Design Of Joints

Design of Machine Element

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

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Design of Machine Element

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

π

=

4

×d2

∴ Tearing strength of the rods

π

=

4

× d 2 ×σ t

Equating this load (P), we have P=

π 4

× d 2 ×σ t

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

π

=

4

(d 2 ) 2 − d 2 .t

Hence, tearing strength of the spigot across the slot π  =  (d 2 ) 2 − d 2 .t σ t 4  Equating this to load (P), we have

[

]

P = π / 4(d 2 ) 2 − d 2 . t σ t

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

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Design of Machine Element

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. =

π

[(d ) 4 1

2

]

− ( d 2 ) 2 − ( d 1 − d 2 ) t.

∴Tearing strength of the socket across the slot

π  =  (d1 ) 2 − (d 2 ) 2 − (d1 − d 2 )t σ t 4 

[

]

Equating this to load (P), we have π  P =  (d1 ) 2 − (d 2 ) 2 − (d1 − d 2 )t σ t 4 

[

]

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 Equating this to load (P) we have 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

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Design of Machine Element

= 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 = 2.d2 .τ x a.

Hence, shear strength of the rod end

P = 2 ad2 . τ

Equating this to load (P), we have

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 =

π

[(d ) 4

2

3

− (d 2 ) 2

]

Crushing strength of the collar =

π 4

[(d )

2

3

]

− (d 2 ) 2 σ c

Equating this to load (P), we have P=

π 4

[(d 3 ) 2 − (d 2 ) 2 ]σ c

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 Equating this to load (P) we have P = π . d2 . t1 . τ From this equation, the thickness of spigot collar ( t1 ) may be obtained. 2003-04

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Design of Machine Element

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 d − d2 d2 P d2  P + − ×  1 / 3 × 4 2 2 2 2 4  P  d − d2 d2 d2  P  d4 − d2 d2  =  4 + − =  +  2 6 2 4  2 6 4  M max =

We know that section modulus of the cotter, Z = t.b2/6 ∴Bending stress induced in the cotter,

σb =

M max Z

P  d4 − d2 d2  +   2 6 4  P(d 4 + 0.5d 2 ) = = 2 t .b /σ 2 .t . b2

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

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Design of Machine Element

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=

π 4

× d 2 ×σ t

π

90 x 103 =

4

d2 =

× d 2 × 90

d = 35.6824 mm

90 × 10 3 x 4 π × 90

[ 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. t = 0.4 d = 0.4 × 40 = 16mm

P=

[π / 4 (d

2

)2 − d 2 . t ]σ t

Substituting values, π  2 90 x 103 =  (d 2 ) − d 2 (16) 90 4  1000 =

π 4

∴1000 = 0.7853 (d 2 ) − 16d 2

(d 2 )2 − 16d 2

2

1273.2395 = (d 2 ) − 20.3718d 2 2

(d 2 )2 − 20.3718 d 2 − 1273.2395 = 0

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Design of Machine Element

20.7318 ± (20.7318) 2 + 4 × 1 ×1 × 1273.2395 d2 = 2 20.7318 ± 74.3153 = [ Taking only positive value] 2 = 47.5235 mm or 50 mm

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 :

 π P =  (d1 ) 2 − (d 2 ) 2 − (d1 − d 2 ) t σ t  4

[

]

 π 90 x 103 =  (d1 ) 2 − (50) 2 − (d1 − 50) 16 90  4

[

]

π [ (d1 ) 2 − (50) 2 ] − (d1 − 50) 16 = 1000 4 (d1 ) 2 − 20.3743 d - 2754.9922 20.3743 ± (20.3743) 2 + 4 × 1 × 2754.9922 d1 = 2 = 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= 90 × 10 3 b= 2 × 16 × 60

b = 46.875 mm.

b = 47 mm. The width of the cotter is 47mm. Step 6 : Failure of the socket collar in crushing :

P = (d 4 − d 2 ) . t . σ c

2003-04

P 2 . t .τ

7

Design of Machine Element

d4 − d2 =

P t .σ c

d4 =

P 90 × 10 3 + d2 = + 50 t.σ c 16 × 130

d4 = 93.2692 or

d4 = 94 mm.

