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A local principal stress model associated with Ernest Masson Anderson in which one principal stress is vertical, as suggested by negligible shear stress on a flat free surface. With compression positive and ${\sigma }_1\ge {\sigma }_2\ge {\sigma }_3$, vertical ${\sigma }_1$ gives normal faulting, vertical ${\sigma }_2$ gives strike-slip faulting, and vertical ${\sigma }_3$ gives reverse faulting. The other two principal stresses are horizontal. Topography, mechanical layering and local fault geometry can violate the idealization.

The water pressure at a glacier-bed interface affects the effective normal pressure $N$, the difference between ice overburden pressure $p_i$ and water pressure $p_w$. Higher water pressure at fixed overburden reduces effective compression of the bed contact and can alter cavitation and basal resistance. It is not identical to basal shear stress, and a sliding law depending only on shear suppresses this independent hydraulic influence.

A quadratic boundary-drag model relates shear stress to mean flow velocity, with dimensionless resistance coefficient $c_d>0$. Uniform steady flow down slope $S$ balances $\rho gAS=\tau \ell$, giving $|u|=\sqrt{gRS/c_d}$. Its sign is chosen so drag opposes the motion.

For the convention $e^{i\omega t}$, complex viscosity is the ratio of oscillatory shear stress amplitude to shear rate amplitude. A causal relaxation modulus gives the displayed transform. Its associated complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, where $G^{\prime }$ is the storage modulus and $G^{\prime \prime }$ the loss modulus.

One of a pair of fault orientations that can form under a common principal stress field with opposite senses of resolved shear stress. For newly formed, optimally oriented faults obeying the same Mohr-Coulomb failure criterion, their acute intersection angle is $90^{\circ }-\varphi$, where $\varphi$ is the friction angle. The maximum compressive principal stress bisects that acute angle. This permits an estimate $\mu =\tan \varphi$ only under the common-formation and optimal-orientation assumptions, rather than for arbitrary inherited faults.

For a stationary, horizontally homogeneous mean flow without a horizontal pressure gradient, the averaged horizontal Navier-Stokes equation is $\nu U_{zz}-(\overline{u^{\prime }{w}^{\prime }}{)}_z=0$. Integrating gives the displayed physical shear stress. The derivative of the Reynolds covariance belongs in the stress-divergence equation, not inside the stress itself. The kinematic stress is ${\tau }_d/\rho$, which must be used when an eddy viscosity with units of kinematic viscosity multiplies the mean gradient.

Consider steady flow of a Newtonian fluid in incompressible flow with constant dynamic viscosity $\mu >0$ between parallel plates at $z=0$ and $z=H$. The bottom plate is stationary and obeys the Navier slip boundary condition with slip length $b$. The top plate moves at velocity $U>0$ and obeys the no-slip boundary condition. There is no streamwise pressure gradient or body force. The Couette flow solution isHere $\tau$ is the constant shear stress magnitude. The lower wall's Navier slip boundary condition holds exactly because $b{u}^{\prime }(0)=bU/(H+b)=u(0)$. At the top, $u(H)=U$ is exactly the prescribed no-slip boundary condition; no finite upper-wall slip length is imposed in this example.

For a numerical example, stipulate $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $U=1\mathrm{mm}{\mathrm{s}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. The bottom velocity is $0.167\mathrm{mm}{\mathrm{s}}^{-1}$ and the shear stress is $0.0833\mathrm{Pa}$, compared with $0.100\mathrm{Pa}$ when both plates obey the no-slip boundary condition. The extrapolated profile reaches zero at $z=-2\mu \mathrm{m}$.

For fixed $H$, $U$ and $\mu$, $b\to 0$ recovers $u=Uz/H$. As $b\to \infty$, $u\to U$ throughout the gap and $\tau \to 0$: the moving upper wall selects the limiting uniform velocity, while the lower wall approaches a stress-free boundary condition. The dynamic viscosity stays positive; the disappearing shear stress comes from the disappearing velocity gradient.

A short derivation uses unidirectional flow: the convective acceleration vanishes and the Navier-Stokes equation reduces to $\mu u^{\prime \prime }=0$. Write $u=Az+B$. The bottom condition gives $B=bA$ and the top condition gives $A(H+b)=U$.

A small step of shear stress $s_{0}H(t)$ produces shear strain $\gamma (t)=s_{0}J(t)$ from a relaxed state. This time-dependent deformation under held stress is creep. The causal constitutive convolution gives $\hat{s}=p\hat{G}\hat{\gamma }$, proving the displayed Laplace transform relation. For a Maxwell fluid, $J(t)=1/G_0+t/(G_0\tau )$: an elastic jump followed by viscous flow.

