creating estimated s-n diagram - uprmacademic.uprm.edu/pcaceres/courses/inme4011/md-6b.pdfcreating...

Creating Estimated S-N Diagram

These equations are used to determine the corrected endurance limit Seor the corrected fatigue strength Sf at a particular number of cycles N in the high-cycle region of the S-N diagram. For the low-cycle region the following estimates are found:

utm

utm

SSloadingaxialSSbending

75.0.........:_9.0....................:

==

Sm material strength at 103 cycles

The equation of the line from Sm to Sf (or Se) is: bn aNS =

for materials that possess an endurance limit the coefficients (a, b) can be calculated from the following two points:

for materials that do not possess an endurance limit, use:

Note: NbaSn loglog)log( +=

Boundary conditions:

82

62

31

105__

10__

10__

×===

===

===

NNatSS

NNatSS

NNatSS

fn

en

mn

Material with endurance limit

Material without endurance limit

21 loglog............log1 NNzwhereSS

zb

e

m −=⎟⎟⎠

⎞⎜⎜⎝

⎛=

Material with endurance limit N1=103 and N2=106

N2 (106) z1.0 -3.0005.0 -3.69910.0 -4.00050.0 -4.699100.0 -5.000500.0 -5.699

bn aNS =

Problem: Create an estimated S-N diagram for a steel bar and define its equations. How many cycles of life can be expected if the alternating stress is 100MPa.

Given: The Sut has been tested at 600MPa. The bar is 150mm square and has a hot rolled finish. The operating temperature is 500oC maximum. The loading will be fully reverse bending.

Assumptions: Infinite life is required (ductile steel with endurance limit). A reliability factor of 99.9% will be used.

Solution:

MPaSS ute 3005.0' ==

Loading factor is bending

0.1=loadk

The part is not round:

mmmmAd

mmmmmmbhA

equiv 2.1210766.0

11250766.0

112515015005.005.02

95

295

===

=××==

The size factor:

747.0)2.121(189.1

189.1.........:2508_097.0

097.0

==

=≤≤−

−

size

size

k

dkmmdmmfor equivalent

Surface Factor:

584.0)600(7.57 718.0 =×=×= −butSurface SAk

Temperature Factor: ( ) ( ) 71.04505000058.014500058.01 =−−=−−= TkTemp

Reliability Factor 753.0=yreliabilitk

Corrected Endurance limit:

MPaS

SkkkkkS

e

eyreliabilitTempsurfsizeloade

70300753.071.0584.0747.00.1

'

=×××××=

×××××=

Creating the S-N diagram

MPaSSbending utm 5406009.09.0....................: =×==

295765.070

540log31log1

−=⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

e

m

SS

zb

7.4165)295765.0(3)540log(3)log()log(

=−×−=−=

abSa m

NbSaNbaS

n

n

log)log(logloglog)log(

−=+=

cyclesNNMPa

5

295765.0

100.3)7.4165(100

×=

×= −

N2 (106) z1.0 -3.0005.0 -3.69910.0 -4.00050.0 -4.699100.0 -5.000500.0 -5.699

bn aNS =

Kt is the theoretical stress concentration factor.Notch sensitivity – q . Materials have different sensitivity to stress concentrations, referred to as the notch sensitivity of the material. In general, the more ductile a material is, the less sensitivity to notches it is.It depends on notch radius, the smaller the radius, the less sensitive the material is

Notches and Stress Concentrations

where a is Neubers constant and it is solely a function of the material, and r is the notch radius (both expressed in inches).

Neuber, Kuhn and Peterson have developed an approach to notch sensitivityPeterson’s equation

Problem: A rectangular step steel bar (as shown below) is to be loaded in bending. Determine the stress-concentration factor for the given dimensions.

Given: H=2in, h=1.8in, r=0.25in. The material has Sut=100kpsi.

