Shaft,Keys and Coupling

8/10/2019 Shaft,Keys and Coupling

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CHAPTER 04 Shaft, Keys & coupling

Introduction, Application , Numericaletc.


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SHAFT

! In machinery, the general term shaft refers to a member, usually

of circular cross-section, which supports gears, sprockets, wheels,

rotors, etc.! Transmission shaft refers to rotating machine element, which

support gear, pulley, and sprocket, and transmit power.

! Shaft are usually cylindrical, may be square or cross-shaped in

section.

! The shaft is always stepped with maximum diameter in the middle

& minimum at end to mount the bearing

! Fillet provided to avoid stress concentration due to abrupt change in

cross section


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SHAFT

" An axleis a non-rotating member that used to supports rotating

member wheels, pulleys which is fitted o housing by bearing and

carries no torque.

" A spindle is a short rotating shaft. Used in al machine tools,

Terms such as line shaft, head shaft, transmission shaft,countershaft, are names associated with special usage.

"

Counter shaft secondary shaft, driven by the main shaft

" Jackshaft Intermediate shaft between two shaft, function

same as countershaft

"

Line shaft Consist no. of shaft, connected in axial direction by

coupling, used in group drive


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!A good practice for Material selection

!

Start with an inexpensive, low or medium carbon steel for the firsttime through the design calculations.

! Applications requiring greater strength often specify alloy steels,

Nickel, chromium corrosion applications call for brass, stainless,Ti, or others.

!

The cost of the material and its processing must be weighed

against the need for smaller shaft diameters.

! Manufacturing of Shafts

!

For low production, turning is the usual primary shaping process.

An economic viewpoint may require removing the least material.

!

High production may permit a volume conservative shaping

method (hot or cold forming, Forging)

!

Shoulder! It allows precise positioning

!

Support to minimize deflection.

! In cases where the loads are small, positioning is not very

important, shoulders can be eliminated.


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ELEMENTSATTACHEDTOASHAFT

Shoulders provide axial positioning location, & allow for larger center shaft diameter

where bending stress is highest.


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SHAFTDESIGNFORSTRESS

! It is not necessary to evaluate the stresses in a shaft at every

point; a few potentially critical locationswill suffice. Criticallocations will usually be on the outer surface.

! Possible Critical Locations, axial locations where:

! The bending moment is large and/or

! The torque is present, and/or

! Stress concentrations exist.

! Stresses due to

! Shear stress (Transmission torque)

! Bending stress (machine element)

! Combined Torsional & Bending loads


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STRESSDUETOTORSION

! Stress should not exceed yield point due to twisting moment

!

Diameter of shaft obtained by Torsion equation

T

J

r

=

Where,

! T = Twisting moment

! J = Polar moment of inertia = ( / 32) x d4

!

= Torsional shear stress! r = Distance from NA to outer most fiber = d / 2

T( / 32) x d4

d / 2

=

T = ( / 16) xx d3

Torque can be obtained by Power formula


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SUM1

!A line shaft at 200rpm is to transmit 20kw. Shaft is

made of up mild steel of shear stress 42MPa. Determinethe diameter of shaft by torsion formula

! Solution

! Calculate T = torque

! Power = (2

NT / 60 )

! Calculate d = Shaft Dia


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STRESSDUETOBENDINGMOMENT

! Stress should not exceed yield point due to twisting moment

!

Diameter of shaft obtained by pure bending equation

M

I

b

y=

Where,

! M = Bending moment

! I = Moment of inertia = ( / 64) x d4

!

b= Bending stress! y = Distance from NA to outer most fiber = d / 2

M( / 64) x d4

bd / 2

=

M = ( / 32) xbx d3

S C &


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STRESSDUETOCOMBINEDTORSION& BENDING

MOMENT

! When shaft subject to two moment simultaneously various theories

are suggested for elastic failure of the material!

