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The principle of transmissibility states that moving a force along its line of action preserves its external mechanical effect on an ideal rigid body. The force vector is unchanged, and its torque about every reference point is unchanged. The statement concerns external rigid-body motion or equilibrium; it does not assert unchanged internal deformation.

For an original proof, a nonzero force $\mathbf{F}$ is applied at $\mathbf{r}$. Move it to $\mathbf{r}+s\hat{\mathbf{F}}$, where $\hat{\mathbf{F}}=\mathbf{F}/\Vert \mathbf{F}\Vert$ and $s$ is a signed distance. The new moment isThe last cross product vanishes because its vectors are parallel. This is the same moment geometry used in the primary torque definition.^[ref]

For a numerical example, let $\mathbf{F}=(0,10,0)\mathrm{N}$ act at $(0.30,0.20,0)\mathrm{m}$. Its moment is $(0,0,3)\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80,0)$ preserves that moment because the displacement follows the force direction. Moving it instead to $(0.50,0.20,0)$ gives $(0,0,5)\mathrm{N}\mathrm{m}$. The second move needs an additional compensating couple to preserve the original wrench.

A one-dimensional elastic counterexample shows the rigid-body limitation. Consider a uniform bar along $0\le x\le L$, fixed at the left end, with a rightward axial force $F$ applied to a cross-section at $x=a$. In a small-strain axial model, internal force is $F$ on $0<x<a$ and zero on $a<x<L$. Its right-end displacement iswhere $E$ is Young's modulus and $A$ the section area. Applying the same force at $a=L$ instead gives $FL/(EA)$, although both applied forces share one line of action and give the same external resultant and moment.

With stipulated $F=100\mathrm{N}$, $L=1\mathrm{m}$, $E=200\mathrm{GPa}$ and $A=1\mathrm{m}{\mathrm{m}}^2$, the end displacement is $0.25\mathrm{mm}$ for $a=L/2$ and $0.50\mathrm{mm}$ for $a=L$. The linear stress–strain relation underlying that calculation is described in the primary elasticity section.^[ref]

A proposed comparison therefore measures support reaction and body deformation separately when changing load position along a common line. Equal reactions and moments can coexist with different local stress and displacement. The external free-body diagram may be equivalent while the internal loading model differs.

For a spherical nonrotating tracer with luminosity density $\lambda$, radial velocity dispersion ${\sigma }_r$, enclosed luminosity $L(r)$, and constant mass-to-light ratio ${\rm Y}$, let $q=\lambda {\sigma }_r^2$ and $\gamma =G{\rm Y}$. DefineThe Spherical Jeans equation and the projected second stellar velocity moment give $p=2A+R{A}^{\prime }$. To differentiate without a singular lower-end contribution, write $A(R)={\int }_0^{\infty }q(\sqrt{R^2+z^2})dz$. If $R^{2}A$ vanishes at both endpoints, ${\int }_0^{\infty }Rp(R)dR=0$. A finite outer aperture instead retains its boundary stress term.

The proportional limit bounds the loading interval over which stress is proportional to strain. It does not by itself establish the last load that permits complete recovery of elastic deformation.

A two-dimensional incompressible line source with area flux $Q$ has radial velocity $Q/(2\pi r)$, obtained from flux conservation around a circle. It is irrotational away from the source, with potential $(Q/(2\pi ))\ln r$. For a circular expanding interface, $Q=2\pi a{a}_t$ and $u_r=a{a}_t/r$, whose radial strain is $-a{a}_t/r^2$. That strain supplies a normal viscous stress even though the velocity is a potential flow.

On positive-determinant deformation gradients, the first Piola-Kirchhoff stress is $\mu (F-J^{-1}{F}^{-T})$. Its zero set is $SO(3)$: multiplication by $F^T$ gives $F{F}^T=J^{-1}I$, and determinants give $J^5=1$. These are also the global minimizers, since $|F{|}^2\ge 3J^{2/3}$ and $3J^{2/3}/2+J^{-1}$ has its unique minimum $5/2$ at $J=1$. The minimum energy is $5\mu /2$ rather than zero; an additive constant changes no stress.