Solution:

11.18.1

2

1388.08.125.0

==

==

hHhr

Kt=1.56 89.0

25.0062.01

1

1

1062.0

=+

=+

=

=

ra

q

a

( )( )

50.1

156.189.01

11

=

−+=

−+=

f

f

tf

K

K

KqK

Design of a Cantilever Bracket for Fully Reversed BendingProblem: Design a cantilever bracket to support a fully reversed bending load of 500-lb amplitude for 109 cycles with no failure. Its dynamic deflection can not exceed 0.01in.. Calculate the safety factor.

Given: Beam width (b) = 2in ; beam depth over length (d) = 1in ; beam depth in wall (D) = 1.125in ; fillet radius (r) = 0.3in ; applied load amplitude at point (F) = 500-lb ; beam length (l) = 6in ; distance to load (a) = 5in ; distance for deflection calculation (lx) = 6in ; Modulus of elasticity (E) = 3x107psi ; ultimate tensile strength (Sut) = 80kpsi for a steel. The cantilever has been machined and operates at a temperature of 120oF

Solution:

5.021

2

1667.012

1212

2500500

33

===

=×

==

−=×===

dc

bdI

inlbaFMlbFR

a

a

Bending at the root

3.013.0

125.11125.1

==

==

drdD

33.1=tK

psiI

Mcnoma 7500

1667.05.02500

, =×

==σ

lbF

lbF

mean

amplitude

0

500

=

=

( )( )

29.1

133.187.01

11

=

−+=

−+=

f

f

tf

K

K

KqK

87.0

30.0080.01

1

1

1080.0

=+

=+

=

=

ra

q

a

psiK nomafa 9675750029.1, =×== σσ

The stress is a principal stress as there is no shear stress at the top surface and there are no other stresses present.

psipsipsi

psi

aVM

aa

aab

96750,9675,

48322

96752

,

31

==

===

σσσ

στ

psiSS ute 40000800005.05.0' =×==

Creating the S-N diagram.

inAd

indbA

equivalent 14.10766.0

1.02105.005.0

95

295

==

=××==

Size factor:

inAd

indbA


1.02105.005.0

95

295

==

=××==

( ) 86.014.1869.0

869.0.........:.103.0_097.0

097.0

=×=

=≤≤−

−

size

size

k

dkindinfor

Load Factor: Pure bending 1=loadk

Surface factor:

( ) 845.0807.2 265.0 =×== −butsurf ASk

Temperature factor: 0.1=Tempk

Reliability factor: As only ten of these parts are required, a high % reliability is chosen (99.9%) 753.0=yreliabilitk

Corrected endurance limit:

psiSpsiS

SkkkkkS

e

e


2184340000753.01845.086.01

'

=×××××=

=

Predicted Safety factor: 26.2967521843

,

===aVM

esf

Snσ

The beam deflection y is calculated

( )[ ] ( )( ) ( )[ ] inaxaxxEIFy 005.0566536

1667.010365003

6323

7323 −=−−××−

×=−−−=

Many repeating and fluctuating stresses have non-zero mean components and these must be taken into account.

Designing for fluctuating Uniaxial Stresses

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

ut

mea S

S σσ 1Modified-Goodman line (for design)

Gerber Parabola (for analysis of failed parts)

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

2

1ut

m

SSea

σσ

Design of a Cantilever Bracket for Fluctuating BendingProblem: Design a cantilever bracket to support a fluctuating bending load of 100 to 1100-lb amplitude for 109 cycles with no failure. Its dynamic deflection can not exceed 0.02in.. Calculate the safety factor.

