18 Beam Bending

18.1 Definition of a beam

18.1.1 Definition of a beam

Beam of arbitrary cross section

Assumptions that underlie beam theory
- A beam is much longer than its characteristic cross section
- A beam carries:
  - Axial forces (P)
  - Shear forces (S)
  - Distributed loads (p)
  - Torques (T)
  - Bending moments (M)
- The reference line for a beam is the line of the centroids
- The reference line is often assumed to be straight

18.2 Kinematics of an Euler-Bernoulli beam

18.2.1 Definition of an Euler-Bernoulli beam

An Euler-Bernoulli beam also assumes:
- Plane sections remain plane under load
- Plane sections remain perpendicular to the reference line
Therefore, the axial deformation is a linear function of the position in the plane of the cross section \[w(x,y,z) = w_0(z) + \alpha(z) x + \beta(z) y\]

18.2.2 Kinematics of an Euler-Bernoulli beam

Beam kinematics

Therefore: \[w(x,y,z) = w_0(z) - x {u {}_{,z}} - y {v {}_{,z}}\] Now, we can write strain: \[\begin{align} {\varepsilon_{zz}}=& \, {\frac{\partial w}{\partial z}} \\ {\gamma_{zy}}=& \, {\frac{\partial v}{\partial z}} + {\frac{\partial w}{\partial y}} = {\frac{\partial v_0}{\partial z}} + \beta = 0\\ {\gamma_{zx}}=& \, {\frac{\partial u}{\partial z}} + {\frac{\partial w}{\partial x}} = {\frac{\partial u_0}{\partial z}} + \alpha = 0\\ {\varepsilon_{zz}}=& \, {w_0 {}_{,z}} - x {\alpha {}_{,z}} - y {\beta {}_{,z}} \\ \end{align}\]

When the assumptions of an Euler-Bernoulli beam hold, shear deformation is assumed negligible: \[\begin{align} {\gamma_{zy}}=& \, 0 = {\frac{\partial v_0}{\partial z}} + \beta \\ {\gamma_{zx}}=& \, 0 = {\frac{\partial u_0}{\partial z}} + \alpha \\ \end{align}\] thus: \[\begin{align} \alpha(z) =& \, - {u {}_{,z}} \\ \beta(z) =& \, - {v {}_{,z}} \\ \end{align}\] \[\begin{align} {\varepsilon_{zz}}=& \, {w_0 {}_{,z}} - x {\alpha {}_{,z}} - y {\beta {}_{,z}} \\ \end{align}\]

18.2.3 Strain in a beam

\[\begin{align} {\varepsilon_{zz}}=& \, {w_0 {}_{,z}} - x {u {}_{,zz}} - y {v {}_{,zz}}\\ {\varepsilon_{zz}}=& \, {\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}} \end{align}\]

The symbol is the axial strain of the reference line.
The symbols and are the curvatures of the reference line in the \(x-z\) and \(y-z\) planes.

18.2.4 Stress in a beam

The stress can now be written: \[\begin{align} {\sigma_{zz}}=& \, E {\varepsilon_{zz}}\\ {\sigma_{zz}}=& \, E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \end{align}\]

18.3 Beam loads

18.3.1 Axial load in a beam

The axial load in a beam can be expressed as: \[P = \int_A {\sigma_{zz}}\, dA\] \[dA = dx \cdot dy\] Therefore: \[\begin{align} P =& \, E \int_A \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \, dA\\ =& \, E \left({\varepsilon_{zz 0}}\int_A \, dA - {u {}_{,zz}} \int_A x \, dA - {v {}_{,zz}} \int_A y \, dA \right) \end{align}\]

18.3.2 Definitions

Lets define some terms: \[\begin{align} \int_A \, dA =& \, A \\ \int_A x \, dA =& \, x_c A \\ \int_A y \, dA =& \, y_c A \\ \end{align}\] Now we have: \[\begin{align} P =& \, E \left(A {\varepsilon_{zz 0}}- x_c A {u {}_{,zz}} - y_c A {v {}_{,zz}} \right) \\ \end{align}\]

18.3.3 Choice of coordinate system

We choose our coordinate system so that: \[\begin{align} x_c = 0 \\ y_c = 0 \\ \end{align}\] Which leads to: \[\begin{align} P =& \, E A \, {\varepsilon_{zz 0}}\\ \end{align}\]

This means that the axial force in the beam is due to the elongation of the centroidal axis and is not related to the bending of the beam.

