FFF Solutions

Solutions Manualto problems in Failure, Fracture, Fatigue An Introduction by Tore Dahlberg Anders Ekberg Studentlitteratur, Lund 2002, ISBN 91-44-02096-1. This manual contains solutions to problems in the textbook Failure, Fracture, Fatigue - An Introduction Studentlitteratur, Lund 2002, ISBN 91-44-02096-1 At present, solutions to all problems given in Chapters 1 to 6 and Chapters 8 and 9 are available in this document (Chapter 7 does not contain any problems and solutions to the problems in Chapter 10 will be available later). It is our hope that these solutions will be to some help for teachers and students when studying the topics covered in the textbook. Samples of examinations are available (2010) on the home page of Solid Mechanics, IEI, at Linköping University, Sweden, address: www.solid.iei.liu.se under the heading "Examinations" and course "Damage mechanics and life analysis TMHL61". Linköping and Gothenburg in February 2010 The authors Address to Tore Dahlberg after May 2010: Åsmark 6, 563 92 Gränna, Sweden Page 1:2 Solutions to problems in T Dahlberg and A Ekberg: Failure, Fracture, Fatigue - An Introduction. Studentlitteratur, Lund 2002, ISBN 91-44-02096-1. Chapter 1 Introduction, failure mechanisms Problems with solutions Elastic deformations 1/1. Side view : P L Top view : P Cross section: B t H H >> t B >> t A cantilever beam, length L, carries a force P at its free end. The beam cross section is an ideal I profile, i.e., the area of the flanges is width B by height t each (B >> t), and the area of the web can be neglected. The height is H, where H >> t, see figure. When the beam is loaded by the force P, the deflection δ1 becomes P L3 ⎛ H ⎞ 2 Bt H 2 δ1 = where I = 2 Bt ⎜ ⎟ = 3 EI ⎝ 2⎠ 2 Thus, δ1 = 2P L 3 / 3E BtH 2 = 0.67P L 3 / E BtH 2. Side view : It has been found that this deflection is too large. Therefore, without increasing the weight (the mass) of the beam, another form of the cantilever beam cross section is tried. Material P is “moved” from the flanges at the free end and “placed” at the flanges at the fixed end to give a beam cross section (still an ideal I profile) with a width varying with the coordinate x, see figure. The height of the profile is still H, but H >> t the width of the flanges is now B(x) = B >> t 2B(1 − x /L). What deflection δ2 will now be obtained when the load is applied at the free end? P L Top view : x Cross section : 2 B (1 - x / L) t H SOLCHAP1.DOC/2010-01-29/TD Chapter 1 Page 1:1 and the second moment of area becomes ⎛ x⎞ ⎛ x ⎞ ⎛ H⎞2 I = 2 × 2B ⎜1 − ⎟ t ⎜ ⎟ = Bt H 2 ⎜1 − ⎟ ⎝ ⎝ L⎠ ⎝ 2 ⎠ L⎠ (b) The direction of the moment M(x) has been selected to give a positive deflection w(x) downwards. Page 1:2 Chapter 1 .Solution: Use the differential equation − EI(x) w’’(x) = M(x) (a) to determine the deflection w(x) of the beam. Entering M(x) and I(x) in the differential equation gives PL − M(x) PL (1 − x/L) = w’’(x) = = EI(x) E Bt H 2(1 − x/L) E Bt H 2 (c) Integration gives w’(x) = and PL x + C1 E Bt H 2 PL x 2 w(x) = + C1 x + C2 E Bt H 2 2 Boundary conditions at x = 0 give (notation BC for boundary condition) BC 1: w(0) = 0 gives C2 = 0 BC 2: w’(0) = 0 gives C1 = 0 (d) (e) (f) (g) The deflection w(x) then becomes PL x 2 w(x) = E Bt H 2 2 (h) and at the free end (at the force P) one obtains PL L 2 PL 3 δ2 = w(L) = = E Bt H 2 2 2E Bt H 2 (i) It is concluded that the rearrangement of the width of the flanges makes the deflection at the free end to decrease from 0. Answer: The rearrangement of the width of the flanges causes the deflection at the free end to decrease from 0.67 PL3/EBtH2 to 0.5 PL3/EBtH2. the deflection decreases by 25 per cent.67 PL3/EBtH2 to 0. The bending moment M(x) in the beam becomes M(x) = − PL (1 − x /L). i..5 PL3/EBtH2.e. 2 (MPa) ⎜ ⎝ 2 ⎠ and do not forget the third principal stress: σ3 = 0.5 on the yield limit gives σY 148. σy = 80 MPa. In the xy-plane one obtains σx + σy σ1. 2 = ± 2 ⎯⎯⎯⎯⎯ √ ⎛ σx − σy ⎞ 2 2 ⎟ + τxy = 100 ± 63.5 with respect to yielding is required. according to the Tresca criterion. Chapter 1 Page 1:3 . For this structure a safety factor s = 1.3 = which gives σY = 222 MPa 1. Answer: The yield strength of the material should be.3 MPa ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ e =√ Again. The stress components in a part of a structure have been calculated to σx = 120 MPa.5 on the yield limit gives σY 163.2 MPa.5 The von Mises equivalent stress is σvM σ2x + σ2y − σx σy + 3τ2xy = 148.2 MPa and the smallest principal stress is σ3 = 0. using safety factor s = 1. Using safety factor s = 1. (a) Thus. and τxy = 60 MPa (all other stress components are zero).5 (b) (c) Thus. σY = 245 MPa (and the von Mises criterion gives σY = 222 MPa). The effective stress according to Tresca then is σeT = σ1 − σ3 = 163. What yield strength σY should the material have to fulfil this condition? Investigate both the Tresca and the von Mises yield criteria.2 = (a) which gives σY = 245 MPa 1.Yielding 1/2. the largest principal stress is σ1 = 163. according to the more conservative criterion (the Tresca criterion) the yield strength of the material should be σY = 245 MPa. Solution: First. determine the principal stresses in the material. Assume that the material is linearly elastic.. the elastic solution giving Pelast will be determined.Yielding and plastic collapse 1/3. The stress in the beam is obtained as Mb σ= (a) Wb where Mb = Pelast L. (b) Pmax = σY bh 2 / 4L. P L h Material: stress b strain (b) Determine the maximum force Pmax the beam may be loaded with. ideally plastic with yield limit σY. Pelast is the maximum load not giving any yielding in the beam). (c) The ratio Pmax /Pelast becomes Pmax /Pelast = 3/2. A cantilever beam of length L and with a rectangular cross section (base b. and Wb is the section modulus: Wb = bh2/6. (a) Determine the load P = Pelast the beam may be loaded with without inducing any plastic deformation in the beam (i. Solution: (a) To start with.e. When σ = σY one obtains Mb σY b h 2 = Pelast = L 6L (b) Y x z (b) The maximum force that may be applied to the beam will produce a plastic hinge at the support of the beam. Answer: (a) Pelast = σY bh 2 / 6L. and (c) Pmax /Pelast = 3/2. Page 1:4 Chapter 1 . determine the ratio of the load Pmax to load Pelast. height h) is loaded by the force P at the free end. The bending moment Mf needed to produce the hinge is Y 0 Mf = ⌠ σ z dA = ⌠ − σY z b dz + ⌠ ⌡ ⌡− h /2 ⌡0 h /2 σY b h 2 σY z b dz = 4 (c) Thus Pmax = Mf /L = σYbh2/4L. (c) Finally. only numerical values of the moments will be discussed below. the moment MA at support A as RC MA = 5PL /27. (For collapse. half the cross-sectional area A (A = b × h) is yielding in tension and half of the area A is yielding in compression. (a) Determine the force P = P0 giving one plastic hinge on the beam (to give a plastic hinge. When a plastic hinge appears. MA > MB.5 PL /27 M 8 PL /81 First determine the distribution of the bending moment M along the beam when the beam is C fully elastic. The beam has length L and a rectangular cross section with base b and height h. Next. This gives (see Problem 1/3) Mf = ⌠ σx z dA = 2 ⌠ ⌡A ⌡0 h /2 bh 2 1 ⎛ h⎞2 σY z bdz = 2 σY b ⎜ ⎟ = σY 2 ⎝ 2⎠ 4 Chapter 1 (a) Page 1:5 . a cross section must be fully plastic). see figure. determine the bending moment M = Mf needed to create a plastic hinge on the beam. ideally plastic with yield limit σY.Plastic collapse 1/4. Handbook formulae give. Note also that. The reaction force at support C becomes RC = P /3 − 5P /27 = 4P /27.) The material is linear elastic. but in most cases only the numerical value of the moment (and not the direction) is of interest. numerically. and it is loaded at x = L /3 by a force P. Therefore. implying that plastic yielding will start at point A. Here they will appear at A and B. The x bending moment MB at B becomes MB = RC 2L /3 = 8PL /81. (b) Determine the force P = Pf giving collapse of the structure. for an elastic beam. two plastic hinges are needed. Note that MA and MB have different directions. According to the conventions used here the moment MA is negative. Solution: A L /3 P 2 L /3 B MA . P A 2 L /3 L /3 x z B C The beam AC is rigidly supported (clamped) at end A and simply supported at end C. and the moment MB at B becomes MB = 2RC L /3 = 2σY bh2/15.(a) Question (a) can now be answered. because the part AC of the beam is still elastic.e.c) Thus. i. confirm that MA = Mf > MB when P = P0.. This moment is less that Mf. the load distribution in the structure may change.e.35 σY bh2/L one plastic hinge has developed (at A) in the beam. When MA = Mf the beam AC acts as a simply supported beam. and loaded with a moment Mf at A and a force P (> P0) at B. Collapse of the beam will appear when the bending moment at B reaches Mf. the support reaction RC becomes RC = P0 /3 − Mf /L. the structure will collapse. beyond P0. thus giving MB = Mf. confirm that MA is larger than MB also when the hinge appears at A. When MA = Mf and P = P0. after yielding has started at point A.. P = P0 2 L /3 MB RC P > P0 A L /3 2 L /3 MA= Mf plastic hinge MA = Mf A P = Pf plastic hinges Collapse mode P = Pf C RC C During the loading of the beam. The first plastic hinge will appear in the beam when the largest moment (here MA) reaches the value Mf. i. (b) The force P can be further increased. i. Therefore. The force P = Pf that gives a hinge also at B can be determined in the following way: the bending moment Page 1:6 Chapter 1 . When the load P > P0 has increased so much that it produces a hinge also at B. MA = Mf gives (numerically) bh 2 5PL Mf = σY = 4 27 27 bh 2 which gives P = P0 = σ 20 L Y (b.) Thus. which implies that the beam is still elastic at B when the hinge appears at A. because the moment at MB is not large enough to create a plastic hinge there. at least part of the cross section at B is elastic. (It is assumed that MA > MB is valid also during yielding at A. when a plastic hinge is obtained also at B.e.. This implies that the ratio MA /MB of the bending moments may change during yielding at A. at load P = P0 = 1. supported at A and C. Answer: P0 = 1. The force P = Pf is still unknown. the force P may be increased to P = Pf = 1. at which load two plastic hinges are obtained. see figure.at B is MB = Mf = RCf (2L /3). which gives RC = RCf = MB /(2L /3) = 3Mf /2L. Mf Chapter 1 Page 1:7 . and Pf = 1.875 σY bh2/L. The loading of the full beam AC then is MA = Mf at A. see figure below. One obtains L L 3Mf MA = Mf = Pf − RCf L = Pf − (d) 3 3 2 from which is solved B 3 15 bh 2 5 Pf = Mf = σY (e) R C f = 3 Mf / 2 L 2 L 8 L Thus. P = Pf at B.35 σY bh2/L. and the structure will collapse. giving one plastic hinge at A. and RC = RCf = 3Mf /2L at C. Mf Pf 2 L /3 R C f = 3 Mf / 2 L Pf A L /3 2 L /3 C Moment equilibrium with respect to point A gives an equation containing the unknown force P = Pf.35 σY bh2/L gives one plastic hinge at A.875 σY bh2/L gives plastic collapse (two hinges). after the first hinge has appeared at load P = P0 = 1. Also.Plastic instability 1/5. It gives dP = σ ⋅ dA + A ⋅ dσ (b) (c) (d) dV = A ⋅ dL + L ⋅ dA = 0 (e) 1 dL dL ⋅ = L/L0 L0 L (f) dε = Plastic instability occurs when the load cannot be increased any more. see Section 1. L P Determine the maximum tensional force Pmax the bar in the figure (see also Section 1. L0 P A.4.4 in the textbook. Solution: As in the example on plastic instability.4) may be subjected to.e. and L depends on P. one has equilibrium: P = σ⋅A (a) V = A0 ⋅ L 0 = A ⋅ L the volume is constant: large strain: ε = ln L L0 Differentiate (a). Assume that the material is hardening due to plastic deformation and use the stress-strain relationship σ = σ0 ⋅ εm Material: stress strain Use true stress and natural (logarithmic) strain. i.e.e. and A depends on P. L > L0 when P > 0. A < A0 when P > 0. Page 1:8 Chapter 1 . The logarithmic strain is defined as ε = ln L L0 where L is the length of the bar when loaded. remember that plastic deformation will take place with no change of volume. A0 .4. (b) and (c). i. i. The true stress is defined as P σ= A where A is the (contracted) cross-sectional area obtained after loading. when dP = 0. It gives σ ⋅ dA + A ⋅ dσ = 0 (g) A ⋅ dL + L ⋅ dA = 0 (h) dL = L dε (i) Eliminate A and L from (g). Then L0 Pmax = σ ⋅ A = σ0 m m ⋅ A = σ0 m m ⋅ A0 (l) L The ratio L0 /L is obtained from equation (c). Chapter 1 Page 1:9 . has been determined. By use of ε = m. One has dP = σ0 A0 {εm ⋅ ( − 1) e − ε + mεm − 1 e − ε } = 0 dε (n) (o) (p) from which ε = m is obtained. Entering ε = m into (o) gives P = Pmax = σ0 m m ⋅ A0 e − m (q) (Verify that a maximum value. (h) and (i). (m) = e −m ε = m = ln L0 L0 L Finally. It gives A L dσ dσ σ=− dσ = dσ = . It gives σ0 ⋅ εm = σ0 ⋅ m ⋅ εm − 1 giving ε = m (k) Thus. thus σ = dA dL dε dε (j) Introduce the material relationship σ = σ0 ⋅ εm into (j). or. the loading force P reaches its maximum value P = Pmax when ε = m.) Answer: Plastic instability will occur when Pmax = σ0 m m e − m A 0. by use of (a) and (c). L0 P = σ ⋅ A = σ0 εm ⋅ A0 = σ0 εm ⋅ A0 e − ε L Determine the maximum value of P. entering (m) into (l) gives Pmax = σ0 m m ⋅ A0 e − m Alternative solution: The loading force P may be written. and not a minimum value. one obtains L0 L L giving em = . Stationary creep 1/6. (a) Calculate the stresses in the two cylinders immediately after the pressure p has been applied to the inner cylinder (no longitudinal stresses will appear. The circumferential stress σi in the inner cylinder becomes σi = (p − q) r /h and in the outer cylinder the circumferential stress σo becomes σo = qr /h (index i for the inner cylinder and index o for the outer). sgn σ means “sign of σ” and is +1 if σ is positive and − 1 if σ is negative. at time t = 0. before the loading has been applied the inner cylinder fits exactly in the outer one. (b) Investigate how the stresses will change with time in the two cylinders. Thus. h h r p p o i A tube subjected to an internal pressure p is made of two circular cylinders of different materials. and t* is a reference time. a contact pressure q will arise between the inner and the outer cylinder. only circumferential). Chapter 1 . Solution: q o o q p i Page 1:10 i (a) When. wall thickness h. The material equation of the inner cylinder reads ˙ ⎛ | σ | ⎞ n sgn σ ˙ε = σ + ⎜ ⎟ Ei ⎝ σn ⎠ t* where Ei is the elastic modulus of the material of the inner cylinder (Ei = 69 GPa). The contact pressure q is unknown. see figure. the pressure p is applied to the inner cylinder. The outer cylinder has inner radius r. and is made of a Norton material. The inner cylinder has outer radius r. σn and n are material parameters (n > 1).e. linear elastic material with constitutive equation ε = σ/Eo). and is made of a Hooke material with modulus of elasticity Eo = 207 GPa (i. wall thickness h (h << r). ui = uo. Chapter 1 Page 1:11 . Thus.18b) in Section 1. k = 1 / τσnn). it is “easier” for the inner cylinder to expand. and in the outer cylinder the circumferential stress is σo = qr /h = 3pr /4h. It gives σ˙ i σ˙ o + kσni = u˙ i = u˙ o giving Ei Eo (d. the outer cylinder carries most of the load (q > p − q) and σo > σi). at time t = 0. (b) How will the circumferential stresses σi and σo develop with time? The deformation condition ui = uo is valid also at times t > 0. thus q = q(t).2. namely q(0) = 3p /4. Due to creep. at time t = 0. the circumferential stress in the inner cylinder is σi = (p − 3p /4) r /h = pr /4h. was determined above.6 in the textbook) where u is the radial displacement (radial expansion) of the thin-walled cylinder. For circular symmetry the circumferential strain ε equals ε = u /r (see equation (1. This follows from the fact that the outer cylinder is stiffer than the inner one (Eo > Ei). Because of the lower modulus of elasticity of the inner cylinder. One obtains ui = r εi = r εo = uo (a) This gives. σi σo = Ei Eo (b) Entering σi = (p − q) r /h and σo = qr /h in (b) gives Ei p −q =q Eo (c) Using the numerical values of Ei and Eo one obtains q = 3p /4 (at t = 0). this stress distribution will be changed as time goes on. implying that the outer cylinder (which is less prone to expansion) has to carry most of the load.The contact pressure q is obtained from the deformation condition that the two cylinders have the same radial expansion u. at time t = 0. It is seen that when the pressure is applied (at t = 0). The dot ( ˙ ) indicates differentiation with respect to time t. The initial condition on q. and σi > 0 gives sgn σi = 1. Thus. The contact pressure q between the inner and outer cylinder will now depend on time t. immediately after the pressure p has been applied.e) where k is a constant replacing 1 / τσnn in the material equation (thus. for n > 1. The homogeneous part of equation (h) reads q˙ + C1 q n = 0 (i) This equation may be separated into one part containing q only and one part containing the time t. One obtains 3p q(0) = {( − n + 1) C0 } 1 / ( − n + 1) + p = (m) 4 giving Page 1:12 1 ⎛ 3p 1 ⎛ − p ⎞ ( − n + 1) ⎞ ( − n + 1) = ⎜ − p⎟ ⎜ ⎟ C0 = −n +1⎝ 4 ⎠ −n +1⎝ 4 ⎠ Chapter 1 (n) . a relation that will be used below.Entering σi = (p − q) r /h and σo = qr /h into (e) gives ⎧ r ⎫n r r + k ⎨ (p − q) ⎬ = q˙ ( p˙ − q) ˙ h Ei h⎭ h Eo ⎩ (f) Here p˙ = 0. The solution to (i) becomes. The particular solution to (h) becomes ⎧ C2 ⎫ 1 / n q = qpart = ⎨ ⎬ = p ⎩ C1 ⎭ (k) q = qhom + qpart = { ( − n + 1) ( − C1 t + C0 )} 1 / ( − n + 1) + p (l) and the complete solution is The boundary (initial) condition q(0) = 3p /4 gives the constant C0. q = qhom = {( − n + 1) ( − C1 t + C0 )} 1 / ( − n + 1) (j) where C0 is an integration constant to be determined from a boundary (initial) condition. because p is the constant pressure in the cylinder. Rearranging the expression (f) gives ⎧ r ⎫n ⎧ p r ⎫n r⎧1 1⎫ ⎨ ⎬ ⎨ ⎬ ⎬ + +k q =k ⎨ (q) ˙ h ⎩ Ei Eo ⎭ ⎩ h⎭ ⎩ h ⎭ or where q˙ + C1 q n = C2 C1 = k { r/h } n r { 1/Ei + 1/Eo }/h and C2 = (g) (h) k{ p r/h } n r { 1/Ei + 1/Eo }/h One notices that C2 /C1 = p n . to protect the outer cylinder from chemical reactions. and in the outer cylinder the stress is σo = 3pr /4h. and the stresses tend to σi = 0 and σo = pr /h. Answer: (a) At time t = 0. Chapter 1 Page 1:13 . After some time the inner cylinder does not carry any load at all (or it carries a very small part of the load). The factor in the curly brackets will then be proportional to t − 1 / (n − 1). and the corresponding term in (q). q ⎧ =⎨(−n ⎩ ⎧ = ⎨ (n ⎩ 1 / ( − n + 1) ⎛ 1 ⎛ − p ⎞ ( − n + 1)⎞ ⎫ ⎟⎬ + 1) ⎜ − C1 t + +p ⎜ ⎟ ⎠⎭ ⎝ −n +1⎝ 4 ⎠ ⎛ − p ⎞ ( − n + 1) ⎫ ⎬ − 1) C1 t + ⎜ ⎟ ⎝ 4 ⎠ ⎭ 1 / ( − n + 1) +p (o) Finally. one obtains r r σi(t) = (p − q) = − h h ⎧ ⎨ (n ⎩ ⎛ − p ⎞ ( − n + 1) ⎫ ⎬ − 1) C1 t + ⎜ ⎟ ⎝ 4 ⎠ ⎭ 1 / ( − n + 1) (p) and 1 / ( − n + 1) qr r ⎧ r ⎛ − p ⎞ ( − n + 1) ⎫ ⎬ σo(t) = = ⎨ (n − 1) C1 t + ⎜ + p ⎟ h h⎩ h ⎝ 4 ⎠ ⎭ (q) One notices that for t = 0 the initial stresses σi = pr /4h and σo = 3pr /4h are obtained (as one should). For large values of time t. will tend to zero. The function of the inner cylinder could then be. or something else. the stress in the inner cylinder is σi = pr /4h. which tends to zero for large values of time t. the term containing the time t will dominate over the second term in the factor enclosed by the curly brackets in (p) and (q). the creep in the inner cylinder force the outer cylinder to carry a larger part of the load. Thus. This implies that after a long time (for large values of time t) the expression giving σi in (p). respectively.Thus. 1 / ( − n + 1) (b) r r⎧ ⎛ − p ⎞ ( − n + 1) ⎫ ⎬ σi(t) = (p − q) = − ⎨ (n − 1) C1 t + ⎜ ⎟ h h⎩ ⎝ 4 ⎠ ⎭ and qr r ⎧ ⎛ − p ⎞ ( − n + 1) ⎫ ⎬ σo(t) = = ⎨ (n − 1) C1 t + ⎜ ⎟ h h⎩ ⎝ 4 ⎠ ⎭ 1 / ( − n + 1) + r p h (where σi tends to zero and σo tends to pr /h when time t becomes large). for example. The outer cylinder will then carry the main part of the load. and τxy = 60 MPa (all other stress components are zero). Page 1:14 Chapter 1 . One obtains.Fracture. The stress components in a part of a structure have been calculated to σxx = 120 MPa. Failure is expected. whereas in tension the strength of the material is σUt = 150 MPa. since in tension the ultimate strength of the material is σUt = 150 MPa only. only. 2 = ± 2 ⎯⎯⎯⎯⎯ √ ⎛ σxx − σyy ⎞ 2 2 ⎟ + τxy = 100 ± 63. in the xy-plane. Answer: Yes.2 MPa and the smallest principal stress is σ3 = 0 (= σzz). The material is brittle and it has the ultimate strength σUt = 150 MPa in tension and the ultimate strength σUc = 200 MPa in compression.2 (MPa) ⎜ ⎝ 2 ⎠ (a) Thus. Using the maximum normal stress criterion. σxx + σyy σ1. the largest principal stress is σ1 = 163. investigate material failure. maximum normal stress 1/7. determine the principal stresses in the material (and don’t forget the third principal stress). Solution: First. σyy = 80 MPa. failure is expected because the maximum principal stress in the material is σ1 = 163 MPa. 2σY = 3σY √ 1.1142σY). When that stress is reached the fracture will occur. A stress test specimen of this material is subjected to a torque.1142 σY Thus. which is increased until fracture occurs. Will there be any plastic deformation of the specimen prior to the fracture? Use the Tresca yield criterion. shear stress The Mohr criterion for brittle fracture is. Chapter 1 Page 1:15 .e.Fracture. Thus. determine the radius of the circle. Answer: Plastic deformation will occur because τ = τfra = 1. Solution: 1. at fracture one has τfra = Rfra = 1. This means that the material must be deformation hardening so that shear stress τ = 1.22 +⎯32 σY ⎯⎯⎯⎯⎯ which gives Rfra = 1.2 Y C B A Rfra 3 O 1 3 Y Pure shear gives Mohr’s circle as shown in the figure (centre of the circle at the origine O).114σY at fracture whereas τ = τyield = 0. At which shear stress τ = τyield will yielding occur? Yielding will occur when σeTresca = σY = σ1 − σ3 = τyield − (− τyield) = 2τyield. Triangles OAB and OAC give Rfra 1. for the material studied here. at yielding one has (according to Tresca) τyield = σY /2.1142σY can be reached. Determine the shear stress giving fracture.5σY and fracture when τ = 1. It is concluded that yielding will occur before fracture (yielding when τ = 0.5σY at yielding. formulated as (see figure) ± τ + 0.4 σ = 1.1142 σY. i.2σY normal where σY is the yield limit of the material. Mohr’s failure criterion 1/8. Thus. use factor m = 6 in the Weibull relationship. and in compression. σzz = τxz = τyz = 0) are given by σxx + σyy σ1.) Assume that the ultimate strength of a material of volume V0 is found to be σ0. σyy = 0. τxy = τ = T/2πr2h. σxx = 0. and wall thickness h = 5 mm. The shear stress in the circular thin-walled cylinder is T T τ= = Wv 2π r 2h (a) Denote this stress τxy. Assume that the Mohr criterion for brittle fracture is given by straight lines in the shear stress versus normal stress diagram (as in Problem 1/8). According to Weibull. Mohr’s failure criterion 1/9. The size effect according to Weibull should be taken into account. here.Fracture. would be ⎛ V0 ⎞ 1 / m σ = σ0 ⎜ ⎟ ⎝V⎠ For the material above. This material is used in a circular thin-walled cylinder of length L = 1 m. but of volume V. σUt = 100 MPa. Solution: The ultimate strength in tension. σUc = 400 MPa. For a brittle material the ultimate strength was determined to be 100 MPa in tension and 400 MPa in compression. (One reason for this could be that there is a larger probability of finding a flaw in a larger volume than in a smaller volume. The volume of the test specimens used to determine the ultimate strengths was 104 mm3. According to Weibull the ultimate strength of a large volume of material is less than the ultimate strength of a small volume of the same material. With respect to brittle fracture. The cylinder is loaded with a torque T. Page 1:16 Chapter 1 (b) . The principal stresses (at plane stress. determine the maximum allowable torque Tmax the cylinder may be subjected to. give the fracture limit curve according to the figure below. and τxy = τ. The brittle fracture may start at the flaw. diameter d = 200 mm. 2 = ± 2 ⎯⎯⎯⎯⎯ √ ⎛ σxx − σyy ⎞ 2 2 ⎟ + τxy ⎜ ⎝ 2 ⎠ where. Within these limits the material is linearly elastic. the ultimate strength of the same material. T O’’ Uc The principal stresses thus are (re-numbered) h σ1 = τ. σ2 = 0 and σ3 = − τ (c) T Mohr’s circle. is entered into the diagram. however. (With no reduction due to volume effects.6 kNm (and Tmax would have been 25 kNm. the allowable torque T would have been T = τWv = 25 kNm.7 MPa giving τreduced = τ ⎜ (g) ⎝ 2π100 ⋅ 5 ⋅ 1000 ⎠ The allowable torque T becomes Tallow = τreducedWv = 9. Ut From triangles O’BC and O’’BE one obtains OB − 50 50 = (d) 200 OB + 200 giving OB = 133 MPa. O O’ Determine first. Triangles O’BC and OBD give (with OD = σ1) 50 OB − 50 = σ1 OB (e) from which σ1 = 80 MPa is solved. The radius of the circle is obtained from triangles OBD and D C B O’BC. the distance OB. Fracture is expected when the circle E R= 1 reaches the fracture limit curve.6 kNm. approximately.) Reduce the allowable stress with respect to the volume effect. Answer: Tmax = 9. Thus τ = σ1 = 80 MPa before any reduction of the stress with respect to volume effects has been done. Weibull gives 1/m σ ⎛ V0 ⎞ =⎜ ⎟ σ0 ⎝ V ⎠ Thus. Chapter 1 Page 1:17 . if the size effect had been neglected). (f) τreduced ⎛ V0⎞ 1 / 6 =⎜ ⎟ τ ⎝V⎠ ⎞1/6 ⎛ 104 ⎟ = 0. The new circle will have the radius σ1 (= OD in the OD = figure).3835τ = 30. as given by the principal stresses r σ1 and σ3. 