Bone Tissue Mechanics
Transcrição
Bone Tissue Mechanics
Bone Tissue Mechanics João Folgado Paulo R. Fernandes Instituto Superior Técnico, 2013 PART 1 and 2 Biomecânica dos Tecidos, MEBiom, IST Introduction The objective of this course is to study basic concepts on hard tissue mechanics. Hard tissue is the structural material of the skeleton, mainly bone and cartilage. In this course the focus will be on bone biomechanics. The skeleton is a mechanical organ. Its primary functions are to transmit forces from one part of the body to another and protect certain organs from mechanical forces that could damage them. Biomecânica dos Tecidos, MEBiom, IST Introduction To study the effect of loads on the skeleton, and in particular in bone we have to know: Which loads are applied to bone? •Basically loads are transmitted by joint, so the question is how to know the forces in joints. •It is possible to obtain an order of magnitude of this loads using free body diagrams and static analysis. What is the effect of these load in bones? • Concept of mechanical stress and strain. Bone as a deformable body. How bone support these loads? •Bone as a structural material. •Mechanical properties of Bone •Bone adaptation to mechanical loads. Biomecânica dos Tecidos, MEBiom, IST Forces in the Hip Joint Modelling assumptions: “single leg stance phase” of gait. two-dimensional analysis. P – Abductor muscles; F – Joint reaction force acting in the middle of the acetabulum.; B – weight of the body on the leg. W – Body weight. Because each lower member is about (1/6)W, B=(5/6)W Biomecânica dos Tecidos, MEBiom, IST Forces in the Hip Joint The lengths b and c can be estimated from X-ray. It was found that: Assuming 5 =0 6 θ is the angle between the abductor muscle line and the y-axis. =0 − − =0⟺ −2 cos − Assuming Remark: The ratio b/c is critical for the hip load magnitude. Biomecânica dos Tecidos, MEBiom, IST Forces in the Elbow Joint W – Weight in the hand; J – reaction in the joint; B – biceps (and brachial) force If θ =75º; w = 0.35 m and b = 0.04 m thus: and orientation is: Biomecânica dos Tecidos, MEBiom, IST Stress in bending Problem (4 point bending) A bone sample, with outer diameter de=32 mm and inner diameter di=16 mm, is subject to a four-point test (see figure, F=1 KN).Determine the maximum bending normal stress . F F de=2r 70mm 40mm 70mm Biomecânica dos Tecidos, MEBiom, IST Note: Solution: Reactions: Static equilibrium: Biomecânica dos Tecidos, MEBiom, IST Shear and Bending Moment diagram: Biomecânica dos Tecidos, MEBiom, IST Note: for the sample is subject to pure bending Biomecânica dos Tecidos, MEBiom, IST Bending normal stress The maximum stress occurs in the section where the absolute value of the bending moment is maximum at the points where the distance to the neutral axis is maximum. Moment of inertia maximum stress Biomecânica dos Tecidos, MEBiom, IST Stress ΔF σ Q = lim ΔA → 0 Δ A x2 x3 - Stress is a measure of the internal forces associated to the plane of interest. - In general every plane containing the point Q has a normal and a shearing stress component. σ2 σ 21 σyz σ12 σ2 σ32 σ31 σ13 σ3 - The general state of stress is described by the components in a x1, x2, x3 reference system. - Only six components because the tensor is symmetric. x1 σ11, σ22, σ33 – normal stress σ12, σ13, σ23 – shearing stress Biomecânica dos Tecidos, MEBiom, IST Stress - Stress components depend on the reference system. - The same state of stress is represented by a different set of components if axes are rotated. Beer & Johnston (McGraw Hill) Biomecânica dos Tecidos, MEBiom, IST 2D example Transformation of coordinates: Problem 1 Assume the plane stress state given by its components in the x-y system (x is the horizontal axis and y is the vertical one): 6 3 [σ ] = − 3 2 Write the components of this stress tensor in the reference system which makes with the previous one: a) 90º b) 18,4º Biomecânica dos Tecidos, MEBiom, IST Transformation of coordinates For an angle θ (and 2D) Biomecânica dos Tecidos, MEBiom, IST Biomecânica dos Tecidos, MEBiom, IST Biomecânica dos Tecidos, MEBiom, IST Biomecânica dos Tecidos, MEBiom, IST Transformation of coordinates: Problem 2 (Using the Mohr’s Circle) Assume the plane stress state given by its components in the x-y system (x is the horizontal axis and y is the vertical one): 6 3 [σ ] = − 3 2 Draw the Mohr’s circle for this stress state. Biomecânica