Given: Beam width (b) = 2in ; beam depth over length (d) = 1in ; beam depth in wall (D) = 1.125in ; fillet radius (r) = 0.3in ; maximum applied load amplitude at point (F) = 1100-lb ; minimum applied load amplitude at point (F) = 100-lb ; beam length (l) = 6in ; distance to load (a) = 5in ; distance for deflection calculation (lx) = 6in ; Modulus of elasticity (E) = 3x107psi ; ultimate tensile strength (Sut) = 80kpsi for a steel; yield strength (Sy) = 60kpsi. The cantilever has been machined and operates at a temperature of 120oF

Solution:

inlbinlbaFMinlbinlbaFMinlbinlbaFM

lbFRlbFRlbFR

lbFFF

lbFFF

MaxMax

mm

aa

MaxMaxmmaa

MinMaxa

MinMaxm

−=×==−=×==−=×==

======

=−

=−

=

=+

=+

=

5500511003000560025005500

1100.....600.......500

5002

10011002

6002

10011002

5.021

2

1667.012

1212

33

===

=×

==

dc

bdI

psiI

cM

psiI

cM

mnomm

anoma

90001667.0

5.03000

75001667.0

5.02500

,

,

=×

==

=×

==

σ

σ

3.013.0

125.11125.1

==

==

drdD

33.1=tK

Verify that the stress concentration are below yield strength of the material.

psipsiK

psipsiK

psiSpsiI

cMKK

KKthenSKif

nommmfm

nomafa

yMax

fMaxf

ffmyMaxf

11610900029.1

9675750029.1

60000212811667.0

5.0550029.1

..............

,,

,

=×==

=×==

=<=×

==

=<

σσ

σσ

σ

σ

( )( )

29.1

133.187.01

11

=

−+=

−+=

f

f

tf

K

K

KqK

87.0

30.0080.01

1

1

1080.0

=+

=+

=

=

ra

q

a

Calculate the Von Mises for the alternating stress and the mean stress

psi

psi

mxymymxmymxm

axyayaxayaxa

11610030116100116103

967503096750967532222

,,,2,

2,

'

2222,,,

2,

2,

'

=×+×++=+−+=

=×+×++=+−+=

τσσσσσ

τσσσσσ

Generating a S-N : Size factor:

inAd

indbA


1.02105.005.0

95

295

==

=××==

( ) 86.014.1869.0

869.0.........:.103.0_097.0

097.0

=×=

=≤≤−

−

size

size

k

dkindinfor

Load Factor: Pure bending 1=loadk

Surface factor: ( ) 845.0807.2 265.0 =×== −butsurf ASk

Temperature factor: 0.1=TempkReliability factor: As only ten of these parts are required, a high % reliability is chosen (99.9%)

753.0=yreliabilitk


psiSpsiS

SkkkkkS

e

e


2184340000753.01845.086.01

'

=×××××=

=

Various safety factors can be calculated:It assumes that the alternate and mean component will have a constant ratio.

70.1

8000011610

218839675

1=

+=fn

The beam deflection y is calculated

( )[ ] ( )( ) ( )[ ] inaxaxxEI

Fy Max 012.05665361667.01036

110036

3237

323 −=−−××−×

=−−−=

82.2116109675

60000=

+=

+=

ma

yy

Sn

σσ

Applying Stress-Concentration Effects with Fluctuating Stresses

0........2....

...........

..............

,,

,

,

=>−

−=>

=<

fmynomMinnomMaxf

nomm

nomafyfmyMaxf

ffmyMaxf

KthenSKif

KSKthenSKif

KKthenSKif

σσ

σσ

σ

σ

Multiaxial Fluctuating Stresses

Problem: Determine the safety factors for the brackets tube shown.

Given:Material: 2024-T4 aluminum with Sy=47ksi and Sut=68ksi.Tube length (l) = 6in. ; arm (a) = 8in. ; Tube outside diameter (OD) = 2in ; Tube inside diameter (ID) = 1.5in The applied load varies sinusoidally from F=340 to -200lb

Assume finite life design 6x107 cycles.The component has been machined. It operates at RT and with 99.9% reliability.

