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CHAPTER 1

Introduction to Mechanical

Engineering Design

CONTENTS 1.0 Design 2.0 Mechanical Engineering Design 3.0 Standards and Codes 4.0 Units 5.0 Specifications 6.0 Safety 7.0 Etc 8.0 Examples 1 9.0 Examples 2 10.0 Question & Answer

Chapter 6 Fatigue Failure Resulting from

Variable Loading

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Introduction

Non rotating shaft

F

Under static failure, the stress on the member is constant Problem: static failure, a very large deformation will occur on the structure or machine members. .

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Rotating shaft

F

Non rotating shaft

- F to F

F

F

-F

Alternating or fluctuating stresses on member will cause the member subjected to fatigue failure.

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R. R Moore Rotating Beam: Figure 6-9 pp 274

87 mm

7.6 mm

250 mm

Testing specimen Subject to pure bending (no traverse shear) Very well machined and polished No circumferential scratches

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Fatigue Strength and Cycle Graph (S-N Graph): Figure 6-10 pp274 F

atig

ue

Str

en

gth

S f

Endurance

limit Se

Cycle

Life

High CycleLow Cycle

Infinite LifeFinite Life

N

A

A

Cycle, N

103

S e

100 101 102 103 104 105 106 107 108

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Terminology

N : Cycle : one rotation of the specimen = 1 cycle of alternating stress

Sf : fatigue strength is the limit of strength where failure occurs when the alternating

stress above the fatigue strength. However, when cycle is larger than 106, the fatigue

strength is constant to a value Se. This value is called Endurance Limit

Cycle: Cycle is related to the cycle of the specimen: boundary is 103

Life: The life is related to the stress of the specimen: boundary is Se.

Significance to design Designers have a choice to set the design life whether finite life or infinite life. Designers can predict the failure to occur (finite life): i.e. when the stress of the

member is sA the predicted life will be NA cycles.

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Endurance limit

Steel

MPaSMPa

MPaSSS

ut

utut

e1400700

14005.0'

Eq 6.8 pp 282

kpsiSkpsi

kpsiSSS

ut

utut

e200100

2005.0'

Note: Sut = 3.41 HB Mpa Eq 2.17 pp 37 Notes: Various class of cast iron, polished and machined: Table A-24 pp 1046 Aluminum alloy: does not have an endurance limit, fatigue strength from Table A-24 is set to 5(108) cycles

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Relationship between fatigue strength Sf and cycle N

ba

b

f

aN

aNS

1

Eq 6.13 and 6.14 pp 285

where a: slope of the curve (log vs log)a1

b: Sf @ 103

e

ut

e

ut

S

Sfa

S

Sfb

2)(

)log(3

1

Eq 6.14 and 6.15 pp 285

Note: f is fatigue strength factor @103

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How to determine fatigue strength factor f @103 1. From Figure 6.18 pp 285 2. Calculation

B

ut

F

Sf )102(

' 3

Eq 6-10 pp 284

where )N2log(

)S/'log(B

e

eF

Eq 6.12 pp 284

Mpa345S' utF Eq 6.11 pp 284

kpsiSutF 50'

*Please differentiate between b (6.15) and B (6.12) General equation for fatigue strength

B

Ff NS )2(' Eq 6-9 pp 284

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ENDURANCE LIMIT (Se)

eedcbae SkkkkkS '

where S’e : endurance limit based on R.R. Moore experiment ka: surface factor kb: size factor kc: loading factor kd: temperature factor ke:miscellaneous-effects factor

Discuss: Why do we need these variables? What do you expect the value of these variables and why?

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Example Calculate the endurance limit Se, if Material AISI 1010 HR, surface is machined until D = 30 mm, d = 20mm and r (fillet radius) = 2 mm is achieved.

d D

r

Surface Factor ka

b

uta aSk Eq 6.19 pp 287

a and b: Table 6-2 pp 280 Discuss: should the factor a and b be hot rolled or machined.

