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Exercises on Fracture Mechanics
Jan Zuidema and Michael Janssen
These exercises are made available free of charge. Details of the accompanying
textbook Fracture Mechanics are available at http://www.sponpress.com and can be
ordered from [email protected] or phone: +44 (0) 1264 343071
© 2002 Jan Zuidema and Michael Janssen
All rights reserved. No part of this book may be reprinted or reproduced or utilized in any
form or by any electronic, mechanical, or other means, now known or hereafter invented,
including photocopying and recording, or in any information storage or retrieval system,
except for the downloading and printing of a single copy from the website of the publisher,
without permission in writing from the publishers.
Publisher's note
This material has been produced from camera ready copy provided by the authors
1 Linear-Elastic Fracture Mechanics
2 Elastic-Plastic Fracture Mechanics
3 Fatigue and Stress Corrosion Crack Growth and Dynamic Effects
4 Combinations & Others
5 Answers
Exercises are presented in an order which is roughly determined by the book “Fracture Mechanics” used in the course with the same name. The exercises contain a rather arbitrary mixture ranging from relatively simple theoretical questions to more complex problems for which a
large part of the knowledge of the book is required. To facilitate self-study, the answers to
most of the exercises are given at the end.
1 Linear-Elastic Fracture Mechanics
1. What is meant by:
a) the stress concentration factor?
b) the stress intensity factor?
c) the critical stress intensity factor?
2. Calculations based on the cohesion force suggest that the tensile strength of glass should be
10 GPa. However, a tensile strength of only 1.5 % of this value is found experimentally.
Griffith supposed that this low value was due to the presence of cracks in the glass. Calculate
the size 2a of a crack normal to the tensile direction in a plate.
Given: Young’s modulus E = 70 GPa
surface tension g = 0.5 J/m
3. From the energy balance the following fracture criterion can be derived:
(F - Ua) > (Ug)
Is this criterion valid for:
a) Linear-elastic material behaviour?
b) Non-linear elastic material behaviour?
c) Elastic-plastic material behaviour?
4. A plate of maraging steel has a tensile strength of 1900 MPa. Calculate the reduction in
strength caused by a crack in this plate with a length 2a = 3 mm oriented normal to the tensile
Given: Young’s modulus E = 200 GPa
surface tension ge = 2 J/m
plastic energy per unit crack surface area gp = 2×104 J/m2
critical stress intensity factor Kc = sc pa
5. Describe the effect of yield strength on fracture behaviour.
6. Consider a plate with an edge crack (see figure). The plate thickness is such that a plane strain
condition is present.
W = 1000 mm
stress intensity factor KI = Cs pa where C = 1.12
(N/mm )
tensile strength suts
(N/mm )
plane strain fracture toughness
KIc (N/mm )
Steel 4340
Maraging steel
Al 7075 -T6
yield strength sys
Answer the next questions for the three materials given in the table above:
a) Does fracture occur at a stress s = 2§3 sys and a crack length a = 1 mm?
b) What is the critical defect size at a stress s = 2§3 sys?
c) What is the maximum stress for a crack length a = 1 mm without permanent consequences?
Figure for exercise 6
7. In a mould a wide and 3 mm thick plastic plate is cast. In this plate a central crack is created
with a length 2a = 50 mm. The plate is then mechanically tested in the length direction normal
to the crack.
a) If the plate fails at a stress s = 5 N/mm , what then is the Kc value of the material? Corrections for finite specimen size need not be taken into account.
b) In a similar plate copper wires with a cross section of 2 mm are introduced in the length
direction to act as reinforcements. These wires have a relative distance of 20 mm, while one
of the wires crosses the central crack exactly through the middle (see figure). By testing the
plate, tensile stresses develop in the parts of the 3 wires located between the crack flanks.
These stresses are found to be equal to 24, 36 and 24 N/mm respectively at a test stress
s = 5 N/mm2. How high is KI at this moment if the stresses in the wires are assumed to load
the crack flanks over the whole plate thickness?
c) To what level can the test stress s be increased before failure occurs? Assume that the
stresses in the wires are proportional to s.
Figure for exercise 7
8. Consider a strip-shaped structural part with a width of 200 mm and a hole in the centre with a
diameter 2R = 10 mm. In service the part is subjected to a tensile load in the length direction of
no more than 450.3 kN, while no load is applied directly to the hole.
Material properties: plane strain fracture toughness KIc = 2500 N/mm3/2
maximum fracture toughness Kc max = 4000 N/mm3/2
yield strength sys = 790 N/mm2
a) During the design a safety factor with respect to the yield strength is adopted equal to 2.
Furthermore, the presence of the hole is taken into account, but not the stress concentration
that is caused by it. What thickness will the structural part be given?
b) Does this plate thickness lead to KIc-behaviour according to the ASTM criteria?
c) Estimate Kc for the plate thickness that is to be applied, using the linear interpolation
method proposed by Anderson.
d) On both sides of the hole cracks may develop due to fatigue. Plot the residual strength as a
function of 2(R+a), where a is the length of the individual cracks. Assume that KI =
s p(R+a), where s is the nominal stress.
e) Determine the critical crack length 2(R+a).
9. For the design of a polycarbonate pane the following data are available:
∑ at a thickness of 5 mm Kc = 4 MPa÷m;
∑ for thicknesses B larger than required for a plane stress state, the value for Kc (in MPa÷m) is
given as a function of B (in m), i.e.
