Design of Machine Elements

Design of Machine Element

2003-04 1

DESIGN OF SOCKET AND SPIGOT COTTOR JOINT

Spigot and Socket Cotter Joint:

The end of the rod, which goes into the socket, is called the spigot. A spigot and

socket cotter joint is shown in Fig. The design of various parts may be accomplished as

discussed below.

Fig: Spigot and Socket Cotter Joint

Design of Socket and Spigot Cotter Joint :

The socket and spigot cotter joint is as shown in Fig.

Let, P = Load carried by the rods,

d = Diameter of the rods,

d1 = Outside diameter of socket,

d2 = Diameter of spigot or inside diameter of socket,

d3 = Outside diameter of spigot collar,

d4 = Diameter of socket collar,

t1 = Thickness of spigot 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 the rod,

σt = Permissible tensile stress for the rod material,

τ = Permissible shear stress for the cotter material, and

σc = Permissible crushing stress for the cotter material.


2003-04 2

The dimensions for a socket and spigot cotter joint can be found by considering the

various modes of failure as discussed below :

Step 1 : Failure of the rods in tension :

The rods may fail in tension due to the tensile load P.

Area resisting tearing

= 2

4d×π

∴ Tearing strength of the rods

= td σπ×× 2

4

Equating this load (P), we have

P = td σπ×× 2

4

From this equation, diameter of the rods (d) may be determined.

Step 2 : Failure spigot in tension across the weakest section (or slot) :

Note : Thickness of the cotter is generally taken as 0.4d, hence t = 0.4d.

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

= tdd .)(4 2

22 −

π

Hence, tearing strength of the spigot across the slot

= ttdd σπ

− .)(

4 22

2

Equating this to load (P), we have

P = [ ] ttdd σπ . )(4/ 22

2 −

From this equation, the diameter of spigot or inside diameter of socket ( d2 ) may be

determined.

Step 3 : Failure of the rod or cotter in crushing :

The area that resists crushing of a rod or cotter = d2. t

∴ Crushing strength = d2. t. σc

Equating this to load (P) we have


2003-04 3

P = d2. t. σc

From this equation, the induced crushing stress may be checked.

Step 4 : Failure of the socket in tension across the slot :

The resisting area of the socket across the slot is as shown in Fig.

= [ ] . )()()(4 21

22

21 tdddd −−−

π

∴Tearing strength of the socket across the slot

= [ ] ttdddd σπ

−−− )()()(

4 212

22

1


P = [ ] ttdddd σπ

−−− )()()(

4 212

22

1

From this equation the outside diameter of socket ( d1 ) may be determined.

Step 5 : Failure of cotter in shear :

Considering the failure of cotter in shear as shown in Fig.

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.τ

Equating this to the load (P), we have

P = 2 b.t.τ

From this equation, width of the cotter (b) can be determined.

Step 6 : Failure of the socket collar in crushing :

Considering the failure of socket color in crushing as shown in Fig.

We know that the area that resists crushing of

socket collar. = ( d4 – d2 ) t

and crushing strength = ( d4 – d2 ) . t . σc


P = ( d4 – d2 ) . t . σc

From this equation, the diameter of socket collar ( d4 ) may be obtained.

Step 7 : Failure of socket end in shearing :

The socket end is in double shear, therefore area that resists shearing of socket collar.


2003-04 4

= 2 ( d4 – d2 ) c

Shearing strength of socket collar = 2 ( d4 – d2 ) c x τ

Equating this to load (P), we have P = 2 ( d4 – d2 ) c x τ

From this equation, the thickness of socket collar ( c ) may be obtained.

Step 8 : Failure of rod end in shear :

The rod end is in double shear, therefore the area resisting shear of the rod end.

= 2 ad2

Hence, shear strength of the rod end = 2.d2 .τ x a.

Equating this to load (P), we have P = 2 ad2 . τ

From this equation, the distance from the end of the slot to the end of the rod (a) may

be obtained.

Step 9 : Failure of spigot collar in crushing :

Consider the failure of the spigot collar in crushing as shown in Fig.

Area that resists crushing of the collar

= [ ]22

23 )()(

4dd −

π

Crushing strength of the collar

= [ ] cdd σπ 22

23 )()(

4−


P = cdd σπ ])()[(4

22

23 −

From this equation, the diameter of the spigot collar ( d3 ) may be obtained.

