Avd Hållfasthetslära, IKP, Linköpings Universitet Tentamen/Examination TMHL61 Tentamen i Skademekanik och livslängdsanalys TMHL61 lördagen den 14/10 2000, kl 8-12 Solid Mechanics, IKP, Linköping University Examination in Damage Mechanics and Life Analysis TMHL61 Saturday, October 14, 2000, time: 8-12 Teacher: Tore Dahlberg, 013-28 1116, 070-66 51103 Examination Part I Part II Literature Part II Language Solutions Results Student s inspection Marking Part I (yellow papers): No books or other external facilities are allowed. The questions should be answered directly on the yellow papers. After having handed in Part I (the four yellow papers), Part II (white papers) is received. The following literature may then be used: 1. T Dahlberg: Fracture Mechanics, LiTH-IKP-S-470. 2. G Hedner (ed): KTH Formelsamling i Hållfasthetslära, or distributed sections from this handbook (for English-speaking students). 3. Tore Dahlberg: Material Fatigue, LiTH-IKP-S-459. 4. Handbooks and mathematical tables, dictionaries, glossary. 5. Any small, handhold calculator in one single unit and without any external communication (as printer, tape-recorder or others). In the literature minor notations "in the margin" are allowed, but no problems, solutions, or answers to problems are allowed. Solutions can be given in Swedish or English. Solutions will be posted on the Solid Mechanics (Hållfasthetsläras) notice-board, entrance 15 or 17, corridor B, immediately after the examination. They will also be made available on the www, address: http://www.solid.ikp.liu.se, under the heading "Examinations" Examination results will be posted on the same notice-board not later that October 31, 2000. The correction of the examination papers is open for inspection until December 15, 2000, whereafter they are distributed in the students "pigeonholes". To obtain a mark (grade), the examination and the laboratory work should be passed Points 0 to 5 6 to 8 9 to 11 12 to 16 Mark failed 3 4 5 NOTE! This page need not be handed in. Keep it to remember the dates and information given. Tore Dahlberg, Hållfasthetslära, tel 013-28 1116 2000-10-23
Name: Examination in Damage Mechanics and Life Analysis (TMHL61) Part 1, page 1 of 4 1. (1 point) Write the equations given by tensor notations below in full (i.e. what would the equations look like if tensor notations were not used)? σ i j, j + k i = ρ 2 u i t 2 1 2000-10-23/TD
Name: Examination in Damage Mechanics and Life Analysis (TMHL61) Part 1, page 2 of 4 2. (1 point) In case of small scale yielding, the Irwin model (or the Irwin correction) is sometimes used. Explain the main features of (huvuddragen - the main ideas behind) the Irwin model. 2 2000-10-23/TD
Name: Examination in Damage Mechanics and Life Analysis (TMHL61) Part 1, page 3 of 4 3. (1 point) Explain Mohr s failure criterion 3 2000-10-23/TD
Name: Examination in Damage Mechanics and Life Analysis (TMHL61) Part 1, page 4 of 4 4. (1 point) Give the equation of the Neuber hyperbola, define the parameters involved, and explain how the hyperbola is used. 4 2000-10-23/TD
Part II, page 1 of 2 0 2 a 5. (3 points) Consider a through-thickness crack of length 2a in a large plane plate subjected to a uniaxial stress σ 0. The plane of the crack is tilted an angle α with respect to the stress direction, see figure. (a) In which Mode(s) will the crack be loaded? (b) Determine K I for the crack. Fig 5 0 6. (3 points) Linus (or if it was Linnéa, I really don t know) has glues the two ends of his/her ruler (linjal) to the table. The ruler has the overall length L, of which the length l at each end of the ruler is glued to the table, see figure. The glue is rigid, so the ruler can be considered fixed at the ends along the parts with length l. Thus, the portion of the ruler that is not glued to the table is 2a = L 2l. The ruler (think of it as a beam of length 2a fixed at the ends) is loaded with a force P at its mid point (at the symmetry line). Assume that the space between the ruler and the table is a crack, and calculate the energy release rate G for the structure (for example by use of the compliance method). The material is linear elastic with modulus of elasticity E. The cross-sectional area of the ruler (beam) is b h. h P glue (rigid) l a a l 5 2000-10-23/TD
