bone tissue mechanics - fenixedu · 2016-04-01 · mohr’s circle for a 3d state of stress • the...

Biomecânica dos Tecidos, MEBiom, IST

Bone Tissue Mechanics

João FolgadoPaulo R. Fernandes

Instituto Superior Técnico, 2016

PART 1 and 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.


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.

Introduction


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


Forces in the Hip Joint

The lengths b and c can be estimated from X-ray. It was found that:

Assuming

θ is the angle between the abductor muscle line and the y-axis.

Assuming

= 0 − − = 0 ⟺ − 2 cos − 56 = 0

Remark: The ratio b/c is critical for the hip load magnitude.


Forces in the Elbow JointW – Weight in the hand; J – reaction in the joint; B – biceps (and brachial) force

If θ =75º; a = 0.35 m and b = 0.04 m thus:

and orientation is:

a

a


Stress

σ2

σ2

σ3

σ13

σ12

σ21σyz

σ31σ32

x1

x2

x3

0limQ A

FA

σΔ →

Δ=Δ - In general every plane containing the

point Q has a normal and a shearing stress component.

- Stress is a measure of the internal forces associated to the plane of interest.

- 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. σ11, σ22, σ33 – normal stressσ12, σ13, σ23 – shearing stress


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)


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):

Write the components of this stress tensor in the reference system which makes with the previous one:

a) 90º b) 18,4º

[ ] 6 33 2

σ = −

2D example


Transformation of coordinates

For an angle θ (and 2D)


Draw the Mohr’s circle for this stress state.

[ ] 6 33 2

σ = −

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):


Mohr’s circle (2D)


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.

- 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.



Koch (1917)

Principal Stresses in the Femur


Principal Stresses in the Femur

Fernandes, Rodrigues and Jacobs (1999)


Proposed Problem:

For the given state of plan stress:

Determine the principal stresses and principal directions.

[ ] 6 33 2

σ = −

Principal Stresses for a 2D state of stress


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 stateis represented by:


Principal directions - eigenvectors


Mohr’s circle for a 3D state of stress

• The three circles represent the normal and shearing stresses for rotation around each principal axis.

• Points A, B, and C represent the principal stresses on the principal planes (shearing stress is zero)

minmaxmax 21 σστ −=

• Radius of the largest circle yields the maximum shearing stress.



Failure Criteria

• Failure of a component subjected to uniaxial stress is directly predicted from an equivalent tensile test

• 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



Elastic strain energy density

= ijijdU εσ0

For a stress state of a single normal stress σxx

[ ] [ ]2222220 2

1)(221

zxyzxyxxzzzzyyyyxxzzyyxx GEU τττσσσσσσνσσσ +++++−++=

EU xx

xxxx 221 2

0σεσ ==

For linear elastic and isotropic materials, subjected to a generalized stress state

ijijU εσ21

0 =


Strain energy of distortion

Average normal stress 1 2 31 ( )3aσ σ σ σ= + +

Deviatoric stress [ ] [ ] [ ]IS aijij σσ −=

−−

−=

azz

yzayy

xzxyaxx

zz

yzyy

xzxyxx

SSSSSS

σστσσττσσ

That is, a stress state can be represented as the sum of two states: a hydrostaticstate (in which shear stress are zero and σ1=σ2=σ3=σa) and a deviatoric stress state

[ ] [ ] [ ]ijaij SI += σσ

- The change of volume is related with the hydrostatic state- The change of shape is related with the deviatoric state


The strain energy of distortion (general case) can be written as

Note: For a case of uniaxial stress, σe=σxx and for a hydrostatic case σe=0

2 2 2 2 2 20

1 ( ) ( ) ( ) 6( )12

σ σ σ σ σ σ τ τ τ = − + − + − + + + d xx yy yy zz zz xx xy yz zxUG

An alternative expressions for the strain energy of distortion (general case) is,

[ ] 2/1222222 )(6)()()(2

1xzyzxyxxzzzzyyyyxxe τττσσσσσσσ +++−+−+−=

Where σe is the Von Mises stress

Strain energy of distortion

For a stress state of a single normal stress, σxx, the strain energy of distortion is

2 2 20 0

1 112 6d xx xx d xxU U

G Gσ σ σ = + =

2 2 2 2 2 2 20

1 1( ) ( ) ( ) 6( )12 6d xx yy yy zz zz xx xy yz zx eU

G Gσ σ σ σ σ σ τ τ τ σ = − + − + − + + + =


Von Mises criterionYielding of an isotropic material (plasticity) begins when the strain energyof distortion reach a limiting value,

Taken into account the definition of the Von Mises stress and that for anuniaxial test σe=σxx=σY

0

0

1 yielding (plasticity)d

d Y

UU

≥

Thus the Von Mises criterion can be written as

2

2

2

0

0

61

61

==Y

e

Y

e

Yd

d

G

GUU

σσ

σ

σ

1 yielding (plasticity)e

Y

σσ

≥


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.


Loading

Cross-section properties


Bending

Normal stress function of y State of stress where the bending stress is maximum:


Torsion

Shear stress function of r State of stress where the shear is maximum:


Bending + Torsion (combined where both shear and normal stress have the maximum values)

2D state of stress (plane stress)


State of stress at A

State of stress at B


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


For the proposed problem

Comparing with the yield stress given for the material

The sample is safe. Yield does not occur.


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


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 MPaand failure compressive stress of σcf=195 Mpa.

Verify if under these conditions the material fails.


Problem:


Loading

Cross-section properties


Bending

Normal stress function of y State of stress where the bending stress is maximum:


Torsion

Shear stress function of r State of stress where the shear is maximum:


Bending + Torsion (combined where both shear and normal stress have the maximum values)

2D state of stress (plane stress)


State of stress at A

State of stress at B



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


Principal stresses for the proposed problem


Mohr’s criterion – Point A

Mohr’s criterion – Point B

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.

Thus, there is no fail

Thus, there is no fail


Bibliography

Skeletal Tissue Mechanics , R. Bruce Martin, David B. Burr, Neil A.Sharkey, Springer Verlag,1998.

Orthopaedic Biomechanics, Mechanics and Design inMusculeskeletal Systems, D. Bartel, D. Davy, T. Keaveny, PearsonPrentice 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

bone tissue mechanics - fenixedu · 2016-04-01 · mohr’s circle for a 3d state of stress • the...

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