Step 10 : Failure of the spigot collar in shearing :

Consider failure of the spigot collar in shearing as

shown in Fig. Area that resists shearing of

the collar. = π . d2 . t1

and shearing strength of the collar = π . d2 . t1 . τ


P = π . d2 . t1 . τ

From this equation, the thickness of spigot collar ( t1 )

may be obtained.


2003-04 5

Step 11 : Failure of cotter in bending :

In all the above relations, it is assumed that the load is uniformly distributed over the

various cross-sections of the joint. But in actual practice, this does not happen and the cotter

is subjected to bending. In order to find out in the bending stress induced it is assumed that

the load on the cotter in the rod end is uniformly distributed while in the socket end it varies

from zero at the outer diameter ( d4 ) and maximum at the inner diameter ( d2 ), as shown in

Fig. The maximum bending moment occurs at the center of the cotter and is given by

+

−=

−+

−=

×−+

−×=

4624262

42223/1

2

2242224

2224max

dddPddddP

dPdddPM

We know that section modulus of the cotter,

Z = t.b2/6

∴Bending stress induced in the cotter,

224

2

224

max

..2)5.0(

/.462

btddP

bt

dddP

ZM

b+

=

+

−

==σ

σ

This bending stress induced in the cotter should be less than allowable bending stress

of the cotter.

Step 12 : To find the length of cotter (l) :

The length of the cotter (l) is taken as 4d

Hence l = 4d


2003-04 6

Example 1

Design a cotter joint as shown in Fig. to transmit a load of 90 kN in tension or

compression. Assume the following stresses for socket, spigot and cotter.

Allowable tensile stress = 90 Mpa

Allowable crushing stress = 120 Mpa

Allowable shear stress = 60 Mpa.

Solution :

Given :

P = 90kN = 90 x 103 N σt = 90Mpa = 90 N/mm2

σc = 120Mpa = 120 N/mm2 τ = 60Mpa = 60 N/mm2

Referring from Fig.

Step 1 : Failure of the rods in tension :

The rods may fail in tension due to the tensile load P.

∴ P = td σπ×× 2

4

90 x 103 = 904

2 ×× dπ 90

41090 32

××

=π

xd

d = 35.6824 mm [ d = 40mm ]

The diameter of the rod is 40mm.

Step 2 : Failure of spigot in tension across the weakest section (or slot) :

Thickness of the cotter is generally taken as 0.4d.

mmdt 16404.0 4.0 =×==

P = ( )[ ] ttdd σπ . 4/ 22

2 −

Substituting values,

90 x 103 = ( ) 90 )16(4 2

22

− ddπ

1000 = ( ) 22

2 164

dd −π ∴1000 = 0.7853 ( ) 2

22 16dd −

1273.2395 = ( ) 22

2 3718.20 dd −

( ) 02395.1273d 3718.20 22

2 =−−d


2003-04 7

mm50or mm5235.47

value]positiveonly Taking [ 2

3153.747318.20

22395.1273114)7318.20(7318.20 2

2

=

±=

×××+±=d

The diameter of spigot or inside diameter of socket is 50 mm.

Step 3 : Failure of the rod or cotter in crushing :

P = d2 . t . σc

∴90 x 103 = 50 x 16.σc

σc = 112.5 N/mm2 < 120N/mm2.

The induced crushing stress is thus checked.

Step 4 : Failure of the socket in tension across the slot :

P = [ ] ttdddd σπ

−−− )()()(

4 212

22

1

90 x 103 = [ ] 90 16 )50()50()(4 1

221

−−− ddπ

[ ] 100016 )50()50()(4 1

221 =−−− ddπ

2754.9922-d 3743.20)( 21 −d

2

9922.275414)3743.20(3743.20 2

1

××+±=d

= 62.67516 or 65mm. [ Taking only positive value ]

The outside diameter of socket is 65mm.

Step 5 : Failure of cotter in shear :

P = 2 . b . t . τ b = τ..2 t

P

60162

1090 3

×××

=b b = 46.875 mm.

b = 47 mm.

The width of the cotter is 47mm.

Step 6 : Failure of the socket collar in crushing :

P = ( ) c24 . t . σdd −


2003-04 8

ct

Pddσ. 24 =−

50130161090

.

3

24 +××

=+= dt

Pdcσ

d4 = 93.2692 or d4 = 94 mm.

The diameter of socket collar is 94mm.

Step 7 : Failure of rod end in shear :

P = 2 a d2 . t τ××

=22 d

Pa

60502

1090 3

×××

=a a = 15 mm.

