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KALLAM HARANADHAREDDY INSTITUTE OF TECHNOLOGY
Mechanical Engineering Department, Guntur, A.P
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DMM-I UNIT-IV
KEYS, COTTERS, KNUCKLE JOINTS AND SHAFTS
Introduction:- (What is the key? State its functions)
“A key is a device which is used for connecting two machine parts (Ex.
Shaft with pulley gear or crank) for preventing relative motion of rotation with
respect to each other.” In other words, key is used to transmit torque from a shaft
to a gear or pulley.
The primary function of key is to prevent the relative rotation between the
shaft and mating member. The secondary function is to prevent the relative
movements in the axial direction of the shaft. Thus, the connected parts act as a
single unit.
It is always inserted parallel to the axis of the shaft. Keys are used as
temporary fastenings and are subjected to considerable crushing and shearing
stresses.
Keys are generally made from cold rolled mild steel. Keys are designed
based on diameter of shaft.
Types of Keys ( Explain different types of keys with sketches)
The following types of keys are important:
1. Saddle keys, 2.Sunk keys, 3.Gib head tapered key, 4. Feather key, 5.
Woodruff key, 6. Round key or Pin, and 7.Splines.
1. Saddle keys:- The saddle keys are of the following two types:
a).Flat saddle key, and b).Hollow saddle key.
A flat saddle key : Bottom of the key is flat and the shaft is flattened to match as
shown in fig. It is likely to slip round the shaft under load. Therefore it is used for
comparatively light loads.
Saddle key.
A hollow saddle key : Bottom of the key is machined to have a curved surface.
2. Sunk keys:-The sunk keys are provided half in the keyway of the shaft and half
in the keyway of the hub of the pulley.
Rectangular sunk key. A rectangular sunk key is shown in Fig.
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Rectangular sunk key
The usual proportions of this key are :
Width of key, w = d / 4 ; and thickness of key, t = 2w / 3 = d / 6
Where d = Diameter of the shaft or diameter of the hole in the hub.
The key has taper 1 in 100 on the top side only.
3.Gib-head tapered key. It is a rectangular sunk key with a head at one end
known as gib head. It is usually provided to facilitate the removal of key. A gib
head key is shown in Fig(a) and its use in shown in Fig(b).
Gib-head key.
The usual proportions of the gib head key are :
Width, w = d / 4 ; and thickness at large end, t = 2w / 3 = d / 6
4.Feather key. A key attached to one member of a pair and which permits
relative axial movement is known as feather key.
Feather key.
The feather key may be screwed to the shaft as shown in Fig(a) or it may have
double gib heads as shown in Fig.(b). The various proportions of a feather key
are same as that of rectangular sunk key and gib head key.
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5.Woodruff key. The woodruff key is an easily adjustable key. It is a piece from a
cylindrical disc having segmental cross-section in front view as shown in Fig. A
woodruff key is capable of tilting in a recess milled out in the shaft by a cutter
having the same curvature as the disc from which the key is made. This key is
largely used in machine tool and automobile construction.
6. Round key or Pin:- The round keys, as shown in Fig (a), are circular in section
and fit into holes drilled partly in the shaft and partly in the hub. They have the
advantage that their keyways may be drilled and reamed after the mating parts
have been assembled. Round keys are usually low power drives.
Round keys.
Sometimes the tapered pin, as shown in Fig(b), is held in place by the friction
between the pin and the reamed tapered holes.
7.Splines:- Splines are multiple parallel keys integral with the shaft or the hub
Such shafts are known as splined shafts as shown in Fig. These shafts usually have
four, six, ten or sixteen splines. The splined shafts are relatively stronger than
shafts having a single keyway.
The splined shafts are used for automobile transmission and sliding gear
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transmissions. By using splined shafts, we obtain axial movement as well as
positive drive.
Forces acting on a Sunk Key
When a key is used in transmitting torque from a shaft to a rotor or hub, the
following two types of forces act on the key :
1. Forces (F1) due to fit of the key in its keyway, as in a tight fitting straight key or
in a tapered key driven in place. These forces produce compressive stresses in
the key which are difficult to determine in magnitude.
2. Forces (F) due to the torque transmitted by the shaft. These forces produce
shearing and compressive (or crushing) stresses in the key.
The distribution of the forces along the length of the key is not uniform
because the forces are concentrated near the torque-input end. The non-
uniformity of distribution is caused by the twisting of the shaft within the hub.
The forces acting on a key for a clockwise torque being transmitted from a
shaft to a hub are shown in Fig.
Forces acting on a Sunk Key
Strength of a Sunk Key
A key connecting the shaft and hub is shown in Fig.
Let T = Torque transmitted by the shaft,
F = Tangential force acting at the circumference of the shaft,
d = Diameter of shaft, l = Length of key, w = Width of key.
t = Thickness of key, and
= Shear and crushing stresses for the material of key.
Due to the power transmitted by the shaft, the key may fail due to shearing or
crushing.
Considering shearing of the key, the tangential shearing force acting at the
circumference of the shaft,
F = Area resisting shearing × Shear stress = l × w ×
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…………...(i)
Considering crushing of the key, the tangential crushing force acting at the
circumference of the shaft,
…………...(ii)
The key is equally strong in shearing and crushing, if
………..[Equating equations (i) and (ii)]
……………….(iii)
The permissible crushing stress for the usual key material is at least twice the
permissible shearing stress. Therefore from equation (iii), we have w = t. In other
words, a square key is equally strong in shearing and crushing.
In order to find the length of the key to transmit full power of the shaft, the
shearing strength of the key is equal to the torsional shear strength of the shaft.
We know that the shearing strength of key,
……………….(iv)
and torsional shear strength of the shaft,
……………....(v)
...(Taking 1 = Shear stress for the shaft material)
From equations (iv) and (v), we have
... (Taking w = d/4) …...(vi)
When the key material is same as that of the shaft, then = 1
l = 1.571 d …………….. [From equation (vi)]
Problem(1):- Design the rectangular key for a shaft of 50 mm diameter. The
shearing and crushing stresses for the key material are 42 MPa and 70 MPa. Assume
Width of key, w = 16 mm,and thickness of key, t = 10 mm
Given data : d = 50 mm ; = 42 MPa = 42 N/mm2 ; c= 70 MPa = 70 N/mm2 ,
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w = 16 mm, t = 10 mm
The rectangular key is designed as discussed below:
The length of key is obtained by considering the key in shearing and crushing.
Let l = Length of key.
Considering shearing of the key. We know that shearing strength (or torque
transmitted) of the key,
…………....(i)
and torsional shearing strength (or torque transmitted) of the shaft,
……………(ii)
From equations (i) and (ii), we have
l = 1.03 × 106 / 16 800 = 61.31 mm
Now considering crushing of the key. We know that crushing strength (or torque
transmitted) of the key,
…………....(iii)
From equations (ii) and (iii), we have
l = 1.03 × 106 / 8750 = 117.7 mm
Taking larger of the two values, we have length of key,
l = 117.7 say 120 mm Ans.
Problem(2):- A 45 mm diameter shaft is made of steel with a yield strength of 400
MPa. A parallel key of size 14 mm wide and 9 mm thick made of steel with a yield
strength of 340 MPa is to be used. Find the required length of key, if the shaft is
loaded to transmit the maximum permissible torque. Use maximum shear stress
theory and assume a factor of safety of 2.
Given data: d = 45 mm; ytfor shaft = 400 MPa = 400 N/mm2 ; w = 14 mm ;
t = 9 mm; ytfor key = 340 MPa = 340 N/mm2; F.S. = 2
Let l = Length of key.
According to maximum shear stress theory, the maximum shear stress for the
shaft,
and maximum shear stress for the key,
We know that the maximum torque transmitted by the shaft and key,
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First of all, let us consider the failure of key due to shearing. We know that the
maximum torque transmitted ( T ),
Now considering the failure of key due to crushing. We know that the maximum
torque transmitted by the shaft and key (T ),
l = 1.8 × 106 / 17 213 = 104.6 mm
Taking the larger of the two values, we have
l = 104.6 say 105 mm Ans.
Effect of Keyways:-
A little consideration will show that the keyway cut into the shaft reduces
the load carrying capacity of the shaft. This is due to the stress concentration near
the corners of the keyway and reduction in the cross-sectional area of the shaft.
It other words, the torsional strength of the shaft is reduced. The following
relation for the weakening effect of the keyway is based on the experimental
results by H.F. Moore.
Where e = Shaft strength factor. It is the ratio of the strength of the shaft with
Keyway to the strength of the same shaft without keyway,
w = Width of keyway, d = Diameter of shaft, and
It is usually assumed that the strength of the keyed shaft is 75% of the solid shaft,
which is somewhat higher than the value obtained by the above relation.
In case the keyway is too long and the key is of sliding type, then the angle of
twist is increased in the ratio k as given by the following relation:
Where k = Reduction factor for angular twist.
