--- title: Stress and Elasticity module: Gravitation and Matter moduleNumber: 6 lessonNumber: 9 order: 609 summary: > Rigid bodies are a fiction; every real material stretches, shears, or squeezes under load, and the useful question is how much. We define stress as force per area and strain as fractional deformation, then find that for small deformations the two are simply proportional โ€” Hooke's law โ€” with Young's, shear, and bulk moduli as the constants for stretch, twist, and volume change. From these we compute extensions, torsional twist, and stored elastic energy, and read a tensile curve for the yield, ultimate, and fracture points where linear elasticity ends. We also mark the practical limits: stress concentrations, fatigue, and the multiaxial states a single uniaxial modulus cannot capture. topics: [Gravitation and Matter] draft: false sources: - book: Tipler & Mosca ref: "Ch. 12 โ€” Static Equilibrium and Elasticity; ยง12-7" --- ## Stress, strain, and tensile response An externally loaded solid transmits force through its interior. Cut an imagined surface through the body and isolate either side. The material removed by the cut exerts a distributed contact force on the retained side. At a small patch of area $\d A$ with unit normal $\hat n$, the contact force is written $$ \d\vec F=\vec t(\hat n)\,\d A, $$ where $\vec t(\hat n)$ is the traction vector. Its component normal to the cut produces extension or compression; its tangential component produces shear. The microscopic forces are carried by atomic bonds, grain contacts, polymer chains, or other material structure. Mechanics replaces that irregular structure with a smooth force-per-area description when the region of interest is much larger than the microscopic spacing and much smaller than the specimen dimensions. The scale of the chosen measurement sets what a reported stress or strain can mean. An extensometer averages deformation over its gauge length. A strain gauge averages over its bonded grid. Digital image correlation averages displacements over an optical subset. None of these readings directly reports the singularly high local strain at an ideal sharp crack, and none replaces a full displacement map when deformation localizes. Select a gauge length that samples the intended uniform region, then report it with the result. Changing gauge length can change measured fracture strain and apparent scatter even when specimens share the same material. Similarly, a nominal stress is often the right quantity for comparing standardized coupons or establishing a simple load path. Local stress is needed near contact edges, holes, changes of section, and cracks. The two quantities answer different questions. Treating a nominal value as a local peak can overstate the certainty of the calculation; treating a local finite-element peak as a uniform specimen stress can overstate the severity. The physical location, averaging area, and load history should accompany each reported stress. Linear elastic equations also support an inverse use of data. A measured extension under a known force can estimate modulus, a measured force under prescribed extension can estimate a spring-like axial compliance, and a measured lateral contraction can estimate Poisson ratio. Each inversion magnifies uncertainty in the denominator when the corresponding deformation is small. A long, slender, well-instrumented specimen provides a larger elastic extension and often gives a more reliable modulus estimate than a short, thick coupon, provided it remains uniformly loaded and does not buckle or exceed the elastic range. **Engineering stress and strain conventions.** The standard uniaxial tensile reduction uses the original gauge geometry: $$ \sigma_{\rm eng}=\frac{F}{A_0}, \qquad \varepsilon_{\rm eng}=\frac{L-L_0}{L_0} =\frac{\Delta L}{L_0}. $$ The numerator of engineering stress is the axial force recorded by the load cell. The denominator is the original gauge-section area, obtained from diameter for a circular coupon or width times thickness for a rectangular coupon. Engineering strain is a fractional change in gauge length and has no unit. A reported extension in millimetres becomes meaningful only after division by the stated initial gauge length. Tensile values are positive under the usual sign convention, while compression gives negative normal stress and negative normal strain. These conventions preserve one reference geometry throughout a test, permitting comparison among standardized coupons. Large plastic deformation changes the actual cross-section and makes the distinction between engineering and instantaneous measures important. The later true-stress treatment addresses that case. For the initial elastic range, engineering and local measures differ by a small amount, and the original geometry provides a clear basis for modulus fitting. **Tensile curves, yielding, strength, and fracture.** A tensile test produces a sequence of paired observations $(F,\Delta L)$. The converted $(\sigma_{\rm eng},\varepsilon_{\rm eng})$ curve contains several material regimes. The idealized curve below labels the proportional limit, elastic limit, yield region, ultimate tensile strength, and fracture. Actual curves vary with alloy, heat treatment, specimen geometry, surface condition, strain rate, and the test standard. The named points are measured features of a particular protocol, rather than universal constants attached to a chemical element. $$ % caption: Engineering stress--strain curve for a ductile tensile coupon. The straight % initial segment gives Young's modulus; the yield point begins permanent deformation, % the highest engineering stress is the ultimate strength, and the falling branch % reflects necking up to fracture. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(6.6,0) node[right] {strain}; \draw[->,black] (0,0)--(0,3.5) node[above] {stress}; \draw[acc,very thick] (0,0)--(1.7,2.0); \draw[very thick,smooth] plot coordinates {(1.7,2.0)(2.1,2.28)(2.9,2.66)(3.9,2.88)(4.7,2.74)(5.4,2.2)}; \draw[fill=white,draw=black] (1.7,2.0) circle (1.7pt); \draw[fill=white,draw=black] (3.9,2.88) circle (1.7pt); \draw[fill=white,draw=black] (5.4,2.2) circle (1.9pt); \draw[black,dashed] (1.7,0)--(1.7,2.0); \draw[black,dashed] (3.9,0)--(3.9,2.88); \node[black] at (0.9,0.4) {elastic}; \node[above left] at (1.7,2.0) {yield}; \node[above] at (3.9,3.0) {ultimate}; \node[right] at (5.4,2.2) {break}; \end{tikzpicture} $$ The proportional limit ends where stress is no longer proportional to strain. The elastic limit is the greatest stress from which unloading restores the original gauge length to the resolution of the experiment. Those points may be close but need not coincide. Engineering practice commonly defines a proof stress when a curve has no sharp yield point. A line parallel to the initial elastic slope is drawn at a prescribed offset strain, often $0.002$ for metals. Its intersection with the curve reports a repeatable offset yield strength. The value has meaning only with the offset convention and test method stated alongside it. The stress needed to yield is not Young's modulus. A material can have a high modulus and a low yield stress, or a low modulus and a high yield stress. Modulus sets elastic deformation under a given working stress. Yield strength sets one limit on the stress permitted before permanent shape change. A stiff material can therefore be unsuitable when a component must absorb deformation without cracking, and a strong material can be unsuitable when a component must remain dimensionally stable under a modest load. $$ % caption: Offset-yield construction on a smooth tensile curve. A line parallel to the % initial elastic slope starts at a fixed strain offset (commonly $0.2\%$ for metals); % its intersection with the measured curve fixes a repeatable proof stress when no sharp % yield point exists. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(6.4,0) node[right] {strain}; \draw[->,black] (0,0)--(0,3.3) node[above] {stress}; \draw[very thick,smooth] plot coordinates {(0,0)(1.1,1.5)(1.9,2.12)(2.9,2.55)(4.1,2.85)(5.2,2.7)}; \draw[black,thick,dashed] (0,0)--(1.55,2.52); \draw[acc,thick,dashed] (0.55,0)--(2.05,2.44); \draw[fill=acc!12,draw=acc] (1.78,2.0) circle (1.9pt); \draw[black,dashed] (1.78,0)--(1.78,2.0); \node[below] at (0.55,0) {of{}fset}; \node[below right] at (1.85,1.98) {proof stress}; \node[black,above left] at (1.55,2.52) {elastic slope}; \node[above] at (4.2,2.85) {measured}; \end{tikzpicture} $$ Permanent deformation in a metal generally involves irreversible motion of defects within its crystal structure. An introductory stress--strain calculation does not need a microscopic model to use the measured yield point, but it must distinguish elastic strain from plastic strain. After loading past yield and returning to zero force, the total strain decomposes as $$ \varepsilon_{\rm total} =\varepsilon_{\rm elastic}+\varepsilon_{\rm plastic}. $$ The elastic part recovers on unloading. The residual plastic part remains as a permanent extension. A load--unload loop therefore separates recoverable strain energy from permanent set. Its exact shape can include hysteresis and hardening, especially in polymers, soils, biological materials, and metals that have already undergone plastic strain. The maximum engineering stress on a ductile curve occurs before fracture because the force falls after a neck develops. The local cross-section inside that neck shrinks faster than the force, so an instantaneous stress based on local area can continue to rise while engineering stress falls. The distinction motivates true stress and true strain for large deformation: $$ \sigma_{\rm true}=\frac{F}{A}, \qquad \varepsilon_{\rm true}=\ln\!\left(\frac{L}{L_0}\right). $$ The logarithmic strain adds across successive small extensions. Before necking, uniform deformation and approximate volume conservation relate engineering and true measures. After necking, a single gauge-average strain does not characterize the neck; local optical or diameter measurement is required. $$ % caption: A ductile bar passes from nearly uniform extension to a localized neck. % Engineering stress keeps the original area $A_0$ in its denominator, while true stress % uses the shrinking neck area $A$; the two measures separate once the neck dominates the % deformation. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (0.4,0.7) rectangle (2.7,1.5); \draw[fill=acc!10,draw=acc,thick] (3.7,0.7) .. controls (4.3,0.7) and (4.5,0.95) .. (4.85,0.95) .. controls (5.2,0.95) and (5.4,0.7) .. (6.0,0.7) -- (6.0,1.5) .. controls (5.4,1.5) and (5.2,1.25) .. (4.85,1.25) .. controls (4.5,1.25) and (4.3,1.5) .. (3.7,1.5) -- cycle; \node[above] at (1.55,1.5) {uniform}; \node[above] at (4.85,1.5) {neck}; \node[below] at (1.55,0.7) {$A_0$}; \node[below] at (4.85,0.7) {$A$}; \end{tikzpicture} $$ Ductility and toughness are separate measurements. Ductility describes the amount of plastic strain before fracture, commonly reported as percent elongation or reduction in area. Toughness is energy absorbed per volume up to fracture, the area under the full stress--strain curve under stated conditions. A brittle glass-like response can have a steep elastic slope but a small fracture strain and small area under the curve. A ductile metal can reach a comparable peak stress while absorbing much more energy through plastic deformation. $$ % caption: Idealized brittle and ductile tensile curves on one axis scale. The ductile % curve reaches much larger strain and encloses far more area, absorbing more energy per % unit volume before fracture; the near-origin slopes set the modulus comparison. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(6.4,0) node[right] {strain}; \draw[->,black] (0,0)--(0,3.5) node[above] {stress}; \fill[acc!10] (0,0)--(1.2,1.7)--(2.1,2.15)--(3.3,2.5)--(4.4,2.6)--(5.5,1.6)--(5.5,0)--cycle; \draw[very thick] (0,0)--(1.55,2.95); \draw[fill=white,draw=black] (1.55,2.95) circle (1.9pt); \draw[very thick,smooth] plot coordinates {(0,0)(1.2,1.7)(2.1,2.15)(3.3,2.5)(4.4,2.6)(5.5,1.6)}; \draw[fill=white,draw=black] (5.5,1.6) circle (1.9pt); \node[above] at (1.55,3.0) {brittle}; \node[right] at (5.55,1.6) {ductile}; \node[acc] at (3.0,1.0) {absorbed energy}; \end{tikzpicture} $$ Nominal tensile strength, yield strength, fracture strain, elastic modulus, and toughness should appear as distinct columns in a material comparison. A table that lists only "strength" suppresses the failure mode and test condition. Compressive strength can greatly exceed tensile strength for concrete, rock, bone, and many ceramics. Fibre-reinforced composites often have high strength along fibres but markedly lower transverse strength. Design data must match the load direction, environment, and manufacturing state of the intended component. ## Compression and normal deformation Compression uses the same signed normal-strain definition as tension. A prismatic specimen under axial compression has $\sigma<0$ and $\varepsilon<0$ under the usual convention. The initial magnitude of the axial slope still gives Young's modulus for a linear isotropic solid. The experiment becomes less simple at larger strain: friction between specimen ends and loading platens restrains lateral expansion, creating nonuniform stress and a barrel-shaped profile. Misalignment introduces bending, so one side reaches a higher compressive stress than the average $F/A$. Some materials fracture under compressive loading; others yield and spread laterally. Slender compression members may become unstable by lateral buckling before the material reaches its compressive yield stress. Buckling is a structural-stability problem and requires geometry, end restraint, and bending analysis beyond this lesson. A material test reports constitutive response of a specified coupon; it does not, by itself, establish the load capacity of an arbitrary column. Bearing stress is an average contact pressure used in elementary joint calculations: $$ \sigma_{\rm bearing}=\frac{F}{A_{\rm projected}}. $$ In a pin-through-plate joint, a simple projected area is pin diameter times plate thickness. The result is an average guide to local crushing. Real contact pressure is concentrated around the pin and depends on clearance, surface hardness, and deformation. A joint can also fail by net-section tension, shear-out, bending, or fatigue, each with a different critical section. Stress is a local quantity. A tensile machine reports an applied force, while a material law relates stress at a material point to deformation near that point. Uniform stress is an appropriate approximation in a long, straight prismatic specimen loaded through well-aligned grips. It deteriorates near a grip, a hole, a sharp corner, a bonded interface, or a rapidly changing cross-section. A calculation using $F/A$ therefore carries an implicit statement about where the section is taken and how evenly the load is distributed over it. On a plane normal to the $x$ direction, compact introductory notation separates normal and shear components: $$ \sigma_{xx}=\frac{F_x}{A},\qquad \tau_{xy}=\frac{F_y}{A},\qquad \tau_{xz}=\frac{F_z}{A}. $$ The first subscript identifies the cut normal and the second identifies the force direction. Full continuum mechanics collects these components into a stress tensor. At the present level, the tensor functions as a bookkeeping device: a plane can carry one normal stress and two independent shear stresses, and changing the plane changes the components. A single scalar $F/A$ describes only a specially chosen loading direction. $$ % caption: Traction on a small element cut normal to the $x$ axis: one normal component % along $x$ and a tangential (shear) component in the plane of the face. A different cut % orientation gives different normal and shear components, so a single $F/A$ describes % only one chosen plane. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (2.0,0.6) rectangle (4.4,2.6); \draw[->,acc,very thick] (4.4,1.3)--(5.35,1.3); \draw[->,black,thick] (4.4,1.85)--(4.4,2.85); \draw[->,acc,very thick] (2.0,1.3)--(1.05,1.3); \draw[->,black,thick] (2.0,1.35)--(2.0,0.35); \draw[->,black] (0.7,0.2)--(1.7,0.2) node[right] {$x$}; \draw[->,black] (0.5,0.4)--(0.5,1.4) node[above] {$y$}; \node[acc,right] at (5.35,1.3) {normal}; \node[above] at (4.4,2.85) {shear}; \end{tikzpicture} $$ The SI unit of normal stress and shear stress is the pascal: $$ 1\ \mathrm{Pa}=1\ \mathrm{N\,m^{-2}}. $$ Engineering data often use $\mathrm{MPa}$ or $\mathrm{GPa}$ because ordinary structural stresses are far larger than $1\ \mathrm{Pa}$. Area must be converted before dividing. A diameter in millimetres gives an area in $\mathrm{mm^2}$; using newtons per square millimetre produces $\mathrm{N\,mm^{-2}}$, numerically equal to $\mathrm{MPa}$. That equivalence is convenient, provided the unit conversion is shown rather than assumed. **Normal strain and a uniaxial tensile specimen.** Strain measures deformation relative to the original local geometry. For a gauge segment of initial length $L_0$ and measured length $L$, the engineering normal strain is $$ \varepsilon=\frac{L-L_0}{L_0}=\frac{\Delta L}{L_0}. $$ Strain has no SI unit. Tensile strain is positive under the usual sign convention; compressive strain is negative when $L **Definition (Elastic modulus).** A modulus is the coefficient relating a stated > stress component to a stated strain component in a specified linear regime. Its > numerical value depends on temperature, material orientation, loading rate, and > the chosen constitutive model when those effects are appreciable. ## Lateral, shear, and bulk response Tension along one direction commonly narrows a solid across the loading direction. In a small uniaxial test of an isotropic material, Poisson ratio is defined by $$ \nu=-\frac{\varepsilon_{\rm transverse}}{\varepsilon_{\rm axial}}. $$ The negative sign makes $\nu$ positive for ordinary tensile behavior: axial strain is positive and transverse strain is negative. A specimen with $\nu=0.30$ and $\varepsilon_{\rm axial}=1.0\times10^{-3}$ has transverse engineering strain $-3.0\times10^{-4}$ in each perpendicular direction when the lateral surfaces are free of traction. $$ % caption: Axial tension lengthens a bar and narrows it laterally. The positive axial % strain and the negative transverse strain form the ratio that defines Poisson's ratio; % lateral change is read in the uniform gauge region, away from grips and necks. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,dashed] (1.4,0.9) rectangle (4.6,2.1); \draw[fill=acc!10,draw=acc,thick] (0.9,1.15) rectangle (5.1,1.85); \draw[->,thick] (0.9,1.5)--(0.15,1.5); \draw[->,thick] (5.1,1.5)--(5.85,1.5); \node[below] at (3.0,0.9) {unloaded}; \node[above] at (3.0,2.1) {loaded and narrower}; \end{tikzpicture} $$ In a small rectangular element with axial strain $\varepsilon_x$ and equal free lateral response in the $y$ and $z$ directions, $$ \varepsilon_y=\varepsilon_z=-\nu\varepsilon_x. $$ The small fractional volume change is the sum of the three normal strains: $$ \frac{\Delta V}{V}\approx \varepsilon_x+\varepsilon_y+\varepsilon_z =(1-2\nu)\varepsilon_x. $$ An incompressible small-strain idealization has $\Delta V/V=0$ and therefore $\nu=1/2$. Rubber-like solids can approach that value during rapid deformation. Metals often have smaller values, so a uniaxial tensile test changes volume slightly before plastic necking. Material symmetry matters. Wood, composites, and rolled sheet can have different lateral responses along different material axes; one scalar Poisson ratio then gives only a limited description. **Shear stress, shear strain, and torsion.** Shear loading changes shape by sliding neighboring material layers parallel to a surface. A rectangular block of height $L$ and loaded-face area $A$ has average shear stress $$ \tau=\frac{F_s}{A}. $$ If the upper face moves sideways by $\Delta x$ while the lower face remains at its original position, the small shear strain is $$ \gamma=\frac{\Delta x}{L}. $$ For small angles, $\gamma=\tan\phi\approx\phi$ when $\phi$ is expressed in radians. Shear strain is dimensionless. It is not an extension divided by the length of the sliding face; the denominator is the perpendicular spacing between the two faces whose relative displacement is being compared. $$ % caption: Simple shear of a block between parallel surfaces. A tangential load moves % the top face sideways by $\Delta x$ relative to the fixed base across separation $L$; % the average shear strain is $\gamma=\Delta x/L$ for the small angle shown. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[fill=black!12,draw=black,thick] (0.8,0.35) rectangle (5.3,0.6); \draw[fill=acc!10,draw=acc,thick] (1.2,0.6)--(4.6,0.6)--(5.2,2.5)--(1.8,2.5)--cycle; \draw[fill=black!12,draw=black,thick] (1.5,2.5) rectangle (6.0,2.75); \draw[->,very thick] (4.6,3.05)--(5.8,3.05); \draw[<->,black] (0.55,0.6)--(0.55,2.5); \draw[<->,black] (1.2,2.95)--(1.8,2.95); \node[left] at (0.55,1.55) {$L$}; \node[above] at (1.5,2.95) {slide}; \node[above] at (5.2,3.05) {shear load}; \node[below] at (3.0,0.35) {base held still}; \end{tikzpicture} $$ The linear elastic shear law is $$ \tau=G\gamma, \qquad G=\frac{\tau}{\gamma}, $$ where $G$ is the shear modulus. Tipler and Mosca use $M_s$ for the same modulus in their discussion of torsion. The SI unit is pascal. A large shear modulus means that a given shear stress produces little angular distortion. A static fluid has zero shear modulus because any sustained tangential stress causes continued flow; an ordinary solid maintains a static shear deformation after the load is applied. Shear stress is direction-sensitive. On a plane with normal in the $x$ direction, $\tau_{xy}$ describes traction along $y$. Local equilibrium requires companion shear components on perpendicular faces. Otherwise the tiny element would acquire an angular acceleration. The paired components have equal magnitude in the classical continuum description: $$ \tau_{xy}=\tau_{yx}. $$ That relation is a local moment balance, not an assertion that every physical face looks the same. A different plane cuts through a different set of material bonds and can have different normal and shear components. Stress transformation and Mohr's circle formalize that dependence in a later mechanics or materials course. Torsion is distributed shear in a member twisted about its longitudinal axis. A circular shaft with one end held and the other rotated through angle $\Theta$ has greater shear strain at larger radius. For a shaft of length $L$ and radius $r$, $$ \gamma(r)=\frac{r\Theta}{L}. $$ The centreline has zero torsional shear strain in this idealized model, and the outer surface has the largest strain. Linear elastic torsion therefore gives $$ \tau(r)=G\frac{r\Theta}{L}. $$ A circular cross-section remains nearly circular under this loading, which makes the radial distribution especially simple. Noncircular bars warp out of their original cross-sectional plane; their torsion analysis needs a more detailed shape solution. $$ % caption: A circular shaft twisted about its axis. A straight surface mark rotates % more at larger radius, so shear strain grows linearly from the centreline to the outer % surface; the twist angle is measured over the shaft length $L$. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (1.2,1.5) ellipse (0.45 and 0.85); \draw[thick] (5.4,1.5) ellipse (0.45 and 0.85); \draw[thick] (1.2,0.65)--(5.4,0.65); \draw[thick] (1.2,2.35)--(5.4,2.35); \draw[black,dashed] (1.2,1.5)--(5.4,1.5); \draw[black,thick] (1.2,1.5)--(1.2,2.35); \draw[black,thick] (5.4,1.5)--(5.75,2.22); \draw[->,acc,very thick] (6.05,2.15) arc (35:315:0.55); \draw[<->,black] (1.2,0.2)--(5.4,0.2); \node[below] at (3.3,0.2) {$L$}; \node[above] at (5.7,2.55) {$T$}; \node[above left] at (1.2,2.35) {surface mark}; \end{tikzpicture} $$ The torque required to produce a twist depends on the polar second moment of area $J$ of the cross-section. For a circular shaft, $$ T=\frac{GJ}{L}\Theta, \qquad J_{\rm solid}=\frac{\pi R^4}{2}, \qquad J_{\rm tube}=\frac{\pi(R_o^4-R_i^4)}{2}. $$ The fourth-power radius dependence is a geometric result. Material placed far from the axis is much more effective against twist than the same area near the centre. A hollow tube can therefore retain a large fraction of the torsional rigidity of a solid bar while using substantially less material. The formula assumes small twist, linear elasticity, a uniform circular shaft, and torque transmitted without slip. A torsion test can measure $G$ if torque, twist, length, and geometry are known. Angular displacement measured at the grips includes compliance of the fixtures, which biases the inferred modulus downward if treated as shaft twist. A calibration shaft of known dimensions and modulus can estimate that apparatus compliance. Surface scratches and keyways also change the local stress distribution; the simple $\tau_{\max}=TR/J$ result gives the nominal outer-surface stress away from those features. **Bulk response, compressibility, and relations among moduli.