The distance from the end of the slot to the end of the rod is 15 mm.

Step 9 : Failure of spigot collar is crushing :

P = [ ] cdd σπ )()(4

22

23 −

23

22 )()(4 ddP

c

=+Π×

×σ

23

23

)()50(120

41090 d=+Π×××

954.9296 + (50)2 = (d3)2

58.7732 = d3

d3 = 58.7732

The diameter of the spigot collar is 58.7732

Step 10 : Failure of the spigot collar in shearing :

P = π x d2 x t1x τ

τπ ××

=2

1 dPt

60501090 3

1 ×××

=π

t

mm 5492.91 =t or t1 = 10 mm.

The thickness of the spigot collar is 10mm.

Step 11 : Failure of cotter in bending :

+

−×=+

−

=4

506

509421090

462

3224

maxdddPM


2003-04 9

mmNM −×=+×= 53max 109249.8]5.123333.7[1045

224

2

224

max

b ..t 2d 5.0(

6/b.46

2/

2+

=

+

−

==dP

t

dddP

Mbσ

90 = 2

3

162))50(5.094(1090

b××+×

2880 b2 = 10.71x106 ∴ b = 60.9815

b = 61mm.

Considering bending of cotter, b = 61mm.

Considering shearing of cotter, b = 47mm

Selecting larger of two values [b = 61mm ]

Step 12 : The length of the cotter (l)

l= 4d

l = 4 x 40

l= 169 mm.

The length of the cotter is 160 mm.


2003-04 10

DESIGN OF KNUCKLE JOINT Knuckle Joint :

A knuckle joint is used to connect two rods subjected to tensile load only.

At the end of one rod an eye is forged and at the other end of the other rod a fork.

The eye and the fork are connected by means of a pin. The joint provides flexibility in one

plane and provides a quick means of connecting and disconnecting the joint. The rods may

not be truly axial and may have angular misalignment. Fig. shows a knuckle joint most

commonly used. It is generally made from mild steel or wrought iron.

If d is the diameter of rod, then diameter of pin, d1 = d

Outer diameter of eye d2 = 2d

Diameter of knuckle pin head and collar, d3 = 1.5d

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.


2003-04 11

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 the eye

t2 = Thickness of pin head and collar

t = Thickness of single eye

t1 = Thickness of fork.

σt, τ, σc = 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.

Due to these assumptions the strengths are approximate, however they serve to

indicate a well proportioned joint. Following are the methods of failure of the joint :

Step 1 : Failure of the solid rod in tension :

Since the rods are subjected to direct tensile load, therefore tensile strength of the rod,

= td σπ×× 2

4

Equating this to the load (P) acting on the rod, we have

P = td σπ×× 2

4

From this equation, diameter of the rod (d) is obtained.

Step : 2 Failure of the knuckle pin in shear :

Since the pin is in double shear, therefore

cross-sectional area of the pin under shearing

21 )(

4 . 2 dπ×=

and the shear of the pin


2003-04 12

τπ 21 )(

42 d×=

Equating this to the load (P) acting on the rod, we have P = τπ 21 )(

42 d×

From this equation τ induced in kunckle pin is obtained. If it is less than

permissible shear stress then the pin is safe in shear.

Step 3 : 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 = ( )tdd 21 −

∴Shearing strength of single eye or rod end

( ) τ.12 tdd −=


( ) τ . t 12 ddP −=

From this equation, the outer diameter of the eye (d2)

may be obtained.

Step 4 : Failure of the single eye or rod end in tension :

The single eye or rod end may tear off due to tensile load. We know that area

resisting tearing = ( ) tdd 12 −

∴ Tearing strength of single eye or rod end ( ) t12 . t σdd −=

Equating this to the load (P), we have ( ) t12 . t σddP −=

From this equation, the induced tensile stress σt 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).


2003-04 13

Step 5 : Failure of the single eye or rod end in crushing :

The single eye or pin may fail in crushing due to the tensile load. We know that area

resisting crushing = d1t

∴Crushing strength of single eye or rod end

= d1 . t . σc


P = d1 . t . σc

From this equation, the induced crushing stress σc for the single eye or pin may be checked.

In case the induced crushing stress is more than the allowable working stress, then increase

the thickness of the single eye.

Step 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) 2t1

∴Tearing strength of the forked end

= (d2 – d1) 2t1 x σt


P = (d2 – d1) x 2t1 x σt

From this equation, the induced tensile stress may be checked.