Problem(3):- A 15 kW, 960 r.p.m. motor has a mild steel shaft of 40 mm diameter
and the extension being 75 mm. The permissible shear and crushing stresses for the
mild steel key are 56 MPa and 112 MPa. Design the keyway in the motor shaft
extension. Check the shear strength of the key against the normal strength of the
shaft.
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Given data : P = 15 kW = 15 × 103 W ; N = 960 r.p.m. ; d = 40 mm ; l = 75 mm ;
= 56 MPa = 56 N/mm2 ; c = 112 MPa = 112 N/mm2
We know that the torque transmitted by the motor,
Let w = Width of keyway or key.
Considering the key in shearing. We know that the torque transmitted (T),
This width of keyway is too small. The width of keyway should be at least d / 4.
Since c = 2 , therefore a square key of w = 10 mm and t = 10 mm is adopted.
According to H.F. Moore, the shaft strength factor,
Strength of the shaft with keyway,
And shear strength of the key
COTTER JOINTS Introduction
A cotter is a flat wedge shaped piece of rectangular cross-section and its
width is tapered (either on one side or both sides) from one end to another for an
easy adjustment. The taper varies from 1 in 48 to 1 in 24.It is usually made of mild
steel or wrought iron.
A cotter joint is a temporary fastening and it is used to connect rigidly two
co-axial rods or bars which are subjected to axial tensile or compressive forces.
It is usually used in connecting a piston rod to the crosshead of a
reciprocating steam engine, a piston rod and its extension as a tail or pump rod,
strap end of connecting rod.
Difference between keys and cotters:- Keys are usually driven parallel to the
axis of the shafts which are subjected to torsional or twisting stresses.
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Cotters are normally driven at right angle to the axes of connected parts, which
are subjected to tensile or compressive stresses.
Why a single taper is provide in cotter and not an both sides?
The taper is provided to ensure the tightness of the joint and to facilitate easy
withdrawal of cotter from the joint. In cotter, the taper is provided only on one
side because machining a taper on two sides of a machine part is more difficult.
In addition, there is no specific advantage in providing taper on both sides.
Types of Cotter Joints
Following are the three commonly used cotter joints to connect two rods by a
cotter :
1. Socket and spigot cotter joint, 2. Sleeve and cotter joint, and 3. Gib and
cotter joint.
Socket and Spigot Cotter Joint
In a socket and spigot cotter joint, one end of the rods (say A) is provided
with a socket type of end as shown in Fig. and the other end of the other rod (say
B) is inserted into a socket. The end of the rod which goes into a socket is also
called spigot.
A rectangular hole is made in the socket and spigot. A cotter is then driven
tightly through a hole in order to make the temporary connection between the
two rods. The load is usually acting axially, but it changes its direction and hence
the cotter joint must be designed to carry both the tensile and compressive
loads. The compressive load is taken up by the collar on the spigot.
Design of Socket and Spigot Cotter Joint
The socket and spigot cotter joint is shown in Fig.
Let P = Load carried by the rods,
d = Diameter of the rods,
d1 = Outside diameter of socket,
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d2 = Diameter of spigot or inside diameter of socket,
d3 = Outside diameter of spigot collar,
t1 = Thickness of spigot collar,
d4 = Diameter of socket collar,
c = Thickness of socket collar,
b = Mean width of cotter,
t = Thickness of cotter,
l = Length of cotter,
a = Distance from the end of the slot to the end of rod,
t= Permissible tensile stress for the rods material,
= Permissible shear stress for the cotter material, and
c = Permissible crushing stress for the cotter material.
1. Failure of the rods in tension
The rods may fail in tension due to the tensile load P. We know that
Area resisting tearing
∴ Tearing strength of the rods,
Equating this to load (P), we have
From this equation, diameter of the rods ( d ) may be determined.
2. Failure of spigot in tension across the weakest section (or slot)
Since the weakest section of the spigot is that section which has a slot in it for the
cotter, as shown in Fig.
therefore Area resisting tearing of the spigot across the slot
and tearing strength of the spigot across the slot
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Equating this to load (P), we have
From this equation, the diameter of spigot or inside diameter of socket (d2) may
be determined.
Note : In actual practice, the thickness of cotter is usually taken as d2 / 4.
3. Failure of the rod or cotter in crushing
We know that the area that resists crushing of a rod or cotter = d2 × t
∴ Crushing strength = d2 × t × c
Equating this to load (P), we have
P =d2 × t × c
From this equation, the induced crushing stress may be checked.
4. Failure of the socket in tension across the slot
We know that the resisting area of the socket across the slot, as shown in Fig.
∴ Tearing strength of the socket across the slot
Equating this to load (P), we have
From this equation, outside diameter of socket (d1) may be determined.
5. Failure of cotter in shear
Considering the failure of cotter in shear as shown in Fig.
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Since the cotter is in double shear, therefore shearing area of the cotter
= 2 b × t
and shearing strength of the cotter
=2 b × t ×
Equating this to load (P), we have
P =2 b × t ×
From this equation, width of cotter (b) is determined.
6. Failure of the socket collar in crushing
Considering the failure of socket collar in crushing as shown in Fig.
We know that area that resists crushing of socket collar = (d4 – d2) t
and crushing strength =(d4 – d2) t × c
Equating this to load (P), we have
P =(d4 – d2) t × c
From this equation, the diameter of socket collar (d4) may be obtained.
7. Failure of socket end in shearing
Since the socket end is in double shear, therefore area that resists shearing of
socket collar = 2 (d4 – d2) c
and shearing strength of socket collar
=2 (d4 – d2) c ×
Equating this to load (P), we have
P =2 (d4 – d2) c ×
From this equation, the thickness of socket collar (c) may be obtained.
8. Failure of rod end in shear
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Since the rod end is in double shear, therefore the area resisting shear of the rod
end = 2 a × d2
and shear strength of the rod end = 2 a × d2 ×
Equating this to load (P), we have
P = 2 a × d2 ×
From this equation, the distance from the end of the slot to the end of the rod (a)
may be obtained.
9. Failure of spigot collar in crushing
Considering the failure of the spigot collar in crushing as shown in Fig.
We know that area that resists crushing of the collar
and crushing strength of the collar
Equating this to load (P), we have
From this equation, the diameter of the spigot collar (d3) may be obtained.
10. Failure of the spigot collar in shearing
Considering the failure of the spigot collar in shearing as shown in Fig.
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We know that area that resists shearing of the collar
= π d2 × t1
and shearing strength of the collar,
= π d2 × t1 ×
Equating this to load (P) we have
P = π d2 × t1 ×
From this equation, the thickness of spigot collar (t1) may be obtained.
Problem(1):- Design and draw a cotter joint to support a load varying from 30 kN in
compression to 30 kN in tension. The material used is carbon steel for which the
following allowable stresses may be used. The load is applied statically.
Tensile stress = compressive stress = 50 MPa ; shear stress = 35 MPa and crushing
stress = 90 MPa.
Given data : P = 30 kN = 30 × 103 N ; t= 50 MPa = 50 N / mm2 ; = 35 MPa = 35
N / mm2 ; c = 90 MPa = 90 N/mm2
The cotter joint is shown in Fig. The joint is designed as discussed below :
1. Diameter of the rods
Let d = Diameter of the rods.
Considering the failure of the rod in tension. We know that load (P),
∴ d 2 = 30 × 103 / 39.3 = 763 or d = 27.6 say 28 mm Ans.
2. Diameter of spigot and thickness of cotter
Let d2 = Diameter of spigot or inside diameter of socket, and
t = Thickness of cotter. It may be taken as d2 / 4.
Considering the failure of spigot in tension across the weakest section. We know
that load (P),
∴ (d2)2 = 30 × 103 / 26.8 = 1119.4 or d2 = 33.4 say 34 mm
Let us now check the induced crushing stress. We know that load (P),
Since this value of c is more than the given value of c
= 90 N/mm2, therefore
the dimensions d2 = 34 mm and t = 8.5 mm are not safe. Now let us find the values
of d2 and t by substituting the value of c = 90 N/mm2 in the above expression,
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∴ (d2)2 = 30 × 103 / 22.5 = 1333 or d2 = 36.5 say 40 mm Ans.
and t = d2 / 4 = 40 / 4 = 10 mm Ans.
3. Outside diameter of socket
Let d1 = Outside diameter of socket.
Consider the failure of the socket in tension across the slot. We know that load (P)
30 × 103/50 = 0.7854 (d1)2 – 1256.6 – 10 d1 + 400
or (d1)2 – 12.7 d1 – 1854.6 = 0
∴ d1 = 49.9 say 50 mm Ans. ................(Taking +ve sign)
4. Width of cotter
Let b = Width of cotter.
Considering the failure of the cotter in shear. Since the cotter is in double
shear, therefore load (P),
30 × 103 = 2 b × t × = 2 b × 10 × 35 = 700 b
∴ b = 30 × 103 / 700 = 43 mm Ans.
5. Diameter of socket collar
Let d4 = Diameter of socket collar.
Considering the failure of the socket collar and cotter in crushing. We know that
load (P),
30 × 103 = (d4 – d2) t × c = (d4 – 40)10 × 90 = (d4 – 40) 900
∴ d4 – 40 = 30 × 103 / 900 = 33.3 or d4 = 33.3 + 40 = 73.3 say 75 mm Ans.
