𝟐𝐌
𝐒𝐛 =
𝛑𝐃𝟑
𝐒𝐛 =
𝟑𝟐𝐌𝐃𝐨
𝛑 𝐃𝟒𝐨 − 𝐃𝟒𝐢
Axial stress (axial loads)
F
Sa =
A
𝐅𝐚
𝐒𝐚 = 𝛑
𝟐
𝐃
𝟒
𝐅𝐚
𝐒𝐚 = 𝛑
𝟐
𝟐
𝐃
−
𝐃
𝐢
𝟒 𝐨
STRESSES ON SHAFTS
Combined Torsion and Bending Stresses
Ductile materials (maximum shear stress)
𝐒𝐬−𝐦𝐚𝐱 =
𝟏
𝐒𝐛
𝟐
𝟐
+
𝐒𝐬−𝐦𝐚𝐱
𝐒𝐬𝟐
𝟏𝟔
𝟐 + 𝐓𝟐
=
𝐌
𝛑𝐃𝟑
Brittle materials (maximum normal stress)
𝟏
𝐒𝐛
𝟐
𝟐
𝐒𝐭−𝐦𝐚𝐱
𝟏
= 𝐒𝐛 +
𝟐
𝐒𝐭−𝐦𝐚𝐱
𝟏𝟔
𝟐 + 𝐓𝟐
=
𝐌
+
𝐌
𝛑𝐃𝟑
+ 𝐒𝐬𝟐
Note: if factor of Safety, N,
is considered.
𝑺𝒚𝒔
𝑺𝒔−𝒎𝒂𝒙 =
𝑵
𝑺𝒚
𝑺𝒕−𝒎𝒂𝒙 =
𝑵
Where:
𝑺𝒚𝒔 = 𝟎. 𝟓𝑺𝒚
STRESSES ON SHAFTS
Combined Torsion and Bending with shock & fatigue factors
Ductile materials (maximum shear stress)
𝐒𝐬𝐦𝐚𝐱 =
𝐒𝐬𝐦𝐚𝐱
𝟏
𝐤 𝐛 𝐒𝐛
𝟐
𝟏𝟔
=
𝛑𝐃𝟑
𝟐
+ 𝐤 𝐬 𝐒𝐬
𝐤𝐛𝐌
𝟐
𝟐
+ 𝐤𝐬𝐓
𝟐
Brittle materials (maximum normal stress)
𝐒𝐭−𝐦𝐚𝐱
𝐒𝐭−𝐦𝐚𝐱
𝟏
= 𝐤 𝐛 𝐒𝐛 +
𝟐
𝟏
𝐤 𝐛 𝐒𝐛
𝟐
𝟏𝟔
=
𝐤𝐛𝐌 +
𝟑
𝛑𝐃
𝟐
𝐤𝐛𝐌
+ 𝐤 𝐬 𝐒𝐬
𝟐
+ 𝐤𝐬𝐓
𝟐
𝟐
SHAFTINGS
Power Transmitted by Shaft
𝐏 = 𝟐𝛑𝐓𝐍
From Machinery’s Handbook:
Allowable twisting moment (of any cross-section)
𝟔𝟑, 𝟎𝟎𝟎 𝑷
𝑻 = 𝑺𝒔 𝒁𝑷 =
𝑵
Note: units
Diameter of Solid Circular Shaft
𝑫=
𝟑
𝟓. 𝟏𝑻
=
𝑺𝒔
𝟑
𝟑𝟐𝟏, 𝟎𝟎𝟎 𝑷
𝑵𝑺𝒔
Ss (shear stress) – psi
Zp (polar section modulus) – in3
P (power transmitted) – hp
T (torque transmitted) – in-lbf
N (rotative speed) – rpm
D (shaft diameter) – in
SHAFTINGS
From Machinery’s Handbook:
Power Transmitted by Shaft
Note: units
Main Transmitting Shaft (Ss = 4000 psi) P (power transmitted) – hp
N (rotative speed) – rpm
𝑫𝟑 𝑵
𝑷=
D (shaft diameter) – in
𝟖𝟎
Line shafts carrying pulleys (Ss = 6000 psi)
𝑫𝟑 𝑵
𝑷=
𝟓𝟑. 𝟓
Small, Short shafts, countershafts (Ss = 8500 psi)
𝑫𝟑 𝑵
𝑷=
𝟑𝟖
SHAFTINGS
Maximum Vertical Shear Stress on Shaft
𝟒𝐕
𝟏𝟔𝐕
(for “Circular” cross-section)
𝐒𝐯 =
=
𝟐
𝟑𝐀 𝟑𝛑𝐃
𝟑𝐕
(for “rectangular/square” cross-section)
𝐒𝐯 =
𝟐𝐀
Shafts Angular Deformation
𝑻𝑳
𝛉=
𝑱𝑮
Note:
V – maximum shear force
A – cross-sectional area
J – polar moment of inertia
G – rigidity modulus
SHAFTINGS
From Machinery’s Handbook:
Angle of twist “not exceeding 0.08° per foot length of shaft”
𝟒
𝑫 = 𝟎. 𝟐𝟗 𝑻 = 𝟒. 𝟔
𝟒
𝑷
𝑵
Note: units
Shaft deflection “not exceeding 1° in a LP =(power
