BME 281: Mechanics of Deformable Solids
Naveen Chandrashekar
Estimated study time: 1 hr 41 min
Table of contents
Sources and References
Primary textbook — R.C. Hibbeler, Mechanics of Materials, 11th ed., Pearson, 2014 (ISBN 9780137605613).
Supplementary sources — MIT OpenCourseWare 2.002 Mechanics and Materials II; Y.C. Fung, Biomechanics: Mechanical Properties of Living Tissues, 2nd ed., Springer, 1993; S.C. Cowin & S.B. Doty, Tissue Mechanics, Springer, 2007; G.A. Holzapfel, Nonlinear Solid Mechanics, Wiley, 2000; Stanford ME 337 Biomechanics lecture notes; Duke BME 271 course materials.
Chapter 1: Stress
1.1 The Concept of Stress
When an external force is applied to a deformable body, internal forces are generated within the material to maintain equilibrium. The notion of stress formalizes how these internal forces are distributed over a cross-sectional area. Unlike the simpler concept of force, stress has both a magnitude and an orientation, since it depends on both the intensity of the internal force and the orientation of the surface element on which it acts.
Consider a body in equilibrium under a set of external loads. Pass an imaginary cutting plane through the body and isolate one portion. The interaction of the removed portion on the remaining body is represented by a distributed force field acting on the cut surface. The intensity of this distribution — force per unit area — is the stress at a point.
The traction vector depends on both the location of the point and the orientation of the surface element, described by its outward unit normal \(\hat{n}\).
1.2 Normal Stress
The component of the traction vector acting perpendicular to the cut surface is the normal stress.
For a prismatic member carrying an axial resultant force \(N\) over a cross-sectional area \(A\) with uniform distribution, this reduces to
\[ \sigma = \frac{N}{A} \]Normal stress is positive (tensile) when it tends to pull material apart, and negative (compressive) when it pushes material together.
The SI unit of stress is the Pascal: \(1\ \text{Pa} = 1\ \text{N/m}^2\). Engineering applications commonly use MPa (\(10^6\ \text{Pa}\)) or GPa (\(10^9\ \text{Pa}\)).
1.3 Shear Stress
The components of the traction vector lying in the plane of the cut surface constitute the shear stress.
Shear stress tends to cause sliding (tangential displacement) between material planes.
1.3.1 Complementary Shear Stress
A fundamental result of moment equilibrium on an infinitesimal element is that shear stresses occur in complementary pairs: if \(\tau_{xy}\) acts on a face with normal in the \(x\)-direction, then an equal shear stress \(\tau_{yx} = \tau_{xy}\) acts on the face with normal in the \(y\)-direction. This symmetry of the stress tensor is a consequence of rotational equilibrium.
1.4 Bearing Stress
In mechanical connections such as bolted or pinned joints, the pin presses against the material of the hole, producing a bearing stress on the projected area of contact.
1.5 Factor of Safety
Structural designs must account for uncertainties in loading, material properties, and manufacturing tolerances. The factor of safety (FS) compares the failure load (or stress) to the allowable load (or stress).
For ductile materials, \(\sigma_{\text{failure}}\) is taken as the yield strength \(\sigma_Y\). For brittle materials, it is the ultimate strength \(\sigma_u\). Typical FS values range from 1.5 (well-characterized loads and materials) to 4 or higher (uncertain conditions or safety-critical applications).
A lap joint connects two steel plates using a 16 mm diameter bolt. The joint carries a tensile load of \(P = 12\ \text{kN}\). The plate thickness is \(t = 10\ \text{mm}\). Find: (a) the average shear stress in the bolt, (b) the bearing stress in the plate, and (c) the FS against bolt shear failure if the bolt shear strength is \(\tau_u = 300\ \text{MPa}\).
Solution.
(a) The bolt is in single shear. The shear area is the bolt cross-section:
\[ A_{\text{bolt}} = \frac{\pi d^2}{4} = \frac{\pi (0.016)^2}{4} = 2.011 \times 10^{-4}\ \text{m}^2 \]\[ \tau_{\text{avg}} = \frac{P}{A_{\text{bolt}}} = \frac{12\,000}{2.011 \times 10^{-4}} = 59.7\ \text{MPa} \](b) Bearing stress on projected area \(A_b = t \cdot d = (0.010)(0.016) = 1.6 \times 10^{-4}\ \text{m}^2\):
\[ \sigma_b = \frac{P}{A_b} = \frac{12\,000}{1.6 \times 10^{-4}} = 75.0\ \text{MPa} \](c) Factor of safety against bolt shear:
\[ \text{FS} = \frac{\tau_u}{\tau_{\text{avg}}} = \frac{300}{59.7} \approx 5.03 \]Chapter 2: Strain and Material Properties
2.1 Normal Strain
Stress describes the intensity of internal forces; strain describes the geometric deformation of the body. The two are linked by the constitutive (material) law.
where \(\delta = L - L_0\) is the elongation. Strain is dimensionless; it is often expressed as mm/mm, m/m, or as a percentage. Tensile strain is positive; compressive strain is negative.
At a point, normal strain is defined as the limiting ratio of deformation to original length as the gauge length shrinks to zero, giving the local stretch at a material point.
2.2 Shear Strain
Shear strain is measured in radians and is dimensionless.
2.3 Small-Deformation Assumption
Most structural analyses assume that displacements are infinitesimally small compared to the geometry of the structure. This linearization:
- Allows the undeformed geometry to be used for equilibrium equations.
- Makes the strain-displacement relations linear (linear kinematics).
- Justifies superposition of load effects.
For biological soft tissues under physiological loading, strains can exceed 20–50%, violating the small-deformation assumption. Nonlinear kinematics and large-deformation continuum mechanics are then required (see Chapter 7).
2.4 The Stress–Strain Diagram
The stress–strain diagram obtained from a uniaxial tensile test is the primary experimental characterization of a material. For a ductile metal such as structural steel:
- Proportional limit (\(\sigma_p\)): The upper limit of linearity. Below this, \(\sigma = E\varepsilon\) (Hooke’s law).
- Elastic limit: The stress below which no permanent deformation occurs upon unloading.
- Yield point (\(\sigma_Y\)): Pronounced in low-carbon steel; at this stress, large plastic strains occur with no increase in load (yielding plateau).
