PA-Solutions 4012.pdf

MAAE 4102 - Strength and Fract ure

CARLETON UNIVERSITY Department of Mechanical and Aerospace Engineering MAAE 4012 - Strength and Fracture

Stress Life

1.

A meth method od use used d to pre prese sent nt mean mean stre stress ss fat fatig igue ue dat dataa is to gene genera rate te a fam famil ily y of of curv curves es on on an S-N plot, with each curve representing a different stress stress ratio, R. Generate the curves for R values of -1, 0 and 0.5 for a steel with with an ultimate strength of 100 ksi. For this example, use the Gerber relationship relationship to generate these curves. Use Eqn. 1 to estimate estimate the 3 6 fully reversed reversed (R = -1) fatigue fatigue behaviour behaviour between between 10 10 and 10 cycles. cycles. (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 (ó = 0)

@ @ @ @

N = 103 N = 104 N = 105 N = 106

S N S N S N S N

= = = =

R = -1

óa = 90 ksi óa = 74 ksi óa = 61 ksi óa = 50 ksi

Case 2

R=0 This case is the same as maximum loading where óa = óm . Substituting into the Gerber equation

Problem Set Solutio ns Page 1 of 31


If values values of S N are substitu substituted ted the equation equation can be solved solved for óa R = -1 R= 0 3 óa = 58.8 ksi @ N = 10 S N = 90 ksi 4 óa = 53.1 ksi @ N = 10 S N = 74 ksi 5 óa = 47.3 ksi @ N = 10 S N = 61 ksi 6 óa = 41.4 ksi @ N = 10 S N = 50 ksi Case 3

R = 0.5 From the definition of stress ratio ómin = 0.5 ómax This gives the following relationships:

Therefore óa = 3 óm Substituting in the Gerber equation and solving for óa R= -1 R = 0.5 3 óa = 27.7 ksi @ N = 10 S N = 90 ksi 4 óa = 26.7 ksi @ N = 10 S N = 74 ksi 5 óa = 25.4 ksi @ N = 10 S N = 61 ksi 6 óa = 24.0 ksi @ N = 10 S N = 50 ksi The óa values for various R cases can be plotted on an S - N curve.



2.

Anothe Anotherr meth method od used used to to pres present ent mean mean stre stress ss fatigu fatiguee data data is to genera generate te a fami family ly of curves curves on an S-N plot, with each curve representing a different mean stress value, óm. Generate the curves for mean stress values of 0, 20 and 4 0 ksi for a steel with an ultimate strength of 100 ksi. For this example, use the goodman relationship to generate these curves. Again use eqn. (1) to estimate the fully reversed (ómean = 0) fatigue behaviour.

SOLUTION The Goodman relationship is: The alternating stress level for a given life can be determined from:

Case 1

óm = 0 (s (same as fully reve eversed lo loadi ading) ng)

@ @ @ @

Case 2

N = 103 N = 104 N = 105 N = 106

S N S N S N S N

= = = =

R = -1

óa = 90 ksi óa = 74 ksi óa = 61 ksi óa = 50 ksi

óm = 20 ksi

Substituting this value of óm into the Goodman Goodman equation for for different different value value of SN gives:

@ @ @ @ Case 3

3

N = 10 N = 104 N = 105 N = 106

R = -1 óm = 20 ksi S N = 90 ksi óa = 72 ksi óa = 59 ksi S N = 74 ksi óa = 49 ksi S N = 61 ksi óa = 40 ksi S N = 50 ksi

óm = 40 ksi

Substituting this value of óm into the Goodman Goodman equation for for different different value value of SN gives:

@ @ @ @

3

N = 10 N = 104 N = 105 N = 106

S N S N S N S N

R = -1 óm = 40 ksi = 90 ksi óa = 54 ksi óa = 44 ksi = 74 ksi óa = 37 ksi = 61 ksi óa = 30 ksi = 50 ksi

The óa for various óm can now be plotted on a S-N curve





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 103 , 104 , 105 and 106 cycles. Make predictions using the Goodman, Gerber and Morrow relationships.

