machine design md by s k mondal t&q

Design of Joint S K Mondals Chapter 1

1. Design of Joint

Theory at a glance (IES, IAS, GATE & PSU)

Cotters In machinery, the general term shaft refers to a member, usually of circular cross-

section, which supports gears, sprockets, wheels, rotors, etc., and which is subjected to torsion and to transverse or axial loads acting singly or in combination.

An axle is a non-rotating member that supports wheels, pulleys, and carries no torque.

A spindle is a short shaft. Terms such as line-shaft, head-shaft, stub shaft, transmission shaft, countershaft, and flexible shaft are names associated with special usage.

A cotter is a flat wedge-shaped piece of steel as shown in figure below. This is used to connect rigidly two rods which transmit motion in the axial direction, without rotation. These joints may be subjected to tensile or compressive forces along the axes of the rods. Examples of cotter joint connections are: connection of piston rod to the crosshead of a steam engine, valve rod and its stem etc.

Figure- A typical cotter with a taper on one side only

A typical cotter joint is as shown in figure below. One of the rods has a socket end into which the other rod is inserted and the cotter is driven into a slot, made in both the socket and the rod. The cotter tapers in width (usually 1:24) on one side only and when this is driven in, the rod is forced into the socket. However, if the taper is provided on both the edges it must be less than the sum of the friction angles for both the edges to make it self locking i.e. + < +1 2 1 2 where 1 2, are the angles of taper on the rod edge and socket edge of the cotter respectively and 1 2, are the corresponding angles of friction. This also means that if taper is given on

Page 1 of 263


one side only then < +1 2 for self locking. Clearances between the cotter and slots in the rod end and socket allows the driven cotter to draw together the two parts of the joint until the socket end comes in contact with the cotter on the rod end.

t

3d 1d 2dd

Figure- Cross-sectional views of a typical cotter joint

Figure- An isometric view of a typical cotter joint

Design of a cotter joint If the allowable stresses in tension, compression and shear for the socket, rod and cotter be

,t c and respectively, assuming that they are all made of the same material, we may write the following failure criteria:

1. Tension failure of rod at diameter d

=24 t

d P

Fig. Tension failure of the rod

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2. Tension failure of rod across slot

=

21 14 t

d d t P

Fig. Tension failure of rod across slot

3. Tensile failure of socket across slot

( ) ( ) = 2 22 1 2 14 td d d d t P

Fig. Tensile failure of socket across slot

4. Shear failure of cotter

=2bt P

Fig. Shear failure of cotter

5. Shear failure of rod end

=1 12l d P

Fig. Shear failure of rod end

6. Shear failure of socket end ( ) =3 12l d d P

Fig. Shear failure of socket end

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7. Crushing failure of rod or cotter

=1 cd t P

Fig. Crushing failure of rod or cotter

8. Crushing failure of socket or rod ( ) =3 1 cd d t P

Fig. Crushing failure of socket or rod

9. Crushing failure of collar

( ) =2 24 14 cd d P

Fig. Crushing failure of collar

10. Shear failure of collar

=1 1d t P

Fig. Shear failure of collar

Cotters may bend when driven into position. When this occurs, the bending moment cannot be correctly estimated since the pressure distribution is not known. However, if we assume a triangular pressure distribution over the rod, as shown in figure below, we may approximate the loading as shown in figure below.

Page 4 of 263


Figure-Bending of the cotter

This gives maximum bending moment = +

3 1 1

2 6 4d d dP and

The bending stress, ( ) + + = =

3 1 3 11 1

3 2

32 6 4 2 6 4

12b

d d d dd dP b P

tb tb

Tightening of cotter introduces initial stresses which are again difficult to estimate. Sometimes therefore it is necessary to use empirical proportions to design the joint. Some typical proportions are given below:

A design based on empirical relation may be checked using the formulae based on failure Mechanisms. Question: Design a typical cotter joint to transmit a load of 50 kN in tension or compression.

Consider that the rod, socket and cotter are all made of a material with the following allowable stresses:

Allowable tensile stress ( y )= 150 MPa

Allowable crushing stress (c) = 110 MPa

Allowable shear stress (y) = 110 MPa.

Answer: Refer to figures

d1 = 1.21.d d4 = 1.5.d

l = l1 = 0.75d

d2 = 1.75.d

t = 0.31d

t1 = 0.45d

d3 = 2.4 d

b = 1.6d

c = clearance

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t

3d 1d 2dd

Axial load ( ) = 24 yP d . On substitution this gives d=20 mm. In general Standard shaft size in mm is 6 mm to 22 mm diameter 2 mm in increment 25 mm to 60 mm diameter 5 mm in increment 60 mm to 110 mm diameter 10 mm in increment 110 mm to 140 mm diameter 15 mm in increment 140 mm to 160 mm diameter 20 mm in increment 500 mm to 600 mm diameter 30 mm in increment We therefore choose a suitable rod size to be 25 mm. Refer to figure

For tension failure across slot =

214 y

d d t P . This gives

d1t = 1.58x10-4m2. From empirical relations we may take t=0.4d i.e. 10 mm and this gives d1= 15.8 mm. Maintaining the proportion let d1= 1.2 d = 30 mm.

Refer to figure

The tensile failure of socket across slot ( ) = 2 22 1 2 14 y

d d d d t P

This gives d2 = 37 mm. Let d2 = 40 mm Refer to figure above For shear failure of cotter 2bt = P. On substitution this gives b = 22.72 mm. Let b = 25 mm. Refer to figure For shear failure of rod end 2l1d1 = P and this gives l1 = 7.57 mm. Let l1 = 10 mm. Refer to figure For shear failure of socket end 2l(d2-d1) = P and this gives l = 22.72 mm. Let l = 25 mm.

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Refer to figure For crushing failure of socket or rod (d3-d1)tc = P. This gives d3 = 75.5 mm. Let d3 = 77 mm.

Refer to figure

For crushing failure of collar ( ) =2 24 14 cd d P . On substitution this gives d4 = 38.4 mm. Let d4 = 40 mm. Refer to figure For shear failure of collar d1t1 = P which gives t1 = 4.8 mm. Let t1 = 5 mm. Therefore the final chosen values of dimensions are d = 25 mm; d1 = 30 mm; d2 = 40 mm; d3 = 77 mm; d4 = 40 mm; t = 10 mm; t1 = 5 mm; l = 25 mm; l1 = 10 mm; b = 27 mm.

Knuckle Joint A knuckle joint (as shown in figure below) is used to connect two rods under tensile load. This joint permits angular misalignment of the rods and may take compressive load if it is guided.

d

2t

3d2d

1d

t1t

1t 2td

Figure- A typical knuckle joint

These joints are used for different types of connections e.g. tie rods, tension links in bridge structure. In this, one of the rods has an eye at the rod end and the other one is forked with eyes at both the legs. A pin (knuckle pin) is inserted through the rod-end eye and fork-end eyes and is secured by a collar and a split pin. Normally, empirical relations are available to find different dimensions of the joint and they are safe from design point of view. The proportions are given in the figure above. d = diameter of rod

d1 = d t = 1.25d

d2 = 2d t1 = 0.75d d3 = 1.5.d t2 = 0.5d

Mean diameter of the split pin = 0.25 d

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However, failures analysis may be carried out for checking. The analyses are shown below assuming the same materials for the rods and pins and the yield stresses in tension, compression and shear are given by t, c and . 1. Failure of rod in tension:

=24 t

d P 2. Failure of knuckle pin in double sheer:

=212 4 d P 3. Failure of knuckle pin in bending (if the pin is loose in the fork)

Assuming a triangular pressure distribution on the pin, the loading on the pin is shown in figure below.

