machine design md by s k mondal t&q.pdf
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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 shaftrefers 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 notorque. 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
where1 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
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Design of Joint
S K Mondals Chapter 1one 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 1
d 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
=2
4td P
Fig. Tension failure of the rod
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Design of Joint
S K Mondals Chapter 12. Tension failure of rod across
slot
=
2
1 14 td d t P
Fig. Tension failure of rod across slot
3. Tensile failure of socket
across slot
( ) ( )
=
2 2
2 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 1
2l 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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Design of Joint
S K Mondals Chapter 17. Crushing failure of rod or
cotter
=1 c
d 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 1
4cd 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 theloading as shown in figure below.
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Design of Joint
S K Mondals Chapter 1
Figure-Bending of the cotter
This gives maximum bending moment
= +
3 1 1
2 6 4
d d dPand
The bending stress, ( )
+ +
= =
3 1 3 11 1
3 2
32 6 4 2 6 4
12
b
d d d dd dP bP
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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Design of Joint
S K Mondals Chapter 1
t
3d 1
d 2dd
Axial load( )
= 2
4
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 = 2
14
yd 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 2
2 1 2 14
yd 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 = Pand this gives l1= 7.57 mm. Let l1= 10 mm.
Refer to figure
For shear failure of socket end2l(d2-d1) = Pand this gives l = 22.72 mm. Let l = 25
mm.
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Design of Joint
S K Mondals Chapter 1Refer to figure
For crushing failure of socket or rod (d3-d1)tc = P. This gives d3= 75.5 mm. Let d3= 77mm.
Refer to figure
For crushing failure of collar ( )
=2 24 1
4c
d 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
3d
2d
1d
t
1t
1t
2t
d
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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Design of Joint
S K Mondals Chapter 1However, 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, cand .
1. Failure of rod in tension:
=2
4t
d P
2. Failure of knuckle pin in double sheer:
=2
12
4d 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
3
1
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:
d1t c= P
6. Failure of rod eye in tension:
(d2 - d1)t t= P
7. Failure of forked end in shear:
2(d2 - d1)t1 = P
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Design of Joint
S K Mondals Chapter 18. Failure of forked end in tension:
2(d2 - d1)t1 t= 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
=2
1
2P
d. We may now put value of d1 from
empirical relation and then find FACTOR OF SAFTEY,(F.S.)=
ywhich should be more
than one.
Q. Two mild steel rods are connected by a knuckle joint to transmit an axialforce 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
3d
2d
1d
t
1t
1t
2t
d
Refer to figure above
For failure of rod in tension,
= 24
yP 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:
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Design of Joint
S K Mondals Chapter 1
1. Failure of knuckle pin in shear:
=
2
124
yP d ,which gives y= 39.8
MPa. This is less than the yield shear stress.
2. For failure of knuckle pin in bending:
+
=
1
3
1
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 d 2should be at least 85.8
mm. Let d2be 90 mm.
4. for failure of rod eye in crushing: d1 t c= Pwhich 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= Pwhich 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 theWeakest so that it is cheap and easy
to replace in case of failure.
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S K
F
r
g
r
Recta
1. Re
pr
W
Th
ond
igure- (a)
ounded; als
ib head. (
ectangular
etric sizes.
ngular
ctangular
portions of
Width
Thick
ere d = Di
e key has t
ls
quare or
o available
) Woodru
key; = hu
(f)Tapere
and sq
sunk key:
this key ar
of key, (w
ess of ke
meter of th
per 1 in 1
De
ectangular
with both
ff key; als
b length, h
gib-head
uare k
A rectang
:
, (t)
e shaft or d
00on the t
ign o
key. (b) S
ends roun
o availabl
= height;
ey; dimens
y
lar sunk
= d/ 4
= 2w/ 3 =
iameter of
p side only
f Joi
uare or re
ed. (c) Squ
with fla
aper is 18
ons and ta
ey is show
d/ 6
he hole in
.
t
ctangular
are or rect
tened bot
in for 12 i
er same as
in figure
he hub.
Chapt
ey with o
angular ke
om. (e) T
or 1 for 1
in (e).
below. The
er 1
e end
with
pered
00 for
usual
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2.
k
3.
is
a
T
If
o
Q
S
K M
Square s
y is that it
Gib-head
usually pr
d its use i
he usual pr
a square k
8x
rl
=
. A heat
50 mmSelect
lution:C
figure b
ndal
nk key: T
s width an
key: It is
ovided to f
shown in
oportions o
idth, (
hicknes
ey of sides
2 [W
treated st
. The shaan approp
nsider a r
elow. The
sFig
he only dif
thickness
rectangul
cilitate th
igure belo
the gib he
)
s at lar
/4 is used t
herex
the
eel shaft
t rotatesriate key
ctangular
ey may fail
Desi
ure-. Recta
erence bet
are equal, i.
w = t = d/
r sunk ke
removal
.
