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Beam cross-section rotation is the signed change in orientation of a material section in the bending plane. The Euler-Bernoulli beam equation assumes that sections remain plane and perpendicular to the deformed centreline. With a consistent angle convention, the section rotation therefore agrees with the centreline tangent angle; for a graph $w(x)$,The last step is a small-angle approximation, with the angle measured in radians. This kinematic relation underlies the axial bending strain, rather than being an extra condition imposed by every support.^[ref]

For an original geometric profile, take $w(x)=a{x}^2$ with $a=0.10{\mathrm{m}}^{-1}$. At $x=0.50\mathrm{m}$, the slope is $0.10$, while the exact tangent angle is $\arctan (0.10)\approx 0.09967$ radians. Replacing it by $0.10$ overestimates the angle by about 0.33 percent. This is a chosen shape illustrating the approximation, not a measured equilibrium profile.

The beam shear angle records a possible difference between centreline slope and section rotation in a model that permits transverse shear strain. The Euler-Bernoulli kinematics set that difference to zero; a short or shear-compliant specimen can require a different model. A simply supported beam leaves its end section rotation free, whereas a clamped boundary condition prescribes it. Neither zero transverse displacement alone nor zero endpoint bending moment requires zero rotation.

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

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

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

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

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

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

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

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

Bending changes the curvature of an elastic beam, rod or sheet. In the elementary beam model, material on opposite sides of the neutral axis stretches and shortens. Their longitudinal normal stresses can balance to give zero resultant force while still transmitting a nonzero bending moment. A bent ruler illustrates why resistance to changing shape cannot be described by axial stretching alone.

For a homogeneous slender elastic beam in small-deflection pure bending, take $x$ along the beam and $y$ perpendicular to its centroidal neutral axis. Let $\kappa$ be the signed centreline curvature. The plane-section kinematics and Hooke's law giveHere $E$ is Young's modulus, taken constant over the section. Choosing the positive bending moment as $M=-{\int }_{A}y{\sigma }_{xx}dA$ yieldsThus the second moment of area $I$, rather than just the cross-sectional area, controls bending stress. The beam flexural rigidity is $EI$. Zero axial resultant requires ${\int }_{A}ydA=0$, locating the neutral axis at the section centroid under these assumptions.^[ref]

A four-point bending test makes the constant-moment region experimentally accessible. Two outer supports carry a specimen while two inner contacts apply the load. The Illinois teaching laboratory uses adjustable inner and outer spans on an Instron 4400 frame and records load and deflection electronically; it also tests notched specimens to examine stress concentration.^[ref]

For an original symmetric loading example, place the outer supports at $x=0,L$ and apply total downward force $P$ equally at $x=a,L-a$, with $0<a<L/2$. Each support reaction is $P/2$. Between the inner contacts the transverse shear force is zero and the bending moment is constant, $M=Pa/2$. This central interval realizes the ideal beam conditions for pure bending away from contact disturbances. A three-point bending test instead has its maximum moment at the single central loading contact.

Choose $L=160\mathrm{mm}$, $a=40\mathrm{mm}$, $P=40\mathrm{N}$, rectangular width $b=10\mathrm{mm}$ and thickness $h=4\mathrm{mm}$. Then $I=b{h}^3/12=53.333\mathrm{m}{\mathrm{m}}^4$ and $M=0.800\mathrm{N}\mathrm{m}$. With stipulated $E=70\mathrm{GPa}$, the model givesThese independently chosen numbers describe an assumed elastic specimen, rather than measurements of a particular material.

The same relation can infer a flexural modulus from a measured bending moment and curvature, provided the geometry and beam assumptions are adequate. Added axial force contributes a uniform longitudinal normal stress and can shift the zero-stress line. Short beams require attention to transverse shear strain; large rotations require a geometrically nonlinear description. The small-deflection Euler-Bernoulli beam equation and the more general Euler elastica cover different levels of that modeling.

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.

A fluid element is a small material parcel followed as it moves with a velocity field. It contains the same particles throughout its motion when no molecular exchange is included in the idealization. The continuum approximation requires it to be large enough to average microscopic structure while small enough to assign approximately local fields.

Its centre follows a Lagrangian trajectory, and its shape can stretch and rotate. For its volume $\delta V$, $D(\delta V)/Dt=(\nabla \cdot \mathbf{v})\delta V$. Mass conservation keeps $\rho \delta V$ constant, so expansion decreases the parcel's mass density. An incompressible flow preserves the parcel's volume, but can still distort it strongly through shear strain. A fixed box in space is a different object: material can enter and leave that box, requiring boundary fluxes in its balance laws.

