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A. F. Bower is the author of Applied Mechanics of Solids by A. F. Bower, an online text whose title page gives his name as Allan F. Bower. The text connects elasticity, plasticity in continuum mechanics and fracture mechanics to explicit models of how solids deform. Its treatment of single-crystal plastic deformation follows slip systems and the rotation of the tensile axis, showing why a single crystal's response depends on its orientation.^[ref]

The beam shear angle is the small-angle difference between centreline slope and material-section beam cross-section rotation:With transverse coordinate $z$ and kinematics $u_x=-z\theta (x)$, $u_z=w(x)$, the engineering shear strain is ${\partial }_z{u}_x+{\partial }_x{u}_z=\gamma$. The Euler-Bernoulli beam equation imposes $\gamma =0$; a shear-deformable model permits $w$ and $\theta$ to vary independently. The primary beam account distinguishes these kinematics and relates transverse force to the resulting shear measure.^[ref]

For an original local example, suppose shape measurements give $w_x=0.015$, while section markers give $\theta =0.010$ radians in the same sign convention. Then $\gamma =0.005$, about $0.286$ degrees. Setting the section rotation equal to the observed slope would erase this measured discrepancy by assumption.

In a simple linear shear model, the transverse force resultant is $V=S\gamma$, with effective beam shear rigidity $S=k_{s}GA$. Here $G$ is the shear modulus, $A$ the material area, and $k_s$ a section correction coefficient. Stipulating $G=1.0\mathrm{GPa}$, $A=20\mathrm{m}{\mathrm{m}}^2$ and $k_s=5/6$ gives $S=16667\mathrm{N}$ and $V\approx 83.3\mathrm{N}$. The correction coefficient accounts for using a reduced section model; it is not universally fixed at $5/6$.

A physical test must measure both the centreline shape and section orientation, for example by tracking markers during loading. Shape alone determines $w_x$ but cannot independently determine $\gamma$ without section-rotation information or an additional mechanical model. The comparison also requires compatible spatial resolution and angular measurement calibration.

The beam shear correction coefficient $k_s$ adjusts a reduced beam model's shear response for the nonuniform shear stress over a material section. With shear modulus $G$, area $A$, transverse force resultant $V$ and generalized beam shear angle $\gamma$,where $S$ is effective beam shear rigidity. Bower's primary account calls this the Timoshenko shear coefficient and explains that its value depends on geometry, Poisson's ratio and how the approximate theory is matched to the underlying solid.^[ref]

An original energy-matching derivation makes the correction precise for one approximation. Assume uniform $G$ and a chosen shear-stress field $\tau$ with resultant $V={\int }_A\tau dA$. The shear elastic energy per unit beam length and its reduced representation areThusThe Cauchy-Schwarz inequality gives $V^2\le A{\int }_A{\tau }^{2}dA$, hence $k_s\le 1$ for this definition. Equality requires uniform stress. This bound belongs to the stated energy-matching definition rather than every convention called a shear correction coefficient.

For the parabolic rectangular stress approximation used in effective beam shear rigidity, the squared-stress integral is $6V^2/(5A)$, giving $k_s=5/6$. Choose $G=80\mathrm{GPa}$, $A=100\mathrm{m}{\mathrm{m}}^2$ and $V=1000\mathrm{N}$. Then $S=6.667\times 10^6\mathrm{N}$ and $\gamma =1.5\times 10^{-4}$ radians. Omitting the correction gives $1.25\times 10^{-4}$, underestimating this generalized shear angle by one sixth.

The generalized angle also differs from the peak local shear strain. Under the same parabolic approximation, peak stress is $3V/(2A)$, giving peak strain $1.875\times 10^{-4}$. The values describe different reductions of the same chosen stress field. Identifying $k_s$ from a bending test therefore requires the section model and displacement readout, not merely a material modulus. For a three-point bending test, the shear deflection contribution varies as $PL/(4k_{s}GA)$, whereas bending varies as $P{L}^3/(48B)$; a span comparison can help separate them.