6. Thickness of socket collar
Let c = Thickness of socket collar.
Considering the failure of the socket end in shearing. Since the socket end is in
double shear, therefore load (P),
30 × 103 = 2(d4 – d2) c × = 2 (75 – 40 ) c × 35 = 2450 c
∴ c = 30 × 103 / 2450 = 12 mm Ans.
7. Distance from the end of the slot to the end of the rod
Let a = Distance from the end of slot to the end of the rod.
Considering the failure of the rod end in shear. Since the rod end is in double
shear, therefore load (P),
30 × 103 = 2 a × d2 × = 2a × 40 × 35 = 2800 a
∴ a = 30 × 103 / 2800 = 10.7 say 11 mm Ans.
8. Diameter of spigot collar
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Let d3 = Diameter of spigot collar.
Considering the failure of spigot collar in crushing. We know that load (P),
∴ (d3)2 = 424 + (40)2 = 2024 or d3 = 45 mm Ans.
9. Thickness of spigot collar
Let t1 = Thickness of spigot collar.
Considering the failure of spigot collar in shearing. We know that load (P),
30 × 103 = d2 × t1 × = × 40 × t1 × 35 = 4400 t1
∴ t1 = 30 × 103 / 4400 = 6.8 say 8 mm Ans.
10. The length of cotter ( l ) is taken as 4 d.
∴ l = 4 d = 4 × 28 = 112 mm Ans.
11. The dimension e is taken as 1.2 d.
∴ e = 1.2 × 28 = 33.6 say 34 mm Ans.
Sleeve and Cotter Joint:(Write about working principle of sleeve and cotter
joint)
Sleeve and cotter joint is type of joint, which is used to connect two similar
coaxial cylindrical rods or two tie rods or two pipes or two tubes. These
cylindrical rods are connected together by a common sleeve and two wedge
shaped tapered cotters. Appropriate slots are cut in the sleeve and in the
cylindrical rods. The cotters are assembled into these slots.
Further,
In this type of joint, a sleeve or muff is used over the two rods and then two
cotters are driven in the holes.
The taper of cotter is usually 1 in 24.
The taper sides of the two cotter should face each other.
The various proportions for the sleeve and cotter joint in terms of the
diameter of rod (d ) are as follows :
Outside diameter of sleeve, d1 = 2.5 d
Diameter of enlarged end of rod, d2 = Inside diameter of sleeve = 1.25 d
Length of sleeve, L = 8 d
Thickness of cotter, t = d2/4 or 0.31 d
Width of cotter, b = 1.25 d
Length of cotter, l = 4 d
Distance of the rod end (a) from the beginning to the cotter hole (inside the
sleeve end)
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= Distance of the rod end (c) from its end to the cotter hole
= 1.25 d
Design of Sleeve and Cotter Joint
The sleeve and cotter joint is shown in above Fig.
Let P = Load carried by the rods,
d = Diameter of the rods,
d1 = Outside diameter of sleeve,
d2 = Diameter of the enlarged end of rod,
t = Thickness of cotter,
l = Length of cotter,
b = Width of cotter,
a = Distance of the rod end from the beginning to the cotter hole (inside the
sleeve end),
c = Distance of the rod end from its end to the cotter hole,
t , and c
= Permissible tensile, shear and crushing stresses
respectively for the material of the rods and cotter.
1. Failure of the rods in tension
The rods may fail in tension due to the tensile load P. We know that
Equating this to load (P), we have
From this equation, diameter of the rods (d) may be obtained.
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2. Failure of the rod in tension across the weakest section (i.e. slot)
Since the weakest section is that section of the rod which has a slot in it for the
cotter, therefore area resisting tearing of the rod across the slot
and tearing strength of the rod across the slot
Equating this to load (P), we have
From this equation, the diameter of enlarged end of the rod (d2) may be
obtained.
Note: The thickness of cotter is usually taken as d2 / 4.
3. Failure of the rod or cotter in crushing
We know that the area that resists crushing of a rod or cotter
= d2 × t
∴ Crushing strength = d2 × t × c
Equating this to load (P), we have
P = d2 × t × c
From this equation, the induced crushing stress may be checked.
4. Failure of sleeve in tension across the slot
We know that the resisting area of sleeve across the slot
Tearing strength of the sleeve across the slot
Equating this to load (P), we have
From this equation, the outside diameter of sleeve (d1) may be obtained.
5. Failure of cotter in shear
Since the cotter is in double shear, therefore shearing area of the cotter
= 2b × t
and shear strength of the cotter = 2b × t ×
Equating this to load (P), we have
P = 2b × t ×
From this equation, width of cotter (b) may be determined.
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6. Failure of rod end in shear
Since the rod end is in double shear, therefore area resisting shear of the rod end
= 2 a × d2
and shear strength of the rod end = 2 a × d2 × Equating this to load (P), we have P = 2 a × d2 ×
From this equation, distance (a) may be determined.
7. Failure of sleeve end in shear
Since the sleeve end is in double shear, therefore the area resisting shear of the
sleeve end = 2 (d1 – d2) c
and shear strength of the sleeve end = 2 (d1 – d2 ) c ×
Equating this to load (P), we have P = 2 (d1 – d2 ) c ×
From this equation, distance (c) may be determined.
Problem(2):-Design a sleeve and cotter joint to resist a tensile load of 60 kN. All
parts of the joint are made of the same material with the following allowable
stresses : t
= 60 MPa ; = 70 MPa ; and c = 125 MPa.
Given data : P = 60 kN = 60 × 103 N ; t = 60 MPa = 60 N/mm2 ;
= 70 MPa = 70 N/mm2 ; c = 125 MPa = 125 N/mm2
1. Diameter of the rods
Let d = Diameter of the rods.
Considering the failure of the rods in tension. We know that load (P),
∴ d 2 = 60 × 103 / 47.13 = 1273 or d = 35.7 say 36 mm Ans.
2. Diameter of enlarged end of rod and thickness of cotter
Let d2 = Diameter of enlarged end of rod, and
t = Thickness of cotter. It may be taken as d2 / 4.
Considering the failure of the rod in tension across the weakest section (i.e. slot).
We know that load (P),
∴ (d2)2 = 60 × 103 / 32.13 = 1867 or d2 = 43.2 say 44 mm Ans.
and thickness of cotter,
Let us now check the induced crushing stress in the rod or cotter. We know that
load (P), 60 × 103 = d2 × t × c = 44 × 11 × c
= 484 c
∴ c = 60 × 103 / 484 = 124 N/mm2
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Since the induced crushing stress is less than the given value of 125 N/mm2,
therefore the dimensions d2 and t are within safe limits.
3. Outside diameter of sleeve
Let d1 = Outside diameter of sleeve.
Considering the failure of sleeve in tension across the slot. We know that load (P)
or (d1)2 – 14 d1 – 2593 = 0
d1= 58.4 say 60 mm Ans. ..............(Taking +ve sign)
4. Width of cotter
Let b = Width of cotter.
Considering the failure of cotter in shear. Since the cotter is in double shear,
therefore load (P),
60 × 103 = 2 b × t × = 2 × b × 11 × 70 = 1540 b
∴ b = 60 × 103 / 1540 = 38.96 say 40 mm Ans.
5. Distance of the rod from the beginning to the cotter hole (inside the sleeve
end)
Let a = Required distance.
Considering the failure of the rod end in shear. Since the rod end is in double
shear, therefore load (P),
60 × 103 = 2 a × d2 × = 2 a × 44 × 70 = 6160 a
∴ a = 60 × 103 / 6160 = 9.74 say 10 mm Ans.
6. Distance of the rod end from its end to the cotter hole
Let c = Required distance.
Considering the failure of the sleeve end in shear. Since the sleeve end is in
double shear, therefore load (P),
60 × 103 = 2 (d1 – d2) c × = 2 (60 – 44) c × 70 = 2240 c
∴ c = 60 × 103 / 2240 = 26.78 say 28 mm Ans.
Gib and Cotter Joint: (Describe the purpose of gib in cotter joint. What are the
applications of cotter joints?)
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Gib and cotter joint is used for joining two square rods. One end of the rod
is forged in the shape of a fork while the other rod is pushed into the fork. Slots
are provided in the fork and the rod to accommodate the gib and cotter while
assembling the parts. The gib is inserted first so that the straight surface touches
the slot of the fork and then cotter is hammered into the rest of the slot. Care
should be taken to ensure that the tapered side of the gib and cotter should be in
contact face to face with each other.
Applications of cotter joints:-
Used to connect bicycle pedal to sprocket wheel
Used to connect piston rod in cross head
Used to connect piston rod with its extension
Used to connect two halves of a flywheel
Used for joining tail rod with piston rod of a wet air pump.
Design of Gib and Cotter Joint for Square Rods:- (What are the
design procedure for gib and cotter joint for square rods)
Consider a gib and cotter joint for square rods as shown in Fig. The rods may be
subjected to a tensile or compressive load. All components of the joint are
assumed to be of the same material.