Kc = KIc +
where Kc,max and KIc are the critical plane stress and plane strain KI values respectively.
a) What is the maximum thickness if it is required that fracture is ductile?
b) What happens if a 10× thicker pane is chosen?
c) Compare the ratio of the critical loads (in terms of forces) for the two cases, while assuming
a plate geometry containing a central crack with a given size.
plane strain fracture toughness KIc = 2.2 MPa÷m
yield strength sys = 64 MPa
10. Consider a 25 mm thick steel specimen with an edge crack. A test is performed in which the
specimen is crack line loaded with 10 000 N at the edge of the crack. While artificially increasing the crack size, the crack opening displacement d at the edge is measured. The following relation between crack length a and displacement d (both in mm) is found:
d = 8×10-7a3 .
What is the maximum crack length a, if KIc for this steel is 1785 N/mm ?
Material data: Young’s modulus E = 210 000 N/mm
Poisson’s ratio n = 0.28
yield strength sys = 650 N/mm
11. Which failure mechanism (besides crack
growth) can occur in a thin uni-axially loaded
centre cracked plate with a large crack length
2a? Explain this phenomenon.
12. From a small loaded hole cracks have
developed (see figure). The hole and the
cracks are also subjected to an internal pressure P. Calculate the K-solution, assuming a
geometry factor equal to 1.
13. Consider a plate with a width W = 300 mm
and a thickness B = 2 mm containing a central crack with a length 2a = 40 mm. The
following material data are known:
∑ Young’s modulus E = 210 000 MPa
∑ yield strength sys = 375 MPa
First the plate is loaded with a force of 100 kN.
Figure for exercise 12
a) How high will KI become at least?
At this load the plate does not fail. However the crack appears to have grown to 2a = 44 mm.
b) How is this phenomenon called and why has the plate not failed?
c) How high has KI become?
d) Is the fracture toughness of the material at the current thickness, Kc, higher or lower than
KI? Is the plane strain fracture toughness KIc higher or lower than KI?
It is decided to increase the load until failure, which eventually occurs at a force of 120 kN.
Using a high-speed movie camera, it is found that just before failure the crack had grown to a
length 2a = 46 mm.
e) Is this a purely fracture mechanical failure, when using the Feddersen approach? Motivate
your answer.
f) Using the available data, which critical K values can be determined for the material under
consideration and what are their values?
14. In service a construction part made of a high strength steel is subjected to a constant stress of
1200 MPa. However, in course of time the part fails. Inspection of the fracture surface points
to non-stable crack extension from an embedded circular crack, normal to the load direction
and with a diameter of 100 mm.
a) How high is the stress intensity factor KI for this defect as a result of the externally applied
It is suspected that the failure is due to hydrogen-induced cracking, i.e. a high hydrogen pressure which has developed inside the crack.
b) Using the superposition principle, derive which formula should be used to calculate KI for
this defect due to an internal pressure only.
The fracture toughness for this steel is known: KIc = 27.5 MPa÷m.
c) How high was the hydrogen pressure inside the crack at the moment of failure? For this
situation it may be assumed that plane strain conditions are present.
15. Through a hole in a large 10 mm thick plate of polycarbonate a steel nail with a diameter of 3
mm was driven. Afterwards it became obvious that the hole was too small for the nail because
cracks developed on opposite sides, each with a length of 1 mm (see figure A). The fracture
toughness of the polycarbonate is known: KIc = 30 Nmm .
Figure A for exercise 15
a) How high is KI for these cracks?
b) How high is the load that is exerted by the nail onto the plate? Consider this load to consist
of two point forces F (see figure A). Furthermore, assume that these forces cause the same
stress intensity as if they were acting on the flanks of a central crack with the same total
length, i.e. 5 mm.
In spite of the cracks the plate is subjected to a load s of 5 MPa in the direction normal to the
cracks. After loading it is found that both cracks have grown to a size of 3½ mm (see figure B).
Figure B for exercise 15
c) How high is KI now?
d) Is the nail still attached to the plate?
16. The total energy U of a loaded elastic body containing a crack is given by:
U = Uo + Ua + Ug - F .
a) What do the terms mean?
b) What is the energy available for crack growth?
c) What is the energy required for crack growth?
17. According to Irwin plastic deformation at the crack tip results in an apparently longer crack,
i.e. KI = s p(a + ry), where ry is the radius of the plastic zone.
What will the KI solution become if we realise that KI affects the size of ry, while ry also affects
the size of KI?
18. Plot how the critical stress intensity Kc depends on the thickness and explain this.
19. For a double cantilever beam specimen loaded by a force P, the displacement
can be written
a = crack length,
E = Young’s modulus,
I = moment of inertia.
A constant displacement is applied to the specimen. Derive whether KI increases or decreases
when the crack grows.
Figure for exercise 19
20. Consider a plate with a width W = 300 mm and a thickness B = 12.5 mm containing a central
crack with a size 2a = 50 mm. In a test the load at failure is found to be 450 kN.
a) How high is the fracture toughness Ke in MPa÷m? Is this a valid value?
b) What is the minimum plate width for a valid Ke determination?
c) Calculate the residual strength (in MPa) for a plate with the same crack length and thickness
but with a width W = 100 mm.
d) If this was the result of a test, what would Ke be? Is this a valid value?
geometry factor f(a/W) = 1
yield strength sys = 500 MPa
21. In a Kc (Ke) test the plate fails at a stress sc higher than 2§3·sys. If this stress is used to calculate
Kc or Ke, will the result be too low or too high? Explain the answer.