Part II, page 2 of 2 7. (3 points) One type of test specimen to determine the fracture toughness of a material is the double p cantilever beam, see figure. The height is 2h = 30 mm (the width t is large). At plane strain, the stress intensity factor can be written p a 2 h K I = 2 3 1 ν 2 p a h 3/2 where ν is the Poisson ratio, ν = 0.3, and p is the loading per unit of the width of the test specimen. Paris law for the material reads da dn = 5.1 10 12 ( K I ) 3 m per cycle where K I has the unit MPa m 1/2. The fracture toughness of the material is (in this example) determined in the following way: First a crack of length a 0 = 20 mm is created by machining. Then, to obtain a sharp crack tip, the test specimen is subjected to a cyclic loading such that the load p is cycled between 0 och 1 MN/m. This is done for N = 12 000 cycles. When this is done the cyclic loading is stopped and the load p is increased monotonicly until fracture occurs. Determine the fracture toughness of the material if failure occurs at p = 1.8 MN/m. The yield limit of the material is σ s = σ Y = 1000 MPa. 8. A flat bar of a linear elastic, ideally plastic material with a rectangular cross section has a small circular hole at its centre axis. The stress concentration factor at the hole is K t = 3.0 and the fatigue notch factor is K f = 2.8. stress Y E strain The bar is subjected to an alternating stress σ = ± 200 MPa. Estimate the number of cycles to fatigue failure by use of Neuber s method. Note that the material is linearly elastic, ideally plastic as shown in the figure. Use the elastic modulus E = 200 GPa and the yield strength σ Y = 430 MPa. Further, σ f = 1200 MPa, ε f = 1.0, b = 0.1, and c = 0.62. 6 2000-10-23/TD
Part 1 1. (1 point) Write the equations given by tensor notations below in full (i.e. what would the equations look like if tensor notations were not used)? σ i j, j + k i = ρ 2 u i t 2 σ xx x + σ x y y + σ xz z + k x σ yx x + σ y y y + σ yz z + k y = ρ 2 u x t 2 = ρ 2 u y t 2 σ zx x + σ z y y + σ zz z + k z = ρ 2 u z t 2 i.e. the equations of equilibrium, including mass intertia. Here σ ij are stresses, k i are loads per unit volume, ρ is density, and u i are displacements. 2. (1 point) In case of small scale yielding, the Irwin model (or the Irwin correction) is sometimes used. Explain the main features of (huvuddragen - the main ideas behind) the Irwin model. In case of small scale yielding there is a small plastic zone in front of the crack tip. In Irwin s model a fictitious crack length (a + r 1 ) is introduced so that the fictitious crack tip falls in (the middle of) the plastic zone. It is assumed that the elastic stress field in front of the plastic zone is similar to the elastic stress field one would have had in front of the crack if the plastic zone was not there, but the elastic stress field is moved away from the crack tip a certain distance (r 1 ) and stresses above the yield limit are removed. In the plastic region between the crack tip and the moved-away elastic stress field (with stresses below the yield limit) the stresses are set to the yield limit of the material. The distance r 1 is selected so that the force transmitted over the plane in front of the crack is the same in the two cases, i.e. σyy(x) dx 0 is the same in the two cases. Thus, in the Irwin model σ yy is the plastic-elastic stress field in front of the crack (with elastic stresses below the yield limit), and in the fully elastic case σ yy is the elastic stress field in front of the crack, with stresses going to infinity at the crack tip. 7 2000-10-23/TD
Part 1 3. (1 point) Explain Mohr s failure criterion If the largest of the three Mohr s circles is large enough to reach (to contact) a limiting border in the stress diagram (σ-τ diagram), failure (often brittle) is expected. In the figure, the left circle indicates that failure is not expected (the circle does not touch the borders), whereas the right circle crosses the borders, and failure is expected. shear stress normal stress no failure failure 4. (1 point) Give the equation of the Neuber hyperbola, define the variables and parameters involved, and explain how the hyperbola is used. The Neuber hyperbola reads σ ε = ( K σ σ K ε ε = ) (K f σ ) 2 E where σ and ε are stress and strain, respectively, K f is the fatigue notch factor, σ is the remote stress (far away from the stress concentration), and E is the modulus of elasticity. The Neuber hyperbola is used to determine the stress and the strain at a point in the material where stress concentration is present: the intersection of the Neuber hyperbola and the cyclic stress/strain relationship gives the stress and the strain at the point of stress concentration. The fatigue notch factor at the point with stress concentration is K f. 8 2000-10-23/TD
0 2 a Part 2 5. (3 points) Consider a through-thickness crack of length 2a in a large plane plate subjected to a uniaxial stress σ 0. The plane of the crack is tilted an angle α with respect to the stress direction, see figure. (a) In which Mode(s) will the crack be loaded? (b) Determine K I for the crack. Fig 5 0 Solution: (a) The crack will be loaded in Mode I (tension) and in Mode II (shear). n (b) First, determine the stress σ n perpendicular to the crack plane. Let the cross-sectional area of the plate be A 0. Equilibrium in the direction of the normal stress σ n gives, see figure 0 A 0 σ n sin α σ 0 A 0 sin α = 0 giving σ n = σ 0 sin 2 α The stress intensity factor K I becomes (case 1, a small crack in a large plate) K I = σ n πa 1 = σ 0 (sin 2 α ) πa 9 2000-10-23/TD