20 D”transmitted) – hp
N (rotative speed) – rpm
D (shaft diameter) – in
𝟑 𝑷
𝟑
T (transmitted torque) – in-lbf
𝑫 = 𝟎. 𝟏 𝑻 = 𝟒. 𝟎
𝑵
L (shaft length) – ft
Shaft Linear Deflection
𝑳 = 𝟖. 𝟗𝟓
𝟑
𝟑
𝑫𝟐
𝑳 = 𝟓. 𝟐 𝑫𝟐
(for shafting subject to no bending action
except its own weight)
(for shafting subject to bending action of pulleys)
SHAFTINGS
Problem 1
Determine the torque that can be applied
to a 1-in diameter circular shaft if the
shearing stress is not to exceed 8000 psi
A. 1570 in-lb
C. 2750 in-lb
B. 1750 in-lb
D. 3560 in-lb
Ans:
𝟏𝟓𝟕𝟎. 𝟖 𝐢𝐧 − 𝐥𝐛
SHAFTINGS
Problem 2
A steel shaft is subjected to a constant torque of
2,260 N-m. The ultimate strength and yield
strength of the shafting material are 668 MPa
and 400 MPa respectively. Assume a factor of
safety 2 based on the yield point and endurance
strength in shear, determine the diameter of the
shaft in inches.
A. 1.521 in
C. 2.321 in
B. 1.915 in
D. 2.417 in
Ans:
𝟏. 𝟗𝟏𝟓 𝐢𝐧
SHAFTINGS
Problem 3
What would be the diameter in millimeters
of a main power transmitting steel shaft
SAE 1040 to transmit 150 kW at 900 rpm?
A. 2.3
C. 66.4
B. 2.6
D. 76.5
Ans:
𝟔𝟔. 𝟒𝟏 𝐦𝐦
SHAFTINGS
Problem 4
A 3 ft length of commercial steel shafting is
to transmit 50 hp at 3600 rpm through
flexible coupling from an AC motor to a DC
generator. What is the nearest standard
shaft size? Take allowable shearing stress
for commercial shafting with keyway to be
6 ksi and the combined shock and fatigue
factor applied to torsional moment is unity.
Ans:
𝟏𝟓
𝟏𝟔
𝐢𝐧
SHAFTINGS
Problem 5
A solid steel shaft having a diameter of 3
in twists through an angle of 5 deg in 20 ft
of length because of the action of a torque.
Determine the maximum shearing stress
in the shaft.
A. 3550 psi
C. 6550 psi
B. 4550 psi
D. 8550 psi
Ans:
𝟔𝟓𝟒𝟓 𝐩𝐬𝐢
SHAFTINGS
Problem 6
Determine the angular deflection in
degrees of a SAE 1040 steel shaft in a
length of 1/2 meter. The shear stress is 69
MPa, shaft diameter is 62 mm and steel
modulus of elasticity is 79.3 GPa.
