pa-solutions 4012.pdf
TRANSCRIPT
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MAAE 4102 - Strength and Fracture
Problem Set Solutions
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CARLETON UNIVERSITY
Department of Mechanical and Aerospace Engineering
MAAE 4012 - Strength and Fracture
Stress Life
1. A method used to present mean stress fatigue data is to generate a family of curves on an
S-N plot, with each curve representing a different stress ratio, R. Generate the curves for
R values of -1, 0 and 0.5 for a steel with an ultimate strength of 100 ksi. For this
example, use the Gerber relationship to generate these curves. Use Eqn. 1 to estimate the
fully reversed (R = -1) fatigue behaviour between 10 and 10 cycles.3 6
(1)
Solution:
The Gerber relationship in general form is
The alternating stress level can be obtained using equation 1 above
Case 1 R = -1
This case is the same as fully reversed (= 0) R = -1
N a@ N = 10 S = = 90 ksi3
N a@ N = 10 S = = 74 ksi4
N a@ N = 10 S = = 61 ksi5
N a@ N = 10 S = = 50 ksi6
Case 2 R = 0
a mThis case is the same as maximum loading where = . Substituting into the Gerber
equation
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N aIf values of S are substituted the equation can be solved for
R = -1 R= 0
N a@ N = 10 S = 90 ksi = 58.8 ksi3
N a@ N = 10 S = 74 ksi = 53.1 ksi4
N a@ N = 10 S = 61 ksi = 47.3 ksi5
N a@ N = 10 S = 50 ksi = 41.4 ksi6
Case 3 R = 0.5
min maxFrom the definition of stress ratio = 0.5
This gives the following relationships:
a m aTherefore = 3 Substituting in the Gerber equation and solving for
R= -1 R = 0.5N a@ N = 10 S = 90 ksi = 27.7 ksi
3
N a@ N = 10 S = 74 ksi = 26.7 ksi4
N a@ N = 10 S = 61 ksi = 25.4 ksi5
N a@ N = 10 S = 50 ksi = 24.0 ksi6
aThe values for various R cases can be plotted on an S - N curve.
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Problem Set Solutions
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2. Another method used to present mean stress fatigue data is to generate a family of curves
mon an S-N plot, with each curve representing a different mean stress value, . Generate
the curves for mean stress values of 0, 20 and 40 ksi for a steel with an ultimate strength
of 100 ksi. For this example, use the goodman relationship to generate these curves.
meanAgain use eqn. (1) to estimate the fully reversed ( = 0) fatigue behaviour.
SOLUTION
The Goodman relationship is:
The alternating stress level for a given life can be determined from:
mCase 1 = 0 (same as fully reversed loading) R = -1
N a@ N = 10 S = = 90 ksi3
N a@ N = 10 S = = 74 ksi4
N a@ N = 10 S = = 61 ksi5
N a@ N = 10 S = = 50 ksi6
mCase 2 = 20 ksi
m NSubstituting this value of into the Goodman equation for different value of S gives:
m R = -1 = 20 ksi
N a@ N = 10 S = 90 ksi = 72 ksi3
N a@ N = 10 S = 74 ksi = 59 ksi4
N a@ N = 10 S = 61 ksi = 49 ksi5
N a@ N = 10 S = 50 ksi = 40 ksi6
mCase 3 = 40 ksi
m NSubstituting this value of into the Goodman equation for different value of S gives:
m R = -1 = 40 ksi
N a@ N = 10 S = 90 ksi = 54 ksi3
N a@ N = 10 S = 74 ksi = 44 ksi4
N a@ N = 10 S = 61 ksi = 37 ksi5
N a@ N = 10 S = 50 ksi = 30 ksi6
a mThe for various can now be plotted on a S-N curve
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Problem Set Solutions
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Problem Set Solutions
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3. Given a material with an ultimate strength of 70 ksi, an endurance limit of 33 ksi, and a
true fracture strength of 115 ksi, determine the allowable zero to maximum (R = 0) stress
which can be applied for 10 , 10 , 10 and 10 cycles. Make predictions using the3 4 5 6
Goodman, Gerber and Morrow relationships.
