Fatigue Assessment

UTS Fracture and Fatigue

Steven TRAN- 11228184

Neslon TRAN- 11025513

Danny DU- 11227966

Fracture and Fatigue 2013

TABLE OF CONTENT

Problem 1:............................................................................................................................3

Step 1: Estimating the actual endurance limit(S ' e):.........................................................4

Step 2: Determine nominal bending stress.......................................................................5

Step 3: Fatigue factor of safety and check whether the infinite life is predicted..............7

Step 4: calculate the cycle to failure (finite life)................................................................7

Problem 2:............................................................................................................................9

Step 1: Estimating the actual endurance limit(S ' e):.......................................................10

Step 2: Determine the applied normal stresses..............................................................11

Step 3: Fatigue factor of safety and check whether the infinite life is predicted............11

Step 4: Calculate the cycle to failure (finite life)..............................................................12

Appendix................................................................................................................................13

2


Problem 1:The shaft shown in the figure is machined from AISI 1040 CD steel. The shaft rotates at 1600-rpm and is supported in rolling bearings at A and B. the applied forces are F1=10kN andF2=4 kN . Determine the minimum fatigue factor of safety based on achieving infinite life. If infinite life is not predicted, estimate the number of cycles to failure. Also check for yielding.

All calculations were taken to 4 decimal places to prevent error when rounding off.

For all figures and tables mentioned, please refer to the appendix

Given:Material: Cold-Drawn, AISI 1040 steel

Type of stress: Bending stress

Forces: F1=10kN , F2=4kN

3


Step 1: Estimating the actual endurance limit(S'e):

Using the formula: S 'e=Se(C surf )(Cm)(C st)(CR)(CS)(CT )

From table A-20 material property, AISI 1040 Cold-Drawn steel:

Sut=590MPa , Sy=490MPa

For actual endurance (Se ):

Using the property: Se={ 0.5Sut , Sut≤1400MPa700Mpa,Sut>1400MPa

Se=0.5×590=295MPa , as Sut=590MPa<1400MPa

For surface factor (C surf ):

Using the formula: C surf=a(Sut)b

And the property for a machined surface finishes:

Factor (a) = 4.51 and Exponent (b) = - 0.265

C surf=4.51(590)−0.265=0.8316

For material factor (Cm):

Cm=1 , as the material is steel

For stress factor (C st):

Using the property: C st={ 1.0 ,Bending0.8 , axial loading0.59 , torsion

C st=1.0 , as it’s bending stress

For reliability factor (CR):

CR=1 , as reliability was not mentioned

For size factor (CS ) :

Using the property for circular section in rotating bending:

For size range 7.62<D≤50 where D is in mm, CS=( D70.62

)−0.11

CS=( 4070.62

)−0.11

=0.8333 , where D = 40mm as it’s the smallest diameter of the shaft

For temperature factor (CT ):

4


CT=1 , as temperature was not mentioned

Therefore actual endurance limit is:

S 'e=Se(C surf )(Cm)(C st)(CR)(CS)(CT )

S 'e=295×0.8316×1×1×1×0.8333×1

S 'e=204.4268MPa

Step 2: Determine nominal bending stress Reaction Force calculation

∑F y=0; F A−F1−F2+FB=0

F A−10−4+FB=0 (1)

∑M A=0 ; −F1[0.192]−F2[0.384 ]+FB[0.576]=0

FB [0.576 ]=10 [0.192 ]+4 [0.384] (2)

Solve (1) and (2)

FB=6kN , F A=8kN

Shear Force calculation

V A=8kN

V C=8−10=−2kN

V E=8−10−4=−6kN

V B=8−10−4+6=0kN

Bending Moment calculation (LHS)

MC=−8 [0.192 ]=−1.536 kN .m (Hogging)

MD=−8 [0.252 ]+10 [0.252−0.192 ]=−1.416 kN .m (Hogging)

ME=−8 [0.384 ]+10 [0.192 ]=−1.152kN .m (Hogging)

MF=−8 [0.492 ]+10 [0.492−0.192 ]=−0.504kN .m (Hogging)

MD=−8 [0.576 ]+10 [0.384 ]+4 [0.192 ]=0kN .m

Check RHS

MF=−6 [0.084 ]=−0.504 kN .m (Check OK)

5


8

-2

-6

0

Shear Force Diagram (V, kN)

C DA E F B

0

1.536 1.416

1.152

0.504

0

Bending Moment Diagram (M, kN.m)

A C D E F B

The critical location occurs at point D, (the shoulder fillet) as this is where the bending moment is large, the diameter is small and the stress concentration exists.

Determine the nominal bending stress at plane D.

σ o=32M

π d3

σ o , D=32 [1.416 ] [106]

π 403=225.3634MPa

Yielding is not predicted as σ o , D<SY

6


225.3634MPa<490MPa

Step 3: Fatigue factor of safety and check whether the infinite life is predicted

Using the formula n f=

1σa

S ' e+σ m

Sut

where σ a=k f σao and K f=1+q (K t−1)

The notch sensitivity factor (q):

Using figure 6-20 notch sensitivity chart for steel:

ForSut=590Mpa,r=1.5mm an estimated value q=0.76

For stress concentration factor (K t):

Using figure A-15-9 stress concentration factor K t under bending:

At rd=1.540

=0.037 andDd

=4540

=1.125, It is estimated that K t=1.95

From this K f=1+0.76 (1.95−1 )=1.722

σ a=1.722×225.3634=388.0758MPa

Therefore the fatigue factor of safety:

n f=204.4268388.0758

=0.5268 , as σ m=0

Fatigue factor of safetyn f<1 , thus infinite life doesn’t exist.