SOLUTION: For R = 0 and óm = óa the mean stress equations can be written as:

Goodman:

Gerber:

Morrow:

When Su = 70 ksi and óf = 115 ksi Find S N from

Given 

Se = 33 ksi and using C = 2.08

S1000 = 0.9 Su = 63 ksi

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: Note: at R = 0 i.e

ómax = 2 óa

at 103

Goodman

Gives óa = 33 ksi



R=0

103 cycles

104 cycles

105 cycles

106 cycles

óa (ksi)

ómax (ksi)

óa (ksi)

ómax (ksi

óa (ksi)

ómax (ksi

óa (ksi)

ómax (ksi

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

Goodman



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: ómax = 75 ksi

ómin = - 5 ksi

Mean Stress

Alternating Stress

Stress ratio

Amplitude ratio Using the Goodman relationship

Given

óa = 40 ksi

Solving for S N Estimating

óm = 35 ksi

and Su = 100 ksi

SN = 61.5 ksi Se as 0.5 Su , S1000 as 0.9 Su

and substituting in

N = 8.9 x 104 cycles



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) aluminum alloy. Esteel = 208 GN/m2 , E Aluminum = 79 GN/m2 .

SOLUTION For a cantilever

Factored endurance from figure

= 5 x 100,000

=

5 x 105 cycles

for mild steel For Aluminum

for steel

for aluminum (x) ()

Aluminum Alloy 24S-T3 reversed axial stress Mild Steel reversed axial stress



6.

A pressure vessel support bracket is to be designed so that it can withstand a tensile loading cycle of 0-500 MN/m2 once every day for 25 years. Which of the following steels would have the greater tolerance to intrinsic defects in this application: (i) a maraging steel (KIC = 82 MN m -(3/2) , C = 0.15 x 10-11 , m = 4.1), or (ii) a medium-strength steel (K IC = 50 MN m-(3/2) , C = 0.24 x 10-11 , m = 3.3)? For the loading situation a geometry factor of 1.12 may be assumed.

SOLUTION The number of cycles in 25 years = Nf = 1 x 365 x 25 = 9125

( i)

(ii)

In a similar way for medium strength steel

Which is more damage tolerant ?



7.

A series of crack growth tests on a moulding grade of polymethyl methacrylate gave the following results: da/dN (m/cycle) Ä K (MN m-3/2)

2.25 x 10-7 0.42

4 x 10-7 0.53

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 ìm long, (2a), estimate the maximum repeated tensile stress which could be applied to this material for at least 106 cycles without causing fatigue failure.

From the graph:

so

c = 2 x 10-6

m = 2.513

ó = 2.13 MN/m2 Problem Set Solutio ns Page 10 of 31

6.2 x 10-7 0.63

17 x 10-7 0.94

29 x 10-7 1.17


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) Fatigue endurance (MN/m2 )

No hole 600

5 250

10 270

20 320

Solution: The Kt 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 Kt = 2.84 Also Kf = 600/250 = 2.4

Therefore

Similarly for other holes: d (mm)

Kt

Kf

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.


25 370


9.

The fatigue endurances from the S-N curve for a certain steel are:

Stress (MN/m2)

Fatigue endurance (cycles)

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/m2 , how many cycles can the material be expected to withstand at 410 MN/m2 before fatigue failure occurs, assuming that Miner’s cumulative damage theory applies? Solution:

Using cumulative damage theory



Strain Life

1.