Let P = Load carried by the rods,
x = Each side of the rod,
B = Total width of gib and cotter,
B1 = Width of the strap,
t = Thickness of cotter,
t1 = Thickness of the strap, and
t , and c
= Permissible tensile, shear and crushing stresses.
.
Gib and cotter joint for square rods.
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1. Failure of the rod in tension
The rod may fail in tension due to the tensile load P. We know that
Area resisting tearing = x × x = x2
∴ Tearing strength of the rod = x2 × t
Equating this to the load (P), we have P = x2 × t
From this equation, the side of the square rod (x) may be determined. The other
dimensions are fixed as under :
Width of strap, B1 = Side of the square rod = x
Thickness of cotter,
Thickness of gib = Thickness of cotter (t)
Height (t2) and length of gib head (l4) = Thickness of cotter (t)
2. Failure of the gib and cotter in shearing
Since the gib and cotter are in double shear, therefore,
Area resisting failure = 2 B × t
and resisting strength = 2 B × t ×
Equating this to the load (P), we have P = 2B × t ×
From this equation, the width of gib and cotter (B) may be obtained. In the joint,
as shown in Fig. one gib is used, the proportions of which are
Width of gib, b1 = 0.55 B ; and width of cotter, b = 0.45 B
In case two gibs are used, then
Width of each gib = 0.3 B ; and width of cotter = 0.4 B
3. Failure of the strap end in tension at the location of gib and cotter
Area resisting failure = 2 [B1 × t1 – t1 × t] = 2 [x × t1 – t1 × t] ... ( B1 = x)
∴ Resisting strength = 2 [ x × t1 – t1 × t] t
Equating this to the load (P), we have
P = 2 [x × t1 – t1 × t] t
From this equation, the thickness of strap (t1) may be determined.
4. Failure of the strap or gib in crushing
The strap or gib (at the strap hole) may fail due to crushing.
Area resisting failure = 2 t1 × t
∴ Resisting strength = 2 t1 × t × c Equating this to the load (P), we have
P = 2 t1 × t × c From this equation, the induced crushing stress may be checked.
5. Failure of the rod end in shearing
Since the rod is in double shear, therefore
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Area resisting failure = 2 l1 × x
∴ Resisting strength = 2 l1 × x ×
Equating this to the load (P), we have
P = 2 l1 × x ×
From this equation, the dimension l1 may be determined.
6. Failure of the strap end in shearing
Since the length of rod (l2) is in double shearing, therefore
Area resisting failure = 2 × 2 l2 × t1
∴ Resisting strength = 2 × 2 l2 × t1 ×
Equating this to the load (P), we have
P = 2 × 2 l2 × t1 ×
From this equation, the length of rod (l2) may be determined. The length l3 of the
strap end is proportioned as 2/3 rd of side of the rod. The clearance is usually
kept 3 mm. The length of cotter is generally taken as 4 times the side of the rod.
Problem(3):- Design a gib and cotter joint as shown in Fig. to carry a maximum
load of 35 kN. Assuming that the gib, cotter and rod are of same material and have
the following allowable stresses :
t = 20 MPa ; = 15 MPa ; and c
= 50 MPa
Given data : P = 35 kN = 35 000 N ; t = 20 MPa = 20 N/mm2 ; = 15 MPa =
15 N/mm2 ; c = 50 MPa = 50 N/mm2
1. Side of the square rod
Let x = Each side of the square rod.
Considering the failure of the rod in tension. We know that load (P),
35 000 = x2 × t = x2 × 20 = 20 x2
∴ x2 = 35 000 / 20 = 1750 or x = 41.8 say 42 mm Ans.
Other dimensions are fixed as follows :
Width of strap, B1 = x = 42 mm Ans.
Thickness of cotter,
Thickness of gib = Thickness of cotter = 12 mm Ans.
Height (t2) and length of gib head (l4) = Thickness of cotter = 12 mm Ans.
2. Width of gib and cotter
Let B = Width of gib and cotter.
Consider the failure of the gib and cotter in double shear. We know that load (P),
35 000 = 2 B × t × = 2 B × 12 × 15 = 360 B
∴ B = 35 000 / 360 = 97.2 say 100 mm Ans.
Since one gib is used, therefore
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Width of gib, b1 = 0.55 B = 0.55 × 100 = 55 mm Ans.
and width of cotter, b = 0.45 B = 0.45 × 100 = 45 mm Ans.
3. Thickness of strap
Let t1 = Thickness of strap.
Considering the failure of the strap end in tension at the location of the gib and
cotter. We know that load (P),
35 000 = 2 (x × t1 – t1 × t) t = 2 (42 × t1 – t1 × 12) 20 = 1200 t1
∴ t1 = 35 000 / 1200 = 29.1 say 30 mm Ans.
Now the induced crushing stress may be checked by considering the failure of
the strap or gib in crushing. We know that load (P),
35 000 = 2 t1 × t × c = 2 × 30 × 12 × c
= 720 c
∴ c= 35 000 / 720 = 48.6 N/mm2
Since the induced crushing stress is less than the given crushing stress,
therefore the joint is safe.
4. Length (l1) of the rod
Considering the failure of the rod end in shearing. Since the rod is in double
shear, therefore load (P),
35 000 = 2 l1 × x × = 2 l1 × 42 × 15 = 1260 l1
∴ l1 = 35 000 / 1260 = 27.7 say 28 mm Ans.
5. Length (l2 ) of the rod
Considering the failure of the strap end in shearing. Since the length of the rod
(l2) is in double shear, therefore load (P),
35 000 = 2 × 2 l2 × t1 × = 2 × 2 l2 × 30 × 15 = 1800 l2
∴ l2 = 35 000 / 1800 = 19.4 say 20 mm Ans.
Length (l3) of the strap end
and length of cotter = 4 x = 4 × 42 = 168 mm Ans.
Knuckle Joint:- (what is the Knuckle joint?) A knuckle joint is used to connect two rods which are under the action of
tensile loads. However, if the joint is guided, the rods may support a compressive
load. A knuckle joint may be readily disconnected for adjustments or repairs.
Its use may be found in the link of a cycle chain, tie rod joint for roof truss,
valve rod joint with eccentric rod, pump rod joint, tension link in bridge structure
and lever and rod connections of various types.
In knuckle joint is shown in Fig, one end of one of the rods is made into an
eye and the end of the other rod is formed into a fork with an eye in each of the
fork leg. The knuckle pin passes through both the eye hole and the
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fork holes and may be secured by means of a collar and taper pin or spilt pin.
The knuckle pin may be prevented from rotating in the fork by means of a small
stop, pin, peg. The material used for the joint may be steel or wrought iron.
Dimensions of Various Parts of the Knuckle Joint:-
Knuckle joint
If d is the diameter of rod, then diameter of pin,
d1 = d
Outer diameter of eye, d2 = 2 d
Diameter of knuckle pin head and collar, d3 = 1.5 d
Thickness of single eye or rod end, t = 1.25 d
Thickness of fork, t1 = 0.75 d
Thickness of pin head, t2 = 0.5 d
Other dimensions of the joint are shown in Fig.
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Methods of Failure of Knuckle Joint:-
Consider a knuckle joint as shown in Fig.
Let P = Tensile load acting on the rod,
d = Diameter of the rod,
d1 = Diameter of the pin,
d2 = Outer diameter of eye,
t = Thickness of single eye,
t1 = Thickness of fork.
t , and c
= Permissible stresses for the joint material in tension, shear and
crushing respectively.
In determining the strength of the joint for the various methods of failure, it is
assumed that
1. There is no stress concentration, and
2. The load is uniformly distributed over each part of the joint.
1. Failure of the solid rod in tension
Since the rods are subjected to direct tensile load, therefore tensile strength of
the rod,
Equating this to the load (P) acting on the rod, we have
From this equation, diameter of the rod ( d ) is obtained.
2. Failure of the knuckle pin in shear
Since the pin is in double shear, therefore cross-sectional area of the pin under
shearing
and the shear strength of the pin
Equating this to the load (P) acting on the rod, we have
From this equation, diameter of the knuckle pin (d1) is obtained.
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Maximum bending (tensile) stress,
From this expression, the value of d1 may be obtained.
3. Failure of the single eye or rod end in tension
The single eye or rod end may tear off due to the tensile load. We know that area
resisting tearing = (d2 – d1) t
∴ Tearing strength of single eye or rod end = (d2 – d1) t × t
Equating this to the load (P) we have P = (d2 – d1) t × t
From this equation, the induced tensile stress ( t) for the single eye or rod end
may be checked. In case the induced tensile stress is more than the allowable
working stress, then increase the outer diameter of the eye (d2).
4. Failure of the single eye or rod end in shearing
The single eye or rod end may fail in shearing due to tensile load. We know that
area resisting shearing = (d2 – d1) t
∴ Shearing strength of single eye or rod end = (d2 – d1) t ×
Equating this to the load (P), we have P = (d2 – d1) t ×
From this equation, the induced shear stress ( ) for the single eye or rod
end may be checked.