SOLUTION:m aFor R = 0 and = the mean stress equations can be written as:
Goodman:
Gerber:
Morrow:
u fWhen S = 70 ksi and = 115 ksi
NFind S from
e 1000 uGiven S = 33 ksi and using S = 0.9 S = 63 ksi
C = 2.08 b = -0.094
Substituting in this equation for each value of cycles and then in the Goodman, Gerber and
Morrow equations for the effect of mean stress we get them following values:
max aNote: at R = 0 = 2
i.e at 103
aGoodman Gives = 33 ksi
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Problem Set Solutions
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R = 0 10 cycles 10 cycles 10 cycles 10 cycles3 4 5 6
a
(ksi)
max
(ksi)
a
(ksi)
max
(ksi
a
(ksi)
max
(ksi
a
(ksi)
max
(ksi
Goodman 33 66 29.4 58.8 25.8 51.6 22.4 44.8
Gerber 41.2 82.4 36.7 73.4 32.2 64.4 27.8 55.6
Morrow 40.7 81.4 35.1 70.2 30.1 60.2 25.6 51.2
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Problem Set Solutions
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4. A component undergoes a cyclic stress with a maximum value of 75 ksi and a minimum
value of -5 ksi. Determine the mean stress, stress range, stress amplitude, stress ratio and
amplitude ratio. If the component is made from a steel with an ultimate strength of 100
ksi, estimate its life using the Goodman relationship.
SOLUTION:
max min = 75 ksi = - 5 ksi
Mean Stress
Alternating Stress
Stress ratio
Amplitude ratio
Using the Goodman relationship
a m uGiven = 40 ksi = 35 ksi and S = 100 ksi
N NSolving for S S = 61.5 ksi
e u 1000 uEstimating S as 0.5 S , S as 0.9 S and substituting in
N = 8.9 x 10 cycles4
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Problem Set Solutions
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5. A switching device consists of a rectangular cross-section metal cantilever 200 mm in
length and 30 mm in width. The required operating displacement at the free end is 2.7
mm and the service life is to be 100,000 cycles. To allow for scatter in life performance a
factor of 5 is employed on endurance. Using the fatigue curves given in Figure,
determine the required thickness of the cantilever if made in (a) mild steel, (b) aluminumsteel Aluminumalloy. E = 208 GN/m , E = 79 GN/m .
2 2
SOLUTION
For a cantilever
Factored endurance = 5 x 100,000 = 5 x 10 cycles5
from figure for mild steel
For Aluminum
for steel
for aluminum
(x) Aluminum Alloy 24S-T3 reversed axial stress
() Mild Steel reversed axial stress
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Problem Set Solutions
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6. A pressure vessel support bracket is to be designed so that it can withstand a tensile
loading cycle of 0-500 MN/m once every day for 25 years. Which of the following steels2
would have the greater tolerance to intrinsic defects in this application: (i) a maraging
IC ICsteel (K = 82 MN m , C = 0.15 x 10 , m = 4.1), or (ii) a medium-strength steel (K-(3/2) -11
= 50 MN m , C = 0.24 x 10 , m = 3.3)? For the loading situation a geometry factor of-(3/2) -11
1.12 may be assumed.
SOLUTION
fThe number of cycles in 25 years = N = 1 x 365 x 25 = 9125
( i)
(ii) In a similar way for medium strength steel
Which is more damage tolerant ?
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Problem Set Solutions
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7. A series of crack growth tests on a moulding grade of polymethyl methacrylate gave the
following results:
da/dN(m/cycle) 2.25 x 10 4 x 10 6.2 x 10 17 x 10 29 x 10-7 -7 -7 -7 -7
K(MN m ) 0.42 0.53 0.63 0.94 1.17-3/2
If the material has a critical stress
intensity factor of 1.8 MN m-3/2
and it is known that the moulding
process produces defects 40 m
long, (2a), estimate the maximum
repeated tensile stress which
could be applied to this material
for at least 10 cycles without6
causing fatigue failure.
