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Bearing thermal stiffness identification uses measured force changes along a specified thermal path, together with independently known expansion behavior, to infer the axial restraint of an assembly. The observed bearing thermal reaction slope is $q=dF/dT$ inN/K. Mechanical stiffness is force per displacement, so converting $q$ into stiffness requires the free relative-expansion rate $\beta$ in displacement per temperature. In the constant-compliance engaged model,Here $C_b$ and $C_s$ represent the bearing and supporting structure's axial mechanical compliance contributions.

For an original identification example, suppose equilibrated engaged measurements give forces160,320 and480N at temperature increases50,60 and70K. A linear regression gives slope16N/K and, within this model, an extrapolated zero-force engagement increase of40K. If an independent free-expansion measurement gives $\beta =2\mu m/K$, the total compliance is $0.125\mu m/N$ and total stiffness is $8N/\mu \mathrm{m}=8\times 10^{6}N/m$. A16N/K thermal slope is therefore not itself a16N/m spring stiffness.

The measurements have an identifiability limit: the slope determines $C_b+C_s$ when $\beta$ is known, rather than determining both contributions separately. A known supporting-structure compliance $C_s=0.100\mu m/N$ would imply $C_b=0.025\mu m/N$. Without that extra information, many pairs give the same total. If $\beta$ is also unknown, scaling both $\beta$ and total compliance by the same positive factor leaves the measured slope unchanged.

A bench experiment can measure axial force with a load cell, track shaft and housing temperatures, wait for equilibrated states and fit only measurements from the engaged regime. A separate low-restraint expansion measurement can estimate $\beta$. Repeating heating and cooling helps test whether friction, changing contact or thermal gradients invalidate the assumed single slope. These are proposed measurement steps, not a claimed manufacturer's experiment. The physical motivation is the need to accommodate relative thermal displacement in a locating/non-locating bearing arrangement.^[ref]

For independent $\beta$ and $q$, first-order propagation givesFor $\beta =2.0$ with standard uncertainty0.1µm/K and $q=16.0$ with standard uncertainty0.8N/K, the inferred total compliance has standard uncertainty about0.00884µm/N. Correlated temperature references require the corresponding covariance contribution rather than this independent-input expression.^[ref]

A biomedical engineer develops, investigates or evaluates systems at the intersection of engineering, biology and medicine. Work can include instrument design, biomechanics, biomaterials, data analysis and technical support for medical equipment. The role includes constructing models and experiments, not only assembling a finished device.^[ref]

An original load cell example makes the measurement work concrete. Suppose reference loads $F=2\mathrm{N}$ and $6\mathrm{N}$ produce outputs $V=0.30\mathrm{V}$ and $0.70\mathrm{V}$. Fitting the stipulated linear measurement model $V=kF+b$ givesA later output of $0.50\mathrm{V}$ then corresponds to $4\mathrm{N}$. Ignoring the offset would instead report $5\mathrm{N}$. With the calibration coefficients held fixed, an output uncertainty of $0.01\mathrm{V}$ contributes $0.10\mathrm{N}$ to force uncertainty; uncertainty in the reference loads and fitted coefficients adds further terms.

The engineer must also determine whether that model remains useful. A third reference load tests interpolation; repeated loading and unloading can reveal hysteresis; a temperature change can expose drift. Mounting the sensor on a biological specimen can alter the load path compared with the calibration fixture. Thus a satisfactory calibration curve is one part of demonstrating fitness for the intended measurement. Collaborating with users and reporting the conditions and limitations connects engineering performance to the biological question.

The center of pressure summarizes where a distributed normal pressure acts. On a planar contact region $A$ in $z=0$, take a nonnegative pressure field $p(x,y)$ and positive total normal force $N$. Its center of pressure is the pressure-weighted positionIt is a continuous weighted mean, rather than necessarily the point of largest pressure. For discrete normal forces $N_i\ge 0$ at $(x_i,y_i)$, replace the integrals by sums. The resulting convex combination places the center of pressure inside the contacts' convex hull, or support polygon.^[ref]

This definition preserves the torque as well as the resultant normal force. About the origin in the contact plane,A single vertical force $(0,0,N)$ applied at $(x_P,y_P,0)$ gives the same torque. Thus its line of action passes through the center of pressure. Conversely, force-and-torque measurements reconstruct the point throughThe origin and sign convention are part of these equations; moving the moment reference above the contact plane requires the corresponding torque translation.

