Part 1 1. (1p) Define the Kronecker delta and explain its use. The Kronecker delta δ ij is defined as δ ij = 0 if i j 1 if i = j and it is used in tensor equations to include (δ ij = 1) or "sort out" (δ ij = 0) terms.. (1 p) Give a short explanation, including figure, of the Mohr fracture criterion. shear If the largest of the three Mohr stress circles is large stress enough to reach (to contact) a limiting border in the stress diagram (σ-τ diagram), failure (often brittle) is normal expected. In the figure, the left circle indicates that stress failure is not expected (the circle does not touch the borders), whereas the right circle crosses the borders, no failure failure and failure is expected. 3. (1 p) In a case where the fracture toughness depends on crack length a (we use the material resistance curve R(a) to describe that), give the conditions that must be fulfilled for a crack of length a 0 shall have an unstable propagation. The energy release rate G must, at a = a 0, fulfill the following conditions G(a)=R(a) and dg da > dr da 4. (1 p) Give the equation of the Neuber hyperbola, define the parameters involved, and explain how the hyperbola is used. The Neuber hyperbola reads ε σ= (K f σ ) 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. 7 007-10-/TD
Part 5. (3 points) A long thin-walled cylindrical pressure vessel of metal contains gas with pressure p. The necessary wall thickness t of the pressure vessel shall be determined by the "leak-before-break"-criterion. This criterion says that the vessel must be able to withstand a crack with any orientation and with length a up to double the thickness of the wall without breaking. (The idea of this type of criterion is that a crack of this size will cause leakage that leads to decreased pressure in the vessel and detection of the crack.) In this particular case, a safety factor s = agains fracture is required, i.e., K I K Ic /. It is assumed that the radius of curvature of the wall is much larger than the wall thickness (r >> t) so that the stress intensity factor can be determined as for a flat sheet. (a) Assuming that LEFM can be used, determine the required wall thickness of the pressure vessel. (b) Knowing the wall thickness, what can be said about the LEFM assumption? Numerical data: p = 3 kpa, vessel radius r = 00 mm, and K Ic = 50 knm -3/. The most dangerous crack is perpendicular to the largest principal stress in the wall. This stress is σ = pr/t. The stress intensity factor is K I =σ πa f, where, here, f =1. The fracture criterion K = K Ic / gives K I =σ πa = pr t πa = K Ic giving 3 0. t πt = 50 giving t = 1.81 mm (and a = 3,6 mm). (b) Check the LEFM condition. One has (t is smallest of t, a and W a) t.5 K Ic sσ Y giving σ Y 99 MPa. Thus, the yield limit should be larger than this. 8 007-10-/TD
6. (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 h = 30 mm (the width t is large). At plane strain, the stress intensity factor can h be written a where ν is the Poisson ratio, ν = 0,3, and p is p loading per unit of the width of the test specimen. Paris law for the material reads da dn = 5.1 10 1 (Δ K I ) 3 m per cycle where ΔK I has the unit MPa m 1/. The fracture toughness of the material is (in this example) determined in the following way: First a crack of length a 0 = 0 mm is created by machining. Then, to obtain a sharp crack tip, the test specimen is subjected to a loading p such that the load p is cycled between 0 och 1 MN/m. This is done for N = 0 000 cycles. When this is done the cycling is stopped and the load p is increased monotonicly until fracture occurs. Determine the fracture toughness of the material if failure occurs at p = 1.5 MN/m. The yield limit of the material is σ s = σ Y = 1000 MPa. First determine the crack length after the cyclic loading. Paris law gives Which gives da dn = 5.1 10 1 3 0, 91 1 from which is solved a final = 38. mm. At the final, monotonicly increasing loading the critical crack length is a final = 38. mm. The fracture toughness is obtained from Thus K Ic = 99 MPa m 1/. Here linear elastic fracture mechanics (LEFM) was used. Is it allowed? One has.5 K Ic =.5 99 = 0.045 < a σ Y 1000 final Thus, LEFM may be used for the final failure. a final 3 1 a 0.015 3/ m per cycle K I = 3 1 ν pa h 3/ 1 1 a = 0.045374 N = 907.5 0 (1/m ) K I = K Ic = 3 p failure a final = 3 3 0.038 = 99 MPa m 1 ν h 3/ 0.91 0.015 3/ 9 007-10-/TD
Part 7. (3 points) To determine the fatigue limit of a material a number of experiments were performed according to the "staircase" method. Calculate an estimation of the mean value and the standard deviation of the fatigue limit of the material if the following test series were obtained (stresses in MPa) σ a = 60, 66, 7, 78, 84, 78, 7, 78, 84, 78, 84, 90, 84, 78, 7, 78, 84, 78, 84, 78, 7, 78 MPa, where the last test (78 MPa) became a run-out. The fatigue limit is supposed to have a normal distribution. From the load levels used in the test it is concluded that the test specimens failed (indicated by an F) or did not fail (became run-outs, indicated by RO) as follows: σ a = 60(RO), 66(RO), 7(RO), 78(RO), 84(F), 78(F), 7(RO), 78(RO), 84(F), 78(RO), 84(RO), 90(F), 84(F), 78(F), 7(RO), 78(RO), 84(F), 78(RO), 84(F), 78(F), 7(RO), 78(RO) MPa. The test series seems to have started on a too low stress level. Therefore, from the tests the first three tests are discarded (the first change from run-out to failure came at stress level 78 MPa). From the remaining 19 tests 10 are run-outs and 9 are failures. The less frequent event should be analysed. Thus, analyse the failures. The failures occurred at σ a = 84(B), 78(B), 84(B), 90(B), 84(B), 78(B), 84(B), 84(B), 78(B) MPa. These events give the following table (see compendium) I II III IV V 78 0 3 0 0 84 1 5 5 5 90 1 4 N =9 A =7 B =9 The mean value of the fatigue limit is estimated to (step d = 6 MPa) S m = S 0 + d A N 1 = 78 + 6 7 9 1 = 79.7 = 80 MPa Further, one obtains NB A = 9 9 7 = 0.395 N 9 The standard deviation of the fatigue limit is estimated to s = 1.6d NB A + 0.09 = 4.1 MPa N 10 007-10-/TD
Part 0,003 0,001-0,005 töjning/ strain tid/time 8. A material is subjected to a load sequence giving the strain shown in the figure. Use the Morrow relationship to determine the expected number of load sequences to fatigue failure. The influence of the mean stress can be neglected. Material properties: E = 00 GPa, ν = 0.3, σ U = 700 MPa, Ψ = 0.65, σ f = 900 MPa, ε f = 0.6, b = 0.095, and c = 0.47. The diagram gives five cycles: - One cycle with strain range Δε = 0,0055, giving strain amplitude ε a = 0,0075. According to Morrow one obtains (neglecting the mean stress) ε a = σ f E (N)b +ε f (N) c giving 0, 0075 = 900 00 000 (N) 0,095 + 0, 6 (N) 0,47 Solving for N gives N = 44 550 cycles (N is load reversals to failure). - Two cycles with strain range Δε = 0,005, giving ε a = 0,0015. Morrow gives 0, 0015 = 900 00 000 (N) 0,095 + 0, 6 (N) 0,47 Solving for N gives N = 180 000 cycles - Two cycles with strain range Δε = 0,003, giving ε a = 0,0015. Morrow gives 0, 0015 = 900 00 000 (N) 0,095 + 0, 6 (N) 0,47 Solving for N gives N = 70 000 cycles. Palmgren-Miner now gives 1 D = 44 550 + 70 000 + 180 000 = 1 38 50 Expected number of sequences to failure is 38 000. 11 007-10-/TD