Shafting(Lec15)

Shafting

ME 147P

• Shaft: is a rotating machine element which is used to transmit power from one place to another and subjected to torsion or a combination of torsion, bending and axial loading.

• In order to transfer the power from one shaft to another, the various members such as pulleys, gears etc., are

mounted on it.

• Shaft is used for the transmission of torque and bending moment.

• The various members are mounted on the shaft by means of keys or splines.

• The shafts are usually cylindrical, but may be square or cross-shaped in section. They are solid in cross-section but sometimes hollow shafts are also used.

• An axle, though similar in shape to the shaft, is a stationary machine element and is used for the transmission of bending moment only.

• It simply acts as a support for some rotating body such as hoisting drum, a car wheel or a rope sheave.

• A spindle is a short shaft that imparts motion either to a cutting tool (e.g. drill press spindles) or to a work piece (e.g. lathe spindles).

Types of Shafts

• 1. Transmission shafts. These shafts transmit power between the source and the machines absorbing power. The counter shafts, line shafts, over head shafts and all factory shafts are transmission shafts. Since these shafts carry machine parts such as pulleys, gears etc., therefore they are subjected to bending in addition to twisting.

• 2. Machine shafts. These shafts form an integral part of the machine itself. The crank shaft is an example of machine shaft.

Standard Sizes of Transmission Shafts

• For SI Units: 25 mm to 60 mm with 5 mm steps; 60 mm to

110 mm with 10 mm steps ; 110 mm to 140 mm with 15 mm steps ; and 140 mm to 500 mm with 20 mm steps.

• For English Units: Use the preferred sizes in our previous

Stresses in Shafts

• Pure Torsion • Pure Bending



16



hollow shaft shaft solid 16 4 4 3       i o o s s s D D TD S D T S J Tr S  



32



hollow shaft shaft solid 32 4 4 3       i o o f f f D D MD S D M S I Mc S  

Deflections of Shafts

• Angular Deflection





hollow shaft 32 shaft solid 32 4 4 4       G D D TL G D TL JG TL i o     

• Lateral Deflection

• For Simply Supported Shaft:

• For Cantilever: EI FL y 48 3  EI FL y 3 3 

Combined Loads on Shafts

• Axial and Flexural Loads:

            8 32 4 32 3 2 3 max max FD M D D F D M S N S or N S S A F I Mc S _allow y u   

• Axial and Torsional Loads:

 

 

                                              2 2 3 2 2 max 2 2 3 2 2 3 2 8 8 16 2 2 : Theory Stress Normal Max. on Based 8 16 2 : Theory Stress Shear Max. on Based 16 ; 4 max FD T FD D S S S S FD T D S S S D T J Tr S D F A F S s s s s    

• Flexural and Torsional Loads:

 

 



2 2



3 2 2 max 2 2 3 2 2 3 3 16 2 2 : Theory Stress Normal Max. on Based 16 2 : Theory Stress Shear Max. on Based 16 ; 32 max M T M D S S S S M T D S S S D T J Tr S D M I Mc S s s s s                              

• Axial, Flexural and Torsional Loads

 

 

              _                    _                   _    2 2 3 2 2 max 2 2 3 2 2 3 3 8 8 16 2 2 : Theory Stress Normal Max. on Based 8 16 2 : Theory Stress Shear Max. on Based 16 ; 8 32 max FD M T FD M D S S S S FD M T D S S S D T J Tr S FD M D A F I Mc S s s s s    

• Maximum Shear Stress Theory (Guest’s Theory)

• This theory is applicable for ductile materials.

• The equation shows the maximum shear stress that would result in a combination of shear stress and normal stress applied to a machine member.

• This is also based on the assumption that a ductile material is most probably going to fail by shearing.

Maximum Stress Theories

 

2 2

2

max

















S

S

S

_{s s}

For design considerations, Ss_max is equated to the allowable stress.