From the graph:
so c = 2 x 10 m = 2.513-6
= 2.13 MN/m2
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Problem Set Solutions
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8. A series of tensile fatigue tests on stainless steel strips containing a central through hole
gave the following values for the fatigue endurance of the steel. If the steel strips were
100 mm wide, comment on the notch sensitivity of the steel.
Hole diameter (mm) No hole 5 10 20 25Fatigue endurance (MN/m ) 600 250 270 320 3702
Solution:
tThe K values for a strip with a central hole may be obtained from Peterson ( see figure):
The un-notched fatigue endurance is 600
MN/m2
For the 5mm hole -
tK = 2.84fAlso K = 600/250 = 2.4
Therefore
Similarly for other holes:
d(mm)t fK K q
5 2.84 2.4 0.76
10 2.74 2.22 0.7
20 2.5 1.875 0.58
25 2.44 1.622 0.43
The high values of q indicate that the steel is notch sensitive.
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Problem Set Solutions
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9. The fatigue endurances from the S-N curve for a certain steel are:
Stress (MN/m ) Fatigue endurance (cycles)2
350
380
410
2,000,000
500,000
125,000
If a component manufactured from this steel is subjected to 600,000 cycles at 350 MN/m2
and 150,000 cycles at 380 MN/m , how many cycles can the material be expected to2
withstand at 410 MN/m before fatigue failure occurs, assuming that Miners cumulative2
damage theory applies?
Solution:
Using cumulative damage theory
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Problem Set Solutions
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Strain Life
1. It has been determined that a certain steel (E = 30 x 10 ksi) follows the following true3
pstress, , true plastic strain, , relation:
The true plastic strain at fracture was found to be 0.48. Determine:
fa) True fracture strength,
b) Total true strain at fracture
c) Strength coefficient, K
d) Strain hardening exponent, n
ye) Strength at 0.2% offset, S
f) Percent reduction in area, % RA
fg) True fracture ductility,
SOLUTION
f(a) True fracture strength,
b) Total true strain at fracture
c) Strength coefficient, K
K = 360 ksi
d) Strain hardening exponent, n
n = 0.11
ye) Strength at 0.2% offset, S
This is the stress corresponding to a plastic strain of 0.002
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f) Percent reduction in area, % RA
RA = 0.381 or 38 %
fg) True fracture ductility,
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Problem Set Solutions
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2. The following stress-strain and strain-life properties are given for a steel:
E = 30 x 10 ksi K = 137 ksi n= 0.22
b = -0.11
c = -0.64
SOLUTION
a)
Elastic strain-life
fintercept at 2N = 1 of
Plastic strain-life
fintercept at 2N = 1 of
tDetermine the transition life (2N). From graph 2Nt = 30,000 reversals
using equation
2Nt = 30,366 reversals
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Problem Set Solutions
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b) Draw the hysteresis loops corresponding to strain amplitude (/2) values of 0.05,
0.00125 and 0.0007. Determine the fatigue life in reversals at these three strain levels.
by iteration
f/2 = 0.05 /2 = 70.1 ksi 2N = 100
f/2 = 0.0125 /2 = 24.18 ksi 2N = 3 x 105
f/2 = 0.0007 /2 = 18 ksi 2N = 1 x 107
For /2 = 0.05
For /2 = 0.0125
The hysteresis loop for 0.0007 is almost a straight line
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Problem Set Solutions
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Using the computer program the life can be calculated
Modulus of Elasticity = 30000
Fatigue Strength Coefficient = 120.0 Fatigue Strength Exponent = -0.11
Fatigue Ductility Coefficient = 0.95
Fatigue Ductility Exponent = -0.64
Cyclic Strength Coefficient (K) = 137.0 input
Cyclic Strain Hardening Exp. (N) = 0.22 input
_______________________________________________________
Strain Mean Max. Life In Reversals
Amplitude Stress Stress _____________________________
Morrow Man-Hal SWT________________________________________________________
.05000 0.00 70.1 1.075E+02 1.075E+02 1.109E+02
.00125 0.00 24.8 3.514E+05 3.514E+05 5.992E+05
.00070 0.00 18.0 1.122E+07 1.112E+07 1.775E+07
.00800 0.00 45.2 2.527E+03 2.527E+03 3.024E+03
fc) Determine the elastic, plastic and total strain amplitude for a life (2N ) of 2 x 106
reversals.