Step 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) x 2t1

∴ shearing strength of the forked end

= (d2 – d1) x 2t1 x τ

Equating this to the load (P) we have

P = (d2 – d1) 2t1 x τ

From this equation the induced shear stress may be checked. In case, the

induced shear stress is more than the allowable working stress, then thickness of

the fork is increased.

Step 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 x 2t1


2003-04 14

∴crushing strength of the forked end

= d1 x 2t1 x σc

Equating this to the load (P) we have

P = d1 x 2t1 x σc

From this equation, the induced crushing stress may be checked.

Example 3 :

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 N/mm2. The ultimate tensile and shearing strength of the pin material

are 510 N/mm2 and 396 N/mm2 respectively. Determine the tie rod section and

pin section. Take factor of safety = 6.

Solution :

Data : P = 70 kN = 70000 N σtu for rod = 420 N/mm2

σtu for pin = 510 N/mm2 σtu = 396 N/mm2

F.S.= 6.

We know that the permissible tensile stress for the rod material,

6

420F.S.

==

rodfortut

σσ

= 70 N/mm2

and permissible shear stress for the pin material,

2N/mm 666

396.

===SF

suστ

We shall now consider the various methods of failure of the joint as discussed below

Step 1 : Failure of the rod in tension

Let d = Diameter of the rod we know that the load (p),

70,000 = 222 d 557044

=××=×× dd tπσπ

d2 = 70,000/55 = 1273

or d = 35.7 say 36 mm.

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 = 2d = 2 x 36 = 72 mm


2003-04 15

Diameter of knuckle pin head and collar, d3 = 1.5d = 1.5 x 36 = 54 mm

Thickness of single eye or rod end, t = 1.25 d = 1.25 x 36 = 45 mm

Thickness of fork, t1= 0.75d = 0.75 x 36 = 27 mm

t2 = 0.5d = 0.5 x 36 = 18 mm

Now we shall check for the induced stresses as discussed below :

Step 2 : Failure of the knuckle pin in shear

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

70,000 = ττπτπ 2036)36(4

2)(4

2 221 =×=× d

∴ τ = 70,000 / 2036 = 34.4 N/mm2

Since 34.4 < 66 N/mm2 pin is safe in shear

Step 3 : Failure of the single eye or rod end in shearing

P = (d2 – d1).t.τ

70000 = ( d2 – 36 ) x 45 x 66 = 59.5690 mm

Hence taking the bigger value as found from the standard formula we take d2 as 72mm.

Step 4 : Failure of the single eye or rod in tension

The single eye or rod end may fail in tension due to the load.

We know that load (P) is

70,000 = (d2 – d1) x t x σt

= (72 – 36) x 45 x σt = 1629 σt

σt = 2/ 2.43162070000 mmN=

Step 5 : Failure of single eye or rod end in crushing :

P = d1 x t x σc

2

1

N/mm 2.434536

70000=

×=

×=

tdP

cσ

Hence the induced crushing stress σc is 43.2 N/mm2.

Step 6 : Failure of the forked end in tension :

P = (d2 – d1) x 2 t1 x σt

70000 = ( 72 – 36 ) x 2 x 27 x σt

σt = 36.0082 N/mm2


2003-04 16

Hence the induced stresses are less than the given values hence the joint is safe in

tension.

Step 7 : Failure of the forked end in shear :

P = (d2 – d1) 2t1 x τ

70000 = ( 72 – 36 ) x 2 x 27 x τ

τ = 36.0082 N/mm2

Hence the induced stresses are less than the given values hence the joint is safe in shear.

Step 8 : Failure of the forked end in crushing :

P = d1 x 2t1 x σc

70,000 = 36 x 2 x 27 x σc

σc = 36.0082 N/mm2

Hence, as the induced stresses are less than the given permissible stresses the joint is

safe in crushing.


2003-04 17

DESIGN OF LEVER SAFETY VALVE

A lever safety valve is used to maintain a constant pressure inside the boiler. When

press inside boiler increase excess steam blows off through the value until press falls to

required limit.

Design of lever :

Step 1 : Finding value of W :

From the pressure given (guage pressure) find value of W using relation

W = pressure x area

Or W = 2

4DP π

×

From above find W.

Step 2 : Finding equilibrium conditions :

Considering equilibrium of lever i.e. ∑ Fx , ∑Fy and M, find forces at points B and F.