Equating the maximum bending stress to tensile or compressive yield stress we have

+ =1

31

163 4

t

t tP

d

Figure- Bending of a knuckle pin

4. Failure of rod eye in shear:

(d2 - d1) t = P 5. Failure of rod eye in crushing:

d1tc = P 6. Failure of rod eye in tension:

(d2 - d1)tt = P 7. Failure of forked end in shear:

2(d2 - d1)t1 = P

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8. Failure of forked end in tension: 2(d2 - d1)t1t = P

9. Failure of forked end in crushing:

2d1t1c = P The design may be carried out using the empirical proportions and then the analytical relations may be used as checks.

For example using the 2nd equation we have = 212Pd

. We may now put value of d1 from

empirical relation and then find FACTOR OF SAFTEY, (F.S.) =

y which should be more

than one. Q. Two mild steel rods are connected by a knuckle joint to transmit an axial

force of 100 kN. Design the joint completely assuming the working stresses for both the pin and rod materials to be 100 MPa in tension, 65 MPa in shear and 150 MPa in crushing.

d

2t

3d2d

1d

t1t

1t 2t

d

Refer to figure above

For failure of rod in tension, = 24 y

P d . On substituting P = 100 kN, y = 100 MPa we have d= 35.6 mm. Let us choose the rod diameter d = 40 mm which is the next standard size.

We may now use the empirical relations to find the necessary dimensions and then check the failure criteria.

d1= 40 mm t= 50 mm d2 = 80 mm t1= 30 mm; d3 = 60 mm t2= 20 mm; Split pin diameter = 0.25 d1 = 10 mm To check the failure modes:

Page 9 of 263


1. Failure of knuckle pin in shear: =

212 4 y

P d ,which gives y = 39.8

MPa. This is less than the yield shear stress.

2. For failure of knuckle pin in bending:

+ =1

31

163 4

y

t tP

d. On substitution

this gives y = 179 MPa which is more than the allowable tensile yield stress of 100 MPa. We therefore increase the knuckle pin diameter to 55 mm which gives y = 69 MPa that is well within the tensile yield stress.

3. For failure of rod eye in shear: (d2 - d1)t = P. On substitution d1 = 55mm = 80

MPa which exceeds the yield shear stress of 65 MPa. So d2 should be at least 85.8 mm. Let d2 be 90 mm.

4. for failure of rod eye in crushing: d1 t c = P which gives c = 36.36 MPa that is

well within the crushing strength of 150 MPa.

5. Failure of rod eye in tension: (d2 - d1)tt = P. Tensile stress developed at the rod eye is then t = 57.14 MPa which is safe.

6. Failure of forked end in shear: 2(d2-d1)t1 = P. Thus shear stress developed in

the forked end is = 47.61 MPa which is safe.

7. Failure of forked end in tension: 2(d2 -d1)t1y = P. Tensile strength developed in the forked end is then y= 47.61 MPa which is safe.

8. Failure of forked end in crushing: 2d1t1c = P which gives the crushing stress

developed in the forked end as c = 42 MPa. This is well within the crushing strength of 150 MPa.

Therefore the final chosen values of dimensions are:

d1 = 55 mm t = 50 mm d2 = 90 mm t1= 30 mm; and d = 40 mm d3 = 60 mm t2= 20 mm;

Keys In the assembly of pulley, key and shaft, Key is made the Weakest so that it is cheap and easy to replace in case of failure.

Page 10 of 263

S K M

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Page 11 of 263

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Page 12 of 263

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Page 13 of 263

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Feather feather key

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Page 14 of 263


1. It accommodates itself to any taper in the hub or boss of the mating piece. 2. It is useful on tapering shaft ends. Its extra depth in the shaft prevents any tendency to turn over in its keyway.

The main dis-advantages of a woodruff key are as follows: 1. The depth of the keyway weakens the shaft. 2. It can not be used as a feather.

Circular (Pin) Keys

Significantly lower stress concentration factors result from this type of key as compared to parallel or tapered keys. A ball end mill can be used to make the circular key seat.

Splines Splines are essentially stub by gear teeth formed on the outside of the shaft and on the inside of the hub of the load-transmitting component. Splines are generally much more expensive to manufacture than keys, and are usually not necessary for simple torque transmission. They are typically used to transfer high torques. One feature of a spline is that it can be made with a reasonably loose slip fit to allow for large axial motion between the shaft and component while still transmitting torque. This is useful for connecting two shafts where relative motion between them is common, such as in connecting a power takeoff (PTO) shaft of a tractor to an implement. Stress concentration factors are greatest where the spline ends and blends into the shaft, but are generally quite moderate.

Page 15 of 263

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Page 16 of 263


Spline Manufacturing Methods Splines are either cut (machined) or rolled. Rolled splines are stronger than cut splines due to the cold working of the metal. Nitriding is common to achieve very hard surfaces which reduce wear.

Welded joints A welded joint is a permanent joint which is obtained by the fusion of the edges of the two parts to be joined together, with or without the application of pressure and a filler material. The heat required for the fusion of the material may be obtained by burning of gas (in case of gas welding) or by an electric arc (in case of electric arc welding). The latter method is extensively used because of greater speed of welding. Welding is extensively used in fabrication as an alternative method for casting or forging and as a replacement for bolted and riveted joints. It is also used as a repair medium e.g. to reunite metal at a crack, to build up a small part that has broken off such as gear tooth or to repair a worn surface such as a bearing surface.

Types of welded joints: Welded joints are primarily of two kinds: (a) Lap or fillet joint Obtained by overlapping the plates and welding their edges. The fillet joints may be single transverse fillet, double transverse fillet or parallel fillet joints (see figure below).

Page 17 of 263

S

(b

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Page 18 of 263

S K M

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Page 19 of 263

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xample: If with convex c

Design he main faiutt joint is

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Page 20 of 263

S K M

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Page 21 of 263

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n designing bove designgure below)

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Page 22 of 263

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2d

e selected fr

max

to be wee subjectedear strengt

ing are to bs in the par

Hence the mi

ngth of the

A usual allhe plate thi

ck are to re subjectetensile stre

ransverse) he throat ars half of tenarries a load

635 10

nt

rom the equ

lded to and to a loadth to be 56

be provided.rallel fillet w

inimum len

weld is to

owance of ickness)

be weldeed to a loaess 70 MPa

joint the wrea. Since tensile strengd of 35 kN a

Chapte

ation

nother plad of 50 kN. MPa.

Each line sweld is p

lt, w

ngth of the w

o be provid

12.5 mm is

d by meaad of 70 kNa.

eld is desensile stren

gth, i.e. , sand the size

er 1

ate by . Find

shares where t

weld is

ded as

s kept.

ans of N, find

signed ngth is = 35 of the

Page 23 of 263


Adding an allowance of 12.5 mm for stopping and starting of the bead, the length of the weld should be 154 mm.

Q. A 50 mm diameter solid shaft is to be welded to a flat plate and is required to

carry a torque of 1500 Nm. If fillet joint is used foe welding what will be the minimum size of the weld when working shear stress is 56 MPa.