22. Show that for crack growth under constant load conditions half of the work performed on a
centre cracked plate is used for an increase of the elastic energy of the plate, while the other
half is used for crack extension.
23. The through-thickness yielding criterion is: KIc > sys B
a) What is the use of this criterion?
b) Explain why at 30 mm thickness 10Ni steel fails in a tougher way than the alloy Ti-6Al-V.
24. The R-curve for a certain material can be expressed as:
+ 3(a-ao) ,
with R in MN/m, a and ao in meters, KIc = 150 MPa÷m and E = 210 000 MPa.
A wide centre cracked plate of this material has a crack length 2ao = 60 mm.
a) Show that this plate allows a maximum stable crack growth of 3.1 mm at both tips.
b) Calculate the critical stress (sc) and the fracture toughness (Kc)?
25. On the fracture surface of a plate-shaped part of aluminium alloy 7075-T6 that has failed due
to an overload, shear lips are found. The shear lips on each side have a width of 1 mm and so
in total 2 mm of the fracture surface in the thickness direction is slant. The thickness of the
part is 4 mm.
Calculate Kc for this material at this thickness using the results of Knott’s analysis of the model
of Krafft et al. (Young’s modulus E = 70 000 N/mm )?
26. There is a choice between two materials, designated as A en B. A construction part is being
designed with a safety factor 3 with respect to the yield strength. Which material should be
chosen if the maximum allowable crack size is to be as large as possible?
material A: sys = 350 MPa ; KIc = 50 MPa÷m
material B: sys = 500 MPa ; KIc = 60 MPa÷m
Figure for exercise 27
27. For a through-thickness crack (length 2a) in a wide plate KI = s pa, whereas for a row of
collinear cracks with interspacing l (see figure)
KI = s
l tan
In guidelines for the safety of welded constructions, requirements are set for plane crack-like
welding defects with respect to the distance s (see figure). If s is smaller than the defect size
2a, two neighbouring defects should be considered as one larger defect (length s + 4a). In this
case the first K-formula is used (KI = s pa).
a) Determine the ratio between the two KI values for s = 2a.
b) Is this guideline conservative, i.e. on the safe side, for s = 2a?
c) Calculate the ratio between the two KI values for a very small s value, i.e. s = a/10, and explain the result.
28. A tool (pick-axe) is used to break up old roads (see figure).
a) What force can be allowed without a crack? Use the fact that stresses acting in the same direction on the same plane may be superposed. Take the yield strength as a failure criterion.
Perform the calculation for the part where a crack is drawn in the figure.
b) A crack develops, but it is only detected when it reaches a size of 5 mm. What is the critical
fracture force at that moment?
fracture toughness KIc = 59 MPa÷m
yield strength sys = 1500 MPa
correction for a free surface: tensile 1.12, bending 1.03
correction for finite dimensions is not necessary
bending stress sb = 6M/Bh2, where M is the bending moment
Figure for exercise 28
29. In a plate of aluminium alloy 7075 a central defect is present with a total length of 3 mm.
a) Calculate the critical stress for fracture using the Griffith criterion and using KIc.
b) Explain possible differences.
surface tension ge = 1.14×10 J/mm
Young’s modulus E = 7×10 N/mm
fracture toughness KIc = 1040 N/mm
2 Elastic-Plastic Fracture Mechanics
30. A JIc test is performed on steel with the following properties:
E = 207 GPa; sys = 360 MPa en suts = 560 MPa ; n = 0.28.
For this purpose a 3-point bend specimen is used with the following dimensions:
W = 50 mm; B = 20 mm; a = 30 mm.
The load is found to increase linearly with displacement. At the onset of crack extension the
load is 25 kN, while the displacement is 4 mm.
a) What value follows for JIc, if this is defined as J at the onset of crack extension?
b) Is this value acceptable according to ASTM standard E 813?
c) What is the corresponding value for KIc?
d) What thickness should the specimen be for a valid KIc determination?
31. For a large welded steel vessel the following data are available:
service stress s = 200 N/mm
yield strength weld metal sys = 450 N/mm2
Young’s modulus weld metal E = 205 000 N/mm2
Due to shrinkage after the welding process, residual stresses develop in the weld seam which
can be as high as the yield strength of the weld metal. By means of a heat treatment these
stresses can be reduced. It is required that the vessel can withstand a crack in the weld metal
with half crack size a = 60 mm. Furthermore the CTOD value in this case may not exceed 0.5
To what level should the residual stresses be reduced to comply with these requirements?
32. A JIc test is performed on HY130 steel. The results, measured on SENB specimens, are:
U [J]
Da [mm]
Specimen dimensions: span L = 4W = 200 mm
width W = 50 mm
thickness B = 0.5 W= 25 mm
crack length a = 30 mm
Material properties:
yield strength sys = 925 MPa
tensile strength suts = 953 MPa
Young’s modulus E = 210 000 MPa
Poisson’s ratio n = 0.28
mass density = 7.8×10 kg/m
a) Approximately (!) determine JIc using a spreadsheet or on paper. Assume that conditions
which can not be checked but are necessary for a valid determination are fulfilled.
b) How high is KIc for this steel?
c) At least how many kilograms of weight would be saved by determining KIc through JIc with
this test? Assume that the ratios of the dimensions of the KIc specimen are equal to those
listed above.