For an original discrete example, four vertical reactions of $100$, $200$, $300$ and $200\mathrm{N}$ act at the successive corners $(0,0)$, $(0.40,0)$, $(0.40,0.30)$ and $(0,0.30)$ in metres. Their total normal force is $800\mathrm{N}$, with $M_x=150\mathrm{N}\mathrm{m}$ and $M_y=-200\mathrm{N}\mathrm{m}$. Hence the center of pressure is $(0.250,0.1875)\mathrm{m}$. A rigid plate on four load cells provides a simple force platform implementation: calibrate the force channels and positions, then test known vertical loads at known locations. Subtracting the unloaded plate's own force and moment removes its contribution when reconstructing the added load. This is a proposed measurement test, not experimental data.

A continuous example shows the inverse problem hidden by this useful compression. On a rectangle of length $L$ and width $B$, put $u=x/L$ and choose profiles independent of $y$:Both profiles are positive on $0\le u\le 1$. The added polynomial has zero integral and zero first weighted integral:Thus both profiles give $N=p_{0}LB$ and the same center of pressure $(0.60L,B/2)$, while their peak pressures are $1.6p_0$ and $1.9p_0$. These are pressure distributions with identical resultant force and center of pressure. For $L=0.40\mathrm{m}$, $B=0.30\mathrm{m}$ and $N=800\mathrm{N}$, both give $(0.240,0.150)\mathrm{m}$ despite different local loads. A force platform alone therefore does not identify every local pressure.

Two limits matter. If $N=0$, the normal-pressure center of pressure is undefined. At fixed $N$, an absolute moment error $\delta M_y$ changes the reconstructed $x_P$ by $-\delta M_y/N$, so fixed absolute measurement uncertainty is amplified as a contact unloads. Also, tangential contact forces can leave a residual contact torque at the center of pressure: with full resultant $\mathbf{F}$,Selecting the normal-pressure centroid does not force this yaw torque to vanish.

Under the gravity-only, resting assumptions of support polygon, the center of pressure coincides with the horizontal center of mass projection. During motion it need not. For a planar model driven only by gravity and ground contact forces, at fixed center of mass height $h$ and with constant angular momentum about that centre, force and torque balance give $x_P=x_{\mathrm{CM}}-(h/g){\ddot{x}}_{\mathrm{CM}}$. The acceleration term explains why the center of pressure is a contact-force measurement rather than a general substitute for center of mass position.^[ref]

Effective beam shear rigidity is the coefficient $S$ relating a transverse shear-force resultant $V$ to the beam shear angle $\gamma$ in a reduced linear beam model:Here $G$ is shear modulus, $A$ is material area and $k_s$ is a beam shear correction coefficient. The coefficient $S$ has units of force, whereas beam flexural rigidity has units $\mathrm{N}{\mathrm{m}}^2$. A shear-deformable beam permits section rotation to differ from centreline slope; Bower gives the corresponding force relation and explains the correction coefficient's dependence on section geometry and the approximation used.^[ref]

An original energy calculation explains one familiar correction. For a homogeneous rectangle of width $b$, thickness $h$ and area $A=bh$, adopt the classical parabolic shear-stress approximationIt has resultant ${\int }_A\tau dA=V$. Its shear elastic energy per unit beam length isEquating this to the reduced-model energy $V^2/(2S)$ gives $S=5GA/6$. Thus $k_s=5/6$ follows from this particular energy-matching approximation; it is not a universal material constant or an exact coefficient for every beam theory.

The effect on a three-point bending test can be derived from that reduced energy. For a simply supported beam under central force $P$, the magnitude of shear force is $P/2$ on each half. Hence $U_s=P^{2}L/(8S)$ and its contribution to central displacement is $\partial U_s/\partial P=PL/(4S)$. Combining it with the bending response givesThe shear contribution becomes relatively more important as the span decreases.

For stipulated $B=2\mathrm{N}{\mathrm{m}}^2$, $S=20000\mathrm{N}$, $L=0.10\mathrm{m}$ and $P=10\mathrm{N}$, the two contributions are $0.1042$ and $0.0125\mathrm{mm}$. Shear adds twelve percent of the bending deflection. A span series with a load cell and direct midspan displacement readout can distinguish cubic bending dependence from linear shear dependence, provided contact and fixture contributions are also controlled. Inferring $G$ from the response additionally requires the section geometry and correction model.