Solution. According to the procedure for calculating strength in the weld joint,

= sthroatT

t d22

Where the symbols have usual significance. For given data, the throat thickness is 6.8 mm. Assuming equal base and height of the fillet the minimum size is 9.6 mm. Therefore a fillet weld of size 10 mm will have to be used.

Q. A strap of mild steel is welded to a plate as shown in the following figure.

Check whether the weld size is safe or not when the joint is subjected to completely reversed load of 5 kN.

Fig.

Solution. As shown in the figure the joint is a parallel fillet joint with leg size as 9 mm and

the welding is done on both sides of the strap. Hence the total weld length is 2(50) = 100 mm.

In order to calculate the design stress the following data are used

k1 = 2.7 (parallel fillet joint, refer table 3) (there is required a table to solve the problem)

w = 0.9 cm K = -1 for completely reversed loading The value of the allowable fatigue stress (assuming the weld to be a line) is then

1358 0.9 214.8 kgf/cm = 214800N/m

1.5 = = (approx).The design stress is Therefore

Page 24 of 263

S K M =1, 2142.dfluctuating

Thread Bothseop Scinin

Bolts

Machi

Monda=800 79556

.7g load allow

ded fas

olt - Threahrough holeecured by tpposite the h

crew - Thrnserted thronto a threade

ne Scre

als N/m . Since

wable for the

steners

aded fastens in matingtightening head of the

readed fastough a holeed hole in a

ews

Des

e the total

e joint is 795

ner designeg members a nut frombolt.

tener desige in one mea mating me

sign o

l length of

55.6 N. The

ed to pass and to be

m the end

gned to be ember and ember.

of Join

f the weld

e joint is the

nt

d is 0.1 m

erefore safe.

Fig.

Fig.

Chaptem, the max

er 1 ximum

Page 25 of 263

S

S

T

ThT Tnu

S K Mo

Sheet M

Thread

he pitch linThread

hread Seriumber of th

ondal

Metal an

F

Profiles

e or diametSeriesies - groups

hreads per in

s

nd Lag S

Figure-Sheet

s

er is located

s of diametench applied

Desig

Screws

t metal scre

d at the h

er-pitch com to a specific

gn of

s

ews are often

height of the

mbinations dc diameter.

Joint

n self-tappi

e theoretical

distinguishe

C

ng.

l sharp v-th

ed from eac

Chapter

hread profile

ch other by

r 1

e.

the

Page 26 of 263


M-Series Metric system of diameters, pitches, and tolerance/allowances.

Tensile Stress Area The average axial stress in a fastener is computed using a tensile stress area.

=avet

FA

+ = 2

4 2r p

t

D DA

F Axial Force Dr Root Diameter Dp Pitch Diameter At Tensile Stress Area ave Average axial stress

Length of Engagement (Equal Strength Materials) If the internal thread and external thread material have the same strength, then

Tensile Strength (External Thread)

maxt

t

FA

=

Shear Strength (Internal Thread)

max

,0.5 t

s i e

FA L

=

Where

max ,0.5t t t s i eF A A L = = =

,

2 te

s i

ALA

Bolt/Nut Design Philosophy ANSI standard bolts and nuts of equal grades are designed to have the bolt fail before the threads in the nut are stripped. The engineer designing a machine element is responsible for determining how something should fail taking into account the safety of the operators and public. Length of engagement is an important consideration in designing machine elements with machine screws.

Objective Questions (For GATE, IES & IAS)

Page 27 of 263


Previous 20-Years GATE Questions Keys GATE-1. Square key of side "d/4" each and length l is used to transmit torque "T"

from the shaft of diameter "d" to the hub of a pulley. Assuming the length of the key to be equal to the thickness of the pulley, the average shear stress developed in the key is given by [GATE-2003]

2 2 34T 16T 8T 16T(a) (b) (c) (d)ld ld ld d

GATE-1. Ans. (c) If a square key of sides d/4 is used then. In that case, for shear failure we

have xd dl T4 2

=

x 28Torld

= [Where x is the yield stress in shear and l is the key length.] GATE-2. A key connecting a flange coupling to a shaft is likely to fail in[GATE-1995]

(a) Shear (b) tension (c) torsion (d) bending GATE-2. Ans. (a) Shear is the dominant stress on the key

Welded joints GATE-3. A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN

along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is [GATE-2006] (a) 2.4 (b) 3.4 (c) 4.8 (d) 6.8

GATE-3. Ans. (b) Strength of materialFactorofsafety

Actual load or strength onmaterial=

3

6o

200(in MPa) 200(in MPa) 3.458.91(in MPa)15 10

660 10 (in MPa)cos45

= =

Threaded fasteners

Page 28 of 263


GATE-4. A threaded nut of M16, ISO metric type, having 2 mm pitch with a pitch diameter of 14.701 mm is to be checked for its pitch diameter using two or three numbers of balls or rollers of the following sizes [GATE-2003] (a) Rollers of 2 mm (b) Rollers of 1.155 mm (c) Balls of 2 mm (d) Balls of 1.155 mm

GATE-4. Ans. (b)

Previous 20-Years IES Questions

Cotters IES-1. Assertion (A): A cotter joint is used to rigidly connect two coaxial rods carrying

tensile load. Reason (R): Taper in the cotter is provided to facilitate its removal when it fails due to shear. [IES-2008] (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-1. Ans. (b) 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 and it may be increased up to 1 in 8, if a locking device is provided. The locking device may be a taper pin or a set screw used on the lower end of the cotter. The cotter is usually made of mild steel or wrought iron. A cotter joint is a temporary fastening and is used to connect rigidly two co-axial rods or bars which are subjected to axial tensile or compressive forces.

IES-2. Match List I with List II and select the correct answer using the code given

below the Lists: [IES 2007] List I List II

(Application) (Joint) A. Boiler shell 1. Cotter joint B. Marine shaft coupling 2. Knuckle joint C. Crosshead and piston road 3. Riveted joint D. Automobile gear box 4. Splines

(gears to shaft) 5. Bolted Joint Code: A B C D A B C D

(a) 1 4 2 5 (b) 3 5 1 4 (c) 1 5 2 4 (d) 3 4 1 5 IES-2. Ans. (b) IES-3. Match List-I (Parts to be joined) with List-II (Type of Joint) and select the

correct answer using the code given below: [IES-2006] List-I List -II

A. Two rods having relative axial motion 1. Pin Joint B. Strap end of the connecting rod 2. Knuckle Joint C. Piston rod and cross head 3. Gib and Cotter Joint D. Links of four-bar chain 4. Cotter Joint

A B C D A B C D (a) 1 3 4 2 (b) 2 4 3 1

Page 29 of 263

SIE IE

IE IE

IE

IE

IE IE

IE

S K Mo(c)

ES-3. Ans. (

ES-4. MaLiA. B.C.D.

Code(a)(c)

ES-4. Ans. (

ES-5. In(a)(c)

ES-5. Ans. (

ES-6. MacoLiA.

B.C D.