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Size Factor kb

mmdd

mmddkb

2545151.1

5179.224.1157.0

107.0

Eq 6.20 pp 288

indd

inddkb

10291.0

211.0879.0157.0

107.0

Note: only applicable to round rotating shaft due to torsion or bending

Other conditions i. Axial load

kb = 1 (for axial load) * Please refer to kc

ii. Round non-rotating shaft

95% stress area is applied 2

95.0 d0766.0A

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Dd

Dd

DA

e

e

37.0

0105.00766.0

0105.0

22

2

95.0

iii. For non circular non rotating beam 5.0)(808.0 hbde

iv. Common non-rotating structural *Refer to Table 6-3 pp 290 for common non rotating structural Loading factor kc Bending kc = 1 Axial kc = 0.85 Eq 6.26 pp 290

Temperature Factor

Refer to Table 6-4 pp 291

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Reliability Factor (ke)

Refer to Table 6-5 pp 293

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Stress concentration and notch sensitivity See Fig 6.3, 6,4 : fatigue failure originates from stress concentration Table A-13: Charts for theoretical Stress Concentration Factor (Kt) For the example above the suitable chart is Figure A-13-9

0 0.05 0.10 0.15 0.20

r/d

0.25 0.301.0

1.4

1.8

2.2

2.6

3.0

D/d=1.5

1.54

Figure A-13-9

geometrical property to be used

in calculating the stress.

o

4

Roundshaft withshoulder fillet

Mcinbending, ,where

I

d dc ,andI

2 64

K

t

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Fatigue stress concentration factor (Kf)

)1K(q1K tf Eq 6.32 pp 295

Kts: shear stress concentration factor Kfs: fatigue shear stress concentration factor

)1K(q1K tsshearf s Eq 6.32 pp 295

q, qshear : notch sensitivity from Figure 6-20 and 6-21 respectively.

If the value of notch sensitivity can not be retrieved from the respective figures, the most conservative assumption is that Kf = Kt or Kfs = Kts

Or using calculation

r

a

KK T

f

1

11

Eq 6-33 pp 296

And Refer to 6-15 pp 335

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Therefore the maximum stress will be

0fmax K or 0fmax K Eq 6.30 pp 285

Studio: example 6-17

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Combination of loading modes Combination purely alternating bending, purely alternating axial and alternating torsional stresses

Alternating axial stress )axial(0)axial(f)axial( K

Alternating bending stress )bend(0)bend(f)bend( K

Alternating torsional stress )torsion(0)torsion(f s)torsion( K

Total alternating stress (DET)

2

)torsion(

2

)axial(

)bend(t )(385.0

*Please note that kc for axial and bending are 0.85 and 1 respectively

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Application of S-N curve to fatigue design. Infinite life

Te

alln

S

Finite life

b

1

a

b

f

aN

aNS

where nT

a

Discuss Example 6-8 and 6-9 Question 6-11

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Contoh

A Z

X

Y

Rajah 1

Rajah 1 menunjukkan sebahagian dari pemegang untuk satu mesin. Pemegang merupakan rod yang dikimpal di A pada mesin

tersebut. Rod ini diperbuat dari AISI 1020 HR dan berdiameter 15mm.

Jika momen pada A menghasilkan tegasan lenturan ulang-alik a =30 MPa. Daya paksi di A menghasilkan tegasan paksi ulang-alik

(avea 20MPa.

Berdasar kepada data di atas jawab soalan berikut

a) Kirakan ketahanan lesu (endurance limit) Se bahan tersebut

b) Kirakan faktor keselamatan jika tumpuan tegasan yang disebabkan oleh kimpalan ialah Kf untuk lenturan = 1.7 dan Kf untuk

daya paksi = 1.2.

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Fluctuating Stresses

a : amplitudestress

Str

ess

m : midrangestres

max : maximumstress

min : minimumstress

Time

Midrange Stress 2

minmaxm

Alternating Stress 2

minmax

a

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Similarly for the fluctuating forces

Midrange Force 2

FFF minmax

m

Alternating Force 2

FFF minmax

a

These forces will generate respective midrange and alternating stresses.

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How to determine the midrange or alternating stress? Moment on rotating shaft.

MAI

4

32 S

tre

ss

Time1

2

3

4

1

2

3

4

1

When the shaft rotates in ccw direction, element A will rotate from position 1, 2, 3 and 4

for one full cycle. At the same time, the stresses of A will fluctuate from 0, smax, 0 and

smin. Therefore, moment on rotating shaft will generate sa = Mc/I and sm = 0.

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Torsion on rotating shaft

AI

4

32 S

tre

ss

Time

A

ta = 0 and tm = Tr/J.

