pa-solutions 4012.pdf

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MAAE 4102 - Strength and Fracture

Problem Set Solutions

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CARLETON UNIVERSITY

Department of Mechanical and Aerospace Engineering


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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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%

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