5. Failure of the single eye or rod end in crushing
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The single eye or pin may fail in crushing due to the tensile load. We know that
area resisting crushing = d1 × t
∴ Crushing strength of single eye or rod end = d1 × t × c
Equating this to the load (P), we have P = d1 × t × c
From this equation, the induced crushing stress ( c) for the single eye or pin
may be checked. In case the induced crushing stress in more than the allowable
working stress, then increase the thickness of the single eye (t).
6. Failure of the forked end in tension
The forked end or double eye may fail in tension due to the tensile load. We
know that area resisting tearing = (d2 – d1) × 2 t1
∴ Tearing strength of the forked end = (d2 – d1) × 2 t1 × t
Equating this to the load (P), we have P = (d2 – d1) × 2t1 × t
From this equation, the induced tensile stress for the forked end may be
checked.
7. Failure of the forked end in shear
The forked end may fail in shearing due to the tensile load. We know that area
resisting shearing = (d2 – d1) × 2t1
∴ Shearing strength of the forked end = (d2 – d1) × 2t1 × Equating this to the load (P), we have P = (d2 – d1) × 2t1 ×
From this equation, the induced shear stress for the forked end may be checked.
In case, the induced shear stress is more than the allowable working stress, then
thickness of the fork (t1) is increased.
8. Failure of the forked end in crushing
The forked end or pin may fail in crushing due to the tensile load. We know that
area resisting crushing = d1 × 2 t1
∴ Crushing strength of the forked end = d1 × 2 t1 × c
Equating this to the load (P), we have P = d1 × 2 t1 × c
From this equation, the induced crushing stress for the forked end may be
checked.
Problem(4):- Design a knuckle joint to transmit 150 kN. The design stresses may be
taken as 75 MPa in tension, 60 MPa in shear and 150 MPa in crushing.
Given data : P = 150 kN = 150 × 103 N ; t = 75 MPa = 75 N/mm2 ; = 60 MPa =
60 N/mm2 ; c = 150 MPa = 150 N/mm2
The knuckle joint is shown in Fig. The joint is designed by considering the
various methods of failure as discussed below :
1. Failure of the solid rod in tension
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Let d = Diameter of the rod.
We know that the load transmitted (P),
∴ d2 = 150 × 103 / 59 = 2540 or d = 50.4 say 52 mm Ans.
Now the various dimensions are fixed as follows :
Diameter of knuckle pin, d1 = d = 52 mm
Outer diameter of eye, d2 = 2 d = 2 × 52 = 104 mm
Diameter of knuckle pin head and collar, d3 = 1.5 d = 1.5 × 52 = 78 mm
Thickness of single eye or rod end, t = 1.25 d = 1.25 × 52 = 65 mm
Thickness of fork, t1 = 0.75 d = 0.75 × 52 = 39 say 40 mm
Thickness of pin head, t2 = 0.5 d = 0.5 × 52 = 26 mm
2. Failure of the knuckle pin in shear
Since the knuckle pin is in double shear, therefore load (P),
= 150 × 103 / 4248 = 35.3 N/mm2 = 35.3 MPa
3. Failure of the single eye or rod end in tension
The single eye or rod end may fail in tension due to the load. We know that load
(P), 150 × 103 = (d2 – d1) t × t = (104 – 52) 65 × t
= 3380 t
∴ t = 150 × 103 / 3380 = 44.4 N / mm2 = 44.4 MPa
4. Failure of the single eye or rod end in shearing
The single eye or rod end may fail in shearing due to the load. We know that load
(P), 150 × 103 = (d2 – d1) t × = (104 – 52) 65 × = 3380
= 150 × 103 / 3380 = 44.4 N/mm2 = 44.4 MPa
5. Failure of the single eye or rod end in crushing
The single eye or rod end may fail in crushing due to the load. We know that load
(P), 150 × 103 = d1 × t × c = 52 × 65 × c
= 3380 c
∴ σ c = 150 × 103 / 3380 = 44.4 N/mm2 = 44.4 MPa
6. Failure of the forked end in tension
The forked end may fail in tension due to the load. We know that load (P),
150 × 103 = (d2 – d1) 2 t1 × t = (104 – 52) 2 × 40 × t
= 4160 t
∴ t
= 150 × 103 / 4160 = 36 N/mm2 = 36 MPa
7. Failure of the forked end in shear
The forked end may fail in shearing due to the load. We know that load (P),
150 × 103 = (d2 – d1) 2 t1 × = (104 – 52) 2 × 40 × = 4160
= 150 × 103 / 4160 = 36 N/mm2 = 36 MPa
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8. Failure of the forked end in crushing
The forked end may fail in crushing due to the load. We know that load (P),
150 × 103 = d1 × 2 t1 × c = 52 × 2 × 40 × c
= 4160 c
∴ c = 150 × 103 / 4180 = 36 N/mm2 = 36 MPa
From above, we see that the induced stresses are less than the given design
stresses, therefore the joint is safe.
Problem(6):- Design a knuckle joint for a tie rod of a circular section to sustain a
maximum pull of 70 kN. The ultimate strength of the material of the rod against
tearing is 420 MPa. The ultimate tensile and shearing strength of the pin material
are 510 MPa and 396 MPa respectively. Determine the tie rod section and pin
section. Take factor of safety = 6.
Given data : P = 70 kN = 70 000 N ; tu for rod = 420 MPa ; tu
for pin = 510 MPa ;
u = 396 MPa ; F.S. = 6
We know that the permissible tensile stress for the rod material,
and permissible shear stress for the pin material,
We shall now consider the various methods of failure of the joint as discussed
below:
1. Failure of the rod in tension
Let d = Diameter of the rod.
We know that the load (P),
∴ d 2 = 70 000 / 55 = 1273 or d = 35.7 say 36 mm Ans.
The other dimensions of the joint are fixed as given below :
Diameter of the knuckle pin, d1 = d = 36 mm
Outer diameter of the eye, d2 = 2 d = 2 × 36 = 72 mm
Diameter of knuckle pin head and collar, d3 = 1.5 d = 1.5 × 36 = 54 mm
Thickness of single eye or rod end, t = 1.25 d = 1.25 × 36 = 45 mm
Thickness of fork, t1 = 0.75 d = 0.75 × 36 = 27 mm
Now we shall check for the induced stresses as discussed below :
2. Failure of the knuckle pin in shear
Since the knuckle pin is in double shear, therefore load (P),
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3. Failure of the single eye or rod end in tension
The single eye or rod end may fail in tension due to load. We know that load (P),
70 000 = (d2 – d1) t × t = (72 – 36) 45 t
= 1620 t
∴ t
= 70 000 / 1620 = 43.2 N/mm2
4. Failure of the forked end in tension
The forked end may fail in tension due to the load. We know that load (P),
70 000 = (d2 – d1) 2 t1 × t= (72 – 36) × 2 × 27 × t
= 1944 t
∴ t = 70 000 / 1944 = 36 N/mm2
From above we see that the induced stresses are less than given permissible
stresses, therefore the joint is safe.
Prepared By Srinivasulureddy.Dorasila M.Tech; M.I.S.T.E; (Ph.D) ; Associate professor; Mechanical engg.
Dept; K.H.I.T; Guntur.
SHAFTS
Introduction:-A shaft is a rotating machine element which transmit power from
one place to another. Shaft are subjected to tensile, bending or torsional stresses
or to a combination of these stresses.
A transmitting shaft is circular is cross section, which supports transmission
elements like pulleys, gears and sprockets.
The design of transmission shaft consists of determining the corret shaft diameter
based on 1). Strength, 2). Rigidity and Stiffness.
Axle: It is a non rotating shaft, which supports the rotating components of the
machine. It does not transmit a useful torque. Axle is subjected to only bending.
Spindle: It is a short rotating shaft in case of drilling machine, lathe spindles.
Line shaft or transmission shaft: It is a comparatively long shaft which is driven
by a motor. The line shaft transmits motion to various machines through counter
shafts. The counter shaft is an intermediate shaft placed between the line shaft
and various driven machines.
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Stub axle: It is short axle capable of small angular motion about the pivots. Front
wheels of rear wheel drive vehicles are supported on stub axles.
Material Used for Shafts
The material used for shafts should have the following properties :
1. It should have high strength.
2. It should have good machinability.
3. It should have low notch sensitivity factor.
4. It should have good heat treatment properties.
5. It should have high wear resistant properties.
The material used for ordinary shafts is carbon steel of grades 40 C 8, 45 C 8, 50
C 4 and 50 C 12.
When a shaft of high strength is required, then an alloy steel such as nickel,
nickel-chromium or chrome-vanadium steel is used.
Types of Shafts
The following two types of shafts are important from the subject point of view :
1. Transmission shafts. These shafts transmit power between the source and the
machines absorbing power. The counter shafts, line shafts, over head shafts and
all factory shafts are transmission shafts. Since these shafts carry machine parts
such as pulleys, gears etc., therefore they are subjected to bending in addition to
twisting.