- Ultimate strength (\(\sigma_u\)): The maximum engineering stress.
- Fracture stress (\(\sigma_f\)): Stress at rupture (lower than \(\sigma_u\) due to necking).
The proportionality constant \(E\) is called Young’s modulus (or the modulus of elasticity). It has units of pressure (Pa) and represents the stiffness of the material. For structural steel, \(E \approx 200\ \text{GPa}\); for aluminum, \(E \approx 70\ \text{GPa}\); for cortical bone, \(E \approx 15\text{–}25\ \text{GPa}\) along the longitudinal axis.
For most engineering materials, \(0 < \nu < 0.5\). Incompressible materials (rubber, soft tissue at short time scales) have \(\nu \to 0.5\).
where \(G\) is the shear modulus (modulus of rigidity). It is related to \(E\) and \(\nu\) by
\[ G = \frac{E}{2(1+\nu)} \]2.5 Energy-Based Material Descriptors
It measures the ability of a material to absorb energy without permanent deformation.
or by the percent reduction in area at the fracture neck.
2.6 True Stress and True Strain
Engineering stress and strain use the original cross-sectional area and gauge length. As the specimen necks, the actual (true) values diverge.
True stress always exceeds engineering stress after necking begins. For large-deformation problems (including soft tissue mechanics), true stress and logarithmic (Hencky) strain are the appropriate measures.
A titanium alloy bone pin (\(E = 110\ \text{GPa}\), \(\nu = 0.34\)) has diameter \(d = 4.5\ \text{mm}\) and length \(L_0 = 80\ \text{mm}\). An axial compressive force of \(F = 3.2\ \text{kN}\) is applied. Find: (a) the axial stress, (b) the axial strain, (c) the lateral strain, (d) the change in diameter.
Solution.
(a) Cross-sectional area:
\[ A = \frac{\pi (0.0045)^2}{4} = 1.590 \times 10^{-5}\ \text{m}^2 \]\[ \sigma = \frac{-F}{A} = \frac{-3200}{1.590 \times 10^{-5}} = -201.3\ \text{MPa} \quad (\text{compressive}) \](b) Axial strain (Hooke’s law):
\[ \varepsilon_{\text{axial}} = \frac{\sigma}{E} = \frac{-201.3 \times 10^6}{110 \times 10^9} = -1.830 \times 10^{-3} \](c) Lateral strain:
\[ \varepsilon_{\text{lat}} = -\nu\,\varepsilon_{\text{axial}} = -(0.34)(-1.830 \times 10^{-3}) = +6.22 \times 10^{-4} \](d) Change in diameter:
\[ \Delta d = \varepsilon_{\text{lat}} \times d = (6.22 \times 10^{-4})(4.5) = +2.80 \times 10^{-3}\ \text{mm} \]The pin bulges slightly outward (as expected for a compressed cylinder).
Chapter 3: Axial Loading, Torsion, and Bending
3.1 Axial Deformation
For a prismatic member of cross-sectional area \(A\), length \(L\), and modulus \(E\), subject to an axial force \(N\):
where \(\delta\) is the elongation (positive for tension). If \(N\), \(A\), or \(E\) vary along the length, the formula becomes
\[ \delta = \int_0^L \frac{N(x)}{A(x)\,E(x)}\,dx \]For a member assembled from \(n\) segments each with constant \(N_i\), \(A_i\), \(E_i\), \(L_i\):
\[ \delta_{\text{total}} = \sum_{i=1}^{n} \frac{N_i L_i}{A_i E_i} \]3.1.1 Stress Concentrations
At geometric discontinuities — holes, notches, fillets, keyways — the local stress exceeds the nominal (average) stress by a factor \(K\):
The factor \(K > 1\) depends on the geometry (shape of the discontinuity) and is determined from charts (Hibbeler Appendix B, Peterson’s charts). For a circular hole in an infinite plate under remote tension, \(K = 3\) at the hole edge.
3.2 Torsion
3.2.1 Torsion Formula
Consider a circular shaft of radius \(c\) and length \(L\) subject to a twisting moment (torque) \(T\). Assuming plane cross-sections remain plane, radii remain straight, and shear strain varies linearly with radius:
The shear strain at radius \(\rho\) is \(\gamma = \rho\phi/L\), where \(\phi\) is the angle of twist. For a linearly elastic material, \(\tau = G\gamma = G\rho\phi/L\). The maximum shear stress occurs at the outer surface (\(\rho = c\)):
\[ \tau_{\max} = \frac{Gc\phi}{L} \]The torque is the resultant moment of all shear stresses about the axis:
\[ T = \int_A \rho\,\tau\,dA = \int_A \rho \cdot \frac{G\phi}{L}\rho\,dA = \frac{G\phi}{L} \int_A \rho^2\,dA = \frac{G\phi}{L} J \]where \(J = \int_A \rho^2\,dA\) is the polar moment of inertia. Solving for \(\phi\):
\[ \phi = \frac{TL}{GJ} \]Substituting back to express \(\tau\) in terms of \(T\):
\[ \tau = \frac{T\rho}{J} \]where \(c\) is the outer radius, \(J\) is the polar moment of inertia, \(G\) is the shear modulus, and \(\phi\) is the angle of twist (radians).
For a solid circular cross-section: \(J = \pi c^4 / 2\).
For a hollow circular cross-section with inner radius \(c_i\) and outer radius \(c_o\): \(J = \pi(c_o^4 - c_i^4)/2\).
3.2.2 Power Transmission
Rotating shafts transmit power. The relationship between power \(P\), torque \(T\), and angular velocity \(\omega\) (rad/s) or speed \(n\) (rpm) is:
\[ P = T\omega = T \cdot \frac{2\pi n}{60} \]A solid stainless steel surgical drill shaft (\(G = 75\ \text{GPa}\)) has diameter \(d = 6\ \text{mm}\) and transmits a torque of \(T = 0.8\ \text{N·m}\) over a length of \(L = 200\ \text{mm}\). Find: (a) the maximum shear stress, (b) the angle of twist.
Solution.