The Favimat+ fibre testing machine combines measurements on individual material fibres, including a single-fibre tensile test and vibroscopic estimation of linear mass density. Its useful feature is the association of the force-extension record with a measurement of the same strand, rather than with a batch-average geometry.

In the reported glass- and carbon-fibre study, a magazine held the prepared strands and a robotic device transferred each to the grips. The built-in vibroscope used acoustic excitation and a frequency sweep to identify resonance; a load cell measured tension and grip separation supplied the suspended length. Tensile measurements followed vibroscope measurements on the same still-gripped fibre.^[ref]

An original pairing example shows why strand identity matters. Suppose two strands have independently established material areas $50$ and $100\mu {\mathrm{m}}^2$, and each carries $0.050\mathrm{N}$. Their engineering stresses areUsing the common mean area $75\mu {\mathrm{m}}^2$ would assign both $0.667\mathrm{GPa}$, concealing the difference. These are chosen values, not measured instrument data.

Frequency first supports a mass-per-length estimate under the vibration model; converting it into area also needs compatible mass density and deformation conventions. The vibroscope area estimator explains that inference and its limitations. Force and displacement readout resolution are likewise not complete statements of measurement uncertainty or specimen extension: machine compliance and grip motion must be assessed when extracting a modulus.

A force-displacement curve records how a measured force varies with a specified displacement during loading or unloading. The choice of displacement matters: it can describe specimen extension, motion of a machine crosshead or another reference. A curve can reveal resistance to motion, nonlinear response and differences between loading and unloading.

A concrete apparatus uses a moving crosshead, specimen fixtures and a load cell. The University of Illinois describes an Instron 4483 frame whose ball screws move the crosshead, whose load cell measures force, and whose encoder records crosshead position; an extensometer can separately measure specimen length change. The acquisition system converts those signals into engineering units and records them.^[ref]

Those displacement channels need not agree. Crosshead motion can include mechanical compliance of the machine, grip seating and deformation elsewhere in the load train. A specimen extensometer therefore answers a different geometrical question from the crosshead encoder.^[ref]

For a monotonic loading branch, the local slope $dF/dx$ is a tangent stiffness, while the ratio $F/x$ from a chosen zero point is a secant stiffness. These coincide for the linear model $F=kx$ but can differ for a nonlinear curve. The force here is the applied loading force in the positive displacement direction; a spring's restoring force has the opposite sign.

The work supplied along this one-dimensional branch isFor an own linear example, $k=200\mathrm{N}{\mathrm{m}}^{-1}$ and $x_1=0.05\mathrm{m}$ give final force $10\mathrm{N}$ and workThe nonlinear model $F=c{x}^3$, with $c=80000\mathrm{N}{\mathrm{m}}^{-3}$, reaches the same final force but givesIts final secant stiffness is $200\mathrm{N}{\mathrm{m}}^{-1}$, while its tangent stiffness is $3c{x}_1^2=600\mathrm{N}{\mathrm{m}}^{-1}$. A final force and displacement therefore do not determine either the full curve or its work integral.

When loading and unloading follow different branches, a single-valued relation between force and displacement may be insufficient. Hysteresis records that path dependence; relating a cycle's work to dissipation also requires an energy balance and the relevant state assumptions. Report the measured displacement channel, sign convention and loading path before treating a curve as a material property independent of its apparatus.

A load cell is a calibrated device for measuring force. In a tensile modulus experiment its output supplies the axial load used to calculate engineering stress.

A load receptor is the part of a weighing scale intended to receive the load, such as the platform supporting a parcel. The instrument can transmit that load's force to a separate measuring element, for example a load cell.^[ref]

A simple static model explains why the physical support path matters. Suppose a $2\mathrm{kg}$ object rests entirely on a horizontal load receptor, with local gravitational acceleration $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$. Neglecting air buoyancy, its gravitational weight is $19.6\mathrm{N}$. The instrument can convert the measured supporting force into a mass indication through its calibration.

Now arrange a separate overhead support to carry $4.9\mathrm{N}$ of the same object's gravitational weight, while the rest still reaches the load receptor. The scale then supports $14.7\mathrm{N}$ and, under the same ideal calibration, indicates $1.5\mathrm{kg}$. The object's mass remains $2\mathrm{kg}$: the indication changed because the support path changed. This is an illustrative static equilibrium experiment, rather than a claim about any instrument's specified error.