A longitudinal wave has oscillating displacement parallel to its propagation direction. Push and pull one end of a long spring along its axis: compressed and expanded regions travel along the spring while individual turns move back and forth. In a small-amplitude acoustic wave in a uniform fluid, the corresponding pressure and mass density variations drive neighbouring fluid elements through pressure gradients.

For a plane acoustic wave, $u(x,t)=u_0\cos (kx-\omega t)$ can describe the particle velocity along $x$. The separation between a particle's small excursion and the distance travelled by the disturbance is essential: the speed of sound is the propagation rate, not the particle velocity. Compressional waves in an solid with elasticity provide another example; their speed depends on both the bulk modulus and resistance to shear strain. A longitudinal polarization leaves no independent choice of transverse direction of the kind available to a transverse wave.

stress describes internal force per area. A cut with unit normal $\hat{\mathbf{n}}$ has traction $\mathbf{t}=\mathbf{\sigma }\hat{\mathbf{n}}$; its normal component is $\hat{\mathbf{n}}\cdot \mathbf{\sigma }\hat{\mathbf{n}}$ and its remainder is shear traction. force balance constrains the stress divergence, while ordinary local angular-momentum balance makes the Cauchy stress tensor symmetric in a medium without body couples.

strain describes deformation rather than force. For small displacement gradients,The antisymmetric displacement-gradient part is a rigid rotation to first order and contributes no infinitesimal strain. Engineering shear strain is ${\gamma }_{xy}=2{ϵ}_{xy}$, so it must not be confused with the tensor component. stress has units of pressure; strain is dimensionless.

An isotropic linear elastic solid obeysTwo independent moduli are required. The shear modulus $G$ and bulk modulus $K=\lambda +2G/3$ must be positive for stability. Young's modulus and Poisson ratio satisfyFor a tensile specimen with free side faces, ${\sigma }_{xx}=F/A$ gives ${ϵ}_{xx}={\sigma }_{xx}/E$ and ${ϵ}_{yy}={ϵ}_{zz}=-\nu {\sigma }_{xx}/E$. Pure shear gives $\tau =G\gamma$; hydrostatic loading gives $\Delta V/V=-p/K$. The incompressible limit has $\nu \to 1/2$ at finite shear stiffness, not vanishing shear resistance. Crystals can be anisotropic and need a fourth-rank elastic stiffness tensor instead of the isotropic two-constant law.

elastic deformation is recoverable on unloading; it need not remain linear at large strain. Beyond yield, irreversible rearrangements such as dislocation motion cause plastic strain, often followed by hardening. Ductile samples can neck before fracture; brittle failure has much less plastic deformation. Engineering stress uses initial area, while true stress uses the changing instantaneous area. A decreasing engineering stress during necking therefore need not mean that local true stress decreases. stress concentrations, flaws and loading geometry can determine failure before a uniform-material strength is reached. Viscoelastic materials also show rate dependence, creep and stress relaxation; their losses cannot be described by a static Hooke law alone.

A tensile test should measure force and gauge-length extension independently, correct machine compliance, align the grips, and control strain rate and temperature. Unloading distinguishes recoverable strain from a residual plastic extension. Separate shear or hydrostatic tests determine moduli that one axial measurement does not fix. Compression specimens need friction and buckling checks. Work per volume is $\int {\sigma }_{ij}d{ϵ}_{ij}$; the recoverable part is elastic energy, while a dissipative cycle produces a loss rather than creating energy. The original loading/unloading curve below is a chosen constitutive sketch, not measured material data.

Use positive normal stress for tension. On a small rectangular material element, ${\tau }_i$ is the normal force per unit area on the face perpendicular to axis $i$, and $e_i=\Delta {\ell }_i/{\ell }_i$ is the corresponding normal strain. Young's modulus $E$ is the ratio of axial normal stress to axial strain in uniaxial stress; Poisson's ratio $\sigma$ is minus the transverse-to-axial strain ratio in the same test. Thus isotropic linear elasticity gives $E{e}_1={\tau }_1-\sigma {\tau }_2-\sigma {\tau }_3$ and cyclic permutations. The stress symbols here follow the question; $\sigma$ denotes Poisson's ratio, not a stress component.

For $e_2=e_3=0$, the two transverse compliance equations imply ${\tau }_2={\tau }_3=\sigma {\tau }_1/(1-\sigma )$. Substitution into the first givesThis uniaxial strain modulus is greater than the uniaxial-stress Young's modulus for the positive Poisson's ratio of interest, because the lateral constraints prevent free Poisson contraction.

For a small twist $\varphi$, the bonded faces' relative tangential displacement at radius $r$ is $r\varphi$. The layer's shear strain is $r\varphi /h$, so the shear stress is $Gr\varphi /h$. Integrating the torque givesThe minus sign denotes a restoring torque; the question quotes its magnitude. This uses the disk's polar second moment of area.