Outer radius \(c = 3\ \text{mm} = 0.003\ \text{m}\). Polar moment of inertia:
\[ J = \frac{\pi c^4}{2} = \frac{\pi (0.003)^4}{2} = 1.272 \times 10^{-10}\ \text{m}^4 \](a) Maximum shear stress:
\[ \tau_{\max} = \frac{Tc}{J} = \frac{(0.8)(0.003)}{1.272 \times 10^{-10}} = 18.9\ \text{MPa} \](b) Angle of twist:
\[ \phi = \frac{TL}{GJ} = \frac{(0.8)(0.200)}{(75 \times 10^9)(1.272 \times 10^{-10})} = 0.01678\ \text{rad} = 0.961° \]3.3 Bending
3.3.1 Kinematics of Pure Bending
When a beam is subjected to a bending moment \(M\), cross-sections rotate and the beam curves. The key kinematic assumption (Euler-Bernoulli hypothesis) is that plane cross-sections remain plane and perpendicular to the deformed neutral axis. This implies:
- Longitudinal fibers above the neutral axis are compressed.
- Longitudinal fibers below the neutral axis are in tension (for a positive moment per convention).
- There exists a neutral axis where \(\varepsilon = 0\).
Let \(\rho\) be the radius of curvature of the neutral axis. A fiber at distance \(y\) from the neutral axis has length \(\rho - y\) times the same subtended angle as the neutral axis (length \(\rho\)). The longitudinal strain at height \(y\) is:
\[ \varepsilon = \frac{(\rho - y)\,d\theta - \rho\,d\theta}{\rho\,d\theta} = -\frac{y}{\rho} \]For a linearly elastic material: \(\sigma = E\varepsilon = -Ey/\rho\).
The neutral axis location is found from the condition of zero net axial force:
\[ \int_A \sigma\,dA = -\frac{E}{\rho}\int_A y\,dA = 0 \implies \int_A y\,dA = 0 \]This means the neutral axis passes through the centroid of the cross-section.
The bending moment is the resultant couple of the stress distribution:
\[ M = -\int_A y\,\sigma\,dA = \frac{E}{\rho}\int_A y^2\,dA = \frac{EI}{\rho} \]where \(I = \int_A y^2\,dA\) is the second moment of area (moment of inertia) about the centroidal axis. Hence \(1/\rho = M/(EI)\). Substituting into the stress expression:
\[ \sigma = -\frac{Ey}{\rho} = -\frac{M y}{I} \]The maximum bending stress occurs at the outermost fiber at distance \(c\) from the neutral axis:
\[ \sigma_{\max} = \frac{Mc}{I} = \frac{M}{S} \]where \(S = I/c\) is the section modulus. The sign convention: positive \(M\) causes compression at the top (\(y > 0\)) and tension at the bottom (\(y < 0\)).
3.3.2 Section Properties
For common cross-sections:
- Rectangular (\(b \times h\)): \(I = bh^3/12\), \(c = h/2\), \(S = bh^2/6\)
- Circular (radius \(r\)): \(I = \pi r^4/4\), \(c = r\), \(S = \pi r^3/4\)
- Hollow circular (inner \(r_i\), outer \(r_o\)): \(I = \pi(r_o^4 - r_i^4)/4\)
- I-section: Use parallel-axis theorem for flanges and web
3.3.3 Composite Beams
A composite beam consists of two or more materials bonded together. The analysis uses the transformed section method: convert all materials to an equivalent section of one reference material.
A simplified femur mid-shaft is modeled as a hollow circular cross-section with outer radius \(r_o = 14\ \text{mm}\) and inner radius \(r_i = 9\ \text{mm}\). During stair climbing, the peak bending moment is estimated at \(M = 120\ \text{N·m}\). Find the maximum tensile and compressive bending stresses.
Solution.
\[ I = \frac{\pi}{4}\left(r_o^4 - r_i^4\right) = \frac{\pi}{4}\left((0.014)^4 - (0.009)^4\right) = \frac{\pi}{4}\left(3.842 \times 10^{-8} - 6.561 \times 10^{-9}\right) \]\[ I = \frac{\pi}{4}(3.186 \times 10^{-8}) = 2.503 \times 10^{-8}\ \text{m}^4 \]Maximum stress (at \(c = r_o = 0.014\ \text{m}\)):
\[ \sigma_{\max} = \frac{Mc}{I} = \frac{(120)(0.014)}{2.503 \times 10^{-8}} = 67.2\ \text{MPa} \]The cortex on the tension side experiences \(+67.2\ \text{MPa}\) and the compression side \(-67.2\ \text{MPa}\). This is well within the tensile (\(\approx 130\ \text{MPa}\)) and compressive (\(\approx 190\ \text{MPa}\)) strengths of cortical bone under quasi-static loading, providing a safety factor of approximately 1.9 against tensile fracture.
Chapter 4: Transverse Shear and Combined Loading
4.1 Shear Stress Due to Transverse Loading
When a beam carries transverse loads, the internal cross-sections experience both a bending moment \(M\) and a shear force \(V\). Although the flexure formula gives the normal stress, the shear force produces shear stresses on horizontal planes within the beam.
Consider two adjacent cross-sections separated by \(dx\), with moments \(M\) and \(M + dM\). The difference in bending stress resultants on the sub-area above height \(y'\) creates a net horizontal force that must be balanced by a shear stress \(\tau\) on the horizontal cut at \(y'\):
\[ \tau \cdot b\,dx = \frac{dM}{I}\int_{y'}^{c} y\,dA = \frac{V\,dx}{I} Q \]where \(Q = \int_{y'}^{c} y\,dA\) is the first moment of the area above the cut about the neutral axis, and \(b\) is the width of the section at the cut.
where \(V\) is the transverse shear force, \(Q\) is the first moment of the area above (or below) the point of interest about the neutral axis, \(I\) is the second moment of area of the entire cross-section, and \(b\) is the width at the point of interest.
For a solid rectangular cross-section (\(b \times h\)):
\[ \tau(y) = \frac{V}{2I}\left(\frac{h^2}{4} - y^2\right), \qquad \tau_{\max} = \frac{3V}{2bh} = \frac{3V}{2A} \]The shear stress is parabolic, zero at the top and bottom surfaces, and maximum at the neutral axis.
4.2 Shear Flow
Shear flow is useful for thin-walled members and for computing the shear in connectors (nails, welds, adhesive bonds) joining parts of a built-up beam.