A removable container placed on the platform contributes tare mass; it is distinct from the instrument's own load receptor. To measure the net mass of the contents, the tare device changes the indicated reference while the receptor continues to carry both container and contents. Contact with an external support can therefore defeat an otherwise correct subtraction.

Machine compliance is the extra load-path displacement per applied force attributed to the apparatus in a specified measurement model. Its units are $\mathrm{m}{\mathrm{N}}^{-1}$, the reciprocal units of a spring constant. Here $C_m$ includes the frame, load cell, adapters and grips insofar as their deformation contributes to the recorded crosshead displacement $x$. It is a property of that configuration and readout, rather than a universal constant of the testing machine.

Assume a uniform specimen with initial length $L$, constant $A_0$ and Young's modulus $E$, small axial elastic deformation, quasistatic loading, no slip, and a linear, length-independent apparatus response over the chosen force window. Equilibrium makes the same force act through specimen and apparatus, while their extensions add. The two springs in series therefore giveThe offset $x_0$ describes the displacement origin after seating; the slope $q$ describes total compliance. Fitting an intercept avoids forcing an arbitrary zero through the origin, but cannot repair slip or a curved response.

If all of $x-x_0$ is incorrectly attributed to specimen extension, the inferred modulus isFor positive $C_m$, this is too small. The fractional bias is most serious for a short, thick, stiff specimen because its own compliance $L/(E{A}_0)$ is small. A local extensometer instead has $d(\Delta L_g)/dF=L_g/(E{A}_0)$ in this ideal model, excluding deformation outside its contacts.

A shaped specimen requires more care. With varying initial cross-sectional area $A(s)$ and uniform $E$, its axial compliance is $\int ds/[EA(s)]$ in the one-dimensional approximation. Shoulder and end deformation must be included explicitly or remain identical between tests and be absorbed into an effective intercept. Changing only the extensometer gauge length on an unchanged specimen does not change its total compliance and cannot perform a two-length machine compliance separation.

A measuring chain is one successive signal path through a measuring system, beginning at a sensor and ending at an output element. The VIM illustrates this with both an electro-acoustic chain and a mechanical chain; a measuring chain need not be electronic.^[ref]

Consider an original force measurement: a load cell produces an electric potential difference, an amplifier scales that signal, an analog-to-digital converter encodes it, and software converts the code to a force value. This is one measuring chain within the larger measuring system, which also includes its excitation supply, references and mounting arrangements. The load cell acts as a measuring transducer.

In an ideal linear model, let sensor response be $S=2\mathrm{mV}{\mathrm{N}}^{-1}$, amplifier gain $G=100$, and converter step $q=1\mathrm{mV}$ per code increment. If the additive output offset is $n_0=10$ increments, the numerical output isThus $n=610$ implies $F=(610-10)/200=3\mathrm{N}$. These are stipulated teaching values, not measured equipment performance.

The example exposes why measurement calibration must describe the relevant path. Changing amplifier gain changes the relation between output and force even if the load cell is untouched. A saturated stage can also discard information that later conversion cannot restore. Gain uncertainty, offsets, instrumental drift and finite conversion steps contribute at different points; treating the final display's increment as the whole measurement uncertainty overlooks those contributions.

Mechanical compliance describes how much displacement a mechanical system produces per applied force. For a one-dimensional linear response measured from an unloaded reference,where $k$ is the stiffness. Translational mechanical compliance has units $\mathrm{m}{\mathrm{N}}^{-1}$. A larger value means a larger displacement under the same force. For a nonlinear force-displacement curve, the ratio $x/F$ and the local derivative $dx/dF$ need not agree; the latter is reciprocal to tangent stiffness where the inverse relation exists.

For a uniform bar in axial linear elasticity,where $L$ is its length, $A$ its cross-sectional area and $E$ its Young's modulus. Thus doubling the length doubles mechanical compliance, while doubling the area halves it, with material and other assumptions fixed.^[ref]

The same idea matters when measuring a specimen in a testing machine. The machine frame, load cell and grips also deform, so a crosshead's total movement can exceed the specimen's extension.^[ref] In an illustrative slowly loaded series model, each part carries the same force and their displacements add:Choose $C_{\mathrm{specimen}}=0.020\mathrm{mm}{\mathrm{N}}^{-1}$ and $C_{\mathrm{machine}}=0.005\mathrm{mm}{\mathrm{N}}^{-1}$. At $100\mathrm{N}$, total movement is $2.5\mathrm{mm}$ but specimen extension is $2.0\mathrm{mm}$. Treating the total as specimen extension would give an apparent stiffness of $40\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$ instead of $50\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$.