** Hydrostatic loading applies equal normal stress from every direction. For an initial volume $V_0$ changing by $\Delta V$, the bulk modulus is $$ K=-\frac{\Delta p}{\Delta V/V_0}. $$ Pressure increase $\Delta p>0$ produces volume decrease $\Delta V<0$, which explains the minus sign. Some texts use $B$ rather than $K$ for bulk modulus. A large bulk modulus means low compressibility. Its reciprocal, $$ \kappa=\frac{1}{K}, $$ is the isothermal or adiabatic compressibility only after the thermal condition has been specified. A compressed gas changes temperature unless heat transfer holds it near a chosen temperature, so its pressure--volume response cannot be treated as one universal material constant over all processes. For small isotropic elastic deformation, the three common elastic moduli and Poisson ratio are linked: $$ Y=2G(1+\nu), \qquad K=\frac{Y}{3(1-2\nu)}. $$ Only two of $Y$, $G$, $K$, and $\nu$ are independent under the assumptions behind these equations. The relations apply to a homogeneous, isotropic material with small reversible strain. They should not be used blindly for a fibre composite, layered biological tissue, wood, a porous foam, a granular packing, or a material near a phase transition. Such systems can possess direction-dependent moduli, nonlinear response, time dependence, or a separate pore-fluid contribution. $$ % caption: Three elementary loading modes isolate different deformations: uniaxial load % changes length, simple shear changes angle, and hydrostatic pressure changes volume. % An isotropic linear solid links their slopes through $Y$, $G$, $K$, and Poisson's ratio. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (0.5,1.0) rectangle (1.9,2.1); \draw[->,thick] (0.5,1.55)--(0.0,1.55); \draw[->,thick] (1.9,1.55)--(2.4,1.55); \node[below] at (1.2,1.0) {normal}; \node[above] at (1.2,2.1) {$Y$}; \draw[thick] (3.3,1.0)--(4.7,1.0)--(5.0,2.1)--(3.6,2.1)--cycle; \draw[->,thick] (3.85,2.45)--(5.05,2.45); \node[below] at (4.15,1.0) {shear}; \node[above] at (4.5,2.55) {$G$}; \draw[thick] (6.2,1.0) rectangle (7.6,2.1); \draw[->,thick] (5.7,1.55)--(6.18,1.55); \draw[->,thick] (8.1,1.55)--(7.62,1.55); \draw[->,thick] (6.9,2.6)--(6.9,2.12); \draw[->,thick] (6.9,0.5)--(6.9,0.98); \node[below] at (6.9,0.42) {volume}; \node[above right] at (7.6,2.1) {$K$}; \end{tikzpicture} $$ The pressure in a liquid at rest is nearly hydrostatic at a point, and a liquid's large bulk modulus explains why its density changes little under ordinary pressure differences. Its shear modulus at long time is effectively zero because a static tangential load produces flow. A solid rubber ball has both a measurable shear modulus and a large bulk modulus; squeezing its sides chiefly changes shape before it substantially changes volume. A foam can have a low apparent bulk modulus because gas-filled pores collapse, even when its solid skeleton has a much larger material modulus. The pressure--volume slope need not be constant across a wide compression range. A tangent bulk modulus uses the local derivative $$ K_{\rm tangent}=-V\frac{\d p}{\d V}. $$ An average secant modulus uses two finite states. The distinction resembles the initial-slope and secant treatment of a nonlinear tensile curve. Both descriptions can be valid when their strain or pressure ranges are stated. A single modulus without a range can conceal substantial nonlinearity. ## Elastic energy and material testing Loading a linear elastic specimen stores mechanical energy. For a uniform axial bar with force increasing from zero to $F$, the force--extension relation is $$ F=\frac{A_0Y}{L_0}\Delta L. $$ The work supplied quasistatically is the area below the force--extension line: $$ U=\int_0^{\Delta L}F\,\d(\Delta L') =\frac12F\Delta L =\frac{F^2L_0}{2A_0Y}. $$ The primed integration variable prevents the upper-limit extension from being mistaken for a constant during the integration. The result applies while the bar remains in its linear elastic regime and the load is introduced slowly enough that kinetic energy and wave propagation are negligible at the measurement scale. Dividing by initial volume $A_0L_0$ gives the uniaxial elastic energy density: $$ u=\frac{U}{A_0L_0} =\frac12\sigma\varepsilon =\frac{\sigma^2}{2Y} =\frac12Y\varepsilon^2. $$ The energy density is a local statement for uniform uniaxial stress. In a bar with changing area or changing axial force, integrate over volume: $$ U=\int_V\frac{\sigma^2}{2Y}\,\d V. $$ An axial member with position-dependent area $A(x)$ and constant tensile force $F$, this becomes $$ U=\int_0^{L_0}\frac{F^2}{2A(x)Y}\,\d x. $$ Thin regions store more energy per length because their stress is higher. That observation matters in a tapered spring or compliant mechanism, but it also identifies locations where yield and fatigue need attention. The linear elastic energy density under simple shear has the analogous form $$ u_{\rm shear}=\frac12\tau\gamma=\frac{\tau^2}{2G}. $$ For hydrostatic compression, the energy density associated with a small fractional volume change is $$ u_{\rm bulk}=\frac12K\left(\frac{\Delta V}{V_0}\right)^2. $$ These expressions describe recoverable energy. A plastic load cycle converts part of the input work into permanent microstructural change and heat, leaving a loop between loading and unloading curves. Material resilience is the maximum elastic energy density before yield in a specified loading mode. It differs from toughness, which includes energy absorbed through plastic deformation up to fracture. Energy methods require care near fracture. A material with high elastic energy density may release that energy suddenly when a crack grows. A long loaded steel wire, a pressurized vessel, and a stretched elastomer can store enough energy for rapid motion after failure. A quasistatic stress calculation identifies a nominal stress state; it does not predict the subsequent dynamics of a released load. **Material-testing apparatus, calibration, and uncertainty.