33. On a bend specimen a single-specimen Jc test is performed. Using a non-destructive technique
crack initiation is observed at a load P = 9315 N. During the test the load is recorded as a
function of the load-point displacement V. A polynominal fit of the P-V curve resulted in:
= 10-5P + 10-14P4 ,
where V is expressed in mm and P in N.
a) Determine the critical J value at initiation.
b) Is the plate thickness sufficient for a valid JIc test?
width W = 50 mm
crack length a = 12 mm
thickness B = 28 mm
flow stress so = 1000 MPa
34. Derive: J = 2soDa
How is this J versus Da line called and what is its purpose?
35. For materials with a moderate toughness (e.g. aluminium alloys) KIc can be determined from
JIc. Express the minimum required thickness for the JIc test (Bj) in terms of the minimum required thickness for the KIc test (Bk).
Young’s modulus E = 70 000 MPa
yield strength sys = 345 MPa
tensile strength suts = 500 MPa
E' for plane strain = E (for plane stress)
36. For the derivation of the J-integral the so-called deformation theory of plasticity is used.
What does this theory mean and which important condition (limitation) results from this?
37. Show that the occurrence of plastic constraint in actual structural parts leads to more safety
when using the COD design curve.
38. A (non-linear) elastic plate is loaded, leading to the
load-displacement diagram shown in the figure.
a) Assume that no crack growth occurs during the
loading process and indicate graphically the part
of the area representing the change in potential
energy of:
∑ the specimen,
∑ the combination of the specimen and the
loading system.
b) Now assume that crack growth occurs under
constant load conditions. Answer the previous
question but now for the change in potential energy due to crack growth.
Figure for exercise 38
39. Show that in the linear elastic regime the COD design curve predicts a maximum permissible
crack length (amax) equal to half the critical crack length calculated with LEFM.
40. In relation with the J integral the expressions dx1 = -n2ds and dx2 = n1ds are used. What is the
meaning of these quantities? Derive these formulas.
41. Explain why in case of short cracks or of very long cracks the linear elastic fracture criteria are
no longer valid, even for relatively brittle materials.
42. The value of the J integral is independent of
the exact path followed surrounding the crack
tip in counterclockwise direction, starting on
the lower and ending on the upper crack
a) What is J for a closed contour, i.e. one not
surrounding the crack tip singularity?
Figure for exercise 42
b) Indicate what is wrong in the following
Along the closed contour ABPA shown in the figure, the J integral is zero. Along the flanks
AP and BP J is zero too. Consequently J must be zero along the contour surrounding the
crack tip, A Æ B.
3 Fatigue and Stress Corrosion Crack Growth
and Dynamic Effects
43. Describe the possible effects of a peak load on the crack growth behaviour during variable
amplitude fatigue.
44. A crack with a depth of 2 cm is found in a cast iron rod of an old steam engine. Every weekend the machine is used for 8 hours for demonstration purposes, during which it is run at 15
rpm and the load in the rod varies from +6.4×10 N to -6.4×10 N. It may be assumed that the
crack is completely closed under compressive load but that under tensile load crack closure is
negligible. The Paris relation is sufficiently accurate until failure.
a) Is it safe to use this machine for demonstrations and for how long?
b) Answer these questions assuming a load variation from +0.92 MN to -0.92 MN.
Kc = 16 MPa÷m
DKth = 5 MPa÷m
cross section rod = 0.04 m
factor in the Paris relation: C = 4.3×10 m(MPa÷m)
KI = 1.12 s pa
45. Which of the two cases (A or B) drawn in the figure gives more retardation of the fatigue crack
growth and why is this?
case A
case B
Figure for exercise 45
46. On a material a constant amplitude fatigue load with smax = 100 MN/m2 and R = 0 is applied.
An edge crack is growing in the material. At a crack length of 16 mm there is a single overload
with smax = 200 MN/m .
Until what crack length does Wheeler’s model predict an effect on da/dn due to the overload?
yield strength sys = 400 MN/m
stress intensity factor KI = s pa
plastic constraint factor C = 1
47. Divers detect a through-thickness crack with length 2a = 70 mm in a support of an oil-winning
platform in the North Sea. The crack is situated along a weld below the water-line. Relevant
dimensions are given in the figure.
The constant amplitude loading exerted by the water waves is Ds = 15 N/mm , with a stress
ratio R = 0 and a frequency f = 1/6 Hz. For this geometry it may be assumed that K = s pa.
The steel of the support (BS 4360-50D) permits a maximum crack size of 2a = 150 mm at the
maximum load of the waves. Further material data are:
fatigue threshold DKth = 60 N/mm
Paris relation: da/dn = 5.1×10-11 DK 2.53 mm/cycle with DK in N/mm3/2
Figure for exercise 47
Only in the summer period, from April 1 to October 1, it is allowed to dive in the North Sea. It
is June 1st when the crack is detected. Calculate if it is necessary to repair the crack before
October or if it is possible to wait until next year.
48. During stress corrosion tests with decreasing K specimens corrosion products can lead to
apparently lower KIth values.
a) Explain this phenomenon.
b) How can we measure if it has had an effect on the measured KIth?