One apparatus-based correction loads a very stiff dummy specimen through the same load path, then subtracts its measured movement at matching forces. Its usefulness depends on reproducing the load path and making the dummy's own deformation sufficiently small.^[ref] An extensometer instead measures extension directly over its chosen specimen length.

Mechanical driving-point impedance describes the relation between a driving force and motion at the same mechanical port. Here the convention iswhere $F,V$ are Laplace transforms with zero initial conditions, and $v$ is the signed relative velocity of the two terminals. Choose signs so $Fv$ is entering power. The units of this mechanical impedance are $\mathrm{N}s/m$. Its inverse is the mechanical driving-point mobility. Conventions must be checked explicitly: in The Inerter: A Retrospective by Malcolm C. Smith, the force–current analogy calls $V/F$ the mechanical impedance and $F/V$ the admittance.^[ref]

For a spring, dashpot and inerter sharing the same terminals, relative velocities agree and forces add. HenceAt a sinusoidal angular frequency $\omega$, using complex amplitudes proportional to $e^{i\omega t}$,The real part determines mean absorbed power, while the imaginary part records the out-of-phase reactive response. With peak velocity amplitude $|V|$, the mean entering power is $\frac{1}{2}\mathrm{Re}(Z)|V{|}^2$.

For an original example, choose $b=10\mathrm{kg}$, $c=30\mathrm{N}s/m$ and $k=1000N/m$. At $\omega =10rad/s$, the inertial and elastic terms cancel, giving $Z=30\mathrm{N}s/m$. A relative velocity amplitude $0.020m/s$ requires force amplitude $0.60\mathrm{N}$, and mean absorbed power is $0.006\mathrm{W}$. The individual spring and inerter still store and exchange energy.

A mechanical impedance measurement drives the chosen port and records synchronized force and motion there. A load cell measures force; displacement or an accelerometer can supply motion. For an acceleration complex amplitude $A$, $V=A/(i\omega )$, so the phase as well as magnitude matters. Fixture inertia, sensor measurement calibration and the second terminal's motion belong in the measurement model. Using a response at a different point instead defines a transfer relation, not this driving-point relation. Initial stored motion also prevents interpreting an arbitrary transient quotient as a zero-initial-state transfer function.

A simply supported beam is prevented from moving transversely at its two supports while its ends remain free to rotate. In the small-deflection Euler-Bernoulli beam equation, a uniform span $0\le x\le L$ with transverse displacement $w(x)$ and positive beam flexural rigidity $B=EI$ has the support boundary conditionsHere $E$ is Young's modulus and $I$ is the second moment of area about the bending axis. The second pair expresses zero applied endpoint bending moment, since $M=-B{w}^{\prime \prime }$ in the convention used here. It does not set the endpoint beam cross-section rotation or slope to 0. A clamped boundary condition instead constrains both displacement and slope.

For an original derivation, apply a downward central point force $P$ and take $w$ positive downward. Assume small elastic deflections, constant $B$ and no axial preload. Symmetry and force balance give an upward reaction $P/2$ at each support. On the left half, the bending moment is $M(x)=Px/2$, so $w^{\prime \prime }=-Px/(2B)$. Integrate twice, using $w(0)=0$ and the symmetry condition $w^{\prime }(L/2)=0$, to obtainThus the midspan compliance of a simply supported beam isThe endpoint slopes are $w^{\prime }(0)=P{L}^2/(16EI)$ and $w^{\prime }(L)=-P{L}^2/(16EI)$: they are generally nonzero. This support geometry and central-load formula are used to infer material stiffness from a three-point bending test.^[ref]

The moment-free natural boundary condition for a beam also follows from the principle of virtual work. For a central load, the total potential energy is $\Pi [w]=\frac{1}{2}{\int }_0^{L}B(w^{\prime \prime }{)}^{2}dx-Pw(L/2)$. An admissible virtual displacement $\eta$ satisfies $\eta (0)=\eta (L)=0$, while its endpoint first derivatives remain free. Two integrations by parts produce the boundary contributionThe terms multiplying $\eta$ vanish because the endpoint displacements are prescribed. Arbitrary endpoint values of ${\eta }^{\prime }$ then require $B{w}^{\prime \prime }=0$ there. The free end rotation and zero end moment are conjugate parts of the same variational boundary condition.

