2. (25 pts) One common graft for surgical reconstruction of a damaged anterior c

Question

2. (25 pts) One common graft for surgical reconstruction of a damaged anterior cruciate ligament (ACL) is a central portion of the patient's patellar tendon (PT) as shown Figure 1. The dimensions of the PT are 5 mm thick x 24 mm wide x 50 mm long and it has a tensile modulus of 220 MPa. The original ACL has a length and cross-sectional area of 27 mm and 44 mm^2, respectively and a tensile modulus of 110 Mpa.Determine the percentage of the PT width that must be taken to match the structural properties of the original ACL. (Assume that both tissues are linear elastic.)

Explanation / Answer

Let y be the Tensile modulus.

y=F*l/(A*x)
where F=force
l=length
A=cross sectional area
x=change in length.

to match structural properties change in length per unit force shoulld be same.

hence
x1/F1=x2/F2

x/F=L/(y*A)

hence
or L1/(y1*A1)=L2/(y2*A2)
now lets n be the fraction of PT width .Hence PT width be n*24.
putting in the equation.
we get
L1=27, L2=50
y1=110, y2=220
A1=44 ,A2n*24*5




27/(110*44)=50/(220*n*24*5)
179.2592=n*528
or n=0.3395
in percentage n=33.95% nearly equals 40%

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