Figure- G
d key are:
e end, (
hen. In tha
4d
l
yield is str
f tensile
at 1000 ror the ge
key of wid
(a) in shea
n of
gular sun
een a rect
.e.
4
with a he
f key. A gi
b-head key
= d
t ) =2
t case, for s
=2
x
dT
ss in shear
ield stren
m and trr.
h (w), thic
r or (b) in c
Joint
key.
angular su
d at one e
head key
.
/ 4
/ 3 = d/
hear failur
and l is th
gth of 350
ansmits 1
ness (t) a
ushing.
k key and
d known a
is shown i
6
we have
key length
MPa has
0 kW thr
d length (
hapte
a square s
sgib hea
n figure be
.]
diamete
ough a g
) as show
1
nk
. It
low
of
ar.
in
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S K
Sh
W
Nan
yie4m
Cr
Flat k
A flat ke
friction fotight fit. T
Saddl
A saddle
is concave
used for h
stress con
Tang
ond
ear failur
ere torqu
eing in rpyis the y
ld stress an2.
shing fai
y
, as shown
r the grip.heses keys
key
key, show
to fit the
eavy loads.
centration
nt Key
ls
: The fail
e transmit
, w, L andield stress
d substitut
ure =
in figure
he sideshave about
Flat key
in figure
haft surfa
A simple
ccurs in th
De
Figure
re criteri
ted is, (
d are then shear of
ing the val
2 2
t L
elowis us
f these kehalf the th
bove, is ve
e. These k
in can be
e shaft in t
ign o
n is =
) =Pow
idth, lengtthe key ma
es in abov
d for ligh
s are paralickness of s
ry similar t
ys also ha
sed as a k
ese cases.
f Joi
yw L
2/
60r
h and diamterial. Taki
equations
t loadbec
lel but theunk keys.
Saddl
o a flat ke
e friction
y to trans
his is sho
t
2
d
eter of theng yto be
and we ha
use they d
top is slig
key
except th
rip and th
it large to
n in figure
Chapt
shaft respehalf of the
e wL = 2.1
epend entir
tly tapere
t the botto
erefore can
rques. Ver
above.
er 1
tivelyensile
9 x 10-
ely on
for a
side
not be
little
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A
t
A
s
k
u
d
T
K M
eather
feather ke
the hub or
oodru
woodruff
own in fig
yway is cu
ed in mac
pth.
he main a
ndal
key
is used w
the shaft a
ff key
key is a f
re below.
t in the sh
hine tools
vantages
s
en one com
nd the key
orm of sun
t is usuall
aft using a
nd autom
of a wood
Desi
Figure-T
ponent sli
ay usually
Figure-f
k key whe
used for
milling cu
biles due
Figure-
ruff key a
n of
angent Key
es over ano
has a slidi
eather key
e the key
hafts less
ter, as sho
to the extr
oodruff ke
e as follo
Joint
ther. The k
g fit.
hape is th
han about
wn in the
advantag
s:
ey may be
at of a tru
60 mm dia
igure belo
e derived
hapte
astened eit
cated disc,
meter and
. It is wi
rom the e
1
her
as
the
ely
tra
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Design of Joint
S K Mondals Chapter 11. 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
Splinesare 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 tomanufacture 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.
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K M
ndal
s
Desi
n of
Joint
Splines c
as a s
keyways
keys ma
shaft.
There are
of spli
industry:
splines,
splines.
Splines
uniform
transfershaft tha
n be thou
eries of
with m
chined on
two major
es used
1) straight-
nd 2) in
rovide a
circumfer
f torque ta key.
hapte
ht of
axial
ating
to a
types
in
sided
olute
more
ntial
o the
1
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Design of Joint
S K Mondals Chapter 1
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 jointsA welded joint is a permanent joint which is obtained by the fusion of the edges of the twoparts 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).
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(
T
K M
) Butt
Formed
(for thick
butt joint Squa Singl Singl Singl Singl
hese are sc
ndal
Joints
y placing t
plates) on
s may be ofe butt joint
V-butt joi
U-butt joi
J-butt join
bevel-butt
ematically
s
Figur
he plates e
the edges
different t
t, double
t, double
t, double J-
joint, doub
shown in f
Desi
e- Differen
dge to edg
efore wel
pes, e.g.
-butt joint
-butt joint
butt joint
le bevel bu
gure below
n of
types of la
and weldi
ing. Accor
t joint.