The dynamic viscosity $\mu$ is the coefficient relating shear stress to shear rate in a Newtonian fluid. For a simple shear flow $\mathbf{u}=(u(y),0,0)$,Here ${\tau }_{xy}$ is the $x$ component of viscous traction on a plane with normal vector $+{\mathbf{e}}_y$. The shear rate $\dot{\gamma }=du/dy$ has units ${\mathrm{s}}^{-1}$. Thus dynamic viscosity measures the stress required to sustain a given rate of deformation. It is the shear coefficient; the separate coefficient for volume change is volume viscosity. ^[ref]

For a worked Couette flow example, place a Newtonian fluid of constant dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$ between parallel plates separated by $h=0.0010\mathrm{m}$. Hold the lower plate fixed and move the upper plate at $U=0.020\mathrm{m}{\mathrm{s}}^{-1}$. Assume steady flow, no streamwise pressure gradient, and the no-slip boundary condition. ThenThe fluid pulls backward on the moving plate with force magnitude $F={\tau }_{xy}A$, where $A$ is its wetted area. For $A=0.010{\mathrm{m}}^2$ this is $0.020\mathrm{N}$. Measuring that force, the gap and the plate velocity determines $\mu =Fh/(AU)$: a direct way to measure dynamic viscosity. The linear profile and opposing wall tractions are the standard plane Couette flow solution. ^[ref]

The velocity gradient matters, rather than the velocity itself. Adding the same constant velocity to both plates and the fluid leaves $du/dy$, and hence shear stress, unchanged. In uniform flow, every layer translates together and $du/dy=0$. More generally, even a nonzero velocity gradient need not generate viscous stress: solid-body rotation has zero strain-rate tensor, as the viscous stress tensor example shows. Thin gaps amplify shear stress for a fixed plate-speed difference, making dynamic viscosity useful for predicting the force needed to move lubricated surfaces.

The equilibrium sediment flux is the transport supported locally by a prescribed bed shear stress. A common idealization is $q_{\mathrm{sat}}={\varphi }_b\chi (\tau -{\tau }_{\mathrm{th}}{)}_{+}^{\gamma }$. Linearization above the sediment entrainment threshold requires perturbations small compared with the positive excess stress.

The locus of critical combinations of normal stress and shear stress. Contact of a Mohr circle with this boundary identifies an ideal failure threshold; it does not by itself describe post-failure dynamics.

A horizontally compressive force per width $F$, carried uniformly through thickness $a$, gives additional horizontal stress $F/a$. Resolving its traction on a fault dipping at angle $\delta$ gives the displayed mean shear stress. An isotropic background pressure adds only normal traction. The integrated force alone does not determine the actual depth-dependent stress distribution or whether a fault fails.

A fluid is a material that flows under sustained shear stress and cannot maintain a static shear stress in its fluid state. Liquids and gases are the usual examples. A Newtonian fluid has deviatoric stress proportional to its rate-of-strain tensor, whereas more general fluid constitutive equations can be nonlinear or have memory. In a porous medium, the fluid occupies connected voids within a load-bearing solid framework. The important distinction is the response to a continuing tangential force: an ordinary elastic solid can settle into a fixed deformation, while an ordinary fluid continues to deform. A fluid can nevertheless be at rest when no unbalanced shear stress drives it. ^[ref]

For a worked example, place a Newtonian fluid between parallel plates separated by $h=1\mathrm{mm}$, keep the lower plate fixed and move the upper one at $U=0.10\mathrm{m}{\mathrm{s}}^{-1}$. With no-slip boundary conditions, steady flow, no imposed pressure gradient and negligible edge effects, Couette flow has $u(y)=Uy/h$. For dynamic viscosity $\mu =0.001\mathrm{Pa}\mathrm{s}$, the shear stress magnitude isThe velocity gradient describes neighbouring layers sliding at different speeds. Doubling $U$ doubles the required shear stress; maintaining the same velocity keeps producing deformation, rather than fixing a final shape. This gives viscosity a concrete meaning: resistance to a rate of motion, not a prohibition on motion.

A fluid surface has no equilibrium static in-plane shear modulus, because material can rearrange after shear. This does not set its membrane tension or membrane bending modulus to zero. Time-dependent deformation can also produce viscous in-plane shear stress; absence of a static elastic coefficient is not absence of dissipation.

