MECHANICAL PROPERTIES OF
MATERIALS
Prof. Hüseyin ÇİMENOĞLU
Assoc. Prof. Murat BAYDOĞAN
Selma KASAL 060130326
Esin Çağla SEÇER 060130338
Yusuf Tolunay KOZAN 060150729
Zehra Berin ORDU 06013005
Ufuk GÜRER 060140333
Hülya Telli 060130448
Tunahan Çağlar ÜLKER 060130438
Tamer Ortunç SÜVARİ 060130603
Fatigue Fracture Mechanisms
1.A high-strength steel plate(Figure7-21), which has a plane strain fracture toughness of 80
MPa√𝑚, is alternately loaded in tension to 500 MPa and in compression to 60 MPa.The plate
is to survive for 10 years, with the stress being applied at a frequency of once every 5
minutes.Design a manufacturing and testing procedure that the component will serve as
intended.Assume a geometry factor f = 1.0 for all flaws.
Solution
To design our manufacturing and testing capability, we must determine the maximum size of
any flaws that might lead to failure within the 10-year period.The critical crack size (ac),using
the fracture toughness and maximum stress, is:
𝑚𝑚𝑚 = 𝑚𝑚√𝑚𝑚𝑚
80 𝑚𝑚𝑚√𝑚 = (1)(500 𝑚𝑚𝑚)√𝑚𝑚𝑚
ac=0.0081m=8.1mm
The maximum stress is 500 MPa; however, the minimum stress is zero , not 60 MPa in
compression, because cracks do not propagate in compression.Thus, Δσ is
Δσ= σmax- σmin=500-0=500 Mpa
We need to determine the minimum number of cycles that he plate must withstand:
N=(1 cycle/5 min)(60min/h)(24h/d)(365d/y)(10y)
N=1,051,200 cycles
If we assume that f=1 for all crack lengths and note that C=1.62*10-12 and n=3.2 from Figure
7-21 in equation N, then
2[(𝑚𝑚)(2−𝑚)/2 −(𝑚𝑚)(2−𝑚)/2 ]
N=
1,051,200=
(2−𝑚)𝑚𝑚𝑚 ∆𝑚𝑚 𝑚𝑚/2
2[(0.008)(2−3.2)/2 −(𝑚𝑚)(2−3.2)/2 ]
(2−3.2)(1.62∗10−12 )(1)3.2 (500)3.2 𝑚3.2/2
1,051,200=
2[(18−(𝑚𝑚)−0.6 ]
(−1.2)(1.62∗10−12 )(1)(4.332∗108 )(6.244)
ai-0.6=18+2764=2782
ai=1.82*10-6m=0.00182mm for surface flaws
2ai=0.00364mm for internal flaws
The manufacturing process must produce surface flaws smaller than 0.00182mm in length.We
can conduct a similar calculation for specifying a limit on edge cracks.In addition,
nondestructive tests must be available to assure that cracks exceeding this length are not
present.
2.A solid shaft for a cement kiln produces from tool steel must be 240cm long and must survive
continuous operation for one year with an applied load of 55,600 N.The shaft makes one
revolution per minute during operation.Design a shaft that will satisfy these requirements.The
S-N curve for the tool steel is shown in Figure 7-19
Solution
The fatigue life required for our design is the total number of cycles N that the shaft will
experience in one year:
N=(1 cycle/ min)(60min/h)(24h/d)(365d/y)
N=5.256*105 cycles/y
From Figure 7-19, the applied stress therefore must be less than about 496 MPa
±σ=16FL/(πd^3 )=5.09FL/(d^3 )
(5.09)(240𝑚𝑚)(55,600)𝑚)
496MPa =
(𝑚^3 )
d=11.1cm
A shaft with a diameter of 11.1 cm should operate for one year under these conditions.However,
a significant margin of safety might be incorporated in the design in addition, we might
consider producing a shaft that would never fail.If we assume the factor of safety to be 2 (we
will assume that the maximum allowed stress level will be 496/2=248MPa).The minimum
diameter required to prevent failure would now be :
(5.09)(249𝑚𝑚)(55,600)𝑚)
248MPa=
(𝑚^3 )
d=14 cm
Selection of a larger shaft reduces the stress level and makes fatigue less likely to occur or
delays the failure.Other considerations might, of course, be important.High temperature and
corrosive environment, fatigue is accelerated.Thus, in the applications involving fatigue of
components regular inspections of the components go a long way toward avoiding a
catastrophic failure.