The diameter of socket collar is 94mm. Step 7 : Failure of rod end in shear :

P 2 × d2 ×τ

P = 2 a d2 . t

a=

90 × 10 3 a= 2 × 50 × 60

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=

π [ (d 3 ) 2 − (d 2 ) 2 ]σ c 4

P×4 + (d 2 ) 2 = (d 3 ) 2 σc ×Π 90 × 10 3 × 4 + (50) 2 = (d 3 ) 2 120 × 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 t1 =

P π × d2 ×τ

t1 = 9.5492 mm or

t1 =

90 × 10 3 π × 50 × 60

t1 = 10 mm.

The thickness of the spigot collar is 10mm. Step 11 : Failure of cotter in bending :

M max =

2003-04

P  d 4 − d 2  d 2 90 × 10 3  94 − 50 50  + =  6 + 4  2  6  4 2 8

Design of Machine Element

M max = 45 × 10 3 [7.3333 + 12.5] = 8.9249 × 10 5 N − mm  d − d2 d2  P / 2 4 +  6 4  P(d 4 + 0.5 d 2  = = 2 t .b /6 2 .t . b 2

σb =

M max 2

90 =

90 × 10 3 (94 + 0.5(50)) 2 × 16 × b 2

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.

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Design of Machine Element

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 d2 = 2d

Outer diameter of eye Diameter of knuckle pin head and collar,

d3 = 1.5d

Thickness of single eye or rod end,

t = 1.25 d t1 = 0.75 d

Thickness of fork, Thickness of pin head,

t2 = 0.5 d

Other dimensions of the joint are shown in Fig. 2003-04

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Design of Machine Element

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, =

π × d 2 ×σ t 4

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

π × d 2 ×σ t 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 = 2 .×

π (d1 ) 2 4

and the shear of the pin

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Design of Machine Element

= 2×

π (d1 ) 2 τ 4 P = 2×

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

π (d1 ) 2 τ 4

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 = (d1 − d 2 )

t

∴Shearing strength of single eye or rod end = (d 2 − d 1 )t.τ

Equating this to the load (P), we have P = (d 2 − d1 ) t . τ

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 = (d 2 − d1 ) t = (d 2 − d1 ) t . σ t

∴ Tearing strength of single eye or rod end

P = (d 2 − d1 ) t . σ t

Equating this to the load (P), we have

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).

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Design of Machine Element

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 Equating this to the load (P), we have 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 Equating this to the load (P), we have 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

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Design of Machine Element

∴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 : σtu for rod = 420 N/mm2

Data : P = 70 kN = 70000 N σtu for pin = 510 N/mm2

σtu = 396 N/mm2

F.S.= 6. We know that the permissible tensile stress for the rod material,

σt =

σ tu for rod F.S.

=

420 6

= 70 N/mm2 and permissible shear stress for the pin material,

τ=

σ su F .S

=

396 = 66 N/mm 2 6

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 =

π π × d 2 × σ t = × d 2 × 70 = 55 d 2 4 4

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, 2003-04

d2 = 2d = 2 x 36 = 72 mm 14

Design of Machine Element

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 t1= 0.75d = 0.75 x 36 = 27 mm

Thickness of fork,

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 = 2 × ∴

π π (d1 ) 2 τ = 2 × (36) 2 τ = 2036 τ 4 4

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 =

70000 = 43.2 N / mm 2 1620

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

P = d1 x t x σc

σc =

P 70000 = = 43.2 N/mm 2 d1 × t 36 × 45

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

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Design of Machine Element

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.

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Design of Machine Element

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 W = P×

Or

π 2 D 4

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

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Design of Machine Element

i.e. if

W= 2 τ

induced

π 2 d p × τ induced 4

< τ

permissible 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

outer diameter of boss   Bending moment M = P  b −  2   I = Using the relation σ =

t 1 (b1 ) 3 12

y = b1/2

M y I

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.

