Unit II
Design for Strength and Rigidity
Syllabus
❑ Design of Cotter Joint
❑ Design of Transmission Shaft
❑ Design of Square and Flat Keys
❑ Design of Coupling
COTTER JOINT
Introduction
⮚ Cotter joint is used to connect two co-axial rods, which are
subjected to either axial tensile force or axial compressive
force.
⮚ It is, also, used to connect rod on one side with some machine
part like crosshead or base plate on the other side.
⮚ It is not used for connecting shafts that rotate and transmit
torque.
Applications of Cotter Joint
⮚ Joint between the piston rod and the crosshead of steam engine
⮚ Joint between the slide spindle and the fork of the valve
mechanism
⮚ Joint between the piston rod and the tail or pump rod.
⮚ Foundation Bolt
Cotter Joint
Cotter Joint
P Tensile force acting on rods (N)
d diameter of each rod (mm)
d1 outside diameter of socket (mm)
d2 diameter of spigot or inside diameter of socket (mm)
d3 diameter of spigot-collar (mm)
d4 diameter of socket-collar (mm)
a distance from end of slot to the end of spigot on rod-B (mm)
b mean width of cotter (mm)
c axial distance from slot to end of socket collar (mm)
t thickness of cotter (mm)
t1 thickness of spigot-collar (mm)
l length of cotter (mm)
FBD of Cotter Joint
Assumptions
⮚ The rods are subjected to axial tensile force.
⮚ The effect of stress concentration due to slot are neglected.
⮚ The stresses due to initial tightening of cotter are neglected.
1. Tensile Failure of Rods
Each rod of diameter d is subjected to
tensile force P. This equation is used
to find the diameter of rod d , from the
permissible tensile stress of the rod.
2. Tensile Failure of Spigot
This is used to determine the
diameter of spigot or inner
diameter of socket (d2).
Weakest Cross section X-X
The thickness of cotter is usually
determined by empirical
relationship
3. Tensile Failure of Socket
This equation is used to find out the
outside diameter of socket ‘d1’.
Weakest Cross section Y-Y
4. Shear Failure of Cotter
The cotter is subjected to double
shear.
The mean width of cotter ‘b’ can
be determined.
5. Shear Failure of Spigot End
This equation is used to
determine the distance ‘a’ .
Spigot end is subjected to
double shear as shown.
6. Shear Failure of Socket End
Socket end is also
subjected to double shear
as shown in fig.
This failure is used to
determine the dimension
‘C’
7. Crushing Failure of Spigot End
• The force ‘P’ causes a narrow
rectangular area of thickness ‘t’ and
width ‘d2’ .
• This equation is used for checking
8. Crushing Failure of Socket End
• As shown in Fig. the force ‘P’
causes compressive stress on a
narrow rectangular area of
thickness ‘t’.
9. Bending Failure of Cotter
SHAFT
Introduction
⮚ A shaft is a rotating member usually of circular cross-section, which
supports transmission elements like gears, pulleys and Sprockets.
⮚ Used to transmit power or motion.
⮚ Specific Categories of Transmission Shafts.
❖ Axle: An axle is a non-rotating member used to support rotating
members such as wheels, pulleys and it do not transmit torque.
e.g. Rear axle of railway Wagon.
❖ Spindle: Spindle is a short rotating shaft. Ex. Machine Tools.
❖ Countershaft: It is secondary shaft, which is driven by the
main shaft. e.g. Multi-stage Gearboxes.
❖ Line Shaft: A line shaft consists of a number of shafts, which
are connected in axial direction by means of couplings. e.g.
Workshops.
Shaft Materials
⮚ Shafts are made of medium carbon steels with a carbon
content from 0.15 to 0.40 per cent such as 30C8 or 40C8
⮚ For High strength is required, high carbon steels such as45C8
or 50C8 or alloy steels are used.
⮚ Alloy steels include nickel, nickel-chromium and molybdenum
steels
⮚ Grades of alloy steels are 16Mn5Cr4, 40Cr4Mo2, 16Ni3Cr2,
35Ni5Cr2
⮚ Alloy steels are costly compared with plain carbon steels
⮚ Alloy steels have higher strength, hardness, higher resistance
to corrosion
Shaft Design on Strength Basis
• When the shaft is subjected to axial tensile force
• When the shaft is subjected to pure bending moment
• When the shaft is subjected to pure torsional moment
When the shaft is subjected to combination of loads,
the principal stress and principal shear stress are
obtained by constructing Mohr’s circle.
The normal stress is denoted by σx while the shear stress, by τ.
Two cases for calculating the
value of σx
• Case I
When shaft is subjected to a combination of
axial force, bending moment and torsional
moment.
• Case II
When shaft is subjected to a combination of
bending and torsional moments without any
axial force
• The principal stress σ1
• The principal shear stress τmax
• The shaft can be designed on the basis of maximum principal stress
theory or maximum shear stress theory.