(a)(c)

ES-6. Ans. (

ES-7. Inthsh(a)

ES-7. Ans. (

ondal) 1 4 (d)

atch List Iist I (Types Riveted jo Welded jo Bolted joi. Knuckle js: A B ) 4 3 ) 2 4 (c)

n a gib and ) Single she) Single shea(d)

atch List Iorrect answist I Bolts in b

cylinder Cotters inRivets in l. Bolts hold

a flange cA B

) 4 1 ) 3 1 (a)

n a cotter hickness ishearing str) 120 N/mm(d) It i

Shear stre

s 3

I with List s of joints) oint oint int joint

C 2 3

cotter joinar only ar and crush

I (Items inwer using t

bolted joincover plat

n cotter joilap joints ding two flcoupling

C 3 4

joint, the s 12 mm. Tress develo

m2 is a case of d

ess Load2 Area

=

Desig

2 (d)

II and sele List 1. P 2. St 3. Lo 4. FD 1 (b) 1 (d)

nt, the gib hing

n joints) wthe codes g

nts of enginte int langes in

D 2 (b) 2 (d)

width of The load aoped in the(b) 100 N/mdouble shea

360 10a 2 50 12

=

gn of

2 3

ect the cort II (An elein trap ock washeillet

A B2 32 4

and cotter(b) doubl(d) doubl

with List IIgiven below

Lne 1

2 3 4

A B4 2 3 2

the cotteracting on te cotter? mm2 (car.

250N / mm2

=

Joint 4

rrect answement of th

er

B C 4 1

r are subjele shear onlle shear and

I (Type of w the Lists

List II .Doubletra

. Torsiona Single tra. Tension

B C 3 4

r at the cethe cotter c) 75 N/mm2

2

C1

er. he joint)

D 1 3

ected to ly d crushing

failure) ans:

ansverse sh

l shear ansverse sh

D 1 1

entre is 50r is 60 kN. 2 (d)

Chapter

[IES-199

[IES-200

nd select t [IES-20

hear

hears

0 mm and . What is t [IES-200) 50 N/mm2

r 1

94]

06]

the 04]

its the 04]

Page 30 of 263


IES-8. The spigot of a cotter joint has a diameter D and carries a slot for cotter. The permissible crushing stress is x times the permissible tensile stress for the material of spigot where x > 1. The joint carries an axial load P. Which one of the following equations will give the diameter of the spigot?

[IES-2001]

(a) t

P x 1D 2x= (b) t

P x 1D 2x+= (c) t

2 P x 1Dx+= (d) t

2PD x 1= + IES-8. Ans. (b) IES-9. Match List-l (Machine element) with List-II (Cause of failure) and select

the correct answer using the codes given below the lists: [IES-1998] List-I List-II

A. Axle 1. Shear stress B. Cotter 2. Tensile/compressive stress C. Connecting rod 3. Wear D. Journal bearing 4. Bending stress Code: A B C D A B C D

(a) 1 4 2 3 (b) 4 1 2 3 (c) 4 1 3 2 (d) 1 4 3 2

IES-9. Ans. (b) In machinery, the general term shaft refers to a member, usually of circular

cross-section, which supports gears, sprockets, wheels, rotors, etc., and which is subjected to torsion and to transverse or axial loads acting singly or in combination.

An axle is a non-rotating member that supports wheels, pulleys, and carries no torque.

A spindle is a short shaft. Terms such as line-shaft, head-shaft, stub shaft, transmission shaft, countershaft, and flexible shaft are names associated with special usage.

IES-10. The piston rod and the crosshead in a steam engine are usually connected

by means of [IES-2003] (a) Cotter joint (b) Knuckle joint (c) Ball joint (d) Universal joint

IES-10. Ans. (a) IES-11. A cotter joint is used when no relative motion is permitted between the

rods joined by the cotter. It is capable of transmitting [IES-2002] (a) Twisting moment (b) an axial tensile as well as compressive load (c) The bending moment (d) only compressive axial load

IES-11. Ans. (b) IES-12. Match List I with List II and select the correct answer using the codes

given below the lists: [IES-1995] List I List II

(Different types of detachable joints) (Specific use of these detachable joints) A. Cotter joint 1. Tie rod of a wall crane B. Knuckle joint 2. Suspension bridges C. Suspension link joint 3. Diagonal stays in boiler D. Turn buckle (adjustable joint) 4. Cross-head of a steam engine Codes: A B C D A B C D

(a) 4 2 3 1 (b) 4 3 2 1 (c) 3 2 1 4 (d) 2 1 4 3

Page 31 of 263

SIE IE

IE

IE

IE

KIE

S K MoES-12. Ans.

ES-13. Magiv

A. B.C.D.

1. 2. 3. 4. 5. Co

ES-13. AnstwmaCoetc

ES-14. AsthupReloa(a)(b)(c)(d)

ES-14. Ans.

Keys ES-15. In

(a)

ondal. (a)

atch List ven below

List I (Ty Cotter joi Knuckle j Turn buck. Riveted jo

List II (M Connects Rigidly co Connects Permanen Connects odes: A

(a) 5 (c) 5

. (b) A cottwo rods whicay be subjeconnection ofc are examp

ssertion (Ae connectin

pon the direceason (R): ading and re) Both A and) Both A and) A is true b) A is false b. (b)

n the assem) pulley is m

s

I with Lis the lists:

ype of jointnt oint kle oint

Mode of jointwo rods o

onnects twtwo rods h

nt fluid-tigtwo shafts

B 1 3

er is a flat ch transmit cted to tensif piston rod ples of cotter

A): When thng rods eithction of rota A turn bucequiring subd R are indid R are indiut R is falsebut R is true

mbly of pulmade the we

Desig

st II and s t)

nting memor bars pero members

having threht joint be

s and transC D 3 2 2 4 wedge-shap motion in tile or compr to the crossr joint.

he coupler oher move clation of the ckle is usedbsequent adividually truividually true e

Fig. Tu

ley, key aneakest

gn of

select the

mbers) rmitting sms eaded end

etween twosmits torqu

A(b) 2 (d) 2

ped piece of the axial dirressive forces-head of a

of a turn buoser or mov coupler. d to connectdjustment foue and R is ue but R is n

urnbuckle

nd shaft (b) key is

Joint

correct an

mall amoun

s o flat pieceue

A B 1 3 steel. This

rection, withes along thesteam engin

uckle is turve away fro t two roundor tightenin the correct not the corr

s made the

C

nswer usin

nt of flexib

es

C D 3 4 1 4

is used to chout rotatioe axes of thene, valve ro

ned in one om each oth

d rods subjeng or looseni explanationrect explana

[IEweakest

Chapter

ng the cod [IES-19

bility

connect rigin. These joi

e rods. od and its st

direction bher depend [IES-199

ected to tening. n of A ation of A

S-1993; 199

r 1

des 93]

idly ints

tem

both ding 96] sile

98]

Page 32 of 263

S K M

IES-15. A

IES-16.

IES-16. A

IES-17.

IES-17. A IES-18.

IES-18. A IES-19.

IES-19. A

Monda(c) Key is m

Ans. (b) Keyfailure.

Match Licorrect anList-I A. WoodruB. KennedC. FeatheD. Flat keCode: A

(a) (c)

Ans. (b) A febe fastened

Match Ligiven beloList-I (KeyA. Gib heaB. WoodruC. ParalleD. SplinesCode: A

(a) (c)

Ans. (c)

A spur gerectangul(a) Shear s(c) Both sh

Ans. (c) Key

Assertioncapacity anReason (R(a) Both A (b) Both A (c) A is tru(d) A is fals

Ans. (d)

als made the str

y is made t

st-I (Typenswer usin uff key dy key r key

ey A B 2 3 2 3

eather key isd either to t

st-I with ow the listy/splines) ad key uff key el key s A B 1 2 2 1

ear transmlar sectionstress alone hear and beay develops bo

(A): The nd to increaR): Highly lo and R are i and R are ie but R is fase but R is t

Des

rongest

the weakest

of keys) ng the code

L 1 2 3 4C D1 4 4 1

s used whenhe hub or th

Fig.