3.The daily, multi-stress loading cycling encountered by a structural element fabricated from
SAE/AISI 2330 Steel is summarized in Table 1 S-N data for the hardened steel is presented
below in figure1. In addition to the data shown, the tensile strength was found to be σuts=250
ksi. Using this loading history and S-N data, determine how many days the structural element
will survive.
Loading Condition
σmax(ksi)
σmin(ksi)
ni(cycles)
1
50
-50
1,000,000
2
175
75
60
3
196
4
4
4
190
-190
0.5
Table 1 General stress history with variations in cyclic amplitude.
Figure 1 S-N data for SAE/AISI 2330 hardened steel.
Solution
In order to determine the component life it is first necessary to find Ni for each condition or
i=1, 2, 3, and 4. The following table summarizes the result of the necessary calculations and
the taking of endurance values, Ni, from the S-N curve.
Loading
Condition
σmax(ksi)
σa(ksi)
σao(ksi)
Ni (from SN)
di=ni/Ni
1
0
50
50
∞
0
2
125
50
100
6*104
10-3
3
100
96
160
4*103
10-3
4
0
190
190
103
0.5*10-3
Table 2 Summary of stresses and associated cycles to failure.
It should be noted that the completely reversed, stress amplitudes were calculated using
𝑚𝑚
𝑚𝑚𝑚 =
𝑚𝑚
(1 − 𝑚𝑚𝑚𝑚)
equation sine the mean stress,σm≠0.Per Miner’s rule, the damage accumulated in one day will
be the ratio of the individual(daily)cycles, ni to the total number of cycles, Ni for a given stress
level.Using this construct, the accumulated damage is:
D=0+0.001+0.001+0.0005=0.0025
Since damage msut equal unity at failure, the total numver of days to failure times the daily
damage must equal the same or:
D(0.0025)=1.0
Solving for the number of days until failure results in D=400 days.
4.A crankshaft in a diesel engine fails. Examination of the crankshaft reveals no plastic
deformation. The fracture surface is smooth. In addition, several other cracks appear at other
locations in the crankshaft. What type of failure mechanism occured?
Since the crankshaft is aa rotating part, the surface experiences cyclical loading. We
should immediately suspect fatigue. The absence of plastic deformation supports our suspicion.
Furthermore, the presence of other cracsis consistent with fatigue; the other cracks did not have
time to grow to the size that produced catastrophic failure. Examination of the fracture surface
will probably reveal beach marks or fatigue striations.
5. A component of a machine is subjected to a repeated loading sequence. The loading
sequence is repeated once a day. The loading is alternating, and one sequence contains the
following loading amplitudes (i.e. stress amplitudes the mean value of the stress is zero)
σa = 180, 160, 140 and 100 MPa
The number of loading cycles at each stress level is
n = 15, 20, 150, and 500, respectively
The Wöhler curve of the material is, in this stress range, given by the relation
σa = 50logN + 400 MPa
Determine the damage accumulation D due to one loading sequence, and then, determine
how many loading sequences (how many days) the component might be used before fatigue
failure is expected.
Solution: For the different stress levels, the expected fatigue life N is obtained from the
Wöhler curve. One obtains
N = 10(400−σ a)/50
giving, respectively,
N = 25 119, 63 096, 158 489, and 1 000 000
The accumulated damage due to one loading sequence is, by use of the Palmgren-Miner
damage accumulation rule,
D = 15 /25119 + 20/ 63096 + 150 /158489 + 500/ 1000000 = 0.00236= 1/424
Finally, the number of sequences Ns to expected fatigue failure is
Ns = 1/D ≅ 424
Thus, the component can be used in 424 days, approximately.