The Clausius-Duhem inequality is a local expression of the Second law of thermodynamics for a simple continuum whose thermodynamic entropy flux is heat flux divided by positive absolute temperature. With specific entropy $s$, outward heat flux $\mathbf{q}$ and heat supply $r$ per unit mass, it readsCombining this with the local internal energy balance and writing the specific Helmholtz free energy as $\psi =e-Ts$ givesHere $D$ is the rate-of-strain tensor and the dot denotes a material derivative.^[ref]

The inequality constrains constitutive equations, including both mechanical dissipation and thermal conduction. For Fourier's law $\mathbf{q}=-k\nabla T$ with $k\ge 0$, the thermal contribution is $k|\nabla T{|}^2/T\ge 0$. This direct sign check explains why passive thermal conduction carries heat down a temperature gradient. More general continua can require a different thermodynamic entropy flux, so the stated assumption matters.

Effective beam shear rigidity is the coefficient $S$ relating a transverse shear-force resultant $V$ to the beam shear angle $\gamma$ in a reduced linear beam model:Here $G$ is shear modulus, $A$ is material area and $k_s$ is a beam shear correction coefficient. The coefficient $S$ has units of force, whereas beam flexural rigidity has units $\mathrm{N}{\mathrm{m}}^2$. A shear-deformable beam permits section rotation to differ from centreline slope; Bower gives the corresponding force relation and explains the correction coefficient's dependence on section geometry and the approximation used.^[ref]

An original energy calculation explains one familiar correction. For a homogeneous rectangle of width $b$, thickness $h$ and area $A=bh$, adopt the classical parabolic shear-stress approximationIt has resultant ${\int }_A\tau dA=V$. Its shear elastic energy per unit beam length isEquating this to the reduced-model energy $V^2/(2S)$ gives $S=5GA/6$. Thus $k_s=5/6$ follows from this particular energy-matching approximation; it is not a universal material constant or an exact coefficient for every beam theory.

The effect on a three-point bending test can be derived from that reduced energy. For a simply supported beam under central force $P$, the magnitude of shear force is $P/2$ on each half. Hence $U_s=P^{2}L/(8S)$ and its contribution to central displacement is $\partial U_s/\partial P=PL/(4S)$. Combining it with the bending response givesThe shear contribution becomes relatively more important as the span decreases.

For stipulated $B=2\mathrm{N}{\mathrm{m}}^2$, $S=20000\mathrm{N}$, $L=0.10\mathrm{m}$ and $P=10\mathrm{N}$, the two contributions are $0.1042$ and $0.0125\mathrm{mm}$. Shear adds twelve percent of the bending deflection. A span series with a load cell and direct midspan displacement readout can distinguish cubic bending dependence from linear shear dependence, provided contact and fixture contributions are also controlled. Inferring $G$ from the response additionally requires the section geometry and correction model.

Fracture mechanics studies the initiation and growth of cracks under loading. It connects a defect's geometry with the local driving force for crack extension and the material's resistance. A component's average stress alone cannot describe the concentration of deformation near a crack tip.^[ref]

In linear elastic fracture mechanics, a mode-I stress intensity factor often has the form $K_I=Y\sigma \sqrt{\pi a}$, where $a$ is a specified crack dimension and $Y$ accounts for geometry. Doubling $a$ increases $K_I$ by $\sqrt{2}$ at fixed nominal stress. Comparing the appropriate intensity with measured fracture toughness is meaningful only when the loading, geometry and small-scale yielding assumptions of the model apply.

For a small displacement field $\mathbf{u}$, define $G_{ij}={\partial }_i{u}_j$ and the infinitesimal strain tensor $\epsilon =(G+G^T)/2$. The isotropic linear elasticity energy density isHere $\lambda ,\mu$ are Lamé parameters and the colon is the Frobenius inner product. The energy depends on symmetric strain, so an infinitesimal rigid rotation costs no elastic energy. A term involving only $G:G^T$ does not have this property. ^[ref]

Differentiation gives the Cauchy stress tensorVarying $\int WdV$ with fixed boundary displacement and no body force yields $\nabla \cdot \sigma =0$, orThese are the static Navier-Cauchy equations.

In two dimensions, separating trace and deviatoric strain givesThus $\mu >0$ and $\lambda +\mu >0$ make every nonzero two-dimensional strain cost positive energy. Rigid translations and infinitesimal rigid rotations remain zero-strain motions.