Maximum shear stress theory or Guests Theory Ductile Material

! Maximum normal stress theory or Ranking's Theory Brittle material

"

Maximum shear stress theory or Guests Theory

- Max shear stress at shaft

!max= (1 / 2)x [(b)2+ 4(!)2]1/2

!max= (1 / 2) x [(32M/ d3)2+ 4 (16T/ d3)2]1/2

!max= (16 / d3) x [ M2+

T2 ]1/2

[ M2+ T2 ]1/2 = Equivalent twisting moment = Denoted by Te


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"

Maximum normal stress theory or Ranking's Theory

b(max)= (1 / 2)b+ {(1 / 2)x [(b)2+

4(!)2

]1/2

}

b(max)= (32/ d3

) x { (1/2) M + [M2

+ T2

]1/2

{(1/2) M +[ (M2+ T2)!]} = Equivalent Bending moment

= Denoted by Me

M = ( / 32) xbx d3

Puttingb &!In above equation

T = (

/ 16) xx d3

We get,


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STRESSDUETOAXIAL, COMBINEDTOTORSION&

BENDINGLOAD

! Shaft is subjected to axial load (tensile or compressive)

P = F / A = F / [(

/ 4) x d2]

! Resultant Stress

3 2

32 4

x

M F

d d

!

" "

= +


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Static or Quasi-Static Loading on

Shaft

The stress at an element located

on the surface of a solid roundshaft of diameter dsubjected to

bending, axial loading, and

twisting is

3 232 4

xM F

d d!

" "

= +

3

16

xy

T

d!

"

=

Normal stress

Shear stress

Non-zero principal

stresses

1 22

2,

2 2

x y x y

A B xy

! ! ! !

! ! "

# $+ %& ' & '= +( )* + * +

( ), - , -. /

CH-18

LEC

29

Slide13


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keys

! A key is the piece inserted in an axial direction between a shaft and

hub of the mounted machine element such as pulley or gear etc.

! A key is used for temporary fastening, Key joint consist of shaft, hub,and key

! The shaft and rotating machine element must have akeyway, also

known as akey seat, which is a slot or pocket the key fits in. The

whole system is called akeyed joint

! 2 basic function of key

!

Primary function is to transmit the torque from shaft to the hub ofmating element and vice versa

! Secondary function is to prevent rotational motion between the shaft and

the hub like gear or pulley.


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! Slot machined on the shaft & hub called keyway, usually cut by

milling cutter

! Drawback keyway result in stress concentration in the shaft & part

becomes weak

! Material usually plain carbon steel to withstand shear andcompressive stresses

! Types of keys are

! Saddle key & sunk key

!

Square key & flat key

! Key with & without Gib head

! Special types


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Square & rectangular key! The sunk keys are provided half in the keyway of the shaft and

half in the keyway of the hub or boss of the pulley.

1.

Rectangular sunk key

- Also called flat key

-

More stability than square key

-

Used in machine tool

2.

Square sunk key

- Used in general industrial machinery! The only difference between a rectangular sunk key and a square

sunk key is that its width and thickness are equal, i.e. b = h

! Selecting key without stress analysis Thumb Rule

! Square, b = h = (d / 4), l = 1.5 d

! Rectangular b = d / 4, h = (2b / 3) = (d / 6), l = 1.5 d

Where, d = Diameter of shaft (mm)

b = Width of key (mm)

h = Height of key (mm)

l = Length of key (mm)


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Failure of key in shear: occurs at Plane AB

! Plane AB

P P

b x l

2Mt

d x b x l=

Where, d = Diameter of shaft (mm)b = Width of key (mm)

h = Height of key(mm)l = Length of key (mm)

Strength of a Sunk Key

Considering shearing of thekey, the tangential shearingforce acting at the

circumference of the shaft:

P = Area resisting shearing #Shear stress


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Failure of key in Compressive

P"

c

Surface AC

P

(h/2) x l

2Mt

(h/2) x l xd

Crushing between shaft and key : Considering crushing of the key, thetangential crushing force acting at the circumference of the shaft:

P = Area resisting crushing #Crushing stress

Square Key h = b

4Mt

h x l x d

4Mt

b x l x d

"c


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! The permissible crushing stress for the usual keymaterial is at least twice the permissible shearing stress

!

From above equation, We get

!

4Mt

d x b x l"c

2Mt

d x b x l=

"c = 2 !Sum 1

required to design a square key for fixing gear on a shaft of 25mm diameter.