925s. Answer: Material B should be selected because it has the largest ultimate strength in compression: σUcompB = 1.809s.85s. with the Which one of the materials is the better if brittle fracture (in compression) of the bar is considered? Solution: A rB rA A B B s 2s The fracture limit of material A intersects the τ axis at τA = s tanϕ = s / 2. materials are available. Similar triangles give rA s /2 (a) = 2 2 rA + s s √ 1 + (1 / 2) ⎯⎯⎯⎯⎯⎯⎯ giving rA = 0. Similar triangles give rB 2s /3 (b) = rB + 2s s √ 22 + (2 / 3)2 ⎯⎯⎯⎯⎯⎯⎯ giving rB = 0.85s. see following data e Material σe tan ϕ A s 1/2 B 2s 1/3 material will be Two different materials have figure.62s). This gives σUcompB = 2rB = 1. The fracture limit of material B intersects the τ axis at τA = 2s tanϕ = 2s / 3.85s (whereas σUcompA = 1. Both linear fracture limit curves. Let the radius of circle A be rA. A circular bar made of a brittle loaded in axial compression. Page 1:18 Chapter 1 . This gives σUcompA = 2rA = 1.Fracture.62s. Mohr’s failure criterion 1/10. Let the radius of circle B be rB. This material should be selected. Material B has the largest fracture strength in (uni-axial) compression: σUcompB = 2rB = 1. Answer: For the ultimate strengths in compression one has (numerically) σUcompC < σUcompD. When loaded in tension.Fracture. Chapter 1 Page 1:19 . and the smaller will the circle to the left of the τ-axis be. Thus. have linear fracture limit curves. The two materials have the same ultimate strength τU when loaded in shear. What can be said about the two materials’ ultimate strengths in compression? Solution: U Ut s From the figure it can be seen that the larger the ultimate strength in tension σUt is. Mohr’s failure criterion 1/11. It can also be shown that (see below) 1 1 1 = + τU σUt σUcomp (a) which gives the above result. as σUtC > σUtD one will obtain σUcompC < σUcompD. C and D. thus σUtC > σUtD. Similar triangles give τU σUt / 2 1 1 1 = = giving − s s − σUt / 2 τU σUt / 2 s and σUcomp / 2 τU = s s + σUcomp / 2 giving 1 1 1 = + τU σUcomp / 2 s (b) (c) Summing (b) and (c) gives (a). the more horizontal will the fracture limit curve be. however. material C has a larger ultimate strength than material D. Two materials. 140 and 100 MPa The number of loading cycles at each stress level is n = 15.00236 = (b) 25 119 63 096 158 489 1 000 000 424 Finally. 158 489. determine how many loading sequences (how many days) the component might be used before fatigue failure is expected. by use of the Palmgren-Miner damage accumulation rule. The loading is alternating. Solution: For the different stress levels. the expected fatigue life N is obtained from the Wöhler curve. and one sequence contains the following loading amplitudes (i. 15 20 150 500 1 D= + + + = 0. the number of sequences Ns to expected fatigue failure is Ns = 1 / D ≅ 424 (c) Thus.36 ⋅ 10 − 3 25 119 63 096 158 489 10 Expected number of sequences (days) Ns to fatigue failure is Ns = 1 / D ≅ 424. and 1 000 000 The accumulated damage due to one loading sequence is. 160. approximately. the component can be used in 424 days. respectively. respectively The Wöhler curve of the material is. and then. 20.e. A component of a machine is subjected to a repeated loading sequence. 63 096. stress amplitudes the mean value of the stress is zero) σa = 180. Page 1:20 Chapter 1 . Answer: Damage D due to one sequence is 15 20 150 500 D= + + + 6 = 2. in this stress range. The loading sequence is repeated once a day. and 500. given by the relation σa = − 50logN + 400 MPa Determine the damage accumulation D due to one loading sequence. One obtains (400 − σa) / 50 N = 10 (a) giving.Fatigue 1/12. 150. N = 25 119. to the second critical load of the unconstrained beam. and in figure (c) the second eigenmode w2(x) of the unconstrained beam (a) is sketched.Elastic instability 1/13. Solution: The problem will be solved by use of the differential equation for an axially loaded beam.e. to obtain the largest possible critical load of the constrained beam shown in figure (d). and thereby the critical load P = Pcrit of the constrained beam (with the extra support) equals the second critical load of the unconstrained beam. The solution to this equation is w(x) = C1 + C2 px + C3 sin px + C4 cos px (b) P/EI where p = √ ⎯⎯⎯ ⎯ .9 (the example on instability). EI (a) First eigenmode: x Pcrit1 (b) Second eigenmode: Pcrit2 x (c) Constrained beam : Pcrit2 x (d) It is possible to increase the buckling load of a beam by introducing extra supports along the beam. Thus.4.e. It can be shown that by use of one extra support the buckling load of a beam at most can be increase to the second eigenvalue of the original beam without the extra support. see Section 1. EI is the bending stiffness. at the point of zero deflection of the second eigenmode). Determine where to place one extra support on the beam that is pinned (simply supported) at one end and fixed (rigidly supported) at the other end. determine the coordinate x where the mode w2(x) has zero deflection. The differential equation reads EI w IV(x) + P w’’(x) = 0 (a) where w(x) is the deflection of the beam. Chapter 1 Page 1:21 . as shown in figure (a). The extra support should then be placed at the node of the second eigenmode of the unconstrained beam (i. i. P x L. then the beam is constrained to buckle according to the second eigenmode. and P is the force loading the beam in compression. In figure (b) the first eigenmode w1(x) of the pinned-fixed beam is sketched. If the extra support is placed at that node. Boundary conditions give the constants C 1 to C 4.e.f) equal to zero.4934 p2 L = 7. one must have the determinant of the system of equations (e. the forms of the deflection of the buckling beam. (h) The first root p1 L gives p1 L = √ P /⎯EI ⋅ L = 4. i.7252 ≈ 5π /2 p3 L = 7π /2 and so on. One obtains p1 L = 4.7252 √⎯⎯⎯ 2 which gives 7.49342EI 2.72522 EI P = Pcrit2 = L2 (j) The eigenmodes. At the beam end x = L one has BC 3: w(L) = 0 giving C2 pL + C3 sin pL = 0 (e) BC 4: w’(L) = 0 giving C2 p + C3 p cos pL = 0 (f) To obtain a solution with at least one of C 2 and C 3 not equal to zero. From the figure it is concluded that the first root is close to 3π / 2. which is used in the following boundary conditions. become (equation (f) gives C 2 = − C 3 cos pL) for p1 L: Page 1:22 w1(x) = C3 { − p1x cos p1L + sin p1x } Chapter 1 (k) .05 π2EI P = Pcrit1 = = L2 L2 (i) The second root p2 L gives p L = ⎯ P /⎯EI ⋅L = 7. It gives ⎢ pL sin pL ⎥ ⎢ ⎥ = 0 or tan pL = pL (g) ⎢ 1 cos pL⎥ tan pL pL pL 2 3 2 5 2 The roots to equation (g) are determined numerically. the second root very close to 5π / 2.4934 ⎯⎯⎯⎯ which gives 4. One has (BC for boundary condition) BC 1: w(0) = 0 gives C1 + C4 = 0 (c) BC 2: M(0) = 0 gives − EIw’’(0) = 0 which gives − C3 p 2 sin p ⋅ 0 − C4 p 2 cos p ⋅ 0 = 0 (d) These two equations give C 4 = 0 and C 1 = 0. and so on. by use of p2 L (p2 L in radians).36 L. One obtains ⎧ x ⎛ x⎞⎫ (n) w2(x) = C3 ⎨ − 7. the critical load of the pinned-clamped beam increases from Pcrit1 = 2.7252 + sin ⎜7.72522 EI / L 2 = 59. ⎧ x ⎛ x⎞⎫ w2(x) = C3 ⎨ − 7.7252 + sin ⎜7. Answer: The support should be placed at x = 0.36 L.05 π2EI / L 2 = 20.68EI / L 2 for one optimally placed support.for p2 L: w2(x) = C3 { − p2x cos p2L + sin p2x } (l) where C 3 is undetemined (and different in the different modes). Thus.23EI / L 2 (no extra support) to Pcrit2 = 7.7252 ⎟ ⎬ L ⎝ L⎠ ⎭ ⎩ (m) Now the value of x giving a zero crossing of w2(x) can be determined.7252 ⎟ ⎬ = 0 L ⎝ L⎠ ⎭ ⎩ which gives x = 0. The second eigenmode becomes.7252 cos 7.36 L. Chapter 1 Page 1:23 . if an extra support is placed at x = 0.7252 cos 7. Ma 2 Solution: Separate the flywheels from the axle and enter the torque Ma on the axle 2 J M( t ) and on the flywheels. 4J 4J L. One obtains M M M sinΩt ¨ a L + a + a = M(t) = 0 (d) M GK 4J 2J 2J 2J Page 1:24 Chapter 1 . GK Ma 1 2J L. What happens if Ω = ωe? The moments of inertia of the two flywheels are 4J and 2J respectively.) Introduce the rotations ϕ1 and ϕ2 of the two flywheels. GK M ( x) A shaft carries two flywheels according to the figure. and the inertia of the shaft can be neglected. That gives a differential equation in the unknown moment Ma. One of the flywheels is loaded with a torque M(t) = M0 sin Ωt. Therefore the torque is the same at the two ends of the axle. The moment-deformation relationship for the axle is Ma L φ2 − φ1 = GK (c) Eliminate ϕ1 and ϕ2 from the equations (a) to (c). where ωe is the eigenfrequency of the structure. is supposed to negligible here. For the flywheel 1 the equation of motion becomes 4J φ¨ 1 = Ma (a) For the flywheel 2 the equation of motion becomes 2J φ¨ 2 = M(t) − Ma (b) The torsion of the axle is ϕ2 ϕ1. (The inertia of the axle. compared to the flywheels. Determine the maximum torque in the shaft if Ω = 0.75ωe and Ω = 0. The directions of ϕ1 and ϕ2 are shown in the figure.90ωe.Resonance 1/14. and Ω = ωe.51 M0 sinΩt 57 0 (k) Ma = Ma = Finally. One obtains. It gives an equation determining the unknown amplitude A.52 M0 sinΩt 21 0 (j) 200 M sinΩt = 3. for Ω = 0. It is concluded that an excitation frequency close to a resonance frequency of the structure might be very dangerous. Answer: Normalised with respect to the load amplitude M0.Assume a particular solution to the differential equation (d) on the form Ma = A sin Ωt (e) Enter the assumption (e) into the differential equation (d). One obtains ωe = ⎯⎯ √ 3GK 4JL (h) The torque in the axle can now be calculated. if Ω = ωe. Ω = 0. Chapter 1 Page 1:25 .75ωe. 3. respectively. The eigenfrequency is the frequency Ω = ωe giving that the amplitude A in (g) tends to infinity (which is equivalent to that the “system determinant” in (f) becomes zero). the torque Ma in the axle tends to infinity.51 and ∞ .52.90ωe one obtains 32 M sinΩt = 1. One obtains M0 L ⎛ 3⎞ ⎜ − Ω2 + ⎟A= (f) ⎝ GK 4J ⎠ 2J from which is solved A= M0 GK (g) 2JL ( − Ω2 + 3GK / 4JL ) The (angular) eigenfrequency ωe can now be determined. the amplitude Ma of the torque in the shaft is Ma /M0 = Mshaft /M0 = 1.75ωe one obtains For Ω = 0.90ωe. from (e) and (g) 2M0 M0 GK sinΩt = sinΩt (i) Ma = 2JL ( − Ω2 + 3GK / 4JL ) 3 (1 − Ω2/ω2e) For Ω = 0. Page 1:26 Chapter 1 .Empty page. Solutions to problems in T Dahlberg and A Ekberg: Failure. radius a. Two points have stress σA and two points have stress σB.b) h 2h Study the stress concentration at the hole. Lund 2002. see figure. Chapter 2 Stresses. stress concentration. There are four critical points at the hole where high stress may occur: points A and B. The vessel is then loaded with a pressure p. stress intensity factor. ISBN 91-44-02096-1. FMCHAP2P. Determine the maximum stress at the hole due to the pressure if the influence of the tube connection is disregarded. and fracture criteria Problems with solutions Stress concentration 2/1. stresses at crack tip. and wall thickness h. Answer: The largest stress at the hole is σmax = 5pa / 2h. One obtains the stresses σA = 3 σlong − σtan = 3 p a a a −p =p 2h h 2h (c) σB = 3 σtan − σlong = 3 p a a 5 a −p = p h 2h 2 h (d) The stress around the hole varies between these two values. Fracture. For a tube connection a small circular hole is opened in the wall (and the tube is mounted). Solution: tan long A A B B tan and long The stresses in the pressure vessel wall are (tangentially and longitudinally) a a σtan = p and σlong = p (a.An Introduction.DOC/2010-01-15/TD Chapter 2 Page 2:1 . A thin-walled circular cylindrical pressure vessel has length L. The largest stress at the hole thus is σmax = σB = 5pa / 2h. Fatigue . Studentlitteratur. Page 2:2 Chapter 2 . By taking stress concentration into account. Answer: The screws should be mounted as far away from the stress concentration as possible. see Figure (a). A thin-walled circular cylindrical pressure vessel has a small circular hole in the wall. (b) major axis of the ellipse. one obtains ⎛ 2⎞ (b) σmax = ⎜1 + 2 ⎟ σnom = 5σnom ⎝ 1⎠ (b) If the load is parallel to the major axis. The hole will be covered with a square plate. Determine the stress concentration factor Kt when the plate is loaded in parallel to the (a) minor axis. The ratio of the axes (major to minor axis) of the elliptical hole is 2 to 1. the plate being screwed to the wall with four screws. here it will be at 45o counted from the longitudinal axis of the pressure vessel.2/2. one obtains ⎛ 1⎞ (c) σmax = ⎜1 + 2 ⎟ σnom = 2σnom ⎝ 2⎠ Answer: Stress concentration factor is Kt = 5 and 2. one in each corner. 2/3.) Solution: See answer. respectively. where do you want to place the screws in the pressure vessel wall? (This is perhaps a good example to try to understand stress concentration. see Figure (b). but it is not a good way to fasten the plate you should never make a hole in a region where you already have stress concentration. Solution: max nom nom (a) max nom nom (b) The maximum stress at an elliptical hole is given by ⎛ a⎞ σmax = ⎜1 + 2 ⎟ σnom (a) ⎝ b⎠ where a and b are the half-axes of the ellipse. (a) If the load is parallel to the minor axis. A large plate has a small elliptical hole in it. then the largest stress will appear at point B. It is seen that if half-axis b is large (b >> a). if a >> b. The plate is supported at two opposite sides. one obtains σyy = ν σxx (b) Stresses σxx (= σ∞) and σyy = νσxx give rise to stresses at the elliptical hole. y A b a x B A plate contains an elliptical hole as shown in the figure. then the largest stress at the hole will be at point A. stress σyy will appear in that direction. Hooke’s law gives 1 εyy = 0 = [σyy − ν (σxx + σzz ) ] (a) E Knowing that σzz = 0. It gives ⎛ b⎞ ⎛ a⎞ ⎜1 + 2 ⎟ σxx − ν σxx = ⎜1 + 2 ⎟ ν σxx − σxx ⎝ ⎝ a⎠ b⎠ Chapter 2 (d) (e) (f) Page 2:3 .2/4. respectively? The material is linearly elastic (with Young’s modulus E and Poisson’s ratio ν) up to its brittle fracture at the ultimate strength σU. Let σxx = σ∞. Contrary. The plate is loaded with an uni-axial stress σ∞ along the two other sides. the support being such that all motion in the y-direction is prevented whereas motion in the x-direction is possible. Under which condition (ratio b / a) will fracture start at point A and at point B. Solution: yy y A b a x B yy Due to the prevented contraction in the y direction (strain εyy = 0). Using the elementary case for uni-axial stress: ⎛ a⎞ σmax = ⎜1 + 2 ⎟ σnom and σmin = − σnom (c) ⎝ b⎠ one obtains (σyy = νσxx is used) ⎛ b⎞ σA = ⎜1 + 2 ⎟ σxx − ν σxx ⎝ a⎠ and ⎛ a⎞ σB = ⎜1 + 2 ⎟ ν σxx − σxx ⎝ b⎠ Set σA = σB. Determine the ratio a / b that gives the same stress at point A as at point B. and show in a graph. then the stress at point A is the largest. one has σA = σB. at plane deformation. i. when β = ν. whereas b < νa gives that the stress at point B is the largest. It gives β2 + (1 − ν) β − ν = 0 (h) Knowing that β > 0. If b > νa. when b = νa. The material is linearly elastic with Poisson’s ratio ν = 0. one obtains the solution β = ν. Stresses at crack tip 2/5. the distribution of the maximum shear stress close to a crack tip loaded in Mode I. at plane stress Page 2:4 σzz = ν (σxx + σyy ) σzz = 0 Chapter 2 . y Calculate. r x z The stress components close to the crack tip are KI ⎡ Θ 3Θ⎞⎤ Θ⎛ σxx = ⎢cos ⎜ 1 − sin sin ⎟⎥ 2⎝ 2 2 ⎠⎦ 2⎯ πr ⎣ ⎯⎯⎯ √ σyy = Θ 3Θ⎞⎤ ⎡ Θ⎛ ⎢cos ⎜ 1 + sin sin ⎟⎥ 2⎝ 2 2 ⎠⎦ 2⎯ πr ⎣ ⎯⎯⎯ √ KI τxy = Θ 3Θ⎤ ⎡ Θ ⎢cos sin cos ⎥ 2 2 2⎦ 2⎯ πr ⎣ ⎯⎯⎯ √ KI τyz = τzx = 0 and. Thus. Answer: Ratio b / a > ν gives fracture (largest stress) at point A and b / a < ν gives fracture at point B.Simplifying (f) gives 1−ν+ b a −ν =0 a b (g) Let b = βa.3.e. and σ3 thus become. 2 σxx + σyy = ± 2 ⎯⎯⎯⎯⎯ √ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ √ ⎛ σxx − σyy ⎞ 2 2 ⎟ + τxy ⎜ ⎝ 2 ⎠ ⎧ Θ ⎨ cos ± = 2 2⎯ πr ⎩ ⎯⎯⎯ √ KI = = KI 2⎯ πr ⎯√⎯⎯ KI 2⎯ πr ⎯⎯⎯ √ ⎧ Θ ⎨ cos ± ⎩ 2 cos ⎛ Θ Θ Θ 3 Θ ⎫⎬ Θ 3 Θ⎞ 2 ⎜cos sin sin ⎟ + cos2 sin2 cos2 ⎝ 2 2 2 ⎠ 2 2 2 ⎭ ⎫ Θ 2Θ ⎛ 23 Θ 2 3 Θ⎞ ⎬ cos sin ⎜sin + cos ⎟ 2 2⎝ 2 2 ⎠ ⎭ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ √ 2 Θ⎞ Θ⎛ ⎜ 1 ± sin ⎟ 2⎝ 2⎠ (d) The principal stresses σ1. They are σ1 . the z direction is one principal direction and the stress component σzz is one of the principal stresses. called σ1 and σ2.Solution: Determine the maximum shear stress by use of the Tresca criterion. In this problem we have a plane state (either plane stress or plane deformation) in the xy plane. with ν = 0. The maximum shear stress is 1 prs τmax = (σprs (a) max − σmin) 2 where superscript “prs” stands for “principal stress”. The principal stresses are then needed. Chapter 2 Page 2:5 . σ2. In case of plane stress one has the principal stress in the z direction σzz = 0 (b) and in the case of plane deformation (plane strain) one obtains the principal stress KI Θ σzz = ν (σxx + σyy ) = 2 ν (c) cos 2 2 π r ⎯⎯⎯⎯ √ Next. Therefore.3. determine the two other principal stresses. But which one of σ2 and σ3 is the smallest? To investigate this. One has KI (k) τmax( 60o ) = 0. Thus. By solving d τmax =0 (j) dΘ one finds that τmax has its largest value when Θ = π / 3 = 60 degrees = 60o (and the smallest value of τmax occurs when Θ = ± π and it is then τmax = 0). which gives if Θ < 47o then σ2 > σ3 and Page 2:6 if Θ > 47o then σ2 < σ3 Chapter 2 (l) .65 2⎯ πr ⎯⎯⎯ √ Plane deformation At plane deformation one still has that stress σ1 is the largest principal stress. The largest principal stress prs then is σprs max = σ1 and the smallest principal stress is σmin = σ3 = 0. set σ2 = σ3. at plane stress the maximum shear stress will be found in the direction Θ = 60 degrees from the x axis. Find the value of Θ that gives the largest τmax.σ1 = σ2 = KI 2⎯ πr ⎯⎯⎯ √ KI 2⎯ πr ⎯⎯⎯ √ σ3 = 0 σ3 = 0. The maximum shear stress then becomes σ1 1 KI Θ⎞ Θ⎛ (i) τmax = = cos ⎜ 1 + sin ⎟ 2 2√ 2⎝ 2⎠ 2⎯ πr ⎯⎯⎯ It is seen that τmax is a function of Θ.6 cos Θ⎞ Θ⎛ ⎜ 1 + sin ⎟ 2⎝ 2⎠ (e) cos Θ⎞ Θ⎛ ⎜ 1 − sin ⎟ 2⎝ 2⎠ (f) if plane stress KI 2⎯ πr ⎯⎯⎯ √ cos Θ if plane deformation 2 (g) (h) Plane stress Now the plane stress situation will be investigated. 5 (o) 2⎯ πr ⎯⎯⎯ √ Answer: The maximum shear stress is at plane stress: at plane deformation: τmax( 60o ) = 0. This is now in agreement with the assumption that Θ should be larger than 47o. and one obtains 1 1 KI Θ Θ τmax = (σ1 − σ2) = cos ⋅ 2 ⋅sin 2 2√ 2 2 2⎯ πr ⎯⎯⎯ (n) Again. Then σ2 is the smallest principal stress.65 τmax(90o) = 0.8 degrees. namely that Θ should be smaller than 47o. we have to repeat these calculations once again with a new assumption: Assume that the maximum value of τmax will be obtained for Θ > 47o. By solving d τmax =0 (k) dΘ one finds that τmax has its largest value when Θ = 75. Thus. determine the value of Θ that gives the largest value of τmax. By solving d τmax / d Θ = 0 one finds that τmax has its largest value when Θ = π / 2 = 90 degrees.5 KI 2⎯ πr ⎯⎯⎯ √ KI 2⎯ πr ⎯⎯⎯ √ Relations (i) and (n) give the graphs asked for (not shown here).6 cos ⎟ 2√ 2⎝ 2⎠ 2⎠ 2⎯ πr ⎝ ⎯⎯⎯ (m) Next. Chapter 2 Page 2:7 .Assume that the maximum value of τmax will be obtained for Θ < 47o. One obtains 1 τmax = (σ1 − σ3) 2 = 1 KI ⎛ Θ ⎛ Θ⎞ Θ⎞ ⎜cos ⎜ 1 + sin ⎟ − 0. But this is not in agreement with the assumption made. Then τmax will be given by the principal stresses σ1 and σ3. find the value of Θ that gives the largest value of τmax in (m). It is then concluded that KI τmax(90o) = 0. 2/6. Answer: The shear stress τ is the same in all directions. Page 2:8 Chapter 2 . y The stress components in front of a crack tip loaded in Mode III are − KIII Θ τxz = sin 2 2⎯ πr ⎯⎯⎯ √ r x z τyz = KIII 2⎯ πr ⎯⎯⎯ √ cos Θ 2 σxx = σyy = σzz = τxy = 0 In which direction Θ will the shear stress have its maximum? Solution: yz y xz The shear stress τ on a surface in the xy-plane is KIII Θ Θ 2 τ =√ τ + τ = sin2 + cos2 ⎯⎯⎯⎯ xz ⎯ yz 2 2 2⎯ πr ⎯⎯⎯ √ ⎯⎯⎯⎯⎯ √ r x z = KIII 2⎯ πr ⎯⎯⎯ √ = τmax (a) Thus. the shear stress τ (= τmax) is the same in all directions (thus independent of the direction Θ). 467.397 ⎜ ⎟ ⎝ σ0⎠ where ⎛ KI ⎞ 2 f8 = 0. (b) 0. (c) 0.0 gives a = π ⎜ σ ⎟ = 0.318 ⎜ σ ⎟ ⎝ 0⎠ ⎝ 0⎠ (b) ⎛ KI ⎞ 2 ⎛a h⎞ KI = σ0 √ ⎯⎯πa f1 ⎜⎝ W . (c) Edge crack of length a in a large plate. (b) Central crack of length 2a in a quadratical plate with side length 4a.34 gives a = 0. a deep and 4a (= 2c) long.254.177. Study the following crack geometries (a) to (f) and determine the characteristic lengths a of the cracks so that the probability of crack growth is the same for all cracks. Solution: Handbook formulae give stress intensity factors KI (functions fi refer to Appendix 3 in the textbook) 2 (a) ⎛ KI ⎞ 2 ⎛a h⎞ 1 ⎛ KI ⎞ KI = σ0 √ ⎯⎯πa f1 ⎜⎝ W . (e) 0. (a) Central crack of length 2a in a large plate.12 where ⎛ KI ⎞ 2 f4 = 1.177 ⎜ σ ⎟ ⎝ 0⎠ (c) (d) (e) (f) ⎛a⎞ KI = σ0 √ ⎯⎯πa f5 ⎜⎝ W ⎟⎠ ⎛a⎞ KI = σ0 √ ⎯⎯πa f4 ⎜⎝ W ⎟⎠ ⎛a⎞ KI = σ0 √ ⎯⎯πa f7 ⎜⎝ c ⎟⎠ ⎛a⎞ KI = σ0 √ ⎯⎯πa f8 ⎜⎝ c ⎟⎠ ⎛ KI ⎞ 2 gives a = 0. (f) Elliptical crack with principal axes a and 2a embedded in a thick plate. Chapter 2 Page 2:9 .826 gives a = 0.896 gives a = 0.Stress intensity factor 2/7. The material is linear elastic. (d) 0. (d) Two opposite edge cracks of length a in a rectangular plate with side length (width) 2W = 4a and height h >> W. in a thick plate.233 ⎜ ⎟ ⎝ σ0⎠ where ⎛ KI ⎞ 2 f7 = 0. (f) 0.233.254 ⎜ ⎟ ⎝ σ0⎠ where f5 = 1. (e) Half-elliptical surface crack. W ⎟⎠ where f1 = 1. W ⎟⎠ where f1 = 1.467 ⎜ ⎟ ⎝ σ0⎠ Answer: Ratio a / (KI / σ0)2 is (a) 0.397.318. All cracks are loaded in Mode I with a remote stress σ0.17 gives a = 0. N and ϕ so that the crack is loaded in Mode I only.Fracture modes 2/8.b) σxx = and τxy = 2πRt 2πR 2t The stress σyy is zero (σyy = 0). respectively. Solution: area = 1 n xy y n sin cos xy x yy xx The axial tensile force N gives the normal stress σxx in the longitudinal direction. Equilibrium equations in the directions of σn and τn give (in the equations below the area is written after the multiplication sign ×. R The plane of the crack is inclined an angle φ (= N T ϕ) with respect to the longitudinal axis of the cylinder. and the torque T gives shear stress τxy. and in Mode II only. the area equals 1 for the largest area and sin φ and cos φ . A thin-walled circular cylinder contains a crack. i. which gives T N T R τn = 0 = cos 2φ giving sin 2φ + = − tan 2φ 4πRt N 2 2πR 2t Loading of the crack is in Mode II if σn = 0 and τn ≠ 0. t T N Determine a relationship between T. for the two smaller areas) σn × 1 = σxx sin φ × sin φ + 0 − τxy sin φ × cos φ − τxy cos φ × sin φ ⎤ 2T ⎡ N cos φ sin φ + ⎥ = σxx sin2 φ − 2τxy cos φ sin φ = sin φ ⎢ ⎦ ⎣ 2πRt 2πR 2t and τn × 1 = σx sin φ × cos φ − 0 + τxy sin φ × sin φ − τxy cos φ × cos φ = N T cos 2φ sin 2φ + 4πRt 2πR 2t The loading of the crack is in Mode I if σn ≠ 0 and τn = 0. see figure. Assume that R >> t so that the curvature of the cylinder may be neglected when the stress intensity factor is determined.e. which gives. either Page 2:10 (c) Chapter 2 (d) (e) . where N −T (a. The cylinder is loaded by a torque T and an axial tensile force N. 015 m.15 m. Linear elastic fracture mechanics (LEFM) 2/9. What would a more accurate analysis give? Could the pressure vessel be used without danger? The vessel is manufactured of steel. Solution: The stresses in the wall are σrr = 0 . σΘΘ = pR . The conditions for linear elastic fracture mechanics are assumed to be fulfilled. The stress intensity factor KI was then determined for a crack in a plane.4 m. and all τij = 0 2t (i ≠ j) (a) The requirements for linear elastic fracture mechanics (LEFM) to be valid are assumed to be fulfilled. and ratio T / N = ( − R / 2) tan φ gives Mode II loading. In an approximate l estimation of the stress intensity factor the curvature of the pressure vessel wall was disregarded. infinitely large plate. overall length of pressure vessel is l = 5 m.σn = 0 = sin φ giving φ = 0 (f) or ⎤ 2T ⎡ N cos φ sin φ + ⎥ σn = 0 = sin φ ⎢ ⎦ ⎣ 2πRt 2πR 2t giving T R = − tan φ N 2 (g) Answer: Load ratio T / N = ( − R / 2) tan 2φ gives Mode I loading of the crack. and it is loaded with a pressure p. Chapter 2 Page 2:11 . for example). and the wall thickness is t = 0. Numerical data: pressure vessel radius is R = 0. In a thin-walled pressure vessel there is a risk that a 300 mm (= 2a) long longitudinal through2R 2a thickness crack will develop almost immediately as the pressure vessel is loaded (due to a bad t welding. where a = 0. t σzz = pR . crack length is 2a. and it was found that the stress intensity factor so determined was only half of the fracture toughness KIc of the material. 02 m (crack length 2a).343. plate thickness t = 0. implying that failure will occur. Determine λ. The following conditions must be fulfilled: R >> t.015 ⎯⎯Rt √ √ ⎯⎯⎯⎯⎯⎯⎯ This gives f14(1. M0 2a M0 A through-thickness crack of length 2a has been found in a large plate. Case B.343 (d) pR ⎯⎯πa √ t (e) which is larger than the fracture toughness KIc and fracture will be expected! Answer: Exact analysis (taking curvature into account) gives KI ≈ 1.936) = 2. yield strength σY = 1300 MPa. a >> t and L >> R. They are all fulfilled. Determine at which moment M0max crack growth will occur.Case A.17KIc. One has λ= a = 0.936 KI = 2. Study a crack in a large flat plate. The stress intensity factor then is pR KI = σ0 √ (b) ⎯⎯πa f1(0.15 0. 2/10. 1) = σ0 √ ⎯⎯πa ⋅1 = t √ ⎯⎯πa KI = But KIc 2 gives KIc = 2 pR ⎯⎯πa √ t (c) The fracture toughness KIc of the material has now been determined. Page 2:12 Chapter 2 . t Numerical data: a = 0. Study a crack in a cylindrical wall.4 ⋅ 0. The plate is subjected to a bending moment M0 (Nm/m) per unit length. Thus = 1. and fracture toughness KIc = 110 MN/m3/2.03 m. The stress intensity factor will now be obtained from a handbook (Case 14 in Appendix 3 of the textbook). 02 ⎯⎯⎯⎯ ⎯ = 0.06 m. t and W − a.032 6√ π 0. l = 3.0658 MNm m (e) This value of the moment M0 gives the stress σ in the plate. t b 2a l A cantilever beam contains a through-thickness crack. giving maximum remote stress σ = 0. 2/11. Thus KI = 6M0 t2 (a) ⎯⎯πa √ (b) Crack propagation is expected when KI = KIc if linear elastic fracture mechanics (LEFM) can be used. beam (plate) thickness t = 0. One obtains 6M0 σ = 2 = 439 MPa = 0.02 m (crack length 2a). The beam is loaded by a system of forces 11 P d (11P and 9P) according to the figure. One obtains M0 = KIc t 2 6√ ⎯⎯πa = 110 ⋅ 0. Thus LEFM can be used. and fracture toughness KIc = 50 MN/m3/2.08 m.5 ⎜ ⎟ = 2.03 m.2 m. The fracture criterion KI = KIc gives 6M0 t2 ⎯⎯πa = KIc √ (d) from which M0 is solved.34 σY in the plate. yield strength σY = 600 MPa.0179 m 2. Chapter 2 Page 2:13 .Solution: Case 13 in the Appendix 3 of the textbook gives 6M0 ⎛a⎞ KI = σ0√ ⎯⎯πa f13 ⎜⎝ W ⎟⎠ where σ0 = 2 and M0 in Nm/m t Ratio a / W = 0 gives f13 = 1. d = 0. 9P Numerical data: a = 0.5 ⎜ ⎝ 1300 ⎠ ⎝ σY ⎠ (c) which is smaller than a.34σY t (f) Answer: Crack growth is expected when bending moment of the plate is M0max = 66 kNm/m. Determine d at which value of P failure may be expected. plate width b = 0. One has ⎛ KIc ⎞ 2 ⎛ 110 ⎞ 2 ⎟ = 0. 