Notch radius at the wall = 0.25inStress concentration factor for bending = 1.7For shear = 1.35

Solution:Aluminum does not have an endurance limit. As Sut > 48kpsi then Sf=19kpsi at 5x108 cyclesCalculation of the correction factors:

Load Factor: Bending 1=loadk097.0869.0.........:.103.0_ −=≤≤ equivalentsize dkindinforSize factor:

895.0)74.0(869.0

74.00766.0

010462.00766.0

010462.0

097.0

295

295

=×=

===

=

−ink

indAd

dA

size

equivalent

This value is used despite the fact that both bending and torsion are present.

Surface factor: (machined) ( ) 883.0687.2 265.0 =×== −butsurf ASk

Temperature factor: (room temperature) 0.1=Tempk

Reliability factor: As only ten of these parts are required, a high % reliability is chosen (99.9%)

753.0=yreliabilitk


psiS

psiS

SkkkkkS

f

f

eyreliabilitTempsurfsizeloadf

11299

19000753.01883.0895.01

'

=

×××××=

=

At 5x108 cycles

The problem calls for a life of 6x107 cycles, so the strength value at that life must be estimated.

ksiSSbending utm 2.61680009.09.0.....: =×==

1287.01129961200log

699.51log1

−=⎟⎠⎞

⎜⎝⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛=

f

m

SS

zb

148929)1287.0(3)61200log(3)log()log(

=−×−=−=

abSa m

( )psiS

NS

n

n

14846106)148929()148929( 1287.071287.0

=××=×=

−−bn aNS =

Notch sensitivity (for hardened aluminum Sut=68000psi)

( )( )

541.1

17.1773.01

11

=

−+=

−+=

f

f

tf

K

K

KqK

773.0

25.0148.01

1

1

1148.0

=+

=+

=

=

ra

q

a

Bending Torsion

( )( )

270.1

135.1773.01

11

,

,

,,

=

−+=

−+=

sf

sf

stsf

K

K

KqK

The bracket tube is loaded in bending (as a cantilever) and in torsion. All the loading are maximum at the wall. Need to find the alternating and mean components of the applied force, moment andtorque at the wall.

inlbinlbaFinlbinlbaF

inlbinlblFMinlbinlblFM

lbFFF

lbFFF

mm

aa

mm

aa

MinMaxm

MinMaxa

−=×==Τ−=×==Τ−=×==−=×==

=−+

=+

=

=−−

=−

=

5608702160827042067016206270

702

)200(3402

2702

)200(3402

The fatigue stress concentration factor for the mean stresses depends on the relationship between the maximum local stress in the notch and the yield strength

270.1.........

541.1.......

4700058555369.0

16340541.1

..............

==

==

=<=××

==

=<

ffm

ffm

yMax

fMaxf

ffmyMaxf

KKtorsion

KKbending

psiSpsiI

lcFKK

KKthenSKif

σ

σ

The largest tensile bending stress will be at the top or bottom of the cantilever. The largest torsional shear stress will be all around the outer circumference of the tube.

psiJ

rTK

psiI

cMK

psiJrT

K

psiI

cMK

mmeanshearFtorsionm

mfmm

ashearFtorsiona

afa

663074.1

1560270.1

12055369.0

1420541.1

2556074.1

12160270.1

46495369.0

11620541.1

,,,

,,

=×

==

=×

==

=×

==

=×

==

τ

σ

τ

σFinding the amplitude and mean components due to bending and torsion.

Finding the alternating/amplitude and mean Von Mises stresses at A.

psi

psi

m

mxymymxmymxm

a

axyayaxayaxa

166466330120501205

3

6419255630464904649

3

222'

2,,,

2,

2,

'

222'

2,,,

2,

2,

'

=×+×−+=

+×−+=

=×+×−+=

+×−+=

σ

τσσσσσ

σ

τσσσσσ

Various safety factors can be calculated:It assumes that the alternate and mean component will have a constant ratio.