Answer: Damage D due to one sequence is
D = 15/ 25119 + 20/63096 + 150/158489 + 500/ 106 =2.36*10-3
Expected number of sequences (days) Ns to fatigue failure is Ns = 1/D ≅ 424
6. In A514 Steel, σmax = 49 kp / mm2 and KIC = 500 kp / mm3/2. On the material, an
edge crack of a0 = 7.6 mm was encountered. This material is subjected to repetitive loading
with σmax = 25 kp / mm2 σmin = 15 kp / mm2. Calculate material life.
(𝑚 = 1.12 × 𝑚𝑚𝑚𝑚 × √𝑚𝑚
𝑚𝑚
, 𝑚𝑚 = 2.3 × 10
−10
)
𝑚 = 1.12 × 𝑚𝑚𝑚𝑚 × √𝑚𝑚𝑚𝑚
500 = 1.12 × 25 × √𝜋𝑚𝑚𝑚
𝑚𝑚𝑚 = 100mm
𝑚=
1
2.3 × 10−10 × (1.123 ) × 𝑚3/2 × 10
3×
7. 6(1−3/2) − 100
3
2−1
N=289000 cycle
7)A large steel cranks (Kc = 45 MN m-3/2) undergoes cyclic tensile (225 MPa) and
compressive (60 MPa) stresses during use
Prior to use, it was inspected using ultrasonic techniques, from which the largest surface
crack found was 2.5 mm in length
For the steel in question, the Paris Law constants are
A = 1.5 x 10-12 m/(MN m-3/2) m per cycle
m = 2.5
Solution
Kc = 45MN m-3/2
σmax = 225MPa (tensile)
σmin = 60MPa (compressive)
a0 = 2.5mm
A = 1.5 x 10-12 m/(MN m-3/2) m per cycle,
m = 2.5
·
Calculate critical crack length (ac = af ):
�� = (assume Y = 1)
45MN m-3/2=1*225Mpa
ac = 12.7 mm
2) A relatively large sheet of steel is to be exposed to cyclic tensile and compressive stresses
of magnitudes 100 MPa and 50 MPa, respectively. Prior to testing it has been determined that
the length of the largest surface crack is 2.0 mm (2*10-3). Estimate the fatigue life of this
sheet if its plane strain fracture toughness is 25 MPa√𝑚 and the values of m and A in are 3.0
and 1.0 ×10-12 respectively, for 𝛥𝜎 in MPa and a in m. Assume that the parameter Y is
independent of crack length and has a value of 1.0.
Solution:
The value of is just MPa, the magnitude of tensile stress, since 𝜎min is compressive.
Therefore, integration yields,
9) Using the Larson- Miller data for S-590 iron shown in figure 1, predict the time up to
rupture for a component that is subjected to a stress of 140 MPa (20,000 psi at 800 °C (1073
K).
Figure 1: Logarithm stress versus the Larson-Miller parameter for an S-590 iron
Solution
From figure 1, at 140 MPa at 140 MPa (20,000 psi) the value of the Larson- Miller
parameter is 103, for T in K and tr in h; therefore,
103 = T(20+ log tr )
And solving fort he time,
22.37= 20+ log tr
tr = 233 h(9.7 days)
10.A 4340 steel bar is subjected to a fluctuating axial load that varies from a maximum of
330 kN tension to a minimum of 110 kN compression. The mechanical properties of the steel
are: σu = 1090 MPa, σo = 1010 MPa, σe = 510 MPa Determine the bar diameter to give
infinite fatigue life based on a safety factor of 2.5.
Cylindrical cross section of the bar = A, the variation of stress will be
From the Equation
11.For the cyclic stress-strain curve, σB =75 MPa and εB = 0.000645. If εf = 0.30 and E =
22x104 MPa. Determine
(a) ∆εe and ∆εp
(b) The number of cycles to failure.
12
A mild steel plate is subjected to constant amplitude uniaxial fatigue loads to produce
stresses varying from σmax = 180 MPa to σmin = -40 MPa. The static properties of the steel
are σo = 500 MPa, σu = 600 MPa, E = 207 MPa, and Kc = 100 MPa.m1/2. If the plate
contains an initial through thickness edge crack of 0.5 mm, how many fatigue cycles will be
required to break the plate?