4.2.1 Shear Center
For non-symmetric cross-sections (channels, angles, thin-walled open sections), the shear stress resultant generally acts at a point that is not the centroid. If the load does not pass through this point — the shear center — the beam will twist in addition to bending.
A wooden beam (\(E = 12\ \text{GPa}\)) has a rectangular cross-section \(100\ \text{mm} \times 200\ \text{mm}\) (\(b \times h\)). It carries a shear force \(V = 25\ \text{kN}\). Find: (a) the maximum shear stress, (b) the shear stress at \(y = 50\ \text{mm}\) from the neutral axis.
Solution.
\[ I = \frac{bh^3}{12} = \frac{(0.1)(0.2)^3}{12} = 6.667 \times 10^{-5}\ \text{m}^4, \qquad A = (0.1)(0.2) = 0.02\ \text{m}^2 \](a) Maximum shear stress at neutral axis:
\[ \tau_{\max} = \frac{3V}{2A} = \frac{3(25\,000)}{2(0.02)} = 1.875\ \text{MPa} \](b) At \(y = 50\ \text{mm} = 0.05\ \text{m}\):
\[ Q = \int_{0.05}^{0.1} b\,y\,dy = b \cdot \frac{c^2 - y'^2}{2} = (0.1)\frac{(0.1)^2 - (0.05)^2}{2} = (0.1)(3.75 \times 10^{-3}) = 3.75 \times 10^{-4}\ \text{m}^3 \]\[ \tau = \frac{VQ}{Ib} = \frac{(25\,000)(3.75 \times 10^{-4})}{(6.667 \times 10^{-5})(0.1)} = 1.406\ \text{MPa} \]4.3 Combined Loading
In practice, structural members rarely carry a single type of load. A general cross-section may simultaneously experience:
- Axial force \(N\) → normal stress \(\sigma_{\text{axial}} = N/A\)
- Bending moments \(M_y\), \(M_z\) → normal stresses \(\sigma_{\text{bend}}\)
- Torque \(T\) → shear stresses \(\tau_T\)
- Shear forces \(V_y\), \(V_z\) → shear stresses \(\tau_V\)
The critical cross-section is the location where combined stresses are most severe. Superposition is valid only in the linear elastic regime.
4.3.1 Identifying Critical Cross-Sections
The procedure for combined loading:
- Draw free-body diagrams and compute the internal force and moment diagrams (\(N(x)\), \(V(x)\), \(M(x)\), \(T(x)\)).
- Identify candidate critical sections (locations of maximum \(|M|\), maximum \(|V|\), or geometric discontinuities).
- At each candidate section, identify the critical point on the cross-section (where \(\sigma\) and \(\tau\) are largest in the appropriate sense).
- Construct the complete stress state at each critical point and apply the governing failure criterion.
A cortical bone screw is subjected simultaneously to an axial pull-out force \(N = 800\ \text{N}\) and a transverse shear load \(V = 200\ \text{N}\). The screw minor (root) diameter is \(d = 2.0\ \text{mm}\). Find the maximum normal and shear stresses at the root cross-section.
Solution.
Cross-sectional area: \(A = \pi(0.001)^2 = 3.142 \times 10^{-6}\ \text{m}^2\).
Axial normal stress: \(\sigma_{\text{axial}} = N/A = 800/(3.142 \times 10^{-6}) = 254.6\ \text{MPa}\).
Average transverse shear stress: \(\tau_V = V/A = 200/(3.142 \times 10^{-6}) = 63.7\ \text{MPa}\).
These are the dominant stress components at the root. The combined stress state must be evaluated using a failure criterion (see Chapter 5) to assess whether the screw is safe.
Chapter 5: Stress Transformation and Failure Theories
5.1 Plane Stress and Stress Transformation
At a point in a body, the stress state generally has six independent components. Under plane stress, the stress components in one direction (say \(z\)) are zero: \(\sigma_z = \tau_{xz} = \tau_{yz} = 0\). This is a good approximation for thin plates and the free surfaces of most structural members.
The stress components on a face inclined at angle \(\theta\) to the \(x\)-axis are found by equilibrium:
5.2 Principal Stresses
The maximum in-plane shear stress is:
\[ \tau_{\max,\text{in-plane}} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} = \frac{\sigma_1 - \sigma_2}{2} \]The absolute maximum shear stress (including out-of-plane) is \(\tau_{\text{abs}} = \max\!\left(\frac{\sigma_1 - \sigma_2}{2},\, \frac{|\sigma_1|}{2},\, \frac{|\sigma_2|}{2}\right)\).
5.3 Mohr’s Circle
Mohr’s circle is a graphical representation of the stress transformation equations. It allows rapid determination of principal stresses, principal planes, and maximum shear stresses.
Rewrite the transformation equations in the form:
\[ \left(\sigma_{x'} - \frac{\sigma_x + \sigma_y}{2}\right)^2 + \tau_{x'y'}^2 = \left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2 \]This is the equation of a circle in the \((\sigma, \tau)\) plane with:
Center: \(C = \left(\dfrac{\sigma_x + \sigma_y}{2},\ 0\right)\)
Radius: \(R = \sqrt{\left(\dfrac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}\)
The two reference points on the circle are \(X = (\sigma_x, \tau_{xy})\) and \(Y = (\sigma_y, -\tau_{xy})\). As \(\theta\) increases by \(\Delta\theta\), the point on the circle rotates by \(2\Delta\theta\).
The principal stresses are the horizontal intercepts: \(\sigma_1 = C + R\) and \(\sigma_2 = C - R\). The maximum in-plane shear stress is the topmost (or bottommost) point of the circle: \(\tau_{\max} = R\), and it occurs on planes rotated \(45°\) from the principal planes.
At a critical point on a rotating shaft, the stress state is \(\sigma_x = 80\ \text{MPa}\), \(\sigma_y = -20\ \text{MPa}\), \(\tau_{xy} = 40\ \text{MPa}\). Find: (a) the principal stresses, (b) the maximum in-plane shear stress, (c) the orientation of the principal planes.
Solution.
Center: \(C = (80 + (-20))/2 = 30\ \text{MPa}\).
Radius: \(R = \sqrt{((80-(-20))/2)^2 + 40^2} = \sqrt{50^2 + 40^2} = \sqrt{2500 + 1600} = \sqrt{4100} = 64.03\ \text{MPa}\).