The shaft transmitting 15kw power at 720rpm to the gear. Ultimate tensile

stress is 460MPa, factor of safety 3. Determine the dimensions of the key.

Solution :1. find out permissible stress

2. Find out torque (Mt) by power formula

3. Calculate b = h = (d / 4)

4. Calculate lby shearing

5.

Check the crushing stress by putting above value


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Coupling

! All manufacturing plants have at least 2 things in common. These

are:

!

Electric motors and.

!

Couplings - There are many different types, sizes and styles.

! All couplings;

1. Connect 2 shafts together 2. Require safety guards.

! Most couplings;

1. Do more than 1 function at a time.

2. Connect 2 shafts together.

3. Allow for minor misalignment.

4. Allow for axial movement.


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!

Two major classification

! Solid couplings (also called rigid or sleeve couplings)

! Flexible couplings

" Rigid coupling : It is used to connect two shafts which are

perfectly aligned.

Types of rigid coupling are

# Sleeve or muff coupling.

# Clamp or split-muff or compression coupling,# Flange coupling

" Flexible coupling : It is used to connect two shafts having both

lateral and angular misalignment.

Types of flexible coupling are

# Bushed pin type coupling,

# Universal coupling, and

#

Oldham coupling

M ff C li


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Muff Coupling

!Also called sleeve or box coupling

! It is the simplest type of rigid coupling, made of cast iron.

! It consist of hollow cylinder or sleeve, shaft (input & output) & key

! Sleeve inner diameter is the same as that of the shaft diameter.

! It is fitted over the ends of the two shafts by means of a sunk key, the

power is transmitted from one shaft to the other shaft by means of a

key and a sleeve.

! Simple in design & manufacture.

D i d


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Design procedure

1. Calculated the diameter of shaft (d) : Shaft subjected to pure Torsionalmoment,

2. Usually defined on shop floor. Empirical relationship were developed

by engineer on the basis their past experience then check for

shearing D = 2d + 13 L = 3.5d

Where, D = Outer diameter of the sleeve

L = axial length of sleeve

3. Determine the dimension of keyl = L / 2 each shaft

4. Check shear & compressive stress in the key

4Mt

d x b x l

"c

2Mt

d x b x l=!

Mtx r

J=!

(D4 d4)

32J =

16 Mt

d3=!

D

2=r


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SUM1

! Design muff coupling to connect two steel shaft transmitting

25kw power at 360rpm. The shaft and key are made of plaincarbon steel of ultimate yield stress 400MPa. Factor of safety for

shaft & key 4. the sleeve made of grey cast iron, ultimate yield

stress 200MPa. Factor of safety for shaft & key 6.

! Solution :

! Find out permissible stress

!

Diameter of shaft

! Dimensions of sleeve

!

Dimensions of key!

Key for stresses in key

C C


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CLAMPCOUPLING

! Second type of rigid coupling, Also called split muff coupling or

compression coupling

! In this, sleeve is made of two halves, which are split along a planepassing trough axes of shaft

! The two halves of the sleeve are clamped together by means of bolts

! No. of bolts can be four or eight

! It is easy to assemble & dismantle

D i d


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Design procedure

1. Calculated the diameter of shaft (d) : Shaft subjected to pure Torsionalmoment,

2. Usually defined on shop floor. Empirical relationship were developedby engineer on the basis their past experience

D = 2.5 d L = 3.5 d

Where, D = Outer diameter of the sleeve

L = axial length of sleeve

3. Determine the dimension of key

l = L / 2 each shaft

4. Check shear & compressive stress in the key

4Mt

d x b x l"

c

2Mt

d x b x l=!

Mtx r

J

=!


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! Calculate the diameter of clamping bolts by using, (d1)

1. By empirical,

when d < 55 mm ------- d = diameter of shaft

d1= 0.2d + 10 mmd > 55 mm

d1= 0.15d + 15 mm

4. Clamping force (P1) : Each boltP1= ( / 4) d1

2x!t

2. Clamping force (N) : Each Shaft (half bolts)

N = (P1 x n) / 2

where n = Total no. of bolts

3. Frictional force : between sleeve halves & shaft

2 x Mt

f x d x n

=P1=Mtfx d x P1 x n

2


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Shaft,Keys and Coupling

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