For a nonrotating force-free sphere moving with speed $U$ in a stationary tube, use its moving frame: the tube wall has velocity $-U$. In the parabolic lubrication gap $h=h_0+x^2/(2a)$, $h_0=ϵa(1+\lambda \cos \theta )$, moving-boundary lubrication flux gives $q=-h^3{p}_x/(12\mu )-Uh/2$. The wall-shear and pressure-drop balance in a tube rules out an $O({ϵ}^{-3/2})$ net pressure jump. Applying this common pressure recovery to each angular strip gives $q=-2U{h}_0/3$. The next balance yields the displayed positive localized pressure drop. Meanwhile the integrated leading shear stress on the sphere vanishes in each strip; this is consistent with negligible leading torque, and does not mean the local shear or pressure force vanishes.

For a two-dimensional decaying Stokes flow in $y>0$, prescribe boundary Fourier mode $(U,V)e^{ikx}$. The boundary shear stress and normal stress are the displayed diagonal map, with the common exponential understood. For $k>0$ and $V=0$, the Unscaled Papkovich–Neuber representation gives $u=U(1-ky)e^{ikx-ky}$, $v=-ikUy{e}^{ikx-ky}$ and $p=-2i\mu kU{e}^{ikx-ky}$. Differentiation proves the tangential traction and vanishing normal stress. The normal-velocity calculation supplies the second diagonal entry. The map converts a viscous exterior into a nonlocal boundary resistance proportional to $|k|$.

The hydraulic radius is cross-sectional flow area divided by wetted perimeter. A perimeter-averaged boundary shear stress $\tau$ exerts drag per unit fluid volume $\tau \ell /A=\tau /R$. It differs from both actual depth and the radius of a noncircular channel.

For circular Couette flow with $v=Ar+B/r$, the rotating-frame reduction of circular viscoelastic shear gives shear rate $g=-2B/r^2$. The total shear stress is $\mu g$, and the azimuthal minus radial normal-stress difference is $2(\mu -{\mu }_r)\tau g^2$ in the stress convention where ${\sigma }_{rr}=-p$. Radial force balance without inertia is $p^{\prime }=-2(\mu -{\mu }_r)\tau g^2/r$, which integrates to the displayed pressure. The pressure constant is undetermined by these equations.

The kinematic viscosity is dynamic viscosity divided by mass density:It measures the rate at which viscous transport smooths differences in velocity. For constant $\mu$ and $\rho$ in incompressible flow, dividing the Navier-Stokes equation by mass density makes the viscous acceleration $\nu {\nabla }^2\mathbf{u}$, where ${\nabla }^2$ is the Laplacian. ^[ref]

For example, stipulate two Newtonian fluids with the same dynamic viscosity $0.10\mathrm{Pa}\mathrm{s}$ but different mass densities. In the same steady Couette flow as above, both have shear stress $2.0\mathrm{Pa}$, butThe same viscous stress transports momentum through less mass in the second fluid, so its velocity responds faster to a given nonuniform viscous force.

To see the diffusion equation directly, consider an unsteady parallel velocity field $\mathbf{u}=(u(y,t),0,0)$ with constant material coefficients and no streamwise pressure gradient or body force. Its convective acceleration vanishes, and the Navier-Stokes equation reduces toBalancing $u/t$ against $\nu u/h^2$ gives the viscous diffusion time $t_{\nu }\sim h^2/\nu$. For the $1.0\mathrm{mm}$ gap, the two fluids have scales $0.010\mathrm{s}$ and $0.0050\mathrm{s}$. This is a characteristic scale, not an exact settling time. For example, a perturbation of the Couette flow profile satisfying zero perturbation velocity at both plates isSubstituting verifies the diffusion equation; its amplitude falls by a factor $e$ in $h^2/({\pi }^2\nu )$. Kinematic viscosity is therefore useful for predicting how quickly a moving wall influences the interior, while dynamic viscosity gives the shear stress at a prescribed shear rate.

Lift is the component of the fluid force perpendicular to the far-field flow relative to the body. It need not point vertically: a tilted flow changes the lift direction, and a banked wing can produce a sideways component. In a planar problem choose a unit vector ${\mathbf{e}}_D$ along the incoming relative velocity ${\mathbf{U}}_{\infty }$ and a perpendicular unit vector ${\mathbf{e}}_L$ pointing towards the chosen positive-lift side. ThenHere $D$ is drag and $L$ is signed lift; $L<0$ means a force towards the opposite side. In three dimensions the plane perpendicular to ${\mathbf{U}}_{\infty }$ also admits a separate side-force direction, so a lift axis must be specified. Lift is also distinct from buoyancy, which can exist without a relative flow. ^[ref]