Step 3 : Design of pins :

The pins are designed from bearing consideration and checked for shearing.

Let dp = diameter of pin

lp = length of pin [if nothing

specified assume]

If not specified in question,

lp = 1.25 dp

Bearing area of pin = dp x lp (Projected area)

∴ W = dp x lp x bearing pressure

From above we find dp and lp , now, pin is checked in double shear.


2003-04 18

i.e. W = inducedpd τπ×2

42

if τinduced < τpermissible pin safe in shear.

Step 4 : Design of fulcrum pin :

Since force at fulcrum is almost same as W, hence we take same pin at fulcrum.

Assuming a 2mm thick gunmetal bush is provided in pin holes to reduce wear

and to increase life of lever.

∴ diameter of hole = dp + 2 x thickness of bush

outer diameter of boss = 2 x diameter of hole

Step 5 : Design of cross section of lever

The c/s is designed for bending. If nothing is specified, assume

b1 = 4t

Bending moment M = P

−

2boss ofdiameter outerb

I = tb 31 )(

121 y = b1/2

Using the relation σ = I

yM

We find ( b1 and t )

The lever dimensions are check for shearing and bending.

Example :

A lever loaded safety value is 70 mm in diameter and is to be designed for a boiler to blow

off at a pressure of 1 N/mm2 (guage). Design a suitable mild steel lever using following

data.

Tensile stress = 70 Mpa.

Shear stress = 50 Mpa.

Bearing pressure = 25 Mpa.

Pin is also made of mild steel. Distance of fulcrum to weight on lever is 880 mm, and

distance between fulcrum and pin connecting the valve spindle links to the lever is 80mm.


2003-04 19

Solution :

Step 1 : To find W :

From the pressure we find W.

W = pressure x area W = 2)70(4

1 π×

W = 3848.451 N

Step 2 : To Find reaction :

Consider equilibrium of lever to calculate forces.

Σ Fy = 0

Rf – P = - 3848.451

Mf = 0

- 3848.451 x 80 + P x 880 = 0

P = 349.85 N

Rf = -3498.601 N

Step 3 : Design of pin :

dp = diameter of pin


2003-04 20

lp = length of pin = 1.25 dp

W = dp x lp x bearing pressure

∴ 3848.451 = dp x 1.25 dp x 25

dp =11.0973 mm dp = 12 mm

lp = 15 mm

Checking in double shear :

W = 2 induced2

4τπ

pd

3848.451 = 2)12(4

2π τinduced

τinduced = 17.0138 Mpa

as τinduced < τpermissible pin safe in shearing

Step 4 : Design of fulcrum pin :

We take same dimensions of pin as discussed above.

Diameter of hole = dp + 2 x thickness of bush

= 12 + 2 x 2

= 16 mm

outer diameter of boss = 2 x diameter of hole

= 32 mm

Step 5 : Design of c/s of lever :

Assuming b1 = 4t

M = P

−

2boss ofdiameter outerb

= 349.88

−

232600

M = 274282.4

I = tb 31 )(

121

= 43 333.5)4(121 ttt =

FyM .

=σ


2003-04 21

tby 22== t

t2

333.54.274282

4 ×=σ

70 = 3

9.102855t

t = 11.368

t = 12 mm b1 = 48 mm

Step 6 : Design check for c/s of lever :

Cross section of dimension are checked for shearing and bending.

a) Check for shearing :

average shear stress induced < permissible shear stress

lever safe in shearing

b) Check for bending :

Checking for bending stress induced is done

at section passing through center of hole at A.

i.e. σinduced = MIy

where M = P x b

= 349.854 x 800

y = b1/2 = 12/2 = 6

I = 15)16(12115)32(

121 33 ×−×

+ 2 x

××+× 212 )20(81212)8(121

= 113664 mm4

σinduced = MPa 09.59113664

6800854.349=

××

as σinduced < σpermissible ; hence lever safe in bending.


2003-04 22

DESIGN OF RIGID FLANGE COUPLING Flange Coupling :

It is a rigid type of coupling. The flange coupling consists of two cast iron flanges,

keyed to the shaft ends and bolted together. To ensure proper alignment, the end of one

shaft may enter into the recess provided in the flange attached to the other shaft. Protected

type of shafts is often used to provide safety to the operator. These flanges project beyond

the heads of the bolts and nuts. The flanges are generally made from cast iron by casting or

steel by a forging process. They are generally preferred for transmitting heavy torques. The

various parts of the flange coupling as shown in Fig. may be designed as explained below :


2003-04 23

Steps to Design Flange Coupling :

Step 1 : On the basis of torque to be transmitted, calculate the shaft diameter d using

the relation.