03 51. then failure will occur when (b) KI = KIc Can LEFM be used? ⎛ KIc ⎞ 2 ⎛ 50 ⎞ 2 ⎟ = 0.04⎠ ⎭ P√ ⎯⎯π⎯ a = 51.6 ⎞⎟ + 12Pd f ⎛⎜ 0. so LEFM can used. LEFM. The stress intensity factor KI is obtained by superposition of the two loading cases.3 kN Page 2:14 Chapter 2 (e) . and the distance b / 2 − a from the crack tip to the edge of the plate. Thus.Solution: The cantilever beam (or plate) will be loaded in tension and in bending.3 kN. Answer: Failure is expected at critical load Pcrit = 9.162 ⎞⎟ bt ⎝ t ⎠ bt ⎝ 0. plate thickness t. The axial force becomes N = 20P loading the plate in tension. and the bending moment becomes M = 2Pd.04 0.6 P √ ⎯⎯π⎯ a bt bt (a) If the conditions for linear elastic fracture mechanics.02 ⎯⎯⎯⎯ ⎯ = 0.89) t ⎝ 0.02 .08 ⋅ 0.6 √ ⎯⎯π⎯ a = 50 ⋅ 0.72 + 27.0174 m) is less than the crack (half-)length a.5 ⎜ ⎝ 600 ⎠ ⎝ σY ⎠ (c) This number (0. ⎟ + 2 √ π⎯ a f13 ⎜ ⎟ ⎯ ⎯ ⎯ ⎯ ⎝W W⎠ b t ⎝W⎠ bt = ⎯⎯π⎯ a ⎧⎨ 20P f ⎛⎜ 0.6 √ π 0.02⎞⎟ ⎫⎬ √ 1 13 = P√ ⎯⎯π⎯ a ⎛⎜20 ⋅ 1.186 + 12 d 1.5 ⎜ ⎟ = 2. KI = 51.6 P√ ⎯⎯π⎯ a = K = 50 MN/m3 / 2 Ic bt (d) giving P = Pcrit = KIc bt 51.04⎠ ⎩ = (23.00928 MN = 9300 N Failure is expected at P = Pcrit = 9.0174 m 2. One obtains (Cases 1 and 13 in Appendix 3 of the textbook) ⎛ a h ⎞ 6M ⎛a⎞ N KI = KIN + KIM = √ π⎯ a f1 ⎜ . 1. at plane strain are fulfilled. determine at which loadings P0 and M0 fracture may be expected. Chapter 2 Page 2:15 . M0 t An edge crack has been discovered in a long beam. can be used.2 (√ ⎯⎯m) 2. This condition is fulfilled if temperature T is T ≤ − 3oC. W − a) = a ⎝ σY ⎠ which gives ⎛ KIc ⎞ ⎜ ⎟≤ ⎝ σY ⎠ ⎯ √ a = 0. Numerical data: crack length a = 0. see figure.5 m. one should have KIc < 0. one obtains KIc = 100 MN/m3/2 and σY = 500 MPa). t. numerically. The fracture toughness KIc and the yield strength σY of the material depend on temperature as given below.5 (a) Thus. approximately.2 m. beam width t = 0.2/12. K Ic MN/m 3/2 160 140 120 100 80 60 40 Y MPa 700 K Ic Y 600 500 400 -200 -100 0 Temperature T oC 100 Figure. One has ⎛ KIc ⎞ 2 2. Fracture toughness KIc and yield strength σY as function of temperature T Solution: (a) First. see diagram (at T = − 3oC.1 m. investigate if linear elastic fracture mechanics.2σY. beam height W = 0.5 ⎜ ⎟ ≤ minimum of (a. LEFM. M0 a P0 W P0 (a) Over which temperature range may linear elastic fracture mechanics be used? (b) For the upper limit of this temperature range. 1⎞ 6M0 ⎛ 0.44M0c = 13.1⎞ ⎫ P0 f5 ⎜ ⎟ + f ⎜ ⎟⎬ tW ⎩ ⎝ 0.5⎠ W 6 ⎝ 0. Thus.90M ) = K 0 0 Ic (d) tW (1.1 MN (f) where P0 should be entered in MN and M0 in MNm. Answer: Temperature T should be T ≤ − 3oC.37P0 + 12. at this temperature fracture will occur when KI = giving ⎯√⎯π⎯ a (1. Page 2:16 Chapter 2 . At temperature T = − 3oC the yield strength is σY = 500 MPa and the fracture toughness is KIc = 100 MN/m3/2 giving the maximum load P0 + 9.366P + 12.5 π 0. Thus.1 MN. where P0 should be entered in MN and M0 in MNm. at fracture one has P0c + 9.44M0 = 13.1 ⎯⎯⎯ √ ⎯ = 17.2 ⋅ 0.90M0) = KIc tW ⎯⎯π⎯ a √ = 100 ⋅ 0.902M ) 0 0 ⎛ 0.37P + 12.5⎠ ⎭ (b) tW Fracture will occur when KI = KIc (c) At temperature T = − 3oC the fracture toughness KIc is KIc = 100 MN/m3/2.84 MN (e) with P0 in MN and M0 in MNm.(b) Superposition of the two loading cases tension and bending gives (Case 5 and 6 in the Appendix 3 of the textbook) P0 ⎛ a ⎞ 6M0 ⎛a⎞ ⎜ ⎟ KI = KIP + KIM = √ π⎯ a f5 ⎜ ⎟ + π a f ⎯ ⎯ ⎯ √ ⎯ ⎯ 6 ⎝ W ⎠ t W2 ⎝W⎠ bt = ⎯⎯π⎯ a ⎧⎨ √ = ⎯√⎯π⎯ a (1. the radius R is very large compared to crack length a and wall thickness t). be used? One has ⎛ KIc ⎞ 2 ⎛ 50 ⎞ 2 (a) ⎟ = 0.005 m. Determine at which torque T crack propagation will be expected. A thin-walled circular cylinder contains a crack o T forming an angle 30 to the longitudinal axis of the cylinder. The stress τxy gives a normal stress σn across the crack.00434 m 2. the curvature of the wall may be neglected when the stress intensity factor is estimated (i. Chapter 2 Page 2:17 . LEFM.) Solution: Can linear elastic fracture mechanics. see figure.e.5 ⎜ ⎟ = 2. In this case.00434 m (and W − a is larger). T 30 o R t The equivalent stress intensity factor Ke is Ke2 = KI2 + KII2 + 4 2 KIII κ+1 Numerical data: radius R = 0.2/13. where (here) g = 1. and a shear stress τnφ. so LEFM can be used. The stress on the element is T T (b) τxy = = Wv 2πR 2t where Wv is the section modulus in torsion of the cylinder. crack (half-)length a and wall thickness t are a = t = 0. and fracture toughness KIc = 50 MN/m3/2. loading the crack in Mode II.2 m. yield limit σY = 1200 MPa.005 m > 0. (For a small crack in a large plate the stress intensity factor in Mode II is KII = τx y∞ √ ⎯⎯πa g . see figure. xy y n n xy 60 o 30 x o Consider an element surrounding the crack. loading the crack in Mode I..5 ⎜ ⎝ 1200 ⎠ ⎝ σY ⎠ Here a = t = 0. ) Answer: Crack propagation is expected when torque T = 500 kNm.005 √⎯⎯⎯⎯⎯ ⎯ = 0.501 MNm (g) (g) (h) At torque T = 500 kNm crack propagation is expected. with φ = 60o. σn = σxx cos 2φ + σyy sin2 φ + 2τxy cos φ sin φ = √3 τ ⎯ 2 xy 1 τnφ = − (σxx − σyy ) sin φ cos φ + τxy (cos2 φ − sin2 φ) = − τxy 2 (c) (d) In Mode I the stress intensity factor KI becomes (for a small crack in a large plate the factor f1 becomes f1 = 1) KI = σn √ ⎯⎯π⎯ a f1 = ⎯3 τ √ 2 xy √⎯π⎯ a ⎯ (e) In Mode II the stress intensity factor KII becomes (for a small crack in a large plate the factor g becomes g = 1) 1 KII = τnφ √ (f) ⎯⎯π⎯ a g = 2 τxy √ ⎯⎯π⎯ a The equivalent stress intensity factor Ke becomes ⎛ 3 1⎞ Ke2 = KI2 + KII2 = τ2xy π a ⎜ + ⎟ = τ2xy π a ⎝ 4 4⎠ Fracture is expected when Ke = KIc (if LEFM can be used). Page 2:18 Chapter 2 .The stresses σn and τnφ become.005 π ⋅ 0.22 ⋅ 0. This gives T Ke = τxy √ ⎯⎯π⎯ a = ⎯⎯π⎯ a = KIc √ 2πR 2t Thus. (The LEFM conditions were checked above. T= KIc 2πR 2t ⎯√⎯π⎯ a = 50 ⋅ 2π 0. Fracture.00034 m 6π ⎝ σY ⎠ 6π ⎝ 500 ⎠ FFFCH3P. Determine the maximum remote stress (the critical stress) the plate may be loaded with. r a a At a circular hole (radius r = 10 mm) in a large plate two cracks (length a = 5 mm) have been discovered. the correction at critical stress. use the Irwin correction of the crack length. Studentlitteratur.Solutions to problems in T Dahlberg and A Ekberg: Failure. is 2 1 ⎛ KIc ⎞ 1 ⎛ 40 ⎞ 2 r1 = ⎜ ⎟ = ⎜ ⎟ = 0.DOC/2010-01-15/TD Chapter 3 (b) Page 3:1 . Lund 2002. when KI = KIc. Data: plate thickness t = 20 mm. i. For plane strain (t is large enough). Chapter 3 Small plastic zone at crack tip Problems with solutions The Irwin correction 3/1. fracture toughness KIc = 40 MN/m3/2.e. Fatigue . but the crack length a is too short. and yield limit σY = 500 MPa.5 ⎜ ⎟ = 2. see figure. ISBN 91-44-02096-1.016 m ⎝ 500 ⎠ ⎝ σY ⎠ The plate thickness t is large enough to ensure plane strain.An Introduction.5 ⎜ (a) ⎟ = 0. Therefore. Solution: Can linear elastic fracture mechanics be used? ⎛ KIc ⎞ 2 ⎛ 40 ⎞ 2 2. Page 3:2 Chapter 3 . is loaded with a bending moment M as shown in the figure. approximately. σ∞ = σcrit = 173 MPa. A beam with rectangular cross section.00534 ⋅1. 3/2.78 ⎯⎯⎯⎯⎯⎯1 3⎜⎝ r + a + r ⎟⎠ = σ∞ √⎯⎯⎯⎯⎯⎯⎯ 1 = KIc = 40 ⋅ 106 N/m3 / 2 (c) from which σ∞ = σcrit = 173 MPa is solved. and its depth is now 3 mm. the material around the crack is removed. After the grinding the beam height thus is 31 mm. The crack. is still there. removed M W a 9 mm M (a) How much has the ultimate bending moment been increased by the grinding? (Thus. It is decided that 9 mm of the beam height should be removed by grinding. calculate the ultimate load M of the beam before grinding and after the grinding. however.Correction of LEFM now gives ⎛ a + r1 ⎞ KI = σ∞ √ π (a + r ) ⋅f π 0. In order to increase the load-carrying capacity of the cracked beam.) (b) The purpose of the grinding was perhaps to remove all the crack.e. Answer: The critical remote stress σ∞ is. What bending moment can be put on the beam if 12 mm were removed (i. Material properties: KIc = 40 MN/m3/2 and σY = 900 MPa. A 12 mm deep crack (a = 12 mm) has appeared in the beam (see figure). even the crack tip has been removed)? In this case the yield limit of the material limits the loading. t = 20 mm and W = 40 mm. Use the Irwin correction of the crack length. and one obtains the ultimate load M = M1 = 985 Nm. then fracture mechanics theory needs not be used at all. i.5 ⎜ ⎟ = 2.031 ⎠ ⎝ W ⎠ 0.020 ⋅ 0. the ultimate bending moment for the beam with cross section 20 mm by 40 mm and crack depth 12 mm: Case 6 in Appendix 3 of the textbook gives KI = ⎛a⎞ ⎛ 0. and W − a (= 28 mm). Determine the ultimate load before grinding. In that case.94 mm.003105⎞ 6M ⎜ ⎟ = ⎜ π a f π 0. if plastic deformation of the beam should be avoided.003105 f ⎟ KIeff = ⎯ √ ⎯⎯⎯⎯⎯⎯⎯ eff ⎯ √ ⎯⎯ 6 6 ⎝ 0. giving M = M2 = 1227 Nm. (b) If all the material surrounding the crack had been removed.000105 m 6π ⎝ σY ⎠ (c) Case 6 gives ⎛ aeff⎞ 6M ⎛ 0. i. when the stress intensity factor reaches KIc. One obtains Chapter 3 Page 3:3 .0312 tW 2 = KIc = 40 ⋅ 106 ⋅ N/m3 / 2 (d) Diagram gives f6 = 1.040 = KIc = 40 ⋅ 106 N/m3 / 2 (b) Diagram gives f6 = 1. and crack depth is 3 mm. is less than t (= 20 mm). LEFM can not be used any longer.012 ⎯⎯π⎯ a f6 ⎜⎝ W ⎟⎠ = ⎯⎯⎯⎯⎯ ⎯ f6 ⎜⎝ 0.94 ⋅ 10−3 m = 4. Thus. a (= 12 mm). 4.115.e.057.94 mm ⎝ 900 ⎠ ⎝ σs ⎠ This length.040⎟⎠ 2√ 2√ tW 0. it is OK to use LEFM for the first crack length (12 mm) but not for the second (length 3 mm). if 12 mm of the beam height were taken away (this was perhaps the original purpose of the grinding operation). One obtains for plane strain. 2 1 ⎛ KIc ⎞ r1 = ⎜ ⎟ = 0.e.020 ⋅ 0. determine the bending moment the beam can be loaded with.012⎞ 6M 6M π 0.Solution: Can LEFM (linear elastic fracture mechanics) be used? ⎛ KIc ⎞ 2 ⎛ 40 ⎞ 2 2.5 ⎜ (a) ⎟ = 4. After the grinding the cross section becomes 20 mm by 31 mm. respectively. The Dugdale model 3/3.020 ⋅ 0. (a) y x 2a A large steel plate of an elastic.028 / 2 = σY = 900 MPa = I 0.0283 / 12 (e) which gives M = 2352 Nm. (a) At which remote stress σ∞ would a 150 mm long crack start to grow? Use the Dugdale model of crack tip opening displacement δ(a) as criterion for crack growth initiation (CTOD criterion).σmax = M ⋅W /2 M ⋅ 0. calculate the size (length) ρ of the plastic zone at the crack tip when the stress is such that the crack starts to propagate. if you try to improve the load-carrying capacity of a structure by removing a crack. One has ⎫ 8 a σY ⎧ 1 ⎬ δ(a) = ln ⎨ πE ⎩ cos (πσ∞ / 2σY) ⎭ (b) For the two cases 2a = 50 mm and 2a = 150 mm. (b) The moment can be increased to M = 2352 Nm if the full crack (with crack tip included) is removed. One obtains Page 3:4 Chapter 3 .) Answer: (a) The maximum bending moment M increases from 985 Nm to 1227 Nm. thus by 25 per cent. at crack length 2a = 50 mm. Solution: The first measurement. ideally plastic material (σY = 500 MPa) contains a throughthickness crack of total length 50 mm (= 2a). (Conclusion: make sure that also the crack tip is removed. At a tension test of the plate the crack was found to start growing at the remote stress σ∞ = 300 MPa. gives the critical value of the crack tip opening displacement (CTOD) δ(a)crit. and at crack length 150 mm the length of the plastic zone is 14. ⎫ σY 8 ⋅ 0.025 ⎜ − 1⎟ = 0. (b) The length of the plastic zone is obtained from 1 ⎞ ⎛ − 1⎟ ρ=a⎜ ⎝ cos (π σ∞ / 2 σY) ⎠ (c) For 2a = 50 mm. with remote stress σ∞ = 300 MPa. with δ(a)crit just calculated in (a).075 ⋅ σY ⎧ 1 ⎬ ln ⎨ δ(a)crit = 0.5 mm ⎝ cos (π ⋅ 184 / 2 ⋅ 500) ⎠ (d) (e) Answer: (a) At crack length 150 mm the critical stress is 184 MPa.δ(a)crit = = ⎫ 8 acrit σY ⎧ 1 ⎬ ln ⎨ πE ⎩ cos (πσ∞ crit / 2σY) ⎭ ⎫ 8 ⋅ 0. with remote stress σ∞ = 184 MPa.0175325 m = 17.5 mm ⎝ cos (π ⋅ 300 / 2 ⋅ 500) ⎠ For 2a = 150 mm. Chapter 3 Page 3:5 .025 σY ⎧ σY 1 ⎬ = 0. one obtains.03382957 = E πE ⎩ cos (π ⋅ σ∞ / 2 ⋅ 500) ⎭ (a) (b) Solving for σ∞ gives σ∞ = 184 MPa.5 mm (for critical remote stress 300 MPa).03382957 ln ⎨ πE E ⎩ cos (π ⋅ 300 / 2 ⋅ 500) ⎭ (a) At which stress would a 150 mm long crack start to grow? The CTOD criterion gives.075 ⎜ − 1⎟ = 0.5 mm (for critical stress σ∞ = 184 MPa). (b) at crack length 50 mm the length of the plastic zone is 17.0145492 m = 14. ⎛ 1 ⎞ ρ = 0. one obtains. ⎛ 1 ⎞ ρ = 0. ⎬ 24 σ πa ⎩ ⎭ ⎯⎯ ⎠ ⎝ Y√ Solution: (a) Using Dugdale’s model. the crack opening displacement δ(0) becomes δ(0) 8σY 1 + sin(π σ∞ / 2 σY) = ln a πE cos(π σ∞ / 2 σY) = σ∞ 8 ⋅ 2σ∞ 1 + sin(π σ∞ / 4 σ∞) ln = 4. The crack opening displacement (COD) is δ(0) 8σY 1 + sin(π σ∞ / 2σY) = ln a πE cos(π σ∞ / 2σY) and if σ∞ << σY one obtains 4 KI δ(0) = a E√ ⎯⎯πa 2 ⎧ ⎫ π2 ⎛ KI ⎞ ⎨1 + ⎜ ⎟ + …. Plane stress is at hand so that the Dugdale model may be used. the two crack tips move the same distance). The material is linearly elastic.e. (0) During a tension test of a large plate with a central crack the crack opening δ(0) was recorded as a function of the load. i. ideally plastic with yield strength σY. The recorded curve deviated from a straight line. Calculate the crack propagation during the loading (assume that symmetry is maintained.e. (b) The deviation from linearity of the recorded curve appeared to be 30 per cent. if σY is very large) one has Page 3:6 Chapter 3 (a) . The deviation from linearity may be explained partly by the appearance of a plastic zone at the crack tip and partly by crack growth during the loading. y x 2a (a) How large deviation from linearity will be expected due to the plastic zone at the crack tip? The remote stress σ∞ is σY / 2.3/4..4888 πE cos(π σ∞ / 4σ∞) E If σ∞ << σY (i. The plastic deformation at the crack tips gives 12. so the remaining deviation up to 30 per cent is explained by crack growth. ⎬ a E√ ⎯⎯πa ⎩ 24 ⎝ σY √ ⎭ ⎯⎯π⎯ a ⎠ (b) Using KI = σ∞√ ⎯⎯πa and σ∞ << σY in (b). the crack has propagated during the loading of the structure.2 ⎫ 4 KI ⎧ δ(0) π2 ⎛ KI ⎞ ⎨1 + = ⎜ ⎟ + ….) This gives σ∞ ⎯⎯π⎯ a δ(0) 4 σ∞ √ = (c) { 1 + 0} = 4 a E E√ ⎯⎯πa The difference between the displacements in (a) and (c) comes from the plastic deformation at the crack tip.3 ⋅ 4 E πE cos(π σ∞ / 4 σ∞) from which is solved afinal = 1. (If σY tends to infinity.2 per cent..158a during the loading.2 per cent (d) 4 (b) At the experiment. the Dugdale model gives σ∞ a 8 ⋅ 2σ∞ afinal 1 + sin(π σ∞ / 4 σ∞) = ln (e) δ(0) = 1. crack has grown to final length afinal = 1. Chapter 3 Page 3:7 . The deviation from linearity is 4.4888 (f) Answer: (a) Deviation from linearity due to plastic zones is 12.1222 = 12. and the solution obtained is for the fully elastic case. and (b).4888 − 4 = 0. the deviation from linearity was 30 per cent. where a is the original crack length in the unloaded structure.158 a 4. no yielding will occur. Using δ(0) = 1.3 δelastic.2 per cent deviation from linearity only. where δelastic was calculated in (c). one obtains the linear elastic solution..3 ⋅ 4 a = 1. Calculate the final crack length afinal. i.e. fracture occurred when Pc = 0. Solution: At fracture the remote tensile stress is σ∞ c = Pc 2Wt The crack opening displacement (COD) at fracture is 8 σY ac 1 + sin (πσ∞ /2σY) ln δc(0) = πE cos (πσ∞ /2σY) (a) (b) which gives ⎧ 1 + sin (πPc /4Wt σY)⎫ − 1 ⎬ ac = π E δc(0) ⋅ ⎨8 σY ln cos (πPc /4Wt σY) ⎭ ⎩ (c) According to the diagram. ideally plastic with yield strength σY.6 mm. the crack has grown the amount 2Δa = 2ac − 2a0 = 0.28 MN and δc(0) = 0. which gives ac = 0. Thus. yield limit σY = 600 MPa. Page 3:8 Chapter 3 .8 mm at each crack tip). plate thickness t = 0.18 mm The material is linearly elastic.0200 m = 0. Numerical data: crack length a0 = 0. and modulus of elasticity E = 200 GPa.0436 − 0. Calculate the crack growth during the loading.28 P MN (0 ) 0. the Dugdale model may be used.0436 m (d) During the loading of the plate.6 mm (thus. (e) Answer: The crack growth is 2Δa = 23. Symmetric stable crack growth was x 2 h observed during the loading of the plate.0218 m. y (0) 2a 2W P 0.0236 m = 23.002 m.2 m.18⋅10 − 3 m.3/5. The plate is thin so that the stress state is plane (PS). During a tension test of a large plate with a central crack the crack opening δ(0) was recorded as a function of the load P.01 m. the crack has propagated 11. The recorded P curve deviated from a straight line as given in the figure. giving 2ac = 0. plate size h = W = 0. where E is the modulus of elasticity of the material.An Introduction Studentlitteratur. The surface energy ws and the cohesive strength σc of a material may (approximately) be x determined from the following simple model of i the inter-atomic forces: Assume that the force σi per unit area between i atomic planes may be represented by the c function x 5x0 x0 x − x0 σi = σc sin α π 4 x0 where x0 is the distance between the planes of the atoms in the unloaded state and x is the atomic separation when the material is loaded.DOC/2010-01-15/TD Chapter 4 Page 4:1 . stresses at crack tip 4/1.Solutions to problems in T Dahlberg and A Ekberg: Failure. FFFCH4P. ISBN 91-44-02096-1 Chapter 4 Energy considerations Problems with solutions Surface energy. σc and ws in terms of E and x0. Use a small through-thickness crack in a large plate and compare the remote stresses σ∞ giving crack growth in the two cases. i. assume that the fracture limit is obtained when the strain is 25 per cent.e. when x = 5x0 /4. The factor σc is a material parameter: the bond strength. Fracture. (b) Compare the fracture criterion by Griffith (released energy equals surface energy) with a stress criterion for fracture of a material with a crack. i (a) Determine the parameters α. Lund 2002. Fatigue . Also. ) Solution: (a) At small displacements (i. for x = x0.For the stress criterion. one has. It gives the cohesive stress (cohesive strength) E σc = 2π Page 4:2 (d) Chapter 4 (h) . (The stress state may be considered as plane. d σi =0 dx which gives Thus. Determine how far away in front of the crack tip (measured in number of atomic planes) the mean stress must be equal to the cohesive strength of the material in order to make the stress criterion equivalent to the energy criterion by Griffith. assume that the mean stress over n planes of atoms in front of the crack tip is a measure of the loading of the material.. Hooke’ s law gives x − x0 σi = E ε = E (a) x0 d σi E = d x x0 which gives From the equation given in the problem formulation (b) σi = σc sin α π x − x0 x0 d σi x − x0 απ cos α π = σc dx x0 x0 one obtains Equations (c) and (b) give. dσi(x = x0) απ E cos 0 = = σc dx x0 x0 giving σc = (c) E απ From the problem formulation.e. when x is close to x0). (e) d σi(x = 5x0 /4) 5 x0 /4 − x0 απ cos α π =0 = σc dx x0 x0 (f) 1 π which gives α π = and α = 2 4 2 (g) 1 cos α π = 0 4 Enter α = 2 into (d). see figure above. when x = 5x0 /4. see equation (4. and μ is the shear modulus of the material: μ = E / 2(1 + ν ). this crack propagation criterion can be written. which means that in practice. Thus E x0 [Nm / m2 ] (j) ws = 2 4π (b) The Griffith criterion According to Griffith. Expression (k) is thus given for the new crack area tΔa created. The surface energy may now be determined from the area below the curve σi(x). κ+1 2 (k) K t ⋅ Δa = 2ws t ⋅ Δa 8μ I where κ = (3 − ν) / (1 + ν ) at plane stress. The ultimate strength σU of a very high-strength steel is in the order of E / 100. where Kc = √ 2w⎯s E ⎯⎯⎯ Chapter 4 (l) (m) Page 4:3 . fracture will occur due to other mechanisms than separation of the atomic planes. has thickness t. The crack growth is Δa (the crack length a has increased by Δa) and the plate.34) in the textbook. One obtains 3x0 /2 2ws = ⌠ ⌡x0 3x0 /2 σi(x) dx = ⌠ ⌡x0 x − x0 E dx sin 2π 2π x0 3x0 /2 x − x0 ⎤ E ⎡ x0 = ⎢ − cos 2π ⎥ 2π ⎣ 2π x0 ⎦ x = 0 E x0 2 π2 (i) where the factor 2 in front of ws is there because two new surfaces are created when the crack appears.It is noted that this stress is very high: in the order of E / 10. fracture will occur when ⎯⎯⎯⎯⎯ √ 2ws 8E 2(1 + ν) ⎛⎝ 1 + ν + 1 ⎞⎠ 3−ν =√ 2w⎯s E ⎯⎯⎯ KI = Kc. the energy that is released when a crack propagates should be equal to the energy needed to create the new crack surfaces. Solving (k) for KI gives KI = ⎯⎯ √ 2ws 8μ = κ+1 Thus. By use of the stress intensity factor KI. in which the (through-thickness) crack is situated. according to the Griffith criterion the crack will propagate when the remote stress is σG∞ = ⎯⎯ √ √ ⎯⎯⎯ 2E E x0 E = πa 4π2 π 2ws E = πa ⎯⎯ √ x0 2πa (p) The stress criterion y yy = We now turn to the stress criterion. the stress intensity factor KI is KI = σG∞ √ ⎯⎯πa (n) Crack propagation will occur (for plane stress conditions) when KI = Kc. determine the remote stress σG∞ (superscript G for Griffith) leading to failure. between the atomic planes. The mean value σyy of σyy = σyy(r) over n atomic planes is σyy 1 ⌠ = n x0 ⌡0 = n x0 1 ⌠ σyy (r) dr = n x0 ⌡0 n x0 ⎯a 1 σ∞ √ r ] [ 2⎯ √ 0 = σ∞ n x0 √ 2 ⎯ n x0 σ∞ ⎯⎯ √ ⎯ √ 2a n x0 a dr 2r (r) At fracture the mean value σyy should reach the cohesive strength σc of the material.Now. or x0. The coordinate r is used here in order to avoid confusion with the distance x. The stress σyy in front of the crack tip is KI 2 r σyy = r KI ⎯⎯2πr √ ⎯ σ∞ √ ⎯⎯πa = ⎯⎯2πr √ ⎯ ⎯ √ = σ∞ a 2r (q) where σ∞ is the remote stress (far away from the crack). giving KI = σG∞ √ 2w⎯s E ⎯⎯πa = Kc = √ ⎯⎯⎯ (o) Thus. For a small through-thickness crack in a large plate. This gives σyy = σ∞ Page 4:4 ⎯⎯ √ 2a E = σc = n x0 2π Chapter 4 (s) . The material properties should then be as given in the equation in the problem: σi = σc sin α π (x − x0) / x0. with parameter values as calculated in problem (a). ws = E x0 /4π2. This gives σG∞ = E π ⎯⎯ √ x0 E = 2πa 2π which gives n. We have now determined the stress σG∞ giving failure according to the Griffith criterion and the stress σS∞ giving failure according to the stress criterion. In the case that these two criteria give failure at the same remote stress. if the mean stress σyy of σyy(r) reaches the cohesive strength σc over n = 4/π atomic planes. cos 2 2 2⎯ πr 2⎯ πr ⎯⎯⎯ √ ⎯⎯⎯ √ and σxx = σyy = σzz = τxy = 0 Chapter 4 Page 4:5 . strain energy 4/2. Tensor notations.and σ∞ = σS∞ = E 2π ⎯⎯ √ n x0 2a (t) where superscript S stands for “stress criterion”. σij εij u= 2 The stress components in front of a crack tip loaded in mode III are − KIII KIII Θ Θ τyz = τxz = sin . σc = E /2π. then the stress criterion (as formulated here) and the Griffith criterion will predict crack propagation at the same remote stress σ∞. Thus n = 4/π = 1. y r Determine the elastic strain energy density in front of a crack tip loaded in mode III. one has σG∞ = σS∞. Answer: (a) α = 2.27. for linearly elastic material. (b) n = 4 /π = 1. ⎯⎯ √ n x0 = σS∞ 2a (u) In conclusion. In which x direction Θ has the strain energy its maximum? z The elastic strain energy density u is.27. (a) εij = 1 τ 2μ ij (b) where μ is the shear modulus. for i ≠ j. and the strains calculated in (b) into the expression (a). ideally plastic up to deformation δU and yield limit σY). One has. P a Y U P Two rectangular bars (same cross section b by h) of linearly elastic material (modulus of h elasticity E) are glued together along a portion h of its length. μ is the shear modulus. (b) Solution: Two solutions will be given. see figure (a).Solution: Use u= σij εij 2 Calculate the strains. Energy balance 4/3. The glue has σ-δ b (a) characteristics according to figure (b) (rigid. One has a >> h and a >> b. It gives 2 τzx τyz τzy ⎫ 1 ⎛ KIII ⎞ ⎧ Θ Θ⎫ 1 ⎧ τxz ⎜ ⎟ ⎨2 sin2 + 2 cos2 ⎬ u = ⎨τxz + τzx + τyz + τzy ⎬ = 2 ⎩ 2μ 2 2⎭ 2μ 2μ 2μ⎭ 4μ ⎝ √ 2⎯ π r⎠ ⎩ ⎯⎯⎯ 2 1 ⎛ KIII ⎞ 1 2 ⎜ ⎟ = = KIII 2μ ⎝ √ 2⎯ π r ⎠ 4π μ r ⎯⎯⎯ (c) Answer: The elastic strain energy u is u = KIII2/4πμr (thus independent on Θ). Determine the critical load P = Pcrit. Enter the stresses given in the problem. (a) Load control gives the potential energy Π = U − PΔ where which gives Page 4:6 (a) σij εij PΔ U =⌠ dV = ⌡Vol 2 2 (b) Π = − PΔ / 2 (c) Chapter 4 . Assume that the two rectangular bars may be considered to be cantilever beams of length a. the work required to overcome the glue strength. is Ws = A ⋅ σY δU ( = A ⋅ 2ws) (g) d Ws = σY δU dA which gives d Π d Ws = . ⎯⎯⎯⎯ √ EI b σY δU a2 gives = Pcrit = Chapter 4 ∂Ws δa = b σY δU δa ∂a bh a P2 a2 = b σY δU EI E h σY δU ⎯√⎯⎯⎯⎯⎯ bh 2a √ ⎯3 (m.n) 2√ ⎯3 (o) (p) E σY δ⎯ Uh ⎯√⎯⎯⎯⎯⎯ Page 4:7 . namely that gives P = Pcrit = (h) gives ⎯⎯⎯⎯⎯ ⎯⎯⎯⎯ √ √ EI b σY δU a 2 = P2 a2 = σY δU EI b E b h 3 b σY δU 12 a 2 = bh a (i) E h σY δU ⎯⎯⎯⎯⎯⎯ √ 2√ ⎯3 (j) (b) Alternative solution From P2 a3 Π=− 3 EI Ws = b ⋅ a σY δU and (k. − dA dA Energy balance. for a crack growth δa. i. namely Thus P = Pcrit = Answer: Critical load is and δWs = − δΠ = δWs . ∂Π P2 3 a2 δΠ = δa = − δa ∂a 3 EI Energy balance.e.l) one obtaines. The displacement Δ /2 of the cantilever beam end due to a force P then is Δ P a3 2 P a3 = giving Δ = (d) 2 3 EI 3 EI The potential energy of the structure becomes P2 a3 PΔ =− Π=− 2 3 EI (e) The energy release due to a change of the crack area is ⎛ − P 2 (b ⋅ a)3⎞ − P 2 3(b ⋅ a)2 − P 2 a 2 dΠ d dΠ = = = ⎜ ⎟= d A d (b ⋅ a) d (b ⋅ a) ⎝ 3EI b 3 ⎠ 3 EI EI b b3 (f) The work required to create the crack surface. was manufactured from the material to be investigated. (a) Load control gives the potential energy PΔ PΔ Π = U − PΔ = − PΔ = − 2 2 /2 P (a) Assume that the rectangular bars may be approximated to two cantilever beams of length a. see figure. Determine the surface energy of b the material. The modulus of elasticity is E. linearly elastic material a test specimen. When loading the test specimen 2 h crack propagation was obtained when the force P was P = Pmax. P a P To determine the surface energy ws of a brittle. the cross section is b by 2h. The displacement Δ /2 of the beam end due to a force P is a Δ P a3 2 P a3 = giving Δ = 2 3 EI 3 EI (b) The potential energy of the structure is P 2 (ba)3 PΔ =− Π=− 2 3 EI b 3 (c) The energy release due to a change of the crack area A = b a is dΠ P2 a2 dΠ P 2 3(ba)2 =− = =− d A d (ba) 3 EI b 3 EI b (d) The work required to create the new surfaces is Ws = 2ws ⋅ A Energy balance which gives Page 4:8 (G =)− which gives d Π d Ws = dA dA d Ws = 2 ws dA gives P2 a2 = 2ws EI b 2 2 P2 a2 P 2 a 2 12 6 Pmax a = = ws = 2 b EI 2 b E bh 3 E b 2 h 3 Chapter 4 (e) (f) (g) . Solution: Two solutions will be given.4/4. and the crack length is a where a >> h and a >> b. Thus, the surface energy ws is ws = 6Pmax2 a 2 / E b 2 h 3. (b) Alternative solution P2 a3 Π=− 3 EI From and W s = 2 b ⋅ a ws (h,i) is obtained, for a crack growth δa, ∂Ws ∂Π P2 3 a2 δΠ = δa = − δa and δWs = δa = 2 b ws δa ∂a 3 EI ∂a Energy balance, − δΠ = δWs, gives P2 a2 = 2 b ws EI Thus ws = (j,k) P2 a2 2 b EI (l) Answer: The surface energy ws is ws = 6Pmax2 a 2 / E b 2 h 3. 