Latent hardening is increased resistance on one crystal slip system caused by plastic slip on another slip system. The word latent emphasizes that the affected system need not have been active during the earlier loading. It is therefore a statement about interactions between deformation mechanisms, rather than simply about the total amount of deformation.

A simple crystal-plasticity model writes ${\dot{g}}_{\alpha }={\sum }_{\beta }{h}_{\alpha \beta }|{\dot{\gamma }}_{\beta }|$, where $g_{\alpha }$ is the critical resolved shear stress for system $\alpha$ and ${\gamma }_{\beta }$ is slip on system $\beta$. Off-diagonal coefficients represent cross-system hardening, while diagonal coefficients describe self-hardening.^[ref] Changing loading direction after predeformation can activate previously less-favoured slip systems and reveal this history dependence. Predicting the next system to move therefore requires both resolved stress geometry and the evolving resistances.

A second slip system begins to operate as the tensile-axis rotation makes its resolved shear stress comparable to that on the primary system. Dislocations on intersecting planes interact much more strongly, giving the transition from easy single slip to rapid work hardening. The standard geometric estimate neglects differences between the evolving critical stresses of the two systems.

The primary direction is $\mathbf{s}\parallel [101]$. Single-slip shear carries the line joining the specimen ends, expressed in crystal coordinates, alongThe initial fundamental sector has $x<y<z$. Its first reached boundary is $x=y$, at $t=1$, giving $\mathbf{r}=[225]$. At that orientation $(\overline{1}11)[101]$ and $(1\overline{1}1)[011]$ have equal Schmid factors. Since the component perpendicular to the primary slip direction is unchanged by that shear, the length ratio follows directly:ThereforeThe corresponding logarithmic strain is $\ln (L/L_0)=0.2260$. This is an ideal equal-critical-stress geometric onset estimate. Real secondary slip can start earlier or later because of latent hardening, nonuniform deformation and grip constraints; the problem supplies no hardening law that would determine that shift.^[ref]

Single-crystal plastic deformation is irreversible shape change of a single crystal, often through dislocation motion on particular slip systems. A slip system specifies a crystallographic plane and a direction within it. Applied loading resolves into a shear stress that drives slip, making the response depend on crystal orientation.^[ref]

Under uniaxial stress $\sigma$, the resolved stress on a slip system is $\tau =\sigma \cos \varphi \cos \lambda$, where $\varphi$ is the angle between the loading axis and plane normal and $\lambda$ the angle to the slip direction. For compatible $45$-degree angles, $\tau =\sigma /2$. Slip begins when the relevant resistance is reached; subsequent deformation can rotate the lattice and change which systems carry further strain.^[ref]

Slip-line field theory solves certain two-dimensional problems of plastic deformation by constructing two orthogonal families of characteristic curves, called slip lines. Its classical assumptions include plane strain, negligible inertia and body force, and a rigid, perfectly plastic material. The curves are aligned with the directions of maximum in-plane shear stress, rather than the principal stress directions.^[ref]

The curve network encodes both the stress field and constraints on admissible velocity. Indentation of a thick workpiece by a rigid flat punch provides a motivating application: the construction predicts a plastic flow region and the load needed to sustain that flow. It is useful because geometry can turn coupled field equations into relations integrated along curves. A slip line is a continuum characteristic, however, not necessarily a microscopic crystal slip system or the actual path of a moving material particle.

Work hardening is an increase in resistance to further plastic deformation caused by earlier deformation. A useful macroscopic test loads a metal beyond yield, unloads it and reloads in the same direction: further plastic flow then requires a larger stress. In a constitutive equation, this history is represented by evolving internal variables rather than by the current elastic strain alone.^[ref]

In a crystal, multiplication and interaction of dislocations can make later dislocation glide more difficult. Forest hardening and latent hardening describe particular aspects of that obstacle formation. Increased resistance to plastic flow is not the same as a large change in Young's modulus: the elastic unloading slope and the load required to restart permanent deformation measure different properties. Work hardening can delay necking by compensating for the shrinking loaded area.