Shafts and Axles

7/27/2019 Shafts and Axles

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Dr. A. Aziz Bazoune Chapter 18: Axles and Shafts

ME 307MachineDesign I

Dr. A. Aziz Bazoune

King Fahd University of Petroleum & MineralsMechanical Engineering Department

CH-18 LEC 30 Slide 1


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18-1 Introduction .92218-2 Geometric Constraints .92718-3 Strength Constraints .933

18-4 Strength Constraints Additional Methods .94018-5 Shaft Materials .94418-6 Hollow Shafts .94418-7 Critical Speeds (Omitted) .945

18-8 Shaft Design .950



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18-1 Introduction .92218-2 Geometric Constraints .92718-3 Strength Constraints .933



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Neglecting axial loads because they are comparatively very small at critical

locations where bending and torsion dominate. Remember the fluctuatingstresses due to bending and torsion are given by

Fatigue Analysis of shafts

a ma f m f

a ma fs m fs

M C M CK K

I I

T C T CK K

J J

Mm: Midrange bending moment, m: Midrange bending stressMa : alternating bending moment, a: alternating bending stress

Tm: Midrange torque, m: Midrange shear stressTa: alternating torque, m: Midrange shear stressKf: fatigue stress concentration factor for bendingKfs: fatigue stress concentration factor for torsion



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For solid shaft with round cross section, appropriate geometry terms can be

introduced for C, IandJresulting in


32 32

16 16

a ma f m f

a ma fs m fs

M MK K

I I

T TK KJ J

Mm: Midrange bending moment, m: Midrange bending stressMa : alternating bending moment, a: alternating bending stress

Tm: Midrange torque, m: Midrange shear stressTa: alternating torque, m: Midrange shear stressKf: fatigue stress concentration factor for bendingKfs: fatigue stress concentration factor for torsion



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ME 307MachineDesign I Fatigue Analysis of shafts

2 2 1/2 3 2 2 1/2 3

2 2 1/2 3 2 2 1/2 3

' 3 16 / 4( ) 3( ) 16 /

' 3 16 / 4( ) 3( ) 16 /

xya

xym

axa f a fs a

mxm f m fs m

d K M K AT d

d K M K T d B

Combining these stresses in accordance with the DE failure theory the von-

Mises stress for rotating round, solid shaft, neglecting axial loads are given by

2 2 2

3 3

' ' 16 161a m a m

e ut e ut e ut

S S n n nA nB

S S S S d S d S

The Gerber fatigue failure criterion

(18-12

where A and B are defined by the radicals in Eq. (8-12) as

2 2 1/2

2 2 1/2

4( ) 3( )

4( ) 3( )

f a fs a

f m fs m

A K M K T

B K M K T

CH-18 LEC 30 Slide 6CH-18 LEC 30 Slide 6


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ME 307MachineDesign I Fatigue Analysis of shafts

1/31/2

2

281 1 e

e ut

BSnAd

S AS

(18-13

The critical shaft diameter is given by

or, solving for 1/n, the factor of safety is given by

1/22

3 21 8 1 1e

e ut

BSAn d S AS

(18-14



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(18-15)

where


2 2

2 2

4( ) 3( )

4( ) 3( )

'

'

f a fs a

f m fs m

a

m

A K M K T

B K M K T

Ar

B



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2

3

f a

fs m

A K M

B K T

Particular Case

Critical ShaftDiameter



1/31/2

216

1 1 3f a fs m e

e f a ut

nK M K T Sd

S K M S

(18-16)

Safety Factor

1/22

3

1611 1

f a fs m e

e f a ut

K M K T S

n d S K M S

(18-17)

For a rotating shaft with constant bending and torsion, thebending stress is completely reversed and the torsion is

steady. Previous Equations can be simplified by settingMm= 0 andTa = 0, which simply drops out some of the terms.