4/5. M a U U A thin rectangular bar (cross section b by h) of b linearly elastic material (Young’s modulus E) is h glued to a rigid foundation, see figure (a). The (a) glue has a linear σ-δ characteristics according to figure (b) (rupture at stress σU when deformation is δU). The bar is loaded with a bending moment M. Determine the critical load Mcrit. One has a >> b > h. (b) Solution: Two solutions will be given. (a) Load control gives the potential energy Π = U − PΔ, which here gives Mα Mα Π=U −M ⋅α= − Mα = − 2 2 (a) Assume that the free part of the rectangular bar may be approximated to a cantilever beam of length a. The rotation α of the beam end due to the bending moment M is M a M a 12 α= (b) = EI E bh 3 The potential energy of the structure becomes (crack area A is A = b⋅a) Mα M 2 (ba) 6 M2 a 6 Π=− =− =− 2 E bh 3 E b 2h 3 Chapter 4 (where A = ba) (c) Page 4:9 The energy release due to a change of the crack area is dΠ M2 6 =− dA E b 2h 3 (d) The work required to create the new surface is 1 Ws = A ⋅ σU δU 2 d Ws 1 = σ δ dA 2 U U which gives (e) Energy balance, d Π d Ws M2 6 1 = σU δU − = gives dA dA E b 2h 3 2 (f) which gives 2 M 2 = Mcrit = σU δU bh E b 2h 3 and Mcrit = 12 2 ⎯⎯⎯⎯ √ σU δU E h 3 (g) (b) Alternative solution M2 6 a Π=− E b h3 From 1 Ws = ba σU δU 2 and (h,i) is obtained ∂Π 6 M2 δΠ = δa δa = − ∂a E b h3 and δWs = ∂Ws 1 δa = b σU δU δa ∂a 2 (j,k) Energy balance, − δΠ = δWs, gives 6 M2 1 = b σU δU E b h3 2 (l) Thus M =M 2 2 crit ⎫ E σU δU b 2h 3 ⎧ N N 2 3 2 ⎨dimension: 2 2 m m m = (Nm) ⎬ = 12 m m ⎩ ⎭ as obtained in (g). Answer: Critical load Mcrit is Page 4:10 Mcrit = Chapter 4 bh 2√ ⎯3 E h σU⎯δU ⎯√⎯⎯⎯⎯⎯ (m) Solutions to problems in T Dahlberg and A Ekberg: Failure, Fracture, Fatigue - An Introduction. Studentlitteratur, Lund 2002, ISBN 91-44-02096-1. Chapter 5 Determination of stress intensity factor KI Problems with solutions Stress intensity factor 5/1. /2 y /2 x 2h << h An infinitely long strip, height 2h, is split along half its length, see the figure (the strip thus has a semi-infinitely long crack in it). The upper and lower boundary of the strip have been given a prescribed displacement Δ /2 each. Determine the stress intensity factor KI for this configuration. Assume plane stress at the crack tip. Solution: The strategy to solve this problem is the following: the stress intensity factor KI can be determined from the energy release rate G. To find G, an energy consideration is made. When the crack grows (say δa) the strip to the right of the crack tip becomes δa shorter, and the energy in that part (i.e. in a volume δa by 2h by strip thickness t) is released. This energy is used to determine G. First the elastic strain energy density u stored in the strip far away from the crack tip is determined. To do this, we need to know the stresses and the strains in the strip. The strains in the strip (far away to the right of the crack tip) are Δ εyy = (a,b,c) , εxx = 0 , and εzz = − ν εyy 2h The strain εxx is zero because no movements can occur in the x direction far away from the crack tip. Also, all shear strain components are zero because of symmetry. Hooke’s law gives, with σzz = 0, SOLCHAP5.DOC/2010-01-15/TD Chapter 5 5:1 εyy = Δ 1 = {σ − νσxx} 2h E yy εxx = 0 = 1 {σ − νσyy} E xx from which the stresses are solved: E Δ σyy = 1 − ν2 2h and σxx = ν σyy (d) (e) (f,g) The strain energy density u (Nm/m3) becomes 1 1 E ⎛ Δ ⎞2 1 ⎜ ⎟ u = σij εij = (σxx εxx + σyy εyy ) = 2 2 2 1 − ν2 ⎝ 2h ⎠ (h) The energy δU in a section of the strip with length δa is (far away to the right of the crack tip) 1 E ⎛ Δ ⎞2 δU = u ⋅ Volume = u ⋅ 2h δa t = ⎜ ⎟ ⋅ 2h δa t 2 1 − ν2 ⎝ 2h ⎠ (i) where t is the thickness of the strip. The strain energy close to the crack tip is not calculated here, because close to the crack tip the stress field is “complicated”, and, as will be seen, it is not necessary to calculate the strain energy in this region. The energy release rate G is, at displacement control, dU 1 U(a + da) − U(a) 1 U(a) − U(a + da) dΠ =− =− = (j) G =− dA t da t da t da Thus, the energy release when the crack grows a distance δa is δU = U(a) − U(a+δa). The energy release rate G then becomes (at displacement control), by use of δU from (h), 1 δU 1 1 E ⎛ Δ ⎞ 2 ⎜ ⎟ ⋅ 2h t = G= t δa t 2 1 − ν2 ⎝ 2h ⎠ (k) Finally, KI is obtained from KI = √ E’G ⎯⎯⎯ ⎯ , (E’ = E at plane stress) giving KI = √ ⎯⎯EG ⎯ = ⎛ Δ⎞ ⎜ ⎟ ⋅√ ⎯h 2 ⎝ 2h ⎠ 1 − ν ⎯⎯⎯⎯ √ E Answer: The stress intensity factor is KI = EΔ / 2√ h(1 − ⎯ν2) ⎯⎯⎯⎯⎯ 5:2 Chapter 5 (l) one obtains ⎯⎯ √ P 2 a 2 P a 2⎯ √3 = (d) bI bh √ ⎯h It is seen that the stress intensity factor KI increases as the crack grows. (b) Displacement control. P 2 dC P 2 2 a 2 G= = (c) 2 b da 2 b EI In the case of plane stress at the crack tip. Solution: The force versus displacement relationship for a cantilever beam gives the displacement Δ / 2 of the beam end due to the loading force P as Δ P a3 2 P a3 = giving Δ = (a) 2 3 EI 3 EI The compliance C of the specimen is Δ 1 2 P a 3 2a 3 dC 2 a 2 C= = = giving = (b) P P 3 EI 3 EI da EI (a) The energy release rate G due to a crack extension is. for prescribed load P. thickness is b and specimen height is 2h. for prescribed displacement Δ. Assume plane stress conditions. Investigate the two loading cases: (a) Load control. P a P Use the compliance method to determine the energy release rate G and from that. with displacement Δ prescribed. determine the stress intensity factor KI for a double 2 h cantilever beam (DCB) specimen. The material is linear elastic with Young’s modulus E. Crack length is a.5/2. with the loading force P prescribed. KI2 G= G ⋅E = giving KI = √ ⎯⎯⎯⎯ E (b) The energy release rate G due to a crack extension is. b where a >> b and a >> h. implying that unstable crack growth will be expected (at least if Kc does not depend on crack length a). P 2 dC P 2 2 a 2 = G= 2 b da 2 b EI Using P= 3 EI Δ 2 a3 Chapter 5 (e) (f) 5:3 . and (b) 2 KI = √ ⎯ 3 EΔh√ ⎯ h / 4a . The material is linear elastic with Young’s modulus E. One obtains Ma 2M a β= giving α = 2 β = (a) EI EI The compliance C of the specimen is α 2a dC 2 C= = giving = (b) M EI da EI The energy release rate G due to a crack extension is. 5/3. implying that stable crack growth will be expected (at least if Kc does not depend on crack length a). Answer: Stress intensity factors are (a) KI = 2√ ⎯ 3Pa / bh√ ⎯ h .a 2 ⎛ 3 EI Δ⎞ 9 EI Δ2 ⎟ = G= ⎜ b EI ⎝ 2 a 3 ⎠ 4 b a4 2 one obtains (g) In the case of plane stress at the crack tip. 12M 2 M 2 dC M 2 2 = = G= 2 b da 2 b EI E b 2 h 3 Answer: The energy release rate G is G = 12M2 / Eb2h3. one obtains KI2 G ⋅E = giving KI = √ G= ⎯⎯⎯⎯ E ⎯⎯⎯ √ 9 E 2 I Δ2 3 E Δ h √ ⎯h = √ ⎯3 E Δ h √ ⎯h = 4 2 2 4b a 4a ⎯⎯12 ⋅2 a √ (h) Here it is seen that the stress intensity factor KI decreases as the crack grows. thickness is b and specimen height is 2h. Crack length is a. for prescribed load M. 5:4 Chapter 5 (c) . where a >> b and a >> h. M 2h a M b Use the compliance method to determine the energy release rate G for a double cantilever beam (DCB) specimen. The specimen is loaded with bending moments M at the beam ends. Solution: The (generalised) force versus (generalised) displacement relationship (here this becomes moment versus rotation) of a cantilever beam M gives the rotation angle β of the beam end due a to the bending moment M. The crack is assumed to grow symmetrically. The energy release rate G due to a crack extension in both directions then is. Chapter 5 5:5 . i. for a prescribed force P. P 2h P a a Use the compliance method to determine the energy release rate G for a beam with a long central crack.e. two crack fronts of length b each). see figure. One obtains Δ P (2a)3 1 1 P a3 = ⋅ ⋅ giving Δ = (a) 2 3 EI 8 8 12 EI The compliance C of the specimen is Δ a3 dC a2 C= = giving = (b) P 12 EI da 4 EI Note that. which gives (in dΠ/dA) that dA = d(2ab) = 2bda (A is crack area) (or. Solution: The force versus displacement relationship of a fixed-fixed beam (length 2a) gives the displacement Δ / 2 of the beam centre due to the loading force P. The beam is loaded symmetrically with two opposite forces P at the beam centre.5/4. beam thickness is b and beam height is 2h. due to symmetry the crack is assumed to grow in both directions. The crack length is 2a. where a >> b and a >> h. The material is linear elastic with modulus of elasticity E. 3 P2 a2 P 2 dC P 2 a 2 (c) = = G= 2 ⋅ (2b) da 4 b 4 EI 4 E b 2 h 3 Answer: The energy release rate G is G = 3P2a2 / 4Eb2h3. it can be seen as one crack with width 2b. G =− The potential energy is. because here there are two crack tips and each crack font has length b. q ⋅ v dS Π = U − W = ⌠ u dV − ⌠ ⌡boundary ⌡Vol 1 1 = ⌠ q ⋅ v dS − ⌠ q ⋅ v dS = − ⌠ q ⋅ v dS ⌡boundary 2 ⌡boundary 2 ⌡boundary (b) where U = W / 2 has been used (the material is linear elastic). The material is linear elastic with modulus of elasticity E. It gives 2a ⎫ 1 − q2 ⎧ 1 4a 5 ⎨ (2a) − Π = − ⌠ q ⋅ v(x) dx = (2a)4 + 4a 2 (2a)3 ⎬ ⌡0 24EI ⎩ 5 4 3 ⎭ 2 q2 a5 =− 45 EI 5:6 (d) Chapter 5 . see figure. The crack length is 2a. q 2h v(x) y x q a a Use the potential energy Π to determine the energy release rate G for a beam with a long central crack. for 0 ≤ x ≤ 2a. The beam is loaded symmetrically with two opposite distributed loads q (q in N/m) at beam upper and lower surfaces. Expression (b) 2a ⎫ gives 1 ⎧⌠ 2a ⌠ ⎨ q ⋅ v(x) dx + − q ⋅ (− v(x)) dx⎬ (c) Π =− ⌡0 2 ⎩⌡0 ⎭ where the first term refers to the upper surface of the beam (loaded upwards) and the second term refers to the lower surface (loaded downwards).5/5. where a >> b and a >> h. see equation (4.4) in the textbook. q v(x) = (x 4 − 4ax 3 + 4a 2x 2) 24 EI Solution: The energy release rate G for the beam is 1 dΠ (a) 2b da The crack front length (the extension of the crack tip in the z direction) is 2b. Insert v(x) into (c) and integrate. beam thickness is b and beam height is 2h. The deflection of a clamped-clamped beam loaded with a distributed load q is. M 2h a M b Determine the stress intensity factor for the structure in Problem 5/3 by use of the reciprocity relation.10) ∂a (κ + 1) KI(1) ⌡Γ The stress intensity factor in Problem 5/2 is known. In a general case one has Ti(2) (N/m2) times dui(1)/da (m/m) times dΓ (m). giving (N/m2) ⋅ (m/m) ⋅ m = N/m. i.10) has dimension N/m. Ti(2)dΓ becomes force per unit width (width in the z direction) of the structure. the bending moment M. This means that the intergration in (5. When discretizing the integral.10) in the textbook to determine the stress intensity factor. One has ∂ui(1) 4μ (2) (2) ⌠ KI = Ti dΓ (5. Note that the integral in (5. Thus. is applied in two discrete points. Chapter 5 5:7 . Solution: Use equation (5. the factor KI(1) to use in the right hand side of (5.e. In case Ti(2) is a stress (N/m2).Finally the energy release rate G can be calculated: 1 dΠ 1 (− 2 q 2) 4 1 q 2 a 4 4 q 2 a 4 G =− =− 5a = = 2b da 2b 45 EI 9 b EI 3 E b 2h 3 (e) Answer: Energy release rate G is G = 4q2a4 / 3Eb2h3. Use Problem 5/2 as the known Case (1). This means that one needs to know the rotation angle in Case (1) (due to load P) at the end of the cantilever beam (where the load M is applied in Case (2)).10) along the perimeter can be discretized. Reciprocity relation 5/6.10) is (from Problem 5/2) KI(1) = P a 2√ ⎯3 bh √ ⎯h (a) From loading case (1) (Problem 5/2) one needs to know also the (generalized) displacement at the point where the load is going to be applied in the second loading case. One obtains Pa 2 (b) θ= 2EI Also. Ti(2)dΓ will have the dimension N/m. dθ/da is needed. One obtains dθ 2Pa Pa = = da 2EI EI (c) In Problem 5/3 the load. cf.10) becomes. and in front of M / b the factor 2.10). the discretized version of (5. at the end of the expression one has dθ/da from (c). ⎯⎯⎯ √ E 12M 2 2√ ⎯3 M = E b 2h 3 bh √ ⎯h Answer: The stress intensity factor is. Using that μ = E / 2(1 + ν) (the shear modulus). The expression Ti(2)dΓ then must have dimension N (unit of force) to give N/m when multiplying Ti(2)dΓ with dθ/da. ∂ui(1) 4μ M Pa 4μ (2) (2) ⌠ KI = T ⋅ 2 ⋅ dΓ = ⋅ (d) i ∂a b EI (κ + 1) KI(1) ⌡Γ (κ + 1) KI(1) Here. This implies that the loading moment M (given in Nm in the problem) should be applied as moment per unit width (in the z direction) when the moment is entered into (5. next to that the bending moment per unit width M / b. which is there because there are two bending moments. one obtains KI(2) = 4E (1 + ν) bh √ ⎯ h ⋅ 2 ⋅ M ⋅ Pa = 2√ ⎯3 M ⋅ 2(1 + ν) 4 b EI bh √ Pa 2√ ⎯3 ⎯h (e) From Problem 5/3 is obtained KI(1) = √ ⎯⎯EG ⎯ = which is in agreement with (e). when the (generalized) displacement is an angle θ. Thus. in this case. 5:8 Chapter 5 (f) KI = 2√ ⎯3 M bh √ ⎯h . and from equatio (4. answer to Problem 5/3. one obtains the dimension of dθ/da as 1/m.34) that κ = (3 − ν) / (1 + ν) for plane stress conditions at the crack tip.Here. 10) in the textbook to determine the stress intensity factor. and compare with Problem 5/2. Thus. Solution: Use equation (5. q(x) = q0. Use Problem 5/2 as the known Case P (1)..5/7. the factor KI(1) to use in the right hand side of (5. the loading is q0 for 0 < x q ( x) < αa. One obtains. say q0 αa = Q? Assume plane stress conditions.10) is (from Problem 5/2) P a 2√ ⎯3 (a) bh √ ⎯h From loading case (1) (Problem 5/2) one needs to know also the (generalized) displacement at the points where the load is going to be applied in the second loading case. along part of the beams. Pa 3 ⎛ 3x x 3 ⎞ v(x) = ⎜2 − (b) + ⎟ 6EI ⎝ a a3 ⎠ KI(1) = Also.. This means that one needs to know the deflection in Case (1) (due to load P) of the cantilever beam where the load q(x) is applied in Case (2). x (b) the loading q(x) is constant. ∂v/∂a is needed.10) ∂a (κ + 1) KI(1) ⌡Γ The stress intensity factor in Problem 5/2 is known. i.e. the loading is q(x) for 0 < x < q ( x) αa (note that x = 0 at the free surface and x = a Case (2) at the crack tip). One obtains ∂v(x) P = (2 ⋅ 3a 2 − 3x ⋅ 2a + 0 ) ∂a 6EI Chapter 5 (c) 5:9 . One has ∂ui(1) 4μ (2) (2) ⌠ KI = Ti dΓ (5. Case (1) 2 h Investigate some different loading cases of the x structure (structure and loadings are symb metric): a P (a) a general loading q(x) (N/m) along part of the beams. Determine the stress intensity factor for the structure in the figure by use of the reciprocity relation. with ui(1)(x) = v(x).e. a (c) what happens in Case (b) above if the factor α becomes small (α << 1) but the force q0αa at a the beam end is a constant. i. Equation (5.10) now gives αa 4μ ⌠ q(x) P (6a 2 − 6xa) dx KI(2) = 2 b 6EI (κ + 1) KI(1) ⌡0 (d) where μ = E / 2(1 + ν) is the shear modulus, κ = (3 − ν) / (1 + ν) (plane stress), and the factor 2 is there due to the two cantilever beams (integration along two beams). In (5.10) T(2) is stress. Dividing q(x) (N/m) by the thickness b gives the stress on the boundary. (b) Enter q(x) = q0 into (d). It gives αa q a 4μ P 0 (2) ⌠ KI = 2 (a − x) dx (1) ⌡0 b EI (κ + 1) KI = q0 a P ⎛ 4μ 1 2⎞ ⎟ 2 ⎜ a αa − (αa) ⎠ 2 (κ + 1) KI(1) b EI ⎝ Thus, K (2) I (e) P a 2 q0 αa ⎛ 8μ α⎞ = ⎜1 − ⎟ (1) b ⎝ 2⎠ (κ + 1) KI EI (f) (c) Enter q0⋅αa = Q into (f), and then let α go to zero. It gives KI(2) = P a2 Q 8μ 8μ b h √ ⎯h P a 2 Q = (κ + 1) KI(1) EI b (κ + 1) P a 2√ ⎯ 3 EI b 8E (1 + ν) b h ⎯ √h a 12 Q = Qa 2√ ⎯3 2(1 + ν) 4 2√ ⎯ 3 Ebh 3 b bh √ ⎯h as obtained in Case (1) (at plane stress), see expression (a) above. = Answer: (a) KI(2) = αa 4μ ⌠ q(x) Pa (a − x) dx 2 b EI (κ + 1) KI(1) ⌡0 (b) Entering q(x) = q0 into (d) gives KI(2) = P a 2 q0 αa ⎛ 8μ α⎞ ⎜ 1 − ⎟ b ⎝ 2⎠ (κ + 1) KI(1) EI (c) Enter Q = q0⋅αa and α = 0 into (f). It gives as in Case (1) (at plane stress), cf. expression (a). 5:10 Chapter 5 KI(2) = Qa 2√ ⎯3 bh √ ⎯h (g) (d) (f) Superposition 5/8. Demonstrate (no calculations needed) why the two loading cases in the figures have the same stress intensity factor KI, cf. (j) and (k) in Section 5.2.3 in the textbook. The plates are large. In Case (2) the stress σ is a compressive stress acting on the two crack surfaces (i.e., σ is a pressure loading the crack surfaces). Case (2) Case (1) 2a 2a Figure Loading case (1) with remote stress σ and loading case (2) with same stress σ acting on the crack surfaces Solution: In Case (2), replace the compressive stress σ on the crack surfaces with a tension stress σ. Call this loading case for Case (2b). Superimpose loading case (2b) on Case (1). It gives the same stress distribution in the plate (in the plane of the crack) as in a plate with no crack at all, see figure below. This implies that the stress intensity factors in the two cases must cancel each other, and therefore they must be the same. Chapter 5 5:11 Case (2b) Case (1) No crack 2a 2a 2a Figure Loading case (1) with remote stress σ and loading case (2b) with tension stress σ acting on the crack surfaces Or, doing it in anther way: Start with a plate without crack, see the leftmost part of the figure. The stress in the left plate is uniaxial (value σ) and evenly distributed all over the plate. Create loading case (1) by cutting a crack in the first plate. It means that in the second plate the stress σ cannot be transferred from the upper crack surface to the lower crack surface (as was done in the first plate). This gives the loading case (1) with a certain stress intensity factor KI(1). Next, the stress σ that could not be transferred over the crack in loading case (1) is applied on the crack surfaces of another plate. This gives loading case (2b), see the figure. The loading case (2b) gives a stress intensity factor KI(2b). Superimposing Cases (1) and (2b) gives the first plate with no crack. But the first plate has no stress singularity, so one must have KI(1) + KI(2b) = 0 giving KI(1) = − KI(2b) (a) Finally, changing the sign of the stress σ in loading case (2b) gives back the original loading case (2), and one has KI(1) = KI(2) (b) Answer: Replace the compressive stress σ in Case (2) by a tensile stress (of magnitude σ) on the crack surfaces. This gives the Case (2b). Superimpose Case (1) and the new Case (2b). It gives a plate with a homogeneous stress field throughout the plate, without stress concentration. Thus, the two stress intensity factors cancel each other, and it follows that Case (1) and Case (2) must have the same stress intensity factor. 5:12 Chapter 5 using ΔKth = 4 MN/m3/2 and Δσ0 = σY / 2 = 750 MPa.0223 mm. (a) A low-strength steel with ΔKth = 7 MN/m3/2 and σY = 300 MPa. Case 8 in Appendix 3 of the textbook gives (terms containing a / t can be neglected here so that f8 = 0.Solutions to problems in T Dahlberg and A Ekberg: Failure. Fracture.637 ⋅ 150 √ ⎯⎯π⎯ a = 7 MN/m (b) giving a = 1.An Introduction.7 mm. assume that the flaws are penny-shaped (the crack looks like a coin with a sharp egde).5σY (and σY is the yield limit of the material). Lund 2002. Studentlitteratur. The loading is cyclic and the stress range is 0. (b) Similarly. These flaws should be treated as if they were cracks.637 ⋅ 750 √ ⎯⎯π⎯ a = 4 MN/m3 / 2 (c) giving a = 0.DOC/2010-01-15/TD Chapter 6 Page 6:1 . Fatigue . one obtains ΔKI = 0.637 is obtained) KI = 0. FMCHAP6P. Paris’ law 6/1. and (b) a high-strength steel with ΔKth = 4 MN/m3/2 and σY = 1500 MPa. Chapter 6 Crack propagation under cyclic loading Problems with solutions Crack propagation at cyclic loading. ISBN 91-44-02096-1. Using ΔKth = 7 MN/m3/2 and Δσ0 = σY / 2 = 150 MPa. one obtains 3/2 ΔKI = 0. Determine the largest flaw allowable with respect to fatigue crack propagation in the following materials. Crack-like flaws may develop in most materials. Solution: A penny-shaped crack in the interior of the material will give the stress intensity factor KI = σ0 √ ⎯⎯π⎯ a f8(a). Then the stress intensity range ΔKI must be less than (or equal to) the threshold value ΔKth.637 σ0 √ ⎯⎯π⎯ a (a) (a) No crack propagation is allowed. Therefore. where a >> b and a >> h.) Answer: Critical crack length is (a) a = 1. Solution: (a) The stress intensity range becomes (Δp = Pmax /b (N/m) is the load range) ΔKI = 2√ ⎯ 3 Δp a= h√ ⎯h Entering (a) into Paris’ law gives Separating (b) gives 2√ ⎯ 3 Pmax b h√ ⎯h da = C (ΔKI)n = C dN da =C an (a) a n ⎛ 2√ ⎯ 3 Pmax ⎞ ⎜ a⎟ h ⎝ b h√ ⎠ ⎯ n ⎛ 2√ ⎯ 3 Pmax ⎞ ⎜ ⎟ dN ⎝ b h√ ⎯h ⎠ (b) (c) Integrating (c) and solving for N give ⎧ b h√ 1 ⎯h ⎫ 1 − n ⎨ ⎬ (a − ai1 − n ) N= C (1 − n) ⎩ 2√ ⎯ 3 Pmax⎭ n Page 6:2 Chapter 6 (n ≠ 1) (d) . P a P A double cantilever beam (DCB) specimen is subjected to a pulsating loading. The material is 2 h linear elastic with Young’s modulus E. thickness is b and specimen height is 2h. (b) the displacement is prescribed. and (b) a = 0. (One reason is that the high-strength steel is loaded at higher stress. Use Paris’ law da b = C (ΔKI )n dN to determine the number of loading cycles required to make the crack grow from its initial length ai to a final length a. The crack length is a. 6/2.7 mm.022 mm. the displacement varies between 0 and vmax (in the figure one has Δ = 2vmax).It is concluded that the high-strength steel is much more sensitive to cracks than the low-strength steel. if (a) the load is prescribed. the load varies between 0 and Pmax. choose safety factor s = 1. A structure is subjected to a pulsating loading. σmin = 0. The crack diameters are 2a0.e. respectively. The structure is made of a material that follows the crack propagation law da = C (ΔKI )n dN (a) Determine the largest (remote) stress range Δσ0 the structure may be subjected to if crack propagation is not allowed. Therefore. n (a) ⎧ b h√ 1 ⎯h ⎫ 1 − n 1 − n ⎨ ⎬ (a N= − ai ) C (1 − n) ⎩ 2√ ⎯ 3 Pmax⎭ (b) (h) (n ≠ 1) ⎧ 2√ 1 ⎯ h ⎫ 2n + 1 2n + 1 ⎨ ⎬ (a − ai ) N= C (2n + 1) ⎩ √ ⎯ 3 E h 2 vmax⎭ n 6/3. (b) Determine the maximum stress range that may be allowed if the structure is loaded by N = 10 000 cycles. i. σY = 620 MPa.0005 m. ΔKth = 2 MN/m3/2. which is much smaller than the thickness of the material.12⋅10 − 11 m7/MN4. Chapter 6 Page 6:3 . n = 4. Embedded circular (penny-shaped) cracks are discovered in the structure. During the loading the crack is not allowed to grow so large that it becomes unstable. Numerical data: crack length a0 = 0. KIc = 36 MN/m3/2.0 and C = 1.(b) The stress intensity range becomes (here vmax = Δ /2 is the range of the displacement) ⎯ 3 E h 2 vmax 1 √ ΔKI = (e) a2 2√ ⎯h n Entering (e) into Paris’ law gives 2 ⎛ ⎞ 3 E h v da ⎯ √ 1 max = C (ΔKI)n = C ⎜ ⎟ (f) 2 dN a 2√ h ⎝ ⎠ ⎯ Separating (f) gives a N ⎛ ⎯ 3 E h 2 vmax ⎞ n ⌠ a 2n da = ⌠ C ⎜ √ (g) ⎟ dN ⌡ai ⌡0 ⎝ 2√ ⎠ ⎯h Integrating (g) and solving for N give ⎧ 2√ 1 ⎯ h ⎫ 2n + 1 ⎨ ⎬ (a − ai2n + 1) N= 2 C (2n + 1) ⎩ √ ⎯ 3 E h vmax⎭ n Answer: Number of cycles required is.5.. 0⎟⎠ where a = c.637σ0√ ⎯⎯π⎯ac = s giving ac = π ⎜⎝ 0. The stress intensity factor KI then is (Case 8 in Appendix 3 in the textbook.0) = 0. which gives f8(1. One then has K = 0.Solution: (a) No crack propagation If no crack propagation is allowed. Page 6:4 Chapter 6 .637 σ0 ⋅ 1.637. the stress intensity range ΔKI must be less than or equal to the threshold value ΔKth.5⎟⎠ = σ2 0 0 (h) where.637 Δσ0√ ⎯⎯πa ≤ ΔKth Solving for Δσ0 gives ΔKth 2 ⋅ 106 Δσ0 ≤ = = 79 ⋅ 106 N/m2 = 79 MPa 0. which gives 0.637 Δσ0√ ⎯⎯πa (c) No crack propagation is expected if ΔKI ≤ ΔKth. with a << t) ⎛a ⎞ KI = σ0√ (a) ⎯⎯πa f8 ⎜⎝ c . Assume that there exists a crack with the most unfavourable direction. σ0 (in MPa) equals the stress range Δσ0 (in MPa) because σmin = 0 (load is pulsating). Thus KI = 0.637 σ s ⎟⎠ = π ⎜⎝ 0. if Δσ0 ≤ 79 MPa no crack growth is expected.637 √ π ⋅ 0. (d) (e) (b) Crack propagation After N = 104 loading cycles the crack should give a stress intensity factor KImax = KIc / s. Assume that the crack stays circular (penny-shaped) during the propagation. Determine the critical crack length ac giving this stress intensity factor.637 σ √ (f) ⎯⎯πa I 0 Fracture will occur (if LEFM is valid) when KIc KImax = s Equations (f) and (g) give (g) KIc KIc ⎞ 2 1 ⎛ 36 ⎞ 2 452 1⎛ 0.0005 ⎯⎯πa 0.637 √ ⎯⎯⎯⎯⎯⎯⎯ Thus. here.637 σ0√ ⎯⎯πa (b) The stress intensity range ΔKI becomes (Δσ0 is stress range) ΔKI = 0. (b) Δσ0 = 324 MPa.12 ⋅ 10 − 11 (0.637 Δσ0 √ ⎯ ⎯ dN ac which gives N da n ⌠ ⌠ = C (0. n and N into (k) gives −1 −1 − (0.637 Δσ0 √ ⎯ −1 Solving for Δσ0 gives Δσ0 = 324 MPa = σ0. During N cycles the crack will grow from a0 to ac. The crack propagation law (Paris’ law) gives da n πa) (i) = C (ΔKI)n = C(0.5 ⋅ 620 ⎠ ⎝ s ⋅ σY⎠ Thus. LEFM was used when ac was determined.0037 m so LEFM may be used (t and W − a are assumed large enough).5 ⎜ ⎟ = 2.637 Δσ0 √ ⎯ 1 − n/2 Entering the values of a0.Now the initial crack length a0 and the final crack length ac (expressed in the unknown stress range Δσ0) are known. Is LEFM valid? The stress σ0 = 324 MPa (= σmax) gives the critical crack length 452 = 0. Only then LEFM must be valid.5 ⋅ 10 − 3 ) 4 π ) ⋅ 104 = 1. Chapter 6 Page 6:5 . The fracture criterion KImax = KIc / s was used to determine ac.5 ⎜ ⎝ 1.637 Δσ π ) dN (j) ⎯ √ 0 n ⌡a0 (√ ⎯ a ) ⌡0 Evaluation of the integrals gives ac1 − n/2 − a01 − n/2 n π ) ⋅ N (k) = C (0. ac is larger than 0. C. (452/(Δσ0)2) Finally. not during the crack propagation phase. Note that only the crack length ac at fracture is of interest here. ac.0043 m ac = 3242 to be compared with (l) (m) ⎛ KIc ⎞ 2 ⎛ 36 ⎞ 2 (n) ⎟ = 0.0037 m 2. Answer: The stress range should be less than (a) Δσ0 = 79 MPa. 5 MN/m 3 / 2 > ΔKth (c) Case (b): ΔKI = 0.001 m gives the stress intensity range Case (a): π 0. 0⎟⎠ = 0. Solution: For an embedded elliptical crack the stress intensity factor is (Case 8 in Appendix 3.26 MN/m 3 / 2 > ΔKth Thus.826 Δσ∞ √ ⎯⎯π⎯ a = 0.001 m and c0 = 0.001 ⎯⎯π⎯ a = 0. see figure. ratio a / c is constant during the crack propagation).826 Δσ∞ √ ⎯⎯π⎯ a (b) The initial crack length a0 = 0. a << t implies that terms containing a / t may be neglected) ⎛a ⎞ ⎛1 ⎞ KI = σ∞ √ (a) ⎯⎯π⎯ a f8 ⎜⎝ c . Determine the cyclic life (the number of cycles to failure) of the plate if (a) the load varies between 0 and σ∞. (b) the load varies between 0. 