Part II solutions

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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  1m

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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