For tilt about H, choose $y$ in the plate perpendicular to H. The vertical displacement of the bonded top face is $y\theta$ to first order, giving normal strain $y\theta /h$. The thin layer is constrained laterally by bonding, so its leading normal traction is $My\theta /h$, not $Ey\theta /h$. Integrating its moment about H givessince the diametric second moment of area is $\pi a^4/4$. This is the specified thin-layer approximation to the torsion and tilt stiffness of a bonded elastic disk. Relaxation near the free cylindrical edge is neglected; for a nearly incompressible layer it can limit quantitative accuracy unless the thinness is sufficient.

The thin square plate's moment of inertia about V is $I_V=2m{b}^2/3$. Its two perpendicular in-plane central moments are equal and sum to $I_V$, so the diagonal H has $I_H=m{b}^2/3$. Dividing the two angular stiffnesses by the appropriate moments of inertia givesThe geometry factors cancel in their ratio. Using the supplied relation between shear modulus and Young's modulus,The result lies just below the incompressible isotropic limit $1/2$, as expected for a rubber with a large compression stiffness compared with its shear modulus. It is an inference within the bonded-layer model, rather than an independently measured material constant.

Compression and stretch viewed on rotated faces. To isolate shear, take equal small principal strains ${ϵ}_{xx}=-e$, ${ϵ}_{yy}=+e$. A square with sides at $45^{\circ }$ to these directions becomes a rhombus: adjacent side lengths are unchanged to first order, but their right angle changes by $2e$. In axes rotated through $45^{\circ }$, the same strain tensor has zero diagonal entries and off-diagonal entries of magnitude $e$.

The pure shear from opposite principal strains therefore has engineering shear strain $\gamma =2e$, not $e$. Equal opposite principal stresses $(-S,+S)$ give purely tangential traction of magnitude $S$ on those rotated faces. HenceIf the compression and stretch are not equal and opposite, an additional dilatational strain remains; their deviatoric part still supplies shear.

The shaft's torsional stiffness. At radius $r$, an end twist $\varphi$ over length $L$ gives tangential displacement $r\varphi$ and shear angle $r\varphi /L$. The local shear stress is $Gr\varphi /L$. An annular area $dA=2\pi rdr$ contributes torque $r\tau dA$, soThus torsion of a solid circular shaft givesHere $J_p$ is the polar second moment of area, in ${\mathrm{m}}^4$, not a mass moment of inertia. This $C$ is the transmitted torque magnitude; the shaft's restoring couple opposes the twist. The formula assumes small elastic shear and a circular uniform section.

Take downward deflection $y(x)$ positive, with the support level $y=0$. The rectangular section second moment of area about the horizontal neutral axis isFor small slopes the curvature relation from part (b), with the stated sign convention, isIntegrating twice givesThe support boundary conditions $y(0)=y(L)=0$ give $C_2=0$ and $C_1=q{L}^3/24$. HenceThis is a downward displacement; its upward component is $-y(x)$. As checks, reflection symmetry gives $y^{\prime }(L/2)=0$, and the maximum downward deflection isThe units are length: $\rho g/(E{b}^2)$ has units ${\mathrm{m}}^{-3}$. The width $a$ cancels because both weight and bending resistance scale with it. The small-slope approximation requires $\rho g{L}^3/(E{b}^2)\ll 1$; shear strain is also neglected, as appropriate for a slender elastic beam.

In linear viscoelasticity, stress is a linear functional of past strain increments: scaling the deformation scales the response, and responses to small deformations add. For a time-translation-invariant material, Boltzmann superposition writes $\sigma (t)={\int }_{-\infty }^{t}G(t-s)\dot{ϵ}(s)ds$, where $G$ is a relaxation modulus. The material combines reversible storage with delayed or dissipative response; the description is restricted to small deformations about the chosen state.

The Maxwell model places an elastic spring and viscous dashpot in series. They carry the same shear stress $\sigma$, and their shear strains add:Differentiating givesHere $ϵ$ is dimensionless shear strain, $\sigma$ shear stress in pascals, $G_0$ the spring shear modulus, $\eta$ dashpot viscosity in $\mathrm{Pa}\mathrm{s}$, and $\tau$ the relaxation time. The first term describes changing elastic strain; the second is viscous flow. In this model the material flows at long times and has zero static shear modulus.

Taking traces of the constitutive equation givesThe assumed finite limits of stress and strain make the right side bounded. Therefore $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$, and the limiting strain satisfies $\mathrm{tr}e=0$. The permitted interchange of derivatives and limits gives $\mathrm{tr}e=\mathrm{div}u$. DefineThen the limit stress is $S=-pI+2\mu e$. Passing to the limit in force balance and using $2\mathrm{Div}e=\Delta u+\nabla \mathrm{div}u$ givesThe pressure acts as the Lagrange multiplier enforcing the volume constraint. The differential equations determine it up to an additive constant, while a specified absolute limit stress can fix that constant.

