Many solids can absorb energy without permanent deformation. To prevent injury,

Question

Many solids can absorb energy without permanent deformation. To prevent injury, this is a vital property for bones. Bone fractures can be caused by falls over relatively short distances onto hard surfaces, such as concrete. The following calculation will illustrate this point. The figure shows a model of a human tibia (lower leg bone). The central portion of the bone is modelled as a hollow cylinder of cortical (compact) bone of length b = 24.20 cmand cross-sectional area, Ac = 5.887 cm2. The ends of the bone are solid cylinders of trabecular (spongy) bone, with cross-sectional area At = 18.398 cm2 and total length 2a = 15.60 cm.

Which tissue is going to fail first when the bone is stressed as shown in the figure? (Enter C - cortical or T - trabecular bone.) Assume that the tissues are all perfectly elastic up to the ultimate compressive strength (UCS).

Calculate the minimum force required to cause a fracture in the bone.

Calculate the strain energy in the trabecular bone just before fracture. (Don't forget to include both ends of the bone in your calculation.)

Calculate the strain energy in the cortical bone just before fracture.

Consider a 60.5 kg person who falls from a certain height onto concrete, landing on both feet, with legs held stiff and straight. Calculate the minimum height from which the person can fall and suffer a fracture. Assume that the bones behave elastically up to the breaking point.

Tissue UCS (MPa) Y (MPa)
trabecular (spongy) bone 22 360
cortical (compact) bone 162 10600 Many solids can absorb energy without permanent deformation. To prevent injury, this is a vital property for bones. Bone fractures can be caused by falls over relatively short distances onto hard surfaces, such as concrete. The following calculation will illustrate this point. The figure shows a model of a human tibia (lower leg bone). The central portion of the bone is modelled as a hollow cylinder of cortical (compact) bone of length b = 24.20 cmand cross-sectional area, Ac = 5.887 cm2. The ends of the bone are solid cylinders of trabecular (spongy) bone, with cross-sectional area At = 18.398 cm2 and total length 2a = 15.60 cm. Tissue UCS (MPa) Y (MPa) trabecular (spongy) bone 22 360 cortical (compact) bone 162 10600 Which tissue is going to fail first when the bone is stressed as shown in the figure? (Enter C - cortical or T - trabecular bone.) Assume that the tissues are all perfectly elastic up to the ultimate compressive strength (UCS). Calculate the minimum force required to cause a fracture in the bone. Calculate the strain energy in the trabecular bone just before fracture. (Don't forget to include both ends of the bone in your calculation.) Calculate the strain energy in the cortical bone just before fracture. Consider a 60.5 kg person who falls from a certain height onto concrete, landing on both feet, with legs held stiff and straight. Calculate the minimum height from which the person can fall and suffer a fracture. Assume that the bones behave elastically up to the breaking point.

Explanation / Answer

1. 40475.6 N --> UCS*Cross-sectional area [done for T bone since it breaks first]

2. 192.93 J --> strain energy equation is (1/2)* (V/E) * (stress^2)

V=volume and you do area*length of T bone

E is young's modulus for the T bone (360*10^6 Pa)

Stress is Force/Area --> use the force from question 1 and divide it by the area of the T bone

3. 31.77 J

Do the same thing but use the values for the C bone (i.e. the Young's modulus would be 10600*10^6 Pa and the stress would be the force in question 1/area of the C bone)

4. 0.758 m

Add the strain energies in 2 and 3 and divide it by the mass*acceleration due to gravity. then multiply it by 2 because you have two legs.

Cheers

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