• Maximum Normal Stress Theory (Rankine’s Theory)

• This theory is applicable for brittle materials.

• The equation shows the maximum normal stress that

would result in a combination of shear stress and normal stress applied to a machine member.

• This is also based on the assumption that a brittle material is most probably going to fail by normal stress (e.g.

bending)

 

2 2 max

2

2



















S

S

S

• This is based on the two aforementioned stress theories described.

• The ff. expressions are used as design stresses based on ASME Code for Design of Transmission Shafting:

Shaft Design by ASME Code

d d u y d s s u y s

S

S

S

S

S

S

S

S

S

S

d d d

75

.

0

keyway,

a

with

is

shaft

If

smaller.

is

whichever

use

36

.

0

or

6

.

0

:

Stress

Design

Normal

75

.

0

keyway,

a

with

is

shaft

If

smaller.

is

whichever

use

18

.

0

or

3

.

0

:

Stress

Design

Shear

' '













Combined Torsion and Bending on Shaft

 

moment

twisting

equivalent

;

16

2

32

16

2

:

applies

theory

stress

shear

max.

ductile,

is

material

If

32

;

16

2 2 2 2 3 2 3 2 3 2 2 3 3 max max





















































M

T

M

T

D

S

D

M

D

T

S

S

S

D

M

I

Mc

S

D

T

J

Tr

S

s s s f s

















_s

 

_m



_sd s m s

S

M

K

T

K

D

S

K

K











2 2 3

16

bending

for

shear;

for

9.1

Table

from

fatigue

and

shock

combined

to

due

factors

g

Considerin

max



 









equivalent

bending

moment

16

2

32

16

2

32

2

2

:

applies

theory

stress

normal

max.

brittle,

is

material

If

2 2 2 2 3 max 2 3 2 3 3 2 2 max





















































M

T

M

M

T

M

D

S

D

M

D

T

D

M

S

S

S

S

_s















_s

 

_m



_d m m s

S

M

K

T

K

M

K

D

S

K

K











_

_







2 2 3 max

16

bending

for

shear;

for

9.1

Table

from

fatigue

and

shock

combined

to

due

factors

g

Considerin



Table 9.1, Recommended Values of K_m and K_s, from the textbook by Faires is similar to the table below:













































































4 3 4 4 4 4 4 4 3 4 4 4 4 4

1

1

32

32

32

1

1

16

16

let

;

16

j

D

M

D

j

D

MD

S

D

D

MD

I

Mc

S

S

j

D

T

D

j

D

TD

S

jD

D

D

D

j

D

D

TD

J

Tr

S

o o o o i o o f o o o o s o i o i i o o s















 





_s

 

_m



_d m o s m s o s

S

j

M

K

T

K

M

K

D

S

S

j

M

K

T

K

D

S

d

























_

_























4 2 2 3 max 4 2 2 3

1

1

16

:

use

applies,

theory

stress

normal

max.

brittle,

is

material

If

1

1

16

:

use

applies,

theory

stress

shear

max.

ductile,

is

material

If

max





If shaft is solid, the equation above can also be referred to by using D_o=D and j=0 since D_i=0.

Combined Axial and Bending Loads on Shaft









_

_

tensile is load axial if 1, action column for factor shaft hollow for 8 1 1 1 32 shaft solid for 8 32 2 4 3 max 3 max          _               _        d m o d m S j FD M K j D S S FD M K D S







fr.Johnson's formula



4 1 1 , 120 30 For B. formula s Euler' fr. , 120 For A. 2 2 2 2                               E k L S k L E k L S k L e y e e y e    

Column Factor

In case of long shafts (slender shafts) subjected to

compressive loads, a factor known as column

factor (α) must be introduced to take the column

Combined Axial, Torsional and Bending Loads on Shaft













_

_





d o m s o m o s o m s o s S j FD M K T K j FD M K j D S S j FD M K T K j D S d                                     _           2 2 2 2 4 3 max 2 2 2 4 3 8 1 8 1 1 1 16 : theory Stress Normal Max. From 8 1 1 1 16 : theory Stress Shear Max. From max     

If shaft is solid, the equation above can also be referred to by using D_o=D and j=0 since D_i=0.

