Bone Tissue Mechanics


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1 Bone Tissue Mechanics João Folgado Paulo R. Fernandes Instituto Superior Técnico, 2016 PART 1 and 2
2 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.
3 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.
4 Forces in the Hip Joint Modelling assumptions: single leg stance phase of gait. twodimensional 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
5 Forces in the Hip Joint The lengths b and c can be estimated from Xray. It was found that: Assuming = 0 =0 2 cos 5 6 =0 θ is the angle between the abductor muscle line and the yaxis. Assuming Remark: The ratio b/c is critical for the hip load magnitude.
6 Forces in the Elbow Joint W Weight in the hand; J reaction in the joint; B biceps (and brachial) force a a If θ =75º; a = 0.35 m and b = 0.04 m thus: and orientation is:
7 Stress σ Q = lim ΔA 0 ΔF ΔA  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. x 2 σ 2 σ 21 σ yz σ 12 σ 32 σ31 σ 2  The general state of stress is described by the components in a x 1, x 2, x 3 reference system. σ 3 σ 13  Only six components because the tensor is symmetric. x 3 x 1 σ 11, σ 22, σ 33 normal stress σ 12, σ 13, σ 23 shearing stress
8 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)
9 2D example Transformation of coordinates: Problem 1 Assume the plane stress state given by its components in the xy system (x is the horizontal axis and y is the vertical one): [ σ ] 6 3 = 3 2 Write the components of this stress tensor in the reference system which makes with the previous one: a) 90º b) 18,4º
10 Transformation of coordinates For an angle θ (and 2D)
11
12
13
14 Transformation of coordinates: Problem 2 (Using the Mohr s Circle) Assume the plane stress state given by its components in the xy system (x is the horizontal axis and y is the vertical one): [ σ ] 6 3 = 3 2 Draw the Mohr s circle for this stress state.
15 Mohr s circle (2D)
16
17 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.
18 Principal Stresses in the Femur Koch (1917)
19 Principal Stresses in the Femur Fernandes, Rodrigues and Jacobs (1999)
20 Principal Stresses for a 2D state of stress Proposed Problem: For the given state of plan stress: [ σ ] 6 3 = 3 2 Determine the principal stresses and principal directions.
21 Principal stresses and directions are solution of an eigenvalues and eigenvectors problem: Principal stresseseigenvalues Principal stresses In the principal reference system (principal directions) the stress state is represented by:
22 Principal directions  eigenvectors
23
24 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 2 σ max σ min
25 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
26 Elastic strain energy density U 0 = σ dε ij ij For linear elastic and isotropic materials, subjected to a generalized stress state U 0 = 1 2E [ σ + σ + σ 2ν ( σ σ + σ σ + σ σ )] + [ τ + τ + τ ] xx yy zz U xx 0 = yy 1 σ ijε ij 2 yy zz zz xx 2G xy yz zx For a stress state of a single normal stress σ xx U 0 = 2 1 σ xx σ xxε xx = 2 2E
27 Strain energy of distortion 1 Average normal stress σa = ( σ1+ σ2+ σ3 ) 3 Deviatoric stress [ S ] = [ σ ] σ [] I ij ij a S xx S S xy yy S S S xz yz zz = σ xx σ a σ τ yy xy σ a τ xz τ yz σ zz σ a That is, a stress state can be represented as the sum of two states: a hydrostatic state (in which shear stress are zero and σ 1 =σ 2 =σ 3 =σ a ) and a deviatoric stress state [ σ ] = σ [] I + [ S ]  The change of volume is related with the hydrostatic state  The change of shape is related with the deviatoric state ij a ij
28 The strain energy of distortion (general case) can be written as U 1 12G d = ( σxx σ yy) + ( σ yy σzz) + ( σzz σxx) + 6( τxy + τ yz + τzx) An alternative expressions for the strain energy of distortion (general case) is, Where σ e is the Von Mises stress Strain energy of distortion For a stress state of a single normal stress, σ xx, the strain energy of distortion is U U 1 1 = σ σ U σ 12G + = 6G d xx xx 0d xx 1 1 = ( σ σ ) ( σ σ ) ( σ σ ) 6( τ τ τ ) σ 12G = 6G d xx yy yy zz zz xx xy yz zx e [( σ σ ) + ( σ σ ) + ( σ σ ) + 6( τ + τ τ )] 1/ 2 1 σ e = xx yy yy zz zz xx xy yz + 2 Note: For a case of uniaxial stress, σ e =σ xx and for a hydrostatic case σ e =0 xz
29 Von Mises criterion Yielding of an isotropic material (plasticity) begins when the strain energy of distortion reach a limiting value, U U 0d 0dY 1 yielding (plasticity) Taken into account the definition of the Von Mises stress and that for an uniaxial test σ e =σ xx =σ Y U U 0d Y Thus the Von Mises criterion can be written as 0d 1 2 σ e = 6G 1 2 σ Y 6G σ e = σ Y σ e 1 yielding (plasticity) σ Y 2
30 Ductile Material Von Mises criterion Problem: A cylindrical sample with an outer diameter of d e =32 mm and a inner diameter of d i =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 σ e =115 MPa. Verify if under these conditions the material yields.
31 Loading Crosssection properties
32 Bending Normal stress function of y State of stress where the bending stress is maximum:
33 Torsion Shear stress function of r State of stress where the shear is maximum:
34 Bending + Torsion (combined where both shear and normal stress have the maximum values) 2D state of stress (plane stress)
35 State of stress at A State of stress at B
36 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
37 For the proposed problem Comparing with the yield stress given for the material The sample is safe. Yield does not occur.
38 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
39 Brittle Material Mohr criterion Problem: A cylindrical sample with an outer diameter of d e =32 mm and a inner diameter of d i =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 σ tf =133 MPa and failure compressive stress of σ cf =195 Mpa. Verify if under these conditions the material fails.
40 Loading Crosssection properties
41 Bending Normal stress function of y State of stress where the bending stress is maximum:
42 Torsion Shear stress function of r State of stress where the shear is maximum:
43 Bending + Torsion (combined where both shear and normal stress have the maximum values) 2D state of stress (plane stress)
44 State of stress at A State of stress at B
45 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
46 Principal stresses for the proposed problem
47
48 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.
49 Bibliography Skeletal Tissue Mechanics,R.BruceMartin,DavidB.Burr,NeilA. Sharkey, Springer Verlag,1998. Orthopaedic Biomechanics, Mechanics and Design in Musculeskeletal Systems, D. Bartel, D. Davy, T. Keaveny, Pearson Prentice Hall, Bone Mechanics Handbook, 2 nd Edition, S.C.Cowin,CRCPress, 2001 Mechanics of Materials, 5 th Edition, F. Beer, Jr., E. R. Johnston, J. DeWolf, D. Mazurek, McGraw Hill, 2009
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