List-II ands:

C D3 4 3 4

mitting pown. The type aring stresseoth shear an

effect of kese its torsioocalized strendividuallyndividuallyalse true

sign o

(d) astreng

t so that it

with Listes given be

List-II . Loose fitt. Heavy du. Self-align. Normal in

D (b) (d) n one compohe shaft and

feather key

d select th List-I1. Sel2. Fac3. Mo4. Ax

D (b) (d)

wer is con (s) of stres

(b) bees (d) shnd bearing s

eyways on onal rigidityesses occur

y true and Ry true but R

of Join ll the thregth is cheap an

t-II (Charaelow the Li

ting, light uty ning ndustrial u

A B 3 2 3 2

onent slidesd the keywa

y

he correct

II (Applicalf aligning cilitates re

ostly used ial movem

A B 1 2 2 1

nnected tosses develo

earing stresshearing, beastresses.

a shaft is y. at or near t

R is the corre is not the c

nt ee are des

nd easy to r

acteristic) ists:

duty

use C 1 4

s over anothay usually h

t answer

ation) emoval

ment possib C

4 4

o the shaftoped in thes alone ring and be

to reduce

the corners ect explanatcorrect expl

Chaptesigned for

replace in c

and selec [IES-

D 4 1

her. The keyhas a sliding

using the [IES-

ble D 3 3

t with a ke key is far [IES-

ending stres

its load ca [IES-of keyways.tion of A anation of A

er 1 equal

case of

ct the -1997]

y may g fit.

code -2008]

key of re. -1995] ses.

rrying -1994] .

A

Page 33 of 263


IES-20. Which key is preferred for the condition where a large amount of impact torque is to be transmitted in both direction of rotation? [IES-1992] (a) Woodruff key (b) Feather key (c) Gib-head key (d) Tangent key

IES-20. Ans. (d) IES-21. What is sunk key made in the form of a segment of a circular disc of

uniform thickness, known as? [IES-2006] (a) Feather key (b) Kennedy key (c) Woodruff key (d) Saddle key

IES-21. Ans. (c) IES-22. What are the key functions of a master schedule? [IES-2005]

1. To generate material and capacity requirements 2. To maintain valid priorities 3. An effective capacity utilization 4. Planning the quantity and timing of output over the intermediate time horizons Select the correct answer using the code given below: (a) 1, 2 and 3 (b) 2, 3 and 4 (c) 1, 3 and 4 (d) 1, 2 and 4

IES-22. Ans. (b) IES-23. A square key of side d/4 is to be fitted on a shaft of diameter d and in the

hub of a pulley. If the material of the key and shaft is same and the two are to be equally strong in shear, what is the length of the key? [IES-2005]

(a) d2

(b) 2 d3 (c) 3 d

4 (d) 4 d

5

IES-23. Ans. (a) IES-24. Which one of the following statements is correct? [IES-2004]

While designing a parallel sunk key it is assumed that the distribution of force along the length of the key (a) Varies linearly (b) is uniform throughout (c) varies exponentially, being more at the torque input end (d) varies exponentially, being less at torque output end

IES-24. Ans. (c) Parallel sunk key. The parallel sunk keys may be of rectangular or square section uniform in width and thickness throughout. It may be noted that a parallel key is a taperless and is used where the pulley, gear or other mating piece is required to slide along the shaft. In designing a key, forces due to fit of the key are neglected and it is assumed that the distribution of forces along the length of key is uniform.

IES-25. Match List-I (Device) with List-II (Component/Accessory) and select the

correct answer using the codes given below the Lists: [IES-2003] List-I List-II (Device) (Component/Accessory)

A. Lifting machine 1. Idler of Jockey pulley B. Fibre rope drive 2. Sun wheel C. Differential gear 3. Sheave D. Belt drive 4. Power screw Codes: A B C D A B C D

(a) 4 3 1 2 (b) 3 4 1 2 (c) 4 3 2 1 (d) 3 4 2 1

IES-25. Ans. (c)

Page 34 of 263

S K M IES-26.

IES-26. A

IES-27.

IES-27. A

MondaA pulley means of key is takthe key isthe length

(a) 4

Ans. (c)

Shearing s

AssertionReason (Rof the mati(a) Both A (b) Both A (c) A is tru(d) A is fals

Ans. (b)

The main a1. It accom2. It is useftendency toThe main d1. The dept2. It can no

als is conneca rectangu

ken as d/4.s equal to th of the ke

(b)

trength of k

(A): A WooR): The Wooing piece. and R are i and R are ie but R is fase but R is t

advantages mmodates its

ful on tapero turn over dis-advantath of the keyot be used a

Des

cted to a ular sunk k. For full pthe torsiony to the di

2

3

dkey:F .4

Torque(T

Torsiona

or T d

For samed d. .l .4 2lord 2

=

=

=

odruff key isodruff key a ndividuallyndividuallyalse true

of a woodruself to any tring shaft en in its keywa

ages of a wooyway weake

as a feather.

sign o

power trakey of widpower trannal shearinameter of

(c) 2

3

3

.l

dT) =F. .2

l shearing,

16e strengthd d2 16

=

=

s an easily aaccommodat y true and Ry true but R

uff key are aaper in the nds. Its extray. odruff key aens the shaf

of Join

ansmissiondth wand lensmission, ng strengththe shaft (

4

d d.l .4 2T

dd232

=

adjustable ktes itself to

R is the corre is not the c

as follows: hub or bossra depth in t

are as followft.

nt

n shaft of ength l. T the shear

h of the sha(l/d) is

(d)

key. any taper i

ect explanatcorrect expl

s of the matthe shaft pr

ws:

Chapte

diameter The width oring strengaft. The ra [IES-

n the hub o [IES-tion of A anation of A

ing piece. revents any

er 1

d by of the gth of atio of -2003]

or boss -2003]

A

Page 35 of 263


IES-28. The key shown in the above figure is a

(a) Barth key (b) Kennedy key (c) Lewis key (d) Woodruff key

[IES-2000]

IES-28. Ans. (a) IES-29. Match List I (Keys) with List II (Characteristics) and select the correct

answer using the codes given below the Lists: [IES-2000] List I List II

A. Saddle key 1. Strong in shear and crushing B. Woodruff key 2. Withstands tension in one direction C. Tangent key 3. Transmission of power through frictional

resistance D. Kennedy key 4. Semicircular in shape Code: A B C D A B C D

(a) 3 4 1 2 (b) 4 3 2 1 (c) 4 3 1 2 (d) 3 4 2 1

IES-29. Ans. (d) IES-30. Match List-I with List-II and select the correct answer using the code

given below the Lists: [IES-2009] List-I List-II

(Description) (shape) A. Spline 1. Involute B. Roll pin 2. Semicircular C. Gib-headed key 3. Tapered on on side D. Woodruff key 4. Circular Code: A B C D A B C D

(a) 1 3 4 2 (b) 2 3 4 1 (c) 1 4 3 2 (d) 2 4 3 1

IES-30. Ans. (c) IES-31. The shearing area of a key of length 'L', breadth 'b' and depth 'h' is equal to