= 0.0008111 + 0.0000881
= 0.000899
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Problem Set Solutions
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fd) life (2N ) of 500 reversals.
= 0.002109 + 0.017798
= 0.019818
e) Determine the cyclic stress amplitude corresponding to fatigue lives of 500 and 2 x 106
reversals.
Basquin Equation
fat 2N = 500
fat 2N = 2 x 106
f) From computer program at a strain amplitude of 0.008 the life is 2500 reversals. (See
computer results above) This component will not meet the life requirements.
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Problem Set Solutions
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3. Smooth aluminum specimens are subjected to two series of cyclic load-controlled tests.
maxThe first test (level A) varies between a maximum stress value, , of 21.3 ksi and a
minminimum value, , of -30.1 ksi. The second test (level B) varies between 61.5 and 10.1
ksi. Predict the life to failure, in reversals, at the two levels. Use the Morrow, Manson-
Halford and Smith-Watson-Topper relationships for the predictions. Assume that there is
no mean stress relaxation. The material properties for the aluminum are
E = 10.6 x 10 ksi K = 95 ksi n= 0.0653
f = 160 ksi b = -0.124
f = 0.22 c = -0.59
Listed below are actual test results at the two levels. Three tests were run at each of the
levels. Compare the predictions to these values.
fLevel Test Results: Lives in Reversals, 2N
A 5.4 x 10 5.5 x 10 7.2 x 105 5 5
B 5.6 x 10 6.4 X 10 6.9 X 104 4 4
SOLUTION
Using the following equations:
Mean Stress
Stress Range
Stress Amplitude
The strain amplitude can be determined using:
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Problem Set Solutions
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Therefore:
MAX MIN o /2 /2
Case A 21.3 - 30.1 - 4.4 25.7 0.002425
Case B 61.5 10.1 35.8 25.7 0.002425
The results from the life prediction program
Modulus of Elasticity = 10600
Fatigue Strength Coefficient = 160.0
Fatigue Strength Exponent = -0.124
Fatigue Ductility Coefficient = 0.22
Fatigue Ductility Exponent = -0.59
Cyclic Strength Coefficient (K) = 59.0 input
Cyclic Strain Hardening Exp. (N) = 0.065 input
_______________________________________________________
Strain Mean Max. Life In Reversals
Amplitude Stress Stress _____________________________
Morrow Man-Hal SWT
________________________________________________________
.002425 -4.4 21.3 3.498E+06 3.545E+06 5.646E+06
.002425 35.8 61.5 4.626E+05 3.694E+05 9.842E+04
Comparing these predictions with measured result, note that all three methods give non-conservative estimates.
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Problem Set Solutions
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Fracture Mechanics
1. A large plate made of AISI 4340 steel contains an edge crack and is subjected to a tensile
cstress of 40 ksi. The material has an ultimate strength of 260 ksi and aKvalue of 45 ksi
in. Assume that the crack is much smaller than the width of the plate. Determine the
critical crack size.
SOLUTION:
Since the ratio of crack length to plate width, a /b is very small the SIF is given by:
For a KC = 45 ksiin and a stress = 40 ksi
Ca = 0.32 in
y2. A large cylindrical bar made of 4140 steel ( = 90 ksi) contains an embedded circular
(penny shaped) crack with a 0.1 in. diameter. Assume that the crack radius, is much
smaller than the radius of the bar,R, so that the bar may be considered infinitely large
compared to the crack. the bar is subjected to a tensile stress of 50 ksi. Determine the
plastic zone size at the crack tip. Are the basic LEFM assumptions violated?
SOLUTION:
Since the crack size is much small than the radius of the bar, the SIF for a circular
embedded crack in an infinite body, can be used.