0⎟⎠ = σ∞ √ ⎯⎯π⎯ a f8 ⎜⎝ 2.002 m. fracture toughness KIc = 60 MN/m3/2. the crack will grow in both cases. yield limit σY = 1200 MPa. Use safety factor s = 1.6/4.826 ⋅ 200 √ ⎯⎯⎯⎯⎯ ⎯ = 9.826 Δσ∞ √ π 0.4. threshold value ΔKth < 6 MN/m3/2.826 ⋅ 400 √ ⎯⎯⎯⎯⎯ ⎯ = 18. The crack is assumed to keep its form during the crack propagation (i. 2a t 2c A large plate of thickness t contains an embedded elliptical crack. The influence of the mean value of the stress intensity factor is disregarded. Page 6:6 Chapter 6 (d) .. thickness t = 0.10 m.5σ∞ and σ∞. The material follows Paris’ law for crack propagation. and crack propagation parameters n = 4. Numerical data: crack length a0 = 0.0 and C = 5⋅10 − 13 m7/MN4.e.826 σ∞ √ ⎯⎯π⎯ a The stress intensity range ΔKI becomes ΔKI = 0.001 ΔKI = 0. σ∞ = 400 MPa. 826 Δσ √ da = π ) dN ⎯ 0 n ⌡a0 (√ ⌡ 0 ⎯a ) N= Thus acr1 − n / 2 − a01 − n / 2 (1 − n / 2) C (0. may by used) when KIc 60 KImax = π⎯ac = giving 0.0054 − 1 − 0.826 Δσ0 √ ⎯π ) n (h) (i) Case (a) gives: N = (a) 0. determine the critical crack length ac. Chapter 6 Page 6:7 .826 Δσ0 √ ⎯ ⎯ dN (g) Separating and integrating give a N cr 1 n ⌠ ⌠ C (0.001 − 1 ( − 1) 5 ⋅ 10 − 13 (0. Check the conditions for LEFM. approximately.0054 − 1 − 0.4 ⋅ 1200 ⎠ ⎝ s σY⎠ Thus. ac > 0.Next.0054 m at failure (or.5 ⎜ 2.5 ⎜ ⎝ 1. Problem 6/9).001 − 1 ( − 1) 5 ⋅ 10 − 13 = 13 855 cycles (j) = 24 N (a) = 221 680 cycles (k) (0. LEFM.826 ⋅ 200 √ ⎯π ) Thus N(b) = 24 N(a) = 221 600 cycles.826 ⋅ 400 √ (e) ⎯ ⎯ s 1. to be exact. the crack will start to grow rapidly at this crack length and the failure will come after a few more cycles).0032 m ⎟ = 2. One has ⎛ KIc ⎞ 2 ⎛ 60 ⎞ 2 (f) ⎟ = 0.4 which gives ac = 0. and (b) N = 221 600 cycles (cf. 4 Answer: Cyclic life to failure is. Failure will occur (if linear elastic fracture mechanics theory.826 ⋅ 400 √ ⎯π ) 4 Case (b) gives: N (b) = 0. Paris’ law now gives da n πa ) = C (ΔKI)n = C (0.0032 m and LEFM can be used (also t and W − a are large enough). (a) N = 13 850 cycles. The crack is assumed to keep its form during the crack propagation. 12.896 σ∞ √ ⎯ ⎯ dN Separating and integrating give a N final da 3. KIc = 70 MN/m3/2. rewritten. (b) Determine also the stress level that may give unstable crack growth at the different crack depths. t = 0.62.75 logσ∞ + logN = log ⎨ 1 which gives.004 m.75 − N ⎜ 0. σY = 1200 MPa. and 12.875 ⎟ ⎬ 3.10 m.002 m. 15.5. 15 and 20 mm. (a) Determine the relationship between the remote stress σ∞ and the number of loading cycles N for crack propagation to depths a = 5.896 σ∞ √ ⎯⎯π⎯ a Paris’ law then becomes da 3. 1 ⎞⎫ −1 ⎛ 1 − ⎜ 0.75 0.75 π⎯ a ) (b) = C (ΔKI)n = C (0. Solution: Here the stress intensity range becomes (Case 7 in Appendix 3 in the textbook. C = 9.875 a a ⎝ ⎠ final initial C (0. 0⎟⎠ = 0.896 σ π ) dN ⎯ √ ∞ 3. respectively.875 a a ⎝ ⎠⎭ final initial ⎩ C (0. terms containing a / t may be neglected here) ⎛1 ⎞ ΔKI = σ∞ √ (a) ⎯⎯π⎯ a f7 ⎜⎝ 2. 12.64 Page 6:8 Chapter 6 (e) (f) .875 0. The load is cyclic and it varies between 0 and σ∞. The material follows Paris’ crack propagation law. a A large plate of thickness t contains an elliptical surface crack. see figure.875 0. t 2c Numerical data: crack length a0 = 0.875 ⎟ = (σ∞) 3.75 ⌡ainitial (√ ⌡0 ⎯a ) 1 ⎞ −1 ⎛ 1 3.75. ΔKth = 6 MN/m3/2. 10.22⋅10 − 12 m6.75 0.896√ π ) ⎯ Thus 1 (c) (d) or.75 ⌠ ⌠ = C (0.75 logσ∞ + logN = 12. Use safety factor s = 1. and 20 mm. 3.44. 10.6/5. c0 = 0.75. n = 3.57. for a = afinal = 5.896√ π) ⎯ ⎧ 3.625/MN3. 294. thickness t = 0. 240.62.5 KIc = 46. threshold value ΔKth = 4 MN/m3/2. Solution: Between the two forces P. Load amplitude is P0 = 0. 12. fracture toughness KIc = 70 MN/m3/2.05.44. 6/6. 240. The maximum value of the bending moment is P0 l and the minimum value contributing to the stress range is zero (it is assumed that the negative part of the moment will not contribute to the crack growth). and 20 mm.64 for a = 5.10 m. The material follows Paris’ crack propagation law. Use safety factor s = 1.75log σ∞ + log N = 12. Chapter 6 Page 6:9 . the range of the bending moment contributing to the stress range is P0 l.005 m.896 σ∞ √ (g) ⎯⎯π⎯ a = 1.(b) The critical stress at the different crack lengths is. which gives f6(0. and 12.01 m.57. 10. With (g). 10. and (b) σ∞ = 416. see figure. and 208 MPa (i) Answer: (a) 3. The stress intensity factor becomes (Case 6 in Appendix 3 in the textbook) ⎛a⎞ 6Pl ⎜ ⎟ KI = π a f (a) ⎯ √ ⎯ ⎯ 6 ⎝W⎠ t W2 For crack length a = a0 = 0. respectively.005 m one has a / W = a0 / h = 1/20 = 0. 294. Determine a t lower limit of the cyclic life of the beam h (assume that no other cracks are present). the bending moment in the beam is constant and it is Pl. 12. (and if LEFM can be used) 1 KImax = ( ΔKI = ) 0. from (a). height h = 0.05) = 1. l P P 10 l a l A beam with an edge crack is loaded with two forces P = P0 sinωt. respectively. σ∞ = 416. respectively. the four crack lengths 5. and 20 mm give. yield limit σY = 1400 MPa.08. length l = 0.7 MN/m3 / 2 Check for LEFM.5 ⎜ (h) ⎟ = 0.10 m. Thus.5. 15. and crack propagation parameters n = 4 and C = 1⋅10 − 13 m7/MN4. 15. One has ⎛ KIc ⎞ 2 2.042 MN.00378 m ⎝ s ⋅ σY⎠ implying that one may use LEFM. and 208 MPa. Numerical data: crack depth a0 = 0. to be on the safe side. Critical crack length is acrit = 0. Page 6:10 Chapter 6 .06. it is expected that N will be slightly more than 17 900 cycles. Therefore. the critical crack length ac = 0.5 0. As we have used a too large value of f6 this will be a lower limit of the cyclic life of the structure.0097 m. This value of ac gives f6(ac / W) = f6(0.05).0097 ⎛ 6 P0 l ⎞4 da ⌠ =N ⋅C ⎜ (f) ⎯ π 1.08 and 1. Paris’ law now gives 4 ⎛ 6 P0 l da ⎛ a ⎞⎞ n (d) = C (ΔKI) = C ⎜ ⎯⎯π⎯ a f6⎜⎝ W ⎟⎠⎟ √ dN ⎝ t W2 ⎠ Again.e.08/1.102 Solving for ac gives ac = 0. The cyclic life is expected to be larger than the life calculated. for stress and fracture toughness expressed in MPa. i.01 ⋅ 0. see equation (f).1 70 π a 1. Then one obtains. Thus.042 ⋅ 0.0097 m. One obtains 6 P0 l ⎛ ac ⎞ KIc ⎜ ⎟= KI = π a f c ⎯ √ ⎯ ⎯ 6 ⎝W⎠ s t W2 (b) Here f6(ac / W) is unknown so one has to try (to guess) a value of f6. which is the same as the value assumed. select f6 = 1. 6 ⋅ 0.06 = (c) KI = c ⎯ √ ⎯ ⎯ 1.06 to see what happens (we see from the diagram that f6 decreases in the region where a / W = 0. we have the problem of not knowing the value of the function f6 = f6(a) when the crack grows. Now equation (d) will give (with stresses in MPa.005 a 2 ⎝ t W2 ⎠ from which N = 17 900 is solved.097) = 1. From the above calculations (and from the form of the function f6 seen in the diagram) it is concluded that f6 varies between 1. when a changes.06 is used. It gives.0097 is selected for the following calculations.08⎟ dN √ a2 ⎝ t W2 ⎠ or 0. Therefore. as it will give a too large crack propagation rate).) Answer: The fatigue life is expected to be slightly more than N = 17 900 cycles (but less than 19290 cycles).08 for the whole interval (this is a conservative estimation.08⎟ √ ⌡0.Determine the crack length ac at failure. Try f6(ac / W) = 1. Nupper = 17 900 (1. (An upper limit of the fatigue life is estimated if f6 = 1. see the dimension of the constant C) ⎛ 6 P0 l ⎞4 da =C ⎜ (e) ⎯ π 1.06 in the crack length interval obtained here.06)4 = 19 290 cycles. The plate is loaded in cyclic tension with σmin = 0. this is an approximation of function f3 in Appendix 3 of the textbook.6/7. It is assumed that the crack propagation is symmetric with respect to the hole.22175 m a = acr = ⎜ ⎟ =⎜ ⎯ π 60 ⋅ 0.42 KI = σ0√ = 1. σY = 600 MPa.081 / 6 ⎠ ⎝ s 1. so LEFM may be used. The geometry function f used to determine the stress intensity factor may in this example. The critical crack length is acr as calculated in (b). see figure. a 2r a t In a large plate of a ship’s hull two cracks of approximately the same size were discovered at a hole. Determine the remaining life (in hours) of the plate. Use safety factor s = 1.42 ⎜ ⎟ ⎝r⎠ ⎝r⎠ (Thus. and W − a are larger than 0.08 m. approximately.) Numerical data: r = a0 = t = 0.5 ⎜ ⎟ = 0.5.5 ⋅ 1.) Determine the critical crack length acr. The material follows Paris’ law for crack propagation. The loading frequency is 6 cycles per minute. all three measures a. Solution: In this example.025 m. ⎛ KIc ⎞3 ⎛ ⎞3 90 1 ⎟ = 0. Chapter 6 Page 6:11 . Equation (a) gives. and for the crack length used here. KIc = 90 MN/m3/2. ⎛a⎞ 1. t. the stress intensity factor may be written. n = 3.42√ ⎯ π σ∞ r 1 / 6 ⎠ ⎝ 1.42√ (a) ⎯⎯πa f ⎜⎝ r ⎟⎠ = σ0√ ⎯⎯πa ⎯ π σ0 a 1 / 3r 1 / 6 1/6 (a/r) (Note that the function f given here is an approximation of the function f3 given in Appendix 3 in the textbook. C = 1⋅10 − 11 m11/2/MN3.2σ∞ and σmax = σ∞ = 60 MPa. be approximated to ⎛a⎞ ⎛a⎞ −1/6 f ⎜ ⎟ = 1. by use of KImax = KIc / s.025 m ⎝ s ⋅ σY⎠ Yes.42√ (b) May linear elastic fracture mechanics (LEFM) be used here? One has ⎛ KIc ⎞ 2 (c) 2. ΔKth = 6 MN/m3/2. n = 4. Numerical data: a0 = 0.8σ∞ a 1 / 3 r 1 / 6 (d) da 3 π 0. The load t consists of a repeated sequence of two cycles as shown in the figure.90 Δσ0√ ⎯⎯π a Page 6:12 Chapter 6 (b) . KIc = 70 MN/m3/2. 0⎟⎠ = σ0√ ⎯⎯π a 0. It is assumed that the crack keeps its form during the crack propagation. σ1 = 200 MPa and σ2 = 400 MPa .8σ∞ a 1 / 3 r 1 / 6 ) = C (ΔKI)n = C (1. 6/8. ⎛a ⎞ KI = σ0√ (a) ⎯⎯π a f7 ⎜⎝ c .42√ π 0.002 m. Solution: A half-elliptical surface crack (Case 7 in Appendix 3 in the textbook) gives the stress intensity factor. c0 = 0.896 = 0. σY = 1200 MPa.8σ∞ r 1 / 6 ) a0 (g) 1 Remaining fatigue life is t = N / (6⋅ 60) = 568 h.8σ∞ r 1 / 6 ) dN ⎯ ⌡a0 a ⌡0 (f) acr ln = 204 400 cycles 3 C (1.004 m. Determine the expected 2c remaining life of the plate expressed in number of sequences to failure.The stress intensity range becomes Δσ∞ = σmax − σmin = 0.42√ ⎯ π 0.5.90 σ0√ ⎯⎯π a The range of the stress intensity factor becomes ΔKI = 0. time Use safety factor s = 1. for a / c = 0.42√ ⎯ π 0.8σ∞. Answer: Fatigue life N = 204 000 cycles gives that the remaining life is expected to be 568 hours. see figure. a 2 1 A large plate of thickness t contains a halfelliptical surface crack.42√ ⎯ dN (e) Paris’ law gives Thus a giving N= N cr 3 ⌠ da = ⌠ C (1. t = 0. The material follows Paris’ law. C = 1⋅10 − 13 m7/MN4.4. which gives ΔKI = 1.100 m. 0061 m ⎟ = ⎜ π ⎝ s ⋅ 0. It is now known that the crack will grow from a0 to acrit.90 ⋅ 200 √ π ⋅ 0.90 σ2 √ π⋅a ) ⎯⎯⎯ ⎯⎯⎯ (g) ⎬ ⎭ Note that Ns stands for the number of sequences (not number of cycles). Thus. linear elastic fracture mechanics can be used. the crack propagation will be da n = C ∑ (ΔKIi )n = C ∑ (0. and thereby also at the higher stress level σ2. For σ0 = σ2 one obtains from (d) (if linear elastic fracture mechanics is valid) KIc ⎞2 1 ⎛ 1⎛ 70 ⎞2 acrit = ⎜ ⎟ = 0. t. i. giving ΔKI > ΔKth.90 σ2√ ⎯ π ) ⎭ ⌡0 dNs n ⌡a0 (√ ⎯a ) Chapter 6 (h) Page 6:13 .90 ⋅ 400 ⎠ (e) Is linear elastic fracture mechanics valid? ⎛ KIc ⎞ 2 ⎛ 70 ⎞ 2 2.4 ⋅ 1200 ⎠ ⎝ s ⋅ σY⎠ which is less than acrit.3 MN/m3 / 2 ⎯⎯⎯⎯⎯⎯ (c) The threshold value is ΔKth = 6 MN/m3/2. Separation of variables gives N a crit s da ⎧ n n⎫ ⌠ ⌠ ⎨ ⎬ = C ⎩ (0.Will the lower stress level σ1 = 200 MPa contribute to the crack growth? Let a = a0 = 0.5 ⎜ (f) ⎟ = 0.90 ⋅ σ2⎠ π ⎝ 1. Fracture will then occur when KImax = KIc / s. and W − a. Per sequence of two cycles. and both cycles in the sequence will contribute to the crack propagation.90 Δσ0i √ π⋅a ) ⎯ ⎯⎯ dNs ⎧ n n⎫ = C ⎨⎩ (0.4 ⋅ 0.e. This implies that the crack will propagate at the lower stress cycle σ1.5 ⎜ ⎝ 1.002 = 14. It gives KIc KImax = 0.0043 m ⎟ = 2. Determine next the crack length acrit at fracture. Assume that linear elastic fracture mechanics (LEFM) is valid. One then obtains from (b) ΔKI = 0.002 m and Δσ0 = σ1 = 200 MPa.90 σ1 √ π ⋅ a ) + (0.90 σ0√ π ⋅ acrit = (d) ⎯⎯⎯⎯ s Fracture will occur when the highest stress is loading the structure.90 σ1√ ⎯ π ) + (0. when the stress is σ2. γ = 0.826 σ∞ √ ⎯⎯π⎯ a The stress intensity range ΔKI becomes ΔKI = 0.e. σY = 1200 MPa. 6/9. Numerical data: crack length a0 = 0.4..⎛ n⎞ n 1 − n /2 acrit − a01 − n /2 = ⎜1 − ⎟ C (0. ΔKth < 6 MN/m3/2. Solution: Following the solution to Problem 6/4.5σ∞ and σ∞.90 √ π ) { σn1 + σn2 } ⋅ Ns ⎯ ⎝ 2⎠ which gives (i) Entering acrit.826 Δσ∞ √ ⎯⎯π⎯ a Page 6:14 Chapter 6 (b) .002 m. and then solving for Ns. give (j) Ns = 19 080 sequences Answer: Expected fatigue life is Ns = 19 000 sequences.10 m. The influence of the mean value of the stress intensity factor has to be taken into account. 0⎟⎠ = 0. t ⎟⎠ = σ∞ √ ⎯⎯π⎯ a f8 ⎜⎝ 2. c0 = 0.7. and the material parameters C and n in (i). Determine the cyclic life (the number of cycles to failure) of the plate if the load varies between 0. (This is Problem 6/4(b) once again. The crack is assumed to keep its form during the crack propagation (i. n = 4. t = 0. a0. Use safety factor s = 1. see figure.0 and C1 = 5⋅10 − 13 m7/MN4. σ∞ = 400 MPa (it is assumed that the material parameters C and n in Paris’ law are the same here as in Problem 6/4). σ1. σ2. KIc = 60 MN/m3/2. giving N = 38 000 cycles to failure.001 m. one obtains the stress intensity factor for an embedded elliptical crack as ⎛a a⎞ ⎛1 ⎞ KI = σ∞ √ (a) ⎯⎯π⎯ a f8 ⎜⎝ c . 2a t 2c A large plate of thickness t contains an embedded elliptical crack.) The material follows Paris’ law for crack propagation. ratio a / c is constant during the crack growth). but now the influence of the mean value is taken into account. to be compared with N = 221 700 cycles in Problem 6/4. These two numbers differ by the factor (1 − R)n (1 − γ ) = 0.26 MN/m 3 / 2 > ΔKth (c) Thus. Check the conditions for LEFM.001 m gives the stress intensity range ΔKI = 0.001 ⎯⎯π⎯ a = 0. Answer: Expected fatigue life is N = 96 500 cycles.5 ⎜ ⎝ 1. Separating and integrating give N= giving Thus. N = (a) a N cr 1 n ⌠ ⌠ da = C (0.826 ⋅ 400 √ MN/m 3 / 2 ⎯ ⎯ s 1. The factor C becomes C = C1 / (1 − R)n (1 − γ ) = 1.826 ⋅ 200 √ ⎯π ) 4 = 96 500 cycles (i) The expected fatigue life now becomes N = 96 500 cycles.0032 m ⎟ = 2.4 which gives ac = 0. the crack will start to grow rapidly at this crack length.826 Δσ0 √ πa ) (f) = ⎯ ⎯ n(1 − γ) dN (1 − R) where stress ratio R is R = 0. Failure will occur (if linear elastic fracture mechanics theory. may by used) when KIc 60 KImax = π⎯ac = (d) giving 0. to be exact. Paris’ law.826 Δσ0 √ ⎯π ) n 0.1487⋅10 − 12 m7/MN4. Chapter 6 Page 6:15 .001 − 1 ( − 1) 1. One has ⎛ KIc ⎞ 2 ⎛ 60 ⎞ 2 (e) ⎟ = 0.826 Δσ π ) dN (g) ⎯ √ 0 n ⌡a0 (√ ⌡0 ⎯a ) acr1 − n/2 − a01 − n/2 (h) (1 − n/2) C (0. the crack will grow at the stress range given.4 ⋅ 1200 ⎠ ⎝ s σY⎠ Thus. with correction for the mean stress value according to Walker. ac > 0.435.5 ⎜ 2.5. LEFM.0032 m and LEFM can be used (also t and W − a are large enough). now gives C1 da n (ΔKI)n = C (0.826 Δσ∞ √ π 0.0054 − 1 − 0. which means that in this case the mean value plays an important role for the fatigue life of the structure.1487 ⋅ 10 − 12 (0.826 ⋅ 200 √ ⎯⎯⎯⎯⎯ ⎯ = 9.0054 m at failure (or. Determine the critical crack length ac. and the failure will come after a very limited number of loading cycles after this crack length has been reached).The initial crack length a0 = 0. say ai = 1.5⋅10 − 12 (units in MN and meter) and n = 3. Solution: First investigate if crack propagation is obtained for the shortest crack length ai = 1 mm. use that W >> a) ⎛a⎞ ΔKI = Δσ0 √ (a) ⎯⎯π⎯ a f5 ⎜⎝ W ⎟⎠ = 1. for crack length ai = 1 mm.28 MN/m / 2 (b) ⎯⎯π⎯ a f5 ⎜⎝ W ⎟⎠ = 1. 60.001 = 6. taking the closeness to the threshold value into account: da = C {(ΔKI)n − (ΔKth)n } dN Compare the results. It is concluded that the crack will grow for all crack lengths ai. The stress range Δσ0 is Δσ0 = 100 MPa. Assume that the constants C and n are the same in the two formulae: C = 2. 40. 10. The material has the fatigue threshold value ΔKth = 6 MN/m3/2. Separating and solving for N give Page 6:16 Chapter 6 (c) . 0 thickness t a W Determine the number of loading cycles required to make the crack in the figure grow by 1 mm. Use some different initial crack W >> a lengths ai.12 ⋅ 100 √ ⎯⎯⎯⎯⎯⎯ which is larger than the threshold value ΔKth = 6 MN/m3/2. Paris’ law gives da = C (ΔKI)n dN where ΔKI is obtained from (a). For crack growth calculations use both the un-modified Paris law: da = C (ΔKI)n dN and the modified version.12 Δσ0 √ ⎯⎯π⎯ a The stress range Δσ0 = 100 MPa gives.6/10. The structure is loaded with stress cycles of constant amplitude. ⎛a⎞ ΔKI = Δσ0 √ π ⋅ 0. 20. and 80 mm.2. The intensity range for the component with the edge crack is (Case 5 in Appendix 3 in the textbook. 5. 2 (d) 1.12 ⋅ 100 √ ⎯π ) a The modified crack propagation law gives a + 0.9 28.73⋅10 6 cycles. since the crack growth rate tends to zero at the threshold value. ai (m) 0. whereas the modified version predicts 2.N =N (d) a + 0.1 6. It is noted that the fracture toughness KIc of the material must Chapter 6 Page 6:17 . Already at the crack length 5 mm the difference in crack growth rate between the two formulae is small. Table: Number of loading cycles required to make the crack grow 1 mm for different initial lengths ai of the crack. The fatigue life N(d) is calculated with the un-modified Paris law whereas N(e) is calculated with the modified crack propagation law given in the problem.1 In the table is also given the stress intensity range ΔKI(ai) for the initial length ai of the crack. It is noted that only when the stress intensity range is very close to the threshold value there is a large difference between the two results (about one million cycles). only a few per cent.020 0.63⋅10 6 73 628 26 056 8 923 3 001 1 579 1 000 N(e) (cycles) 1. The main part of this difference (one million cycles) depends on the slower crack growth the first millimetre.0 0. whereas the unmodified Paris law predicts a finite crack growth rate also at the threshold value.010 19. A difference was expected. according to the unmodified Paris law. It is seen that the two formulae give different results only when the stress intensity range ΔKI(ai) is very close to the threshold value ΔKth (here ΔKth = 6 MN/m3/2). Making the crack grow from 1 mm to 81 mm would.27 0.001 1⌠ i N =N = C ⌡ai (e) 3.6 56. require 1.7 48.89⋅10 6 cycles.68⋅10 6 78 105 26 592 8 985 3 008 1 581 1 000 ΔKI(ai) (MN/m3/2) 39.005 14.12 ⋅ 100 √ ⎯ π ) a 1.6 da (e) (1.080 N(d) (cycles) 0.040 0.060 0.6 − (ΔKth)3.2 Numerical solution (for example by use of MATLAB) of (d) and (e) gives results according to the table below.001 da i 1 = ⌠ C ⌡ai (1.001 0.2 3. t = 0. (a) Determine the expected number of cycles (twice the number of sequences) required to make the crack grow from a0 to afinal.5) ⎛a⎞ KI = σ0√ (a) ⎯⎯π a f7 ⎜⎝ c ⎟⎠ = 0.006 m before the time crack has to be repaired. σY = 1200 MPa. ΔKrms) were used as an equivalent measure of the stress intensity range? Numerical data: crack depth a0 = 0.002 m.63⋅10 6. 26 600.5 MN/m3/2. n = 4.002 m. The crack is assumed to keep its form during the crack propagation.004 m.e. Use Paris’ law to determine the crack propagation “cycle by cycle”.be so large that after reduction with the safety factor s the fracture toughness should be larger than ΔKI(ai + 0. and the modified formula gives N = 1. respectively. 1000 cycles. Thus. 8 900. KIc = 70 MN/m3/2.100 m.896 σ0√ ⎯⎯π a The range of the stress intensity factor becomes ΔKI = 0. 1 580. σ1 = 200 MPa and σ2 = 400 MPa. a 2 1 A large plate of thickness t contains a halfelliptical surface crack. 9 000.68⋅10 6. The load consists of a repeated sequence of two cycles as t shown in the figure. 26 060. The 2c material follows Paris’ law.001) = ΔKI(0. 3 000.5 MN/m3/2.896 Δσ0√ ⎯⎯π a Page 6:18 Chapter 6 (b) . KIc must be such that KIc > s⋅56. (b) What number of cycles would be obtained if the root mean square value (i. 1 580. Answer: The un-modified propagation law gives N = 0. 6/11. see figure. c0 = 0. Solution: A half-elliptical surface crack (Case 7 in Appendix 3 in the textbook) gives the stress intensity factor (for a / c = 0. 3 000. The crack is detected when the crack length a0 is a0 = 0.081) = 56. 73 600. 78 100. 1000 cycles. C = 1⋅10 − 13 m7/MN4. It is decided that the crack can be allowed to grow to the length afinal = 0. ΔKth = 6 MN/m3/2. per loading sequence the crack growth will be da n = C ∑(ΔKI i )n = C ∑(0.896 √ π ) { σn1 + σn2 } ⋅ Ns afinal ⎯ ⎝ 2⎠ (g) Entering afinal.896 σ2√ π⋅a ) ⎯⎯⎯ ⎯⎯⎯ (e) ⎬ ⎭ Note that Ns here stands for the number of sequences (not number of cycles).896 σ1⎯ dNs √π ) + (0.006 = 49. σ1. This implies that the crack will propagate for the lower stress cycle σ1.896 Δσ0 i √ π⋅a ) ⎯ ⎯⎯ dNs ⎧ n⎫ n = C ⎨⎩ (0. a0.42). and the material parameters C and n in (g). give Ns = 19 265 sequences. One obtains KI = 0. One obtains ΔKI = 0. the stress intensity factor is KI = 49. and then solving for Ns.2 MN/m 3 / 2 ⎯⎯⎯⎯⎯⎯ (d) Thus.896 σ1√ π ⋅ a ) + (0. at the planned repair of the crack. σ2. Both cycles in the sequence will contribute to the crack growth. giving N = 38 500 cycles (h) Chapter 6 Page 6:19 .896 ⋅ 200 √ π ⋅ 0. giving ΔKI > ΔKth.2 MN/m 3 / 2 ⎯⎯⎯⎯⎯⎯ (c) The threshold value is ΔKth = 6 MN/m3/2. Then.002 m and Δσ0 = σ1 = 200 MPa.2 = 1. Separation of variables gives afinal ⌠ ⌡a 0 da Ns = C (0. Check the stress intensity factor for the higher stress σ2 and crack length afinal = 6 mm.002 = 14.896 σ2⎯ √π ) ⌠ n ⌡ 0 (⎯ a ) √ ⎧ ⎨ ⎩ n n⎫ ⎬ ⎭ (f) which gives ⎛ n⎞ n 1−n /2 − a01 − n / 2 = ⎜1 − ⎟ C (0. It is given in the problem that the crack will grow from a0 to afinal. which is 70 per cent of the critical value KIc = 70 MN/m3/2 (the “safety factor” is s = 70 / 49. and thereby also for the higher stress cycle σ2.Will the lower stress level σ1 = 200 MPa contribute to the crack growth? Let a = a0 = 0.896 ⋅ 400 √ π ⋅ 0.2 MN/m3/2. This crack length gives the stress intensity factor. (i) Paris’ law then gives ⎛ n⎞ n ⎧ n⎫ 1 − n /2 afinal − a01 − n /2 = ⎜1 − ⎟ C (0.896√ ⎯⎯π⎯ a ∑ ni ⎯⎯⎯⎯ √ 2002 + 4002 2 = 0.(b) Using the rms value as an equivalent stress intensity range.896 σ2√ ⎯⎯π a = 0.9 MN/m 3 / 2 KI = 0. the equivalent stress range becomes Δσeq(rms) = 316. During the 52 400 − 38 500 ≈ 13 900 extra cycles.23 MN/m 3 / 2 Thus. One obtains ⎛ n⎞ n 1 − n /2 − a01 − n /2 = ⎜1 − ⎟ C (0. one obtains ΔKrms = ⎯⎯⎯⎯ √ ∑{ (ΔKi )2 ni } = 0. Comment: The importance (influence) of the high amplitude loading cycles (here σ2) is diminished when the rms value is used.896 √ π) { σn1 + σn2 } ⋅ 26 200 (k) afinal ⎯ ⎝ 2⎠ giving afinal = 0.896√ ⎯⎯π⎯ a ⋅316. use ⎧ ∑{ (ΔKIi )n ni } ⎫ 1 / n ⎬ ΔKI equivalent = ⎨ ∑ ni ⎩ ⎭ (m) where n is the factor in Paris’ law (and ni is number of cycles at each stress level). (c) Extra problem: Use N = 52 400 cycles (26 200 sequences) to see what crack length would have been obtained if Paris’ law were used as in problem (a). One then obtains the equivalent stress intensity range Page 6:20 Chapter 6 . Conclusion: The conclusion is that the rms value can not be used for crack propagation calculations.23 MPa. π ⋅ 0.92 √ π) ⎨⎩ (Δσ(rms) (j) ) ⎬⎭ ⋅ N ⎯ eq ⎝ 2⎠ giving N = 52 400 cycles.896 ⋅ 400 √ ⎯⎯⎯⎯⎯⎯⎯ (l) which is much larger than the fracture toughness KIc of the material.4 mm.0214 m = 21. Thus. the rms value here predicts 52 400 / 38 500 = 1. Instead.0214 = 92. the crack will grow (far) beyond the 6 mm limit where the crack should have been repaired. at stress σ2. The structure would fail long before the 52 400 cycles had been applied.36 times too many cycles. failure would occur during the loading if the rms value were used in the calculations. 6/12. The irregular loading sequence of a machine component has been analysed and reduced to the 60 stress range cycles given in the table below. 52 400 cycles are obtained.896√ ⎯⎯π⎯ a 341. The maximum permissible crack length in the component is a = 5 mm. as it should. (o) Answer: (a) 38 500 cycles will propagate the crack to length 6 mm.ΔK I equivalent ⎧ ∑{ (ΔKIi )n ni } ⎫ 1 / n ⎬ =⎨ = 0. Thus.54 } ⋅ 38 500 afinal ⎯ ⎝ 2⎠ which gives afinal = 0. That many cycles would propagate the crack to a length that would be longer than the critical value of the crack length. Upon inspection of the component. edge cracks of depth a = 1 mm (or longer) will be detected and eliminated. see figure.006 m. Stress range Δσ0 (MPa) 285 270 200 92 Number of cycles N 3 12 44 0 thickness t a W 1 Determine the maximum number of loading sequences the component may be loaded with between two inspections.896√ ⎯⎯π⎯ a ∑ ni ⎩ ⎭ ⎧ 2004 + 4004 ⎫ 1 / 4 ⎨ ⎬ ⎩ ⎭ 2 3/2 = 0. W >> a The material has the fatigue threshold value ΔKth = 8 MN/m3/2.5 MN/m (n) This value of the stress intensity range gives ⎛ n⎞ n 1 − n /2 − a01 − n /2 = ⎜1 − ⎟ C (0. Chapter 6 Page 6:21 . (b) using the rms value of the stress intensity range. Assume that cracks of length 1 mm may remain in the structure after an inspection. which is 36 per cent too many cycles.896 √ π ) { 341. 001 = 12.5⋅10 − 12 (units in MN and meter) and n = 3.92 mm.92 mm all cycles (all four stress ranges) will be damaging to the structure (i. When the crack has reached 1.92 mm. use that W >> a) ⎛a⎞ ΔKI = Δσ0 √ (a) ⎯⎯π⎯ a f5 ⎜⎝ W ⎟⎠ = 1.77 MN/m / 2 (b) ΔKI = Δσ0 √ ⎯⎯π⎯ a f5 ⎜⎝ W ⎟⎠ = 1.12 ⋅ 92 √ ⎯⎯⎯⎯⎯⎯ which is below the threshold value ΔKth = 8 MN/m3/2.2.12 Δσ0 √ ⎯⎯π⎯ a The stress range Δσ0 = 92 MPa gives ⎛a⎞ π ⋅ 0.92 mm. give crack propagation). only the cycles with stress range 200 MPa and larger will make that the crack grows. One obtains ⎛a⎞ π ⋅ 0.12 ⋅ 200 √ ⎯⎯⎯⎯⎯⎯ This cycle will induce crack propagation at all crack lengths. and then the growth from 1. One has ⎛a⎞ π⎯ ⋅ a1 = 8 MN/m / 2 (d) ΔKI = Δσ0 √ ⎯⎯π⎯ a f5 ⎜⎝ W ⎟⎠ = 1. Paris’ law gives Page 6:22 da = C (ΔKI)n dN Chapter 6 (e) . as long as the crack is shorter than 1. The stress intensity range for the component with the edge crack is (Case 5 in Appendix 3 in the textbook.e.For crack growth calculations Paris’ law may be used: da = C (ΔKI)n dN where C = 2. Solution: First investigate if all stress cycles will contribute to the crack propagation. It is concluded that at least initially the lower stress range cycles Δσ0 = 92 MPa do not contribute to the crack propagation. The crack growth must be calculated i two steps: first the growth from 1 mm to 1. (c) Determine at which crack length the stress range cycles Δσ0 = 92 MPa will start to contribute to the crack propagation.6 MN/m / 2 ΔKI = Δσ0 √ ⎯⎯π⎯ a f5 ⎜⎝ W ⎟⎠ = 1.12 ⋅ 92 √ ⎯⎯⎯ giving a1 = 1. Check also for the next stress range cycle: Δσ0 = 200 MPa.92 mm to 5 mm. Thus.001 = 5. 12√ π ) {1 ⋅ 285n + 3 ⋅ 270n + 12 ⋅ 200n } Ns ⎯ 1.12√ π ) {1 ⋅ 285n + 3 ⋅ 270n + 12 ⋅ 200n } (√ a) ⎯ ⎯ dNs (f) Separating and integrating give 0.12 Δσ0 √ π ⋅ 0. This gives the stress intensity range ΔKI = 1.6 ⌡0. In total. Paris’ law has been used for crack length a = 5 mm. Chapter 6 Page 6:23 . 