Clamping both ends changes the central-load deflection to $P{L}^3/(192EI)$, one quarter of the simply supported value for the same $P,L,E,I$. It introduces endpoint reaction moments; it does not merely rename the same support. The clamped-beam Green function provides the corresponding fourth-order boundary-value solution. The support conditions likewise determine the simply supported beam normal modes, so a static stiffness measurement and a vibration measurement must use the same mechanical support model when they are compared.

A three-point bending test places a specimen on two separated supports and applies a central force. A load cell measures force, and a midspan displacement readout measures response. A direct plunger and extensometer can measure specimen deflection separately from machine crosshead displacement, as in the published Instron fixture arrangement.^[ref] Fitting a proportional small-load segment estimates $E=L^3/(48IC)$, where $C$ is the measured incremental midspan compliance. Record support spacing, load position and cross-section dimensions with that estimate.

For stipulated $L=0.50\mathrm{m}$, rectangular width $b=20\mathrm{mm}$, thickness $h=3\mathrm{mm}$ and $E=200\mathrm{GPa}$, the second moment of area is $I=b{h}^3/12=4.5\times 10^{-11}{\mathrm{m}}^4$ and $EI=9\mathrm{N}{\mathrm{m}}^2$. An incremental $1\mathrm{N}$ central load gives $0.289\mathrm{mm}$ incremental deflection under the simply supported model, versus $0.0723\mathrm{mm}$ for the clamped model. Incremental readings remove an unchanged baseline deflection in this linear model; support settling, contact indentation and large rotations require separate interpretation.

A single-fibre tensile test pulls one material fibre between grips while measuring axial force and extension over a defined gauge length. A load cell supplies force information; a direct specimen-extension measurement or a corrected crosshead displacement supplies extension. With initial area $A_0$ and initial gauge length $L_g$, the conventional engineering stress and engineering strain areThe initial proportional slope estimates Young's modulus; a changing local slope is a tangent modulus. The area and displacement conventions must accompany that result.

The glass- and carbon-fibre study by Rajnish Kumar, Lars Pilgaard Mikkelsen, Hans Lilholt and Bo Madsen used a semi-automated single-fibre machine and compared stress-strain-based modulus determinations.^[ref] Its dataset contains individual tensile curves, gauge-length groups and test-setup compliance calculations.^[ref]

For an original ideal calculation, take a circular strand of diameter $10\mu \mathrm{m}$ and $L_g=20\mathrm{mm}$. Its area is $A_0=7.854\times 10^{-5}\mathrm{m}{\mathrm{m}}^2$. A force increment $0.010\mathrm{N}$ produces measured crosshead travel $0.040\mathrm{mm}$. Suppose independently established machine compliance is $C_m=0.50\mathrm{mm}{\mathrm{N}}^{-1}$, with no grip slip and the same deforming specimen length as $L_g$. ThenUsing the uncorrected travel would instead give $63.7\mathrm{GPa}$. These chosen increments represent a proportional segment, not measured dataset values or a complete uncertainty analysis.

Check grip slip, specimen alignment and the force zero before interpreting the slope. An irregular area cannot automatically be replaced by a circular area from one apparent diameter. A vibroscope offers another route to fibre-area estimation through vibration under tension.^[ref] Preserve the original force and displacement records so corrections and area assumptions remain inspectable.

A support polygon is the convex hull of a body's supporting contact positions, projected onto a horizontal plane. For a rigid body resting on that plane in a uniform gravitational field, with gravity as the only applied load besides support reactions, it describes which horizontal center of mass positions can admit static equilibrium. Widely separated feet can support a center of mass above empty space between them: the footprint's convex hull, rather than the material under every point, matters.^[ref]

The geometry follows directly from force and torque balance. Let the contact positions be $(x_i,y_i,0)$ and their upward normal forces be $N_i\ge 0$. The inequality expresses unilateral contact: an ordinary unbonded support can push the body upward but cannot pull it downward. For total mass $m>0$, gravitational acceleration $g>0$ and horizontal center of mass projection $c=(x_c,y_c)$, balance givesDivide by $mg$ and write ${\lambda }_i=N_i/(mg)$. ThenThus $c$ must be a convex combination of the contacts, which is exactly membership in their convex hull. Conversely, any such convex combination supplies nonnegative reactions $N_i=mg{\lambda }_i$ satisfying this model's balance equations. The center of pressure of these vertical contact forces coincides with $c$. This proves existence of allowable reactions; it does not determine every reaction uniquely when there are more contacts than independent balance equations.

