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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
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Fracture and Fatigue 2013
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
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Fracture and Fatigue 2013
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 ):
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Fracture and Fatigue 2013
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)
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Fracture and Fatigue 2013
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
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Fracture and Fatigue 2013
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 )¿:
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Fracture and Fatigue 2013
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
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Fracture and Fatigue 2013
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
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Fracture and Fatigue 2013
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 )
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Fracture and Fatigue 2013
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:
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Fracture and Fatigue 2013
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 ):
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Fracture and Fatigue 2013
Appendix Material property:
Size factor for circular section in rotating bending:
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Fracture and Fatigue 2013
Stress concentration factors K t under bending:
Notch sensitivity charts for steel:
Fatigue strength fractionf :
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