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In an inviscid barotropic fluid subject to conservative body force, the circulation of a material loop is constant. For constant density, parameterize that loop by $\mathbf{r}(s,t)$ with ${\mathbf{r}}_t=\mathbf{u}$. Differentiating $\Gamma =\int \mathbf{u}\cdot {\mathbf{r}}_{s}ds$ gives an integral of $(D\mathbf{u}/Dt)\cdot d\mathbf{r}$ plus $\int {\partial }_s(|\mathbf{u}{|}^2/2)ds$. The latter vanishes on a closed loop, and the Euler equations make the former the integral of a gradient, also zero. For a barotropic density, $\nabla p/\rho$ is still a gradient of specific enthalpy. Constant circulation is consistent with changing local vorticity when a vortex tube changes its cross-sectional area.

A constant-tension slender viscous thread loses the same cross-sectional area along each material element: $A=A_0-Ft/(3\mu )$. Volume conservation gives material stretch $A_0/A$. Near an isolated quadratic minimum of $A_0$, integration over a shrinking material width proportional to $(t^{\ast }-t{)}^{1/2}$ gives total extension exponent $-1/2$. A uniformly thin thread has the same divergent stretch everywhere and exponent $-1$. The localized coefficient is independent of remote endpoints.

For fixed mean turn length and fixed conductor volume, a coil of $N$ turns has conductor length proportional to $N$ and cross-sectional area proportional to $1/N$. Its electrical resistance is therefore $R_{\mathrm{coil}}=a{N}^2$. With source electromotive force $\mathcal{E}$ and internal resistance $R_s$, suppose the small-deflection driving torque is proportional to $NI=N\mathcal{E}/(R_s+a{N}^2)$. Its derivative has the sign of $R_s-a{N}^2$, so the maximum occurs at $R_{\mathrm{coil}}=R_s$. Geometry, damping and finite response time can alter the practical optimization; this is a conditional winding model, not a universal rule for all galvanometers.

The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

The end displacement relative to the initial unit axial length isBy Hooke's law, the total resisting force supplied by the spring is $F_3=-K\delta$. To convert this resultant into the axial Cauchy stress tensor, divide by the current end cross-sectional area. The transverse total stretches are both ${\alpha }^{-1/2}$, so that area isUniform traction over that face consequently givesEquate this to the expression from part (ii):The extra factor $\alpha$ on the left is essential: the spring law specifies a total force, whereas the material constitutive equation specifies force per current area. The reference end area is one, so the same spring resultant can also be read as a nominal end traction.

At a fixed $z\in [-1,1]$, both cylinder inequalities becomeThus the cross section is a square of side $2\sqrt{1-z^2}$ and area $4(1-z^2)$. Integrating these cross-sectional areas gives the volume of this bicylinder:

The original double integral covers the triangle $0\le y\le x\le 1$. By Fubini's theorem, after reversing the order, $y$ runs from $0$ to $1$ and $x$ from $y$ to $1$. Since a continuous function is integrable on this bounded region,For the square pyramid, a horizontal section at height $z$ is a square of side $2(1-z)$: each vertex moves linearly from the base towards $(1,1,1)$. Hence its cross-sectional area is $4(1-z{)}^2$, and its volume is $4{\int }_0^1(1-z{)}^{2}dz$. To use the identity just proved, choose $f(y)=1-y$:

Assume the rod has constant cross-sectional area $A$, no internal source of heat, no bulk motion, and no lateral heat flux. Assume also that each section has a uniform temperature, so that $T$ depends only on $x,t$, and that Fourier's law applies with the stated constant thermal conductivity $k$. The longitudinal heat flux density is then $q=-k{T}_x$.

By conservation of energy, the rate of increase of internal energy in a slice $[x,x+\Delta x]$ equals incoming minus outgoing heat:Divide by $A\Delta x$ and take the limit $\Delta x\to 0$. For a sufficiently differentiable temperature field this gives $\rho c{T}_t=-q_x=k{T}_{xx}$, and henceHere $\rho$ is mass density, $c$ is specific heat capacity, and $\kappa$ is thermal diffusivity.