For through thickness edge crack, α = 1.12, and for ferritic-pearlitic steels, A = 6.9 x
10-12 MPam1/2 and m = 3.0. σr = (180-0), since compressive stress are ignored, and neglect
the influence of mean stress on the crack growth.
From the Equation
http://www.sut.ac.th/Engineering/metal/pdf/MechMet/12_Fatigue%20of%20metals.pdf
13
A double cantilever beam (DCB) specimen is subjected to a pulsating loading. The crack length is a,
thickness is b and specimen height is 2h, where a >> b and a >> h. The material is linear elastic with
Young’s modulus E. Use Paris’ law
to determine the number of loading cycles required to make the crack grow from its initial length ai
to a final length a, if the load is prescribed; the load varies between 0 and Pmax.
Solution:
The stress intensity range becomes (Δp = Pmax /b (N/m) is the load range)
Entering into Paris’ law gives
Separating
Integrating and solving for N give
14.The mean value of the fatigue limit of a welded 4 mm thick aluminium plate has been
found to be 92 MPa and the standard deviation of the spread of the fatigue limit
measurements is 6 MPa. Determine the stress level at which the probability of fatigue failure
is 0.1 per cent. The spread of the fatigue limit is assumed to have a normal (Gaussian)
distribution. For the normal distribution function Φ(x) one has
Solution:
The mean value of the fatigue limit (here 92 MPa) gives the failure probability 50 per
cent. The standard deviation s is s = 6 MPa. The failure probability 0.1 per cent is
obtained at the stress level
σ = σmean- 3.09⋅s = 92-3.09⋅6 = 73.5 MPa (where 3.09 was
obtained from the table). This implies that in mean one plate out of 1000 plates loaded at
the stress level 73.5 MPa is expected to fail due to fatigue.
15.A generator is rotating with a speed of 3000 rpm (revolutions per minute). Determine the
number of loading cycles N the axle is subjected to due to its own weight. The machine is
operating 7000 hours per year during 20 years.
Solution:
3000 revolutions per minute give 3000 loading cycles per minute. It gives 3000 60 7000
loading cycles per year, giving 3000 60 7000 20 = 25.2 109 loading cycler per 20 years.
Answer:
Number N of loading cycles is N = 25.2 109 cycles (thus, it may be exposed to fatigue).
16.Sketch the Wöhler (SN, Stress-Number) curve for a normalised 37Mn Si5 steel bar with
an ultimate strength of σU = 810 MPa. The bar is subjected to a rotating bending moment. At
an investigation of a similar material the following fatigue lives were found:
σFLR ≅ 0.45σU yields N 10^6 cycles,
σFLR ≅ 0.90σU yields N = 1000 cycles,
and σFLR = σU yields N =1 cycle.
17.Constant-amplitude fatigue strengths for materials subjected to different cyclic loading
conditions are often expressed in Haigh diagrams. In the Haigh diagram given three straight lines (A,
B and C) are shown. Each line corresponds to constant loading conditions. The loading varies
sinusoidally with mean value σm, amplitude σa, maximum value σmax (= σm + σa), minimum value
σmin (= σm σa), and stress ratio R = σmin /σmax. Determine which one of these five loading
variables is constant in the three cases shown in the figure.
Solution: For line A one notices that the sum σm + σa is a constant (= 300 MPa).
Thus, line A gives that stress σmax is constant.
For line B one notices that ratio σa / σm is constant (= 2). Thus, stress ratio R is constant.
For line C one notices that mean stress σm = 200 MPa + σa, giving σm σa = 200 MPa = σmin.
Thus, σmin is constant.
Answer: Line A: stress σmax is constant, line B: stress ratio R is constant, line C: stress σmin is
constant.
18.
19.Crack-like flaws may develop in most materials. These flaws should be treated as if they
were cracks. Therefore, assume that the flaws are penny-shaped (the crack looks like a coin
with a sharp egde). Determine the largest flaw allowable with respect to fatigue crack
propagation in the following materials. The loading is cyclic and the stress range is 0.5σY
(and σY is the yield limit of the material).
(a) A low-strength steel with ΔKth = 7 MN/m3/2 and σY = 300 MPa, and
(b) a high-strength steel with ΔKth = 4 MN/m3/2 and σY = 1500 MPa.