A. 0.08
C. 0.1
B. 0.01
D. 0.8
Ans:
𝟎. 𝟖𝟎𝟒𝐨
SHAFTINGS
Problem 7
The maximum torque that can be applied
to a hollow circular steel shaft of 120 mm
outside diameter and 80 mm inside
diameter without exceeding a shearing
stress of 80 MPa is:
A. 18.46 KN-m
C. 42.83 KN-m
B. 36.24 KN-m
D. 21.78 KN-m
Ans:
𝟐𝟏. 𝟕𝟖𝟐 𝐤𝐍 − 𝐦
SHAFTINGS
Problem 8
A hollow shaft with outside diameter of 14
cm and wall thickness of 0.08 cm
transmits 200 kW at 400 rpm. What must
be the angular deflection of the shaft if the
length is 5 meters? Take G = 12,000,000
psi.
A. 0.019 deg
C. 1.94 deg
B. 1.14 deg
D. 2.44 deg
Ans:
𝟏. 𝟏𝟒𝐨
SHAFTINGS
Problem 9
Determine the thickness of a hollow shaft
having an outside diameter of 100 mm if it
is subjected to a maximum torque of
5,403.58 N-m without exceeding a
shearing stress of 60 MPa or a twist of 0.5
degree per meter length of shaft. G =
83,000 Mpa.
A. 15 mm
C. 16.8 mm
B. 30 mm
D. 14.2 mm
• Based on max shear: t = 7.11 mm
• Based on max twist: t = 15 mm
Ans:
𝟏𝟓 𝐦𝐦
SHAFTINGS
Problem 10
A round steel shaft to a torque 200 rpm
and is subjected to a torque of 226 N-m.
the allowable shearing stress 41.4 Mpa. It
is also subjected to a bending moment of
339 N-m. The allowable tensile stress is
55 MPa. Find the diameter.
A. 37 mm
C. 45 mm
B. 41 mm
D. 51 mm
• Based on allowable shear: d = 36.87 mm
• Based on allowable tensile: d = 41.04 mm
Ans:
𝟒𝟏 𝐦𝐦
SHAFTINGS
Problem 11
The shaft of a 50 hp 850 rpm electric
motor is 31 in from center to center of
bearings and 2.5 in. in diameter. If the
magnetic pull on the armature is 1500 lb
concentrated
midway
between
the
bearings, determine the maximum shear
and the maximum tensile stress in the
shaft.
SHAFTINGS
Problem 11
1500 lb
15.5 in
31 in
2.5 in
750 lb
750 lb
Ans:
𝟑. 𝟗𝟖 𝐤𝐬𝐢; 𝟕. 𝟕𝟕 𝐤𝐬𝐢
SHAFTINGS
Problem 12
A section of commercial shafting 5 ft long
between bearings carries a 200 lb pulley at its
midpoint. The pulley is keyed to the shaft and
receives 20 hp at 150 rpm which is transmitted
to a flexible coupling just outside the right
bearing. The belt drive is horizontal and the sum
of the belt tensions is 1500 lb. Assume Kt = Kb =
1.5. Calculate the necessary shaft diameter and
determine the angle of twist between bearings.
Take G = 12,000,000 psi.
SHAFTINGS
Problem 12
200 lb
T1 + T2 = 1500 lb
T1
2.5 ft
5 ft
T2
100 lb
• ½ in to 2 ½ in by 1/16 in increments
• 2 5/8 in to 4 in by 1/8 in increments
• 4 ¼ in to 6 in by ¼ in increments
100 lb
Ans:
𝟏
𝟑. 𝟏𝟒 𝐢𝐧 ≈ 𝟑 𝟖 ; 𝟎. 𝟏𝟐𝟔°
SHAFTINGS
Problem 13
A solid transmission shaft is 3.5 inches in
diameter. It is desired to replace it with a hollow
shaft of the same material and same torsional
strength but its weight should only be half as
much as the solid shaft. Find the outside
diameter and inside diameter of the hollow shaft
in millimeters.
Ans:
𝟏𝟎𝟕. 𝟑𝟏𝟓 𝐦𝐦 ; 𝟖𝟔. 𝟗𝟕 𝐦𝐦
SHAFTINGS
Problem 14
A 76 mm solid shaft is to be replaced with
a hollow shaft of equal torsional strength.
Find the inside diameter and percentage
of weight saved, if the outside diameter of
the hollow shaft is 100 mm.