Therefore:
Since the crack is embedded in a large cylindrical bar plane strain conditions are
developed and the plastic zone size is given by:
YSince r is very much smaller than a the LEFM assumptions are not violated
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Problem Set Solutions
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Ic y3. A very wide late made from Al 7075-T651 (K = 27 ksi in., = 80 ksi) contains an
edge crack. Plot the allowable nominal stress (ksi) as a function of crack size, (in
inches), if the design requirements specify a factor of safety of 2 on the critical stress
intensity factor. If the plate specifications were changed so that Al 7050-T73651 was
Ic yused (K = 35 ksi in., = 70 ksi), re-plot the curve. For a nominal stress of one-half the
yield stress, determine the increase in allowable flaw size by changing from the Al 7075alloy to the Al 7050 alloy.
SOLUTION:
For a very wide edge cracked plate the SIF is given by:
For the 7075 -T651 Al alloy the material yields at a stress of 80 ksi. Therefore the
smallest crack that can occur while the material remains nominally elastic is:
a = 0.007 in
For the 7050-T73651 Al alloy the smallest crack that can occur while the material
remains nominally elastic is:
a = 0.016 in
The allowable crack size as a function of crack size is shown below:
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Problem Set Solutions
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For a nominal stress of the yield stress the allowable flaw size for
7075 alloy
a = 0.029 in
7050 alloy
a = 0.063 in
The 12.5% reduction in yield strength is overshadowed by the 117% increase in allowable
flaw size.
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Problem Set Solutions
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4. Design a pressure vessel that is capable of withstanding a static pressure of 1000 psi and
that will leak-before-burst. The required material has a fracture toughness of 60 ksi
in. and a yield strength of 85 ksi. The diameter of the vessel is specified to be 4 ft. A
crack with surface length of 1 in. can reliably be detected. Since the cost of the vessel is
related directly to the amount of material used, optimize the design so that the cost is
minimized.
SOLUTION:
The stress intensity factor for a 1 inch crack must be less than the fracture toughness for
I ICthe vessel to leak before break, K < K
The stress intensity factor for a semi-elliptical crack is
The shape parameter Q is found from the graph.
ICGiven: K = 60 ksiin
Y = 85 ksi
dia = 48 in
pressure = 1000 psi
The stress due to the pressure is:
YThe ratio / is:
To leak before break, the crack dimension, a is equal to the wall thickness, t.
The SIF is:
Using the figure solve this equation iteratively
Y It / Q K
0.35 0.81 1.6 63.67 too large
0.4 0.71 1.85 55.39
The optimum design results in a wall thickness of 0.37 in
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5. A component made from 7005-T53 aluminum contains a semi-circular surface crack (a/c
= 1) and is subjected toR = 0.1 loading with a stress range, , of 250 MPa. (Refer to
Example 1 for an expression for the stress intensity range, K.) the following crack
growth data were obtained in laboratory air environment. Using these data:
a) Plot crack length, (mm), versus cycles,Nb) Plot da/dNversus K. Identify the three regions of crack growth.
c) Determine the Paris law constants, Cand m, for the linear region of crack growth.
________________________________________________________
N(cycles) a(mm) da/dN(mm)
_________________________________________________________
95,000 0.244
100,000 0.246 7.00 x 10-7
105,000 0.251 3.920 x 10-6
110,000 0.285 9.665 x 10-6
115,000 0.347 1.053 x 10-5
125,000 0.414 1.230 x 10-5
130,000 0.490 2.063 x 10-5
135,000 0.621 4.661 x 10-5
140,000 0.956 9.565 x 10-5
145,000 1.577 3.964 x 10-4
147,000 2.588 1.105 x 10-3
147,400 3.078 1.554 x 10-3
147,500 3.241 8.758 x 10-3
147,500 3.445
__________________________________________________________
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SOLUTION:
The SIF for a semi-elliptical crack is:
for a semi-circular crack Q is /2 therefore
For a = 0.244 mm
N(cycles) a(mm) K
(MPam)
da/dN(mm) Log(K) Log(da/dN)
95000 0.244 4.93
100000 0.246 4.95 7.00 x 10 0.695 -6.155-7
105000 0.251 5 3.920 x 10 0.699 -5.406-6
110000 0.285 5.33 9.665 x 10 0.727 -5.015-6
115000 0.347 5.89 1.053 x 10 0.77 -4.977-5
125000 0.414 6.43 1.230 x 10 0.808 -4.91-5
130000 0.49 6.99 2.063 x 10 0.845 -4.686-5
135000 0.621 7.87 4.661 x 10 0.896 -4.332-5
140000 0.956 9.77 9.565 x 10 0.99 -4.019-5
145000 1.577 12.55 3.964 x 10 1.099 -3.402-4
147000 2.588 16.07 1.105 x 10 1.206 -2.957-3
147400 3.078 17.53 1.554 x 10 1.244 -2.808-3
147500 3.241 17.99 8.758 x 10 1.255 -2.058-3
147500 3.445 18.54
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The a vs N curve is:
The da/dN plot is:
For all points:
Neglecting points 1to 3 and point11
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6. Calculate the critical defect size for each of the following steels assuming they are each
ysubjected to a stress of 0.5 . Comment on the results obtained.