18.3.4 Bending moments about \(x\)

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\[\begin{align} {\sigma_{zz}}=& \, E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \end{align}\] The resultant bending moment about the centroid is: \[M_x = \int_A y \cdot {\sigma_{zz}}\, dA\]

\[M_x = E \left( {\varepsilon_{zz 0}}\cdot \int_A y \, dA - {u {}_{,zz}} \cdot \int_A x y \, dA - {v {}_{,zz}} \cdot \int_A y^2 \, dA\right)\]

\(\int_A y \, dA = 0\) : By choice of coordinate system
\(\int_A x y \, dA = {I_{xy}}\) : Area product of inertia
\(\int_A y^2 \, dA = {I_{xx}}\) : Area moment of inertia about \(x\)-axis
Note: \({I_{xy}}= 0\) if there is symmetry about the \(x\) or \(x\) axis. (If there is symmetry about another axis, the computation can be transformed to be about the principal axis of the cross section.)
If \(E=E(x,y)\), then it must remain inside the integral and we do not have a clean separation between material and cross-section properties

18.3.5 Bending moments about \(y\)

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\[\begin{align} {\sigma_{zz}}=& E \left({\varepsilon_{zz 0}}- x {u {}_{,zz}} - y {v {}_{,zz}}\right) \end{align}\] The resultant bending moment about the centroid is: \[M_y = - \int_A x \cdot {\sigma_{zz}}\, dA\]

\[M_y = - E \left( {\varepsilon_{zz 0}}\cdot \int_A x \, dA - {u {}_{,zz}} \cdot \int_A x^2 \, dA - {v {}_{,zz}} \cdot \int_A x y \, dA\right)\]

\(\int_A x \, dA = 0\) : By choice of coordinate system
\(\int_A x y \, dA = {I_{xy}}\) : Area product of inertia
\(\int_A x^2 \, dA = {I_{yy}}\) : Area moment of inertia about -axis
Note: \({I_{xy}}= 0\) if there is symmetry about the \(x\) or \(y\) axis (If there is symmetry about another axis, the computation can be transformed to be about the principal axis of the cross section.)

\[\begin{align} M_x =& \, - E{I_{xy}}{u {}_{,zz}} - E{I_{xx}}{v {}_{,zz}} \\ M_y =& \, + E{I_{yy}}{u {}_{,zz}} + E{I_{xy}}{v {}_{,zz}} \\ \frac{P}{A} =& \, E {\varepsilon_{zz 0}}= {\sigma_{zz}}^{\mathrm{ave}} \\ \end{align}\]
These equations are uncoupled due to the centroidal -axis
Note also we can invert and solve

18.4 Differential equations of static equilibrium

18.4.1 Axial static equilbrium

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In the axial direction (\(z\)):
- \(P = P(z)\) – axial force resultant
- \(p_z = p_z(z)\) – distributed axial load [force/length]
  - e.g., \(p_z = \rho A g_z\)
  - Note \(\rho g_z\) is related to the stress equilibrium equations (body force per unit volume) as \(b_z = \rho g_z\)
- \(\displaystyle\sum F_z = 0 = P(z + dz) - P(z) + p_z dz\)
- \(0 = \frac{P(z + dz) - P(z)}{dz} + p_z\)
- \(\frac{d P}{d z} = - p_z(z)\)

For \(P = {\sigma_{zz}}\cdot A = E A \, {\varepsilon_{zz 0}}= E A \, {{w}_0 {}_{,z}}\)
\({\frac{d P(z)}{d z}} = {\frac{d }{d z}} \left[ E A {\frac{d w_0}{d z}} \right] = - p_z(z)\)
For constant \(EA\), \(E A \, {w_0 {}_{,zz}} = - p_z(z)\)
Note, when \(p_z = 0\), the axial force does not vary (for small deflections)

18.4.2 Bending static equilibrium

Beam bending: In the lateral directions (\(x\) & \(y\)):

(\(S_x\) ; \(S_y\)) - shear force resultants
(\(p_x\) ; \(p_y\)) - distributed lateral loads [force/length]
(\(M_x\) ; \(M_y\)) - bending moment resultants

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Summing forces, we find: \[\displaystyle \sum F_x = S_x + p_x dz = 0\] \[{\frac{d S_x(z)}{d z}} = -p_x\]

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Using similar arguments, we can find: \[{\frac{d S_y(z)}{d z}} = -p_y\]

Now use moment equilibrium and including bending moments:

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Consider first the counterclockwise moments about \(y\) axis (using the the center as reference point): \[\begin{align} \left(M_y + d M_y\right) - M_y + (S_x+d S_x) \frac{d z}{2} + (S_x) \frac{d z}{2} =& \; 0\\ d M_y = -(2 S_x +d S_x)\frac{d z}{2} \end{align}\] Because \(d S_x\) is small in the limit as \(d z\) goes to zero. \[\begin{align} \frac{dM_y}{dz} =& -S_x \\ \end{align}\]

The same can be applied to moments about the \(x\) axis:

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Note carefully the difference in the convention for direction of the moments. \[\begin{align} \frac{dM_x}{dz} =& +S_y \\ \end{align}\]

In summary, equilibrium arguments require: \[\begin{align} {\frac{d S_x(z)}{d z}} =& -p_x \\ {\frac{d S_y(z)}{d z}} =& -p_y \\ \frac{d M_x}{dz} =& +S_y \\ \frac{d M_y}{dz} =& -S_x \\ \end{align}\]

Combining these leads to: \[\begin{align} \frac{d^2 M_x}{dz^2} =& -p_y \\ \frac{d^2 M_y}{dz^2} =& +p_x \\ \end{align}\]

Recall also: \[\begin{align} M_x =& \, - E{I_{xy}}{u {}_{,zz}} - E{I_{xx}}{v {}_{,zz}} \\ M_y =& \, + E{I_{yy}}{u {}_{,zz}} + E{I_{xy}}{v {}_{,zz}} \\ \end{align}\]

In matrix form these can be written as: \[\begin{align} \left\{ \begin{array}{c} M_x \\ M_y\\ \end{array} \right\} = \left[ \begin{array}{cc} -EI_{xy} & -EI_{xx} \\ EI_{yy} & EI_{xy} \\ \end{array} \right] \left\{ \begin{array}{c} {u {}_{,zz}} \\ {v {}_{,zz}} \\ \end{array} \right\} \end{align}\]

Recall inversion: \[\begin{pmatrix}a & b \\ c & d \\ \end{pmatrix}^{-1} = \begin{pmatrix}\frac{d}{a\,d-b\,c} & -\frac{b}{a\,d-b\,c} \\ - \frac{c}{a\,d-b\,c} & \frac{a}{a\,d-b\,c} \\ \end{pmatrix}\]

Simple inversion yields the following: \[\begin{align} \left\{ \begin{array}{c} {u {}_{,zz}} \\ {v {}_{,zz}} \\ \end{array} \right\} = \frac{1}{\det EI} \left[ \begin{array}{cc} EI_{xy} & EI_{xx} \\ -EI_{yy} & -EI_{xy} \\ \end{array} \right] \left\{ \begin{array}{c} M_x \\ M_y\\ \end{array} \right\} \end{align}\] where: \[{\det EI} = + EI_{xx} EI_{yy} - (EI_{xy})^2\]

Now, lets examine the stress equation again: \[\begin{align} \tag{18.1} {\sigma_{zz}}=& E \left( -{u {}_{,zz}} x - {v {}_{,zz}} y \right) \\ =& \frac{E}{\det EI} \left[ -x \left(EI_{xy} M_x + EI_{xx} M_y \right) + y \left(EI_{yy} M_x + EI_{xy} M_y\right) \right] \\ \end{align}\]

It is these equations which allow us to compute deflections, strains, and stresses from geometry, material, and loads.

18.5 Example: Uniform beam

Thus: \[\begin{align} \frac{d^2}{d z^2} \left[ E {I_{yy}}{u {}_{,zz}} + E {I_{xy}}{v {}_{,zz}} \right] =& \, p_x \\ \frac{d^2}{d z^2} \left[ E {I_{xy}}{u {}_{,zz}} + E {I_{xx}}{v {}_{,zz}} \right] =& \, p_y \\ \end{align}\]

If symmetry exists across the \(x\) or \(y\) axis, the bending equations are uncoupled. \[\begin{align} \frac{d^2}{d z^2} \left[ E {I_{yy}}{u {}_{,zz}} \right] =& \; p_x \\ \frac{d^2}{d z^2} \left[ E {I_{xx}}{v {}_{,zz}} \right] =& \; p_y \\ \end{align}\] \[\begin{align} M_x =& \, - E{I_{xx}}{v {}_{,zz}} \\ M_y =& \, + E{I_{yy}}{u {}_{,zz}} \\ \end{align}\]