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Design of Machine Element

Solution : Step 1 : To find W :

From the pressure we find W. W = pressure x area 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

19

W = 1×

π (70) 2 4

Design of Machine Element

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

lp

= 15 mm

dp = 12 mm

Checking in double shear : W=2

π 2 d p τ induced 4

3848.451 = 2

π (12) 2 τinduced 4

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 outer diameter of boss   M = P b −  2   32   = 349.88 600 −  2  M = 274282.4 I= =

2003-04

1 (b1 ) 3 t 12 1 5.333 4 (4t ) 3 t = t 12

σ=

20

M .y F

Design of Machine Element

y= 70 =

b = 2t 2

σ=

102855.9 t3

274282.4 × 2t 5.333 t 4

t = 11.368 b1 = 48 mm

t = 12 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 where

=

M

y I

M=Pxb = 349.854 x 800 y = b1/2 = 12/2 = 6 I =

1 1 (32) 3 × 15 − (16) 3 × 15 12 12 1  + 2 x  (8)12 × 12 + 12 × 8 × (20) 2  12 

= 113664 mm4 σinduced

=

349.854 × 800 × 6 = 59.09 MPa 113664

as σinduced < σpermissible ; hence lever safe in bending.

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Design of Machine Element

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 :

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Design of Machine Element

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

the relation. T=

π × d 3 × (τ permissible )shaft 16

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

Step 2 :

(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) =

4 d +3 150

or number of bolts can be taken as. n =3 for shaft dia. upto 40 mm. n =4 for shaft dia. upto 100 mm. n =6 for shaft dia. upto 180 mm. n =8 for shaft dia. upto 230 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

(b) for rectangular key

w=

d 4

t=

d 4

w=

d 4

t=

d 6

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=

  d 4  π × D 3 1 −    × (τ induced )hub.  D  16  

We find (τ induced )hub 2003-04

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Design of Machine Element

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) = (τ induced ) flange × π D × t f ×

∴

T=

D 2

π D2 × t f × (τ induced ) flange 2

Using the above relation we find (τ induced ) flange 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) = w × l ×

d × (τ induced )key 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) =

t d × l × × (σ induced )key 2 2

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.

∴

2003-04

Load on each bolt = τbolt ×

π 2 d1 4

24

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

Design of Machine Element

Total load on all bolts = τbolt × Torque (T) = τ bolt ×

Hence,

π 4

π × d12 × n 4

× d 12 × n ×

D1 2

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. ∴

D1 × (σ induced )bolt 2

T = n n × d1 × t f ×

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) =

P × 60 15 × 10 6 × 60 ∴Tmean = 2 × π × 200 2π n

Tmean = 716197.2439 Nmm. Since maximum Torque exceeds by 25% ∴

Tmax = 1.25 x 716197.2439 = 895246.5549 Nmm.

Now

Tmax. =

∴ ∴ Step II :

(a) 2003-04

π × d 3 × (τ permissible )shaft 16

895246.5549 =

π × d 3 × 40 16

d = 48.486 mm.

d = 50 mm.

Using empirical relations find all other dimensions of the flange coupling. Outside diameter of the hub (D) = 2d = 100 mm. 25

Design of Machine Element

(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, (σ permissible )key

= 2 (τ permissible )key

∴We choose a square key. ∴

∴

w=

d 50 = = 12.5 mm. 4 4

t =

d 50 = = 12.5 mm. 4 4

w = t = 14 mm

Length of key (l ) = length of hub (L) = 75 mm Step IV : (1)

Design check for hub :

The hub is checked considering it as a hollow shaft :   d 4  π 3 × D 1 −    × (τ induced )hub T=  D  16     50  4  π 3 (100) 1 −   (τ induced )hub 895246.5549 = 16   100  

(τ induced )hub

= 4.863 Mpa.

Since τinduced is less than τpermissible hence design of hub is safe. (2)

Design check for flange :

Torque (T) =

πD 2 × t f × (τ induced ) flange 2

895246.5549 =

(τ induced ) flange

π (100)2 × 25 × (τ induced ) flange 2

= 2.2797 Mpa.