dos Tecidos, MEBiom, IST Mohr’s circle (2D) Biomecânica dos Tecidos, MEBiom, IST Biomecânica dos Tecidos, MEBiom, IST Principal Stresses - Structures are often subject to different combined loads. For instance a beam is usually subject to normal stress due to bending and shear stress due to the transverse load. Beer & Johnston (McGraw Hill) - Principal stresses are the stresses in the planes where the shear stress is zero. - The highest principal stress is the maximum normal stress while the lowest is the minimum normal stress. Biomecânica dos Tecidos, MEBiom, IST Principal Stresses in the Femur Koch (1917) Biomecânica dos Tecidos, MEBiom, IST Principal Stresses in the Femur Fernandes, Rodrigues and Jacobs (1999) Biomecânica dos Tecidos, MEBiom, IST Principal Stresses for a 2D state of stress Proposed Problem: For the given state of plan stress: 6 3 [σ ] = − 3 2 Determine the principal stresses and principal directions. Biomecânica dos Tecidos, MEBiom, IST Principal stresses and directions are solution of an eigenvalues and eigenvectors problem: Principal stresses-eigenvalues Principal stresses In the principal reference system (principal directions) the stress state is represented by: Biomecânica dos Tecidos, MEBiom, IST Principal directions - eigenvectors Biomecânica dos Tecidos, MEBiom, IST Biomecânica dos Tecidos, MEBiom, IST Mohr’s circle for a 3D state of stress Beer & Johnston (McGraw Hill) • Points A, B, and C represent the principal stresses on the principal planes (shearing stress is zero) • The three circles represent the normal and shearing stresses for rotation around each principal axis. • Radius of the largest circle yields the maximum shearing stress. τ max = 1 σ max 2 − σ min Biomecânica dos Tecidos, MEBiom, IST Failure Criteria • Failure of a component subjected to uniaxial stress is directly predicted from an equivalent tensile test Beer & Johnston (McGraw Hill) • Failure of a component subjected to a general state of stress cannot be directly predicted from the uniaxial state of stress in a tensile test specimen • Failure criteria are based on the mechanism of failure (ductile vs. brittle materials). Allows comparison of the failure conditions for uniaxial stress tests and multiaxial component loading Biomecânica dos Tecidos, MEBiom, IST Octahedral Stress z y x σ xx τ xy τ xz σ yy τ yz σ zz 3 1 σ 1 0 0 σ2 0 2 σ 3 Octahedral plane – makes equal angles with principal stress directions 1 3 1 3 σ oct = (σ 1 + σ 2 + σ 3 ) = (σ xx + σ yy + σ zz ) τ oct 1 1 2 2 2 1/ 2 = [(σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) ] = ( 2 I12 − 6 I 2 )1/ 2 3 3 1/ 2 1 = [(σ xx − σ yy ) 2 + (σ yy − σ zz ) 2 + (σ zz − σ xx ) 2 + 6(τ xy2 + τ yz2 + τ zx2 )] 3 Biomecânica dos Tecidos, MEBiom, IST Elastic strain energy density U 0 = σ ij dε ij For linear elastic and isotropic materials, subjected to a generalized stress state 1 U 0 = σ ij ε ij 2 U0 = 1 2 1 2 [ [ σ xx + σ yy2 + σ zz2 − 2ν (σ xxσ yy + σ yyσ zz + σ zzσ xx )] + τ xy + τ yz2 + τ zx2 ] 2E 2G For a stress state of a single normal stress σxx σ xx2 1 U 0 = σ xxε xx = 2 2E Biomecânica dos Tecidos, MEBiom, IST Strain energy of distortion 1 3 Average normal stress σ a = (σ 1 + σ 2 + σ 3 ) = σ oct Deviatoric stress [S ] = [σ ]− σ [I ] ij ij S xx a S xz σ xx − σ a τ xy τ xz S yz = σ yy − σ a τ yz S zz σ zz − σ a S xy S yy That is, a stress state can be represented as the sum of two states: a hydrostatic state in which the principal stresses are σ1=σ2=σ3=σa and a state in which all stresses are deviatoric [σ ] = σ [I ] + [S ] ij a ij - No change of shape is produced by the hydrostatic state - No change of volume is produced by the deviatoric state Biomecânica dos Tecidos, MEBiom, IST Strain energy of distortion The strain energy of distortion (general case) can be written as U 0d = 1 3 2 (σ xx − σ yy ) 2 + (σ yy − σ zz ) 2 + (σ zz − σ xx ) 2 + 6(τ xy2 + τ yz2 + τ zx2 ) = τ oct 12G 4G For a stress state of a single normal stress, σxx, the strain energy of distortion is U 0d = 1 1 2 σ xx2 + σ xx2 U 0 d = σ xx 12G 6G An alternative expressions for the strain energy of distortion (general case) is, U 0d = 1 1 2 (σ xx − σ yy ) 2 + (σ yy − σ zz ) 2 + (σ zz − σ xx ) 2 + 6(τ xy2 + τ yz2 + τ zx2 ) = σe 12G 6G Where σe is the Von Mises stress 1/ 2 1 [ σe = (σ xx − σ yy ) 2 + (σ yy − σ zz ) 2 + (σ zz − σ xx ) 2 + 6(τ xy2 + τ yz2 + τ xz2 )] 2 Note: For a case of uniaxial stress, σe=σxx and for a