Medical prosthetic implants such as hip replacements have traditionally been mad
ID: 1842473 • Letter: M
Question
Medical prosthetic implants such as hip replacements have traditionally been made of metals such as stainless steel or titanium. These are not ideal as they are much stiffer than the bone, giving relatively poor transfer of load into the bone. New composite materials are being developed to provide a much closer stiffness match between the implant material and the bone. One possibility uses high-density polyethylene (HDPE) containing particulate hydroxyapatite (HA), the natural mineral in bone, which can be produced artificially. Data for some experimental composites are provided in the table below, and those for the bulk materials are HDPE 0.65 GPa and Hydroxyapatite 80 GPa. (a) Plot the upper and lower bounds for the Young's modulus of HDPE-HA composites against the volume fraction of reinforcement, f (from 0 to 1), together with the experimental data. (b) Which of the bounds is closer to the data for the particulate composite? Use this bound as a guide to extrapolate the experimental data to identify the volume fraction of hydroxyapatite particulate that is required to match the modulus of bone, E = 7 GPa. Is this practical?Explanation / Answer
solution:
1)composite is mixture of two different material in suitable volume fraction and to achieve desire properties for target material.
2)in this case data is given for particulate composite and it is mixture of HDPE and HA and matrix is HDPE and particulate fibre of HA are used.
3)based on given experimental data we can provide upper and lower bounds on youngs modulus of composites and it is as folllows
i)when volumefraction of fibre of HA is zero in HDPE matrix then whole strength is mainly due to HDPE and it is minimum strength of composite and hence lower bound on youngs modulus value is when volume fraction of fibre is zero that is
lower bound, for f=0,Ec=0.65 GPA
ii)when fibre are constiture whole volume of composite then we get maximum value for strength of composite and it is upper bound on youngs modulus value for composites
upper bound,f=1,Ec=80 GPA
4)based on experimental data given as follows we can get equation of interpolation by lagranges interpolation formula and data is as follows
(f,Ec)=(0,.65),(.1,.98),(.2,1.6),(.3,2.73),(.4,4.29),(.45,5.54),(1,80) and we have to find value of f for Ec=7 GPA
so we get interpolation formula for lagranges interpolation as
Ec=Ec0*L0+Ec1*L1+Ec2*L2+Ec3*L3+Ec4*L4+Ec5*L5+Ec6*L6+Ec7*L7
L=lagranges coefficient
Ec=7 GPA
we have to calculate equation for them and it comes as follows
Ec=1961.04f^6+4251.44f^53101.77f^4+1011.36f^3129.477f^2+8.8303f+.65
for Ec=7 gpa we get value as roots of above equations
1961.04f^6+4251.44f^53101.77f^4+1011.36f^3129.477f^2+8.8303f-6.35=0
solution of this equation will gives us value of volume fraction for composite and on solving it come to be
out of six roots ,four are imaginary and remaining two roots are real and they are as follows
f=.48969 or f=1.12253
as we have limit for f as 0<f<1
so we choose f as f=.48969
5)in this way for volume fraction of f=.48969 we will have desirable properties of composite as Ec=7 GPa and it is practical to get this volume fraction for composite as per lagranges interpolation formula within lower and upper bound.