(a) b x h (b) Lx h (c) Lx b (d) Lx (h/2) [IES-1998] IES-31. Ans. (c)

Splines IES-32. Consider the following statements: [IES-1998]

A splined shaft is used for 1. Transmitting power 2. Holding a flywheel rigidly in position 3. Moving axially the gear wheels mounted on it 4. Mounting V-belt pulleys on it. Of these statements (a) 2 and 3 are correct (b) 1 and 4 are correct

Page 36 of 263


(c) 2 and 4 are correct (d) 1 and 3 are correct IES-32. Ans. (d)

Welded joints IES-33. In a fillet welded joint, the weakest area of the weld is [IES-2002]

(a) Toe (b) root (c) throat (d) face IES-33. Ans. (c) IES-34. A single parallel fillet weld of total length L and weld size h subjected to a

tensile load P, will have what design stress? [IES 2007] (a) Tensile and equal to P

0.707Lh (b) Tensile and equal to P

Lh

(c) Shear and equal to P0.707Lh

(d) Shear and equal to PLh

IES-34. Ans. (c)

Throat, t = h cos450 =

2

1 hv

= 0.707h

T = PLt

= P0.707Lh

IES-35. Two metal plates of thickness t and width 'w' are joined by a fillet weld of 45 as shown in given figure.

[IES-1998]

When subjected to a pulling force 'F', the stress induced in the weld will be

(a) oF

wt sin 45 (b) F

wt (c)

oFsin 45wt

(d) 2Fwt

IES-35. Ans. (a) IES-36. A butt welded joint, subjected to

tensile force P is shown in the given figure, l = length of the weld (in mm) h = throat of the butt weld (in mm) and H is the total height of weld including reinforcement. The average tensile stress t, in the weld is given by

[IES-1997]

( ) ( ) ( ) ( )t t t tP P P 2Pa b c d Hl hl 2hl Hl = = = = IES-36. Ans. (b)

Page 37 of 263


IES-37. In the welded joint shown in the given figure, if the weld at B has thicker fillets than that at A, then the load carrying capacity P, of the joint will

(a) increase (b) decrease (c) remain unaffected (d) exactly get doubled

[IES-1997]

IES-37. Ans. (c) IES-38. A double fillet welded joint with parallel fillet weld of length L and leg B is

subjected to a tensile force P. Assuming uniform stress distribution, the shear stress in the weld is given by [IES-1996]

(a) 2P B.L

(b) P 2.B.L

(c) P2.B.L

(d) 2PB.L

IES-38. Ans. (c) IES-39. The following two figures show welded joints (x x x x x indicates welds),

for the same load and same dimensions of plate and weld. [IES-1994]

The joint shown in (a) fig. I is better because the weld is in shear and the principal stress in the weld is not in line with P (b) fig. I is better because the load transfer from the tie bar to the plate is not direct (c) fig. II is better because the weld is in tension and safe stress of weld in tension is greater than that in shear (d) fig. II is better because it has less stress concentration.

IES-39. Ans. (c) Figure II is better because the weld is in tension and safe stress of weld in tension is greater than shear.

IES-40. Assertion (A): In design of double fillet welding of unsymmetrical sections with

plates subjected to axial loads lengths of parallel welds are made unequal. Reason (R): The lengths of parallel welds in fillet welding of an unsymmetrical section with a plate are so proportioned that the sum of the resisting moments of welds about the centre of gravity axis is zero. [IES-2008] (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-40. Ans. (a) Axially loaded unsymmetrical welded joints

Page 38 of 263


1

1

1 1

1 1

2 2

1 1 2 2

1 1 2 2

1 1 2 2

PA

P AP t IP t IP y P ytI y tI y

I y I y

== = =

= =

=

IES-41. Two plates are joined together by means of

single transverse and double parallel fillet welds as shown in figure given above. If the size of fillet is 5 mm and allowable shear load per mm is 300 N, what is the approximate length of each parallel fillet?

(a) 150 mm (b) 200 mm (c) 250 mm (d) 300 mm [IES-2005] IES-41. Ans. (b) ( )300 100 2l 15000 or l 200 + = = IES-42. A circular rod of diameter d is welded to a flat plate along its

circumference by fillet weld of thickness t. Assuming w as the allowable shear stress for the weld material, what is the value of the safe torque that can be transmitted? [IES-2004]

(a) 2 wd .t. (b) 2

wd .t.2

(c) 2

wd .t.

2 2 (d)

2

wd .t.2

IES-42. Ans. (b)

( )

W

W2

W W

Shear stressShear fore dt

d dTorque T dt .t2 2

= =

= =

IES-43. A circular solid rod of diameter d welded to a rigid flat plate by a circular

fillet weld of throat thickness t is subjected to a twisting moment T. The maximum shear stress induced in the weld is [IES-2003] (a) 2

Ttd (b) 2

2Ttd (c) 2

4Ttd (d) 3

2Ttd

IES-43. Ans. (b) 3 2

dT.T.r 2T2J td td

4

= = =

Page 39 of 263


IES-44. The permissible stress in a filled weld is 100 N/mm2. The fillet weld has equal leg lengths of 15 mm each. The allowable shearing load on weldment per cm length of the weld is [IES-1995] (a) 22.5 kN (b) 15.0 kN (c) 10.6 kN (d) 7.5 kN.

IES-44. Ans. (c) Load allowed = 100 x 0.707 x 10 x15 = 10.6 kN

Threaded fasteners IES-45. A force F is to be transmitted through a square-threaded power screw

into a nut. If t is the height of the nut and d is the minor diameter, then which one of the following is the average shear stress over the screw thread? [IES 2007]

(a) 2fdt (b)

Fdt (c)

F2 dt (d)

4Fdt

IES-45. Ans. (b) IES-46. Consider the case of a square-

threaded screw loaded by a nut as shown in the given figure. The value of the average shearing stress of the screw is given by (symbols have the usual meaning)

( )

r r

2F F(a) (b)d h d h

2F F(c) ddh dh

[IES-1997] IES-46. Ans. (b) IES-47. Assertion (A): Uniform-strength bolts are used for resisting impact loads.

Reason (R): The area of cross-section of the threaded and unthreaded parts is made equal. [IES-1994] (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

IES-47. Ans. (c) A is true and R is false. IES-48. How can shock absorbing capacity of a bolt be increased? [IES 2007]

(a) By tightening it property (b) By increasing the shank diameter (c) By grinding the shank (d) By making the shank diameter equal to the core diameter of thread

IES-48. Ans. (d)

IES-49. The number of slots is a 25 mm castle nut is [IES-1992] (a) 2 (b) 4 (c) 6 (d) 8

IES-49. Ans. (c)

Page 40 of 263


Answers with Explanation (Objective)

Page 41 of 263

Design of Friction Drives S K Mondals Chapter 2

2. Design of Friction Drives

Theory at a glance (GATE, IES, IAS & PSU)

Couplings

Introduction Couplings are used to connect two shafts for torque transmission in varied applications. It may be to connect two units such as a motor and a generator or it may be to form a long line shaft by connecting shafts of standard lengths say 6-8m by couplings. Coupling may be rigid or they may provide flexibility and compensate for misalignment. They may also reduce shock loading and vibration. A wide variety of commercial shaft couplings are available ranging from a simple keyed coupling to one which requires a complex design procedure using gears or fluid drives etc. However there are two main types of couplings: Rigid couplings. Flexible couplings.