Step 4: calculate the cycle to failure (finite life)Using Shigley’s S-N equation:

S f=a N b , where a=( f Sut)

2

Se

and b=−13log(

f Sut

Se

)

For fatigue strength factor ( f ):

Using figure 6-18 fatigue strength fraction:

ForSut=590MPa , an estimated value f=0.867

a=(0.867×590)2

204.4268=1279.9835

b=−13log( 0.867×590204.4268 )=−0.1328

For the equivalent fatigue stress corresponding to the applied loading (S¿¿ f )¿:

7


Using Goodman’s equation: σa

Sn∨S+σm

Sut

=1 ≈σa

Sf

+σm

Sut

=1

Rearranging the equation: S f=

σa

1−σ m

Sut

As σ m=0 ∴S f=σa=388.0758MPa

Rearranging Shigley’s equation to obtain the number of cycles (N):

N=( S f

a )−1b =( 388.0758

1279.9835)

−10.1328=7994 , therefore it takes 7994 cycles to fail.

Checking yield factor of safety(nY ):

Using the equation nY=Sy

σo

= 490225.3634

=2.1743

8


Problem 2:The cold drawn AISI 1040 steel bar shown in the figure is subjected to a completely reversed axial loading fluctuating between 28KN in compression to 28KN in torsion. Estimate the fatigue factor of safety based on achieving infinite life, and the yield factor of safety. If infinite life is not predicted, estimate the number of cycles to failure.

All calculations were taken to 4 decimal places to prevent error when rounding off.

For all figures and tables mentioned, please refer to the appendix

Given:Material: Cold-Drawn, AISI 1040 steel

Type of stress: Axial loading

Forces: F=±28kN

9


Step 1: Estimating the actual endurance limit(S'e):

Using the formula: S 'e=Se(C surf )(Cm)(C st)(CR)(CS)(CT )

From table A-20 material property, AISI 1040 Cold-Drawn steel:

Sut=590MPa , Sy=490MPa

For actual endurance (Se ):

Using the property: Se={ 0.5Sut , Sut≤1400MPa700Mpa,Sut>1400MPa

Se=0.5×590=295MPa , as Sut=590MPa<1400MPa

For surface factor (C surf ):

Using the formula: C surf=a(Sut)b

And the property for a machined surface finishes:

Factor (a) = 4.51 and Exponent (b) = - 0.265

C surf=4.51(590)−0.265=0.8316

For material factor (Cm):

Cm=1 , as the material is steel

For stress factor (C st):

Using the property: C st={ 1.0 ,Bending0.85 , axial loading0.59 , torsion

C st=0.85 , as the material is under axial loading

For reliability factor (CR):

CR=1 , as reliability was not mentioned

For size factor (CS ) :

CS=1, as the material is under axial loading

For temperature factor (CT ):

CT=1 , as temperature was not mentioned

Therefore actual endurance limit is:

S 'e=Se(C surf )(Cm)(C st)(CR)(CS)(CT )

10


S 'e=295×0.8316×1×0.85×1×1×1

S 'e=208.5237MPa

Step 2: Determine the applied normal stresses

For maximum stress (σ ¿¿max)¿:

Using the formula σ max=Fmax

A= 28KN190mm2=147.3684MPa

For minimum stress (σ min) :

Using the formula σ min=Fmin

A=−28KN190mm2=−147.3684MPa

For amplitude:

Using the formulaσ ao=σmax−σ min

2

σ ao=147.3684−(−147.3684)

2=147.3684MPa

For mean:

Using the formulaσ m=σmax+σmin

2

σ m=147.3684+(−147.3684)

2=0MPa

Step 3: Fatigue factor of safety and check whether the infinite life is predicted

Using the formula n f=

1σa

S ' e+σ m

Sut

where σ a=k f σao and K f=1+q (K t−1)

The notch sensitivity factor (q):

Using figure 6-20 notch sensitivity chart for steel:

ForSut=590MPa,r=3mm an estimated value q=0.83

For stress concentration factor (K t):

Using figure A-15-1 stress concentration factor K t for bar in tension or simple compression with a transverse hole:

11


At dw

= 625

=0.24It is estimated that K t=2.44

From this K f=1+0.83 (2.44−1 )=2.1952

σ a=2.1952×147.3684=323.5031MPa

Therefore the fatigue factor of safety:

n f=208.5237323.5031

=0.6446 , as σ m=0

Fatigue factor of safetyn f<1, thus infinite life doesn’t exist.

Step 4: Calculate the cycle to failure (finite life)Using Shigley’s S-N equation:

S f=a N b , where a=( f Sut)

2

Se

and b=−13log(

f Sut

Se

)

For fatigue strength factor ( f ):

Using figure 6-18 fatigue strength fraction:

ForSut=590MPa , an estimated value f=0.867

a=(0.867×590)2

208.5237=1254.8355

b=−13log( 0.867×590208.5237 )=−0.1299

For the equivalent fatigue stress corresponding to the applied loading (S¿¿ f )¿:

Using Goodman’s equation: σa

Sn∨S+σm

Sut

=1 ≈σa

Sf

+σm

Sut

=1

Rearranging the equation: S f=

σa

1−σm

Sut

As σ m=0 ∴S f=σa=323.5031MPa

Rearranging Shigley’s equation to get number of cycles (N):

N=( S f

a )−1b =( 323.5031

1254.8355)

−10.1299=34041.75422 , therefore it takes 34041 cycles to fail.

Check for yield factor of safety(nY ):

12


Using the equation nY=S y

σ ao

= 490147.3684

=3.325

13


Appendix Material property:

Size factor for circular section in rotating bending:

14


Stress concentration factors K t under bending:

Notch sensitivity charts for steel:

Fatigue strength fractionf :

15


Theoretical stress concentration factor:

16

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