hydrostatic case σe=0 Biomecânica dos Tecidos, MEBiom, IST Von Mises criterion Yielding of an isotropic material (plasticity) begins when the strain energy of distortion reach a limiting value, U 0d ≥ 1 yielding (plasticity) U 0d Y Taken into account the definition of the Von Mises stress and that for an uniaxial test σe=σxx=σY U 0d U 0d Y 1 2 σ e σ 2 = 6G = e 1 2 σ Y σY 6G Thus the Von Mises criterion can be written as σe ≥ 1 yielding (plasticity) σY Biomecânica dos Tecidos, MEBiom, IST Ductile Material – Von Mises criterion Problem: A cylindrical sample with an outer diameter of de=32 mm and a inner diameter of di=16 mm, is subject to a bending moment M=140 N.m and a torque of T=210 N.m. The material is isotropic with a normal yield stress of σe=115 MPa. Verify if under these conditions the material yields. Biomecânica dos Tecidos, MEBiom, IST Loading Cross-section properties Biomecânica dos Tecidos, MEBiom, IST Bending Normal stress function of y State of stress where the bending stress is maximum: Biomecânica dos Tecidos, MEBiom, IST Torsion Shear stress function of r State of stress where the shear is maximum: Biomecânica dos Tecidos, MEBiom, IST Bending + Torsion (combined where both shear and normal stress have the maximum values) 2D state of stress (plane stress) Biomecânica dos Tecidos, MEBiom, IST State of stress at A State of stress at B Biomecânica dos Tecidos, MEBiom, IST Yield Criterion -Von Mises criterion (Ductile Materials) • Yield occurs when the distortion energy per unit volume is greater than that occurring in a tensile test specimen at yield. • In the distortion energy is possible to identify a term to compare directly to the yield normal stress given by the tensile test. • This term is the equivalent Von Mises Stress Biomecânica dos Tecidos, MEBiom, IST Yield Criterion -Von Mises criterion (Ductile Materials) • In pratice we compare the Von Mises stress with the yield stress of the material. Thus, the failure (yield) occurs when: = Yield stress of the material. Von Mises Stress Biomecânica dos Tecidos, MEBiom, IST For the proposed problem Comparing with the yield stress given for the material The sample is safe. Yield does not occur. Biomecânica dos Tecidos, MEBiom, IST Brittle Material – Mohr criterion Problem: A cylindrical sample with an outer diameter of de=32 mm and a inner diameter of di=16 mm, is subject to a bending moment M=140 N.m and a torque of T=210 N.m. The material is isotropic and brittle with failure tensile stress of σtf=133 MPa and failure compressive stress of σcf=195 Mpa. Verify if under these conditions the material fails. Biomecânica dos Tecidos, MEBiom, IST Loading Cross-section properties Biomecânica dos Tecidos, MEBiom, IST Bending Normal stress function of y State of stress where the bending stress is maximum: Biomecânica dos Tecidos, MEBiom, IST Torsion Shear stress function of r State of stress where the shear is maximum: Biomecânica dos Tecidos, MEBiom, IST Bending + Torsion (combined where both shear and normal stress have the maximum values) 2D state of stress (plane stress) Biomecânica dos Tecidos, MEBiom, IST State of stress at A State of stress at B Biomecânica dos Tecidos, MEBiom, IST Brittle Material – Mohr criterion failure occurs when where are the principal stresses (the highest and the lowest) is the limiting tensile stress (tensile test) is the limiting compressive stress (compression test) Can be positive or negative Biomecânica dos Tecidos, MEBiom, IST Principal stresses for the proposed problem Biomecânica dos Tecidos, MEBiom, IST Biomecânica dos Tecidos, MEBiom, IST Mohr’s criterion – Point A Thus, there is no fail Mohr’s criterion – Point B Thus, there is no fail Remark: At B the risk of the material is bigger (0.58 > 0.47) because the sample is in tension due to bending and the limiting stress in tension is smaller than in compression. Biomecânica dos Tecidos, MEBiom, IST Bibliography Skeletal Tissue Mechanics , R. Bruce Martin, David B. Burr, Neil A. Sharkey, Springer Verlag,1998. Orthopaedic Biomechanics, Mechanics and Design in Musculeskeletal Systems, D. Bartel, D. Davy, T. Keaveny, Pearson Prentice Hall, 2006. Bone Mechanics Handbook, 2nd Edition, S.C. Cowin, CRC Press, 2001 Mechanics of Materials, 5th Edition, F. Beer, Jr., E. R. Johnston , J. DeWolf, D. Mazurek, McGraw Hill, 2009 Biomecânica dos Tecidos, MEBiom, IST
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