Solution: A penny-shaped crack in the interior of the material will give the stress intensity
factor . Case 8 in Appendix 3 of the textbook gives (terms containing a / t can be neglected
here so that f8 = 0.637 is obtained)
(a) No crack propagation is allowed. Then the stress intensity range ΔKI must be less than (or
equal to) the threshold value ΔKth. Using ΔKth = 7 MN/m3/2 and Δσ0 = σY / 2 = 150 MPa,
one obtains
giving a = 1.7 mm.
(b) Similarly, using ΔKth = 4 MN/m3/2 and Δσ0 = σY / 2 = 750 MPa, one obtains
giving a = 0.0223 mm.
Answer: Critical crack length is (a) a = 1.7 mm, and (b) a = 0.022 mm.
20. For the cyclic stress-strain curve, σB =75 MPa and εB = 0.000645. If εf = 0.30 and E =
22x104 MPa.
Determine (a) ∆εe and ∆εp
Answer:
(b)The number of cycles to failure.
Answer:
From the Coffin-Manson relation;
If c=0.6 and ef and e'f is same.
21. A component of a machine is subjected to a repeated loading sequence. The loading
sequence is repeated once a day. The loading is alternating, and one sequence contains the
following loading amplitudes (i.e. stress amplitudes the mean value of the stress is zero)
σa = 180, 160, 140 and 100 MPa
The number of loading cycles at each stress level is
n = 15, 20, 150, and 500, respectively
The Wöhler curve of the material is, in this stress range, given by the relation
σa = 50logN + 400 MPa
Determine the damage accumulation D due to one loading sequence, and then, determine
how many loading sequences (how many days) the component might be used before fatigue
failure is expected.
Solution: For the different stress levels, the expected fatigue life N is obtained from the
Wöhler curve. One obtains
giving respectively,
N = 25 119, 63 096, 158 489, and 1 000 000
The accumulated damage due to one loading sequence is, by use of the Palmgren-Miner
damage accumulation rule,
Finally, the number of sequences Ns to expected fatigue failure is
Thus, the component can be used in 424 days, approximately
22.Estimate the fatigue limit in tension/compression for alternating and pulsating loading of a
steel (approximately SS 2225) with the ultimate strength σU = 810 MPa. Sketch the Haigh
diagram. At an investigation of a similar material the relationships
σFL = 0.468 σU ± 50 MPa and σFLP ≅ 0.85 σFL
were found. Use these relations and other “thumb rules” to estimate the fatigue limits
(alternating and pulsating) of the material.
Solution:
The relationship σFL = 0.468σU 50 MPa gives, for alternating loading: σFL ≅ 0.468⋅810
50 MPa = 380 50 MPa, i.e., the fatigue limit σFL in tension/compression at alternating
loading may, according to this relationship, be expected to fall in the range 330 MPa to
430 MPa. For pulsating loading the relationship between pulsating and alternating
loading, σFLP ≅ 0.85σFL, gives σFLP = 0.85σFL = 0.85 (380 50) MPa = 320 42 MPa.
Thus, the fatigue limit in tension at pulsating loading is expected to fall in the range from
280 280 MPa (from 320 42 MPa) to 360 360 MPa (from 320 + 42 MPa).
Also, compare with other materials.
Material data for SIS 2225-03, -04 and -05 gives, for rotating bending, that σFLB should
be in the range 350 MPa to 460 MPa. The engineering rule σFL ≅ 0.8σFLB gives that
σFL should fall in the range 280 MPa to 370 Mpa
Another engineering rule says that σFLR (= σFLB) decreases from 0.45σU for mild steel
to 0.38σU for the harder materials. The most unfavourable case, σFLR = 0.38σU gives
σFLR = σFLB ≅ 308 MPa, while the case σFLR = 0.45σU gives σFLR = σFLB ≅ 365
MPa. This gives, using σFL ≅ 0.8σFLB, that σFL should be in the range 250 MPa to 290
MPa.
Finally, select, hopefully “on the safe side”, for alternating loading σFL = 280 MPa and for
pulsating loading σFLP = 0.85σFL, which gives σFLP = 240 240 MPa. (In practice, fatigue
limits will probably be higher than the values estimated here.)