Rigid couplings are used for shafts having no misalignment while the flexible couplings can absorb some amount of misalignment in the shafts to be connected. In the next section we shall discuss different types of couplings and their uses under these two broad headings.

Types and uses of shaft couplings

Rigid couplings Since these couplings cannot absorb any misalignment the shafts to be connected by a rigid coupling must have good lateral and angular alignment. The types of misalignments are shown schematically in figure below.

Page 42 of 263


Figure- Types of misalignments in shafts

Sleeve coupling One of the simple types of rigid coupling is a sleeve coupling which consists of a cylindrical sleeve keyed to the shafts to be connected. A typical sleeve coupling is shown in figure below

0d

L

id

Figure- A typical sleeve coupling Normally sunk keys are used and in order to transmit the torque safely it is important to design the sleeve and the key properly. The key design is usually based on shear and bearing stresses. If the torque transmitted is T, the shaft radius is r and a rectangular sunk key of dimension b and length L is used then the induced shear stress (figure below) in the key is given by

2

TLb r

=

And for safety ( )2 / yT bLr <

Page 43 of 263


Where y is the yield stress in shear of the key material. A suitable factor of safety must be used. The induced crushing stress in the key is given as

2 2br

Tb L r

=

And for a safe design 4T /(bLr) < c Where c is the crushing strength of the key material.

Figure- Shear and crushing planes in the key.

The sleeve transmits the torque from one shaft to the other. Therefore if di is the inside diameter of the sleeve which is also close to the shaft diameter d (say) and d0 is outside diameter of the sleeve, the shear stress developed in the sleeve is

( ) = 04 4016

sleevei

Tdd d

and the shear stress in the shaft is given by

= 316

shafti

Td

. Substituting yield shear stresses of the sleeve and shaft materials for

sleeve and shaft both di and d0 may be evaluated. However from the empirical proportions we have: d0 = 2di + 12.5 mm and L=3.5d. These may be used as checks. Sleeve coupling with taper pins Torque transmission from one shaft to another may also be done using pins as shown in figure below.

Page 44 of 263


Figure- A representative sleeve coupling with taper pins

The usual proportions in terms of shaft diameter d for these couplings are: d0 = 1.5d, L = 3d and a = 0.75d. The mean pin diameter dmean = 0.2 to 0.25 d. For small couplings dmean is taken as 0.25d and for large couplings dmean is taken as 0.2d. Once the dimensions are fixed we may check the pin for shear failure using the relation

= 22

4 2meandd T .

Here T is the torque and the shear stress must not exceed the shear yield stress of the pin material. A suitable factor of safety may be used for the shear yield stress.

Clamp coupling A typical clamp coupling is shown in figure below. It essentially consists of two half cylinders which are placed over the ends of the shafts to be coupled and are held together by through bolt.

Figure- A representative clamp coupling

The length of these couplings L usually vary between 3.5 to 5 times the and the outside diameter d0 of the coupling sleeve between 2 to 4 times the shaft diameter d. It is assumed that even with a key the torque is transmitted due to the friction grip. If now the number of bolt on each half is n, its core diameter is dc and the coefficient of friction between the shaft and sleeve material is we may find the torque transmitted T as follows. The clamping pressure between the shaft and the sleeve is given by

Page 45 of 263


( )2 / 22 4 c tnp d dL = Where n is the total number of bolts, the number of effective bolts for each shaft is n/2 and t is the allowable tensile stress in the bolt. The tangential force per unit area in the shaft periphery is F = p. The torque transmitted can therefore be given by

= 2 2dL dT p

Ring compression type couplings The coupling (figure below) consists of two cones which are placed on the shafts to be coupled and a sleeve that fits over the cones. Three bolts are used to draw the cones towards each other and thus wedge them firmly between the shafts and the outer sleeve. The usual proportions for these couplings in terms of shaft diameter d are approximately as follows: d1 = 2d + 15.24 mm L1 = 3d d2 = 2.45d + 27.94 mm L2 = 3.5d + 12.7 mm d3 = 0.23d + 3.17 mm L3 = 1.5d And the taper of the cone is approximately 1 in 4 on diameters.

Figure- A representative ring compression type coupling.

.

Oldham coupling These couplings can accommodate both lateral and angular misalignment to some extent. An Oldham coupling consists of two flanges with slots on the faces and the flanges are keyed or screwed to the shafts. A cylindrical piece, called the disc, has a narrow rectangular raised portion running across each face but at right angle to each other. The disc is placed between the flanges such that the raised portions fit into the slots in the flanges. The disc may be made of flexible materials and this absorbs some misalignment. A schematic representation is shown in figure below.

Page 46 of 263

S K M

UniverThese joinwidely usejoint is scommerciacenter piethe two foroften knowparallel of

Designcoupli Flange It is a verbolted. Th

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Universal jofor rig

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d consists o

ion Dr

n Oldham c

coupling angular ms and autom forms of keyed or scrlly perpendient can be al rigidity n

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Page 47 of 263

S

T(a(b(c

RA flaboprboon Thtocopr (1

W (2

S K Mo

he main fea) Design of b) Design of ) Overall de

Rigid Fl typical rigianges whicholted in the rovided to golt circle diane flange an

he design porque transoupling are rocedure is

1) Shaft diam

Where T is th

2) Hub diam

Dondal

eatures of t bolts. hub. esign and di

ange Cid flange couh are keyed annular spguard the pameter and

nd a recess o

procedure ismission an obtained. given in the

meterd bas

he torque an

meter, d1 =

Desigs

Figu

the design

imensions.

Couplinupling is shd to the shaace between

projecting bd the numbeon the oppos

is generallynd then foCheck for

e following s

sed on torqu

nd y is the y

=1.75d + 6.5

n of F

ure- A typica

n are essen

g hown in Figuafts to be jon the hub anolt heads aer of bolt dsing face is

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different fasteps:

ue transmis

1d =

yield stress

5 mm

Frictio

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tially

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and nuts. Thepends on t provided fo

determininmpirical relailure mode

ssion is give

1/16

s

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in shear.

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If essentiallflanges are ecting flanghe bolts arethe shaft dir ease of as

ng the shaftlations diffees can then

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3

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r 2

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ven the

sign

Page 48 of 263


(3) Hub length, L = 1.5d But the hub length also depends on the length of the key. Therefore this length L must be checked while finding the key dimension based on shear and crushing failure modes. (4) Key dimensions: If a square key of sides b is used then b is commonly taken as

4d . In that case, for shear failure

we have

= k4 2yd dL T

Where y is the yield stress in shear and Lk is the key length.