Joint
p joints
g them. G
ing to the
rooves are
shape of t
hapte
sometimes
e grooves,
1
cut
the
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S K
There ar
Co Ed T-j
Weldi
A welding
1.
2.
3.
4.5.
6.
7.
8.
These wel
ond
other ty
ner joint (s
ge or seal j
oint (see fig
g sym
symbol ha
eference li
rrow.
asic weld
imensionsupplemen
inish sym
ail.
pecificatio
ding symbo
ls
es of wel
ee figure b
int ( see fig
ure below)
bol
following
ne.
ymbols (li
ary symbol
ols
processes
ls are place
De
Figure- Di
ed joints,
low)
ure below)
asic eleme
e fillet, but
s.
d in standa
ign o
ferent type
for examp
nts:
t joints etc.
rd location
f Joi
of butt joi
le,
(see figure
t
ts
below)
Chapt
er 1
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E
w
T
b
F
(s
K M
xample: If
ith convex
esign
he main fai
tt joint is
here t=
t =t
l =le
or a square
ee figure b
ndal
the desire
ontour, th
of a bu
lure mecha
allowable t
ickness of
ngth of the
butt joint
low).
s
weld is a f
weld symb
tt joint
nism of we
ensile stren
the weld
weld.
is equal t
Desi
illet weld o
ol will be a
Figure-
ded butt jo
=Pgth of the
the thickn
n of
f size 10 m
s following
Tee joint
int is tensil
tt
eld materi
ess of the p
Joint
to be don
e failure. T
al.
lates. In g
e on each s
herefore th
neral, this
hapte
de of Tee j
e strength
need not b
1
int
f a
so
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S K
Desig
Consider
the weld
loadP act
a fillet we
throat is
transvers
Wheres
Athroat= th
For a dou
Desig
Consider
to see froarea (try
throat are
ond
n of tra
single tra
etal is ver
s as shear
ld. If the fil
asily seen
fillet weld
= allowable
roat area.
le transver
n of pa
parallel fi
the streno prove it).
a =2
t
lhA .
ls
nsver
nsverse joi
complicat
orce on the
let weld ha
o be2
hl .
is
shear stre
se fillet joi
Fig
rallel fi
let weld as
th of mateThe allow
he total al
De
Figu
e fillet
t as shown
ed. In desi
throat are
s equal bas
ith the ab
P =
s
t the allow
re- Enlarg
llet joi
shown in f
ial approaable load c
lowable loa
ign o
re-butt join
joint
in figure b
n, a simple
a, which is
e and heig
ove conside
s .Athroa
able load is
e view of fil
t
gure below
h that therried by e
d isP = 2s
f Joi
elow. The
procedure
the smalles
t, (h, say),
ration the
t
twice that
let welding
. Each weld
maximumch of the j
At.
t
eneral stre
is used ass
t area of th
then the c
ermissible
of the singl
carries a l
hear occuroint is
s
Chapt
ss distribu
ming that
e cross sec
oss section
load carrie
e fillet join
ad 2P . It i
along thetA ,whe
er 1
ion in
entire
ion in
of the
d by a
.
s easy
throatre the
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I
a
fi
C
b
a
di
t
K M
designing
ove design
gure below)
here
esign
onsider a c
low. If the
in parall
ameter of
rque the m
here
ndal
a weld joi
criteria.
the allowa
At
tA = thro
of circ
ircular sha
shaft is su
l fillet join
he shaft, t
aximum sh
Td= outer
throatt = t
pJ = pola
=
32
s
t the desig
hen a com
le load is
= throat a
t area alon
lar fil l
ft connecte
jected to a
t. Assumin
e maximu
ear stress i
max
torque apiameter of
hroat thick
moment o
( + throat2td
Desi
n variables
ination of
2 sP =ea along t
g the trans
Figuret weld
d to a plat
torque, sh
g that the
shear st
the weld i
2
p
dT
J
+
lied.the shaft
ness
f area of th
) 4 4d
n of
are hand
transverse
t s tA A + e longitudi
verse direc
subje
e by mean
ear stress
weld thick
ess occurs
s
throat
throat sec
Joint
l. They can
and paralle
nal directio
ion.
ted to
of a fillet
evelops in
ness is ver
in the thro
tion.
be selecte
l filled join
n.
torsio
joint as s
the weld in
y small co
at area. Th
hapte
based on
t required
own in fig
a similar
mpared to
us, for a gi
1
the
see
ure
ay
the
en
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S K
When tt
The throa
Q. A pl
mea
the l
Solution.
a loa
= th
25
56
allo
(Not
Q. Two
tran
the
Solution:
assu
speci
MPa
weld
ond
roat
,d
t dimension
ate 50 m
ns of para
ength of t
In a parall
d of 502
P=
oat length
3
3
10 25
2.5 10
=
ance for s
that the a
plates 2
sverse we
ize of the
According
ming maxi
fied the sh
. Assuming
is calculat
ls
=max
4and hence
wide an
llel fillet
e weld. A
el fillet we
N 25kN= .