2867 + 2254 = 5121 sequences.00192 da n ⌠ = C (1.005 = 40 MN/m3 / 2 ⎯⎯π⎯ a = 1.12 π ) {1 ⋅ 285n + 3 ⋅ 270n + 12 ⋅ 200n + 44 ⋅ 92n } Ns ⎯ √ 1.6 ⌡0.Using notation Ns for number of loading sequences.12 ⋅ 285 √ ⎯⎯⎯⎯⎯⎯ (j) It is concluded that after reduction with a safety factor s (s > 1) the fracture toughness KIc of the material must be at least 40 MN/m3/2.92 mm. giving N = N1 = 2867⋅60 = 172 000 cycles (of these 172 000 cycles 2867⋅44 = 126 100 cycles do not contribute to the crack growth). all stress range cycles contribute to the crack propagation. one must have KIc > s⋅40 MN/m3/2.001 a (g) Solving for Ns gives Ns = Ns1 = 2867 sequences. When the crack has reached length 1.12√ ⎯ π ) {1 ⋅ 285n + 3 ⋅ 270n + 12 ⋅ 200n + 44 ⋅ 92n } (√ ⎯a ) Separating and integrating give 0. giving 307 000 cycles may be allowed between two inspections. One obtains da = C ∑(ΔKI i )n dNs n n = C (1. one obtains da n n = C ∑(ΔKI i )n = C (1. The largest stress range cycle is Δσ0 = 285 MPa. Alternative solution: Solve the problem by use of the equivalent stress range Δσeq.00192 a (h) (i) Solving for Ns now gives Ns = 2254 sequences (giving N = N2 = 2254⋅60 = 135 240 cycles). Thus.005 da n ⌠ = C (1. Dividing N by 60 (the number of damaging cycles in a sequence) gives the number of sequences Ns to make the crack grow from length 1.2⎫ 1 / 3.00192 da n ⌠ = C (1.92 mm to length 5 mm.005 da n ⌠ = C (1. Dividing N by 16 (the number of damaging cycles in a sequence) gives the number of sequences Ns to make the crack grow from length 1 mm to length 1.2⎫ 1 / 3. One obtains Ns = Ns1 = 45 867 / 16 = 2867 sequences. For crack length a < 1.For crack length less than 1.12 Δσequivalent √ ⎯ ⎯ dN Separating and integrating give 0.2 + 12 ⋅ 2003. giving 307 000 cycles.00192 (√ ⎯a ) (n) (o) Solving for N gives N = 135 247 cycles.92 mm.85 MPa Paris’ law gives (k) da n πa ) = C (ΔKI)n = C (1. the equivalent stress range becomes Δσequivalent ⎧ ∑{ (Δσ0i )n ni } ⎫ 1 / n ⎬ =⎨ ∑ ni ⎩ ⎭ ⎧ 1 ⋅ 2853.92 mm.001 (√ ⎯a ) (l) (m) Solving for N gives N = 45 867 cycles.2 + 3 ⋅ 2703. which is in agreement with the result obtained above. Ns = 2867 + 2254 = 5121 sequences may be allowed between two inspections. which is in agreement with the result obtained above. Page 6:24 Chapter 6 . the equivalent stress range becomes Δσequivalent ⎧ ∑{ (Δσ0i )n ni } ⎫ 1 / n ⎧ 1 ⋅ 2853.85 √ π) ⋅N ⎯ n ⌡0.2 ⎬ =⎨ = 155.13 π ) ⋅N ⎯ √ n ⌡0.2 + 3 ⋅ 2703.13 MPa ⎩ 60 ⎭ Separating and integrating Paris’ law give 0.12 ⋅ 155.92 mm the stress range cycle Δσ0 = 92 MPa will not contribute to the crack propagation.2 ⎬ ⎬ =⎨ =⎨ ⎩ ⎭ ∑ ni 16 ⎩ ⎭ = 223. For crack length larger than 1.2 + 44 ⋅ 923.92 mm. Answer: In total.2 + 12 ⋅ 2003.12 ⋅ 223. One obtains Ns = Ns2 = 135 247 / 60 = 2254 sequences. Solutions to problems in T Dahlberg and A Ekberg: Failure. Lund 2002. Fatigue . Fracture. Studentlitteratur. ISBN 91-44-02096-1.An Introduction.DOC/2010-01-29 Chapter 7 Page 7:1 . Chapter 7 An introduction to fatigue No problems in this chapter FMCHAP7P. Empty page. Page 7:2 Chapter 7 . according to this relationship. The machine is operating 7000 hours per year during 20 years. Lund 2002. Studentlitteratur. A generator is rotating with a speed of 3000 rpm (revolutions per minute).An Introduction. Determine the number of loading cycles N the axle is subjected to due to its own weight.Solutions to problems in T Dahlberg and A Ekberg: Failure.468 σU ± 50 MPa and σFLP ≅ 0. be expected to fall in the range 330 MPa to 430 MPa. Fracture.85 σFL were found. Solution: 3000 revolutions per minute give 3000 loading cycles per minute. 8/2.. for alternating loading: σFL ≅ 0. ISBN 91-44-02096-1. Chapter 8 Stress-based fatigue design Problems with solutions 8/1.468σU ± 50 MPa gives.e. i.2 ⋅ 109 cycles (thus. Use these relations and other “thumb rules” to estimate the fatigue limits (alternating and pulsating) of the material.468⋅810 ± 50 MPa = 380 ± 50 MPa. At an investigation of a similar material the relationships σFL = 0. Answer: Number N of loading cycles is N = 25. Solution: The relationship σFL = 0. giving 3000 ⋅ 60 ⋅ 7000 ⋅ 20 = 25. Estimate the fatigue limit in tension/compression for alternating and pulsating loading of a steel (approximately SS 2225) with the ultimate strength σU = 810 MPa. the fatigue limit σFL in tension/compression at alternating loading may.2 ⋅ 109 loading cycler per 20 years. Fatigue .DOC/2010-01-29/TD Chapter 8 Page 8:1 . It gives 3000 ⋅ 60 ⋅ 7000 loading cycles per year. Sketch the Haigh diagram. it may be exposed to fatigue). For pulsating loading the relationship between pulsating and alternating FMCHAP8P. using σFL ≅ 0. The bar is subjected to a rotating bending moment. Thus. σFLP ≅ 0. approximately. select.45σU yields N ≥ 106 cycles. At an investigation of a similar material the following fatigue lives were found: σFLR ≅ 0. FLR / U 1. -04 and -05 gives.90σU yields N = 1000 cycles. The most unfavourable case.85 (380 ± 50) MPa = 320 ± 42 MPa. fatigue limits will probably be higher than the values estimated here. and σFLR = σU yields N =1 cycle.) Answer: The fatigue limit σFL at alternating loading is estimated to 280 MPa.45 0 1 Page 8:2 Chapter 8 3 6 log N . and the fatigue limit σFLP at pulsating loading is estimated to 240 ± 240 MPa. for rotating bending. Also. compare with other materials.45σU for mild steel to 0. Sketch the Wöhler (SN. Finally. which gives σFLP = 240 ± 240 MPa. Stress-Number) curve for a normalised 37Mn Si5 steel bar with an ultimate strength of σU = 810 MPa. approximately. Solution and answer: See figure.9 0. the fatigue limit in tension at pulsating loading is expected to fall in the range from 280 ± 280 MPa (from 320 − 42 MPa) to 360 ± 360 MPa (from 320 + 42 MPa). that σFLB should be in the range 350 MPa to 460 MPa.38σU gives σFLR = σFLB ≅ 308 MPa. σFLR ≅ 0. for alternating loading σFL = 280 MPa and for pulsating loading σFLP = 0.85σFL. (In practice. while the case σFLR = 0. Another engineering rule says that σFLR (= σFLB) decreases from 0. 8/3.8σFLB gives that σFL should fall in the range 280 MPa to 370 MPa. Material data for SIS 2225-03. gives σFLP = 0. that σFL should be in the range 250 MPa to 290 MPa.0 0.loading. This gives.85σFL. σFLR = 0.8σFLB.45σU gives σFLR = σFLB ≅ 365 MPa. The engineering rule σFL ≅ 0.38σU for the harder materials.85σFL = 0. hopefully “on the safe side”. contain more information than the Haigh diagram. σFLP = 180 ± 180 MPa. Chapter 8 Page 8:3 . Solution: a (MPa) 200 180 m 180 590 m 310 + a (MPa) - 360 310 200 180 180 310 (a) Material data gives σFL = ± 200 MPa. see figure. from which the Haigh diagram can be constructed. (From m now on. 590 MPa (b) The same data gives the Goodman diagram according to the second figure. It does not. maximum value σmax (= σm + σa). (b) Draw the fatigue diagram according to Goodman for the normalised steel SS 1650-01 subjected to loading in pure tension/compression. It is noted that this diagram is more complicated to draw than the Haigh diagram (one has both mean value σm and amplitude σa on the ordinate). and stress ratio 0 0 100 200 300 400 MPa R = σmin /σmax. Determine which one of these five loading variables is constant in the three cases shown in the figure. Each line C corresponds to constant loading conditions. σY = 310 MPa and σU = 590 MPa. The 200 loading varies sinusoidally with mean value σm.) Answer: See figures.310 8/5. In the Haigh diagram given three A straight lines (A. only the Haigh diagram will be 590 MPa used. amplitude σa. 100 m minimum value σmin (= σm − σa). . B and C) are shown.8/4. however. Constant-amplitude fatigue strengths for materials subjected to different cyclic loading conditions are often expressed in Haigh a B diagrams. (a) Draw the fatigue diagram according to Haigh for the normalised steel SS 1650-01 subjected to loading in pure tension/compression.200 . 72⋅κ The reduced fatigue limits σred FL give Case (b) σred. line A gives that stress σmax is constant. One obtains for Case (a) that λ(a)⋅δ(a)⋅κ(a) = 1. stress ratio R is constant. Thus. Page 8:4 Chapter 8 .72 ⋅ κ ⋅ σFL = 0. due to surface finish: (a) κ(a) = κ.72. and (b) λ(b) ≅ 0.72 1. respectively. Solution: In Case (a). but the same in the two cases. axle of diameter 10 mm. giving σm − σa = 200 MPa = σmin. Thus.Solution: For line A one notices that the sum σm + σa is a constant (= 300 MPa). due to loaded volume: (a) δ(a) = 1. axle of diameter 100 mm. 0. 8/6. the following reduction factors are obtained: due to the size (volume) of the raw material (15 mm and 120 mm.8. Thus. where κ is unknown. and (b) κ(b) = κ. FL Case (a) σred. line B: stress ratio R is constant. The axle diameters are 10 mm and 100 mm. approximately. The loading is rotating bending. For line C one notices that mean stress σm = 200 MPa + σa. FL = 0.0 ⋅ κ ⋅ σFL Answer: The ratio of the two reduced fatigue limits is. Two axles with the same geometry (but of different size) are machined from the normalised steel SS 1650-01 (σU = 590 MPa).0⋅κ.0. and for Case (b) that λ(b)⋅δ(b)⋅κ(b) ≅ 0. line C: stress σmin is constant. and in Case (b). For line B one notices that ratio σa / σm is constant (= 2). respectively): (a) λ(a) = 1.0.9. Answer: Line A: stress σmax is constant. One axle was machined from a raw material with a diameter 15 mm and the other from a raw material with diameter 120 mm. σmin is constant. Estimate the ratio of the reduced fatigue limits of the two axles if they have the same surface finish. and (b) δ(b) ≅ 0. 8/7. for example a polished surface. Solution: Factor κ for reduction of fatigue limit due to surface finish. Assume that the fatigue limits of polished test specimens subjected to alternating loading in tension/compression are given by (ultimate strength is Rm = σU) σFL = Rm 2 σFL = 500 + when Rm is less than 1000 MPa. see Figure 8. where κ = 1/Kr.8⋅250 = 200 0. and Rm − 1000 5 when Rm is greater than 1000 MPa. gives for some values of the surface roughness measure Ra Chapter 8 Page 8:5 . the KTH Handbook.59⋅500 = 295 240 MPa It is concluded that high-strength steels are very sensitive to surface irregularities (surface finish) and notches.68 ⋅ 600 = Machined specimen σFLred = κ ⋅ σFLpol ≅ 0. give Rm = σU = 500 1000 1500 MPa 250 500 600 MPa Polished specimen (κ = 1. Use some different surface roughnesses. and formulae above. a machined surface and a surface with a notch (a standard notch.0) σFLpol = 0.78⋅500 = 390 0. Estimate how the fatigue limits of different materials depend on the ultimate strength.87⋅250 = 220 0.7).4 ⋅ 600 = Notch σFLred = κ ⋅ σFLpol ≅ 410 MPa 0. Sundström (1998). Using the reduction factor Kr. 62⋅250 = 155 0.90 1. The failure probability 0. For the normal distribution function Φ(x) one has Φ(x) = x = 0.1 per cent is obtained at the stress level σ = σmean − 3. It is concluded that high-strength steels are very sensitive to surface roughness and to notches. and the two formulae given in the problem.09⋅6 = 73. The standard deviation s is s = 6 MPa. 8/8. This implies that in mean one plate out of 1000 plates loaded at the stress level 73. Determine the stress level at which the probability of fatigue failure is 0.09 was obtained from the table).33 0.33⋅500 = 167 150 MPa Also here it is noted that the high-strength steels are sensitive to surface irregularities (surface finish) and notches.9999 3.90⋅250 = 225 0.28 0. The spread of the fatigue limit is assumed to have a normal (Gaussian) distribution. Answer: Factor κ for reduction of fatigue limit due to surface finish.5 MPa (where 3.99 2.1 per cent.5 MPa is expected to fail due to fatigue.Rm = σU = 500 1000 1500 MPa 250 500 600 MPa Polished specimen σFLpol = Ra = 10 μm 0.25 ⋅ 600 = σFLred = κ ⋅ σFLpol ≅ 0.09⋅s = 92 − 3.60 ⋅ 600 = σFLred = κ ⋅ σFLpol ≅ 0. Page 8:6 Chapter 8 .72 Solution: The mean value of the fatigue limit (here 92 MPa) gives the failure probability 50 per cent.09 0. give reduced fatigue limits as given in the tables above.50 0 0.999 3. The mean value of the fatigue limit of a welded 4 mm thick aluminium plate has been found to be 92 MPa and the standard deviation of the spread of the fatigue limit measurements is 6 MPa.65⋅500 = 325 Ra = 100 μm 360 MPa 0. 889 9 2.965 3.856 2.621 25 1.885 20 1.006 3. 8/9. and 135 000.353 4. P = 0.1 per cent. The numbers of cycles to failure were 49 000. 110 000. The table gives factors k1 and k2 such that the following kind of statements can be made: At least the proportion P is less than x + k1 s with confidence 0. 61 000. 71 000.393 3. and s is sample standard deviation.546 4.520 3.292 2.99 n k1 k2 k1 k2 k1 k2 6 3.215 2. x is sample mean.492 2. and so on.396 2.241 8 2.746 4. Determine the allowable number of cycles at the stress level given. The table below gives the “k-factors” for different values of n and P and with confidence C = 0.390 3.068 2.760 3.062 5.454 2.158 3.95. if the probability of failure is not allowed to exceed 1 per cent with confidence C = 0. At least the proportion P is greater than x − k1 s with confidence 0.319 2. 88 000.95. Stress range 250 MPa was used. 81 000.566 2.926 2.911 3.) Answer: At stress level 73. Tolerance limits for the normal distribution (from Råde and Westergren (1998)).641 5.355 2.031 3.399 4.582 3.758 7 2.020 4.638 3.95.462 Chapter 8 Page 8:7 .838 2. The fatigue lives of welded beams have been determined at seven (n = 7) identical tests. n is sample size.437 15 2.33⋅s = 78 MPa the failure probability is one per cent.707 4.95 P = 0.422 5.156 3.(At stress lever σmean − 2.733 3. At least the proportion P is between x − k2 s and x + k2 s with confidence 0.295 3.188 3.755 3.633 10 2.95. which means that one plate out of 100 is expected to fail due to fatigue if the plates are loaded at this stress level.986 3.143 4.5 MPa the probability of fatigue failure is expected to be 0.981 4.95.90 P = 0. and 5. 4.01 (= 1 − P) becomes x = xallowable = x − k ⋅ s = 4. Kt = (1 + 2b / a).99. probability P = 0.9074 = 80 789 cycles) and s= ⎯⎯⎯⎯⎯⎯ √ 1 n ∑(xi − x )2 = 0. One obtains n ∑ xi x= = 4.641 Allowable value of x at failure probability 0.13 Assume that xi (i.95: k = 4. Determine the stress concentration factor Kt when the plate is loaded in parallel to the (a) minor axis.9074 n (which gives N = 104.1496 n −1 The probability of fatigue failure should be 1 per cent at confidence level 95 per cent. A large plate has a small elliptical hole in it. 4. for 1 per cent failure probability.04. 4.Solution: First calculate the logarithm of the number of cycles N to fatigue failure. Page 8:8 Chapter 8 . 5. using number of samples n = 7.641 ⋅ 0. respectively.94. Using b / a = 2 and b / a = 1/2 one obtains Kt = 5 and 2.9074 − 4. It gives xi = log Ni = 4.91.2131 Number of cycles Nallowable that can be allowed at failure probability 1 per cent is x Nallowable = 10 allowable = 10 4. log Ni ) has a normal distribution. and confidence level C = 0.e.2131 = 16 334 Answer: Allowable number of cycles is. 4. Solution: Handbook (for example Appendix 2 in the textbook) gives.79. 8/10. This means that it is assumed that the xi:s are samples taken from a normally distributed process. Nallowable = 16 300 cycles.85. Determine the mean value x of xi and the standard deviation s. (b) major axis. The ratio of the axes of the hole is 2 to 1.69.1496 = 4. A table of statistics then gives. 125 and r / d = 0. Answer: Maximum allowable bending moment M is M = 19 kNm. Chapter 8 Page 8:9 . Consequently. Using safety factor sa = 2.e.0 one obtains λδκ λδκ σFLB which gives σnom σ = 0. use large radii at the geometric irregularities). The factors reducing the fatigue limit are λ = 0. 8/11.18 ⋅ 270 MPa Kf σnom allowable = allowable = sa sa Kf FLB Using σnom allowable = 32M = 0.80 (diameter 200 mm). The ultimate strength of the material is σU = 590 MPa.05 where q = 0. The axle has a change of diameter from 180 mm to 160 mm.03).25 − 1) = 2. notches and other geometric irregularities should be placed so that the stress concentration is avoided or made as small as possible (i. Answer: The stress concentration factor Kt is (a) Kt = 5.Comment: A large stress concentration implies a large probability of fatigue failure if the loading varies with time.. see figure.0 with respect to fatigue failure should be used. The fatigue notch factor Kf becomes Kf = 1 + q (Kt − 1) = 1 + 0.87).25 (here D / d = 1.84 has been taken from Figure 8. and κ = 0.11 in the textbook. δ = 1 here.92 (ground axle with ultimate strength σU = 590 MPa). Solution: The material has fatigue limit σFLB = ± 270 MPa when loaded with an alternating bending moment.84 ⋅ (2. one would have had δ = 0.18 ⋅ 270 ⋅ 106 Pa 3 πd one obtains the bending moment M = 19 kNm. Determine the maximum rotating bending moment M the axle may be subjected to if a safety factor of 2. Stress concentration gives Kt = 2. because the fatigue notch factor Kf will be used and this factor takes over the role of δ (else. and (b) Kt = 2. with the shoulder fillet radius r = 5 mm. A ground (surface finish Ra = 3 μm) axle of steel SS 1650-01 (σU = 590 MPa) has been r = 5 mm M M manufactured from a circular bar of 200 mm diameter. 9.5 mm P P 26 30 mm An axle has a groove (ground.8/12.0 should be used? Solution: The shear stress at the shoulder fillet is τshoulder = Kf τnom = Kf 16MT πd 3 Figures 8.94 ⋅ 0. using d = 0. and Kt = 1.85 − 1) = 1. The fatigue notch factor becomes Kf = 1 + q (Kt − 1) ≅ 1 + 0.6 = κ λ δ τpolished = ⋅ 0. see figure.90. The material is SS 1550-01 with an ultimate strength σU = Rm = 490 MPa and a fatigue limit (in torsion) τFLT = ± 140 MPa.94.6 One obtains 16MT 1 1 1. and 11. δ = 1. gives MT ≅ 130 Nm. σFL = 140 MPa and σFLP = 130 ± 130 MPa. The material is 1450-01 with σU = 430 MPa. 8. σY = 250 MPa.90 ⋅ 1 ⋅ 140 ⋅ 106 Pa 3 s 3 πd which. Answer: The maximum allowable torque is MT = 130 Nm. q = 0.70 ⋅ (1. and Appendix 2 in the textbook give κ = 0. Ra = 7 μm) according to the figure.85.7. λ = 0. Determine the maximum alternating torque MT the axle may be subjected to if a factor of safety 3. Determine the safety factor with respect to fatigue failure if the ratio of the mean value and the amplitude of the loading stress is constant. The fillet is ground (R = 5 a 30 40 μm). The axle is subjected to a tensile loading P = 18 ± 12 kN. 8/13. r = 1.70.030 m. Page 8:10 Chapter 8 . An axle has a change of diameter from 40 mm to 30 mm with a shoulder fillet with radius r = r = 1 mm MT MT 1 mm. and − 800 ± 850 MPa.6 − 1) = 2.7 kN. One has 4Pm Kt ⋅ 2 = σm = 180 ⋅ 106 Pa. m Ratio σa / σm becomes σa 2. 8/14. at the notch). The safety factor then is Pallowable 36. But. approximately.95 ⋅ 0.544 σm 2.7 ⋅ (2. the applied load was P = Papplied = 18 kN. 400 ± 560.7 kN πd Without safety factor the mean value of the froce can be Pallowable = 36.6 ⋅ 18 A straight line with this slope is entered into the Haigh diagram.0 Papplied 18 Answer: The safety factor is s = 2.5 mm and it is polished. 130 180 250 430MPa The notch (the groove) gives the stress concentration factor Kt ≅ 2. − 400 ± 860. see figure.12 Stresses in the material at the notch become: mean value σm = 2.95 ⋅ 1 ⋅ 130 = 117 MPa 0. Chapter 8 Page 8:11 .Solution: a 140 126 Reduce the fatigue data. Allowable mean value Pm of the tensile force P can now be calculated. which gives Pm = Pallowable = 36. Draw the Haigh diagram for notched cylindrical specimens of diameter D = 11 mm.544 1 Draw the Haigh diagram. When testing smooth specimens (polished) made from a high strength steel with yield limit σY = 1510 MPa and ultimate strength σU = 1800 MPa.5 mm deep with a radius in the trough of 0.95 ⋅ 0.95 ⋅ 1 ⋅ 140 = 126 MPa (MPa) 117 MPa σred FLP = κ λ δ σFLP = 0.11 in the textbook) Kf = 1 + q (Kt − 1) ≅ 1 + 0.0.12 ⋅ 12 = = 0. 0 ± 720. One obtains σred FL = κ λ δ σFL = 0. according to the problem. The intersection of this line with the reduced fatigue limit of the material gives that the allowable mean stress σm is σm ≅ 180 MPa. The fatigue notch factor Kf becomes (notch sensitivity factor q from Figure 8. The notch is 1. fatigue limits at N = 107 cycles were obtained at the following stress levels: 800 ± 400. d = 26 mm.6 Pm / A and amplitude σa = 2.12 Pa / A (the cross-sectional area A is calculated using the smallest diameter.7 s= = ≅ 2.6. Thus κλδ = 1. Fatigue limits are ± 270 200 mm MPa at alternating bending and 240 ± 240 MPa at pulsating bending. Use safety factor s = 2. Polished surface give κ = 1.7. where Kt = 2. 143 ± 207.38 and ρ / d = r / d = 0. the yield limit 390 MPa and the ultimate D = 50 strength 590 MPa.7 = 286 ± 148 MPa. and − 286 ± 315 MPa. For the points given. Chapter 8 . Fatigue notch factor Kf = 1 + q (Kt − 1) = 1 + 0. r = 10 65 d = 40 200 Page 8:12 The surface of the beam is polished.8 and Kf = 2. If yielding is to be avoided.95 (2. σ′a = σa / Kf and σ′m + σ′a < σY / Kt.0. λ.7.8 − 1) = 2. δ. 8/15. The right end of the cantilever beam (see figure) is moved up and down.0. − 143 ± 318.Solution: Determine the reduction factors κ. the mean value (corresponding to a static load) should be reduced by the factor Kt and the amplitude by the factor Kf.8 ± 400 / 2.0.0. To make the Haigh diagram valid for the notched specimen. No bending or torsion give δ = 1. 0 ± 267. and the fatigue notch factor Kf.06 give Kt = 2. − 143 ± 318. the stress concentration factor Kt. the following fatigue limits are obtained: 286 ± 148. Stress concentration: D / d = 11/8 = 1. Answer: σ′m = σm / Kt.98 ≈ 1.0. Diameter D = 11 mm gives λ ≈ 0. The yield strength of the material is σY = 1510 MPa. Determine the maximum allowable value of a with respect to fatigue failure. The r = 5 (mm) material has the modulus of elasticity E = 206 GPa. which means that no reduction due to surface finish and volume is needed here. 0 ± 267. which give fatigue limits 286 ± 148.5/8 = 0. the displacement upwards is 2a and downwards a. The point of fatigue limit 800 ± 400 then “moves” to 800 / 2. 143 ± 207.8 = 540 MPa both on the abscissa and on the ordinate. the Haigh diagram should be limited at stress σY / Kt = 1510 / 2. and − 286 ± 315 MPa.8. 5 = 2.5 a ± Kfnotch 1. Reduction factors: κ = 1. λ = 0.8 ⋅ 0. shoulder 3EI = 2 (Ktshoulder 0. At the notch the bending moment Mb.88 give Kfshoulder = 1.66 ⋅ 1.5 a ± Kfshoulder 1.Solution: The force P that loads the free (right) end of the beam is obtained from PL 3 3EI δ= which gives P = 3 δ 3EI L Here δ = 0.87⋅240 = 240 ± 210 MPa. The stresses at the notch and at the shoulder may now be determined (as a function of deflection a).5 a) L L Stress concentration at the notch: Ktnotch = 1. 590 This gives a Haigh diagram according to the m 240 390 (MPa) figure. notch = P = 2 δ = 2 (0.77 1. notch and σshoulder = Mb. shoulder L Wb.35.83 give Kfnotch = 1. L = 400 mm and I (= πD4 / 64) is the second moment of area of the circular beam cross section with diameter D = 50 mm (implying that here the influence of the notch on the beam deflection is neglected).red = 0.4 and q = 0.5 Chapter 8 Page 8:13 .red = 240 ± 0. notch is obtained as L 3EI 3EI Mb.66.5 a ± 1.5a ± 1. δ = 1 The reduced fatigue limit at alternating loading a (MPa) becomes 270 210 MPa 235 σFLB.5 a) σnotch = Wb.5 a) 2 2L 2L At the shoulder the bending moment becomes 3EI 3EI Mb.87. shoulder where Wb is the section modulus in bending. Stress concentration at the shoulder: Ktshoulder = 1. notch 3EI = 2 (Ktnotch 0. This gives at the notch: σnotch a σ notch m = 1.5a.5 a) Wb.87⋅270 = 235 MPa and at pulsating loading it becomes m σFLBP. notch 2L Wb.5 a ± 1.8 and q = 0. The loading is such that the ratio of the amplitude to the mean value is constant. shoulder = P L = 2 δ = 2 (0. One obtains Mb. 5 = = 2.8 on the nominal mean stress applied.5 a 3 2 ⎝ s⎠ Wb.77 in the Haigh diagram. Fatigue failure is thus expected when the mean value of the stress is σm = 82 MPa (the stress amplitude is then 2. This line intersects the reduced fatigue limit at σm ≅ 82 MPa.4 σnominal = 1. The mean value σm of the stress at the notch will now be determined from the calculated bending moment. and using the safety factor s = 2 on the reduced fatigue limit σm = 82 MPa determined.4 2 0.35 ⋅ 1.8 σnominal = σm ⋅ which gives ⎜ 1. Using the stress concentration factor Ktshoulder = 1. It is concluded that the notch is the most critical part of the structure.9σm = 226 MPa). Fatigue failure is thus expected when the mean value of the stress is σm = 78 MPa (the stress amplitude is then 2.9 in the Haigh diagram. one obtains ⎛ ⎞ Mb. The line intersects the reduced fatigue limit at σm ≅ 78 MPa.48 mm.4 on the nominal stress applied to the beam. The mean value σm of the stress at the shoulder will now be determined from the calculated bending moment.mean 3EI 32 1 ⎛ 1⎞ shoulder ⎜ 1.00048 m = 0.5 σshoulder m Enter a straight line of slope 2. and using the safety factor s = 2 on the reduced fatigue limit. The stresses at the notch will be limiting for the displacement a.mean 3EI 1 32 notch 6 1 1. Page 8:14 Chapter 8 . shoulder ⎠ L πD ⎝ From this a = 0.77σm = 227 MPa).58 mm is solved. Using the stress concentration factor Ktnotch = 1.9 1. Answer: Maximum allowable displacement a at the free beam end will be a = 0.00058 m = 0.8 = ⎟ 1. notch ⎠ 2L πd ⎝ From this a = 0. one obtains ⎞ ⎛ Mb.Enter a straight line of slope 2. At the shoulder is obtained: σshoulder a 1.5 a 3 = 82 ⋅ 10 ⋅ Pa s 2 Wb.4 ⋅ 0.48 mm is solved.4 = 78 ⋅ 106 ⋅ ⎜ = σm ⋅ ⎟ = ⎟ 1.8 2 0. The fatigue notch factor Kf becomes Kf = 1 + q (Kt − 1) ≅ 1 + 0. the stress σ0 in the lower flange of the beam would have been M σ0 = Wb Due to the holes. cross section: 200 100 200 9 mm 15 holes 1m P 1m A beam with cross section HE200B is made of the material 1312-00.8 ± 2. Each flange has two rows of holes (machined. and it is (use total beam length 2l) P ⋅l l M= = (P0 ± P0 ) where l = 1 m 2 2 The beam HE200B has the section modulus Wb = 570 ⋅ 10 − 6 m3 in bending. Material data: σU = 360 MPa. It is assumed that the holes are so widely separated that when calculating the stress concentration factor at one hole. Using r / a = 4/50 and B / a = 100/50 the stress concentration factor Kt is obtained as Kt ≅ 2. This can be done because of symmetry.8. σFL = 110 MPa and σFLP = 110 ± 110 MPa.44) B − 2r 2Wb where B is half the width of the flange. stress concentration will appear. with respect to fatigue failure of the beam. Determine. the disturbance of the stress distribution from the other holes may be neglected. σY = 240 MPa. thus B = 100 mm.8/16. Solution: The largest bending moment in the beam will appear at the mid-point of the beam.44. One obtains B P0 l σ = (2. This gives the maximum stress σ at a hole situated where the bending moment is the largest.8(2. The beam is simply supported at its ends and loaded by a force P = P0 ± P0 at its centre point. Ra = 20 μm) with diameter 8 mm. Study half of the flange width. The holes are drilled 100 mm from each other. Chapter 8 Page 8:15 . If the holes had not been there.8 − 1) = 2. the maximum allowable value of P0. 