DMM-I UNIT-IV KEYS, COTTERS, KNUCKLE JOINTS AND SHAFTS

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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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


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


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 ×


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


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 ×


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 ×


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


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


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.


d = Diameter of the rods,

d1 = Outside diameter of sleeve,

d2 = Diameter of the enlarged end of rod,


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


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


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


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


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 ×


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


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.


x = Each side of the rod,

B = Total width of gib and cotter,

B1 = Width of the strap,


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



23

Area resisting failure = 2 l1 × x

∴ Resisting strength = 2 l1 × x ×


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 ×


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



24

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


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



25

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.



26

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.



27

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



28

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



29

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


Since the knuckle pin is in double shear, therefore load (P),

= 150 × 103 / 4248 = 35.3 N/mm2 = 35.3 MPa


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


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



30

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 :


Since the knuckle pin is in double shear, therefore load (P),



31


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


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.



32

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,



33

(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



34

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




35

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 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



36

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



37

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



38

...............(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



39

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


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,



40

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


.......(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



41

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



42

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).



43

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



44

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



= 894 × 103 N-mm


∴ 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



45

The space diagram of the shaft is shown in Fig (a).


The torque diagram is shown in Fig (b).



46

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



47

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


We know that the equivalent twisting moment,


∴ 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



48

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.


and maximum bending moment of a simply supported shaft carrying a central

load,


= 1108 × 103 N-mm


∴ d 3 = 1108 × 103 / 8.25 = 134.3 × 103 or d = 51.2 mm

We know that the equivalent bending moment,


∴ d 3 = 835.25 × 103 / 5.5 = 152 × 103 or d = 53.4 mm


d = 53.4 say 55 mm

Size of the shaft when subjected to gradually applied load


Assume for rotating shafts with gradually applied loads,

Km = 1.5 and Kt = 1




49


∴ d 3 = 1274 × 103 / 8.25 = 154.6 × 103 or d = 53.6 mm

We know that the equivalent bending moment,


∴ d 3 = 1059 × 103 / 5.5 = 192.5 × 103 = 57.7 mm


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.



50


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





51

= 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


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



52

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)



53

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.

Documents

DMM-I UNIT-IV KEYS, COTTERS, KNUCKLE JOINTS AND SHAFTS