12.5
2mm .
.5mm . Ho
arting or s
llowance h
0 mm wi
ds at the
weld assu
to the desi
um shear
ear strengt
there are t
d from
35
Or l
De
3
2
throat
dT
t d
weld dime
2
throt
d 12.5 m
elds. The
ssume all
ding two li
Maximum
inces
p
lt
ever som
topping of
s no conne
de and 1
ends. If th
ming the
gn principl
stress occ
h may be c
wo welds, e
3 110 l
=
= 141.42 m
ign o
2
throat
T
t
sion can b
2
td
=
Fig.
thick is
plates ar
wable sh
nes of weld
shear stres
656 10 . H
extra len
the bead.
tion with t
mm thi
e plates a
llowable
of fillet (t
rs along th
alculated a
ach weld c
30 103
2
m.
f Joi
2
selected fr
ax
to be we
subjecte
ar streng
ing are to b
in the par
ence the m
gth of the
usual all
he plate thi
k are to
e subject
ensile str
ransverse)
e throat ar
half of te
rries a loa
65 10
t
om the equ
lded to a
to a loa
h to be 56
e provided.
allel fillet
inimum len
weld is t
owance of
ckness)
be welde
d to a loa
ss 70 MP
joint the w
ea. Since t
sile streng
of 35 kN
Chapt
ation
other pla
of 50 kN.
MPa.
Each line
eld is plt
,
gth of the
be provi
12.5 mm is
d by mea
d of 70 k
.
eld is de
ensile stre
th, i.e. ,s
nd the size
er 1
te by
Find
hares
here t
eld is
ed as
kept.
ns of
, find
igned
gth is
= 35
of the
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Design of Joint
S K Mondals Chapter 1
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,
=s
throat
T
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 theproblem)
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
1
358 0.9214.8 kgf/cm = 214800N/m
1.5
= = (approx).The design stress is Therefore
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S K
=1,214
2.d
fluctuatin
Threa
B
th
se
o
S
in
in
Bolts
Machi
ond
=800
795567
g load allo
ed fa
lt - Thre
rough hole
cured by
posite the
rew - Th
serted thro
to a thread
ne Scr
ls
N/m . Sinc
able for th
teners
ded faste
s in matin
ightening
ead of the
eaded fas
ugh a hole
ed hole in a
ws
De
the tota
joint is 79
er designe
members
a nut fro
bolt.
ener desig
in one m
mating me
ign o
l length o
55.6 N. The
d to pass
and to be
the end
ned to be
mber and
mber.
f Joi
f the wel
joint is the
t
is 0.1 m
refore safe.
Fig.
Fig.
Chapt
, the ma
er 1
imum
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T
T
n
K M
heet
hread
he pitch lin
hread
hread Ser
mber of th
ndal
etal an
F
Profile
e or diamet
Series
es- group
reads per i
s
d Lag
igure-Shee
er is locate
of diamet
ch applied
Desi
Screw
metal scre
at the h
r-pitch co
to a specifi
n of
ws are ofte
eight of th
binations
c diameter.
Joint
n self-tappi
theoretica
istinguish
ng.
l sharp v-th
d from eac
hapte
read profil
h other by
1
.
the
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Design of Joint
S K Mondals Chapter 1
M-SeriesMetric system of diameters, pitches, and tolerance/allowances.
Tensile Stress AreaThe average axial stress in a fastener is computed using a tensile stress area.
=ave
t
F
A
+ =
2
4 2
r p
t
D DA
F Axial Force
DrRoot Diameter
DpPitch Diameter
At Tensile Stress Area
aveAverage 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
F
A =
Shear Strength (Internal
Thread)
max
,
0.5 ts i e
F
A L =
Where
max ,0.5
t t t s i eF A A L = =
=,
2 te
s i
AL
A
Bolt/Nut Design PhilosophyANSI 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)
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Design of Joint
S K Mondals Chapter 1
Previous 20-Years GATE Questions
KeysGATE-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 3
4T 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 d
l T4 2
=
x 2
8Tor
ld = [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)
Strengthof materialFactorofsafety
Actualloadorstrengthonmaterial=
3
6
o
200(in MPa) 200(in MPa)3.4
58.91(in MPa)15 10
660 10 (in MPa)
cos45
= =
Threaded fasteners
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Design of Joint
S K Mondals Chapter 1GATE-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
CottersIES-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
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I
I
I
I
I
I
I
I
I
K M(c)
S-3. Ans.