It has been determined that a certain steel (E = 30 x 103 ksi) follows the following true stress, ó, true plastic strain, å p, relation:

The true plastic strain at fracture was found to be 0.48. Determine: a) True fracture strength, óf b) Total true strain at fracture c) Strength coefficient, K d) Strain hardening exponent, n e) Strength at 0.2% offset, Sy f) Percent reduction in area, % RA g) True fracture ductility, åf SOLUTION (a)

True fracture strength, óf

b)

Total true strain at fracture

c)

Strength coefficient, K

K = 360 ksi d)

Strain hardening exponent, n n = 0.11

e)

Strength at 0.2% offset, Sy This is the stress corresponding to a plastic strain of 0.002



f)

Percent reduction in area, % RA

RA = 0.381 or 38 %

g)

True fracture ductility, åf



2.

The following stress-strain and strain-life properties are given for a steel: E = 30 x 10 ksi

K = 137 ksi b = -0.11 c = -0.64

SOLUTION a) Elastic strain-life

intercept at 2Nf = 1 of Plastic strain-life intercept at 2Nf = 1 of Determine the transition life (2Nt ). From graph 2Nt = 30,000 reversals

using equation

2Nt = 30,366 reversals


n = 0.22


b)

Draw the hysteresis loops corresponding to strain amplitude (Äå/2) values of 0.05, 0.00125 and 0.0007. Determine the fatigue life in reversals at these three strain levels.

by iteration Äå/2 = 0.05 Äó/2 = 70.1 ksi 2Nf = 100 5 Äå/2 = 0.0125 Äó/2 = 24.18 ksi 2Nf = 3 x 10 Äå/2 = 0.0007 Äó/2 = 18 ksi 2Nf = 1 x 107 For

Äå/2

= 0.05

For Äå/2 = 0.0125

The hysteresis loop for 0.0007 is almost a straight line Problem Set Solutio ns Page 16 of 31


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. Amplitude Stress Stress

Life In Reversals _____________________________

Morrow Man-Hal ________________________________________________________

c)

SWT

.05000 .00125 .00070

0.00 0.00 0.00

70.1 24.8 18.0

1.075E+02 3.514E+05 1.122E+07

1.075E+02 3.514E+05 1.112E+07

1.109E+02 5.992E+05 1.775E+07

.00800

0.00

45.2

2.527E+03

2.527E+03

3.024E+03

Determine the elastic, plastic and total strain amplitude for a life (2Nf ) of 2 x 106 reversals.

= 0.0008111 + 0.0000881 = 0.000899



d)

life (2Nf ) 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

at 2Nf = 500

at 2Nf = 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.



3.

Smooth aluminum specimens are subjected to two series of cyclic load-controlled tests. The first test (level A) varies between a maximum stress value, ómax , of 21.3 ksi and a minimum value, ómin , 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, MansonHalford 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 103 ksi

K = 95 ksi

óf = 160 ksi

b = -0.124

åf = 0.22

c = -0.59

n = 0.065

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. Level

Test Results: Lives in Reversals, 2Nf

A

5.4 x 105

5.5 x 105

7.2 x 105

B

5.6 x 104

6.4 X 104

6.9 X 104

SOLUTION Using the following equations:

Mean Stress

Stress Range

Stress Amplitude

The strain amplitude can be determined using:



Therefore: óMAX

óMIN

óo

Äó/2

Äå/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. Amplitude Stress Stress

Life In Reversals _____________________________

Morrow Man-Hal ________________________________________________________ .002425 .002425

-4.4 35.8

21.3 61.5

3.498E+06 4.626E+05

3.545E+06 3.694E+05

SWT

5.646E+06 9.842E+04

Comparing these predictions with measured result, note that all three methods give nonconservative estimates.



Fracture Mechanics

1.

A large plate made of AISI 4340 steel contains an edge crack and is subjected to a tensile stress of 40 ksi. The material has an ultimate strength of 260 ksi and a K value of 45 ksi c 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 ksiin and a stress ó = 40 ksi

aC = 0.32 in

2.

A large cylindrical bar made of 4140 steel (óy = 90 ksi) contains an embedded circular (penny shaped) crack with a 0.1 in. diameter. Assume that the crack radius, á 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:

Since rY is very much smaller than a the LEFM assumptions are not violated Problem Set Solutio ns Page 21 of 31


3.