49. A wide plate with a central crack is subjected to a constant amplitude fatigue load with Ds =
100 MPa and R = 0. At a crack length 2a = 6 mm an overload occurs with a stress of 170 MPa.
How high is the crack growth rate immediately after the overload, after 0.2 mm and after 0.4
mm of crack growth (at each tip)? Assume plane stress conditions and, for convenience, consider K to remain constant during the indicated crack growth intervals.
Wheeler exponent m = 1.5
yield strength sys = 420 MPa
Paris relation: da/dn = 1.5×10
(DK) mm/cycle with DK in MPa÷m
50. Explain why a rapidly propagating crack in a pressurised pipeline will sooner show arrest if the
pipeline is fluid-filled than if it is gas-filled?
51. Consider the following plot of G,R versus crack length.
Figure for exercise 51
Indicate in the graph where the crack will grow accelerated and where it will slow down.
52. The following striation patterns are found on a fatigue fracture surface using a scanning
electron microscope (SEM).
Figure for exercise 52
Sketch a possible K variation with crack length for both tests. The frequency may be assumed
the same for both cases and independent of crack length.
53. In the context of fatigue crack growth, what is meant by delayed retardation?
Describe the conditions for which this occurs.
54. Show that the reversed plastic zone in fatigue is about 1§4 of the normal (monotonic) plastic
zone size. Hint: take R = 0.
55. During stress corrosion crack growth tests using decreasing K specimens, crack tip blunting
and/or crack tip branching leads to apparently higher KIth values. Explain both phenomena.
56. During a fatigue crack growth test on
aluminium a transition takes place in the applied DK, as is indicated in the figure.
a) How high is R (approximately) before and
after the transition?
b) Draw a line in the figure, representing the
value of Kop (both before and after the
transition). Use a linear crack closure relation for this purpose.
c) Indicate qualitatively how the crack
growth rate da/dn depends on time.
Figure for exercise 56
57. A structural part of about 20 cm thickness is subjected to a tensile load that varies 5 times per
minute between 110 and 200 MPa. For the material the following data are known:
∑ KIc = 30 MPa m
∑ DKth = 5 MPa m
∑ Paris curve: da/dn = 2×10-11(DK)3 with da/dn in m/cycle and DK in MPa m
a) Based on these data, what can be concluded about the lifetime of the part?
Investigations show that after the manufacture of this type of parts semi-circular surface cracks
may be present. Using an eddy current technique such cracks can be found if they are at least 2
mm deep.
b) Calculate whether or not there is a risk for crack growth.
c) What is the maximum inspection interval that ensures no failure will occur?
58. During fatigue loading of an aluminium specimen transitions take place as drawn in the
figures, i.e. the frequency remains the same, but Kmin (case A) respectively. Kmax (case B)
Draw qualitatively how da/dn depends on crack length directly after the load transition for
both cases. Take the effect of crack closure into account.
case A
case B
Figure for exercise 58
59. A cylindrical steel pressure vessel, with a diameter of 7.5 m and a wall thickness of 40 mm,
must operate at a pressure of 5.1 MNm . In the design it is assumed, that the vessel fails as a
result of unstable extension of a fatigue crack developing in the axial direction of the vessel. To
avoid such a failure, the number of times the vessel may be loaded from zero to maximum load
is limited to 3000.
and fatigue crack
The maximum allowable K (at failure) in this material is 200 MNm
growth is described by:
da/dn = A(DK)4 , where A = 2.44×10-14 (MNm-3/2)-4m.
Determine the pressure at which the vessel should be tested before use in order to guarantee
that failure does not occur before loading it 3000 times. Assume KI = s pa.
4 Combinations & Others
60. In fracture mechanics inclusions are generally subdivided in three categories classified according to their size. The largest category is that of inclusions ranging from 0.5 to 50 mm.
What role do these inclusions play in the following phenomena?
a) Ductile fracture
b) Initiation of a fatigue crack in:
1. smooth specimens
2. notched specimens
c) Fatigue crack growth at:
1. low crack growth rates
2. high crack growth rates
d) Fracture toughness KIc.
61. A bar containing a crack with a length ao = 0.002 m is subjected to a fatigue load. For this
geometry KI = 0.8s pa for all crack sizes.
a) Calculate the crack length after the bar has been subjected to 6000 cycles between a stress
of 60 and 120 N/mm and next to 3000 cycles between 0 and 120 N/mm .
b) Calculate the crack length if the loading order is reversed.
c) Assume the bar with the original crack was subjected to a static load of 160 MPa in an envi1.5
ronment that leads to a KIscc of 6 MN/m . Would stress corrosion cracking occur?
Paris relation: da/dn = 10 (DK) with a in meters and DK in MN/m
fatigue threshold DKth = 5 MN/m1.5
62. A long cylindrical pressure vessel is being designed. In practise the vessel will experience
internal (gas) pressure variations between 0 and 3.5 MPa. It is recognised that under these circumstances a half circular fatigue crack can extend from the interior to the exterior, after
which leakage will occur.
What is the minimum wall thickness the vessel should have to ensure the crack stops growing
due to the pressure drop caused by leakage before (unstable) failure of the vessel can occur?