Answer:
The fatigue limit σFL at alternating loading is estimated to 280 MPa, approximately, and the
fatigue limit σFLP at pulsating loading is estimated to 240 240 MPa, approximately.
23
24
A fatigue test was conducted in which the stress amplitude was 210 MPa and the mean
stress was 70 MPa.
(a) Compute the maximum and minimum stresses
(b) Compute the stress range
(c) If the A= 7.2 x 10-12 and m = 4 for stress in MPa and crack length in m, determine the
crack growth rate for a surface crack of length 1.2 mm. (Y=1.5) Solution:
a) �� = ����− ���� 2 = 210 �� = ����+ ���� 2 = 70 ���� = 280 MPa , ���� = −140 ���
b) �� = ���� − ���� = 420 MPa
c) �� �� = �(��)� = 7.2 ∗ 10−12 (� ���� √�� − � ���� √�
a)4 With the assumption of ���� = 0 , the crack growth rate is �� �� = 3.18 ∗ 10−6 cycles
25.A structure contains a critical component made from A514 steel. After fabrication of the
structure, a welding defect 7.6 mm deep is discovered in this steel plate. The flaw is
essentially an edge crack under tension loading, and the required cyclic life of the structure is
100 000 cycles. The component is subject to a fluctuating load which causes a stress
variation from 172 MPa to 310 MPa.
Material properties for the A514 steel are: yield stress = 689 MPa, K1C = 165 MPa m½
geometry correction factor Y = 1.12, and the Paris law is:
where da/dN is in m/cycle, and:
Calculate the fatigue life of this component based on attaining a critical defect size for fast
fracture.
SOLUTION:
In order to calculate the cyclic life, we require the critical crack size causing fast fracture.
This can be obtained by substituting the appropriate data into the K calibration equation:
71.9 mm. Hence the limits on the Paris law integration are 7.6 mm and 71.9 mm, while the
stress range is (310 - 172) MPa = 138 MPa. Separating the variables and rearranging the
Paris law gives:
We can perform the integration symbolically and then substitute the actual values into the
equation to obtain the required life.
26.A steel was subjected to two fatigue test at +(-)400MPa and +(-)250 MPa. Failure
occurred after 2x10^4 and 1.2x10^6 cycles, respectively. Estimate the fatigue life at +()300MPa of a part made of this steel.
SOLUTION:
For Nf>10^4 ∆(Nf)a= c1
800 (2x104)^a= c1
500 (1.2x106)^a= c1
a=0.115 c=2498 MPa
At +(-)300 MPa,
600 (Nf)0.115= 2498 Nf=2.45x10^5 cycles
27.An alloy steel plate is subjected to constant-amplitude uniaxial fatigue cyclic tensile and
compressive stresses of magnitudes of 120 and 30 MPa, respectively. The static properties of
the plate are a yield strength of 1400 MPa and a fracture toughness KIC of 45 MPa(m) 1/2. If
the plate contains a uniform through thickness edge crack of 1.0 mm, how many fatigue
cycles are estimated to cause fracture?
SOLUTION:
Use the equation
REFERENCES
[1]Askeland, D. R., Phule, P. P., & Battacharya, D.K.(2009). Essentials of materials science
and engineering(2nd ed.). Australia:Cengage Learning
[2]Queeney, R. A., & Segall A. E.(2008). Mechanical Response of Engineering Materials(7th
ed.). Duboque, IA: Kendall/Hunt Publising Company
[3](Dahlberg, T. and Ekberg, A., ' Solutions Manual to problems in Failure, Fracture,
Fatigue', 2002.)
Ay, İ., ‘Kırılma Mekaniği’, n.d.
[4]W.D.Callister, Jr. Material and Science and Engineering,fourth edition, 1997
[5]Dr Rob Thornton, Fracture and Fatigue Revision Examples, March 2015
[6] Dahlberg and Ekberg.,’Failure, Fracture, Fatigue’2002
[7]Retrieved from;
http://www.tech.plymouth.ac.uk/sme/Tutorials/FMtut/Fatigue/Solutions/Solution3.htm
http://myweb.ncku.edu.tw/~cyylin/97Lecture04_modified.pdf