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

Write the wire diameter as $d_w$ to distinguish it from other geometric labels. Its cross-sectional area is $A_w=\pi d_w^2/4$, and the total wire length is $4a$. From the electrical resistivity, its electrical resistance isWhile the loop straddles the boundary, let $h$ be the height of each vertical side remaining in the magnetic field. The linked magnetic flux has magnitude $Bah$, with $\dot{h}=-v$. Faraday's law therefore gives $|\mathcal{E}|=Bav$. Neglecting the loop's self-inductance, Ohm's law givesThe into-page magnetic flux is decreasing, so Lenz's law requires a clockwise electric current as viewed in the diagram. If the loop is wholly inside a uniform magnetic field, or wholly outside it, its magnetic flux is constant and this induced electromotive force is zero. During a rapid transient one should instead solve $L\dot{I}+RI=Bav$; at the assumed steady speed the same boxed relation holds.

Use the geometric cross-section shown in the source: the outer cylinder has radius $2a$ and centre $O$, and the circular cavity has radius $a$ and centre $C=a\hat{\mathbf{x}}$. Let the electric current be in the $+z$ direction. The conducting cross-sectional area is $3\pi a^2$, so the uniform electric current density is $\mathbf{J}=I\hat{\mathbf{z}}/(3\pi a^2)$.

To obtain the exterior magnetic field of a wire with an eccentric cylindrical cavity, apply the superposition principle to a full cylinder of electric current density $\mathbf{J}$ centred at $O$ and a second cylinder of electric current density $-\mathbf{J}$ centred at $C$. Their electric currents are $4I/3$ and $-I/3$. This reproduces the actual electric current density everywhere, including zero in the cavity. Outside a long circular cylinder, Ampère's law gives $\mathbf{B}={\mu }_0{I}_{\mathrm{cyl}}\hat{\mathbf{z}}\times \mathbf{r}/(2\pi r^2)$. Every point outside the outer boundary is outside both component cylinders. Consequently, writing $\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}$,In Cartesian coordinates, the same magnetic flux density isThe cavity breaks rotational symmetry. Thus the magnetic flux density is generally neither constant in magnitude nor tangential on a circle centred at $O$; replacing its line integral by $2\pi rB$ would be unjustified. Its line integral is nevertheless ${\mu }_{0}I$, as Ampère's law requires. At distances $r\gg a$, the leading term becomes ${\mu }_{0}I\hat{\mathbf{\varphi }}/(2\pi r)$, with corrections of order ${\mu }_{0}Ia/r^2$.

Let $t(n)$ denote traction: the force per unit area exerted across an oriented internal surface with unit normal $n$. Define ${\sigma }_{ij}$ as component $i$ of the traction on a coordinate plane whose normal is $e_j$. Balance of linear momentum on a shrinking tetrahedron makes the surface-force terms, of order area, dominate its inertia and body forces, of order volume. Resolving its oblique face into the three coordinate projected areas gives Cauchy's traction formulaOpposite faces have opposite tractions, as follows independently from a shrinking pillbox. This fixes both the index and normal conventions.

For a material volume $V(t)$, the Reynolds transport theorem says $d{\int }_{V(t)}fdV/dt={\int }_{V(t)}[f_t+\nabla \cdot (fu)]dV$. Apply it componentwise to $f=\rho u_i$ and use mass conservation. With no body force in this calculation, balance of linear momentum and the divergence theorem giveArbitrariness of the material volume yields $\boxed{\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}}$, with $D/Dt={\partial }_t+u_j{\partial }_j$.

For constant dynamic viscosity and incompressible velocity, ${\partial }_j{u}_j=0$, soThus $\boxed{\rho (u_t+u\cdot \nabla u)=-\nabla p+\mu {\nabla }^{2}u}$.

In fully developed steady pipe flow $u=w(x,y)e_z$, the convective acceleration vanishes. The transverse equations give $p_x=p_y=0$ and the axial equation gives $p_z=\mu (w_{xx}+w_{yy})$. One side depends only on $z$ and the other only on $(x,y)$, so both equal a constant. HenceIf $n$ is outward from the fluid cross-section, the wall-on-fluid axial shear per length is $\mu \oint {\partial }_{n}wds=\mu {\int }_A\Delta wdA=A{p}_z$. The fluid-on-pipe force is its negative:For forward pressure-driven flow $p_z<0$, this force points along the flow. Its value depends on cross-sectional area and pressure drop, even though the detailed velocity profile depends on the shape.