S-4. M
Li
A.
B.
C.
D.
Code
(a
(c)
S-4. Ans.
S-5. In
(a
(c)
S-5. Ans.
S-6. Mco
Li
A.
B.
C
D.
(a
(c)
S-6. Ans.
S-7. In
th
sh
(a
S-7. Ans.
ndal 1 4
(d)
atch List I
st I (Type
Riveted j
Welded jo
Bolted joi
Knuckle j
s: A B
4 3
2 4
(c)
a gib and
Single she
Single she
(d)
atch Listrrect ans
st I
Bolts in
cylinder
Cotters i
Rivets in l
Bolts hol
a flange
A B
4 1
3 1
(a)
a cotter
ickness is
earing str
120 N/mm
(d) It i
Shear str
s3
with List
of joints)
int
int
nt
oint
C
2
3
cotter joi
ar only
r and crus
(Items ier using
olted join
cover plat
cotter joi
ap joints
ing two fl
oupling
C
3
4
joint, the
12 mm.
ess develo2
s a case of
ssLoad
2 Are=
Desi
2 (d)
II and sel
List
1. P
2. S
3. L
4. F
D
1 (b)
1 (d)
t, the gib
hing
joints)he codes
ts of engi
e
nt
anges in
D
2 (b)
2 (d)
width of
he load
ped in the
(b) 100 N/
ouble shea360 10
2 50 12
=
n of
2 3
ct the cor
II (An ele
in
rap
ock washe
illet
A
2 3
2 4
and cotte
(b) doub
(d) doub
ith List IIiven belo
L
e 1
2
3
4
A
4 2
3 2
the cotter
cting on
cotter?
m2 (
r.
250N/mm=
Joint
4
rect answ
ment of th
r
C
4
1
are subje
e shear onl
e shear an
(Type ofthe List
ist II
.Doubletr
. Torsiona
Single tr
. Tension
C
3
4
at the c
he cotter
) 75 N/mm
1
er.
e joint)
D
1
3
cted to
y
crushing
failure) a:
nsverse s
l shear
nsverse s
D
1
1
ntre is 5
is 60 kN.
(d)
hapte
[IES-19
[IES-20
d select[IES-20
ear
ears
mm and
What is
[IES-20
50 N/mm2
1
4]
06]
he04]
its
he
04]
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Design of Joint
S K Mondals Chapter 1IES-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 2
x
=
(b)
t
P x 1D 2
x
+=
(c)
t
2 P x 1D
x
+=
(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 circularcross-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 notorque.
A spindle is a short shaft. Terms such as line-shaft, head-shaft, stub shaft,transmission shaft, countershaft, andflexible 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 codesgiven 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
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I
I
I
I
I
I
K MS-12. Ans
S-13. M
gi
A.
B.
C.
D.
1.
2.
3.
4.
5.
C
S-13. Ans
t
m
C
et
S-14. A
th
u
R
lo(a
(b
(c)
(d
S-14. Ans
eysS-15. In
(a
ndal. (a)
atch List
ven below
List I (Ty
Cotter joi
Knuckle j
Turn buc
Riveted j
List II (
Connects
Rigidly co
Connects
Permane
Connects
des: A
(a) 5
(c) 5. (b)A cott
o rods whic
y be subje
nnection o
are exam
sertion (
e connectin
on the dire
ason (R):
ding and rBoth A an
Both A an
A is true b
A is false
. (b)
the asse
pulley is
s
I with Li
the lists:
pe of join
nt
oint
le
int
ode of joi
two rods
nnects tw
two rods
t fluid-tig
two shafts
B
1
3er is a flat
h transmit
ted to tens
piston rod
les of cotte
):When t
g rods eith
ction of rot
A turn bu
quiring sud R are ind
d R are ind
ut R is fals
ut R is tru
bly of pul
ade the we
Desi
t II and
)
ting me
r bars pe
o member
aving thr
ht joint b
and tran
C D
3 2
2 4wedge-sha
motion in t
le or comp
to the cros
joint.
e coupler
er move cl
tion of the
kle is used
sequent avidually tr
vidually tr
e
Fig. Tu
ley, key a
akest
n of
elect the
bers)
mitting s
s
eaded end
tween tw
mits torq
(b) 2
(d) 2ed piece of
he axial dir
essive forc
s-head of a
f a turn b
oser or mo
coupler.
to connect
justment fe and R is
e but R is
rnbuckle
d shaft
(b) key i
Joint
correct a
all amou
s
flat piec
e
B
1
3steel. This
ection, wit
s along the
steam engi
ckle is tur
e away fr
two round
r tightenithe correct
otthe cor
made the
swer usi
t of flexi
s
C D
3 4
1 4is used to
out rotatio
axes of the
ne, valve ro
ned in one
m each ot
rods subje
g or loosenexplanatio
ect explan
[IE
weakest
hapte
g the co
[IES-19
ility
connect rig
n. These joi
rods.
d and its s
direction b
her depend
[IES-19
cted to ten
ing.of A
tion of A
S-1993; 19
1
es
93]
dly
nts
em
oth
ing
96]
sile
8]
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S K
IES-15.