18.5.1 With uniform loading

Cantilever beam uniformly loaded

\[\begin{align} E I_{yy} \; {u {}_{,zzzz}} =& \; p_x \\ E I_{yy} \; {u {}_{,zzz}} =& \; p_x z + C_1 \\ E I_{yy} \; {u {}_{,zz}} =& \; p_x \frac{z^2}{2} + C_1 z + C_2\\ E I_{yy} \; {u {}_{,z}} =& \; p_x \frac{z^3}{6} + C_1 \frac{z^2}{2} + C_2 z +C_3\\ E I_{yy} \; u(z) =& \; p_x \frac{z^4}{24} + C_1 \frac{z^3}{6} + C_2 \frac{z^2}{2} +C_3 z +C_4\\ \end{align}\]

What are some boundary conditions we can use: \[\begin{align} u(0) =& \; 0 \Longrightarrow C_4 = 0\\ u'(0) =& \; 0 \Longrightarrow C_3 = 0\\ \end{align}\] \[\begin{align} M(0) =& \; E I_{yy} {u {}_{,zz}} = \frac{p_x l^2}{2} \Longrightarrow C_2 = \frac{p_x l^2}{2}\\ M(l) =& \; E I_{yy} {u {}_{,zz}} = 0 \Longrightarrow C_1 = - l p_x\\ \end{align}\]

\[E I_{yy} u(z) = p_x \left(\frac{z^4}{24}-\frac{l\,z^3}{6}+\frac{l^2\,z^2}{4}\right)\]

Also at \(z=0\) \[\begin{split} {u {}_{,z}} =& \; 0 \Longrightarrow {u {}_{,z}} = -\frac{z}{270000} \\ {u {}_{,z}} =& \; -\frac{z}{270000} \Longrightarrow u = -\frac{z^2}{540000} + C \\ \end{split}\]
At \(z=0\): \[\begin{split} u =& \; 0 \Longrightarrow u = -\frac{z^2}{540000} \\ {v {}_{,zz}} =& \; \frac{(197 - 2 \cdot z)}{1620000} \\ {v {}_{,z}} =& \; \frac{(197 \cdot z - z^2)}{1620000} \\ v =& \; \left(\frac{197}{2} - \frac{z}{3}\right) \cdot \frac{z^2}{1620000} \\ \end{split}\]

18.5.2 Caution!!!

Plates and screws can often be modeled as symmetric/uniform...
Keep in mind bones are neither symmetric nor uniform
Our approach for modeling biological structures may be guided by engineering theories, but theories must not be blindly accepted as truth

18.5.3 Contact Relationship Between Components

Components	Relationship	Comments
Screw - Bone (i.e. screw threads)	Rigid (tied contact)	Fixed all DOF
Plate - Bone	Contact pair	Friction sliding (\(\mu=0.3\))
Fracture surfaces	Contact pair	Friction sliding (\(\mu=1.0\))
Plate - Screw (Conventional)	Universal joint	Provide a universal connection between the screw control node and nodes on the bearing surface of the plate.
Plate - Screw (Locked)	Rigid	Provide a rigid connection between the screw control node and nodes on the bearing surface of the plate.

Fracture Plate Combinations \[fracture-plate-combinations\]
Fracture type	Construct	Construct	Construct
	Neutralization plate (Conventional screw construct)	Neutralization plate (Fixed angle screw construct)	Lateral periarticular distal fibular plate (Fixed angle screw construct)
Comminuted fracture	Model #1	Model #3	Model #5
Danis-Weber B fracture	Model #2	Model #4	Model #6

Displacement (\(|\Delta \vec{u}|\)) Due to Fibulotalar Reaction Load and External Moment With Danis-Weber B Fracture

18.6 Bartel Chapter 5 Structural Analysis of Musculoskeletal Systems: Beam Theory

@Bartel2006

18.7

@Bartel2006

18.8

@Bartel2006

18.9

@Bartel2006

18.10 Advanced Structural Analysis of Musculoskeletal Systems

18.10.0.1 Beam on elastic foundation

Many engineered structures and orthopaedic structures behave as a beam on an elastic foundation

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18.10.0.1.1 Examples

18.10.0.1.2 Free body diagram

Sum forces in \[\begin{split} V(x+dx) - V(x) +p(x) dx - q(x) dx =0 \\ {\frac{d V}{d x}} = q - p \end{split}\]

Elastic force from the foundation: \(q(x) = k \cdot v(x)\)