Given: external cylinder diameter = 1 m
fracture toughness Kc = 49.4 MPa÷m
Hints: Neglect the wall thickness relative to the diameter of the vessel, assume that the crack
remains half circular during extension until leakage and take KI = s pa (solution for a
through-thickness crack).
63. Cracks develop on opposite sides of an unloaded central hole in a tensile element. The plate
width is 195 mm, the plate thickness is 6 mm and the hole diameter is 15 mm. In service the
load varies 100 times per day between 0 and 351 000 N. The minimum detectable crack length,
a, is 1 mm. For this geometry it is safe to assume that KI = s p(a+r), where r is the radius of
the hole.
The material has a KIc of 2000 N/mm and reaches a maximum Kc of 3400 N/mm at small
thicknesses. Furthermore, for a loading ratio R = 0 the crack growth rate da/dn = 10 (DK)
mm/kilocycle, with DK in N/mm . The yield strength, sys, is 800 N/mm .
a) Estimate Kc for the thickness to be applied using Anderson’s model.
b) In what range of a-values will the plate show Kc behaviour according to Feddersen?
c) Determine the inspection interval if at least three inspections should fall within the crack
growth period.
64. Investigation of a failed structural part leads to fatigue as the cause. The fatigue crack depth is
5 mm, after which failure occurred. On the fracture surface striations are found: the striation
distance at the end of the fatigue crack is 2 mm at a magnification of 2000×.
The fatigue load the structural part has been subjected to, is a constant load amplitude with a
load ratio R = 0, but with an occasional overload cycle with a maximum load of twice the
maximum of the constant amplitude load.
Material tests show that the yield strength sys = 900 MPa and that fatigue crack growth at con-12
stant amplitude and R = 0 is described by da/dn = 6×10 ×DK , with da/dn in m/cycle and DK
in MPa÷m. This relation may be assumed valid until fracture and crack closure can be ignored.
For this structural part KI = s pa.
a) Calculate the fracture toughness Kc (in MPa÷m) of the material.
b) Calculate the maximum stress (in MPa) of the constant amplitude load.
65. Consider a DCB specimen (see figure). With the bolt a displacement is applied of 0.1 mm.
After that the specimen is placed in an environment that causes stress corrosion in the material.
Figure for exercise 65
How high is KIscc if the crack stops growing at a = 50 mm and plane stress conditions prevail?
Young’s modulus E = 70 000 N/mm
height h = 25 mm
66. A 300 mm wide and 11 mm thick plate is subjected to a fatigue load at a frequency of 1 Hz.
The stress varies between smax = 120 MPa and smin = 40 MPa. At a certain moment an edge
crack is found of 4.5 mm.
Calculate how many hours it takes before the plate fails.
Material data: Paris relation: da/dn = 3.5×10 DK m/cycle with DK in MPa÷m
yield strength sys = 381 MPa
plane strain fracture toughness KIc = 35 MPa÷m
fracture toughness for thin plate (ª 1 mm) Kc = 62 MPa÷m
67. Failure analysis showed that a construction part, which was subjected to a constant load, had
failed by stress corrosion cracking. The crack length at failure was 6 mm. The material thickness was sufficient for plane strain. The factory where the parts are made checks every sample
using non-destructive techniques. It is guaranteed that no crack-like defects larger than 1 mm
are present on delivery.
Calculate the percentage with which the thickness of the part should be increased to avoid future failures due to stress corrosion cracking.
KIc = 60 MPa÷m
KIscc = 20 MPa÷m
KI = 1.12 s pa
68. A wide plate contains a central crack. The plate is subjected to constant amplitude fatigue
loading until failure. Analyses of the fracture surface reveals that the striation spacing is
0.5×10 m at 2a = 2.0 cm and 2×10 m at 2a = 4.0 cm crack length. For this material the co-13
efficient C in the crack growth relation of Paris is equal to 4×10 where da/dn is expressed in
m/cycle and DK in MPa÷m.
a) What was the applied stress interval Ds?
b) How many cycles were needed for crack growth from 2 until 4 cm?
69. For a material 4 constant amplitude test results at 4 different R values (R = -0.1, 0.2, 0.5 and
0.8 respectively) have been found (see the logarithmic and linear plots).
a) Find the coefficients c1 and c2 of the linear crack closure relation U = c1 + c2·R. Assume
that DKeff = DK for R = 1.
b) What is the crack growth rate da/dn at DKeff = 10 and 15 MPa÷m?
Figures for exercise 69
70. A cylindrical pressure vessel with a very large diameter has a wall thickness t = 1.25 cm. A
half-circular shaped crack of 0.25 cm depth is found on the inner surface. The orientation of
the crack is normal to the hoop stress in the cylinder wall. There is a risk of stress corrosion
crack growth. For the combination of the material and the environment it is known that KIscc =
10 MPa÷m. For safety reasons it is required that, should crack growth occur, the vessel will
leak before it breaks.
a) For which interval of hoop stresses will this apply, i.e. both crack growth and leak before
break will occur, assuming that for the material Kc = 88 MPa÷m?
b) Answer the same question assuming Kc = 25 MPa÷m.
Hint: Assume that KI cannot become larger than that for a through-thickness crack with a
length of twice the wall thickness.
71. During a JIc determination using a technique similar to that of Begley and Landes a critical
displacement (at the onset of crack growth) is found of 20 mm for an initial crack length of 10
mm. In the figure load displacement curves are given for 3 initial crack lengths (5, 10 and 15
mm). The tests are performed with bend specimens for which the thickness B and the height W
are both equal to 20 mm. The measured displacement is that of the load point.
a) How high is JIc?
b) Approximately determine the value for the critical displacement at an initial crack length of
15 mm.