(a) Principal stresses:
\[ \sigma_1 = 30 + 64.03 = 94.0\ \text{MPa}, \qquad \sigma_2 = 30 - 64.03 = -34.0\ \text{MPa} \](b) Maximum in-plane shear stress: \(\tau_{\max} = R = 64.0\ \text{MPa}\).
(c) Orientation of \(\sigma_1\) plane: \(\tan 2\theta_p = \tau_{xy}/\left[(\sigma_x - \sigma_y)/2\right] = 40/50 = 0.8\), so \(2\theta_p = 38.66°\) and \(\theta_p = 19.3°\) CCW from \(x\).
5.4 Theories of Failure
A failure theory predicts the onset of yielding or fracture in a body under multi-axial stress, based on a uniaxial yield/failure stress measured in a simple tension test.
5.4.1 Maximum Normal Stress Theory (Rankine)
This criterion works well for brittle materials (cast iron, ceramics, glass, cortical bone under tension) but is inadequate for ductile materials.
5.4.2 Maximum Shear Stress Theory (Tresca)
In terms of principal stresses (assuming \(\sigma_1 \geq \sigma_2 \geq \sigma_3\)):
\[ \sigma_1 - \sigma_3 \geq \sigma_Y \]For plane stress (\(\sigma_3 = 0\)), this defines a hexagonal yield surface in \((\sigma_1, \sigma_2)\) space.
5.4.3 Von Mises Distortion Energy Theory
For plane stress (\(\sigma_3 = 0\)):
\[ \sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2 \geq \sigma_Y^2 \]In terms of \(\sigma_x\), \(\sigma_y\), \(\tau_{xy}\):
\[ \sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2 \geq \sigma_Y^2 \]The von Mises criterion is an ellipse in principal stress space, circumscribing the Tresca hexagon. It is the most accurate criterion for ductile metals and is widely used in implant design.
A titanium hip stem has \(\sigma_Y = 795\ \text{MPa}\). At the critical section, the stress state is \(\sigma_x = 320\ \text{MPa}\), \(\sigma_y = 0\), \(\tau_{xy} = 180\ \text{MPa}\). Assess safety using (a) Tresca and (b) von Mises.
Solution.
Principal stresses:
\[ \sigma_{1,2} = \frac{320 + 0}{2} \pm \sqrt{\left(\frac{320}{2}\right)^2 + 180^2} = 160 \pm \sqrt{25600 + 32400} = 160 \pm 240.8 \]\[ \sigma_1 = 400.8\ \text{MPa}, \quad \sigma_2 = -80.8\ \text{MPa} \](a) Tresca: \(\sigma_1 - \sigma_2 = 400.8 - (-80.8) = 481.6\ \text{MPa}\). Since \(481.6 < 795\), the material does not yield by Tresca. \(\text{FS}_T = 795/481.6 = 1.65\).
(b) Von Mises:
\[ \sigma_{VM} = \sqrt{\sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2} = \sqrt{320^2 - 0 + 0 + 3(180)^2} = \sqrt{102400 + 97200} = \sqrt{199600} = 446.8\ \text{MPa} \]Since \(446.8 < 795\), safe by von Mises. \(\text{FS}_{VM} = 795/446.8 = 1.78\).
Both criteria indicate safety, but with limited margin — particularly important given fatigue loading in vivo.
Chapter 6: Beam Deflection and Buckling
6.1 The Elastic Curve Equation
When a beam deflects under transverse loading, the deflection curve — the elastic curve — satisfies a differential equation that relates curvature to the applied moment.
This follows from the curvature–moment relationship \(1/\rho = M/(EI)\) and the small-slope approximation \(1/\rho \approx d^2v/dx^2\).
Successive integration and application of boundary conditions yields the deflection \(v(x)\) and slope \(\theta(x) = dv/dx\).
The boundary conditions for common supports:
- Pin/Roller: \(v = 0\) (no deflection), \(M = 0\) (no moment)
- Fixed (clamped) end: \(v = 0\) and \(dv/dx = 0\) (no deflection, no slope)
- Free end: \(M = 0\) and \(V = 0\)
6.2 Integration Method
A simply supported beam of length \(L\), with bending stiffness \(EI\), carries a uniformly distributed load \(w\) (N/m). Find the deflection curve and the maximum deflection.
Solution.
By symmetry, the reactions are \(R_A = R_B = wL/2\). The internal moment at position \(x\) (using left FBD):
\[ M(x) = \frac{wL}{2}x - \frac{w x^2}{2} \]The elastic curve equation:
\[ EI\,\frac{d^2 v}{dx^2} = \frac{wL}{2}x - \frac{wx^2}{2} \]Integrate once (slope):
\[ EI\,\frac{dv}{dx} = \frac{wLx^2}{4} - \frac{wx^3}{6} + C_1 \]Integrate again (deflection):
\[ EI\,v = \frac{wLx^3}{12} - \frac{wx^4}{24} + C_1 x + C_2 \]Boundary conditions: \(v(0) = 0 \Rightarrow C_2 = 0\); \(v(L) = 0\):
\[ 0 = \frac{wL^4}{12} - \frac{wL^4}{24} + C_1 L \Rightarrow C_1 = -\frac{wL^3}{24} \]Deflection:
\[ v(x) = \frac{w}{24EI}\left(2Lx^3 - x^4 - L^3 x\right) \]Maximum deflection at midspan (\(x = L/2\)):
\[ v_{\max} = \frac{5wL^4}{384\,EI} \]A cantilever beam of length \(L\) and stiffness \(EI\) carries a concentrated load \(P\) at the free end. Find the tip deflection.
Solution.
With the fixed end at \(x = 0\) and the free end at \(x = L\), the bending moment is \(M(x) = -P(L - x)\) (hogging). Integrating twice:
\[ EI\,\frac{d^2v}{dx^2} = -P(L-x) \]\[ EI\,\frac{dv}{dx} = -PLx + \frac{Px^2}{2} + C_1 \]BC: \(dv/dx = 0\) at \(x = 0\) \(\Rightarrow C_1 = 0\).
\[ EI\,v = -\frac{PLx^2}{2} + \frac{Px^3}{6} + C_2 \]BC: \(v = 0\) at \(x = 0\) \(\Rightarrow C_2 = 0\). Tip deflection (\(x = L\)):
\[ v(L) = \frac{P}{EI}\left(-\frac{L^3}{2} + \frac{L^3}{6}\right) = -\frac{PL^3}{3EI} \]The negative sign indicates downward deflection (consistent with convention).