2. Machine shafts. These shafts form an integral part of the machine itself. The
crank shaft is an example of machine shaft.
Standard Sizes of Transmission Shafts
The standard sizes of transmission shafts are : 25 mm to 60 mm with 5 mm steps;
60 mm to 110 mm with 10 mm steps ; 110 mm to 140 mm with 15 mm steps ; and
140 mm to 500 mm with 20 mm steps. The standard length of the shafts are 5 m, 6
m and 7 m.
Stresses in Shafts
The following stresses are induced in the shafts :
1. Shear stresses due to the transmission of torque (i.e. due to torsional load).
2. Bending stresses (tensile or compressive) due to the forces acting upon
machine elements like gears, pulleys etc. as well as due to the weight of the shaft
itself.
3. Stresses due to combined torsional and bending loads.
Design of Shafts
The shafts may be designed on the basis of
1. Strength, and 2. Rigidity and stiffness.
In designing shafts on the basis of strength, the following cases may be
considered :
(a) Shafts subjected to twisting moment or torque only,
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(b) Shafts subjected to bending moment only,
(c) Shafts subjected to combined twisting and bending moments, and
(d) Shafts subjected to axial loads in addition to combined torsional and bending
loads.
Shafts Subjected to Twisting Moment Only
When the shaft is subjected to a twisting moment (or torque) only, then the
diameter of the shaft may be obtained by using the torsion equation. We know
that
.....................(i)
where T = Twisting moment (or torque) acting upon the shaft,
J = Polar moment of inertia of the shaft about the axis of rotation,
= Torsional shear stress, and
r = Distance from neutral axis to the outer most fibre
= d / 2; where d is the diameter of the shaft.
We know that for round solid shaft, polar moment of inertia,
The equation (i) may now be written as
................(ii)
From this equation, we may determine the diameter of round solid shaft ( d ).
We also know that for hollow shaft, polar moment of inertia,
where do and di = Outside and inside diameter of the shaft, and r = do / 2.
Substituting these values in equation (i), we have
.........(iii)
Let k = Ratio of inside diameter and outside diameter of the shaft = di / do
Now the equation (iii) may be written as
...........(iv)
From the equations (iii) or (iv), the outside and inside diameter of a hollow shaft
may be determined. It may be noted that
1. The hollow shafts are usually used in marine work. These shafts are stronger
per kg of material and they may be forged on a mandrel, thus
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making the material more homogeneous than would be possible for a solid shaft.
When a hollow shaft is to be made equal in strength to a solid shaft, the twisting
moment of both the shafts must be same. In other words, for the same material of
both the shafts,
2. The twisting moment (T) may be obtained by using the following relation :
We know that the power transmitted (in watts) by the shaft,
where T = Twisting moment in N-m, and
N = Speed of the shaft in r.p.m.
3. In case of belt drives, the twisting moment ( T ) is given by
T = (T1 – T2 ) R
where T1 and T2 = Tensions in the tight side and slack side of the belt
respectively, and
R = Radius of the pulley.
Problems(1):- A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The shaft may
be assumed to be made of mild steel with an allowable shear stress of 42 MPa.
Determine the diameter of the shaft, neglecting the bending moment on the shaft.
Given data : N = 200 r.p.m. ; P = 20 kW = 20 × 103 W; = 42 MPa = 42 N/mm2
Let d = Diameter of the shaft.
We know that torque transmitted by the shaft,
We also know that torque transmitted by the shaft ( T ),
∴ d3 = 955 × 103 / 8.25 = 115 733 or d = 48.7 say 50 mm
Problem(2):- A solid shaft is transmitting 1 MW at 240 r.p.m. Determine the
diameter of the shaft if the maximum torque transmitted exceeds the mean torque
by 20%. Take the maximum allowable shear stress as 60 MPa.
Given data : P = 1 MW = 1 × 106 W ; N = 240 r.p.m. ; Tmax = 1.2 Tmean ; = 60 MPa
= 60 N/mm2
Let d = Diameter of the shaft.
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We know that mean torque transmitted by the shaft,
∴ Maximum torque transmitted,
Tmax = 1.2 Tmean = 1.2 × 39 784 × 103 = 47 741 × 103 N-mm
We know that maximum torque transmitted (Tmax),
∴ d3 = 47 741 × 103 / 11.78 = 4053 × 103
or d = 159.4 say 160 mm
Problem(3):- Find the diameter of a solid steel shaft to transmit 20 kW at 200 r.p.m.
The ultimate shear stress for the steel may be taken as 360 MPa and a factor of safety
as 8.If a hollow shaft is to be used in place of the solid shaft, find the inside and
outside diameter when the ratio of inside to outside diameters is 0.5.
Given data :P = 20 kW = 20 × 103 W ; N = 200 r.p.m. ; u = 360 MPa = 360 N/mm2
F.S. = 8 ; k = di / do = 0.5
We know that the allowable shear stress,
Diameter of the solid shaft
Let d = Diameter of the solid shaft.
We know that torque transmitted by the shaft,
We also know that torque transmitted by the solid shaft (T),
∴ d3 = 955 × 103 / 8.84 = 108 032 or d = 47.6 say 50 mm
Diameter of hollow shaft
Let di = Inside diameter, and
do = Outside diameter.
We know that the torque transmitted by the hollow shaft ( T ),
∴ (do)3 = 955 × 103 / 8.3 = 115 060 or do = 48.6 say 50 mm
and di = 0.5do = 0.5 × 50 = 25 mm
Shafts Subjected to Bending Moment Only
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When the shaft is subjected to a bending moment only, then the maximum stress
(tensile or compressive) is given by the bending equation. We know that
...................(i)
where M = Bending moment,
I = Moment of inertia of cross-sectional area of the shaft about the axis of
rotation,
b= Bending stress, and
y = Distance from neutral axis to the outer-most fibre.
We know that for a round solid shaft, moment of inertia,
Substituting these values in equation (i), we have
From this equation, diameter of the solid shaft (d) may be obtained.
We also know that for a hollow shaft, moment of inertia,
........(where k = di / do )
and y = do/ 2
Again substituting these values in equation (i), we have
From this equation, the outside diameter of the shaft (do) may be obtained.
Problem(4):- A pair of wheels of a railway wagon carries a load of 50 kN on each
axle box,acting at a distance of 100 mm outside the wheel base. The gauge of the
rails is 1.4 m. Find the diameter of the axle between the wheels, if the stress is not to
exceed 100 MPa.
Given data : W = 50 kN = 50 × 103 N ; L = 100 mm ; x = 1.4 m ; b = 100 MPa =
100 N/mm2
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The axle with wheels is shown in Fig.
A little consideration will show that the maximum bending moment acts on the
wheels at C and D. Therefore maximum bending moment,
M = W.L = 50 × 103 × 100 = 5 × 106 N-mm
Let d = Diameter of the axle.
We know that the maximum bending moment (M),
∴ d 3 = 5 × 106 / 9.82 = 0.51 × 106 or d = 79.8 say 80 mm
Shafts Subjected to Combined Twisting Moment and Bending Moment
When the shaft is subjected to combined twisting moment and bending
moment, then the shaft must be designed on the basis of the two moments
simultaneously. Various theories have been suggested to account for the elastic
failure of the materials when they are subjected to various types of combined
stresses. The following two theories are important from the subject point of view :
1. Maximum shear stress theory or Guest's theory. It is used for ductile materials
such as mild
steel.
2. Maximum normal stress theory or Rankine’s theory. It is used for brittle
materials such as cast iron.
Let = Shear stress induced due to twisting moment, and
b = Bending stress (tensile or compressive) induced due to bending
moment.
According to maximum shear stress theory, the maximum shear stress in the
shaft,
Substituting the values of and b
The maximum B.M. may be obtained as follows :
RC = RD = 50 kN = 50 × 103 N
B.M. at A, MA = 0
B.M. at C, MC = 50 × 103 × 100 = 5 × 106 N-mm
B.M. at D, MD = 50 × 103 × 1500 – 50 × 103 × 1400 = 5 × 106 N-mm
B.M. at B, MB = 0
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...............(i)
The expression is known as equivalent twisting moment and is
denoted by Te. The equivalent twisting moment may be defined as that twisting
moment, which when acting alone, produces the same shear stress ( ) as the
actual twisting moment. By limiting the maximum shear stress ( max) equal to the
allowable shear stress ( ) for the material, the equation (i) may be written as
...............(ii)
From this expression, diameter of the shaft ( d ) may be evaluated.
Now according to maximum normal stress theory, the maximum normal stress in
the shaft,
...............(iii)
....................(iv)
The expression is known as equivalent bending moment and
is denoted by Me. The equivalent bending moment may be defined as that
moment which when acting alone produces the same tensile or compressive
stress ( b) as the actual bending moment. By limiting the maximum normal
stress [ (max)b] equal to the allowable bending stress ( b
), then the equation (iv)
may be written as
....................(v)
From this expression, diameter of the shaft ( d ) may be evaluated.
Notes: 1. In case of a hollow shaft, the equations (ii) and (v) may be written as
2. It is suggested that diameter of the shaft may be obtained by using both the
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theories and the larger of the two values is adopted.