T = ( )shaftepermissibld τπ

×× 3

16

Step 2 : Using empirical relations find all other dimensions of the flange coupling.

(a) Outside diameter of hub (D1) = 2 d. (b) Length of hub (L) = 1.5 d.

(c) P.C.D. of bolts (D1) = 3 d. (d) Outside diameter of flange (D2) = 4 d.

(e) Thickness of flange ( tf ) = 0.5 d. (f) Number of bolts (n) = 3150

4+d

or number of bolts can be taken as.

n =3 for shaft dia. upto 40 mm.




(g) If the coupling is protective type flange coupling then, thickness of protective

circumferential flange ( tp ) = 0.25 d.

Step 3 : On the basis of shaft diameter we decide dimensions of the key.

(a) for square key

w = 4d t =

4d

(b) for rectangular key w = 4d t =

6d

Length of key (l) = Length of hub (L).

Step 4 : Since we have used empirical relations to find various dimensions, now we shall

check each part for safety.

(1) Design check for hub :

The hub is designed by considering it as a hollow shaft transmitting the same torque

T. Using the relation.

T = ( ) .

43 1

16 hubinducedDdD τπ

×

−×

We find ( )hubinducedτ


2003-04 24

If τinduced is less than τpermissible for cast iron then the design of hub is safe.

(2) Design check for Flange :

The flange at the junction of the hub is under shear while transmitting the torque.

Area of flange subjected to shear = π D x tf

Now force at junction (F) = Shear stress x shear area.

And Torque (T) = Force x radius

Hence Torque (T) = ( )2

DtD fflangeinduced ×××πτ

∴ T = ( ) flangeinducedftD τπ××

2 2

Using the above relation we find ( ) flangeinducedτ If τinduced is less than τpermissible for cast

iron then design of flange is safe.

Step 5 : Design check for key :

Dimensions of the key are checked in shear and crushing.

(a) For shear :

Torque (T) = ( )keyinduceddlw τ×××2

From the above equation we find τinduced

If τinduced in less than τpermissible for key material then key is safe in shearing.

(b) For Crushing :

Torque (T) = ( )keyinduceddlt σ×××22

From the above equation we find σinduced

If σinduced is less than σpermissible then key is safe in crushing.

Step 6 : Design for bolts :

The bolts are subjected to shear stress due to the torque transmitted. The bolts are

designed for shear and checked for crushing.

(a) Design for shear :

Load on each bolt = shear stress x shear area.

∴ Load on each bolt = τbolt2

14dπ

× Torque = Force x P.C.D. of bolts


2003-04 25

Total load on all bolts = τbolt nd ××× 214

π

Hence, Torque (T) = 24

121

Dndbolt ××××

πτ

Using the above equation we find diameter of bolt d1.

(b) Check in Crushing :

Area resisting crushing = n x d1 x tf

Also Force = crushing stress x area resisting crushing

and Torque = Force x P.C.D. of bolts.

∴ T = n ( )boltinducedfD

tdn σ××××2

11

From the above equation σinduced is found.

If σinduced is less than σpermissible for bolt then design of bolt is safe.

Problem : Design and draw a flange coupling for a steel shaft transmitting 15 kW at

200 r.p.m. and having allowable shear stress of 40 Mpa. Shearing in bolts should not

exceed 30 Mpa. Assume that same material is used for shaft and key and crushing stress is

twice the value of shearing stress. Maximum torque is 25% greater than full load torque.

The shear stress for Cast Iron is 14 Mpa.

Solution :

Step 1 : To calculate shaft diameter ‘d’

Torque (T) = n

P2

60π× ∴Tmean =

2002601015 6

××××

π

Tmean = 716197.2439 Nmm.

Since maximum Torque exceeds by 25%

∴ Tmax = 1.25 x 716197.2439 = 895246.5549 Nmm.

Now Tmax. = ( )shaftepermissibld τπ

×× 3

16

∴ 895246.5549 = 4016

3 ×× dπ

∴ d = 48.486 mm. d = 50 mm.

Step II : Using empirical relations find all other dimensions of the flange coupling.

(a) Outside diameter of the hub (D) = 2d = 100 mm.