6.3 Euler Column Buckling
Long slender columns can fail by sudden lateral deflection — buckling — at a load well below the material’s yield strength. This is a stability failure, not a strength failure.
Consider a pin-pin column of length \(L\), stiffness \(EI\), subject to axial compressive load \(P\). Assume a slightly buckled configuration with lateral deflection \(v(x)\). The bending moment at position \(x\) is \(M(x) = -P\,v\) (the load \(P\) acts at eccentricity \(v\) from the column axis).
The elastic curve equation becomes:
\[ EI\,\frac{d^2v}{dx^2} = -P\,v \]\[ \frac{d^2v}{dx^2} + \frac{P}{EI}\,v = 0 \]Let \(k^2 = P/(EI)\). The general solution is:
\[ v(x) = A\sin(kx) + B\cos(kx) \]Boundary conditions: \(v(0) = 0 \Rightarrow B = 0\); \(v(L) = 0 \Rightarrow A\sin(kL) = 0\).
Non-trivial solutions require \(\sin(kL) = 0\), so \(kL = n\pi\) (\(n = 1, 2, 3, \ldots\)). The smallest critical load corresponds to \(n = 1\):
\[ k_{\text{cr}} = \frac{\pi}{L} \Rightarrow \frac{P_{\text{cr}}}{EI} = \frac{\pi^2}{L^2} \Rightarrow P_{\text{cr}} = \frac{\pi^2 EI}{L^2} \]The critical stress (buckling stress) is:
\[ \sigma_{\text{cr}} = \frac{P_{\text{cr}}}{A} = \frac{\pi^2 E}{\left(L_e/r\right)^2} \]where \(r = \sqrt{I/A}\) is the radius of gyration and \(L_e/r\) is the slenderness ratio.
Effective length factors \(K\) (so \(L_e = KL\)): pin-pin \(K = 1\); fixed-free (flagpole) \(K = 2\); fixed-pin \(K \approx 0.7\); fixed-fixed \(K = 0.5\).
In biomedical contexts, bone pins and intramedullary nails are long slender rods that can buckle if their effective length exceeds the critical value. An IM nail in the femur must be sized with sufficient diameter to push the slenderness ratio below the critical threshold for the applied compressive forces during weight-bearing.
A solid titanium IM nail (\(E = 110\ \text{GPa}\), \(\sigma_Y = 795\ \text{MPa}\)) has diameter \(d = 11\ \text{mm}\) and effective length \(L_e = 350\ \text{mm}\) (fixed-pin end conditions from locking screws). The axial compressive load during single-leg stance is \(P = 2800\ \text{N}\). Find: (a) the critical buckling load, (b) the critical stress, (c) the FS against buckling.
Solution.
\[ I = \frac{\pi d^4}{64} = \frac{\pi (0.011)^4}{64} = 7.188 \times 10^{-10}\ \text{m}^4 \]\[ A = \frac{\pi d^2}{4} = 9.503 \times 10^{-5}\ \text{m}^2, \qquad r = \sqrt{I/A} = \sqrt{7.188\times10^{-10}/9.503\times10^{-5}} = 2.75 \times 10^{-3}\ \text{m} \](a) Critical load (pin-pin effective length used; fixed-pin gives \(K \approx 0.7\), \(L_e = 0.245\ \text{m}\)):
\[ P_{\text{cr}} = \frac{\pi^2 EI}{L_e^2} = \frac{\pi^2 (110 \times 10^9)(7.188 \times 10^{-10})}{(0.245)^2} = \frac{780\,700}{0.06002} = 1.301 \times 10^6\ \text{N} = 1301\ \text{kN} \](b) Critical stress:
\[ \sigma_{\text{cr}} = \frac{P_{\text{cr}}}{A} = \frac{1.301 \times 10^6}{9.503 \times 10^{-5}} = 1.369\ \text{GPa} \]Since \(\sigma_{\text{cr}} > \sigma_Y\), the nail will yield before it buckles; the Euler formula is not applicable for this geometry. This nail is stocky enough that buckling is not the limiting mode.
(c) FS against yielding (axial compression only): \(\sigma = P/A = 2800/(9.503 \times 10^{-5}) = 29.5\ \text{MPa}\). \(\text{FS} = 795/29.5 = 26.9\) — very safe in pure compression.
Chapter 7: Mechanics of Biological Materials
7.1 Overview: Why Biological Tissues Are Different
Engineering materials such as steel and aluminum have well-defined, reproducible mechanical properties that are essentially time-independent (at room temperature, quasi-static rates), isotropic, and linear elastic up to yielding. Biological tissues deviate from every one of these assumptions:
- They are hierarchically structured across molecular, fibrillar, lamellar, and organ scales.
- Their properties are highly anisotropic, reflecting the directional arrangement of fibers.
- Their stress–strain behavior is nonlinear (J-shaped or S-shaped curves rather than straight lines).
- They exhibit viscoelasticity — their response depends on the rate and history of loading.
- They are living materials that remodel in response to mechanical stimuli (Wolff’s Law for bone; mechanobiology of soft tissue).
- Their properties are highly variable with age, disease, hydration state, and anatomical site.
7.2 Viscoelasticity
Viscoelastic materials exhibit both elastic (spring-like, energy-storing) and viscous (dashpot-like, energy-dissipating) mechanical behavior. The response depends on both the magnitude and the rate of applied deformation.
7.2.1 Creep and Stress Relaxation
Two hallmark phenomena of viscoelastic materials:
7.2.2 The Maxwell Model
The Maxwell model connects a spring (stiffness \(E\)) and a dashpot (viscosity \(\eta\)) in series.
Under stress relaxation (\(\varepsilon = \varepsilon_0 = \text{const}\), \(\dot{\varepsilon} = 0\)):
\[ \sigma(t) = \sigma_0\,e^{-t/\tau_R}, \qquad \tau_R = \frac{\eta}{E} \]where \(\tau_R\) is the relaxation time. The Maxwell model predicts complete stress relaxation as \(t \to \infty\), which is too aggressive for most biological tissues.
