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Total Life Given: A high cycle fatigue power law, with the coefficients A and b obtained from experiments on the alloy. S n A( N f ) b a) Engineers run S-N tests (in repeated loading, periodically from 0 to Sn) to find that Nf=1 x 105 cycles for Sn=800 MPa, Nf=1 x 107 cycles for Sn=510 MPa. What are the coefficients A and b? Nf = # of cycles to failure, Sn= Stress range. Given the equation above, a and b are as follows: 800 MPa A(1E 5)b 510 MPa A(1E 7)b 800 log 2b 510 b 0.0977 800 a (1E 5) 0.0977 a 2463 MPa b) Find the stress life for an unetched neck (no stress concentration). Assume that the neck is cyclically loaded during daily activity to the maximum stress computed in part (b) above. Choose (justify) an appropriate minimum stress. In Part. 1, Static Analysis, you were told that the force is some multiple of the body weight, α. As determined from gait analysis, the maximum load that occurs is about 4x body weight. Hence, you solve for the maximum load (Pmaz) with α=4. The maximum stress at point A and point B are tensile and compressive, respectively. Since the stresses at point B are compressive, we are not concerned with the stress range there. What remains to be determined is the minimum stress at point A. The minimum could be tensile or zero. Generally, during walking α is about zero. I accepted either of these two assumptions, but for simplicity, the solutions here assume α=0 for computed the minimum stresses (α=0). From Part I, ∆σ @ A = 348 MPa. Substituting back into the equation: 348 2463( Nf ) 0.0977 Nf 4.9 E108 Cycles c) If the device is required to survive for 3 x 107 cycles (assuming an unetched surface and no endurance limit), will the device fail? Since the number of cycles to failure is greater than 3 x 107 cycles, the device will not fail. d) As given in the case study, the device failed after 43 months post-operatively. If the patient loaded the implant for 1 x 106 cycles/year, what was the total number of cycles to failure? What would the stress concentration factor from the laser etching have to have been to cause failure at 43 months? Total number of cycles to failure: 43Months cylces cycles (1E 6 ) 3.58 E 6 12months year year Setched 2463(3.58 E 6) 0.0977 564 MPa 564 SCF 1.62 348 e) If laser etching has to be included in the design, how might the neck design (geometry, materials, implantation) be changed to allow the etching but still meet the 3 x 107 cycles life requirement? There were several acceptable answer: (1) Same material, larger x-sectional area →lower stresses. What would be a reasonable x-section? Will it lead to excessive stress shielding? Would material costs be prohibitive? (2) Decrease neck length → lower moment. How short a neck is anatomically possible without compromising range of motions? (3) Different material → enhanced toughness. Will the material yield? (4) Placement of the etching in a place where tensile loading is minimized These are all qualitative suggestions, but I wanted to see at least one concrete calculation proving that the device wouldn’t fail with the design changes. Total life curves are usually generated from smooth, unnotched specimens without any stress concentration. Discuss what the implications are for applying these data to geometries with notches. Is the simple addition of a stress concentration factor due to a flaw sufficient to obtain a reliable result? Total life takes into account both the crack initiation and the propagation, both of which are non-linear processes. Hence, introducing a stress concentration through a flaw essentially changes the crack initiation phase. Thus, the assumption that the propagation phase can be effectively represented by using the total life approach is not necessarily sufficient. However, a number of experiment procedures and pre-factors have been devised that allow one to take a sample with known stress concentration and use existing data from un-notched specimens effectively. Damage Tolerant Given: A crack propagates from the stress concentration due to laser etching as given above. The crack is loaded such that the stress intensity factor KI for the crack is as follows: 1 M KI Pft fb B W W a 2 tan a 3 2W a ft 0.752 2.02 0.37 1 sin a W 2W cos 2W a 6 2 tan 2W 0.923 0.199 1 sin a 4 fb a 2W cos 2W Where P is the axial load and M is the bending moment, W, B = 10 mm. KIC = 60 MPa√m α=4 and d=25 mm. Assume m=100kg. Find: a) If ΔKthreshold is 5 MPa√m, what is the initial crack length required for crack propagation under the given service loading? From the equation for KI given, 1 M o K I Pft W fb . Here, ΔP is equal to the P max- B W Pmin in the axial direction (i.e, the direction along the neck axis only). Again, some component of the load must be tensile for crack propagation to occur, so we are concerned with the stresses at point A only. From calculations in Part. 1: Pmax 3000 N , Pmin 0 N P 3000 N M 0 min 0 Nm M 0 max 63 Nm M 0 63 Nm The initial crack length, a_init, is computed from the matlab code given below. a_ini t= 37 μm. b) What is the critical crack length for this service loading? Catastrophic failure will occur when the maximum value of KI reaches KIC (not when ΔKI reaches KIC. In this particular case, KI,max=ΔKI because KI, min = 0, but this is not typically the case). Using the same matlab code as above, KI,max = KIC = 60 MPa√m: acrit = 4 mm for catastrophic failure. c) Using the data provided in hip.xls, which is fatigue crack propagation data for the alloy in the Paris regime, find the coefficients of the Paris equation. According to the Paris Equation, fatigue crack propagation (change in crack length per cycle) is related to the stress intensity factor via a power law of the form. da C (K ) m dN From the plot given above (plotted on log-log paper and fit to a power-law the Paris Law coefficients are: C=9x10-11, m=2.9676 Matlab Code: d) Suppose an initial sharp flaw exists in the material, prior to implantation, at points A and B. If the number of cycles to failure is known (as computed above), then invert the Paris equation to determine the initial (Mode I) crack size. The inverted equation must be integrated numerically to obtain an answer. Please use the trapezoidal rule with crack length increments of 1 μm, and include your code. Include a plot of the crack length as a function of time. To solve the problem, we needed several pieces of information that have been computed throughout the problem: (1) KI as a function of geometric factors and loads (given in Problem statement) (2) Cyclic Loads (computed in Part 1) (3) Paris Law Coefficients: Computed as: C=9x10-11, m=2.9676 (4) Number of cycles to failure: Nf – from the case study, at 43 months Nf=3.588x106. (5) Final crack length: Since KIC and the specimen geometry is known, we know the critical crack size prior to when catastrophic failure occurred: acrit = 5.042 μm. To find an expression for the initial flaw size: da C (K ) m 9 x10 11 (K ) 2.9676 dN 1 M o 1 1 63 K Pf t fb 3000 f t fb B W W 0.01 0.01 0.01 af Nf 11 1 (9 x10 ) (K ) 2.9676 da dN N f 3.58 x10 6 cycles ai N i 0 Since the expression of initial flaw size is not tractable analytically, a numerical solution is required. According to the trapezoidal rule: af af 1 1.1x10 (K ) 10 2.9676 da f (a)a f (ai ) f (ai 10 m) a ai ai 2 2.9676 10 1 63 f (a) 1.1x10 3000 f t fb 0.1 0.1 0.01 a 1m Please see the attached matlab code for determining the initial flaw size based on the equations above. e) Your supervisor asks you to compare the fatigue performance of this material to that of other orthopedic materials. List the yield strength, elastic modulus, ΔKthreshold and KIC for several other common orthopedic metals in a table, and comment on how they might resist crack propagation and fracture. Are there any apparent trade-offs for performance in these materials? Be sure to include the worked condition (as wrought, cold rolled, hot forged, fully aged, etc.), for each entry in the table, as the material properties depend heavily on that state.

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