IES-16.
IES-16.A
IES-17.
IES-17. A
IES-18.
IES-18. A
IES-19.
IES-19. A
ond(c) Key is
ns. (b)Ke
failure.
Match Li
correct a
List-I
A. Woodr
B. Kenne
C. Feathe
D. Flat ke
Code:
(a)
(c)
ns. (b)A fe
be fastene
Match Li
given bel
List-I (Ke
A. Gib he
B. Woodr
C. Parall
D. Spline
Code:
(a)
(c)
ns. (c)
A spur g
rectangul
(a) Shear s
(c) Both sh
ns. (c)Key
Assertion
capacity a
Reason (
(a) Both A
(b) Both A
(c) A is tru
(d) A is fal
ns. (d)
lsade the st
is made
st-I (Type
swer usi
ff key
y key
r key
y
B
2 3
2 3
ather key i
either to t
st-I with
w the list
/splines)
d key
ff key
l key
B
1 2
2 1
ar trans
ar section
tress alone
ear and be
develops b
(A): The
d to increa
):Highly l
and R are i
and R are i
e but R is f
e but R is
De
ongest
he weakes
of keys)
g the cod
L
1
2
3
4
C
1 4
4 1
s used whe
he hub or t
Fig.
List-II an
s:
C
3 4
3 4
itting po
. The type
ring stress
th shear a
effect of k
se its torsio
calized str
ndividually
ndividually
lse
rue
ign o
(d) a
stren
so that it
with List
s given b
ist-II
. Loose fit
. Heavy d
. Self-alig
. Normal i
(b)
(d)
one comp
e shaft an
feather key
d select t
List-
1. Sel
2. Fa
3. Mo
4. Ax
(b)
(d)
er is co
(s) of stre
(b) be
es (d) sh
d bearing
yways on
nal rigidity
esses occur
true and R
true but R
f Joi
ll the thr
th
is cheap a
-II (Char
low the L
ing, light
ty
ing
dustrial
A B
3 2
3 2
nent slides
the keyw
e correc
I (Applica
f aligning
ilitates r
stly used
ial movem
A B
1 2
2 1
nected t
ses devel
aring stres
earing, bea
stresses.
a shaft is
.
at or near
is the corr
is notthe
t
e are de
d easy to
cteristic)
sts:
duty
se
C
1
4
over anoth
y usually h
answer
tion)
moval
ent possib
C
4
4
the shaf
ped in th
alone
ring and be
to reduce
he corners
ct explana
orrect expl
Chaptigned for
eplace in
and sele
[IES
D
4
1
er. The key
as a slidin
using the
[IES
le
D
3
3
with a
e key is fa
[IES-
nding stres
its load ca
[IES-
of keyways.
tion of A
anation of
er 1equal
ase of
t the
1997]
may
fit.
code
2008]
ey of
e.
1995]
ses.
rrying
1994]
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Design of Joint
S K Mondals Chapter 1IES-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 4IES-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)d
2
(b)
2 d
3
(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)
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S K
IES-26.
IES-26. A
IES-27.
IES-27.A
ond
A pulley
means of
key is ta
the key is
the lengt
(a)4
ns. (c)
Shearings
Assertion
Reason (
of the mati(a) Both A
(b) Both A
(c) A is tru
(d) A is fal
ns. (b)
The main
1. It accom
2. It is use
tendency t
The main
1. The dep
2. It can n
ls
is connec
a rectang
en as d/4.
equal to
of the ke
(b)
trengthof
(A):A Wo
):The Wo
ng piece.and R are i
and R are i
e but R is f
e but R is
dvantages
modates it
ul on taper
turn over
is-advanta
h of the ke
t be used a
De
ted to a
lar sunk
For full
he torsio
to the di
2
dey:F .
4
Torque(
Torsiona
or T d
For sam
d. .l .
4
lor
d 2
=
=
=
druff key i
druff key
ndividually
ndividually
lse
rue
of a woodr
elf to any t
ing shaft e
in its keyw
ges of a wo
yway weak
s a feather.
ign o
power tra
ey of wid
ower tra
al sheari
ameter of
(c)2
3
3
.l
d) =F. .
