A) Assume the PCL is cylinder, with a length of 40 mm, and a diameter of 10 mm.
ID: 3482096 • Letter: A
Question
A) Assume the PCL is cylinder, with a length of 40 mm, and a diameter of 10 mm. How much tensile force is required to stretch it by 0.1%. Assume Y= 10^8 N/m^2.
B) Explaine in no more than 4-5 sentences the compressional performance of this ligament, i.e. what can you expect when you attempt to compress it. Justify your answer with mechanical and/or evolutionary arguments. How much tensile force is required to stretch it by 0.1%. Assume Y 10 pts) xplain in no more than 4 5 sentences the compressional perform igament, i.e. what can you expect when you attempt to compress it. nswer with mechanical and/or evolutionary arguments. (10 pts) Femur (thighbone) Articular cartilage ateral llateral pament (LCL)Patella nterior cruciate Medial collater ligament (MCL ament (ACL) eniscus ads Posterior cruci ligament (PCL Fibula Tibia
Explanation / Answer
Response
The amount of elongation or elongation in under tension is can be calculated with the formula.
Elongation = (1 × F) / (A × E), in
L is length, in meter
F is applied force.
A is cross section, square meter;
E is modulus of elasticity
We have to find the Force so rearrange equation
Force = (A × E)/Elongation× L
As we know our length is given 40 mm
Diameter is 10mm
Radius= diameter/2
Radius =5mm
A=Area of cyclinder
A=2rh+2r2
= 2 (3.144X5X40)+2(3.144+25)
=1413.72mm2
= 1.41372 m2
Modulus of elasticity =108 N/m2
Elongation =0.1 %
= 0.1*40
=4 mm
=0.004 m
Length= 40 mm
=0.04m
Substituting in formula
Force = (A × E)/Elongation
=(1.4132m2 X108N/m2)/0.004m
=1.4132 m2 X108N/0.004 m3
=353.3X108 N /m
=3.533X1010N/m
B)
As the ligament is in cylindrical in shape Cylindrical shapes greater gives uniformity of results and compressive stress is uniformly distributed in horizontal planes in comparison to other shapes.