100 ⋅ P0 ⋅ 1 B P0 l = 2.8⋅94 / 2.90. and κ ≅ 0. The hole is machined (Ra = 7 μm) and it has a diameter of 20 mm. with respect to fatigue failure at hole A. Answer: The force P = P0 ± P0 may be allowed to be.092 ⋅ 2 ⋅ 570 ⋅ 10 which gives P0 ≅ 40 kN.44 −6 B − 2r 2Wb 0. it was correct to use the stress amplitude σa = 94 MPa as the design stress.85σFL ).95 ⋅ 1 ⋅ 0. The reduced fatigue limit at a pulsating load then becomes σred FLP = λ δ κ ⋅ σFLP = 0. Chapter 1. This give λ ≅ 0. the maximum allowable value of P0.44 2. It gives 0. implying that the straight line giving the service stress intersects the horizontal branch of the fatigue limit curve just slightly to the left of the breaking point at 110 MPa. and the surface of the hole is machined.95. δ = 1. Geometrical data: H = 200 mm and h = 20 mm (cf. 8/17. Section 1.44 = 108 MPa) is less than 110 MPa. Determine. where Kf > 1. Thus. The stress amplitude was used as the design stress. Material data (for SS 1510-00): σU = 600 MPa. P A side view 32 H A top view beam cross section 3h A 4H Page 8:16 h H The structure in the figure is loaded by a force P = P0 ± P0. P = 40 ± 40 kN (no safety factor included here).a (MPa) 94 MPa 110 94 2.8 110 240 m Reduction of fatigue limits: The flange thickness is 15 mm.4. Chapter 8 . σFL = 230 MPa and σFLP is unknown (use thumb rule σFLP ≅ 0. σY = 320 MPa. the fatigue notch factor Kf is used. approximately.90 ⋅ 110 MPa = 94 MPa The allowable stress amplitude at the hole is 94 MPa (with no safety factor). The mean stress σmean (= 2.8).44 = 94 ⋅ 10 6 Pa 2. and 3000.90 and δ = 1. 150.0 (here the case “a small hole in a large plate subjected to uniaxial loading” is used). Pulsating loading implies that the fatigue limit σFLP should be used.0 − 1) = 2. 180.e.87⋅(3. stress) amplitudes (the loading is alternating) σa = 200. 150. which gives P0 = 26 kN. The stress concentration factor Kt at the hole is Kt = 3. A component of a road vehicle is subjected to a repeated loading sequence. This gives the fatigue notch factor Kf = 1 + q(Kt − 1) = 1 + 0. Answer: Without safety factor. 20.90⋅1⋅0.74⋅σnom. the force P may be P = 26 ± 26 kN. in this stress range.85 σFL By use of the reduction factors one obtains σFLPred = κλδ⋅σFLP = κλδ⋅0. respectively The Wöhler curve of the material is.74 The factors reducing the fatigue limit due to surface finish and volume become: κ ≅ 0. One sequence contains the following loading (i.85⋅230 MPa = 151 MPa The amplitude P0 of the load gives the stress amplitude Kf⋅σnom = 2. It gives 2. Damage accumulation. Let Kf⋅σnom = σFLPred. and 100 MPa The number of loading cycles at each stress level is n = 15. λ ≅ 0. given by the relation σa = − 55logN + 430 MPa Chapter 8 Page 8:17 .86⋅0. z = H /2 and 4H H 3 (4H − 6h) (H − 2h)3 I= − 12 12 One obtains σnom = 2125P = 2125(P0 ± P0) N/m2 (P in N). counting of load cycles 8/18.Solution: The nominal stress at the hole becomes σnom = Mz / I where M = P⋅32H.85 σFL = 0. One obtains σFLP = 0.86.74⋅2125P0 = 151⋅106. ( 430 − σa ) / 55 One obtains N = 10 At the different stress levels the following damages are obtained σa (MPa) 200 180 150 100 Ni ni damage ni / Ni 15 199 35 112 123 285 106 15 20 150 3 000 15 / 15 199 20 / 35 112 150 / 123 285 3 000 / 106 Damage D due to one sequence is 15 20 150 3 000 D= = 5.77 ⋅ 10 − 3 + + + 6 15 199 35 112 123 285 10 The expected number of sequences Ns to fatigue failure is Ns = 1 / D ≅ 173 Answer: Damage D due to one sequence is D = 5. Solution: The Wöhler curve σa = − 55 logN + 430 of the material gives the fatigue life N at a given stress amplitude σa. One sequence contains the following stress levels and cycles: Page 8:18 Chapter 8 . and expected number of sequences Ns to fatigue failure is Ns ≅ 173. For a welded plate. a number of points on the Wöhler curve have been determined.Determine the damage accumulation D due to one loading sequence. At the probability of failure p = 50 per cent and at the stress ratio R = 0 the following points were obtained: σmax (MPa) Nf (cycles) 189 1⋅104 182 3⋅104 161 1⋅105 131 3⋅105 107 1⋅106 (a) Use linear regression to fit a straight line in a σmax−logN diagram to these points.77 ⋅ 10 − 3. and then determine how many loading sequences the component might resist before fatigue failure. Estimate by use of linear damage accumulation (the Palmgren-Miner rule) how many sequences (as given below) the plate might resist before fatigue failure is expected. 8/19. e.034945 and B = = 368.03 logN + 368. The Wöhler curve of the material thus becomes σmax = − 43. Linear regression gives (n is number of points) n Σx ⋅ y − Σx ⋅ Σ y Σ y − A Σx A= = − 43.78 − σ From this is solved: ) / 43.03 max N = 10 For the different stress levels the fatigue life and the damage are obtained as Chapter 8 Page 8:19 . Let the coordinate y be y = σmax and the coordinate x = logN.47712 5 5.78 MPa ( 368. the relationship σk ⋅ N = constant could give a better result at spectrum loading (i. This relation gives a straight line in a logσmax−logN diagram. This gives x= 4 4. possibly random. loading with different. Solution: (a) Fit a straight line in a σmax−logN diagram to the measured data.σmax (MPa) n (cycles) 120 100 150 50 180 20 160 40 140 60 (b) According to Jarfall (1977). Fit a straight line in a logσmax−logN diagram to the data given and estimate by the formula so obtained the number of sequences to fatigue failure.98 1) that the points fall almost on a straight line.47712 6 y= 189 182 161 131 107 Fit the straight line y = Ax + B to the points. amplitudes). see reference in textbook.78089 2 2 n n Σ x − (Σ x) The coefficient of correlation r becomes r= n Σx ⋅ y − Σx ⋅Σ y [n Σ x − (Σ x) ] [n Σ y − (Σ y) ] ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ √ 2 which means (r close to 2 2 2 = − 0. 8146 (and the coefficient of correlation r now is r = − 0.σmax Ni ni damage ni / Ni 120 150 180 160 140 603 815 121 282 24 360 71 028 207 093 100 50 20 40 60 0.1656136⋅10 − 3 0.12755 and B = 2.8240612⋅10 − 3 0.4882⋅10 − 3.25176⋅10 − 3 .1715581⋅10 − 3 0. Answer: Number of sequences to fatigue failure becomes (a) 444.260 5 2.4933274⋅10 − 3 0. and (b) 402. (b) Fit a straight line in a logσmax−logN diagram to the measured data.276 4. The Wöhler curve gives N = 10 ( B − log σmax ) / − A from which is obtained σmax Ni ni damage ni / Ni 120 150 180 160 140 582893 101353 24270 61108 174077 100 50 20 40 60 0.47712 2.47712 2.8209972⋅10 − 3 0.2897249⋅10 − 3 The damage D due to one sequence becomes D = Σ (ni / N i) = 2.5631617⋅10 − 3 0. Page 8:20 Chapter 8 . Expected number of sequences to fatigue failure is 1 / D = 444. Let y = logσmax and x = logN. one then obtains x= y= 4 2.6545826⋅10 − 3 0.4122625⋅10 − 3 0. approximately.9723).207 5. which gives the expected number of sequences Ns to fatigue failure: Ns = 1 / D ≅ 402.029 This gives the line logσmax = A logN + B.117 6 2.3446760⋅10 − 3 The damage D due to one loading sequence becomes D = Σ (ni / N i) = 2. where A = − 0. the stress amplitude of the first cycle in each sequence is σ0. cf. The number of cycles to failure at the stress amplitude σ0 would be N0. i. log a log 0 0 0 1 m t Figure (a) Figure (b) log N 0 log N Solution: The Wöhler curve is a straight line of slope − 1/m in a log σ a−log N diagram.8/20. equation (8.. assumed to be a straight line in a diagram with logσa on the ordinate (the “y” axis) and logN on the abscissa (the “x” axis). The equation of the curve may be written 1 log σa = − log N + C (a) m The constant C is determined from the condition that the fatigue life is N0 cycles when the stress amplitude is σ0. the equation of the Wöhler curve takes the form σa 1 N0 log = log (c) σ0 m N The stress amplitude of the different stress cycles following the first stress cycle (with amplitude σ0) in each sequence will now be determined together Chapter 8 Page 8:21 . The Wöhler curve of the material is. In the point of the material to be investigated. The logarithmic decrement δ is defined as δ = ln (σ i / σi+1) where σi is the amplitude of the vibration cycle i and σi+1 is the amplitude of the cycle i + 1. the mean value of each stress cycle is zero. Determine the number of vibration sequences the structure may be expected to survive. This gives 1 (b) log σ0 = − log N0 + C m By inserting C as given by relation (b) into the expression (a).2b). The damping of the vibrations is characterized by the logarithmic decrement δ. for stress amplitudes σa < σ0. see figure (a). Each sequence consists of free damped vibrations giving rise to stresses with the stress ratio R = − 1. A part of a structure is subjected to sequences of vibrations. The material has no fatigue limit. The slope of the curve is − 1/m in the diagram.e. The accumulated damage D due to one loading sequence becomes 1 1 1 1 αm α2m α3m 1 D = + + +… = + + + + … = (1 + αm +α2m + α3m + …. σi = αi σ0.05 and m = 8 one obtains I = 1 / D = 0. the stress amplitude σ0 alone would have given a cyclic life of N0 loading cycles. Determine the cyclic life N = Ni at stress level σi = αi σ0. σ2 = ασ1 = α2σ0. It gives α = exp(− δ) (here exp(− δ) stands for e− δ ). The relation (c) gives αi σ0 1 N0 log = log (e) σ0 m Ni This relation gives Ni = N0 / αi ⋅ m .95) and m = 8 one obtains I = 0. Let σ1 = ασ0.with the fatigue life at each stress amplitude. Fatigue failure is expected when the accumulated damage becomes unity. This gives 1 (i) 1=I⋅ N0 (1 − αm ) Solve for I. thus when I ⋅ D = 1.) (f) N0 N1 N2 N0 N0 N0 N0 N0 Use the series expansion 1 when | x| < 1 1−x This gives the damage D of one sequence of vibration as: 1 D= N0 (1 − αm ) 1 + x + x 2 + x 3 + …. . Page 8:22 Chapter 8 . = (g) (h) Let I be the total number of sequences the structure may be subjected to.33 N0 (solution also given in the Example in Section 8.3 in the textbook)... The definition of the logarithmic decrement yields σi σi δ = ln = ln (d) σi + 1 ασi Solve (d) for α. in this case.2. The accumulated damage then is I ⋅ D.. The free damped vibration after the first cycle (with amplitude σ0) reduces. Answer: The number I of sequences to fatigue failure is expected to be 1 I = = N0 (1 − αm ) where α = e − δ D If δ = 0. and so on.05 (giving α ≅ 0.33 N0 I = N0 (1 − αm ) = Hence. the fatigue life of the structure to one third of N0. which gives 1 D Using numerical values δ = 0. see also figure): 4 6 5 7 3 4 2 4 0 4 2 4 3 7 6 8 7 9 6 7 2 6 1 3 0 2 0 2 0 5 2 4 3 7 2 5 3 6 5 7 2 3 1 5 4 (= static level at rest) 9 8 7 6 5 4 3 2 1 0 P (kN) time Determine (a) the distribution of the peaks (number of peaks larger than or equal to a certain level). Solution: See answer. At a random loading of a structure. and (c) the distribution of exceedances (the load spectrum). (b) the distribution of the troughs (number of troughs smaller than or equal to a certain level). the following time sequence was recorded (force in kN.8/21. Answer: 9 8 7 6 5 4 3 2 1 0 P (kN) cumulative distribution of troughs exceedances peaks 5 10 15 Chapter 8 20 25 number Page 8:23 . see also figure in that problem): 4 6 5 7 3 4 2 4 0 4 2 4 3 7 6 8 7 9 6 7 2 6 1 3 0 2 0 2 0 5 2 4 3 7 2 5 3 6 5 7 2 3 1 5 4 (= static level at rest) Identify loading cycles by use of the rain flow count method. Now. but the problem of obtaining half cycles that must be matched to each other. that the solution could be slightly simplified if the sequence is rearranged so that it starts at its largest maximum or its smallest minimum. 9 8 7 6 5 4 3 2 1 0 P (kN) 9 17 15 1 3 16 14 20 5 7 2 31 23 25 27 12 8 10 22 24 4 Page 8:24 21 13 11 6 29 35 39 37 19 26 Chapter 8 28 43 38 41 33 32 36 30 34 42 18 44 40 time .8/22. At a random loading of a structure. see the figure. If this is done. 1 kN. A total of 44 drops are needed. will be avoided. Make a list of those cycles whose amplitude is larger than. Here. fore example. or equal to. the part of the sequence from the starting point at P = 4 kN to the minimum at P = 0 could be cut off and moved to the end of the sequence given. the following time sequence was recorded (force in kN. the final result will be the same. see below. let one drop of rain-water start to run (flow) from each maximum value and each minimum value of the loading sequence (the starting point and the final point here form a minimum). however. same sequence as in Problem 8/21. It should be noted. Solution: This solution is given for the load sequence as it is given in the problem. write down its number (the drop number). 11 12. 6 7. 15 16. 33 34. It gives Drop No starts stops at (kN) at (kN) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 4 6 5 7 3 4 2 4 0 4 2 4 3 7 7 5 6 0 4 3 4 2 9 2 4 3 4 6 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 6 8 7 9 6 7 2 6 1 3 0 2 0 2 0 5 7 7 8 0 7 6 6 2 3 1 2 0 2 0 7 2 31 32 33 34 35 36 37 38 39 40 41 42 43 44 2 4 3 7 2 5 3 6 5 7 2 3 1 5 5 3 4 2 7 3 5 5 6 1 3 2 5 4 Collect pairs of drops so that one pair forms a closed loop. 17 19. 35 36. 31 2 3 6 7 6 2 1 0 0 2 4 4 7 8 7 6 3 2 2 5 Chapter 8 32. 20 21. 22 23.Make a list of the run-ways of all drops. 24 25. 39 (40 41.maximum mum value value (1 2. Write down the two drop numbers. One obtains Drops No mini. 37 38. the minimum value (in kN) of the closed loop and the maximum value (in kN). 13 14. 8 9. its stating point (load level in kN) and its end point (load level in kN). 42 (43 (44 3 2 3 5 7 2 1 5 4 7 5 6 1) 3 5) 4) Page 8:25 . For each drop. 18 4 5 7 3 2 0 7) 6 0 4 4 9 10. 28 30. 29 5. 26 27. 3 4. the numbers of the two drops. The five one-way drop ways above may thus be combined to give two closed loops. initially it will run to 5 kN and then further “on the next roof” to 7 kN. Thus drops 43 and 1b together form half a cycle from 1 kN to 7 kN. The drop No 43 would then start at 1 kN. be grouped into closed loops in the following way: Divide the running way of drop No 1 into two parts. Due to the fact that the sequence does not starts at the largest maximum or the smallest minimum (see the introduction to the solution above) there will be some single drops left: drops No 1. 43. 40 1 Page 8:26 5 7 Chapter 8 . One obtains Drop way minimum maximum No value value 1a. Again. however. The drop 1a now stops at 5 kN and it is seen that this part may be matched to the drop No 44. Also these drops may. 1b. and 44.Here 20 closed loops have been identified. 1a and 1b. 40. the minimum value (in kN) of the closed loop and the maximum value (in kN) are written down. It gives Drop No starts stops 1a 1b 40 43 44 4 5 7 1 5 5 7 1 5 4 If we think of a continuation of the given sequence as a repetition of the first sequence. 44 4 43. This half-cycle may be matched to the drop No 40 to form a full loading cycle (one loop). then the run-way of the drop No 43 would continue the same way as the drop No 1 has from 5 kN and onwards. 4).5). If cycles with an amplitude less than 1 kN are neglected. then 12 loading cycles are left to form the basis of a fatigue damage analysis. 31 35.6). 34 0 0 2 2 3 1 2 2 5 7 5 7 Answer: Twelve cycles with an amplitude larger than or equal to 1 kN are obtained (and ten cycles with an amplitude less than 1 kN).2). 18 10.7). (0. 40 36. (3. Sometimes cycles of a small amplitude do not take part in the fatigue process. 1b.7). Chapter 8 Page 8:27 .9). (2.3).4). 27 7. 22 23. The ranges (twice the amplitude) of he loading cycles are given in the tables above. These are Drops No 4.2). These cycles may now be used in a fatigue analysis.5). Minimum and maximum values of the twelve cycles are: (0. 29 30. 11 21. (2. 24 minimum maximum value value 0 2 0 2 2 1 7 4 9 4 6 3 25. (2.In total 22 loops (cycles) have been obtained. 37 43. (1. (0. (2. (2. and (1. (0. 8 9.7). 26 28. Page 8:28 Chapter 8 .Empty page. 3.b) in the textbook. where the stresses far away from the hole are relatively low (i. Taking this into consideration.An Introduction. Thus.3 in the textbook.2a. the second term in the material relation (9.b) can be disregarded. Chapter 9 Strain-based fatigue design Problems with solutions 9/1. A flat bar of material SAE 1045 with a rectangular cross section has a small circular hole at its longitudinal centre axis.7 was given. The fatigue notch factor Kf is (a) K = 1 + q (K − 1) f t No information for calculation of the notch sensitivity factor q is given here. For low stresses. ISBN 91-44-02096-1.e.7.b) in the textbook then become FMCHAP9P. Lund 2002. Expressions (9. mainly within the elastic range). the material at the hole will yield locally.Solutions to problems in T Dahlberg and A Ekberg: Failure. Assume that Kf = 0. the stress concentration factor Kσ and the strain concentration factor Kε may be determined from equations (9.DOC/2010-01-15/TD Chapter 9 Page 9:1 . The bar is subjected to a remote alternating stress σ∞ = ± 300 MPa. the second term will be small when compared to the first term.9a. then the stress concentration factor would have been Kt = 3. (a) Estimate by use of Neuber’s method and the Morrow equation the number of cycles to fatigue failure. Due to the high stresses close to the hole. use Kf = 2. This means that here. Solution: (Compare with the problem solved in Section 9. Fatigue .9a. Instead. Studentlitteratur.9Kt. Fracture. Kf = 0.) (a) Number of cycles to fatigue failure If the stress state at the small hole in the flat bar were purely elastic.9Kt = 2. (c) Determine the fatigue life if σ∞ = 100 ± 200 MPa. (b) For the maximum stress and strain at the hole draw the hysteresis loop when σ∞ = ± 300 MPa. c) where. as can be seen.e. the upper end point of the hysteresis loop is situated at σ = σmax = 499 MPa and ε = εmax = Page 9:2 Chapter 9 .72 = Kf2.) Now the number of cycles to fatigue crack initiation (or fatigue failure) may be determined.006574.72 ⋅ 3002 ⎪σ ⋅ ε = = = 3. According to Morrow.18 ⎪ σa ⎛ σa ⎞ 1/n’ σa ⎞ ⎛ σ a ⎪ε = + ⎜ ⎟ = ⎟ +⎜ a E ⎝ K’⎠ 200 000 ⎝ 1344 ⎠ ⎩ This system of equations gives σa = σmax = 499 MPa and εa = εmax = 0. equations (9. One obtains ⎧ Kf2σ2∞ 2. one obtains σ’f − σm 1227 εa = (i) ( 2N )b + ε’f ( 2N )c = ( 2N ) − 0.006574. Hooke’s law) in (9.006574 = = 1. giving N = 2300 cycles to fatigue failure. when the remote loading stress is σ∞ = ± 300 MPa In problem (a) above the stress and strain amplitudes at the hole have been calculated when the remote stress amplitude is σ∞ = 300 MPa. and by use of the mean stress σm = 0.2805 ⎪ a a E 200 000 ⎨ (e.f) 1/ 0.0 ( 2N ) − 0. The local stress amplitude σa and strain amplitude εa at the hole are obtained as the intersection point between the Neuber hyperbola and the material stress-strain relation for cyclic loading.663 ⋅ 4. as it should.383 = 7.663 and Kε = = = 4.2a. (b) Display the hysteresis loop for the material at the hole. This implies that when the remote stress has its maximum ( σ∞ = 300 MPa). The stress and strain concentration factors Kσ and Kε then become Kσ = σmax 499 εmax 0.66 E 200 000 Using εa = 0.Kσ = σmax εmax εmax and Kε = = σ∞ ε∞ σ∞ / E (b.h) (Verify: Kσ ⋅ Kε = 1.b) has been used in (c).3828 σ∞ 300 σ∞ / E 300/ 200 000 (g. only the first term (i.4) and (9.29 = 2. The Neuber hyperbola becomes Kf2 σ2∞ σ ⋅ ε = Kσ σ∞ ⋅ Kε ε∞ = E (d) The material data given earlier (see the textbok) for the material SAE 1045 and the nominal stress amplitude σ∞ = 300 MPa now determine the Neuber hyperbola. the fatigue life 2N = 4600 reversals to failure is obtained.7) in the textbook.095 + 1. 004670 ⎪ (n. Choose point A as starting point of the hysteresis loop.e. the left) branch of the hysteresis loop. will pass through the point C.006574 − 0. The coordinates of point D are obtained from ⎧σD = σB + Δσ = − 499 MPa + 703 MPa = 204 MPa ⎪ ⎨εD = εB + Δε = − 0.006574 0.3a). see point B in the figure.004670 and Δσ = 703 MPa are solved. 499 MPa A . see figure below.499 MPa We now turn to the upper (i.006574 + 0. At a change of stress Δσ∞ = 300 MPa the upper branch of the hysteresis loop. These values give the lower end point of the hysteresis loop.0. When the remote stress has its minimum ( σ∞ = − 300 MPa).203 MPa . The lower branch of the hysteresis loop. A change Δσ∞ of the remote stress (the stress far away from the stress concentration) will cause a change of stress Δσ and a change of strain Δε at the stress concentration (the hole). see point A in the figure below. at the hole) is σ = − σmax = − 499 MPa and ε = − εmax = − 0. the stress and strain at the stress concentration (i.006574. starting at point B.0. The co-ordinates of point C are obtained from ⎧σC = σA − Δσ = 499 MPa − 703 MPa = − 204 MPa ⎨ (l. starting at point A.006574 C B 0.0. Chapter 9 Page 9:3 .004670 = 0. Using Δσ∞ = 300 MPa one obtains 2 2 2 2 ⎧ ⎪Δσ ⋅ Δε = Kf (Δσ∞) = 2. The changes Δσ and Δε are obtained by use of the Neuber hyperbola (9.006574.001904 For the student: verify that the change of the remote stress Δσ∞ = 600 MPa will take this branch of the hysteresis loop to the point B.18 ⎪ Δσ ⎛ Δσ ⎞ ⎛ Δσ ⎞ Δσ ⎪Δε = +2⎜ ⎟ = +2⎜ ⎟ ⎩ E ⎝ 2K’⎠ 200 000 ⎝ 2 ⋅ 1344 ⎠ From this system of equations Δε = 0. see figure.7 ⋅ 300 = 3.001912 .k) 1/n’ 1/ 0.e. will pass through point D.o) ⎩ = − 0.13d) and the material relation for changes of stress and strain (9.001904 By use of these values (and perhaps some more) the hysteresis loop may be drawn.m) ⎩εC = εA − Δε = 0.2805 ⎪ E 200 000 ⎨ (j.001912 203 MPa D . but now the bar is loaded to a nominal (remote) stress σ∞ = 100 ± 200 MPa. i. This is the case also at the hole. the residual stress at the hole will become σD (in tension) and the residual strain is εD. For equilibrium to be fulfilled at the cross section of the bar at the hole. once again. the residual stress at the hole is σC (compressive) and the residual strain is εC. On the other hand. (The remote stress σ∞ = 0 indicates that the axial force in the bar is zero. This implies that residual stresses in tension close to the hole will be balanced by residual stresses in compression further away from the hole.006574 F B Page 9:4 2 2 2 2 ⎧ ⎪Δσ ⋅ Δε = Kf (Δσ∞) = 2.18 C ⎪ . This explains why the residual stress is in tension at point D when the remote stress is σ∞ = 0. It gives. and vice versa.458 ⎪ E 200 000 0. The residual stress is compressive because the material has been “stretched out” (by plastic deformation) at the hole and it was “too long” when the bar was unloaded. Determine some more points on the hysteresis loop that is obtained when σ∞ = 100 ± 200 MPa.203 MPa ⎪Δε = Δσ + 2 ⎛⎜ Δσ ⎞⎟ ⎩ 200 000 ⎝ 2 ⋅ 1344 ⎠ .001912 ⎨ 1/ 0. as in (j. the starting point of the loop at point A where σA = 499 MPa and εA = 0. when the bar is unloaded (i.499 MPa Chapter 9 (p. when unloading from the remote stress σ∞ = − 300 MPa to stress σ∞ = 0.7) will be investigated.006574 E 0. there must be residual stresses both in tension and in compression. . The nominal (remote) stress σ∞ will now vary between 300 MPa and − 100 MPa.Comment: It is noticed that when the loop passes the point C the external loading (far away from the hole) will give a nominal stress σ∞ equal to zero. 499 MPa A .e.k) above.001912 203 MPa D Select. Determine the point on the loop (the stress at the hole) when Δσ∞ = 200 MPa.) (c) Fatigue life when σ∞ = 100 ± 200 MPa The same bar as above (Kf = 2. when σ∞ = 0).0.7 ⋅ 200 = 1. Thus. In this case the material was “compressed” at the hole so it was made “too short” when the bar was unloaded from σ∞ = − 300 MPa to σ∞ = 0.006574.e.q) .0. the bar has been loaded a number of times up to stress σ∞ = 300 MPa and then unloaded by the stress Δσ∞ = 300 MPa so that the total stress far away from the stress concentration is σ∞ = 0. This implies that the stress at the hole will vary between σA (when σ∞ = 300 MPa) and a point on the hysteresis loop below σC (the stress at the hole is σC when the remote stress is σ∞ = 0). 003519. determine the turning point of the new loop (the stress at the hole) when Δσ∞ = 400 MPa (the remote stress then is σ∞ = − 100 MPa).000464) / 2 = 0.006574 − 0.5 εa = 0.003764 Finally.u) ⎨ 1/ 0. some points on the upper branch of the hysteresis loop may be determined from the stress changes already calculated. The number of loading cycles to crack initiation is obtained.006574 − 0.007038. The turning point F on the new hysteresis loop is now obtained from ⎧σF = σA − Δσ = 499 MPa − 829 MPa = − 330 MPa ⎨ ⎩εF = εA − Δε = 0. according to Morrow. (This result may be compared with the result obtained in Section 9. By use of these values in (x).3.This system of equations gives Δσ = 519 MPa and Δε = 0. and the strain amplitude will be εa = (εA − εF) / 2 = (0.3 for the same structure. The point E on the hysteresis loop is now obtained from ⎧σE = σA − Δσ = 499 MPa − 519 MPa = − 20 MPa ⎨ (r.000464 (v.002810.72 ⋅ 4002 f ∞ ⎪Δσ ⋅ Δε = = = 5. and by use of the parameter values given.) Chapter 9 Page 9:5 . to fatigue failure is solved.095 + 1.k) above. 2 2 ⎧ (Δσ ) K 2.66 200 000 From this 2N = 23 070 reversals. It gives. giving N = 11 500 cycles.s) ⎩εE = εA − Δε = 0.003519 = (y) ( 2N ) − 0.002810 = 0. as in (j.006574 + 0.18 ⎪ ⎛ Δσ ⎞ Δσ ⎪Δε = +2⎜ ⎟ 200 000 ⎝ 2 ⋅ 1344 ⎠ ⎩ This gives Δσ = 829 MPa and Δε = 0.w) Using point F as starting point. This is left to the student as an exercise. one obtains 1227 − 84.832 ⎪ E 200 000 (t.007038 = − 0. There the remote stress σ∞ = 200 ± 200 MPa gave the fatigue life N = 6 500 cycles to failure.0 ( 2N ) − 0. from σ’f − σm εa = (x) ( 2N )b + ε’f ( 2N )c E The mean value σm of the stress becomes σm = (σA + σF) / 2 = (499 − 330) / 2 = 84.5 MPa. ) Determine the expected number of load cycles to fatigue failure.19.5 N − 0.6N − 0.12 + D 0.65 Page 9:6 Chapter 9 (a) (b) (c) . 