Apply constant stress ${\sigma }_0$ at $t=0$ and remove it at $t=t_h$. For series spring and Voigt-branch creep, the series spring recovers its strain ${\sigma }_0/E_0$ immediately on unloading. The Kelvin-Voigt model branch has retardation time $\tau =\eta /E_1$ and must keep continuous strain because an instantaneous dashpot displacement would require an impulse. Its post-unloading equation is $\eta \dot{\epsilon }+E_1\epsilon =0$. Solving with the branch's strain at $t_h$ gives the title formula. Thus all remaining strain eventually recovers: this standard linear solid model has no permanent set.

The reduced elastic modulus combines the elastic compliance of two contacting isotropic bodies:Here $E_i$ is each body's Young's modulus and ${\nu }_i$ its Poisson's ratio. Both bodies deform under the same contact load, so their contributions to the indentation add. The factors involving Poisson's ratio account for the lateral constraint inherent in the local contact geometry.^[ref]

For identical materials, $E^{\ast }=E/[2(1-{\nu }^2)]$; against an effectively rigid probe, $E^{\ast }=E/(1-{\nu }^2)$ for the specimen. Thus even with a rigid probe, an indentation fit does not directly return Young's modulus. This matters when comparing Hertzian contact measurements with tensile testing, which imposes a different pattern of stress.

The reference configuration is a chosen arrangement used to label a body's material particles and measure their subsequent deformation. A particle at $\mathbf{X}$ in the chosen reference has position $\mathbf{x}=\mathbf{\chi }(\mathbf{X},t)$ in the current configuration. The reference coordinates are Lagrangian coordinates and remain attached to the particles.

The initial shape of a specimen is a convenient reference, but a reference configuration need not be stress-free. A preloaded or residually stressed specimen can also be chosen as the reference. For example, engineering strain compares length with the chosen original gauge length, while the deformation gradient compares infinitesimal material line elements. Current and reference mass densities satisfy $\rho J={\rho }_0$ for conserved mass, with $J$ the local volume ratio. Keeping this choice explicit prevents mixing stress or strain measures based on different areas and lengths.

The relaxation modulus is the shear-stress response at lag s to a unit small step of shear strain. By linear superposition it is the causal memory kernel multiplying shear rate. Its decaying part represents stored stress that relaxes after deformation ceases.

Residual stress is stress present after the external loading that produced it has been removed. Different regions can acquire incompatible preferred shapes through nonuniform plastic deformation or thermal expansion. Because the regions must still fit together, elastic deformation balances those mismatches and leaves an internal stress field.

A free specimen can therefore have zero resultant external force while containing regions in tension and compression. For example, cooling different parts at different rates can leave different permanent strains, even after their temperatures become equal. Cutting the specimen releases some constraints, and the resulting shape change can reveal the stored stress. In a simple traction-free static body with no body force, the residual stress must satisfy internal force balance and zero surface traction. It affects later loading because the applied stress adds to a pre-existing field rather than starting from zero everywhere.

For uniaxial stress $\sigma$, projection onto a slip-plane normal and a slip direction gives the displayed shear stress. The angles $\varphi$ and $\lambda$ are measured from the tensile axis to that normal and direction respectively.

The characteristic time for a delayed strain response to approach equilibrium under a step in stress. For a Kelvin-Voigt model with elastic modulus $E$ and viscosity $\eta$, the equation $\eta \dot{\epsilon }+E\epsilon ={\sigma }_0$ gives $\epsilon =({\sigma }_0/E)(1-e^{-t/{\tau }_r})$, where ${\tau }_r=\eta /E$. A relaxation time instead describes decay of stress after an imposed strain; the two times need not coincide in a standard linear solid model.

The Reuss elastic approximation assumes uniform stress. The local strain is the derivative of the convex conjugate $W^{\ast }(S,x)$; its mean is $D⟨W^{\ast }⟩(S)$. Under invertible differentiable duality, this mean determines $S=D{W}_R(E)$. For positive linear elasticity, $C_R=⟨C^{-1}{⟩}^{-1}$. Uniform stress is generally not accompanied by a compatible local displacement field, so this approximation is distinguished from an exact heterogeneous solution.

Renewed plastic deformation during unloading or reversed loading when the stress reaches the opposite part of the yield surface. For the spherical elastoplastic cavity with maximum pressure $P_m>4{\tau }_Y/3$, elastic unloading first reaches the opposite Tresca yield criterion at the cavity wall after a pressure decrease $8{\tau }_Y/3$. Purely elastic return to zero pressure is possible when $P_m\le 8{\tau }_Y/3$, with equality reaching the yield surface exactly at the endpoint.