This gives =k 28

y

TLd

If Lk determined here is less than hub length L we may assume the key length to be the same as hub length. For crushing failure we have

= k8 2cd dL T Where c is crushing stress induced in the key. This gives

= 2k

16c

TL d

And if c < cy , the bearing strength of the key material ,the key dimensions chosen are in order. (5) Bolt dimensions: The bolts are subjected to shear and bearing stresses while transmitting torque. Considering the shear failure mode we have

cb yb

dn d24 2

= Where n is the number of bolts, db nominal bolt diameter, T is the torque transmitted, yb is the shear yield strength of the bolt material and dc is the bolt circle diameter. The bolt

Page 49 of 263


diameter may now be obtained if n is known. The number of bolts n is often given by the following empirical relation:

4 3150

n d= + Where d is the shaft diameter in mm. The bolt circle diameter must be such that it should provide clearance for socket wrench to be used for the bolts. The empirical relation takes care of this. Considering crushing failure we have

= cb cyb dn d t2. 2 Where 2t is the flange width over which the bolts make contact and cyb is the yield crushing strength of the bolt material. This gives t2. Clearly the bolt length must be more than 2 2t and a suitable standard length for the bolt diameter may be chosen from hand book. (6) A protecting flange is provided as a guard for bolt heads and nuts. The thickness t3 is less than 2 2

t the corners of the flanges should be rounded. (7) The spigot depth is usually taken between 2-3mm. (8) Another check for the shear failure of the hub is to be carried out. For this failure mode we may write

11 2 2yf

dd t = Where d1 is the hub diameter and yf is the shear yield strength of the flange material. Knowing yf we may check if the chosen value of t2 is satisfactory or not. Finally, knowing hub diameter d1, bolt diameter and protective thickness t2 We may decide the overall diameter d3. Flexible rubber bushed couplings Flexible coupling

Page 50 of 263

S K MAs discussvariety of will be dis This is simbelow.

In a rigid the bolts a However iabove) pro Because of

(1) Bea Rubber bubushes aremm thicknbrass sleev Thickness

Where db ithickness

Mondased earlier t flexible coucussed here

mplest type

Fi

coupling thand in this a

in the busheovide flexibi

f the rubber

aring sushings aree mostly avaness for larve

may be tak

is the diamof rubber bu

Desials these coupliuplings aree.

of flexible c

igure- A typ

e torque is arrangemen

ed coupling lity and the

r bushing th

tress e available ailable in thrger bores.

ken to be 1.5

eter of the bushing. We

gn of

ings can acc available c

coupling an

ical flexible

transmittednt shafts nee

the rubberese coupling

he design fo

for differenhickness betBrass sleev

5mm. The o

dr = d

bolt or pin, may now w

Frict commodate commercial

d a typical

e coupling w

d from one hed be aligne

r bushings og can accomm

r pins shoul

nt inside antween 6 mmves are ma

utside diam

db +2 tbr +2

tbr is the thwrite

ion Dr some misally and prin

coupling of

with rubber b

half of the ced very well

over the pinmodate som

ld be consid

nd out side m to 7.5 mmade to suit

meter of rubb

tr

hickness of t

rives ignment an

ncipal featu

this type is

bushings.

coupling to t.

ns (bolts) (asme misalignm

dered carefu

diameters. m for bores u

the require

ber bushing

the brass sl

Chaptend impact. Ares of only

s shown in F

the other th

s shown in Fment.

ully.

However rupto 25 mm ements. How

g dr is given

leeve and tr

er 2 A large a few

Figure

hrough

Figure

rubber and 9 wever,

by

is the

Page 51 of 263


2. 2c

r bdn d t p T=

Where dc is the bolt circle diameter and t2 the flange thickness over the bush contact area. A suitable bearing pressure for rubber is 0.035 N/mm2 and the number of pin is given by

325dn = + where d is in mm.

The dc here is different from what we had for rigid flange bearings. This must be judged considering the hub diameters, out side diameter of the bush and a suitable clearance. A rough drawing is often useful in this regard. From the above torque equation we may obtain bearing pressure developed and compare this with the bearing pressure of rubber for safely.

(2) Shear stress The pins in the coupling are subjected to shear and it is a good practice to ensure that the shear plane avoids the threaded portion of the bolt. Unlike the rigid coupling the shear stress due to torque transmission is given in terms of the tangential force F at the outside diameter of the rubber bush. Shear stress at the neck area is given by

2

2

4

b rb

neck

p t d

d =

Where dneck is bolt diameter at the neck i.e. at the shear plane.

Bending Stress The pin loading is shown in Figure below.

Page 52 of 263

S K M

Clearly thbending of

Knowing tusing appr We may al Hub diam Hub lengt

Pin diame

Monda

he bearing f the pin. Co

the shear anropriate the

lso assume

meter = 2d

th = 1.5d

eter at the n

Desials

Figure- Lo

pressure thonsidering a

nd bending eories of fail

the followin

neck = 0.5n

gn of

oading on a

hat acts asan equivalen

3b =

stresses welure.

ng empirical

5dn

Frict

pin support

s distributent concentra

( 23

32br

F td

e may check

l relations:

ion Dr

ting the bus

d load on rated load F=

)/ 2r

k the pin dia

rives

shings.

rubber bus= pt2d the b

ameter for p

Chapte

h would prbending stre

principal st

er 2

roduce ess is

tresses

Page 53 of 263


Problems with Solution Q. Design a typical rigid flange coupling for connecting a motor and a centrifugal

pump shafts. The coupling needs to transmit 15 KW at 1000 rpm. The allowable shear stresses of the shaft, key and bolt materials are 60 MPa, 50 MPa and 25 MPa respectively. The shear modulus of the shaft material may be taken as 84GPa. The angle of twist of the shaft should be limited to 1 degree in 20 times the shaft diameter.

Solution: The shaft diameter based on strength may be given by

= 316

y

Td Where T is the torque transmitted and y is the allowable yield

stress in shear.

Here T32 15 10Power / 143Nm2 100060

60

N

= = =

And substituting y = 60 x 106 Pa we have. Let us consider a shaft of 25 mm which is a standard size. From the rigidity point of view

T GJ L

=

Substituting T = 143Nm, ( )4 9 4 90.025 38.3 10 , 84 1032J m G Pa = = =

9 9143

38.3 10 84 10L

= = 0.044 radian per meter The limiting twist is 1 degree in 20 times the shaft diameter

Which is 180 0.03520 0.025

= radian per meter

Therefore, the shaft diameter of 25mm is safe. We now consider a typical rigid flange coupling as shown in Figure above Hub

Page 54 of 263


Using empirical relations Hub diameter d1 = 1.75d + 6.5 mm. This gives d1 = 1.75 x 25 + 6.5 = 50.25mm say d1 = 51 mm Hub length L=1.5d. This gives L = 1.5 x 25 = 37.5mm, say L= 38mm. Hub thickness, 11

51 25 132 2

d dt mm = = =

Key Now to avoid the shear failure of the key (refer to Figure above)

4 2k yd dL T = , Where the key width 4

dw = and the key length is Lk

This gives ( ) ( )268 143. . 0.0366 36.6

50 10 0.025i e m mm = = k 2y

8TL = d

The hub length is 37.5 mm. Therefore we take Lk = 37.5mm. To avoid crushing failure of the key (Refer to Figure below)

8 2kd dL = , Where is the crushing stress developed in the key.

This gives 216

k

TL d

= Substituting T = 143Nm, Lk = 37.5 x 10-3 m and d = 0.025 m

( )6

23

16 143 10 97.62MPa37.5 10 0.025

= =

Assuming an allowable crushing stress for the key material to be 100MPa, the key design is safe. Therefore the key size may be taken as: a square key of 6.25 mm size and 37.5 mm long. However keeping in mind that for a shaft of diameter between 22mm and 30 mm a rectangular key of 8mm width, 7mm depth and length between 18mm and 90mm is

Page 55 of 263


recommended. We choose a standard key of 8mm width, 7mm depth and 38mm length which is safe for the present purpose.

Bolts To avoid shear failure of bolts

machine design md by s k mondal t&q

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