A very wide late made from Al 7075-T651 (K Ic = 27 ksi in., óy = 80 ksi) contains an edge crack. Plot the allowable nominal stress (ksi) as a function of crack size, á (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 used ( K Ic = 35 ksi in., óy = 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 7075 alloy 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 occ ur while the material remains nominally elastic is:

a = 0.016 in The allowable crack size as a function of crack size is shown below:



For a nominal stress of ½ 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.



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 the vessel to leak before break, KI < KIC The stress intensity factor for a semi-elliptical crack is

The shape parameter Q is found from the graph. Given: KIC = 60 ksiin óY = 85 ksi dia = 48 in pressure = 1000 psi The stress due to the pressure is:

The ratio ó / óY 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 t

ó / óY

Q

K I

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 Problem Set Solutio ns Page 24 of 31


5.

A component made from 7005-T53 aluminum contains a semi-circular surface crack (a/c = 1) and is subjected to R = 0.1 loading with a stress range, Äó, of 250 MPa. (Refer to Example 1 for an expression for the stress intensity range, Ä K .) the following crack growth data were obtained in laboratory air environment. Using these data: a) b) c)

Plot crack length, á (mm), versus cycles, N Plot da/dN versus Ä K . Identify the three regions of crack growth. Determine the Paris law constants, C and 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 __________________________________________________________



SOLUTION: The SIF for a semi-elliptical crack is: for a semi-circular crack Q is ð/2 therefore

For a = 0.244 mm

N (cycles)

a (mm)

ÄK

da/dN (mm)

Log(ÄK)

Log(da/dN )

(MPam)

95000

0.244

4.93

100000

0.246

4.95

7.00 x 10-7

0.695

-6.155

105000

0.251

5

3.920 x 10-6

0.699

-5.406

110000

0.285

5.33

9.665 x 10-6

0.727

-5.015

115000

0.347

5.89

1.053 x 10-5

0.77

-4.977

125000

0.414

6.43

1.230 x 10-5

0.808

-4.91

130000

0.49

6.99

2.063 x 10-5

0.845

-4.686

135000

0.621

7.87

4.661 x 10-5

0.896

-4.332

140000

0.956

9.77

9.565 x 10-5

0.99

-4.019

145000

1.577

12.55

3.964 x 10-4

1.099

-3.402

147000

2.588

16.07

1.105 x 10-3

1.206

-2.957

147400

3.078

17.53

1.554 x 10-3

1.244

-2.808

147500

3.241

17.99

8.758 x 10-3

1.255

-2.058

147500

3.445

18.54



The a vs N curve is:

The da/dN plot is:

For all points: Neglecting points 1to 3 and point11



6.

Calculate the critical defect size for each of the following steels assuming they are each subjected to a stress of 0.5óy. Comment on the results obtained. Steel

Yield Strength (MN/m2)

óy

Fracture 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 Strength óy (MN/m2)

Fracture Toughness (MN-3/2)

ac (mm)

Critical Defect Size 2ac (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 be cause critical defect size is small, however materials such as mils steel would require very large specimens in order to achieve this critical defect size



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 own weight. For glass KIC = 0.3 (MN-3/2 ) and density = 2600 kg/m3 .

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

Using the Flaw Shape Parameter graph of Question 5 and assuming that ó /óYS = 0 The Shape Parameter Q = 1.62

óc

= 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



8.

The accident report on a steel pressure vessel which fractured in a brittle manner when the internal pressure of 19 MN/m2 had been applied to it shows that the vessel had a longitudinal crack 8 mm long and 3.2 mm deep. A subsequent fracture mechanics test on a sample of the steel showed that it had a KIC value of 75 MN-3/2 . If the vessel diameter 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

Using the Flaw Shape Parameter graph of Question 5 and assuming that ó /óYS = 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


PA-Solutions 4012.pdf

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