Nonlinear elastic incompressibility is $\det Dy=1$. Expanding the determinant for $Dy=I+Du$ gives $\det (I+Du)=1+\mathrm{div}u+O(|Du{|}^2)$, so its first-order form is precisely $\mathrm{div}u=0$. The volumetric energy term $\lambda (\mathrm{div}u{)}^2/2$ becomes increasingly expensive as $\lambda$ grows, while the fixed positive $\mu$ leaves shear strain possible. Thus the infinite-bulk-stiffness limit is the incompressible limit of isotropic linear elasticity, with the same equilibrium operator as the steady Stokes equations.

A surface-reaching normal fault earthquake commonly produces a down-facing step towards the hanging wall, accompanied by extension across the break. A fresh free face can expose bedrock, polish or striations, roots or offset soil. The actual rupture is usually irregular: fault slip changes along strike, separate strands overlap or step, and tension fissures, small grabens and secondary cracks distribute part of the extension.

Exposed competent rock or very thin cover: a relatively narrow, sharp break can closely follow the principal fault plane. A steep free face can be recognizable immediately, then weather and shed debris. The face need not be vertical or maintain the same height along strike.

Loose granular or poorly consolidated cover: the main shear can splay upwards through alluvium or regolith. Open fissures, step-like scarps, rotated blocks, synthetic and antithetic normal faults, and a small intervening graben can form across a wider zone. Collapse and raveling quickly build colluvium at the toe. A single rupture event can therefore create several parallel breaks; counting them does not count earthquakes.

Thick weak or cohesive cover: deeper fault slip can be transferred into distributed shear strain, tilted beds, broad warping or monoclines, with tension cracks above a propagating tip. A discrete break may be attenuated or absent if the fault remains blind. Cover thickness alone is not enough to predict the response: cohesion, grain size, layering, saturation, rupture displacement and fault depth also matter. Saturated loose sediment can additionally develop soil liquefaction, sand blows or lateral spreading; these are secondary shaking effects and do not independently locate the tectonic rupture.

The observed vertical step, crack aperture and total width of disturbed ground should be measured separately. Reconstruct the total displacement across all strands and account for the original ground slope. Landslides or gravitational spreading can imitate normal-fault scarps, so map connections to the main fault and displaced markers rather than assuming every open crack is primary tectonic slip.

The infinitesimal strain tensor $\mathrm{diag}(-e,e)$ compresses one principal direction and stretches the perpendicular one by the same amount. Rotating the coordinate axes through $45^{\circ }$ turns it into a tensor with zero normal strains and off-diagonal entries of magnitude $e$. The engineering shear strain is twice the tensor shear component, so its shear angle has magnitude $\gamma =2e$. Likewise principal stresses $(-S,S)$ give zero normal stress and shear stress of magnitude $S$ on those rotated faces. The shear modulus law $S=G\gamma$ therefore gives $S=2Ge$.

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.

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.

The shear modulus $G$ is the ratio of shear stress to small engineering shear strain in an isotropic linear elastic material:For a slab of thickness $h$ whose upper face moves sideways by $\Delta x$ while the lower face is fixed, $\gamma \simeq \Delta x/h$. A tangential force $F$ distributed over face area $A$ gives $\tau =F/A$. Thus $G=Fh/(A\Delta x)$ in the uniform-shear idealization. Its units are pressure, and the engineering convention gives $\gamma =2{\epsilon }_{xy}$ in terms of the infinitesimal strain tensor.

An experiment on the torsion of a solid circular shaft measures the same parameter through torque and end rotation: for shaft length $L$ and polar second moment of area $J_p$, $\mathcal{T}=G{J}_p\theta /L$. This separates material resistance to shear strain from the specimen's geometric resistance to twisting. In isotropic linear elasticity, $G=E/[2(1+\nu )]$, where $E$ is Young's modulus and $\nu$ is Poisson's ratio.

The shear rate is the scalar rate at which adjacent material layers acquire shear strain. In simple shear $u_x=u_x(y)$, its signed value is $\dot{\gamma }=d{u}_x/dy$.

Engineering shear strain is the first-order change of the right angle between material lines initially directed along two perpendicular axes. For a displacement $\mathbf{u}$, ${\gamma }_{xy}={\partial }_y{u}_x+{\partial }_x{u}_y=2{\epsilon }_{xy}$, where ${\epsilon }_{xy}$ is the tensor strain component. Slender Euler-Bernoulli beam equation models neglect the transverse shear strain contribution to deflection.