** A tensile result begins with specimen identification and a traceable geometry measurement. Record material condition, heat treatment if known, specimen orientation, gauge length, diameter or width and thickness, surface finish, and test temperature. For a circular coupon, measure diameter at several angles and several positions in the gauge region. The area calculation then states whether it uses a mean diameter, a minimum diameter, or a directly measured area. A small diameter error has amplified effect because $A_0$ is proportional to $d_0^2$. The loading system applies a commanded crosshead displacement or load history. A load cell converts elastic deformation of a calibrated sensing element into an electrical signal. A displacement transducer, clip-on extensometer, video gauge, or bonded resistance strain gauge measures deformation. Sampling rate must resolve the imposed loading rate and any transient events. A slow monotonic tensile test does not require the sampling rate of an impact test, but too few data points near yield can obscure the fitted elastic slope and the offset-yield intersection. Load-cell calibration compares indicated force with certified reference loads across the range used in the test. A zero reading alone is insufficient. Calibration should report bias, repeatability, hysteresis between increasing and decreasing loads, and resolution. An apparatus can have a near-zero offset while retaining a scale-factor error. The same principle applies to displacement sensors: compare their indicated travel with known displacements, then determine whether the calibration is valid over the gauge range and at the loading rate used. Machine compliance is a frequent source of modulus error. Let the total measured crosshead extension be $\Delta L_{\rm meas}$. A simple series model gives $$ \Delta L_{\rm meas} =\Delta L_{\rm specimen}+\Delta L_{\rm machine}. $$ The machine term includes frame stretch, grip deformation, wedge seating, and load-train deflection. It may be nearly proportional to force over a limited range. An extensometer removes much of that term by spanning the specimen gauge marks. Alternatively, a compliance calibration using a short, high-modulus reference specimen can estimate the apparatus contribution. The correction should be applied only when the calibration geometry and force path are comparable to the test. Modulus determination needs a stated fitting interval. The first few data points can include slack removal and grip seating. At high stress, nonlinear response, temperature drift, or the onset of yielding can bend the curve. Plot the candidate fit range and report the fitted slope, its standard uncertainty, the strain range, and the method used to estimate area. A high coefficient of determination does not repair systematic error from crosshead compliance or misalignment; it measures agreement with a line, not the physical correctness of the line's axes. An uncertainty calculation begins with the measured quantities. For $$ Y=\frac{FL_0}{A_0\Delta L}, $$ independent small relative uncertainties combine approximately as $$ \left(\frac{u_Y}{Y}\right)^2 \approx \left(\frac{u_F}{F}\right)^2 +\left(\frac{u_{L_0}}{L_0}\right)^2 +\left(\frac{u_{A_0}}{A_0}\right)^2 +\left(\frac{u_{\Delta L}}{\Delta L}\right)^2. $$ If area comes from diameter, $u_{A_0}/A_0\approx2u_d/d_0$ for small diameter uncertainty. Correlated terms require a covariance treatment rather than simple quadrature. Repeating tests measures specimen-to-specimen variation as well as instrument scatter. A single test can establish a worked estimate; it cannot characterize a production material population. Misalignment adds bending stress to the intended axial stress. A small eccentricity $e$ between the load line and specimen centroid generates a moment $M=Fe$. One side of the gauge section then carries higher normal stress and can yield early. Opposed strain gauges or diameter measurements around the circumference can reveal that gradient. A clean tensile specimen shape alone does not prove axial alignment; alignment must be established through grips, fixtures, and measured response. ## Fatigue, concentrations, and constitutive limits A component can fracture after many load cycles at a nominal stress below the monotonic tensile strength. Fatigue testing applies a repeated stress history and counts cycles to a defined failure condition. For a cycle with maximum and minimum normal stresses, $$ \sigma_a=\frac{\sigma_{\max}-\sigma_{\min}}{2}, \qquad \sigma_m=\frac{\sigma_{\max}+\sigma_{\min}}{2}, \qquad R=\frac{\sigma_{\min}}{\sigma_{\max}}. $$ The stress amplitude $\sigma_a$ measures cyclic variation. The mean stress $\sigma_m$ distinguishes fully reversed loading from tensile cycling with a positive mean. The ratio $R$ is another compact description of the same history. A fatigue result must identify which convention was used because two tests with equal maximum stress can have very different amplitudes and mean stresses. An S--N curve plots stress amplitude against number of cycles to failure, commonly with a logarithmic cycle axis. It describes the tested specimen, surface finish, environment, size, frequency, stress ratio, and failure criterion. Some steels show a nearly horizontal long-life region under particular laboratory conditions. Many nonferrous alloys retain a declining curve across the tested range. A claimed \"infinite-life\" stress is therefore a material-and-protocol result, not a general property of every metal or component shape. $$ % caption: An S--N curve plots stress amplitude against cycles to failure on a % logarithmic cycle axis. Scatter between nominally identical coupons is common, so a % design curve sits below the mean trend and states stress ratio, surface, environment, % and target reliability. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(6.4,0) node[right] {cycles (log)}; \draw[->,black] (0,0)--(0,3.3) node[above] {stress amplitude}; \draw[very thick,smooth] plot coordinates {(0.3,2.7)(1.2,2.3)(2.1,1.95)(3.0,1.7)(4.0,1.55)(5.6,1.5)}; \draw[acc,thick,dashed] (0.3,2.35)--(5.6,1.2); \draw[fill=white,draw=black] (0.65,2.55) circle (1.7pt); \draw[fill=white,draw=black] (1.55,2.15) circle (1.7pt); \draw[fill=white,draw=black] (2.5,1.82) circle (1.7pt); \draw[fill=white,draw=black] (3.6,1.6) circle (1.7pt); \draw[fill=white,draw=black] (4.8,1.5) circle (1.7pt); \node[above] at (3.6,1.98) {mean trend}; \node[acc,below] at (4.7,1.28) {design curve}; \end{tikzpicture} $$ Stress concentration increases local stress near a geometric discontinuity. A hole, notch, thread root, sharp shoulder, corrosion pit, or surface scratch diverts load paths through a reduced ligament and changes the local stress field. A concentration factor is often written $$ K_t=\frac{\sigma_{\max,\rm local}}{\sigma_{\rm nominal}}. $$ The nominal stress might be $F/A_{\rm net}$ or $F/A_0$, depending on the stated reference area. The local peak depends on notch radius, width ratio, loading mode, and material response. A tabulated $K_t$ from linear elasticity is valuable for elastic components; yielding can redistribute the local stress, and fatigue response also depends on notch sensitivity, surface finish, and residual stress. $$ % caption: A plate with a circular hole under remote tension. Far from the hole the % nominal stress $F/A$ describes the load; near the curved edge the load paths crowd and % the local tensile stress rises to a peak of several times the nominal value. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (0.5,0.6) rectangle (6.3,2.6); \draw[fill=white,draw=black,thick] (3.4,1.6) circle (0.5); \draw[->,very thick] (0.5,1.6)--(-0.2,1.6); \draw[->,very thick] (6.3,1.6)--(7.0,1.6); \draw[acc,thick] (0.8,1.95) .. controls (2.6,1.95) and (2.9,2.28) .. (3.4,2.28) .. controls (3.9,2.28) and (4.2,1.95) .. (6.0,1.95); \draw[acc,thick] (0.8,1.25) .. controls (2.6,1.25) and (2.9,0.92) .. (3.4,0.92) .. controls (3.9,0.92) and (4.2,1.25) .. (6.0,1.25); \draw[->,black,thick] (4.7,3.0)--(3.78,2.15); \node[above] at (4.7,3.0) {stress peak}; \node[below] at (3.4,0.35) {load paths crowd here}; \end{tikzpicture} $$ Crack growth changes the relevant geometry as loading proceeds. A short crack can initiate at a notch root or surface inclusion, then advance a small amount per cycle. The remaining ligament shrinks and the crack-tip stress field becomes very large. Fracture mechanics uses stress intensity, crack length, and fracture toughness to quantify that stage. The present stress--strain methods supply material modulus and nominal loads, but they do not calculate crack-tip fields or reliable remaining life after a crack is detected. Loading rate and temperature can alter the measured curve. Metals can show rate sensitivity; polymers can shift from compliant to glassy response across a modest temperature range; wood and biological tissue vary with moisture and time. Under constant stress, a time-dependent material can continue to strain through creep. Under fixed strain, its stress can fall through relaxation. A modulus quoted without temperature, rate, and duration may be adequate for a narrow laboratory comparison and inadequate for a long-lived component. Thermal expansion adds strain even before mechanical loading. A uniform temperature change gives free axial strain $$ \varepsilon_{\rm th}=\alpha\Delta T, $$ where $\alpha$ is the coefficient of linear expansion. A fully restrained isotropic bar develops an approximate thermal stress magnitude $$ |\sigma_{\rm th}|=Y\alpha|\Delta T| $$ within the linear elastic regime. Partial restraint, temperature gradients, creep, and an assembled structure require compatibility and heat-transfer analysis. A material test at room temperature cannot establish those service stresses by itself. An isotropic linear-elastic law is an approximation with a clear domain. It treats the specimen as homogeneous, regards stress and strain as small, ignores permanent microstructural change, and uses material constants independent of direction. Composite laminates, welds, additively manufactured parts, textured metals, and natural materials can violate several of those assumptions. A material direction or processing route belongs in the test record. Replacing a missing constitutive model with a single tabulated modulus produces numerical output without resolving the physical uncertainty. ## Worked reductions and multiaxial models > **Worked example.** A circular steel wire has initial length $L_0=1.80\ \mathrm m$, > diameter $d_0=6.00\ \mathrm{mm}$, and carries a tensile load $F=8.00\ \mathrm{kN}$. > With $Y=200\ \mathrm{GPa}$, find the extension and the stored elastic energy. > > Convert the diameter before computing area: > > $$ > A_0=\frac{\pi d_0^2}{4}=\frac{\pi(6.00\times10^{-3}\ \mathrm m)^2}{4} > =2.83\times10^{-5}\ \mathrm{m^2}. > $$ > > The engineering stress and strain follow, then the extension: > > $$ > \sigma_{\rm eng}=\frac{F}{A_0}=\frac{8.00\times10^3}{2.83\times10^{-5}} > =2.83\times10^8\ \mathrm{Pa}, > \qquad > \varepsilon_{\rm eng}=\frac{\sigma_{\rm eng}}{Y}=1.41\times10^{-3}, > $$ > > $$ > \Delta L=\varepsilon_{\rm eng}L_0=2.55\times10^{-3}\ \mathrm m=2.55\ \mathrm{mm}. > $$ > > The stored energy is the triangular area under the linear force--extension line, > > $$ > U=\tfrac12 F\,\Delta L=\tfrac12(8.00\times10^3)(2.55\times10^{-3})=10.2\ \mathrm J. > $$ > > The estimate is elastic only if the wire's yield stress exceeds the working stress > of $283\ \mathrm{MPa}$; a generic $Y$ does not settle that. Note the diameter enters > the area squared, so a diameter error weighs twice as heavily as a length error of > the same relative size. > **Worked example.