• These theories can be applied to transmission shaft subjected to
combined bending and torsional moments.
Maximum Principal Stress Theory
• The maximum principal stress is σ1. Since the shaft is subjected to
bending and torsional moments without any axial force.
• The permissible value of maximum principal stress is
Maximum Shear Stress Theory
• The principal shear stress is τmax
• The permissible value of maximum shear stress is
Shaft Design on Torsional Rigidity Basis
• In some applications, the shafts are designed on the basis of either
torsional rigidity or lateral rigidity.
• A transmission shaft is said to be rigid on the basis of torsional
rigidity, if it does not twist too much under the action of an
external torque.
• The transmission shaft is said to be rigid on the basis of lateral
rigidity, if it does not deflect too much under the action of external
forces and bending moment.
• In certain applications, like machine tool spindles, it is necessary to
design the shaft on the basis of torsional rigidity.
• The angle of twist θr (in radians) is given by,
• Converting θr from radians to degrees (θ)
l = length of shaft subjected to twisting
moment (mm)
Mt = torsional moment (N-mm)
G = modulus of rigidity (N/mm2)
• The permissible angle of twist for machine tool applications is 0.25°
per meter length.
• For line shafts, 3° per meter length is the limiting value.
• Modulus of rigidity for steel is 79 300 N/mm2 or approximately 80
kN/mm2.
ASME Code for Shaft Design
• According to this code, the permissible shear stress τmax for the
shaft without keyways is taken as 30% of yield strength in tension
or 18% of the ultimate tensile strength of the material, whichever is
minimum.
• If keyways are present, the above values are to be reduced by 25%.
ASME Code for Shaft Design
• According to the ASME code, the bending and torsional moments
are to be multiplied by factors kb and kt respectively, to account for
shock and fatigue in operating condition.
• The ASME code is based on maximum shear stress theory of
failure.
Similarly
,
kb = Combined shock and fatigue factor applied to bending moment
kt = Combined shock and fatigue factor applied to torsional moment
Values of shock and fatigue factors kb and kt
Application Kb Kt
Load gradually applied 1.5 1
Load suddenly applied
1.5 - 2.0 1.0 - 1.5
(minor shock)
Load suddenly applied
2.0 – 3.0 1.5 – 3.0
(Heavy shock)
Design of Hollow shaft on Strength Basis
• Compared with solid shaft, hollow shaft offers following
advantages:
1. The stiffness of the hollow shaft is more than that of solid shaft
with same weight.
2. The strength of hollow shaft is more than that of solid shaft with
same weight.
3. The natural frequency of hollow shaft is higher than that of solid
shaft with same weight.
• Compared with solid shaft, hollow shaft has the following
disadvantages:
1. Hollow shaft is costlier than solid shaft.
2. The diameter of hollow shaft is more than that of solid shaft and
requires more space.
• Assume,
where,
di = inside diameter of the hollow shaft (mm)
do = outside diameter of the hollow shaft (mm)
C = ratio of inside diameter to outside diameter.
• When the shaft is subjected to axial tensile force
• When the shaft is subjected to bending moment
For hollow circular cross-section,
• When the shaft is subjected to pure torsional moment
For hollow circular cross-section,
• The principal stress σ1
• The principal shear stress τmax
• The hollow shaft can be designed on the basis of maximum principal
stress theory or maximum shear stress theory.
• These theories can be applied to transmission shaft subjected to
combined bending and torsional moments.
Maximum Principal Stress Theory
Maximum Shear Stress Theory
Shaft Design on Torsional Rigidity Basis
Shaft Design on Torsional Rigidity Basis
l = length of shaft subjected to twisting
moment (mm)
Mt = torsional moment (N-mm)
G = modulus of rigidity (N/mm2)
• The permissible angle of twist for machine tool applications is 0.25°
per meter length.
• For line shafts, 3° per meter length is the limiting value.
• Modulus of rigidity for steel is 79 300 N/mm2 or approximately 80
kN/mm2.
KEY’S
Introduction
⮚ A key can be defined as a machine element which is used to
connect the transmission shaft to rotating machine elements
like pulleys, gears, sprockets or flywheels.
⮚ Functions
1. The primary functions of the key is to transmit the torque
from the shaft to the hub of the mating element and vice
versa.
2. The second function of the key is to prevent relative motion
between the shaft and the joined machine element like gear
or pulley.
Key Joint
Types of Keys
1. Sunk keys and Saddle keys
2. Square Key and Flat Key
3. Taper Key and Parallel Key
4. Key with and without Gib-head
There are special types of keys such as Woodruff key, Kennedy key or
feather key.
Selection of Keys
The selection of the type of key for a given application depends
upon the following factors.
1. Power to be transmitted
2. Tightness of fit
3. Stability of connection
4. Cost
Saddle Keys
• A saddle key is a key which fits in the keyway of the hub only.