Rheology studies how matter deforms and flows under applied stress, including the dependence on loading history and observation time. Its central task is to connect a measured stress to strain, strain rate and relevant internal state through a constitutive equation. A Newtonian fluid has an instantaneous proportional response to strain rate; an elastic solid responds to strain; a viscoelastic material retains information about previous deformation.

These differences can be exposed by changing the experiment rather than only changing the load. Holding stress fixed measures material creep; holding strain fixed measures stress relaxation; small-amplitude oscillatory shear separates storage modulus from loss modulus. A rheometer supplies controlled motion and measures the resulting force or torque. Such measurements help predict whether a paste will remain in place, how much pressure is needed to drive a fluid flow, and whether a material will recover after loading.

A rheometer imposes a controlled stress or deformation and measures the complementary response. In a rotational instrument, a sample fills a gap between surfaces such as concentric cylinders, parallel disks or a cone and plate. The directly controlled or measured quantities include torque, angular displacement and angular velocity; the geometry converts these into shear stress, shear strain and shear rate.^[ref]

For example, a small-angle cone-and-plate rheometer produces approximately uniform shear rate $\dot{\gamma }=\Omega /\alpha$ and shear stress $\tau =3\mathcal{T}/(2\pi R^3)$, with cone angle $\alpha$, radius $R$, rotation rate $\Omega$ and measured torque $\mathcal{T}$. A parallel-plate rheometer instead has a radius-dependent shear rate. The conversion therefore requires the appropriate flow model. Wall slip, edge failure, temperature drift and the instrument's own inertia can mimic a material response, especially for very weak or rapidly varying signals.

Measurement of stress–deformation response using a calibrated rheometer. Geometry, temperature, sample history and possible wall slip must be controlled before interpreting a measured force or torque as a bulk constitutive response.

The ridge-push force in a plate model is the depth integral of the horizontal pressure difference between hot and cold compensated columns. To leading order, the old plate's linear excess density gives $F=g{\int }_0^{a}s\delta \rho (s)ds={\rho }_{m}g\alpha \Delta T{a}^2/6$. Here $s$ is depth below the old seafloor. Retaining the geometrical contribution from its lowered surface adds $g({\rho }_m-{\rho }_w)h^2/2$ within the linear-density model. Force per ridge length has units $\mathrm{N}{\mathrm{m}}^{-1}$, distinct from stress.

R. Marini is a coauthor of the 1999 report Functional arteries grown in vitro, a primary study of engineered vascular tissue. The paper combined cultured-vessel measurements with implantation in an animal model.^[ref]

This work connects two distinct questions in tissue engineering: whether a construct tolerates mechanical loading and whether it functions in a biological setting. Pressure resistance, attachment at a surgical connection and the maintenance of an open flow path are different observations. Success in one test does not logically establish the others.

An illustrative biomechanics calculation makes the distinction concrete. For a thin cylindrical vessel wall, a force balance gives circumferential stress ${\sigma }_{\theta }\approx pr/h$, with internal pressure $p$, radius $r$ and wall thickness $h$. Chosen values $p=100\mathrm{kPa}$, $r=1\mathrm{mm}$ and $h=0.1\mathrm{mm}$ give ${\sigma }_{\theta }\approx 1\mathrm{MPa}$. This is a model calculation, not a reported result or an operating recommendation. Geometry and pressure must be specified before comparing a vessel’s material response.

A depth-dependent estimate of the differential stress a rock column can sustain at a specified strain rate, composition, thermal structure and fluid state. In a simple competing-mechanism model, failure occurs by whichever of frictional sliding and ductile material creep requires the lower stress, giving the displayed lower envelope.

Effective confining pressure generally increases frictional strength with depth. In contrast, a power-law creep model $\dot{ϵ}=A{\sigma }^n\exp [-(Q+P{V}^{\ast })/(RT)]$ gives ${\sigma }_{\mathrm{creep}}=(\dot{ϵ}/A{)}^{1/n}\exp [(Q+P{V}^{\ast })/(nRT)]$. At fixed strain rate, the rise of temperature along a geothermal gradient generally lowers the required creep stress. Their crossing is a brittle-ductile transition, while composition changes at the Moho can produce additional peaks. A rock strength envelope is not uniquely determined by depth or by the labels “continental” and “oceanic”.