In a closed tank of cross-sectional area $S$, entrainment into a uniform upper layer conserves the density deficit $({\rho }_l-{\rho }_u)h$. Thus $B=g^{\prime }h$ is constant. With depth $z$ increasing downward, gravitational potential energy is $-gS{\int }_0^H\rho (z)zdz=E_{\ast }+{\rho }_{l}SBh/2$, so its rate of increase is proportional to the rate of deepening. This includes homogenization throughout the upper layer, not merely lifting a thin parcel across the interface.

The pressure-area law of a collapsible tube specifies transverse mechanical equilibrium of a deformable tube wall. With physical internal and external pressures $P,P_e$ and cross-sectional area $A$, write $P-P_e=\mathcal{P}(A)$. For constant fluid mass density $\rho$,is the squared tube wave speed about an undisturbed area $A_0$.

The kinematic pressure convention used in the parent model instead writes $p=P/\rho$, $p_e=P_e/\rho$ and $p-p_e=\tilde{P}(\alpha )$. It gives the equivalent expression $c^2={\alpha }_0{\tilde{P}}^{\prime }({\alpha }_0)$, whether $\alpha$ denotes area or area divided by a fixed reference area. Linearized area and momentum conservation yield this coefficient of the resulting wave equation. A stable local wall branch has positive derivative; prescribed external pressure is held fixed when taking it. This distinguishes a wall stiffness from a change in applied external loading.

For a long axisymmetric bubble with slowly varying radius $a(z,t)$, the leading normal-stress balance and internal lubrication flux couple capillary pressure to conservation of the cross-sectional area.

For a nutrient-consuming chemotactic travelling band in a tube of cross-sectional area $a$, let $N=a{\int }_{\mathbb{R}}B(z)dz$. Integrating $v{C}^{\prime }=kB$ between depleted nutrient behind and $C_{\infty }$ ahead gives $v{C}_{\infty }=kN/a$, hence $v=Nk/(a{C}_{\infty })$. This is a nutrient budget: advancing by a unit length supplies $a{C}_{\infty }$ nutrient, while the bacteria consume at total rate $kN$. Diffusion and chemotactic sensitivity affect the band shape and admissibility, but not this speed at fixed $N,k,a,C_{\infty }$.

A stress concentration is a localized increase of stress caused by geometry or another disturbance of the load path. Holes, notches and abrupt section changes can make the largest local normal stress much greater than a stress calculated from the total force divided by a reference cross-sectional area. A stress concentration factor $K_t={\sigma }_{\max }/{\sigma }_{\mathrm{ref}}$ is meaningful only after both the stress component and its reference have been specified.

The Kirsch equations give an instructive exact result for a circular hole of radius $a$ in an infinite, isotropic linear elastic elastic plate under remote tensile stress ${\sigma }_{\infty }$ in one direction, using a plane stress model. Measure the polar angle $\theta$ from the loading direction. At the hole boundary,At $\theta =\pi /2,3\pi /2$, the tangential normal stress reaches $3{\sigma }_{\infty }$, giving $K_t=3$ relative to remote stress. At $\theta =0,\pi$, it is $-{\sigma }_{\infty }$. A traction-free hole therefore does not require every component of its stress tensor to vanish: it requires zero traction on the radial boundary.^[ref]

For a finite plate, distinguish gross and net reference areas. A plate of width $W$, thickness $t$ and central hole diameter $d$ transmits force $P$ through gross area $A_g=tW$ and minimum net area $A_n=t(W-d)$. Defining ${\sigma }_g=P/A_g$ and ${\sigma }_n=P/A_n$ gives the exact conversionThis is a change of normalization, not a solution for the finite plate's stress field. For an original numerical example, take $W=40\mathrm{mm}$, $t=2\mathrm{mm}$, $d=10\mathrm{mm}$ and $P=800\mathrm{N}$. The reference stresses are $10$ and $13.333\mathrm{MPa}$. If the local maximum is stipulated as $32\mathrm{MPa}$, the same physical maximum gives $K_{tg}=3.2$ and $K_{tn}=2.4$. Neither number changes the local stress by choosing a different denominator.