2003-04 26

(b) Length of hub (L) = 1.5 d = 75 mm.

(c) P.C.D.of bolts (D1) = 3d = 150 mm.

(d) Outside diameter of flange (D2) = 4 d = 200 mm.

(e) Thickness of flange ( tf ) = 0.5 = 25 mm.

(f) Number of bolts (n) = 4.

Step III : On the basis of shaft diameter we decide dimensions of the key.

Since, ( )keyepermissiblσ = 2 ( )

keyepermissiblτ

∴We choose a square key.

∴ w = mm. 5.124

504

==d

t = mm. 5.124

504

==d

∴ w = t = 14 mm

Length of key (l ) = length of hub (L) = 75 mm

Step IV :


The hub is checked considering it as a hollow shaft :

T = ( )hubinducedDdD τπ

×

−×

43 1

16

895246.5549 = ( ) ( )hubinducedτπ

−

43

100501100

16

( )hubinducedτ = 4.863 Mpa.

Since τinduced is less than τpermissible hence design of hub is safe.

(2) Design check for flange :

Torque (T) = ( ) flangeinducedftD τπ××

2

2

895246.5549 = ( ) ( ) flangeinducedτπ×× 25100

22

( ) flangeinducedτ = 2.2797 Mpa.

Since τinduced is less than τpermissible hence design of flange is safe.

Step V : Design check for key :


2003-04 27


(a) For shear :


895246.5549 = ( )keyinducedτ×××2

507514

( )keyinducedτ = 34.104 Mpa

Since τinduced is less than τpermissible hence key is safe in shearing.

(b) For crushing :


895246.5549 = ( )keyinducedσ×××

4507514

( )keyinducedσ = 68.2092 Mpa.

Since σinduced is less than σpermissible hence key is safe in crushing.

Step VI : Design for bolts :

The bolts are designed for shear and checked for crushing.


(Torque) T = 24

121

Dndbolt ××××

πτ

895246.5549 = 42

1504

30 21 ××× dπ

mm. 65147.12621 =d

d1 = 11.253 mm. = 12 mm.

Hence diameter of the bolt is 12 mm.

(b) Check in crushing :

Since crushing stress for bolt material is not given hence check in crushing is not

needed to be found out.

Fig. Shows the dimensions of the coupling


2003-04 28

Problem : Power of 11 kW at 500 r.p.m. is transmitted to a pump, through a rigid

coupling by an engine. Design a protected type flange coupling with a overload capacity of

25%. Design the flange coupling.

Data given is :

Material C.I flange material M.S.shaft and key

material

Plain carbon steel

for bolt

(1) Allowable

Tensile stress

20 Mpa 100 Mpa 80 Mpa

(2) Allowable

compressive stress

60 Mpa ----- 60 Mpa

(3) Allowable shear

stress

10 Mpa 60 Mpa 40 MPa


2003-04 29

Solution :

Step 1 : To calculate shaft diameter ‘d’

Torque (T) = n260

π×P Tmean =

5002601011 6

××××

π

Tmean = 210084.52

Since maximum Torque exceeds by 25%

T = 1.25 x Tmean = 1.25 x 210084.52 = 262605.6561 Nmm

Now T = ( )shaftepermissibld τπ

×× 3

16

262605.6561 = 6016

3 ×× dπ d = 28.143 mm.

d = 30 mm.

Step II : Using empirical relations find all other dimensions of the flange coupling.

(a) Outside diameter of the hub (D) = 2d = 60 mm.

(b) Length of hub (L) = 1.5 d = 45 mm.

(c) P.C.D. of bolts ( D1) = 3d = 90 mm.

(d) Outside diameter of flange ( D2 ) = 4d = 120 mm.

(e) Thickness of flange ( tf ) = 0.5 d = 15 mm.

(f) Number of bolts (n) = 3

(g) Thickness of protective circumferential flange

tp = 0.25 d = 7.5 mm.

Step III : On the basis of shaft diameter we decide dimensions of the key

Since,

( )keyepermissiblσ < 2 ( )

keyepermissiblτ

∴ We choose a rectangular key

∴ w = mm. 8 mm 5.74

304

===d

t = mm. 56

306

==d

w = 8 mm.

Length of key (l) = Length of hub (L) = 45 mm.