ySteel Yield Strength
(MN/m )2Fracture Toughness
(MN )-3/2
Mild Steel 207 200
Low-alloy Steel 500 160
Medium Carbon Steel 1000 280
High-carbon Steel 1450 70
18% Ni (Maraging)
Steel
1900 75
Tool Steel 1750 30
SOLUTION
Critical defect size
For mild steel
Critical Defect Size 2c = 2 x 1.189 = 2.38 m
Steel Yield
yStrength
(MN/m )2
FractureToughness
(MN )-3/2
ca(mm)
Critical Defect
cSize 2a
(mm)
Mild Steel 207 200 1189 2378
Low-alloy Steel 500 160 130. 260
Medium Carbon Steel 1000 280 100 200
High-carbon Steel 1450 70 2.97 5.9
18% Ni (Maraging) Steel 1900 75 1.98 4.0
Tool Steel 1750 30 0.374 0.75
LEFM can be applied to high strength steels because critical defect size is small, however
materials such as mils steel would require very large specimens in order to achieve this critical
defect size
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7. A sheet of glass 0.5 m wide and 18 mm thick is found to contain a number of surface
cracks 3 mm deep and 10 mm long. If the glass is placed horizontally on two suports,
calculate the maximum spacing of the supports to avoid the fracture of the glass due to its
ICown weight. For glass K = 0.3 (MN ) and density = 2600 kg/m .-3/2 3
SOLUTION
The worst case is when the defect is midway between the supports on the bottom of the plate
The aspect ratio of the defect = a / 2c = 3/10 = 0.3
For a semi-elliptical flaw
YSUsing the Flaw Shape Parameter graph of Question 5 and assuming that / = 0The Shape Parameter Q = 1.62
c = 3.52 MN / m2
From beam theory the stress at the crack is
where
weight per unit length w = 2600 x 0.5 x 0.018 x 9.81 = 230 N /m
L = 1.82 m
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8. The accident report on a steel pressure vessel which fractured in a brittle manner when the
internal pressure of 19 MN/m had been applied to it shows that the vessel had a2
longitudinal crack 8 mm long and 3.2 mm deep. A subsequent fracture mechanics test on
ICa sample of the steel showed that it had a K value of 75 MN . If the vessel diameter-3/2
was 1 m and the thickness was 10 mm, determine whether the data reported are consistent
with the observed failure.
The aspect ratio of the defect = a / 2c = 3.2 / 8 = 0.4
For a semi-elliptical flaw
YSUsing the Flaw Shape Parameter graph of Question 5 and assuming that / = 0
The Shape Parameter Q = 2.0
For a thin walled cylinder
Hence LEFM predicts quite accurately the observed fracture pressure of 10 MN/m2
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MAAE 4012 - Strength and Fracture
9. An aluminum alloy plate with a yield stress of 450 MN/m fails in service at a stress of2
110 MN/m . The conditions are plane stress and there is some indication of ductility at2
the fracture. If a surface crack of 20 mm long is observed at the fracture plane calculatethe size of the plastic zone at the crack tip. Calculate also the percentage error likely if
LEFM was used to obtain the fracture toughness of this material.
SOLUTION
(i) Using LEFM
(ii) Using the plastic zone correction
Hence the % error in using LEFM would be 1.5%