Figure for exercise 71
72. For a certain steel the fatigue crack growth rate da/dn is known as a function of DK (see
figure). From this steel a cylindrical pressure vessel is made with a diameter of 7.5 meter and a
wall thickness of 100 mm. The operational pressure in the vessel is 2 MNm . At the design of
the vessel it was assumed that failure would occur from a fatigue crack in the axial direction
and that leakage would precede failure.
Calculate the number of fatigue cycles (R = 0) between leakage and failure. The Kc value is
200 MNm-3/2. Assume that at the onset of leakage a through thickness crack has developed
with a length equal to twice the wall thickness.
Figure for exercise 72
73. Using a single edge notched bend specimen of nodular cast iron, a JIc test is performed based
on the unloading compliance technique. In figure A the measured load P is plotted as a function of the load displacement V. From these data the slope of the P-V curve during the different
unloading-loading cycles are found as follows:
slope [N/mm]
(no crack growth yet)
For specimens with this geometry and made of this material the relative crack length a/W is
plotted in figure B as a function of the compliance C. Other data are:
Material: Young’s modulus E = 200 000 MPa
yield strength sys = 350 MPa
ultimate tensile strength suts = 450 MPa
Specimen: thickness B = 7 mm
height W = 4×B
a) Roughly indicate the positions of the J-Da points in figure C, as they follow from the experimental results. Be sure to add numerical values to the vertical axis. J may be calculated
without making a distinction between elastic and plastic displacements.
b) Explain how crack tip blunting is quantified for a JIc test. Indicate this in figure C.
The critical J value according to ASTM E 813 is defined as the value after 0.2 mm of stable
crack growth.
c) In figure C, draw the line that represents this stable crack growth.
d) Roughly show in figure C how the critical J-value, JIc, is obtained.
e) What is the JIc value (not a calculation but a graphical estimate)?
f) Is this value valid in view of the size requirements?
Figure A for exercise 73
Figure B for exercise 73
Figure C for exercise 73
74. To determine the susceptibility of a material for stress corrosion cracking, 4 specimens with
initial crack lengths of 12.5 mm (specimens with one crack tip) are tested. The times to failure
are found as 1, 10, 500 and 5000 hours, while the final crack sizes are 20, 30, 50 and 51 mm
Estimate KIscc of the material as accurate as possible.
specimen width W = 100 mm
plane strain fracture toughness KIc = 40 MPa÷m
geometry factor f(a/W) = 1.12 + (a/W)
75. Consider a centre cracked plate subjected to a uniform stress of 130 MPa. The fracture toughness Kc = 50 MPa÷m, the yield strength sys = 420 MPa and the plate width W = 300 mm.
a) What is the maximum allowable crack length?
b) Calculate the maximum allowable crack length if KI is corrected for plasticity.
76. In an elastic finite element analysis the stress component sy for the first 3 elements in front of
the tip of an edge crack are calculated as 170, 98 en 75 MPa respectively. The element size is
2.5 mm and the calculated values may be considered representative for the stress in the centre
of each element. In the calculation the crack length is taken as 25 mm, the plate thickness as 10
mm and the plate width as 150 mm.
a) Estimate KI as accurate as possible.
b) Determine the load applied to the plate.
77. A thin-walled cylindrical pressure vessel is found to contain a serious scratch in the length
direction. Reparation by means of welding is less desirable in view of the nature of the contents
and will only be done should catastrophic failure occur before the moment of leakage. On the
other hand, if the situation remains stable at leakage, the reparation is postponed.
Should reparation take place immediately? Motivate your answer.
vessel diameter = 2500 mm
wall thickness = 25 mm
length of the scratch = 50 mm
maximum pressure = 4 MPa
Kc= 75 MPa÷m (at this thickness)
Figure for exercise 78
78. An R-curve is measured for a thin wide plate made of an aluminium alloy containing a central
crack with length 2a = 10 mm (see figure).
a) Approximately how high is the plane strain fracture toughness KIc of the aluminium?
b) Determine at which KI value (Kc) an identical plate will fail if it is loaded normal to the
c) How high is the (nominal) stress in the plate at that moment?
Young’s modulus E = 7×10 MPa
Poisson’s ratio = 0.33
79. In a very large steel part a crack is found with a length of 2a = 40 mm. The structural part will
be applied under circumstances where stress corrosion plays a role. The incubation time for the
combination of material, thickness and environment can be derived from the plot in figure A.
Information about the stress corrosion crack growth rate is given in figure B.
For the geometry of the part the stress intensity as a function of crack length is given by KI =
s pa. The part will be subjected to a constant load resulting in a nominal stress equal to 100
What is the lifetime of the part if possible crack growth is allowed as long as KI < Kc/4?
Figure A for exercise 79
Figure B for exercise 79
80. After being used for a long time a 7 cm long edge crack oriented normal to the load direction
was found in a wide panel of some alloy. Investigation showed that crack growth started from
a small notch at the edge of the panel. Apparently this notch was present when the panel was
put into operation. The panel was designed for constant amplitude fatigue loads (at R = 0)
lower than 20 % of the yield strength (sys = 345 MPa).