2003-04 30

Step IV :


The hub is checked considering it as a hollow shaft :

T = ( )hubinducedDdD τπ

×

−×

43 1

16

262605.6561 = ( )hubinducedD τπ×

−×

43

60301

16

( )hubinducedτ = 6.6046 Mpa

Since τinduced is less than τpermissible hence design of hub is safe.

(2) Design check for flange :

Torque (T) = ( ) flangeinducedft τπ××

2D 2

262605.6561 = ( ) flangeinducedτπ××

× 152602

( ) flangeinducedτ = 3.0959 Mpa.

Since τinduced is less than τpermissible hence design of flange is safe.

Step V : Design check for key :


(a) For shear :


262605.6561 = ( )keyinducedτ×××2

30845

( )inducedτ = 48.63 Mpa.

Since τinduced is less than τpermissible hence key is safe in shearing.

(b) For crushing :


262605.6561 = ( )keyinducedσ×××2

304525

( )inducedσ = 155.618 Mpa. > 100


2003-04 31

As σinduced is greater than σpermissible thus the key fails in crushing.

Hence finding the new value of ‘t’ put σmax in the equation.

262605.6561 = 1004

3045 ×××t

t = 7.7809 = 8 mm.

Hence the key dimensions are ltw ××

= 4588 ××

Step VI : Design for bolts :

The bolts are designed for shear and checked for crushing.


Torque (T) = 24

121

Dndbolt ××××πτ

262605.6561 = 32

90404

21 ×××× dπ

d1 = 7.8688 mm.

(b) Check in crushing :

T = nDtd f ××××2

11 σ

262605.6561 = 7.8688 inducedσ×××× 32

9015

σinduced = 16.498 Mpa.

As σinduced is less than σpermissible hence bolt is safe in crushing.


2003-04 32

DESIGN OF SHAFT CARRING ONE PULLEY AND SUPPORTED IN

TWO BEARING

A belt pulley is keyed to the shaft, midway between the supporting bearings kept at

1000 mm apart. The shaft transmits 20 KW power at 400 rpm. Pulley has 400 mm

diameter. Angle of wrap of belt on pulley is 1800 and the belt tensions act vertically

downwards. The ratio of belt tensions = 2.5.

The shaft is made of steel having ultimate tensile stress and yield stress of 400 Mpa

and 240 Mpa respectively. Use ASME code to design the diameter of shaft with combined

fatigue and shock factors in bending and torsion as 1.5 and 1.25 respectively.

Solution : We draw a rough diagram.

Step I : Applying ASME code to find τpermissible

τpermissible = 0.3 x Syt

OR

τpermissible = 0.18 x Sut

whichever of the above two is minimum

τpermissible = 0.3 x 240 = 72 Mpa

OR τpermissible = 0.18 x 400 = 72 Mpa

Since, the pulleys are keyed to the shaft therefore, reducing smaller value by 25%.

τpermissible = 0.75 x 72

τpermissible = 54 MPa


2003-04 33

Step II : Torque Transmitted :

Using the relation, Power (P) = 60

NT 2π

(Torque) T = 400142.32601020

260 6

××××

=×

NPπ

T = 477464.829 Nmm

Since torque is transmitted by a belt drive,

∴ Torque = ( T1 – T2 ) r

∴ 477464.829 = (T1 – T2) x 200

∴ T1 – T2 = 2387.3241

also 5.22

1 =TT (given)

∴ (2.5 T2 – T2) = 2387.3241

∴ T2 = 1591.5494 N Hence, T1 = 3978.8735 N

Step III : To find maximum B.M. i.e.; M :

(a) Since belt tensions act vertically downwards hence vertical load at the center of the shaft

becomes (T1 + T2) in the downward direction.

Now, ∑Fy RA + RB = 5570.422

∑MA = 0

5570.422 x 500 – RB x 1000 = 0

RB = 2785.211 N and RA = 2785.211 N


2003-04 34

Bending Moment Calculations :

B.M. at A = 0

B.M. at C = 2785.211 x 500 = 1392605.644 Nmm

B.M. at B = 0

∴ Maximum Bending Moment (Mmax) = 1392605.644 Nmm

Step IV : Using the final formula as per theory mentioned we find shaft diameter ‘d’

i.e. 22 )()( MkTk bt ×+× = epermissibld τπ×3

16

∴ )644.13926055.1()829.47746425.1( 2 ×+× = 5416

3 ×dπ

∴ d3 = 204897.027

∴ d = 58.9538 mm.

…………………

Documents

Design of Machine Elements