Under creep (\(\sigma = \sigma_0 = \text{const}\)):
\[ \varepsilon(t) = \frac{\sigma_0}{E} + \frac{\sigma_0}{\eta}t \]The Maxwell model predicts unbounded creep (viscous flow), which is appropriate for fluids but not for solid-like tissues.
7.2.3 The Kelvin-Voigt Model
The Kelvin-Voigt model connects a spring and dashpot in parallel.
Under creep (\(\sigma = \sigma_0\)):
\[ \varepsilon(t) = \frac{\sigma_0}{E}\left(1 - e^{-t/\tau_C}\right), \qquad \tau_C = \frac{\eta}{E} \]The Kelvin-Voigt model predicts that creep asymptotes to \(\sigma_0/E\) (correct for a solid that returns to zero strain on unloading). However, it cannot model stress relaxation (instantaneous stress application requires infinite force with a rigid dashpot in parallel).
7.2.4 The Standard Linear Solid (Zener Model)
The Standard Linear Solid (SLS) — also called the Zener model — combines a spring in parallel with a Maxwell element. It correctly predicts both creep and stress relaxation.
In compact form, with \(E_1\) the series spring and \(E_\infty\) the parallel spring:
\[ \sigma + \frac{\eta}{E_1}\dot{\sigma} = \left(E_\infty + E_1\right)\varepsilon\bigg|_{\text{instant}} + E_\infty\frac{\eta}{E_1}\dot{\varepsilon} \]Under stress relaxation:
\[ \sigma(t) = \varepsilon_0\left[E_\infty + E_1\,e^{-t/\tau_R}\right] \]As \(t \to \infty\): \(\sigma \to E_\infty \varepsilon_0\) (equilibrium modulus). As \(t \to 0^+\): \(\sigma \to (E_\infty + E_1)\varepsilon_0\) (instantaneous modulus).
7.3 Mechanical Properties of Bone
Bone is a hierarchically organized composite: at the nanoscale, mineralized collagen fibrils (hydroxyapatite crystals in a type-I collagen matrix); at the microscale, osteons (cylindrical lamellar units in cortical bone); at the macroscale, cortical (compact) and cancellous (trabecular) architectures.
7.3.1 Cortical Bone
Cortical bone is stronger in compression than in tension, and stronger longitudinally than transversely. This anisotropy arises from the alignment of osteons along the bone’s long axis — a functional adaptation to the predominant axial loading during locomotion.
Cortical bone is linearly elastic up to approximately 0.6–0.7% strain, after which microcracking and damage occur. Its stress–strain curve shows a well-defined yield point, a short post-yield region, and brittle fracture with little plastic deformation (especially in elderly, more mineralized bone).
7.3.2 Cancellous Bone
Cancellous (trabecular) bone is a porous open-cell foam structure. Its mechanical properties depend strongly on apparent density \(\rho\) and trabecular architecture:
\[ E_{\text{cancellous}} \approx E_0 \left(\frac{\rho}{\rho_0}\right)^n, \quad n \approx 2\text{–}3 \]This power-law relationship (analogous to Gibson-Ashby cellular solid theory) means that a 50% reduction in density leads to a 75–88% reduction in stiffness. In osteoporosis, this dramatic property loss with bone loss explains the increased fracture risk.
Apparent density ranges from \(\sim 0.1\ \text{g/cm}^3\) (very porous) to \(\sim 0.9\ \text{g/cm}^3\) (nearly cortical), with corresponding moduli from \(\sim 1\ \text{MPa}\) to \(\sim 1500\ \text{MPa}\).
7.4 Mechanical Properties of Articular Cartilage
Articular cartilage is a highly hydrated (65–80% water) biphasic material: a solid matrix (type-II collagen and aggrecan proteoglycans) saturated with interstitial fluid. Its mechanical behavior is inherently time-dependent due to fluid flow.
where \(p\) is the interstitial fluid pressure and \(\mathbf{I}\) is the identity tensor. Under instantaneous loading, fluid is trapped (\(\nabla \cdot \mathbf{v} = 0\)) and \(\nu \to 0.5\). Under slow (drained) loading, fluid escapes freely and the intrinsic matrix modulus (\(H_A \approx 0.5\text{–}1.0\ \text{MPa}\)) governs. The hydraulic permeability \(k \sim 10^{-15}\ \text{m}^4/\text{N·s}\) controls the rate of fluid exudation.
The aggregate modulus \(H_A = E(1-\nu)/[(1+\nu)(1-2\nu)]\) is the stiffness under confined compression (no lateral expansion permitted, simulating joint geometry). Values:
- Femoral condyle cartilage: \(H_A \approx 0.70\ \text{MPa}\), \(G \approx 0.25\ \text{MPa}\)
- Patellar cartilage: \(H_A \approx 0.92\ \text{MPa}\)
The dynamic modulus (at physiological loading frequencies of 1 Hz) is 5–20 times the static equilibrium modulus, because fluid pressurization at higher rates bears most of the load — a beautifully efficient biological design.
7.5 Mechanical Properties of Soft Tissues: Tendons and Ligaments
Tendons and ligaments are dense regular connective tissues composed primarily of type-I collagen fibers arranged in a hierarchical crimp pattern. Their nonlinear stress–strain behavior (the characteristic J-curve) arises from progressive fiber recruitment:
- Toe region (low stiffness, \(\varepsilon \lesssim 3\%\)): Crimped fibers straighten, requiring little force. The apparent modulus is low (10–100 MPa).
- Linear region (\(3\% \lesssim \varepsilon \lesssim 6\%\)): All fibers are recruited and loaded in tension. Tangent modulus is 1–2 GPa for tendon.
- Failure region (\(\varepsilon \gtrsim 6\%\)): Progressive fiber failure; ultimate tensile strength of tendon \(\approx 50\text{–}100\ \text{MPa}\).
or the simpler exponential form: \(\sigma = A\left(e^{B\varepsilon} - 1\right)\), where \(A\) and \(B\) are experimentally fitted constants. The tangent stiffness is \(d\sigma/d\varepsilon = AB\,e^{B\varepsilon}\), which increases with strain — reproducing the J-curve stiffening.