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Determine the alternating and midrange stresses for the following cases? Rotating shaft

i. Axial load (P) ii. Combination of T and M

Non rotating shaft Moment (M) between -10Nm and 10Nm 20 Nm and 50 Nm -30 Nm and 10Nm Torsion (T) between 10Nm and 10Nm 20 Nm and 50 Nm -30 Nm and 10Nm

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Stress Ratio max

min

R

Amplitude Ratio m

aA

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Fatigue Failure Criteria: graphical and empirical method

Load line

Sy

Se

Sy Sutm

A

B

C

D

a

a m

a1 m1

a2 m2

a3 m3

A : ( , )

B : ( , )

C : ( , )

D : ( , )

Langer Line

Gerber Line

Modified Goodman Line

Load Line

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Graphical Method

Goodman line m

1m

a

1a SSn

Gerber Failure Theory m

2m

a

2a SSn

Langer Line m

3m

a

3a SSn

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Empirical Method

Goodman Line

Basic line equation 1S

S

S

S

e

a

ut

m

For a known (sa, sm) nSS e

a

ut

m 1

Gerber Failure Locus (Parabolic Equation)

Basic parabolic equation 1

S

S

S

S

e

a

2

ut

m

For a known (sa, sm) 1

S

n

S

n

e

a

2

ut

m

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Langer Line (failure due to yielding)

Basic line equation 1

y

a

y

m

S

S

S

S

For a known (sa, sm) n

S

n

1

SS

ma

y

y

a

y

m

Other failure theories, same principles can be applied.

*In the following discussion, these three theories will be used to demonstrate and

similar principle can be applied to other theories.

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Langer line vs. other theories (i.e. Modified Goodman and Gerber Theory)

Sy

Se

Sy Sut

a

m

Sy

Se

Sy Sut

a

m

If the stresses on the member is plotted on the shaded area, the load line of the member

will intersect with Langer line first. Therefore, failure due to yielding will occur first. This is

type of failure is called Langer-first-cycle failure. Empirically, it can be determined when

nlanger < nf(other).

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Procedure in solving the fatigue problem

1. Determine sa, sm ta and tm.

2. Combine the alternating stresses using DET

3. Combine the mean stresses using DET

4. Using the theories to solve the problem

Question 6-39 (pp 352)

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Sy

Se

Sy Sut

a

m

Sy

Se

Sy Sut

a

m

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100 101 102 103 104 105 106 107 108

Fa

tig

ue

Str

en

gth

Sf

Endurance limit Se

Cycle

Life

High CycleLow Cycle

Infinite LifeFinite Life

N

A

A

Cycle, N

103

Se

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Notes

Purely alternating bending stresses ( 0 )

e

f

dcda

eedcdae

eall

SK

kkkk

SkkkkkS

n

S

'1

'

0

Combination purely alternating bending and purely alternating axial stress

Alternating axial stress )(ao

Alternating bending stress )(bo

Total alternating stress )()( boaoo

0 n

Seall

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What is the value of kc and Kf as both factors have different values Bending : kb (eq) kc = 1 and Kf from bending

Axial : kb,0778.023.1 utc Sk and Kf from axial

Soln : let kb,kc and Kf is common in Se which is equal to 1

Axial stress (assuming n =1)

)(

)(

)(

)(

)(

)()(

0

'111

1

'1

ao

ac

af

eda

b

aoe

af

dacaba

ae

all

k

KSkk

k

SK

kkkk

n

S

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Bending stress

)(

)()(

)(

)(

)(

)()(

'111

'1

bo

bbbc

bf

eda

boe

bf

dbcbba

obe

all

kk

KSkk

SK

kkkk

n

S

Final equation with factor of safety n

)(

)()(

)(

)(

)(

)('bo

bbbc

bf

ao

ac

afeda

kk

K

k

K

n

Skk

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CONTOH

A Z

X

Y

Rajah 1

Soalan

Rajah 1 menunjukkan sebahagian dari pemegang untuk satu mesin. Pemegang merupakan rod yang dikimpal di A pada mesin

tersebut. Rod ini diperbuat dari AISI 1020 HR dan berdiameter 15mm.

Jika momen pada A ialah (30 10%)Nm dan daya paksi ialah (8 10%) kN. Untuk memastikan getaran diambil kira, 10%

ditambah.

Berdasar kepada data di atas jawab soalan berikut

c) Kirakan ketahanan lesu (endurance limit) Se bahan tersebut

d) Tentukan a, m,a dan m jika faktor tumpuan tegasan yang disebabkan oleh kimpalan ialah Kf untuk lenturan = 1.7 dan Kf

untuk daya paksi = 1.2.

Kirakan faktor keselamatan terhadap kegagalan alahan dan kegagalan lesu. Yang mana akan gagal dahulu : alahan atau lesu?