2.2

680001664

148466419

1=

+=fn

82.516646419

47000=

+=

+=

ma

yy

Sn

σσ

We need to check the shear due to transverse loading at point B in the neutral axis where the torsional shear is also maximal.

psipsi

psiAVK

psiAVK

torsionmtransversemtotalm

torsionatransverseatotala

mmeanshearftransversem

ashearftransversea

72966312930552556499

129374.1

702270.12

499374.12702270.12

,,,

,,,

,,,

,,

=+=+=

=+=+=

=×

==

=×

==

ττττττ

τ

τ

IbVQ

=τ

psi

psi

m

mxymymxmymxm

a

axyayaxayaxa

137279230000

3

5291305530000

3

222'

2,,,

2,

2,

'

222'

2,,,

2,

2,

'

=×+×−+=

+×−+=

=×+×−+=

+×−+=

σ

τσσσσσ

σ

τσσσσσ

Safety factors: It assumes that the alternate and mean component will have a constant ratio.

7.2

680001372

148465291

1=

+=fn

05.713725291

47000=

+=

+=

ma

yy

Sn

σσ

Both points A and B are safe against fatigue failure.

We have a rotating shaft, as shown, made of AISI 1040 hot rolled (Sy = 290MPa; Sut = 524MPa; E = 210GPa). Determine the Moment it can support for 100,000 cycle life.

05.042.0

5.146

==

==

drdD

03.2=tK

097.0

097.0

189.1.........:2508_

869.0.........:.103.0_

1.....:)8_(3.0_

−

−

=≤≤

=≤≤

=≤

dkmmdmmfor

dkindinfor

kmmindfor

size

equivalentsize

size

Determine the Fatigue Stress Concentration

( )( )

791.1

103.2768.01

11

=

−×+=

−+=

f

f

tf

K

K

KqK

768.0

4.25/2085.01

1

1

1085.0

=+

=+

=

=

ra

q

a

TS=524MPa=76ksi( )

085.0

080.080768070

080.0093.08070

080.0093.08076080.0

=

+−⎟⎠⎞

⎜⎝⎛

−−

=

−−

=−−

a

a

a

Determine the S-N Diagram

MPaS

S

kkkkkSS

f

f

yreliabilitTempsurfsizeloadef

5.105

753.016437.0831.01262

'

=

×××××=

×××××=

MPaMPaSS UTe 2625245.05.0' =×=×=

1=−bendingloadk

831.0)40(189.1189.1

400766.0

56.1220766.0

56.122400766.00766.0

097.0097.0

95

22295

=×=×=

===

=×==

−− mmdk

mmAd

mmdA

size

equivalent

( ) 6437.05247.57 718.0 =×== −butsurf ASk

0.1=Tempk 753.0=yreliabilitk

0.9x524=471.6MPa

105.5MPa

105

MPaS

S

S

aNbS

f

f

f

f

8.173

24.2)log(

)2108log()10log()10log()10log(

)5.105log()6.471log()log(

)log()log()log(

563

=

=

+×−−

=

+×=

21677.05.1056.471log

31log1

−=⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

e

m

SS

zb

bn aNS =

N2 (106) z

1.0 -3.000

5.0 -3.699

10.0 -4.000

50.0 -4.699

100.0 -5.000

500.0 -5.6992108

)21677.0(3)6.471log(3)log()log(=

−×−=−=a

bSa m

NbSaNbaS

n

n


−=+=

Calculate the stress (amplitude and mean)

M

MMI

My

MMI

My

rI

Amplitude

Mean

Min

Max

××=

=

××−=×

×−=−=

××=×

×==

×=×

==

−

−

−

5

56

56

644

105915.1

0

105915.1101257.002.0

105915.1101257.002.0

101257.04

)02.0(4

σ

σ

σ

σ

ππ

Stress Concentration

mNMoment

MPaS

M

MK

fCorrectedAmplitude

CorrectedAmplitude

AmplitudefCorrectedAmplitude

−=

==

××=

×××=×=

−

−

−

7.609

8.173

1028504.0

105915.1791.16

5

σ

σ

σσ

P

R

The load P varies between Pmax=8000lb and a Pmin=-4000lb

The load R varies between +/-1000lb

Given:

AISI 4340 Oil Quenched at 855oC and Temper at 230oC 4hours (Sy=200ksi; Sut=260ksi)Machined component working at room temperature and a reliability of 99.99%

Problem: Determine the safety factor for the component shown below. Consider all stress concentration factors at the wall equals to 1.0 and all notch radius of 0.25in. At 1in from the wall the stress concentration factors are Kbending=1.8; K torsion=1.6; Kshear=1.3; Kaxial=1.5


KsiS

S

kkkkkSS

f

f


197.36

702.01619.0833.01100

'

=

×××××=

×××××=

KsiSe 100' =

1=−bendingloadk 833.0)53.1(869.0869.0

53.10766.0

18.00766.0

18.04.25.105.0

097.0097.0

95

295

=×=×=

===

=××=

−− mmdk

inAd

inA

size

equivalent

( ) 619.02607.2 265.0 =×== −butsurf ASk

0.1=Tempk

702.0=yreliabilitk


( )( )

( )( ) 296.1..............13.1986.01

493.1..............15.1986.01

788.1...........18.1986.01

11

,,

,,

,,

=−×+=

=−×+=

=−×+=

−+=

shearfshearf

axialfaxialf

bendingfbendingf

tf

KK

KK

KK

KqK

99.0

25.0005.01

1

1

1005.0

=+

=+

=

=

ra

q

a

43

43

2

675.012

5.14.2

728.112

4.25.16.34.25.1

inI

inI

inAArea

z

x

=×

=

=×

=

=×==

PR At the wall:

5×= PM x

25.325.600.66

25.675.1

×⎟⎠⎞

⎜⎝⎛×+×⎟

⎠⎞

⎜⎝⎛×= RRM z

( ) 25.675.16 22 =+

⎟⎠⎞

⎜⎝⎛×=

25.600.6RRy ⎟

⎠⎞

⎜⎝⎛×=

25.675.1RVx

PVz =

Point A:

A

Bx

y

z

RPRPA

R

I

M

x

x

y 266.0472.32666.0728.1

2.1525.66

24.2

−=−××

=⎟⎠⎞

⎜⎝⎛×

−⎟⎠⎞

⎜⎝⎛×

=σ

RR

AVx

yx 0583.06.3225.675.15.1

23

=×

⎟⎠⎞

⎜⎝⎛×

=××

=τ

RPy 266.0472.3 −=σ

Ryx 0583.0=τ

lbPlbPlbPlbP

AmplitudeMean

MinMax

6000........20004000.......8000==

−==

lbRLbRlbRlbR

AmplitudeMean

MinMax

1000..........01000......1000==

−==

psi

psi

Amplitudey

Meany

205661000266.06000472.3

69442000472.3

,

,

=×−×=

=×=

σ

σ

psiAmplitudeyx

Meanyx

3.5810000583.0

0

,

,

=×=

=

τ

τ

Finding the amplitude and mean Von Mises stresses at A.

psi

psi

m

mxymymxmymxm

a

axyayaxayaxa

6944036944069440

3

205663.583205660205660

3

222'

2,,,

2,

2,

'

222'

2,,,

2,

2,

'

=×+×−+=

+×−+=

=×+×−+=

+×−+=

σ

τσσσσσ

σ

τσσσσσ

Safety factors for Point A at the wall:

46.1

2600006944

3119720566

1=

+=fn

27.7694420566

200000=

+=

+=

ma

yy

Sn

σσ

A

B x

y

z

Point B:RRR

A

R

I

M

z

z

y 066.52666.0675.0

75.08.425.66

25.1

=−××

=⎟⎠⎞

⎜⎝⎛×

−⎟⎠⎞

⎜⎝⎛×

=σ

PPA

Vzyz 208.0

6.325.1

23

=××

=××

=τ

Ry 066.5=σ

Pyz 208.0=τ

lbPlbPlbPlbP

AmplitudeMean

MinMax

6000........20004000.......8000==

−==

lbRLbRlbRlbR

AmplitudeMean

MinMax

1000..........01000......1000==

−==

psipsi

Amplitudey

Meany

50661000066.50

,

,

=×=

=

σ

σ

psi

psiP

Amplitudeyz

Meanyz

12486000208.0

416208.0

,

,

=×=

=×=

τ

τ

psi

psi

m

mxymymxmymxm

a

axyayaxayaxa

72041630000

3

5508124835066050660

3

222'

2,,,

2,

2,

'

222'

2,,,

2,

2,

'

=×+×−+=

+×−+=

=×+×−+=

+×−+=

σ

τσσσσσ

σ

τσσσσσ

6.5

260000720

311975508

1=

+=fn

11.327205508

200000=

+=

+=

ma

yy

Sn

σσ

43

43

2

675.012

5.14.2

728.112

4.25.16.34.25.1

inI

inI

inAArea

z

x

=×

=

=×

=

=×==

PR

At 1in from the wall:

4×= PM x

25.325.600.65

25.675.1

×⎟⎠⎞

⎜⎝⎛×+×⎟

⎠⎞

⎜⎝⎛×= RRM z

( ) 25.675.16 22 =+

⎟⎠⎞

⎜⎝⎛×=

25.600.6RRy ⎟

⎠⎞

⎜⎝⎛×=

25.675.1RVx

PVz =

Point A:

A

Bx

y

z

RPRP

RKPK

A

RK

I

MK

y

axialfbendingfy

axialfx

x

bendingfy

397.097.4266.0493.178.27883.1

2666.0728.1

2.14

25.66

24.2

,,

,,

−=×−×=

×−××

×=

⎟⎠⎞

⎜⎝⎛×

×−⎟⎠⎞

⎜⎝⎛×

×=

σ

σ

σ

RR

AVK x

shearfyx 0755.06.3225.675.15.1

296.123

, =×

⎟⎠⎞

⎜⎝⎛×

×=××

×=τ

RPy 397.097.4 −=σ

Ryx 0755.0=τlbPlbP

lbPlbP

AmplitudeMean

MinMax

6000........20004000.......8000==

−==lbRLbR

lbRlbR

AmplitudeMean

MinMax

1000..........01000......1000==

−==

psi

psi

Amplitudey

Meany

294231000397.0600097.4

9940200097.4

,

,

=×−×=

=×=

σ

σ

psiAmplitudeyx

Meanyx

5.7510000755.0

0

,

,

=×=

=

τ

τ

Finding the amplitude and mean Von Mises stresses at A.

psi

psi

m

mxymymxmymxm

a

axyayaxayaxa

9940039940099400

3

294235.753294230294230

3

222'

2,,,

2,

2,

'

222'

2,,,

2,

2,

'

=×+×−+=

+×−+=

=×+×−+=

+×−+=

σ

τσσσσσ

σ

τσσσσσ

Safety factors for Point A at 1 in from the wall:

02.1

2600009940

3119729423

1=

+=fn

08.5994029423

200000=

+=

+=

ma

yy

Sn

σσ

Examples of fatigue Failure

BEACH MARKSSTRIATIONS INDICATING SLOW FATIGUE

CRACK GROWTH

MULTIPLE CRACK ORIGINOVERLOAD FAILURE

Failure of a steam turbine blade from a nuclear power plant due to fatigue

A fatigue crack that started at the site of a lightning strike is shown below.