7.6 Anisotropy and Constitutive Modeling
For a fully anisotropic linear elastic material, the constitutive law (generalized Hooke’s law) requires up to 21 independent elastic constants. Symmetry reduces this number:
- Isotropic: 2 constants (\(E\), \(\nu\))
- Transversely isotropic (one symmetry axis; cortical bone, muscle): 5 constants (\(E_L\), \(E_T\), \(\nu_{LT}\), \(\nu_{TT}\), \(G_{LT}\))
- Orthotropic (three symmetry planes; cancellous bone, myocardium): 9 constants
The full constitutive law in compliance form:
\[ \begin{pmatrix}\varepsilon_1 \\ \varepsilon_2 \\ \varepsilon_3 \\ \gamma_{23} \\ \gamma_{13} \\ \gamma_{12}\end{pmatrix} = \begin{pmatrix}1/E_1 & -\nu_{21}/E_2 & -\nu_{31}/E_3 & 0 & 0 & 0 \\ -\nu_{12}/E_1 & 1/E_2 & -\nu_{32}/E_3 & 0 & 0 & 0 \\ -\nu_{13}/E_1 & -\nu_{23}/E_2 & 1/E_3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/G_{23} & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/G_{13} & 0 \\ 0 & 0 & 0 & 0 & 0 & 1/G_{12}\end{pmatrix} \begin{pmatrix}\sigma_1 \\ \sigma_2 \\ \sigma_3 \\ \tau_{23} \\ \tau_{13} \\ \tau_{12}\end{pmatrix} \]7.7 Nonlinear Kinematics: The Deformation Gradient
For soft tissues undergoing large deformations, linear elasticity is inadequate. The correct kinematic measure is the deformation gradient tensor \(\mathbf{F}\):
\(J = 1\) for incompressible materials (constant volume). The right Cauchy-Green tensor \(\mathbf{C} = \mathbf{F}^T \mathbf{F}\) and the Green-Lagrange strain tensor \(\mathbf{E} = (\mathbf{C} - \mathbf{I})/2\) are the appropriate finite-strain measures.
Hyperelastic constitutive models for soft tissue are expressed via a strain energy function \(W(\mathbf{C})\) or \(W(\mathbf{E})\), and the second Piola-Kirchhoff stress is:
\[ \mathbf{S} = 2\frac{\partial W}{\partial \mathbf{C}} \]Common models include the neo-Hookean, Mooney-Rivlin, and Holzapfel-Ogden (for fiber-reinforced tissues) models.
7.8 Wolff’s Law and Mechanobiology
This principle has profound implications for implant design. A rigid femoral stem (titanium, \(E \approx 110\ \text{GPa}\) vs. bone \(E \approx 17\ \text{GPa}\)) shields the proximal femoral cortex from its normal stress environment. This stress shielding leads to cortical bone resorption, which can loosen the implant over years. Modern implant design strategies to mitigate stress shielding include:
- Using composite (PEEK-carbon fiber) stems with modulus closer to bone.
- Short stems that load the femoral neck.
- Surface coatings (hydroxyapatite) to promote osseointegration.
- Distal fixation designs that preserve proximal bone loading.
An intervertebral disc (IVD) is modeled as a standard linear solid with \(E_\infty = 0.5\ \text{MPa}\), \(E_1 = 1.5\ \text{MPa}\), and relaxation time \(\tau_R = 1500\ \text{s}\) (\(25\ \text{min}\)). Under a sustained compressive stress of \(\sigma_0 = 0.5\ \text{MPa}\) applied at \(t = 0\), find the creep strain \(\varepsilon(t)\) using the SLS model.
Solution.
For the SLS under constant stress (creep), the compliance function \(J(t)\) gives:
\[ \varepsilon(t) = \sigma_0 \cdot J(t) = \sigma_0 \left[\frac{1}{E_\infty + E_1} + \frac{E_1}{E_\infty(E_\infty + E_1)}\left(1 - e^{-t/\tau_C}\right)\right] \]where the creep time constant is \(\tau_C = \tau_R(E_\infty + E_1)/E_\infty = 1500 \times (2.0/0.5) = 6000\ \text{s}\).
At \(t = 0\): \(\varepsilon(0) = \sigma_0/(E_\infty + E_1) = 0.5/2.0 = 0.25\%\).
At equilibrium (\(t \to \infty\)): \(\varepsilon(\infty) = \sigma_0/E_\infty = 0.5/0.5 = 1.0\%\).
At \(t = 7200\ \text{s}\) (2 hours of sitting):
\[ \varepsilon = 0.25\% + \frac{1.5}{0.5 \times 2.0}(0.5\%)\left(1 - e^{-7200/6000}\right) = 0.25\% + 0.75\%\left(1 - e^{-1.2}\right) = 0.25\% + 0.75\%(0.699) = 0.774\% \]This \(\sim 0.77\%\) compressive strain in the disc over 2 hours of sitting corresponds to the experimentally measured 1–2% height loss of the lumbar spine during prolonged sitting — consistent with the well-known diurnal variation in human body height (\(\sim 1.5\ \text{cm}\) shorter in the evening than morning).
A patellar tendon specimen is rapidly stretched to a strain of \(\varepsilon_0 = 3\%\) (into the linear region) and held fixed. The initial stress is \(\sigma_0 = 12\ \text{MPa}\). Using a Maxwell model with \(\tau_R = 600\ \text{s}\), find the stress at \(t = 300\ \text{s}\) and \(t = 1800\ \text{s}\).
Solution.
\[ \sigma(t) = \sigma_0\,e^{-t/\tau_R} = 12\,e^{-t/600}\ \text{MPa} \]At \(t = 300\ \text{s}\): \(\sigma = 12\,e^{-0.5} = 12 \times 0.6065 = 7.28\ \text{MPa}\).
At \(t = 1800\ \text{s}\): \(\sigma = 12\,e^{-3} = 12 \times 0.0498 = 0.60\ \text{MPa}\).
The Maxwell model predicts nearly complete relaxation by 30 minutes. In reality, the tendon retains a significant residual stress (the equilibrium elastic response), so an SLS or Prony series model gives a better fit. This experiment motivates the preconditioning protocol used in tensile testing of soft tissue: samples are cycled several times before data collection to reach a repeatable, preconditioned state.