Since τinduced is less than τpermissible hence design of flange is safe. Step V : 2003-04

Design check for key : 26

Design of Machine Element

Dimensions of the key are checked in shear and crushing. (a) For shear : Torque (T) = w × l ×

d × (τ induced )key 2

895246.5549 = 14 × 75 ×

(τ induced )key

50 × (τ induced )key 2

= 34.104 Mpa

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

For crushing : Torque (T) = 895246.5549 =

(σ induced )key

t d × l × × (σ induced )key 2 2 14 × 75 × 50 × (σ induced )key 4

= 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. (a)

Design for shear : (Torque) T = τ bolt × 895246.5549 = 30 ×

π 4

× d 12 × n ×

D1 2

π 2 150 d1 × ×4 4 2

d12 = 126.65147 mm. 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

27

Design of Machine Element

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 Plain carbon steel material

(1) Allowable

for bolt

20 Mpa

100 Mpa

80 Mpa

60 Mpa

-----

60 Mpa

10 Mpa

60 Mpa

40 MPa

Tensile stress (2) Allowable compressive stress (3) Allowable shear stress

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28

Design of Machine Element

Solution : Step 1 :

To calculate shaft diameter ‘d’ Torque (T) =

P × 60 2π n

Tmean =

11 × 10 6 × 60 2 × π × 500

Tmean = 210084.52 Since maximum Torque exceeds by 25% T = 1.25 x Tmean = 1.25 x 210084.52 = 262605.6561 Nmm Now

π × d 3 × (τ permissible )shaft 16

T= 262605.6561 =

π × d 3 × 60 16

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 :

Since,

∴

On the basis of shaft diameter we decide dimensions of the key

(σ

)

permissible key

< 2 (τ permissible )key

We choose a rectangular key ∴

w=

d 30 = = 7.5 mm = 8 mm. 4 4

t=

d 30 = = 5 mm. 6 6

w = 8 mm. Length of key (l) = Length of hub (L) = 45 mm.

2003-04

29

Design of Machine Element

Step IV : (1)

Design check for hub :

The hub is checked considering it as a hollow shaft : T=

  d 4  π × D 3 1 −    × (τ induced )hub  D  16     30  4  π × D 3 1 −    × (τ induced )hub   60   16  

262605.6561

=

(τ induced )hub

= 6.6046 Mpa

Since τinduced is less than τpermissible hence design of hub is safe. (2)

Design check for flange :

Torque (T) =

π D2 × t f × (τ induced ) flange 2 π × 60 2 × 15 × (τ induced ) flange 2

262605.6561

=

(τ induced ) flange

= 3.0959 Mpa.

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

Design check for key :

Dimensions of the key are checked in shear and crushing. (a) For shear : Torque (T) = w × l × 262605.6561

(τ induced )

d × (τ induced )key 2

= 45 × 8 ×

30 × (τ induced )key 2

= 48.63 Mpa.

Since τinduced is less than τpermissible hence key is safe in shearing. (b) For crushing : Torque (T) = 262605.6561

(σ induced ) 2003-04

=

t d × l × × (σ induced )key 2 2 5 30 × 45 × × (σ induced )key 2 2

= 155.618 Mpa. > 100 30

Design of Machine Element

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 = t × 45 ×

30 × 100 4

t = 7.7809 = 8 mm. Hence the key dimensions are w × t × l = 8 × 8 × 45 Step VI :

Design for bolts :

The bolts are designed for shear and checked for crushing. (a) Design for shear : Torque (T) = τ bolt × 262605.6561 = d1

π 4

× d 12 × n ×

D1 2

90 π × d12 × 40 × × 3 4 2

= 7.8688 mm.

(b) Check in crushing : T = d1 × t f × σ × 262605.6561 = 7.8688 × 15 ×

D1 ×n 2

90 × 3 × σ induced 2

σinduced = 16.498 Mpa. As σinduced is less than σpermissible hence bolt is safe in crushing.

2003-04

31

Design of Machine Element

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

32

Design of Machine Element

Step II : Torque Transmitted :

Using the relation, (Torque) T =

Power (P) =

2π NT 60

P × 60 20 × 10 6 × 60 = 2 π N 2 × 3.142 × 400

T = 477464.829 Nmm

Since torque is transmitted by a belt drive, ∴

Torque = ( T1 – T2 ) r

∴

477464.829 = (T1 – T2) x 200

∴

T1 – T2

also ∴

= 2387.3241

T1 = 2 .5 T2

(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

2003-04

= 2785.211 N

33

and

RA = 2785.211 N

Design of Machine Element

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.

(k t × T ) 2 + (k b × M ) 2 =

π 3 d × τ permissible 16

∴ (1.25 × 477464.829) 2 + (1.5 × 1392605.644) = ∴

d3 = 204897.027

∴

d = 58.9538 mm.

π 16

d 3 × 54

…………………

2003-04

34