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Kinesiology 201 Solutions Tissue Mechanics Tony Leyland School of Kinesiology Simon Fraser University Connective Tissue 1 a) Tensile stress = 2500/2 = 1250 N/cm2 b) The area has been reduced by 10% and therefore is now 1.8 cm2 Tensile stress = 3000/1.8 = 1667 N/cm2 c) The ligament may fail completely. As can be seen from a typical ligamentous tissue stressstrain curve opposite, once the collagen fibres begin to fail, the ligament will tend to fail completely without much further increase in stress (indeed the stress may even reduce but the ligament still will fail). Point of rapid failure Stress 2. A visco-elastic tissue is able to resist deformation with a greater force when loaded quickly. Many fractures occur due to acute loading (except stress fractures) so this is a beneficial property. As can be seen from the graph opposite the bone not only offers greater resistance when loaded quickly it can also withstand greater deformation before failing. [3] 3. Strain is defined as change in length divided by original length. Therefore: 0 Load 5 Strain (percent) 10 Fracture Quick Slow Fracture Deformation Δlength = 0.02 m and original length = 0.1 m. Therefore: strain = 0.2 (or 20%) [1] As strain is a ratio of lengths it has no units. [1] 4. (i) Spreads load. Cartilage can reduce maximum contact stresses by 50% or more. (ii) Reduces friction during movement. 5. a) The table shows that the total mass is 74.4 kg and that 4.4 kg is located in shank and foot (lower = 3.4 kg, foot = 1 kg). So the total mass located above knees is 65.6 kg [74.4 - (4.4 x 2)] representing a force of -644 N. Therefore compressive force on each knee is -322 N. b) If he stands on one foot then the entire body except the mass of one lower and one foot is above support by that knee joint. Therefore, the mass above knee is 70 kg (74.4 - 4.4) representing a force of -687 N. the the the leg the 1 c) Obviously the contact forces would be higher still. If you factored in the muscle forces needed to accelerate and decelerate in activities like running you can imagine how high the loads are on the lower joints (approximately three times body weight in distance running and higher still in sprinting). 6. Total relative mass above L5/S1 level = M (we must account for two arms, forearms and hands) M = 0.073 + 0.507 + (2 x (0.026 + 0.016 + 0.007)) = 0.678 Weight on this disk = = total body mass x relative mass (90 x 0.678) x 9.81 = 598.40 N = 594 N Pressure = Force/area = 598.40/25 = 23.94 N/cm2 (or 239442.48 Pascal [N/m2]) (or 239.44 KiloPascal [kPa]) 7. Ligament Crimp [2] 8. Bone is composed of water (approximately 25-30%) [1], protein (about 5-15% primarily in the form of collagen fibres) [1] and minerals (60-70% primarily calcium phosphate) [1]. Note that giving percentages is more information then was requested in the question. It is termed a two-phase material as it is structured to effectively resist both compression and tension. [1] Anisotropic Characteristics 9. Anisotropic material is a material that does not have the same strength when loaded in different ways. The graph opposite shows the quantitative strength of bone under three loading conditions. [2] Compression is loading that is directed along the long axis of the specimen and tends to crush the specimen. [1] tension is directed along the long axis and tends to pull the specimen apart. [1] and shear is force directed at right angles to the long axis and tends to bend or torque the specimen. [1] Stress to Fracture Compression Tension Shear 2