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SOLUTIONS Mechanics of Biomaterials ENGSCI 274SC 2003 1. Loads Stress Strain Displacements Constitutive Law 2. (i) My / I (ii) I y 2 dA 3. 3wL 3wLx wx 2 V ( x) wx, M ( x) 8 8 2 4. So the neutral axis is 4mm up from the base of the beam. 5. 1 1 1 3 v 2 PLx 6 Px 2 EI 6. z zx yx xx x y 7. (no shear stress anywhere) Could evaluate max. shear stresses in the other two perpendicular planes, but this was not expected (question could have been a bit clearer) 8. homogeneous: material properties are the same everywhere incompressible: volume remains constant 9. 12 means the (negative of the) transverse strain in the “2” direction divided by the strain in the “1” (applied) direction 10. Types of biological materials that can be described by the linear elastic model: hard tissue, e.g. bone, or soft tissue under low stress. Biological materials that cannot be described well by the linear elastic model: soft tissue at higher stress levels 11. The equations are 2 0 The question actually implies that 1 0 , but this doesn’t make sense, since you need a compressive stresses 1 , 2 if the material is pushed up against the constraint. Question a bit ambiguous here. 12. E 13. (i) A biological material which is fairly viscoelastic: Myocardium One which is not very viscoelastic: Bone (ii) Creep: Relaxation: 14. creep compliance: the strain response to a unit load Useful to know because we can then obtain the strain response to any loading (using the hereditary integral condtitutive law) 15. Let pi p, p o 0 and ri ro r (the average radius) pi ri3 p o ro3 ri3 ro3 po pi 1 t r ro3 ri3 2 ro3 ri3 r 3 p 3 ri3 ro3 p 3 3 ri 3 3 r ro ri3 3 2r ro ri3 2 Now ro3 ri3 r t / 2 r t / 2 3r 2t rt 3 / 4 3r 2t for small t 3 3 Thus we have pr t 2t 16. 0 1 1 2 2 2 3 2 3 1 2 2 SECTION B 1. (a) the solution is 27 9 RA P, M A PL 32 64 (b) (i) Consider the equilibrium of a triangular-shaped piece of material: F x : AB x OB cos y OA sin OB sin OA cos x etc. (ii) We have 5 0 x 4 7 0 y 4 3 0 4 (iii) Adding the first two stress transformation equations gives x y x y so that the sum of the stresses in any coordinate is a constant. For 0 o : x y 3 o For 30o : x y (5 / 4) o (7 / 4) o 3 o So they are the same. (iv) Here the principal stresses are simply 1 2 0 , 2 0 (v) 0 2 0 (vi) It’s non-linear (not a straight line) and it’s viscoelastic (if it was not viscoelastic, the load/unload curves would coincide – the response is different on the way down because the material’s response changes with time) (vii) Uniaxial Test – Young’s modulus, info about stress/strain in 1-d Biaxial Test – young’s modulus, Poisson’s ratio, info about anisotropy Shear Test – Shear modulus Compression (confined) Test – Bulk Modulus (viii) The tensorial shear strain is half the value of the engineering shear strain. 2. (a) (i) We have 1 0.75 0 10 9 (ii) The max. shear stresses are 0 1 2 4 2 3 0 0 13 4 So, according to the maximum shear stress theory, failure will occur when Y 0 0 2 Y 24 Since the yield stress is Y 0 / 3 , it does not fail. (b) Stress: 10 kPa . Strain: (t ) 0.011 e t / 100 . Creep Compliance is the response due to a unit stress, so J (t ) 0.01 1 e t / 100 10 10 3 10 6 1 e t / 100 Then, with (t ) t Pa , d / dt 1 , (t ) 10 6 t 100e t / 100 e t / 100 1 t 10 6 100 1 e t / 100