At o degree of knee flexion, the contact area between the tibia and the femur is

Question

At o degree of knee flexion, the contact area between the tibia and the femur is 1.6 cm2. If the compressive force across the tibiofemoral joint is 1350 N, the mechanical stress (in Mpa) would be calculated to be 8.44 84.4 8440 need more information to calculate At o degree knee flexion, the mechanical stress at the tibifemoral join is 843, 75n/cm2. If the compressive force is 1350 N, the joint contact area (in cm2) must be 0.160 1.600 16.00 160.0 need more information to calculate Stiffness is the slope of the curve. stress-strain stress-length force-length force-strain A tendon starts of with an initial length of 10 cm and a cross-sectional of 0.1cm2. A force of 100 N is applied, which streches the ligament to a new length of 11 cm. What is the modulus of Elasticity for this ligament? 0.01mpa 0.1mpa 1 mpa 10 mpa 100 mpa

Explanation / Answer

1. stress = force/area = 1350*10000/1.6 = 8.44 MPa

2. area = force/stress = 1350/843.75 cm^2 = 1.6 cm^2

3. k = stress/ strain => slope of stress - strain curve

4. modulus of elasticity = strss/strain = 100*10*10000/1*0.1 = 100 MPa

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