Since the crack length of 7 cm is unacceptably long, it is decided to replace the panel. Examination of the fracture surface at distances of 1.50 and 6.96 cm from the edge reveals striations
with average widths of 2.16×10 and 2.16×10 m respectively. For the material a crack
growth rate relation is known: da/dn = C(DKeff) with C = 5.566×10 and m = 3, where da/dn
is expressed in m/cycle and DKeff is in MPa÷m. DKeff is defined according to the Elber concept,
i.e. using U = 0.5 + 0.4·R.
a) The fact that the panel is prematurely taken out of operation, is this due to the notch initially
being unnoticed or has there been a fatigue load higher than the above-mentioned 20% of
the yield strength?
b) Was the fatigue load of the constant amplitude type?
c) How long would the remaining lifetime (in cycles) have been, if crack growth was allowed
until KI = Kc/2?
fracture toughness Kc = 64.04 MPa÷m
K solution: KI = 1.12 s pa.
81. The following data are known for a material:
∑ plane stress fracture toughness Kc = 35 MPa÷m
∑ plane strain toughness KIc = 25 MPa÷m
∑ yield stress sys = 375 MPa
Of this material a 25 mm thick double cantilever beam specimen is made, containing only a
machined notch. The specimen is subjected to a displacement that is such that cracking occurs.
a) Is it to be expected that the specimen fails? Motivate your answer.
b) How high will KI eventually become?
One wants to know the threshold value for stress corrosion crack growth in sea water. Literature indicates that this value in any case is lower than 10 MPa÷m.
c) How many percent should the displacement be decreased in order to measure this threshold
value as quickly as possible?
82. Consider the case described in exercise 15 but now for an applied load s that varies between 0
and 5 MPa.
a) How high is Kmax?
b) Immediately after applying the fatigue load, is the nail still attached to the plate?
One has to make the choice whether or not to remove the nail in order to minimise crack
growth by the fatigue load.
c) Indicate what the best choice is and motivate this. It may be assumed that fatigue crack
growth in polycarbonate does not depend on the loading ratio R.
5 Answers
2×10-6 m
yes; yes; no
31.5 %
a) yes, no and no; b) 0.59, 1.6 and 2.47 mm; c) 756, 1461 and 500 N/mm
a) 44.3 N/mm ; b) 35.6 N/mm ; c) 6.2 N/mm
a) 6 mm; b) no; c) 3779 N/mm3/2; e) 2(R+a) = 64.5 mm and thus a = 27.25 mm
a) 2.6 mm; b) Kc = 2.55 MPa÷m; c) Pc,b/Pc,a = 4.4
171 mm
buckling of the crack flanks
P pa + ½ (F/ pa +s pa)
a) 41.8 MPa÷m; c) 43.8 MPa÷m; d) higher, lower; f) Kc = 53.8 MPa÷m and Ke = 50.1 MPa÷m
a) 9.57 MPa÷m; c) 2247 MPa
a) 30 Nmm ; b) 840 N; c) 30 Nmm ; d) yes
KI = s pa/ 1-s2/2sys2
KI decreases
a) 33.6 MPa÷m; b) 19.4 mm; c) 110 MPa; d) 30.8 MPa÷m
too low
b) 1905 MPa and 614 MPa÷m
64.8 MPa÷m
material A
a) Kseparate/Ktogether = 0.65; b) yes; c) Kseparate/Ktogether = 2.09
a) 56 604 N; b) 17 190 N
a) Griffith: sc = 5.8 MPa, KIc: sc = 479 MPa
a) 250 N/mm; c) 237 MPa÷m; d) 1.1 m
0.85×sys = 383 MPa
a) ª 260 N/mm; b) ª 7700 N/mm ; c) 643 kg
a) 1.055 MN/m; b) yes
Bk ª 25 Bj
a) zero
a) yes, unlimited; b) no, lifetime 224 cycles
case B
18.8×10 meter
repair now (1.31×106 cycles to failure)
0.2×(da/dn)ca, 0.45×(da/dn)ca and 1×(da/dn)ca where (da/dn)ca = 1.33×10 mm/cycle
a) nothing; b) yes, there is a risk; c) 131 days
8.98 MPa
a) 4.0 mm; b) 4.6 mm; c) yes
minimum thickness 4 mm
a) 2983 N/mm ; b) 2.46 < a < 25 mm; c) 107/4 days
a) 55 MPa÷m; b) 219 MPa
4.8 MPa÷m
16.7 hours
22 % thicker
a) 188.6 MPa ; b) ± 10 000 cycles
a) c1 = 0.553, c2 = 0.447; b) 7.7×10 and 2.2×10 mm/cycle
a) 151 - 444 MPa; b) in case of crack growth the vessel will break before leakage
a) 1000 N/mm; b) ª 17 mm
ª 910 000 cycles
e) JIc ª 15 N/mm; f) yes
16.3 MPa÷m
a) 94 mm; b) 90 mm
a) 15 MPa÷m; b) 71.7 kN
stable situation, do not repair now
a) KIc ª 69 MPa÷m; b) Kc = 100 MPa÷m; c) 690 MPa
total time is about 30000 seconds
a) due to the notch; b) yes; c) 3267 cycles
a) no; b) 25 MPa÷m; c) 60%
a) 30 Nmm ; b) yes; c) do not remove the nail
Fly UP