2
l shearing,
16
strength
d16
=
=
an easily
ccommoda
true and R
true but R
ff key are
aper in the
ds. Its ext
ay.
druff key
ens the sha
f Joi
nsmission
th wand l
smission,
g strengt
the shaft (
4
d d.l .
4 2
T
dd
232
=
djustable
es itself to
is the corr
is notthe
s follows:
hub or bos
a depth in
re as follo
ft.
t
shaft of
ngth l. T
the shea
of the sh
l/d) is
(d)
ey.
any taper i
ct explana
orrect expl
of the mat
the shaft p
s:
Chapt
diameter
he width
ing stren
aft. The r
[IES
n the hub
[IEStion of A
anation of
ing piece.
events any
er 1
d by
f the
th of
tio of
2003]
r boss
2003]
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Design of Joint
S K Mondals Chapter 1IES-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 frictionalresistance
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. InvoluteB. 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)
SplinesIES-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
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Design of Joint
S K Mondals Chapter 1(c) 2 and 4 are correct (d) 1 and 3 are correct
IES-32. Ans. (d)
Welded jointsIES-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 toP
0.707Lh (b) Tensile and equal to
P
Lh
(c) Shear and equal toP
0.707Lh (d) Shear and equal to
P
Lh
IES-34. Ans. (c)
Throat, t = h cos450=2
1h
v= 0.707h
T =P
Lt=
P
0.707Lh
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)o
F
wtsin45 (b)
F
wt (c)
oFsin45
wt (d)
2F
wt
IES-35. Ans. (a)
IES-36.A butt welded joint, subjected to
tensile force P is shown in the
given figure, l = length of theweld (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 2P
a b c d Hl hl 2hl Hl
= = = =
IES-36. Ans. (b)
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Design of Joint
S K Mondals Chapter 1IES-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)P
2.B.L (d)
2P
B.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 ofwelds 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
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Design of Joint
S K Mondals Chapter 11
1
1 1
1 1
2 2
1 1 2 2
1 1 2 2
1 1 2 2
P
A
P A
P t I
P t I
P y P y
tI 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 was 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
w
d.t.
2
(c)
2
w
d.t.
2 2
(d)
2
w
d.t.
2
IES-42. Ans. (b)
( )
W
W
2
W W
Shear stress
Shear fore dt
d dTorque T dt .t
2 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
T
td (b)
2
2T
td (c)
2
4T
td (d)
3
2T
td
IES-43. Ans. (b)3 2
dT.
T.r 2T2
J td td
4
= = =
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Design of Joint
S K Mondals Chapter 1IES-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 fastenersIES-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)2f
dt (b)
F
dt (c)
F
2 dt (d)
4F
dt
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) d
dh 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 notthe 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)
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Design of Joint
S K Mondals Chapter 1
Answers with Explanation (Objective)
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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 shaftby 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 couplingsare used for shafts having no misalignmentwhile the flexiblecouplings
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 rigidcoupling must have good lateral and angular alignment. The types of misalignments are
shown schematically in figurebelow.
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Design of Friction Drives
S K Mondals Chapter 2
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 bearingstresses. 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
TL
b r
=
And for safety
( )2 / yT bLr <
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Design of Friction Drives
S K Mondals Chapter 2
Where yis 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 2
br
Tb L r =
And for a safe design
4T /(bLr) < c
Where cis 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 d i 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
( )
=
0
4 4
0
16sleeve
i
Td
d d and the shear stress in the shaft is given by
=3
16shaft
i
T
d. Substituting yield shear stresses of the sleeve and shaft materials for
sleeveand shaftboth diand d0may 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 pinsTorque transmission from one shaft to another may also be done using pins as shown in figure
below.
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Design of Friction Drives
S K Mondals Chapter 2
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 dmeanis taken as 0.25d and for
large couplings dmeanis taken as 0.2d. Once the dimensions are fixed we may check the pin forshear failure using the relation
=
22
4 2mean
dd 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 cylinderswhich 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 dcand the coefficient of friction between the shaft
and sleeve material is we may find the torque transmitted T as follows.
The clamping pressurebetween the shaft and the sleeve is given by
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Design of Friction Drives
S K Mondals Chapter 2
( )2 / 22 4
c t
np d dL
=
Where nis the total number of bolts, the number of effective bolts for each shaft is n/2and tis 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 2
dL 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 usualproportions 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 couplingThese couplings can accommodate both lateral and angular misalignmentto 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.