The force comes from adding local surface tractions, rather than assigning a single pressure to each side. Let $\Sigma$ be the wetted body surface and $\mathbf{n}$ its unit normal vector pointing out of the body into the fluid. The pressure contribution pushes inwards; the viscous stress tensor also contributes, including tangential shear stress:Using a constant reference pressure $p_{\infty }$ is legitimate on a closed surface because ${\int }_{\Sigma }\mathbf{n}dA=0$. This surface integral makes the direction explicit: a surface-normal force is generally neither pure lift nor pure drag. Pressure and skin shear stress both enter the resultant, although the pressure term usually dominates lift on an airfoil. ^[ref]

The lift coefficient isolates a useful scale. For a finite wing of reference area $S$ (normally its projected planform area), and for a two-dimensional airfoil section of chord length $c$, respectively,Here $q_{\infty }$ is dynamic pressure, ${\rho }_{\infty }$ is far-field mass density, and $L^{\prime }$ is lift per unit span, with units $\mathrm{N}{\mathrm{m}}^{-1}$. The signs of $C_L$ and $c_l$ follow the chosen lift axis. For stipulated values ${\rho }_{\infty }=1.20\mathrm{kg}{\mathrm{m}}^{-3}$, $U_{\infty }=20\mathrm{m}{\mathrm{s}}^{-1}$, $S=0.50{\mathrm{m}}^2$ and $C_L=0.60$, $q_{\infty }=240\mathrm{Pa}$ and $L=72\mathrm{N}$. Doubling $U_{\infty }$ gives four times the lift only if $C_L$ stays fixed: changes in Reynolds number, Mach number, surface condition or angle of attack can change it. ^[ref]

A motivating experiment is to compare two independently observed quantities: the lift registered by a force balance, and the lift inferred from surface pressures. A low-speed wind tunnel holds the model fixed while air passes it at adjustable speed; a calibrated balance measures forces, while small flush holes on both surfaces connect by tubing to differential-pressure sensors referenced to $p_{\infty }$. Iowa State's GA(W)-1 teaching model has $43$ taps around a $101\mathrm{mm}$ chord; its instructions give their coordinates and close the contour by repeating the first tap. A classroom comparison can combine this pressure-panel method with a balance on the same model; this is an experiment design, not a claim that those instructions report a combined measurement. ^{[ref] [ref]}

Before turning on the wind, apply known positive and negative loads to the balance through its resolving centre; use $mg$ for the known load from a mass. Record all balance channels during loading and unloading, including response to drag and torque loads, to detect cross coupling, hysteresis and zero drift. Record a wind-off tare at each model orientation so its weight does not become an aerodynamic lift reading. Map tunnel flow direction and uniformity, rather than using the fan setting as a speed measurement. Covert's MIT guide describes these calibration, tare and flow-survey requirements. ^[ref]

For each pressure channel, apply zero and several known pressure differences spanning the expected range; determine the voltage-to-pressure relation and recheck it after a run. Inspect tubing for leaks and blocked taps, and wait for readings to settle before averaging: tubing can delay a pressure signal. This calibration principle is used in NASA's electronically scanned pressure systems, where several applied calibration pressures determine each sensor's response and repeated calibration tracks drift. A simple pressure standard can instead use a liquid-column height difference and the hydrostatic pressure law. ^[ref]

Measure far-field static and total pressure outside the model's fluid wake; in low-speed incompressible flow their difference gives $q_{\infty }$ by the Bernoulli equation, so $U_{\infty }=\sqrt{2q_{\infty }/{\rho }_{\infty }}$. Record temperature and ambient pressure to determine mass density using the ideal gas law. At fixed $q_{\infty }$, step the angle of attack, repeat selected settings, and sweep back down to check reproducibility. Record balance lift, each tap's mean pressure, and fluctuation sizes. NASA describes pressure probes connected to differential transducers and emphasizes recording their reference pressure. ^[ref]

Convert each tap to a pressure coefficient $C_{p,i}=(p_i-p_{\infty })/q_{\infty }$. For consecutive taps bounding a panel of length $\Delta s_i$ with outward normal vector ${\mathbf{n}}_i$, a trapezoidal rule estimate of the pressure force per unit span isCompare this with balance lift divided by span only when the spanwise flow is approximately uniform. Agreement tests the pressure-resultant picture; disagreement calls for checking unresolved leading-edge pressure variations, skin shear stress, support loads and three-dimensional end effects before rejecting it. A force balance senses the total load, while taps alone reconstruct the pressure part. Tunnel walls and model blockage alter the effective flow and require corrections; end plates or a model extending between walls approximate two-dimensional flow but do not guarantee it. Record the Reynolds number and Mach number with the result, and do not interpret tunnel lift coefficients as universal wing data. ^{[ref] [ref]}