Part 2 6. (3 points) Linus (or if it was Linnéa, I really don t know) has glues the two ends of his/her ruler (linjal) to the table. The ruler has the overall length L, of which the length l at each end of the ruler is glued to the table, see figure. The glue is rigid, so the ruler can be considered fixed at the ends along the parts with length l. Thus, the portion of the ruler that is not glued to the table is 2a = L 2l. The ruler (think of it as a beam of length 2a fixed at the ends) is loaded with a force P at its mid point (at the symmetry line). Assume that the space between the ruler and the table is a crack, and calculate the energy release rate G for the structure (for example by use of the compliance method). The material is linear elastic with modulus of elasticity E. The cross-sectional area of the ruler (beam) is b h. h P glue (rigid) Solution: Load control gives Elementary case for a fixed-fixed beam gives Two crack edges, each of length b (thus, the total crack edge length is 2b), gives (page 4:7) or (page 4:9) G = 1 Π 2b a = 1 2b l a a l δ = P(2a)3 3 EI a Π = U W = Pδ 2 Pδ 2 Pδ = Pδ 2 (a) 1 1 8 8 = P a 3 24 EI = C P (b) = 1 2b P 2 G = P 2 d C 2 2b d a = P 2 2 2b δ a = P 2 4b 3a 2 24EI = 3P 2 a 2 8Eb 2 h 3 3a 2 24EI = 3P 2 a 2 8Eb 2 h 3 10 2000-10-23/TD
7. (3 points) One type of test specimen to determine the p fracture toughness of a material is the double cantilever beam, see figure. The height is 2h = 30 mm (the width t is large). At plane strain, the stress intensity factor can 2 h be written p a where ν is the Poisson ratio, ν = 0.3, and p is the loading per unit of the width of the test specimen. Paris law for the material reads da dn = 5.1 10 12 ( K I ) 3 m per cycle where K I has the unit MPa m 1/2. The fracture toughness of the material is (in this example) determined in the following way: First a crack of length a 0 = 20 mm is created by machining. Then, to obtain a sharp crack tip, the test specimen is subjected to a cyclic loading such that the load p is cycled between 0 och 1 MN/m. This is done for N = 12 000 cycles. When this is done the cyclic loading is stopped and the load p is increased monotonicly until fracture occurs. Determine the fracture toughness of the material if failure occurs at p = 1.8 MN/m. The yield limit of the material is σ s = σ Y = 1000 MPa. Solution: First determine the crack length after the cyclic loading. Paris law gives Which gives da dn = 5.1 10 12 2 3 0, 91 1 a 3 0.015 3/2 1 2 a final m per cycle K I = 2 3 1 ν 2 p a h 3/2 1 1 2 2 a = 0.045374 N = 544.5 0 (1/m2 ) from which is solved a final = 26.62 mm. At the final, monotonicly increasing loading the critical crack length is a final = 26.62 mm. The fracture toughness is obtained from K I = K Ic = 2 3 p failure a final = 99 MPa m 1 ν 2 h 3/2 Thus K Ic = 99 MPa m 1/2. Here linear elastic fracture mechanics (LEFM) was used. Is it allowed? One has 2.5 2 K Ic = 2.5 2 99 = 0.0245 < a σ Y 1000 final Thus, LEFM may be used for the final failure (t and W a are "large"). 11 2000-10-23/TD
Part 2 8. A flat bar of a linear elastic, ideally plastic material with a rectangular cross section has a small circular hole at its centre axis. The stress concentration factor at the hole is K t = 3.0 and the fatigue notch factor is K f = 2.8. stress Y E strain The bar is subjected to an alternating stress σ = ± 200 MPa. Estimate the number of cycles to fatigue failure by use of Neuber s method. Note that the material is linearly elastic, ideally plastic as shown in the figure. Use the elastic modulus E = 200 GPa and the yield strength σ Y = 430 MPa. Further, σ f = 1200 MPa, ε f = 1.0, b = 0.1, and c = 0.62. Solution: Due to the high stresses near the hole, the material will yield locally. The stress concentration factor and the strain concentration factor may be written K σ (Hooke s law is valid for stresses below the yield strength σ Y ε = σ / E.) The Neuber hyperbola becomes = 430 MPa. Thus, σ ε = ( K σ σ K ε ε = ) K 2 2 f σ (c) E The material data given (E), the fatigue notch factor K f, and the nominal stress amplitude σ = 200 MPa now determine the Neuber hyperbola. The local stress and strain amplitudes at the hole, σ a and ε a, respectively, are obtained as the intersection of the Neuber hyperbola and the material stress/strain relationship. One obtains σ ε = K 2 2 f σ E = 2.82 200 2 200 000 = 1.568 (d,e) σ = σ Y = 430 MPa This gives ε = ε a = ε max = 0.0036465116 (and σ = σ a = 430 MPa). Now the number of cycles to fatigue failure may be determined. According to Morrow, and by use of the mean stress σ m = 0, one obtains ε a = σ f σ m E Using ε a = 0.0036465, the fatigue life N = 17 257 (or 17 000) cycles is obtained. K ε K σ = σ max and K σ ε = ε max = ε max ε σ / E (a,b) ( 2N ) b + ε f ( 2N ) c = 1200 200 000 ( 2N ) 0.1 + 1.0 ( 2N ) 0.62 (f) 12 2000-10-23/TD