Problem(5):- A solid circular shaft is subjected to a bending moment of 3000 N-m
and a torque of 10 000 N-m. The shaft is made of 45 C 8 steel having ultimate tensile
stress of 700 MPa and a ultimate shear stress of 500 MPa. Assuming a factor of safety
as 6, determine the diameter of the shaft.
Given data : M = 3000 N-m = 3 × 106 N-mm ; T = 10 000 N-m = 10 × 106 N-mm ;
tu= 700 MPa = 700 N/mm2 ; u
= 500 MPa = 500 N/mm2
We know that the allowable tensile stress,
and allowable shear stress,
Let d = Diameter of the shaft in mm.
According to maximum shear stress theory, equivalent twisting moment,
We also know that equivalent twisting moment (Te),
∴ d3 = 10.44 × 106 / 16.36 = 0.636 × 106 or d = 86 mm
According to maximum normal stress theory, equivalent bending moment,
We also know that the equivalent bending moment (Me),
∴ d3 = 6.72 × 106 / 11.46 = 0.586 × 106 or d = 83.7 mm
Taking the larger of the two values, we have
d = 86 say 90 mm
Problem(6):- A shaft made of mild steel is required to transmit 100 kW at 300 r.p.m.
The supported length of the shaft is 3 metres. It carries two pulleys each weighing
1500 N supported at a distance of 1 metre from the ends respectively. Assuming the
safe value of stress, determine the diameter of the shaft.
Given data: P = 100 kW = 100 × 103 W ; N = 300 r.p.m. ; L = 3 m ; W = 1500 N
We know that the torque transmitted by the shaft,
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The shaft carrying the two pulleys is like a simply supported beam as shown in
Fig.
The reaction at each support will be 1500 N,
i.e. RA = RB = 1500 N
A little consideration will show that the maximum bending moment lies at each
pulley i.e. at C and D.
∴ Maximum bending moment, M = 1500 × 1 = 1500 N-m
Let d = Diameter of the shaft in mm.
We know that equivalent twisting moment,
= 3519 × 103 N-mm
We also know that equivalent twisting moment (Te),
.......(Assuming τ = 60 N/mm2)
∴ d 3 = 3519 × 103 / 11.8 = 298 × 103 or d = 66.8 say 70 mm
Problem(7):- 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 belt having 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 electric motor and belt, which is
placed horizontally to the right. The angle of contact for both the pulleys is 1800 and
μ = 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 shaft. Assume that the
torque on one pulley is equal to that on the other pulley.
Given data : AB = 1 m ; DC = 600 mm or RC = 300 mm = 0.3 m ; AC = 300 mm =
0.3 m ; T1 = 2.25 kN = 2250 N ; DD = 400 mm or RD = 200 mm = 0.2 m ; BD = 200
mm = 0.2 m ; θ = 180° = π rad ; μ = 0.24 ; σb = 63 MPa = 63 N/mm2 ; τ = 42 MPa =
42 N/mm2
The space diagram of the shaft is shown in Fig (a).
Let T1 = Tension in the tight side of the belt on pulley C = 2250 N ......(Given)
T2 = Tension in the slack side of the belt on pulley C.
We know that
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eTT
2
1
eTT )(24.0
2
1 =
127.22
1 TT
∴ Vertical load acting on the shaft at C,
WC = T1 + T2 = 2250 + 1058 = 3308 N
and vertical load on the shaft at D = 0
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The vertical load diagram is shown in Fig (c).
We know that torque acting on the pulley C,
T = (T1 – T2) RC = (2250 – 1058) 0.3 = 357.6 N-m
The torque diagram is shown in Fig (b).
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Let T3 = Tension in the tight side of the belt on pulley D, and
T4 = Tension in the slack side of the belt on pulley D.
Since the torque on both the pulleys (i.e. C and D) is same, therefore
(T3 – T4) RD = T = 357.6 N-m
................(i)
We know that
................(ii)
From equations (i) and (ii), we find that
T3 = 3376 N, and T4 = 1588 N
∴ Horizontal load acting on the shaft at D,
WD = T3 + T4 = 3376 + 1588 = 4964 N
and horizontal load on the shaft at C = 0
The horizontal load diagram is shown in Fig(d).
Now let us find the maximum bending moment for vertical and horizontal
loading.
First of all, considering the vertical loading at C. Let RAV and RBV be the reactions
at the bearings A and B respectively. We know that
RAV + RBV = 3308 N
Taking moments about A,
RBV × 1 = 3308 × 0.3 or RBV = 992.4 N
and RAV = 3308 – 992.4 = 2315.6 N
We know that B.M. at A and B,
MAV = MBV = 0
B.M. at C, MCV = RAV × 0.3 = 2315.6 × 0.3 = 694.7 N-m
B.M. at D, MDV = RBV × 0.2 = 992.4 × 0.2 = 198.5 N-m
The bending moment diagram for vertical loading in shown in Fig (e).
Now considering horizontal loading at D. Let RAH and RBH be the reactions at the
bearings A and B respectively. We know that
RAH + RBH = 4964 N
Taking moments about A,
RBH × 1 = 4964 × 0.8 or RBH = 3971 N
and RAH = 4964 – 3971 = 993 N
We know that B.M. at A and B,
MAH = MBH = 0
B.M. at C, MCH = RAH × 0.3 = 993 × 0.3 = 297.9 N-m
B.M. at D, MDH = RBH × 0.2 = 3971 × 0.2 = 794.2 N-m
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The bending moment diagram for horizontal loading is shown in Fig( f ).
Resultant B.M. at C,
and resultant B.M. at D,
The resultant bending moment diagram is shown in Fig (g).
We see that bending moment is maximum at D.
∴ Maximum bending moment,
M = MD = 819.2 N-m
Let d = Diameter of the shaft.
We know that equivalent twisting moment,
= 894 × 103 N-mm
We also know that equivalent twisting moment (Te),
∴ d 3 = 894 × 103 / 8.25 = 108 × 103 or d = 47.6 mm
Again we know that equivalent bending moment,
We also know that equivalent bending moment (Me),
∴ d 3 = 856.6 × 103/6.2 = 138.2 × 103 or d = 51.7 mm
Taking larger of the two values, we have
d = 51.7 say 55 mm.
Problem(8):- A steel solid shaft transmitting 15 kW at 200 r.p.m. is supported on
two bearings 750 mm apart and has two gears keyed to it. The pinion having 30
teeth of 5 mm module is located 100 mm to the left of the right hand bearing and
delivers power horizontally to the right. The gear having 100 teeth of 5 mm module
is located 150 mm to the right of the left hand bearing and receives power in a
vertical direction from below. Using an allowable stress of 54 MPa in shear,
determine the diameter of the shaft.
Given data : P = 15 kW = 15 × 103 W ; N = 200 r.p.m. ; AB = 750 mm ; TD = 30 ; mD
= 5 mm ; BD = 100 mm ; TC = 100 ; mC = 5 mm ; AC = 150 mm ; τ = 54 MPa = 54
N/mm2
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The space diagram of the shaft is shown in Fig (a).
We know that the torque transmitted by the shaft,
The torque diagram is shown in Fig (b).
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We know that diameter of gear = No. of teeth on the gear × module
∴ Radius of gear C,
and radius of pinion D,
Assuming that the torque at C and D is same (i.e. 716 × 103 N-mm), therefore
tangential force on the gear C, acting downward,
and tangential force on the pinion D, acting horizontally,
The vertical and horizontal load diagram is shown in Fig (c) and (d) respectively.
Now let us find the maximum bending moment for vertical and horizontal
loading.
First of all, considering the vertical loading at C. Let RAV and RBV be the reactions
at the bearings A and B respectively. We know that
RAV + RBV = 2870 N
Taking moments about A, we get
RBV × 750 = 2870 × 150
RBV = 2870 × 150 / 750 = 574 N
and RAV = 2870 – 574 = 2296 N
We know that
B.M. at A and B, MAV = MBV = 0
B.M. at C, MCV = RAV × 150 = 2296 × 150 = 344 400 N-mm
B.M. at D, MDV = RBV × 100 = 574 × 100 = 57 400 N-mm
The B.M. diagram for vertical loading is shown in Fig (e).
Now considering horizontal loading at D. Let RAH and RBH be the reactions at the
bearings A and B respectively. We know that
RAH + RBH = 9550 N
Taking moments about A, we get
RBH × 750 = 9550 (750 – 100) = 9550 × 650
∴ RBH = 9550 × 650 / 750 = 8277 N
and RAH = 9550 – 8277 = 1273 N
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We know that
B.M. at A and B, MAH = MBH = 0
B.M. at C, MCH = RAH × 150 = 1273 × 150 = 190 950 N-mm
B.M. at D, MDH = RBH × 100 = 8277 × 100 = 827 700 N-mm
The B.M. diagram for horizontal loading is shown in Fig ( f ).
We know that resultant B.M. at C,
= 393 790 N-mm
and resultant B.M. at D,
= 829 690 N-mm
The resultant B.M. diagram is shown in Fig (g). We see that the bending moment
is maximum at D.