A. 56.53 %
C. 48.49 %
B. 67.31%
D. 72.50 %
Ans:
𝟓𝟔. 𝟓𝟒𝟕 %
SHAFTINGS
Problem 15
A 20 feet steel line shaft has no bending
action except its own weight. What power
in HP can the shaft deliver at a speed of
360 rpm. Consider that the torsional
deflection will not exceed 0.08 degree per
ft length.
A. 100
C. 55
B. 120
D. 135
Ans:
𝟏𝟎𝟎. 𝟏𝟐 𝐡𝐩
SHAFTINGS
Problem 16
A 20 ft steel line shaft has bending action
of pulleys. What power in hp can the shaft
deliver at a speed of 360 rpm. Consider
that the torsional deflection will not exceed
0.08 deg. per ft length.
A. 2600
C. 1250
B. 1100
D. 900
Ans:
𝟐𝟔𝟎𝟐. 𝟕𝟒 𝐡𝐩
KEYS and SPLINES
Keys
• Are used to prevent relative motion between a
shaft and the connected member through
which torque is being transmitted.
KEYS and SPLINES
Splines
• Are keys made integral with the shaft and
usually consist of four, six, or ten in members.
ANALYSIS ON KEYS
Shear Stress on key:
𝐒𝐬 =
𝐅𝐭
𝐀 𝐬𝐡𝐞𝐚𝐫
𝐅𝐭
𝟐𝐓
=
=
𝐰𝐋 𝐰𝐋𝐃𝐬
Compressive Stress on key:
𝐅𝐭
𝐅𝐭
𝟒𝐓
𝐒𝐜 =
=
=
𝟏
𝐀𝐜
𝐭𝐋 𝐭𝐋𝐃𝐬
𝟐
Note:
When key and shaft are made of same material
Let:
L = 1.18D
w = D/4
ANALYSIS ON SPLINES
Shear Stress on splines:
𝐅𝐭
𝐅𝐭
𝟐𝐓
𝐒𝐬 =
=
=
𝐀 𝐬𝐡𝐞𝐚𝐫 𝐧𝐬 𝐰𝐋𝐧𝐬 𝐰𝐋𝐝𝐧𝐬
Compression between splines & hub:
𝐅𝐭
𝐅𝐭
𝐓
𝐒𝐜 =
=
=
𝟏
𝐀 𝐜 𝐧𝐬
𝐭𝐋𝐧𝐬 𝐭𝐋𝐑 𝐦 𝐧𝐬
𝟐
𝐑+𝐫 𝐃+𝐝
=
𝟐
𝟒
Total torque transmitted:
𝐓 = 𝐅𝐑 𝐦 = 𝐩𝐀𝐑 𝐦
𝐑𝐦 =
Torque of one spline:
𝐓
𝐓𝐬 = 𝐱 𝟏. 𝟏
𝐧𝐬
Note:
𝐧𝐬 - number of splines
𝑹𝐦 - mean radius
𝐩 – permissible pressure
COUPLINGS
Couplings
• Are used to connect
sections of shafts or
to connect the shaft
of a driving machine
to the shaft of a
driven machine.
ANALYSIS OF COUPLINGS
COUPLING ELEMENT
FAILURE MODE
Shearing of key
KEY
Compression of key
Shearing of bolts
BOLTS
FLANGE
STRESS FORMULA
𝟐𝐓
𝐒𝐬 =
𝐰𝐋𝐃𝐬
𝟒𝐓
𝐒𝐜 =
𝐭𝐋𝐃𝐬
𝟐𝐓
𝐒𝐬 = 𝛑
𝟐
𝐝
𝟒 𝐛 𝐧𝐛 𝐃𝐁𝐂
Compression
between bolt and
flange
𝟐𝐓
𝐒𝐜 =
𝐝𝐛 𝐭 𝐟 𝐧𝐛 𝐃𝐁𝐂
Punching shear
𝟐𝐓
𝐒𝐬 =
𝛑𝐭 𝐟 𝐃𝐡
KEYS & COUPLINGS
Problem 17
A 4 inches shaft using a flat key whose
width is 1 inch is transmitting a torque of
63 000 in-lb. If the design shearing stress
is 5000 psi the safe length is:
A. 6.3 in
C. 4.3 in
B. 5.3 in
D. none of these
Ans:
𝟔. 𝟑 𝐢𝐧
KEYS & COUPLINGS
Problem 18
A 1200 mm cast iron pulley is fastened to
a 112.5 mm shaft by means of a 28.13 mm
square key 175 mm long. What force
acting at the pulley rim will shear the key if
shear stress of the key is 20.67 kg/mm2?