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S K
UniveThese joi
widely us
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T
(
(b
(c
A
fl
b
p
b
o
T
t
c
p
(
(2
K M
he main f
) Design of
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he design
mensions.
ouplinpling is sh
to the sh
ace betwee
rojecting b
the numb
n the oppo
s generall
d then fo
Check for
following
sed on torq
d yis the
1.75d + 6.
n of
re- A typic
are essen
g own in Fig
fts to be jo
the hub a
olt heads a
r of bolt d
sing face is
based on
llowing e
different f
teps:
e transmis
d =
ield stress
mm
ricti
l flange co
tially
re above. I
ined. The f
nd the prot
nd nuts. T
epends on
provided fo
determinin
pirical rel
ilure mod
sion is give
1/
6
s
T
in shear.
n Dri
pling
f essentiall
langes are
cting flang
e bolts ar
he shaft d
r ease of as
g the shaf
ations diff
s can the
n by
3
es
y consists
brought to
e. The prot
placed eq
ameter d.
sembly.
t diameter
rent dime
be carrie
hapte
f two cast i
ether and
ective flang
ui-spaced o
spigot A
d for a gi
nsions of
d out. Des
2
ron
are
e is
n a
on
en
the
ign
Page 48 of 263
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Design of Friction Drives
S K Mondals Chapter 2(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 bis used then b is commonly taken as4
d. In that case, for shear failure
we have
=
k4 2
y
d dL T
Where yis the yield stress in shear and Lkis the key length.
This gives
=k 28TL d
If Lkdetermined 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 2
c
d dL TWhere cis crushing stress induced in the key. This gives
=2
k
16c
T
L 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 failuremode we have
cb yb
dn d2
4 2=
Where n is the number of bolts,dbnominal bolt diameter, Tis the torque transmitted, b is
the shear yield strength of the bolt material and dc is the bolt circle diameter. The bolt
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Design of Friction Drives
S K Mondals Chapter 2diameter may now be obtained if n is known. The number of bolts nis often given by the
following empirical relation:
43150n d= +
Whered 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 failurewe have
= cb cybd
n d t2
.2
Where2
t is the flange width over which the bolts make contact and cybis 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 t3is less
than 22
tthe 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 d1is the hub diameter and yf is the shear yield strength of the flange material.
Knowing yf we may check if the chosen value of t2is 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
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S KAs discus
variety of
will be dis
This is si
below.
In a rigidthe bolts
However i
above) pr
Because o
(1) Be
Rubber b
bushes ar
mm thickbrass slee
Thickness
Where db
thickness
onded earlier t
flexible co
cussed her
plest type
Fi
coupling thnd in this
n the bush
vide flexibi
f the rubbe
aring s
shings are
mostly av
ness for lae
may be ta
is the diam
of rubber b
Desi
lshese coupli
plings are
.
of flexible
gure- A typ
e torque isrrangemen
d coupling
lity and th
bushing t
tress
available
ailable in t
ger bores.
en to be 1.
eter of the
shing. We
gn of
ngs can acc
available
oupling an
ical flexible
transmittet shafts ne
the rubber
se couplin
e design fo
for differen
ickness be
Brass slee
mm. The o
dr =
olt or pin,
may now
Frict
ommodate
commercial
d a typical
coupling w
from oned be aligne
bushings o
can accom
r pins shou
t inside an
tween 6 m
es are ma
utside dia
b+2 tbr+2
tbris the t
rite
ion D
some misal
ly and prin
coupling of
ith rubber
alf of the cd very well
ver the pin
odate so
ld be consid
d out side
to 7.5 mm
de to suit
eter of rub
tr
ickness of
ives
ignment an
cipal featu
this type i
ushings.
oupling to.
s (bolts) (a
e misalign
ered carefu
diameters.
for bores
the require
er bushin
he brass sl
Chaptd impact.
res of only
shown in
he other t
shown in
ent.
lly.
However
pto 25 mm
ments. Ho
dris given
eeve and tr
er 2large
a few
igure
rough
igure
ubber
and 9
ever,
by
is the
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Design of Friction Drives
S K Mondals Chapter 2
2.
2c
r b
dn d t p T =
Where dcis the bolt circle diameter and t2the 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
325
dn= + 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 dneckis bolt diameter at the neck i.e. at the shear plane.
Bending Stress
The pin loading is shown in Figure below.
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S K
Clearly t
bending o
Knowing
using app
We may a
Hub diam
Hub lengt
Pin diam
ond
e bearing
the pin. C
he shear a
ropriate th
lso assume
eter = 2d
h = 1.5d
ter at the
Desi
ls
Figure- L
pressure t
nsidering
d bending
ories of fail
the followi
eck =0.
gn of
ading on a
at acts as
n equivale
b =
stresses w
ure.
g empirica
d
Frict
pin suppor
distribute
nt concentr
( 23
2
br
F t
d
may chec
l relations:
ion D
ing the bus
d load on
ted load F
)/ 2
the pin di
ives
hings.
rubber bus
pt2d the b
ameter fo