• No keyway on the shaft.
The saddle keys are of the following two types
1. Hollow key
2. Flat key
⮚ A flat saddle key is a taper key which fits in a keyway in
the hub and is flat on the shaft.
⮚ It is likely to slip round the shaft under load.
⮚ Used for light loads
⮚ A hollow saddle key is a taper key which fits in a keyway in
the hub and the bottom of the key is shaped to fit the curved
surface of the shaft.
Hollow saddle keys hold on by friction.
⮚ Suitable for light loads.
⮚ It is usually used as a temporary fastening in fixing and
setting eccentrics, cams etc.
Sunk Keys
⮚ A sunk key is a key in which half the thickness of the key fits
into the keyway on the shaft and the remaining half in the
keyway on the hub.
⮚ Keyways are required both on the shaft as well as the hub of the
mating element. So, cost is more than saddle key.
⮚ This is a standard form of key and may be either of rectangular
or square cross-section.
Rectangular Sunk Keys
▣ Width of key, b=d /4;
▣ Thickness of key, h = 2 b / 3 = d / 6
▣ Length of key, l = 1.5 d
▣ The key has taper 1 in 100 on the top side only
Square Sunk Keys
▣The only difference between a rectangular sunk key and a
square sunk key is that its width and thickness are equal,
▣ b=h =d /4
▣ Length of key, l = 1.5 d
Design of Square and Flat Keys
Design of Square and Flat Keys
COUPLINGS
Introduction
Coupling can be defined as a mechanical device that
permanently joins two rotating shafts to each other.
The shafts to be connected by coupling may have collinear axes,
intersecting axes or parallel axes with a small distance in between.
Oldham coupling is used to connect two parallel Shafts.
Hooke’s coupling is used to connect two shafts having intersecting
axes.
When the axes are collinear or in the same line , rigid or flexible
couplings are used.
Requirements of Couplings
1. The coupling should be capable of transmitting torque from the
driving shaft to the driven shaft.
2. The coupling should keep the two shafts in proper alignment.
3. the coupling should be easy to assemble and disassemble for
the purpose of repairs and alterations.
4. The failure of revolving bolt heads, nuts, key heads and other
projecting parts may cause accidents.
They should be covered by giving suitable shape to the flanges or
by providing guards.
Parameter Rigid Coupling Flexible Coupling
Misalignment A rigid coupling cannot A flexible coupling due to provision
tolerate misalignment of flexible elements like bush or disk,
between the axes of can tolerate 0.5° of angular
shafts misalignment and 5 mm of axial
displacement between the shafts
Shock & It can be used only where
Vibration the motion is free from Absorb Shock & Vibration
Shock & Vibration
cost Simple & inexpensive Costlier due to additional parts
Rigid Flange Couplings
Rigid Flange Coupling consists of two flanges one keyed to the driving
shaft and the other to the driven shaft.
The two flanges are connected together by means of four or six bolts
arranged on a circle concentric with the axes of the shafts.
Advantages:-
1. High torque transmitting capacity
2. Easy to assemble and dismantle
3. Rigid coupling has simple construction
4. It is easy to design and manufacture.
Disadvantage:-
1. cannot tolerate misalignment between the axes of two shafts
2. can be used only where the motion is free from shocks and vibrations
3. It requires more radial space
Unprotected Type Flange Coupling
Protected Type Flange Coupling
Rigid Flange Coupling
Standard Proportions for
Various Dimensions of the Flange
i) Outside Diameter of Hub dh = 2d
ii) Length of hub lh = 1.5d
iii) Pitch circle diameter of bolts D = 3d
iv) Thickness of Flange t = 0.5d
v) Thickness of protecting rim t1 = 0.25d
vi) diameter of spigot and recess dr = 1.5d
vii) Outside diameter of Flange DO = (4d + 2t1)
Number of bolts
N = 3 For shafts up to 40 mm diameter
N = 4 For shafts from 40 to 100 mm diameter
N = 6 For shafts from 100 to 180 mm diameter
Case I Bolts Fitted in Reamed and Ground Holes
➢ The forces acting on individual bolts due to transmission of the
torque.
➢ Equating the external torque with the resisting torque,
• It should be noted that the bolts are subjected to direct shear
stress due to the force P and not torsional shear stress.
• No torque is acting about the axis of the bolt. The force P
results in only direct shear stress.
The direct shear stress in the bolt is given by,
Case II Bolts Fitted in Large Clearance Holes
➢ bolts are sufficiently tightened with a preload and the torque is
transmitted from one flange to the other by means of friction between
flanges.
➢ For uniformly distributed pressure, the friction radius Rf is given by,
Shear Failure of flange at the junction of the hub
Bushed-Pin Flexible Coupling
It is similar to the rigid type of flange coupling except for
the provision of rubber bush and pins in place of bolts.