A. 9538 kg
C. 5839 kg
B. 8593 kg
D. 3859 kg
KEYS & COUPLINGS
Problem 18
Key|
Shaft|
Frim
Fkey
Pulley |
Ans:
𝟗𝟓𝟑𝟗 𝐤𝐠
KEYS & COUPLINGS
Problem 19
Determine the required length of a square
key if the key and shaft are to be of the
same material and of equal strength but
due to stress concentration, there is a 25%
torque reduction.
Ans:
𝟏. 𝟏𝟖𝐃
KEYS & COUPLINGS
Problem 20
A line shaft with a power of 100 kW at a
speed of 1200 rpm, had a rectangular key
used in its pulley connection. Consider the
shearing stress of the shaft to be 40 MPa
and the key to be 200 MPa, determine the
width of the rectangular key if it is onefourth of the shaft diameter.
A. 23.65 mm
C. 11.65 mm
B. 14.65 mm
D. 9.65 mm
Ans:
𝟏𝟏. 𝟔𝟓 𝐦𝐦
KEYS & COUPLINGS
Problem 21
A solid steel machine shaft with a safe
shearing stress of 7000 psi transmits a
torque of 10,500 in-lb. A square key is
used whose width is equal to one-fourth
the shaft diameter and whose length is
equal to 1 ½ times the shaft diameter. Find
key dimensions.
A. ½ in and 3 in
C. ¾ in and 2.5 in
B. ½ in and 2.5 in D. ¾ in and 3 in
Ans:
½ 𝐢𝐧 𝐚𝐧𝐝 𝟑 𝐢𝐧
KEYS & COUPLINGS
Problem 22
Two shafts are connected by a flange
coupling. The coupling is secured by 6
bolts, 20 mm in diameter on a pitch circle
diameter of 150 mm. If torque of 120 N-m
is applied, find the shear stress in the
bolts.
A. 0.85 N/mm2
C. 0.85 Pa
B. 0.85 kPa
D. 0.95 Pa
Ans:
𝟎. 𝟖𝟓 𝐌𝐏𝐚
KEYS & COUPLINGS
Problem 23
A flanged coupling has an outside
diameter of 200 mm and connects two 40
mm shafts. There are four 16 mm bolts on
a 140 mm bolt circle. The radial flanged
thickness is 20 mm. If the torsional stress
in the shaft is not to exceed 26 MPa,
determine the shear stress in the bolts if
uniformly distributed
A. 1.2 MPa
C. 4.3 MPa
B. 2.9 MPa
D. 5.8 MPa
Ans:
𝟓. 𝟖 𝐌𝐏𝐚
KEYS & COUPLINGS
Problem 24
Two short shafts having identical
diameters of 38.1 mm and rotating at 400
rpm are connected by a flange coupling
having 4 bolts with a 100 mm bolt circle.
The design shearing stress of the bolt is
12 N/mm2 and the design compressive
stress of the flange is 15 N/mm2 . How
thick should the flange be in mm?
A. 11.51 mm
C. 12.49 mm
B. 13.60 mm
D. 15.65 mm
Ans:
𝟏𝟏. 𝟓𝟏 𝐦𝐦
KEYS & COUPLINGS
Problem 25
A flange coupling is to be designed, using
25-mm diameter bolts at a distance of 152
mm from the center of the shaft. Allowable
shearing stress on the bolt is 103 MPa. If
the shaft is to transmit 5,800 hp at a speed
of 1,200 rpm, how many bolts are needed
in the connection?
A. 2
C. 4
B. 3
D. 5
Ans:
𝟓 𝐛𝐨𝐥𝐭𝐬
“The more I PRACTICE,
the Luckier I get.”
-Gary Player