The distinction appears in real historical measurements. In 1939, C. Dumont loaded a perforated aluminum-alloy plate in out-of-plane bending and measured strain and deflection at two loads. The reported peak tangential stress was 1.59 times the computed net-section stress, but 1.85 times the computed stress for an unperforated plate under the same bending forces. These are bending results for that geometry, rather than the infinite-plate tensile factor three.^[ref]

Modern demonstrations can combine a metal plate instrumented with strain gauges and a transparent polymer viewed in a polariscope. The latter uses stress-induced birefringence; its fringes respond to differences of principal stresses. The MIT laboratory compares local and remote strain measurements with qualitative photoelasticity observations around holes and a slot.^[ref]

The elastic stress concentration factor is dimensionless. It differs from the stress intensity factor used to describe a sharp crack's singular field. Local yielding or a crack changes the mechanical problem; multiplying a remote stress by the elastic $K_t$ does not, by itself, establish a failure load.

A stress-strain curve plots a material's stress response against a specified strain measure during a loading experiment. Its initial slope gives an elastic modulus in a linear regime; yielding, hardening and fracture can produce later changes.

The curve depends on temperature, loading rate, sample history and the definitions of stress and strain. Engineering stress divides force by original cross-sectional area, whereas true stress uses current area. After necking these can differ greatly. Unloading can distinguish recoverable elastic strain from permanent deformation, so one loading curve alone does not specify every constitutive property.

An unbonded material-fibre bundle collects separate material fibres without continuously bonding neighbouring strands together. Contact and friction may still transfer force, so unbonded does not automatically mean freely sliding. The response depends on which strands are gripped, their alignment, slack and whether inter-fibre sliding is permitted.

For straight strands of common length $L$, let both end fixtures impose the same small extension $\Delta L$ on every strand. Under linear elasticity, strand $i$ carries $F_i=E_i{A}_i\Delta L/L$, soThis follows by applying the uniform-bar axial stiffness relation to the strands in parallel.^[ref] Equal extension here is an end-loading condition. It does not require continuous bonding, nor does it describe the separate freely sliding bending experiment in bending rigidity of an unbonded fibre bundle.

The material cross-sectional area $A_m={\sum }_i{A}_i$ excludes gaps. An envelope enclosing the bundle can have larger area $A_e$. Define the bundle material area fraction $\varphi =A_m/A_e$. For identical moduli $E$, inferring a modulus from the envelope and the measured axial stiffness givesAs an original example, four circular strands of radius $0.10\mathrm{mm}$ have $A_m=0.12566\mathrm{m}{\mathrm{m}}^2$. Stipulate an envelope area $0.20\mathrm{m}{\mathrm{m}}^2$, $E=100\mathrm{GPa}$ and $L=50\mathrm{mm}$. Then $\varphi =0.6283$, $K_{\mathrm{ax}}=251.3\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$ and $E_{\mathrm{envelope}}=62.83\mathrm{GPa}$. That lower effective value does not mean the strand material has changed.

A useful tensile comparison counts the engaged strands, measures their areas and checks that the fixtures actually impose the assumed extension. A loose or broken strand changes the load-bearing area even if the outer envelope barely changes. Individual strand properties can be characterized by a single-fibre tensile test; neither envelope dimensions nor a bundle slope alone identifies every strand's Young's modulus.

A periodic slender bubble with $a=f+\sqrt{2}g\sin kz$ and negligible internal pressure variation has exterior radial velocity $u_r=a\dot{a}/r$ and pressure $P=2\mu \dot{a}/a+\gamma /a$. Its conserved mean cross-sectional area gives ${\alpha }_0=f^2+g^2=f(0{)}^2+g(0{)}^2$. Comparing the constant and sinusoidal terms gives the displayed equations. The long-wave growth rate about radius $a_0$ is $\gamma /(2\mu a_0)$, and ${\dot{a}}_{\min }=-\gamma (g^2+\sqrt{2}fg)/(2\mu {\alpha }_0)<0$. Small internal viscosity eventually produces appreciable axial pressure gradients near a neck, so this uniform-pressure reduction cannot persist to pinch-off at fixed positive viscosity ratio.