∴ Maximum bending moment, M = MD = 829 690 N-mm
Let d = Diameter of the shaft.
We know that the equivalent twisting moment,
We also know that equivalent twisting moment (Te),
∴ d3 = 1096 × 103/10.6 = 103.4 × 103
or d = 47 say 50 mm.
Shafts Subjected to Fluctuating Loads:-
In the previous articles we have assumed that the shaft is subjected to constant
torque and bending moment. But in actual practice, the shafts are subjected to
fluctuating torque and bending moments. In order to design such shafts like line
shafts and counter shafts, the combined shock and fatigue factors must be taken
into account for the computed twisting moment (T ) and bending moment (M ).
Thus for a shaft subjected to combined bending and torsion, the equivalent
twisting moment,
and equivalent bending moment,
where Km = Combined shock and fatigue factor for bending, and
Kt = Combined shock and fatigue factor for torsion.
Problem(9):- A mild steel shaft transmits 20 kW at 200 r.p.m. It carries a central
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load of 900 N and is simply supported between the bearings 2.5 metres apart.
Determine the size of the shaft, if the allowable shear stress is 42 MPa and the
maximum tensile or compressive stress is not to exceed 56 MPa. What size of the
shaft will be required, if it is subjected to gradually applied loads?
Given data : P = 20 kW = 20 × 103 W ; N = 200 r.p.m. ; W = 900 N ; L = 2.5 m ;
τ = 42 MPa = 42 N/mm2 ; σb = 56 MPa = 56 N/mm2
Size of the shaft
Let d = Diameter of the shaft, in mm.
We know that torque transmitted by the shaft,
and maximum bending moment of a simply supported shaft carrying a central
load,
We know that the equivalent twisting moment,
= 1108 × 103 N-mm
We also know that equivalent twisting moment (Te),
∴ d 3 = 1108 × 103 / 8.25 = 134.3 × 103 or d = 51.2 mm
We know that the equivalent bending moment,
We also know that equivalent bending moment (Me),
∴ d 3 = 835.25 × 103 / 5.5 = 152 × 103 or d = 53.4 mm
Taking the larger of the two values, we have
d = 53.4 say 55 mm
Size of the shaft when subjected to gradually applied load
Let d = Diameter of the shaft.
Assume for rotating shafts with gradually applied loads,
Km = 1.5 and Kt = 1
We know that equivalent twisting moment,
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We also know that equivalent twisting moment (Te),
∴ d 3 = 1274 × 103 / 8.25 = 154.6 × 103 or d = 53.6 mm
We know that the equivalent bending moment,
We also know that equivalent bending moment (Me),
∴ d 3 = 1059 × 103 / 5.5 = 192.5 × 103 = 57.7 mm
Taking the larger of the two values, we have
d = 57.7 say 60 mm
Problem(10):- A horizontal nickel steel shaft rests on two bearings, A at the left and
B at the right end and carries two gears C and D located at distances of 250 mm and
400 mm respectively from the centre line of the left and right bearings. The pitch
diameter of the gear C is 600 mm and that of gear D is 200 mm. The distance
between the centre line of the bearings is 2400 mm. The shaft transmits 20 kW at
120 r.p.m. The power is delivered to the shaft at gear C and is taken out at gear D in
such a manner that the tooth pressure FtC of the gear C and FtD of the gear D act
vertically downwards.
Find the diameter of the shaft, if the working stress is 100 MPa in tension and
56 MPa in shear. The gears C and D weighs 950 N and 350 N respectively. The
combined shock and fatigue factors for bending and torsion may be taken as 1.5
and 1.2 respectively.
Given data : AC = 250 mm ; BD = 400 mm ; DC = 600 mm or RC = 300 mm ; DD =
200 mm or RD = 100 mm ; AB = 2400 mm ; P = 20 kW = 20 × 103 W ; N = 120 r.p.m ;
t = 100 MPa = 100 N/mm2 ; = 56 MPa = 56 N/mm2 ; WC = 950 N ; WD = 350 N ;
Km = 1.5 ; Kt = 1.2
The shaft supported in bearings and carrying gears is shown in Fig.
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We know that the torque transmitted by the shaft,
Since the torque acting at gears C and D is same as that of the shaft, therefore the
tangential force acting at gear C,
and total load acting downwards on the shaft at C
= FtC + WC = 5300 + 950 = 6250 N
Similarly tangential force acting at gear D,
and total load acting downwards on the shaft at D
= FtD + WD = 15 900 + 350 = 16 250 N
Now assuming the shaft as a simply supported beam as shown in Fig, the
maximum bending moment may be obtained as discussed below :
Let RA and RB = Reactions at A and B respectively.
∴ RA + RB = Total load acting downwards at C and D
= 6250 + 16 250 = 22 500 N
Now taking moments about A,
RB × 2400 = 16 250 × 2000 + 6250 × 250 = 34 062.5 × 103
∴ RB = 34 062.5 × 103 / 2400 = 14 190 N
and RA = 22 500 – 14 190 = 8310 N
A little consideration will show that the maximum bending moment will be either
at C or D. We know that bending moment at C,
MC = RA × 250 = 8310 × 250 = 2077.5 × 103 N-mm
Bending moment at D,
MD = RB × 400 = 14 190 × 400 = 5676 × 103 N-mm
∴ Maximum bending moment transmitted by the shaft,
M = MD = 5676 × 103 N-mm
Let d = Diameter of the shaft.
We know that the equivalent twisting moment,
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= 8725 × 103 N-mm
We also know that the equivalent twisting moment (Te),
∴ d3 = 8725 × 103 / 11 = 793 × 103 or d = 92.5 mm
Again we know that the equivalent bending moment,
We also know that the equivalent bending moment (Me),
∴ d3 = 8620 × 103 / 9.82 = 878 × 103 or d = 95.7 mm
Taking the larger of the two values, we have
d = 95.7 say 100 mm.
Design of Shafts on the basis of Rigidity:-
Sometimes the shafts are to be designed on the basis of rigidity. We shall
consider the following two types of rigidity.
1. Torsional rigidity. The torsional rigidity is important in the case of camshaft of
an I.C. engine where the timing of the valves would be effected. The permissible
amount of twist should not exceed 0.25° per metre length of such shafts. The
torsional deflection may be obtained by using the torsion equation,
where = Torsional deflection or angle of twist in radians,
T = Twisting moment or torque on the shaft,
J = Polar moment of inertia of the cross-sectional area about the axis
of rotation,
G = Modulus of rigidity for the shaft material, and
L = Length of the shaft.
2. Lateral rigidity. It is important in case of transmission shafting and shafts
running at high speed, where small lateral deflection would cause huge out-of-
balance forces. The lateral rigidity is also important for maintaining proper
bearing clearances and for correct gear teeth alignment.
The lateral deflection may be determined from the fundamental equation
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for the elastic curve of a beam, i.e.
Problem(11):- A steel spindle transmits 4 kW at 800 r.p.m. The angular deflection
should not exceed 0.25° per metre of the spindle. If the modulus of rigidity for the
material of the spindle is 84 GPa, find the diameter of the spindle and the shear
stress induced in the spindle.
Given:- P = 4 kW = 4000 W; N = 800 r.p.m. ; = 0.25° = 0.25 X 180
= 0.0044 rad ;
L = 1 m = 1000 mm ; G = 84 GPa = 84 × 109 N/m2 = 84 × 103 N/mm2
Diameter of the spindle
Let d = Diameter of the spindle in mm.
We know that the torque transmitted by the spindle,
We also know that
∴ d 4 = 129 167 × 32 / π = 1.3 × 106 or d = 33.87 say 35 mm
Shear stress induced in the spindle
Let = Shear stress induced in the spindle.
We know that the torque transmitted by the spindle (T),
= 47 740 / 8420 = 5.67 N/mm2 = 5.67 MPa.
Problem(12):- Compare the weight, strength and stiffness of a hollow shaft of the
same external diameter as that of solid shaft. The inside diameter of the hollow shaft
being half the external diameter. Both the shafts have the same material and length.
Given data: do = d ; di = do / 2 or k = di / do = 1 / 2 = 0.5
Comparison of weight
We know that weight of a hollow shaft,
WH = Cross-sectional area × Length × Density
................(i)
and weight of the solid shaft,
................(ii)
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Since both the shafts have the same material and length, therefore by dividing
equation (i) by equation (ii), we get
Comparison of strength
We know that strength of the hollow shaft,
...................(iii)
and strength of the solid shaft,
....................(iv)
Dividing equation (iii) by equation (iv), we get
Comparison of stiffness
We know that stiffness
∴ Stiffness of a hollow shaft,
..................(v)
and stiffness of a solid shaft,
.................(vi)
Dividing equation (v) by equation (vi), we get
= 1 – k4 = 1 – (0.5)4 = 0.9375
Srinivasulureddy.Dorasila M.Tech; M.I.S.T.E; (Ph.D) ; Associate professor; Mechanical engg. Dept; K.H.I.T; Guntur.