Elasticity: Stress, Strain and Compression

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1. The figure below shows the forces acting on a tibia when a person stands on the ball of one foot in equilibrium. As shown, the force of the tibia on the ankle joint for a person (of weight 750 N) standing this way is 2800 N. The tibia has a length of 0.4 m, an average inner diameter of 1.3 cm, and an average outer diameter of 2.5 cm. (The central core of the bone contains marrow that has negligible compressive strength. Take the Young's modulus of bone to be 17.0 x 109 Pa.)

(a) Explain why the force on the tibia isn't equal to 750 N.

The foot is not supported on only on tibia but the tension in fibula and mussels is also there. Thus the net force on tibia is the resultant of the weight of the person and tension in the other parts.

(b) What is the average cross-sectional area of the tibia (you can assume it's a circle)?

The cross-sectional area is the difference of the area of outer and inner circles and thus will be

Or

Or m2

Or m2

(c) What is the compressive stress in the tibia?

The stress in the material is given by the force per unit area and thus stress in tibia is given by

(d) What is the change in length for the tibia due to the compressive forces?

The ratio of the change in length L and length L is called longitudinal strain. Thus strain in the tibia is given by

Now according to the Hook's law within elastic limits, stress is proportional to strain and the constant of proportionality is called modulus of elasticity.

For longitudinal strain the constant is called Young's modulus Y and thus we can write

Or

Or

Or m

Thus compression in the length of tibia is 1.84*10-4 m or 0.184 mm.

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