Sample Problem 1

• A solid circular shaft is to transmit 20 hp at 600 rpm. It also supports a bending load of 2000 in-lbs. The shaft is made of AISI C1020, as rolled steel. A gear is fastened at the midpoint by means of a key. If the load is gradually applied with a

reversed bending, determine the shaft diameter. • Solve the above if shaft is hollow where D_o=2D_i.









psi S psi S lbs in M lbs in N P T u y rpm 65000 48000 steel rolled as C1020, AISI of Properties 2000 . 83 . 2100 600 63025 20 63025        



 















5

.

1

;

0

.

1

:

bending

reversed

with

load

applied

gradually

p.279,

9.1,

Table

From

:

factors

fatigue

and

shock

combined

For

8775

11700

75

.

0

75

.

0

keyway,

with

is

shaft

But

11700psi.

is,

that

smaller,

is

whichever

Use

11700

65000

18

.

0

18

.

0

14400

48000

3

.

0

3

.

0

16

:

theory

stress

shear

max.

ductile,

is

material

If

' ' max 2 2 3





























m s s s u s y s s m s s

K

K

psi

S

S

psi

S

S

psi

S

S

S

M

K

T

K

D

S

d d d d d



 









 





 









 









in

in

D

S

D

S

d s s

8

3

1

286

.

1

8775

2000

5

.

1

83

.

2100

0

.

1

16

8775

2000

5

.

1

83

.

2100

0

.

1

16

:

values

Solving

3 2 2 2 2 3 ' max























 









 







 























 







 









in in D D psi S S psi S S psi S S S D S S M K T K M K D S d d u d y d d d m s m 4 1 1 24 . 1 17550 2000 5 . 1 83 . 2100 0 . 1 2000 5 . 1 16 17550 23400 75 . 0 75 . 0 keyway, with is shaft But 23400psi. is, that smaller, is whichever Use 23400 65000 36 . 0 36 . 0 28800 48000 6 . 0 6 . 0 2000 5 . 1 83 . 2100 0 . 1 2000 5 . 1 16 16 : material brittle For 3 2 2 ' ' 2 2 3 max ' 2 2 3 max       _{ }                _{ }       _{ }    

Sample Problem 2

diametersof shafts x and y : Required keyways. have shafts Both applied. gradually is Load 20 gears of angle Pressure 78000 60000 : properties material Shaft System on Transmissi : Given     psi S psi S u y



 

















 





 

lbs in in lbs L F M lbs F F F lbs F F lbs D T F M lbs in N P T S M K T K D S x R r t R t r x t x s m s s _d                        66 . 2235 4 8 83 . 1117 4 . 83 . 1117 32 . 382 42 . 1050 32 . 382 20 tan 42 . 1050 tan . 42 . 1050 6 25 . 3151 2 2 : For . 25 . 3151 600 63025 30 16 ductile, is material Assume : shaft x 2 2 2 2 2 2 3 ' max  













5

.

1

;

0

.

1

:

bending

reversed

with

load

applied

gradually

p.279,

9.1,

Table

From

:

factors

fatigue

and

shock

combined

For

10530

14040

75

.

0

75

.

0

keyway,

with

is

shaft

But

14040psi.

is,

that

smaller,

is

whichever

Use

14040

78000

18

.

0

18

.

0

18000

60000

3

.

0

3

.

0

:

Stress

Design

'























m s s s u s y s

K

K

psi

S

S

psi

S

S

psi

S

S

d d d d

 









 





 









 









in

in

D

S

D

S

d s s

8

3

1

3

.

1

10530

66

.

2235

5

.

1

25

.

3151

0

.

1

16

10530

66

.

