Bone Tissue Mechanics

Bone Tissue Mechanics
João Folgado
Paulo R. Fernandes
Instituto Superior Técnico, 2013
PART 1 and 2
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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.
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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.
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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
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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.
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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:
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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
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Note:
Solution:
Reactions:
Static equilibrium:
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Shear and Bending Moment
diagram:
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Note: for
the sample is subject
to pure bending
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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
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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
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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)
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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
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Transformation of coordinates
For an angle θ (and 2D)
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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.
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Mohr’s circle (2D)
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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.
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Principal Stresses in the Femur
Koch (1917)
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Principal Stresses in the Femur
Fernandes, Rodrigues
and Jacobs (1999)
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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.
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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:
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Principal directions - eigenvectors
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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
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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
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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
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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
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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
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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
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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
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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.
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Loading
Cross-section properties
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Bending
Normal stress function of y
State of stress where the bending stress is maximum:
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Torsion
Shear stress function of r
State of stress where the shear is maximum:
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Bending + Torsion (combined where both shear and normal stress have the maximum values)
2D state of stress (plane
stress)
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State of stress at A
State of stress at B
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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
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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
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For the proposed problem
Comparing with the yield stress given for the material
The sample is safe. Yield
does not occur.
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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.
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Loading
Cross-section properties
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Bending
Normal stress function of y
State of stress where the bending stress is maximum:
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Torsion
Shear stress function of r
State of stress where the shear is maximum:
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Bending + Torsion (combined where both shear and normal stress have the maximum values)
2D state of stress (plane
stress)
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State of stress at A
State of stress at B
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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
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Principal stresses for the proposed problem
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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.
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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
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