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2'' 3

f aa

m fs m

K MrK T

(18-18)




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Shaft Diameter Equation for the DE-EllipticCriterion

2 2 1/2 3 2 2 1/2 3

2 2 1/2 3 2 2 1/2 3

' 3 16 / 4( ) 3( ) 16 /

' 3 16 / 4( ) 3( ) 16 /

xya

xym

axa f a fs a

mxm f m fs m

d K M K AT d

d K M K T d B

Remember

2 2 22 2 2

3 3

' ' 16 161a m a m

e y e y e y

S S n n nA nB

S S S S d S d S

The Elliptic fatigue-failure criterion is defined by

(18-12

where A and B are defined by

2 2 1/2

2 2 1/2

4( ) 3( )

4( ) 3( )

f a fs a

f m fs m

A K M K T

B K M K T



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Substituting for A and B gives expressions for d, 1/n and r:

(18-19

Critical Shaft Diameter

1/31

2 2 22/2

4 3 316

4f m fsf a fs a

e e

m

y y

K M K T

S S

K M

S

n K T

S

d

Safety Factor

2/2

2 2

12

34

643 3

1 1 f m fs m

y

a

e y

f fs a

e

K M K T

S S

K M K T

d S Sn

(18-20)



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2 2

2 2

4 3'

' 4 3

f a fs aa

mf m fs m

K M K T Ar

B K M K T



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2

3

f a

fs m

A K M

B K T

Particular Case

Critical Shaft Diameter (18-21)

Safety Factor (18-22)

1/31/2

22

46

31 f a f

ye

s mK Tn

dK M

SS

1

2/2

3

2

164

13

fs m

y

f a

e

K M

Sn

T

d

K

S

For a rotating shaft with constant bending and torsion,the bending stress is completely reversed and thetorsion is steady. Previous Equations can be simplified

by settingMm = 0 andTa = 0, which simply drops outsome of the terms.



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2

2

4 2'

' 33

f a f aa

m fs mfs m

K M K MAr

B K TK T

At a shoulderFigs. A-15-8 and A-15-9 provide information about Kt andKts.

For a hole in a solid shaft, Figs. A-15-10 and A-15-11 provide about Kt andKts .

For a hole in a solid shaft, use Table A-16 For grooves use Figs. A-15-14 and A-15-15



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The value of slope at which the load line intersects the junction of

the failure curves is designated rcrit.

It tells whether the threat is from fatigue or first cycle yielding

Ifr> rcrit, the threat is from fatigue

Ifr< rcrit

, the threat is from first cycle yielding.



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For the Gerber-Langer intersection the strength components Sa and

Sm are given in Table 7-10 as

22 2

1 1 12

1

yut e

me ut e

a y m

y m yacrit

m m m

SS SS

S S S

S S S

S S SSrS S S

(18-23)



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For the DE-Elliptic-Langer intersection the strength components Sa

and Sm are given by

2

2 2

2y e

a

e y

m y a

a a

critm y a

S SS

S S

S S S

S S

r S S S

(18-24)



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Note that in an analysis situation in which the diameter is known and the

factor of safety is desired, as an alternative to using the specialized equationsabove, it is always still valid to calculate the alternating and mid-rangestresses using the following Eqs.

and substitute them into the one of the equations for the failure criteria ,

Eqs. (7-48) to (7-51) and solve directly for n.



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In a design situation, however, having the equations pre-solved for

diameter is quite helpful.

It is always necessary to consider the possibility ofstatic failure in

the first load cycle.

The Soderberg criteria inherently guards against yielding, as can

be seen by noting that its failure curve is conservatively within the

yield (Langer) line on Fig. 727, p. 348.

The ASME Elliptic also takes yielding into account, but is not

entirely conservative throughout its range. This is evident by

noting that it crosses the yield line in Fig. 727.

The Gerber and modified Goodman criteria do not guard against

yielding, requiring a separate check for yielding. A von Mises

maximum stress is calculated for this purpose.



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To check for yielding, this von Mises maximum stress is compared to the

yield strength, as usual

For a quick, conservative check, an estimate for maxcan be obtained by

simplyadding a andm . (a + m ) will always be greater than or equal

to max, and will therefore be conservative.



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Example

Solution

7-20

7-20



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Figure 7-20

Figure 7-21



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18-22

18-14

18-24

Solderberg



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D A A i B Ch t 18 A l d Sh ft


C 8 C 30 Slid 36

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Shafts and Axles