A rotating shaft loaded by the 5kN and 10kN forces is represented below. The grinding relief groove at B is 2.5mm deep. The cylindrical surface AB is ground but the groove is machined. The shaft is made of a steel hardened and tempered to SUT=1300MPa and a SYield=1000MPa according to the BS 826M40 (EN26). The shaft transmit a power of 15kW at 900RPM between the loads. A safety factor (on load) of 2 corresponding to a life of 0.35x106 cycles has been used for the shaft design. Is the design correct? If not determine the new life for a S.F.=2. The shaft operates at room temperature with a reliability of 99.9%.

Quiz

Kbending=1.63……KTorsion=1.32


( )( )( ) 305.1..............132.1954.01

601.1...........163.1954.01

11

,,

,,

=−×+=

=−×+=

−+=

shearftorsionf

bendingfbendingf

tf

KK

KK

KqK

954.0

4.2550215.01

1

1

10215.0

=+

=+

=

=

ra

q

a

1300MPa=188.5ksi


MPaS

S

kkkkkSS

f

f


278

753.016745.08422.01650

'

=

×××××=

×××××=

MPaSe 650' = 1=−bendingloadk

( ) 6745.0130051.4 265.0 =×== −butsurf ASk

0.1=Tempk

753.0=yreliabilitk8422.0)35(189.1189.1

350766.0

835.930766.0

835.93350766.00766.0

097.0097.0

95

22295

=×=×=

===

=×==

−− mmdk

mmAd

mmdA

size

equivalent

0.9x1300=1170MPa

278MPa

3.5x105MPaS

S

S

aNbS

f

f

f

f

95.345

539.2)log(

)4.4922log()105.3log()10log()10log()278log()1170log()log(

)log()log()log(

563

=

=

+××−−

=

+×=

208.0278

1170log31log1

−=⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

e

m

SS

zb

bn aNS =

N2 (106) z1.0 -3.0005.0 -3.69910.0 -4.00050.0 -4.699100.0 -5.000500.0 -5.699

4.492269218.3)log()208.0(3)1170log(3)log()log(

=⇒=−×−=−=

aabSa m

NbSaNbaS

n

n


−=+=

345.95MPa

mNRmNRRM A

−=⇒⇒−=×−×+×==∑

428.6571.83501025010050

12

2

642.8N-m

857N-m

MC=214.3N-mMB=749.9N-mMA=160.7N-m

mNT

wattsPmNTrpmnnTP

−=××

=

−=

15.1599002

1500060

)()...()..(....60

2

π

π

T=159.15N-m

Critical point at B:MB=749.9N-mTB=159.15N-m

Bending moment gives the stress amplitude, while the Torque gives the mean stress.

JcTK

IcMK

torquemeantorque

bendingamplitudebending

××=

××=

,

,

τ

σ

( )

( ) 484

484

1073.14035.032

10366.7035.064

mJ

mI

−

−

×==

×==

π

π

6.428kN

1.428kN

8.571kN

( )( )( )

MPamNI

cMK

amp

bendingamplitudebending

23.28510366.7

2035.09.749

601.1 8

,

=×

−×=

××=

−σ

σ

( )( )( )

MPamN

JcTK

mean

torquemeantorque

68.241073.14

2035.015.159

305.1 8

,

=×

−×=

××=

−τ

τ

Finding the Von Mises stresses

( ) MPaMPa

MPa

mVM

aVM

74.4268.243

23.2852

,

,

=×=

=

σ

σ

Modified Goodman

17.1..1300

74.4295.34523.285

..1

=⇒+= FSFS

Langer

05.374.4223.285

1000,,

=+

=+

=MPaMPa

MPaLangerSFampVMmeanVM

Yield

σσσ

Calculating the new life

MPaNN

2.5741300

74.4223.28521

=⇒+= σσ

( )

cyclesN

N

baS

NaNbS ff

30620

486.4208.0

)4.4922log()2.574log()log(

)log()log()log()log()log()log(

=

=−−

=

−=⇒+×=

creating estimated s-n diagram - uprmacademic.uprm.edu/pcaceres/courses/inme4011/md-6b.pdfcreating...

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