9/2. Assume that the material follows the cyclic stress-strain curve σa ⎛ σa ⎞ 1/n ’ ⎛ Δσ ⎛ Δσ ⎞ 1/n’⎞ ⎜and for changes: εa = + ⎜ ⎟ Δε = +2⎜ ⎟ ⎟ E ⎝ K’⎠ ⎝ E ⎝ 2K’⎠ ⎠ where E = 210 GPa. only ranges are of interest here.3.8 ⋅ 300)2 Neuber: Δσ ⋅ Δε = = = 3.0498 Δε = 3.19 Δσ +2⎜ ⎟ = +2⎜ ⎟ Δε = E ⎝ 2 K’⎠ 210 000 ⎝ 2 ⋅ 1200 ⎠ (Kf ⋅ Δσ∞)2 (2. Use fatigue notch factor Kf = Kt.Answer: (Compare with the problem solved in Section 9.36 E 210 000 This system of equations gives Δσ = 650 MPa and Δε = 0. One obtains Δσ ⎛ Δσ ⎞ 1/n’ ⎛ Δσ ⎞ 1/0. respectively. The influence of the mean values is disregarded.6 where D = ln E 1−Ψ 1 − 0.005169. (b) stress and strain at the hole are (end points of the hysteresis loop) σa = ± 499 MPa and εa = ± 0.) (a) At remote stress σ∞ = ± 300 MPa the number of cycles N to fatigue failure is N = 2300. the reduction of the cross-sectional area at rupture is Ψ = 65 per cent.3 in the textbook.006574. approximately. K’ = 1200 MPa and n’ = 0.8 has been found. Coffin-Manson’s fatigue law reads σU 1 1 = ln = 1. (In addition to this stress. and (c) fatigue life is expected to be N = 11 500 cycles to failure (according to Morrow) at remote stress σ∞ = 100 ± 200 MPa. and the ultimate strength of the material is σU = 470 MPa. As we are going to use the Coffin-Manson rule. In a structure a notch with stress concentration factor Kt = 2. Solution: The material relation and the Neuber hyperbola give for stress and strain ranges (stress range is Δσ∞ = 250 − ( − 50) = 300 MPa). The structure is loaded so that the nominal stress some distance away from the notch varies between − 50 MPa and + 250 MPa. the stress concentration will be added at the notch. approximately. n’ = 0. 19 000 cycles. Assume that the material follows the cyclic stress-strain relationships (for amplitudes and changes. of course. σf’ = 1400 MPa. b = − 0. and less than 2.6 N − 0.Numerical values give 470 N − 0. One obtains Kf2 ⋅ σ2∞ 2. The specimen has a notch with stress concentration factor Kt and fatigue notch factor Kf such that Kt = Kf = 2. and c = − 0. σY’ = 850 MPa.005169 = 3. 9/3.65 (a) E 206 000 σa ⎛ σa ⎞ 1/n’ σa ⎛ σa ⎞ 1/0.82 ⋅ 7002 σa ⋅ εa = = = 18.8. Thus.60.11 ⎟ +⎜ εa = + ⎜ ⎟ = E ⎝ K’⎠ 206 000 ⎝ 1750 ⎠ Chapter 9 (b) Page 9:7 .11. failure is expected after. The upper end point of the hysteresis loop is obtained from the intersection of the Neuber hyperbola and the cyclic stress-strain relation. The material has the following cyclic properties: E = 206 GPa. Comment: When σ∞ = 700 MPa.8⋅700 = 1960 MPa.5 (d) Answer: Fatigue failure is expected after N = 19 000 cycles (approximately). Then use the calculated values in a low-cycle fatigue analysis to determine the expected fatigue life N of the test specimen.55. A test specimen is subjected to a remote nominal stress σ∞ = 500 ± 200 MPa. respectively) σa ⎛ σa ⎞ 1/n ’ Δσ ⎛ Δσ ⎞ 1/n’ εa = + ⎜ ⎟ and Δε = +2⎜ ⎟ E ⎝ K’⎠ E ⎝ 2K’⎠ Determine the stress (mean value and amplitude) and the strain (mean value and amplitude) in the material at the place of stress and strain concentration. εf’ = 0. which is the stress one should have had if the material were fully elastic. Solution: Determine the hysteresis loop obtained when the remote stress is varying between 700 MPa and 300 MPa.10. 0.6 210 000 from which N = 19 200 cycles (approximately) is solved. the stress at the stress concentration should be more than 700 MPa. K’ = 1750 MPa.04980.12 + 1. The second term on the right hand side of (h) is thus.j) Comment: One notices that 2.005479 (h) (i.d) The stress concentration factor Kσ and the strain concentration factor Kε may now be determined (if desired). as it should.5386 and Kε = = = 5. and from that starting point the elastic range is twice the yield limit of the material until yielding starts at compression.119 (g) E 206 000 Δσ Δσ ⎛ Δσ ⎞ 1 / n ’ ⎛ Δσ ⎞ 1 / 0. A change Δσ∞ = 400 MPa (i.82 = 7.005479 = 0. One obtains εmax εmax 1077 0. The reason why we have such a large range for the elastic deformation here is that the material is yielding in tension when the change of load is applied.01184 (l) The mean value and the amplitude of the stress and the strain at the notch can now be calculated. much smaller than the first term. The lower end point of the hysteresis loop may now be determined. in this case. One obtains Kf2 ⋅ (Δσ∞)2 2.017319 Kσ = = = 1.8418.8 ⋅ 400 = 1120 MPa. which is the change one would have obtained if the material were fully elastic.From these two equations the upper end point (the turning point) of the hysteresis loop will be found.f) 700 ε∞ σ∞ / E 700 / 206 000 One finds that Kσ⋅Kε = 7. One obtains σmin = σmax − Δσ = 1077 − 1116 MPa = − 39 MPa and (k) εmin = εmax − Δε = 0. which implies that almost all deformation at the notch is elastic at this change of the remote stress.84.11 Δε = = +2⎜ ⎟ +2⎜ ⎟ E ⎝ 2K’⎠ 206 000 ⎝ 2 ⋅ 1750 ⎠ This gives the changes of stress and strain at the stress concentration as Δσ = 1116 MPa and Δε = 0. One obtains Page 9:8 Chapter 9 . and that Kf2 = 2. Here we have got the change Δσ = 1116 MPa in (i).017319 − 0. twice the amplitude 200 MPa) of the remote stress σ∞ causes changes of Δσ and Δε at the stress concentration.017319 (c.e.096 (e.82 ⋅ 4002 Δσ ⋅ Δε = = = 6. The changes Δσ and Δε at the stress concentration are obtained from the intersection of the Neuber hyperbola and the stress-strain relation for changes. One obtains σmax = 1077 MPa and εmax = 0. Answer: Failure is expected after N = 33 400 cycles (approximately). Solution: Use the Coffin-Manson rule to calculate the maximum strain amplitude allowed. the expected number of cycles to fatigue failure is N = 33 400 cycles. 9/4.01184 = = 0.60(2N) − 0. It gives σU εa = 1.6 N − 0. The structure is designed for cyclic loading. Material data: The material can be considered linear elastic. A structure has a notch with the fatigue notch factor Kf.05. The remote loading (the loading far away from the notch) is σ∞ = ± 250 MPa. and (maximum) 10 000 loading cycles is expected during the life of the structure. give Chapter 9 Page 9:9 . Determine the highest value of the fatigue notch factor Kf that can be allowed. (q) (r) Thus. N = 33 400 cycles is solved.0027395 = (2N) − 0.σmean = σmax + σmin 1077 − 39 MPA = = 519 MPa 2 2 (m) σa = σmax − σmin 1077 + 39 MPA = = 558 MPa 2 2 (n) εmean = εa = εmax + εmin 0.6 (a) E Numerical values. ideally plastic with modulus of elasticity E = 210 GPa and yield strength σY = 350 MPa.10 + 0.01184 = = 0.0027395 2 2 (p) The fatigue life N is obtained from the Morrow expression: σf’ − σmean εa = (2N) b + εf’(2N) c E which gives 1400 − 519 0.75 N − 0. as given above.5 D 0. The ultimate strength of the material is σU = 380 MPa and the ductility D = 1.017319 − 0.55 206 000 From this.01450 2 2 (o) εmax − εmin 0.12 + 0.017319 + 0. 6 10 000 − 0. A structure has a notch with the fatigue notch factor Kf. with σmax = σY = 350 MPa. The remote loading (the loading far away from the notch) is σ∞ = ± 200 MPa. as the ultimate strength of the material is 380 MPa.εa = 1. (In fact. Kf2 2502 350 ⋅ 0. The ductility is D = 1. stress U Y E 0. it could be slightly higher than 350 MPa.6 210000 = 0. The modulus of elasticity is E = 210 GPa.6 N − 0. One obtains Kf = 1.5 D 0. at which stress the strain (at failure) is 0. and (maximum) 10 000 loading cycles is expected during the life of the structure.0030983 (b) The stress at the notch can not be higher than σY = 350 MPa. Determine the highest value of the fatigue notch factor Kf that can be allowed. It gives σU εa = 1.12 + 0. the ultimate strength of the material is σU = 380 MPa.001667 strain 0.003098 = (d) 210 000 From this the maximum value of the fatige notch factor is solved.5 ⋅ 1. some deformation hardening could take place in the material. Solution: Use the Coffin-Manson rule to calculate the maximum strain amplitude allowed.75 N − 0.005 Material data: The material behaviour at cyclic loading may be regarded as linear elastic.005.05 0. 9/5.) The Neuber hyperbola Kf2 σ2∞ σ⋅ε = E (c) gives. Thus. deformation hardening (see figure).05.91. but this is not taken into account here. The structure is designed for cyclic loading.6 E Page 9:10 Chapter 9 (a) . the yield strength σY = 350 MPa.91 (e) Answer: Maximum allowable value of the fatige notch factor is Kf = 1.12 + 0.75 380 10 000 − 0. 0 and the fatigue notch factor is Kf = 2. the maximum stress σ = 363 MPa can be allowed at the notch.43. ideally plastic material with a rectangular cross section has a small circular hole at its centre axis. with σmax = 363 MPa. The stress concentration factor at the hole is Kt = 3. Kf2 2002 (d) 363 ⋅ 0.e.5 ⋅ 1. so it is concluded that the Neuber hyperbola must intersect the strain-hardening branch of the stress-strain curve given. σf’= 1200 MPa.1 and c = − 0. Further.05 0.003098 = 210 000 From this the maximum allowable value of the fatige notch factor is solved. The modulus of elasticity is E = 200 GPa and the yield strength is σY = 400 MPa.001667 (the elastic limit. Kf2 σ2∞ σ⋅ε = E The Neuber hyperbola (c) gives.6 10 000 − 0. Use the Neuber method to estimate the stress and the stain at the hole. 9/6. stress Y strain The bar is subjected to an alternating stress σ∞ = ± 200 MPa. give 380 εa = 1. Thus. εf’= 1.003098 into this equation. Chapter 9 Page 9:11 .0030983 (b) This strain is larger than the strain 0.75 10 000 − 0. i.62. The equation of this branch is σ = 335 + 9000 ⋅ ε (stress in MPa) Enter the strain ε = 0. and then use the Morrow equation to estimate the number of cycles to fatigue failure.8. see figure). Note that the material is assumed to be ideally plastic also at cyclic loading. as given above. and less than 0. A flat bar of a linear elastic. It gives stress σ = 363 MPa. One obtains Kf = 2.12 + 0.43 (e) Answer: The largest fatigue notch factor that can be allowed is Kf = 2.6 210000 = 0.005.Numerical values. no deformation hardening or softening is present. b = − 0.0. stress Y strain Page 9:12 Determine the expected number of loading cycles to fatigue failure of the specimen.e) ⎨ E 200 000 ⎪ ⎩εa is unknown. the fatigue life N = 13800 (or 13827) cycles is obtained.0 ( 2N ) − 0. thus ε∞ = σ∞ / E ) The Neuber hyperbola becomes Kf2σ2∞ σ⋅ε = E (c) The material data given and the nominal stress amplitude σ∞ = 200 MPa now determine the Neuber hyperbola. respectively. εa = εmax = 0. Now the number of cycles to fatigue failure may be determined.82 ⋅ 2002 ⎪σa ⋅ εa = f ∞ = = 1.b) σ∞ ε∞ σ∞ / E (Hooke’s law is valid for stresses below 400 MPa. at the hole are obtained as the intersection of the Neuber hyperbola with the material stress-strain relation. but σa = σY = 400 MPa This gives σa = σmax = 400 MPa and. The material is assumed to be linear elastic.5) is subjected to a remote stress σ∞ = ± 300 MPa. One obtains ⎧ K 2σ2 2. Data: Cross-section reduction Ψ at fracture is Ψ = 65 per cent and fracture strength is σU = 650 MPa. Morrow’s formula gives σ’f − σm 1200 (f) ( 2N )b + ε’f ( 2N )c = ( 2N ) − 0. ideally plastic (also at cyclic loading) with modulus of elasticity E = 210 GPa. The local (maximum) stress and strain amplitudes σa and εa.00392.00392. A test specimen with a notch (stress concentration factor Kt = 3.1 + 1. from (d). Using the mean stress σm = 0. εa = Answer: The fatigue life N = 13 800 cycles is expected. see figure.568 (d. Chapter 9 .Solution: Due to the high stresses at the hole. and yield limit σY = 650 MPa. The stress concentration factor Kσ and the strain concentration factor Kε may be written σmax εmax εmax Kσ = and Kε = = (a.62 E 200 000 Using εa = 0.0 and fatigue notch factor Kf = 2. 9/7. the material will yield locally. Solution: The Neuber hyperbola gives (for amplitudes) (Kf σ∞)2 εa ⋅ σa = (a) where σ∞ = 300 MPa E As the material is ideally plastic, it is concluded that the stress cannot exceed σY = 650 MPa at the point where the stress concentration appears. Using this (i.e. σa = σY = 650 MPa) in equation (a), one obtains 2 1 (Kf σ∞) 1 (2.5 ⋅ 300)2 = = 0.00412 (b) εa = σa E 650 210 000 The Coffin-Manson relationship gives, with εa = 0.00412, 650 N − 0.12 + D 0.6N − 0.6 Δε = 2 ⋅ εa = 2 ⋅ 0.00412 = 3.5 210 000 Where D = ln[1/(1 − Ψ)] = 1.04982. Solving (c) for N gives N = 8330 cycles. (c) Answer: Fatigue failure is expected after 8300 cycles, approximately. 9/8. A flat bar of a linear elastic, deformation hardening plastic material has a rectangular cross section. A small circular hole has been drilled in the bar at its centre axis. The stress concentration factor at the hole is Kt = 3.0 and the fatigue notch factor is Kf = 2.8. stress Y E The bar is subjected to an alternating remote stress σ∞ = ± 200 MPa. Estimate the number of cycles to fatigue failure by use of Neuber’s method and Morrow’s equation. Note that the material is assumed to be linearly elastic, linearly deformation k = E /10 hardening also at cyclic loading as shown in the figure. Use modulus of elasticity E = 200 GPa, strain slope k = E /10, and the yield strength σ = 400 Y MPa. Further, σf’= 1200 MPa, εf’= 1.0, b = − 0.1, and c = − 0.62. Chapter 9 Page 9:13 Solution: Due to the high stresses near the hole, the material will yield locally. The stress concentration factor Kσ and the strain concentration factor Kε may be written Kσ = σmax σ∞ and Kε = εmax εmax = ε∞ σ∞ / E (Hooke’s law is valid for stresses below 400 MPa; thus ε∞ = σ∞ / E in (b).) The Neuber hyperbola becomes Kf2σ2∞ σ⋅ε = E (a,b) (c) The material data given and the nominal stress amplitude σ∞ = 200 MPa now determine the Neuber hyperbola. The local stress and strain amplitudes σa and εa, respectively, at the hole are obtained as the intersection of the Neuber hyperbola and the material stress-strain relationship given in the figure. One obtains ⎧ Kf2σ2∞ 2.82 ⋅ 2002 ⎪σa ⋅ εa = = = 1.568 (d,e) ⎨ E 200 000 ⎪ ⎩σa = 360 + k ⋅ εa for σa > σY = 400 MPa This gives σa = σmax = 432.5 MPa and, from (d), εa = εmax = 0.003625. Now the number of cycles to fatigue failure may be determined. According to Morrow, and by use of mean stress σm = 0, one obtains σ’f − σm 1200 (f) ( 2N )b + ε’f ( 2N )c = ( 2N ) − 0.1 + 1.0 ( 2N ) − 0.62 εa = E 200 000 Using εa = 0.003625, the fatigue life N = 17575 (or 17600) cycles is obtained. Answer: The fatigue life N = 17 600 cycles (approximately) is expected. 9/9. A test specimen is loaded with a repeated stress sequence with a nominal stress σ∞ according to the figure. The test specimen contains a notch with stress concentration factor Kt = 2.7 and fatigue notch factor Kf = 2.5. Assume that the material follows the cyclic stress-strain curve σa ⎛ σa ⎞ 1/n ’ ⎛ Δσ ⎛ Δσ ⎞ 1/n’ ⎞ ⎜and at changes: Δε = ⎟ εa = + ⎜ ⎟ +2⎜ ⎟ E ⎝ K’⎠ ⎝ ⎠ E ⎝ 2K’⎠ Page 9:14 Chapter 9 (a) Determine the stresses and strains at the notch needed for a fatigue life analysis (i.e. determine the mean values of the stress cycles and the amplitudes of the strain cycles at the notch). Use the rain-flow count method for the cycle counting. (b) Determine the expected fatigue life (the expected number of loading sequences before failure) of the test specimen. The material has the following cyclic properties: E = 206 GPa, K’ = 1750 MPa, n’ = 0.11, σY = 850 MPa, σf’ = 1500 MPa, εf’ = 0.60, b = − 0.10, and c = − 0.55. Stress (MPa) 600 400 200 0 time Solution: (a) Stresses and strains at the notch The mean value and the amplitude of the stress and strain at the notch are obtained from the turning points of the hysteresis loops. Determine these points. The stress and strain at the notch, when σ∞ = 600 MPa, are obtained from the relationships (the Neuber hyperbola and the cyclic material relation for amplitudes): 2 2 ⎧ K 2(σ )2 ⎪σ ⋅ ε = f ∞ = 2.5 ⋅ 600 = 10.922 ⎪ a a E 206 000 ⎨ (a,b) 1 / n ’ 1 / 0.11 ⎪ σa ⎛ σa ⎞ σa ⎞ ⎛ σ a ⎪ε = + ⎜ ⎟ ⎟ = +⎜ a E ⎝ K’⎠ 206 000 ⎝ 1750 ⎠ ⎩ It is assumed that after a number of loading sequences, the stress and the strain stabilize as given by the Ramberg-Osgood’s stress-strain relation even if the loading is not alternating here. Therefore, the remote stress 600 MPa is used as an amplitude when the upper turning point of the hysteresis loop is determined. From (a,b) the upper turning point of the hysteresis loop is found. This point is marked (1) in the figure to the right below. One obtains σ1 = 999 MPa and ε1 = 0.010943 Thus, each time the remote stress σ∞ reaches the level 600 MPa (points (A) and (C) in the figure to the left), the stress and the strain at the notch become σ1 = 999 MPa and ε1 = 0.010943, respectively (point (1) in the figure to the right). Chapter 9 Page 9:15 006078 − 0. from level (C) to level (D).004865 = 0. Point (3) coincides with point (1).854 ⎪ E 206 000 ⎨ 1/n’ ⎪ Δσ ⎛ Δσ ⎞ 1 / 0.004865 The stress and the stain at the notch.010 (4) Stress and strain at the notch A change Δσ∞ = 400 MPa of the remote stress.52 ⋅ 4002 f ∞ ⎪Δσ ⋅ Δε = = = 4.Stress (MPa) 600 Stress (MPa) (A) (C) 200 (A) (E) (E) 400 (1) (3) 1000 (5) 500 0 (B) (2) time 0 (D) (D) (D) Remote (nominal) stress giving stress concentration at the notch -500 0.004865 = 0. After this the loading stress σ∞ turns upwards the amount Δσ∞ = 400 MPa up to stress level σ∞ = 600 MPa. see point (C). and the strain increases from ε2 to ε3 = 0.010943 = ε1.010943 − 0.006078 This point has been marked as point (2) in the stress-strain diagram. giving point (3) in the diagram. from level (A) to level (B) in the figure. A stress change Δσ∞ = 600 MPa of the remote stress. become σ2 = σ1 − Δσ = 999 − 998 MPa = 1 MPa and ε2 = ε1 − Δε = 0. gives Page 9:16 Chapter 9 . The stress at the notch then increases Δσ = 998 MPa from σ2 to σ3 = 1 MPa + 998 MPa = 999 MPa = σ1. gives 2 2 ⎧ (Δσ ) K 2.d) Δσ = 998 MPa and Δε = 0. when σ∞ = 200 MPa at (B).11 ⎪Δε = Δσ + 2 ⎛⎜ Δσ ⎞⎟ = +2⎜ ⎟ E ⎝ 2 ⋅ K’⎠ 206 000 ⎝ 2 ⋅ 1750 ⎠ ⎩ This gives the stress and strain changes at the notch: (c. 5 ⋅ 600 = 10.008212 The final stress change. At next maximum.2 2 2 2 ⎧ ⎪Δσ ⋅ Δε = Kf (Δσ∞) = 2.007596 = 0. the stress and the strain at the notch become (see point (5) in the figure) σ5 = σ4 + Δσ = − 439 + 998 MPa = 559 MPa and ε5 = ε4 + Δε = 0.004865 = 0.010943 0.005780 0.007596 The stress and the strain at the notch when σ∞ = 0 (σ∞ is situated at point (D)) become (see point (4) in the figure) σ4 = σ3 − Δσ = 999 − 1438 MPa = − 439 MPa and ε4 = ε3 − Δε = 0.003798 2 − 439 559 60 499 0.007145 0.922 ⎪ E 206 000 ⎨ 1/n’ ⎪ Δσ ⎛ Δσ ⎞ 1 / 0. corresponding to cycle (A)-(B)-(C) for the remote stress. i. these values give the following stresses and strains at the notch (not all these values were asked for in the problem): No of cycles σmin σmax σmean 1 1 999 500 499 0.006078 0. Chapter 9 Page 9:17 .two cycles between the points (4) and (5) in the figure. at the notch the following is obtained: .002433 1 − 439 999 280 719 0.002433 σampl εmin εmax εmean εampl Thus. when σ∞ has come to level (E).003347 After this two loops with stress changes Δσ∞ = 400 MPa follow.010943 − 0. To summarize.008510 0.11 ⎪Δε = Δσ + 2 ⎛⎜ Δσ ⎞⎟ = +2⎜ ⎟ E ⎝ 2 ⋅ K’⎠ 206 000 ⎝ 2 ⋅ 1750 ⎠ ⎩ (e.003347 + 0. corresponding to cycle (C)-(D)-(A) for the remote stress.one cycle between the points (3) and (4) in the figure. and .f) This gives the stress and strain changes at the notch: Δσ = 1438 MPa and Δε = 0.008212 0.e. from level (D) back to (A).003347 0.010943 0. .003347 0.one cycle between the points (1) and (2) in the stress-strain diagram. corresponding to cycles (D)-(E)-(D)-(E)-(D) for the remote stress. gives the branch from (4) back to (1) in the hysteresis loop. 55 206 000 giving N1 = 63 650 0. with the corresponding mean value of the stress.55 206 000 giving N2 = 20 355 0.10 + 0.002433 = 1500 − 60 (2 N3) − 0. Page 9:18 Chapter 9 .003798 = 1500 − 280 (2 N2) − 0.002433 = 1500 − 500 (2 N1) − 0.10 + 0.10 + 0.(b) Fatige life The fatigue life Ni at each strain amplitude.60 (2 N1) − 0.60 (2 N3) − 0. Answer: One expects S = 1/D = 13 300 sequences to fatigue failure.55 206 000 giving N3 = 195 400 The accumulated damage from one sequence is (use the Palmgren-Miner damage accumulation rule) 1 2 1 1 + + = (h) D= 63 650 20 355 195 400 13 320 From this. is obtained from the Morrow relationship σf’ − σmi εai = (g) (2 Ni )b + ε’f (2 Ni )c E which gives 0.60 (2 N2) − 0. S = 1/D = 13 320 sequences to fatigue failure is obtained. c) σ∞ ε∞ σ∞ / E The Neuber hyperbola becomes Kf2σ2∞ σ⋅ε = E (d) The material data given earlier for the material SAE 1045 and the nominal stress amplitude σ∞ = 300 MPa now determine the Neuber hyperbola. use Kf = 2.Extra problem. The local stress amplitude σa and strain amplitude εa at the hole are obtained as the Chapter 9 Page 9:19 . the material will yield locally.9a. (c) Determine the fatigue life if σ∞ = 150 ± 150 MPa. Thus. (b) Draw the hysteresis loop for σ∞ = ± 300 MPa. Due to the high stresses close to the hole. Solution: (a) Number of cycles to fatigue crack initiation If the stress state at the small hole in the flat bar was purely elastic. Kf = 0.b) as σmax εmax εmax Kσ = and Kε = = (b. Assume that Kf = 0.7. A flat bar of material SAE 1045 with a rectangular cross section has a small circular hole at its centre axis. (a) Estimate the number of cycles to fatigue failure by use of Neuber’s method.b) is disregarded (for stresses far away from the hole the second term is supposed to be small as compared with the first term) one obtains (9. Instead.9a. a version of 9/1.7 was given.2a. then the stress concentration factor would have been Kt = 3.9Kt. the stress concentration factor Kσ and the strain concentration factor Kε may be determined from equations (9.b). Taking this into consideration. The fatigue notch factor Kf is Kf = 1 + q (Kt − 1) (a) No information for calculation of the notch sensitivity factor q is given here. 9/1x. If the second term in the material relation (9. The bar is subjected to an alternating stress σ∞ = ± 300 MPa.9Kt = 2. 006574. The stress and strain concentration factors Kσ and Kε then become Kσ = σmax 499 εmax 0.663 and Kε = = = 4. as it should. (b) Display the hysteresis loop for the material at the notch. The lower turning point is situated at σ = − σmax = − 499 MPa and ε = − εmax = − 0.383 = 7.7 ⋅ 300 = 3.4) and (9. when the remote loading is σ∞ = ± 300 MPa According to problem (a) above the upper turning point of the hysteresis loop is situated at σ = σmax = 499 MPa and ε = εmax = 0.l) 1/n’ 1/ 0. see point B in the figure.29 = 2.h) (Verify: Kσ ⋅ Kε = 1. giving N = 2300 cycles to fatigue failure. see figure below. (9.663 ⋅ 4.006574.0 ( 2N ) − 0. and by use of mean stress σm = 0. see point A in the figure below.72 ⋅ 3002 ⎪σ ⋅ ε = = = 3.7). One obtains ⎧ Kf2 σ2∞ 2. According to Morrow.18 ⎪ Δσ ⎛ Δσ ⎞ ⎛ Δσ ⎞ Δσ ⎪Δε = +2⎜ ⎟ = +2⎜ ⎟ ⎩ E ⎝ 2K’⎠ 200 000 ⎝ 2 ⋅ 1344 ⎠ Page 9:20 Chapter 9 . Choose the point A as starting point of the hysteresis loop.18 ⎪ σa ⎛ σa ⎞ 1/n’ σa ⎞ ⎛ σ a ⎪ε = + ⎜ ⎟ = ⎟ +⎜ a E ⎝ K’⎠ 200 000 ⎝ 1344 ⎠ ⎩ This system of equations gives σa = σmax = 499 MPa and εa = εmax = 0.2805 ⎪ E 200 000 ⎨ (k.006574. the fatigue life 2N = 4600 reversals to failure is obtained. Using Δσ∞ = 300 MPa one obtains 2 2 2 2 ⎧ ⎪Δσ ⋅ Δε = Kf (Δσ∞) = 2.) Now the number of cycles to fatigue failure may be determined.095 + 1. A nominal change of stress Δσ∞ (far away from the stress concentration) will cause a change of stress Δσ and a change of strain Δε at the stress concentration.3828 σ∞ 300 σ∞ / E 300/ 200 000 (g.006574 = = 1.006574. one obtains σ’f − σm 1227 εa = (j) ( 2N )b + ε’f ( 2N )c = ( 2N ) − 0.intersection point between the Neuber hyperbola and the material stress-strain relation (for cyclic loading).72 = Kf2.66 E 200 000 Using εa = 0.f) 1/ 0.2805 ⎪ a a E 200 000 ⎨ (e. The changes Δσ and Δε are obtained by use of the Neuber hyperbola and the material relation for changes of stress and strain. but now the bar is loaded to a nominal stress σ∞ = 150 ± 150 MPa. starting at point B.006574 + 0. see the figure.006574 C B 0.499 MPa At a change of stress Δσ∞ = 300 MPa the upper branch of the hysteresis loop.0. the bar has been loaded (a number of times) up to stress σ∞ = 300 MPa and then unloaded by the stress Δσ∞ = 300 MPa so that the total stress far away from the stress concentration will be σ∞ = 0.006574 0. At the stress concentration. will pass through the point C.004662 ⎪ ⎩ = − 0.004662 = 0. The lower branch of the hysteresis loop.7) will be investigated. starting at point A.001912 (o. The co-ordinates of point C are obtained from ⎧σC = σA − Δσ = 499 MPa − 702 MPa = − 203 MPa ⎨ (m. will pass through point D. Chapter 9 Page 9:21 . 499 MPa . however. i.001912 Verify that the change of stress Δσ∞ = 600 MPa will take this branch of the hysteresis loop to the point B.From this system of equations Δε = 0. Comment It is noticed that when the loop passes the point C the external loading (far away from the hole) will give a nominal stress σ∞ equal to zero.001912 203 MPa A D .004662 and Δσ = 702 MPa are solved.p) By use of these values (and perhaps some more) the hysteresis loop may now be drawn. This implies that the stress at the hole will vary between σC and σA according to problem (b).006574 − 0.e.203 MPa . The coordinates of point D are obtained from ⎧σD = σB + Δσ = − 499 MPa + 702 MPa = 203 MPa ⎪ ⎨εD = εB + Δε = − 0. The nominal (remote) stress σ∞ will now vary between 0 and 300 MPa.0.001912 . (c) Fatigue life when σ∞ = 150 ± 150 MPa The same bar as above (Kf = 2.n) ⎩εC = εA − Δε = 0. the residual stress σC and a residual strain εC will remain. 0 ( 2N ) − 0.499 MPa This gives Δσ = 519 MPa and Δε = 0. the starting point of the loop at point A where σA = . as in (j.0.As an extra exercise the hysteresis loop between A A and C will be drawn. respectively (c) Nf = 47 800 cycles (according to Morrow). approximately.001912) / 2 = 0.458 ⎪ E 200 000 (q.006574 will be chosen. and the strain amplitude will be εa = (εA − εC) / 2 = (0. to fatigue failure is solved.002810 = 0. The point E on the hysteresis loop is now obtained from ⎧σE = σA − Δσ = 499 MPa − 519 MPa = − 20 MPa ⎨ (s. (b) stress and strain at hole is σa = ± 499 MPa and εa = ± 0.006574 2 2 2 2 ⎧ ⎪Δσ ⋅ Δε = Kf (Δσ∞) = 2.v) ⎩εF = εC + Δε = 0.006574 − 0. Here Δσ∞ = 200 MPa . By use of these values in (v). one obtains 1227 − 148 εa = 0.006574 − 0.002331.006574. according to Morrow. E 0. once again.3.66 200 000 From this 2N = 95 600 reversals.7 ⋅ 200 = 1. obtained in Section 9.001912 + 0. Page 9:22 Chapter 9 . and by use of the given parameter values.002331 = (x) ( 2N ) − 0. giving N = 47 800 cycles.001912 499 MPa and εA = 0. It gives. Select.095 + 1. Determine one 203 MPa D more point on the loop.0.r) 0.k) above.t) ⎩εE = εA − Δε = 0.002810 = 0. from σ’f − σm (w) ( 2N )b + ε’f ( 2N )c εa = E The mean value σm of the stress becomes σm = (σA + σC) / 2 = (499 − 203) / 2 = 148 MPa.006574.203 MPa ⎪Δε = +2⎜ ⎟ 200 000 ⎝ 2 ⋅ 1344 ⎠ ⎩ 499 MPa B .1.004722 The number of loading cycles to fatigue failure is obtained.18 C ⎪ ⎛ Δσ ⎞ Δσ F .002810. (This result may be compared with the result N = 6 500 cycles to fatigue failure obtained when σ∞ = 200 ± 200 MPa.003764 The point F on the loop (branch starting at C. see the figure) gets the coordinates ⎧σF = σC + Δσ = − 203 MPa + 519 MPa = 316 MPa ⎨ (u.) Answer: (Compare with the solution given in example above) (a) Number of cycles N to crack initiation is 2300.001912 ⎨ 1/ 0.