2235

5

.

1

25

.

3151

0

.

1

16

:

values

Solving

3 2 2 2 2 3 ' max























 









 







 























 







 









in in D D psi S S psi S S psi S S S D S S M K T K M K D S d d u d y d d d m s m 4 1 1 24 . 1 21060 66 . 2235 5 . 1 25 . 3151 0 . 1 66 . 2235 5 . 1 16 21060 28080 75 . 0 75 . 0 keyway, with is shaft But 28080psi. is, that smaller, is whichever Use 28080 78000 36 . 0 36 . 0 36000 60000 6 . 0 6 . 0 66 . 2235 5 . 1 25 . 3151 0 . 1 66 . 2235 5 . 1 16 16 : brittle is material If 3 2 2 ' ' 2 2 3 max ' 2 2 3 max       _{ }                _{ }       _{ }    



 



 

 











lbs in in lbs L F M lbs F lbs F lbs F M lbs in T rpm N D D D C D D N N N P T S M K T K D S y R R r t y y y x y x x y y s m s s _d                       49 . 3353 4 12 83 . 1117 4 . 83 . 1117 ; 32 . 382 ; 42 . 1050 : For . 5 . 6302 300 63025 30 300 12 6 600 ; " 12 " 6 " 9 2 2 ; ; 16 ductile, is material Assume : y shaft ' max 2 2 3 













5

.

1

;

0

.

1

:

bending

reversed

with

load

applied

gradually

p.279,

9.1,

Table

From

:

factors

fatigue

and

shock

combined

For

10530

14040

75

.

0

75

.

0

keyway,

with

is

shaft

But

14040psi.

is,

that

smaller,

is

whichever

Use

14040

78000

18

.

0

18

.

0

18000

60000

3

.

0

3

.

0

:

Stress

Design

'























m s s s u s y s

K

K

psi

S

S

psi

S

S

psi

S

S

d d d d

 









 





 









 









in

in

D

S

D

S

d s s

8

5

1

57

.

1

10530

49

.

3353

5

.

1

5

.

6302

0

.

1

16

10530

49

.

3353

5

.

1

5

.

6302

0

.

1

16

:

values

Solving

3 2 2 2 2 3 ' max























 









 







 























 







 









in in D D psi S S psi S S psi S S S D S S M K T K M K D S d d u d y d d d m s m 2 1 1 47 . 1 21060 49 . 3353 5 . 1 5 . 6302 0 . 1 49 . 3353 5 . 1 16 21060 28080 75 . 0 75 . 0 keyway, with is shaft But 28080psi. is, that smaller, is whichever Use 28080 78000 36 . 0 36 . 0 36000 60000 6 . 0 6 . 0 49 . 3353 5 . 1 5 . 6302 0 . 1 49 . 3353 5 . 1 16 16 : brittle is material If 3 2 2 ' ' 2 2 3 max ' 2 2 3 max       _{ }                _{ }       _{ }    

More Problems

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• A shaft supported at the ends by ball bearings carries a straight tooth spur gear at its mid span and is to transmit 7.5 kW at 300 rpm The pitch circle diameter of the gear is 150 mm. The distances between the centre line of bearings and gear are 100 mm each. If the shaft is made of steel and the allowable shear stress is 45 MPa, determine the diameter of the shaft. The pressure angle of the gear may be taken as 20° and the load may be assumed as gradually applied.

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tensions 5.4 kN and 1.8 kN on the tight side and slack side of the belt respectively. Both these tensions may be assumed to be vertical. If the pulley be overhung from the shaft, and the distance of the centre line of the pulley from the centre line of the bearing being 400 mm, find the diameter of the shaft. Assuming that the yield strength of the shaft material is 187 MPa and the load is gradually applied with reversed bending. The pulley is also mounted on the shaft with a key.

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maximum tensile or compressive stress is not to exceed 56 MPa. What size of the shaft will be required, if it is