** A solid circular shaft has radius $R=12.0\ \mathrm{mm}$, length > $L=0.600\ \mathrm m$, and shear modulus $G=26.0\ \mathrm{GPa}$, and carries a torque > $T=120\ \mathrm{N\,m}$. Find the peak surface shear stress and the angle of twist. > > The polar second moment of the circular section is > > $$ > J=\frac{\pi R^4}{2}=\frac{\pi(12.0\times10^{-3})^4}{2}=3.26\times10^{-8}\ \mathrm{m^4}. > $$ > > The shear stress is largest at the outer surface, and the twist follows from > $T=GJ\Theta/L$: > > $$ > \tau_{\max}=\frac{TR}{J}=44.2\ \mathrm{MPa}, > \qquad > \Theta=\frac{TL}{GJ}=8.50\times10^{-2}\ \mathrm{rad}\approx4.87^\circ. > $$ > > The units check: $TR/J$ carries $\mathrm{N\,m}\cdot\mathrm m/\mathrm{m^4}=\mathrm{N\,m^{-2}}$, > a stress, and $TL/(GJ)$ is dimensionless. The result holds for the uniform circular > portion; a keyway or shoulder can lower the allowable torque through stress > concentration. > **Worked example.** A modulus fit uses force with $0.50\%$ relative uncertainty, > gauge length with $0.02\%$, diameter with $0.25\%$, and extension with $1.00\%$. > Estimate the relative uncertainty in $Y$ and identify the dominant term. > > The diameter enters the area squared, so the area carries > $u_{A_0}/A_0\approx2(0.25\%)=0.50\%$. Adding the independent contributions in > quadrature, > > $$ > \frac{u_Y}{Y}\approx > \sqrt{(0.0050)^2+(0.0002)^2+(0.0050)^2+(0.0100)^2}=0.0122, > $$ > > a relative standard uncertainty of $1.22\%$. The extension term, at $1.00\%$, > dominates: sharpening the load cell from $0.50\%$ to $0.25\%$ would help far less > than improving the extension measurement or removing machine compliance. The > ranking comes from the budget, not from the size of the force reading. Report a material-test result with the specimen geometry, loading mode, temperature, strain rate, measurement method, fitted range, stress and strain conventions, number of repeats, and uncertainty basis. State whether values are engineering or true measures, and identify any excluded data with a physical reason such as grip slip or sensor saturation. Those details define the result more completely than a bare modulus or strength value. **Multiaxial elastic response and selection of a model.** Many real parts carry more than one stress component. A pressurized wall has circumferential and axial normal stresses. A rotating shaft carries torsional shear and may also carry bending stress. A bonded joint can combine normal opening and shear. The uniaxial equation $\sigma=Y\varepsilon$ applies to a coupon whose lateral surfaces are free, but it cannot represent those multiaxial states by itself. In a small-strain isotropic elastic solid, the normal components obey $$ \varepsilon_x= \frac{1}{Y}\left[\sigma_x-\nu(\sigma_y+\sigma_z)\right], $$ $$ \varepsilon_y= \frac{1}{Y}\left[\sigma_y-\nu(\sigma_x+\sigma_z)\right], \qquad \varepsilon_z= \frac{1}{Y}\left[\sigma_z-\nu(\sigma_x+\sigma_y)\right]. $$ The shear relations remain $$ \gamma_{xy}=\frac{\tau_{xy}}{G}, \qquad \gamma_{yz}=\frac{\tau_{yz}}{G}, \qquad \gamma_{zx}=\frac{\tau_{zx}}{G}. $$ Poisson coupling appears explicitly: a stress in one direction contributes to strain in the other directions. These equations use a Cartesian coordinate system aligned with the chosen material axes. In an isotropic solid, changing axes changes the listed stress components while preserving the same scalar elastic constants. In an anisotropic laminate, the material law itself depends on orientation and requires additional constants. $$ % caption: A two-dimensional element under combined normal and shear loading. Normal % tractions change the edge lengths while the paired tangential tractions change the % angle; a single uniaxial modulus cannot convert this combined state into the full % strain state. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (2.0,0.7) rectangle (4.6,3.0); \draw[->,acc,very thick] (4.6,1.85)--(5.5,1.85); \draw[->,acc,very thick] (2.0,1.85)--(1.1,1.85); \draw[->,black,thick] (2.6,3.0)--(3.6,3.0); \draw[->,black,thick] (4.0,0.7)--(3.0,0.7); \draw[->,black,thick] (4.6,2.15)--(4.6,2.85); \draw[->,black,thick] (2.0,1.55)--(2.0,0.85); \node[acc,right] at (5.5,1.85) {normal}; \node[acc,left] at (1.1,1.85) {normal}; \node[above] at (3.1,3.0) {shear}; \end{tikzpicture} $$ Plane stress and plane strain are distinct simplifications. A thin sheet loaded in its own plane often has a free thickness direction, giving approximately $\sigma_z=0$. Its thickness can still change through Poisson strain. A very long body constrained against deformation along its length can have $\varepsilon_z\approx0$, while $\sigma_z$ develops as a reaction stress. Using plane stress when the body is actually constrained, or plane strain when the body has free surfaces, produces the wrong apparent stiffness and stress distribution. Principal normal stresses are the normal stresses on specially oriented planes where the shear traction vanishes. For a two-dimensional stress state, they are the eigenvalues of the in-plane stress matrix. Their directions identify material planes that experience pure tension or compression within that plane. A tensile coupon with well-aligned grips approximates one principal direction along the specimen axis. Near a hole or a fillet, the principal directions rotate from point to point, which is one reason simple axial force divided by one gross area becomes inadequate for local failure assessment. The selection sequence for a constitutive calculation is concrete. - **Geometry and boundary conditions:** identify free surfaces, contacts, restraints, and the dimensions that permit a plane-stress, plane-strain, beam, shaft, or three-dimensional model. - **Loading history:** record force, torque, pressure, displacement, temperature, rate, and cycle count rather than reducing every case to a single static force. - **Material description:** select isotropic linear elasticity only after checking that the specimen direction, strain range, temperature, and time scale support that approximation. - **Failure criterion:** compare the relevant stress or strain measure with a measured yield, fracture, fatigue, creep, or buckling limit appropriate to the actual loading mode. These choices occur before numerical substitution. A spreadsheet can evaluate a linear formula accurately while using an unsuitable area, omitted constraint, or wrong material direction. Dimensional consistency catches unit errors; comparison with the specimen shape and test history catches model errors.