Sum moments about \(x+dx\) \[\begin{split} M(x+dx)-M(x) + V(x) dx + q dx \frac{dx}{2} + p dx \frac{dx}{2} =0 \\ {\frac{d M}{d x}} = p - q \end{split}\]

18.10.0.1.3 Governing differential equation

Moment equation \[M = EI \frac{d^2 v}{x^2}\]
Thus \[\frac{d^2}{x^2} \left(EI \frac{d^2 v}{x^2}\right) + k v(x) = p(x)\]
This is an ordinary differential equation that must be solved with boundary conditions
“Relatively easy” for distributed loads p(x) and infinite beams
Challenge for point loads and/or finite beam lengths as in the diagram... multiple equations needed for each “section”
The easiest of these problems don’t exist in orthopaedics

18.10.0.1.4

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18.10.0.1.5

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18.10.0.1.6

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18.10.0.1.7

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18.10.1 Torsion

18.10.1.0.1

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18.10.1.0.2

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18.10.1.0.3

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18.10.1.0.4

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18.10.2 Contact stress analysis

18.10.2.0.1 Contact stress analysis

Joints and joint replacement put surfaces into contact
Most human and engineering joints have interfaces with similar material properties (modulus)
- Bone-to-bone with cartilage bearing surfaces
- Ceramic-to-ceramic for some implant interfaces
Orthopaedic implants are often composed of materials with dissimilar material properties
- Metal-to-polyethylene
When dissimilar, one material may be “rigid” relative to the other

18.10.2.0.2 Conforming vs non-conforming surfaces

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Non-conforming surfaces has surface contours which are not coincident
Conforming surfaces have coincident surfaces
The degree of conformity clearly influences the stress for a given force that is passed
The elastic modulus influences the amount of conformity through deformation of the surfaces

18.10.2.0.3 Contact stress

Wikipedia

The contact stress is not uniform over the contact surface (in most situations–perhaps all joints)
For spherical (or toroidal) surfaces, the contact pressure is largest at the center (or centerline)
For other surfaces, max pressure is often at or near the edge
All contact produces normal and shear stresses below the contact surface (even frictionless)
Appropriate friction and non-linear material properties are necessary for accurate models of joint-replacement failure
Fatigue due to cyclic loading may drive failure (ie gait)

18.10.2.0.4 Hertz theory for contact between elastic bodies

18.10.2.0.4.1 with similar moduli

Good approximate solution method for materials with similar moduli
Provides insights into other contact problems (rigid-compliant)

18.10.2.0.5 Hertz theory for contact between elastic bodies

18.10.2.0.5.1 with similar moduli

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Maximum pressure \[\pmax = \frac{3 P}{2 \pi a^2}\]

Contact radius \[a = 0.721 \left(P C_G C_M\right)^{1/3}\]

Shortening of the distance between spheres \[\delta = 1.04 \left(\frac{P^2 C_M^2}{C_G}\right)^{1/3}\]

\(C_G, C_M\) are geometric and material parameters

Material parameter \[C_M = \frac{1-\nu_1^2}{E_1} + \frac{1-\nu_2^2}{E_2}\]

Geometric parameter for this problem \[C_G = \frac{D_1 D_2}{D_1+D_2}\]

This is a non-linear response, as are all contact problems

Non-linearity in

Deformation
Stress
Material
Geometry

The equations can be used to make general statements about the response when material or geometry changes

Must be aware of the limitations (assumptions)

Contact area is small relative to body
Both surfaces deform
Similar moduli, isotropic material

18.10.2.0.6 Hertz theory for contact between elastic bodies

18.10.2.0.6.1 Maximum normal stress as a function of load

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Warning: other (non-normal) stresses are of interest as well!

18.10.2.0.7

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18.10.2.0.8 Sub-surface stresses in Hertzian contact

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Tensile stress in radial direction around edges
Nearly hydrostatic at the center of the contact area
Maximum shear stress below the center of the contact area (\(z/a=0.51\)) with magnitude \(\tau_\mathrm{max} = 0.31 \pmax\)

18.10.2.0.9 Summary and conclusion of contact stress discussion

Analytical solutions offer insights that numerical solutions cannot
- ie, the non-linear dependence on material properties, geometry, etc
- Analytical solutions guide the thought process for “real” contact problems
Numerical solutions are likely to be used for “accurate solutions to real problems”
Contact problems are non-linear
Contact involves shear and normal stresses, fatigue is an issue

Potentially drop from here to the next section