For an original three-foot example, take contact positions $(0,0)$, $(1,0)$ and $(0,1)$ in metres. The support polygon is the triangle described by $x\ge 0$, $y\ge 0$, $x+y\le 1$. A $10\mathrm{kg}$ body with $c=(0.30,0.20)\mathrm{m}$ requires the convex combination coefficientsWith $g=9.81m/{\mathrm{s}}^2$, the corresponding normal forces are $49.05$, $29.43$ and $19.62\mathrm{N}$. Moving the projection to $(0.70,0.50)\mathrm{m}$ would require ${\lambda }_1=-0.20$, a downward reaction at the first foot. No set of nonnegative normal forces can hold that configuration at rest under the stated loads.

A simple test uses a rigid plate on three small supports, a movable known mass, and three load cells measuring the upward reactions. Include the plate's own mass and center of mass when calculating the combined projection. As that projection approaches the edge $x+y=1$, the first support's reading approaches zero. Beyond the edge, maintaining the same contact configuration would require the forbidden downward reaction; the observed motion or changed contact set marks the failure of the original static equilibrium. This is a proposed apparatus test of the balance equations, not reported experimental data.

The distance from an interior projection to the polygon's nearest edge is a static tipping margin; here it is $\min (0.30,0.20,0.50/\sqrt{2})=0.20\mathrm{m}$. A boundary projection has zero margin: some supports unload, and an arbitrarily small outward shift can make balance impossible. A strictly interior projection gives a positive geometric margin against small horizontal shifts under the same model. It is not a general certificate of dynamical stability for a flexible or articulated system.

This explains a constraint in a quadrupedal robot's gait. Four feet at the corners of a square give a square support polygon. Lifting the corner $(1,1)$ leaves the triangle $x\ge 0$, $y\ge 0$, $x+y\le 1$. A center of mass projection at $(0.60,0.60)$ was inside the square but is outside that triangle, so the body must shift or use dynamics to lift that foot without losing the assumed balance. The contact set must be updated when computing the support polygon.

The assumptions matter when applying the test. Nonhorizontal contacts require the full balance equations and admissible contact forces, including Coulomb friction; a projected hull alone need not decide feasibility on irregular terrain.^[ref] Additional horizontal loads also change the balance problem. For example, a horizontal force $F$ applied at center of mass height $h$ produces an overturning torque $Fh$ about the relevant edge. If the inward horizontal distance to that edge is $d$, incipient tipping of a rigid body satisfies $Fh=mgd$. Whether the body slides first depends on static friction. The gravity-only support polygon contains no information about $h$ or that additional load.

MIT's Spring 2004 laboratory assignment Properties of Heat Treated Steel gives a concrete tensile testing protocol. Record transducer channels and calibration factors; measure specimen diameter and length, and the extensometer gauge length with its locking pin inserted. Zero the load cell, pin the threaded grips into their adapters, and attach the extensometer centrally. Apply the specified seating load, record actuator position, then acquire readings every second while moving at about $0.05\mathrm{cm}\mathrm{mi}{\mathrm{n}}^{-1}$. The handout derives engineering stress from load cell force, engineering strain from extensometer deformation, and Young's modulus by linear regression below the proportional limit; it warns that actuator-derived strain includes apparatus deformation.^[ref]

For a modulus-focused measurement, select a load window after seating, hold temperature and loading rate consistent, and record force, local extension and crosshead displacement together. Fit $F=F_0+k_g\Delta L_g$ within that window and calculate $E=k_g{L}_g/A_0$. Inspect the regression residuals for curvature; repeat at a lower maximum load and include unloading to check elastic deformation. These are diagnostic additions to the cited protocol, whose full test continues through failure. The handout's seating load is specific to its specimens and apparatus, rather than a universal preload.

An extensometer measures the change between its contact points; crosshead displacement measures travel of the machine's moving member. The manufacturer Instron explicitly distinguishes local strain from travel that includes machine deformation and machine compliance.^[ref] The gauge length $L_g$ can be shorter than the uniform length $L$ between grips. Thus even a perfectly rigid machine generally has $\Delta L_g\ne x$: the local instrument sees only its own segment. Do not divide one channel's extension by the other channel's reference length.

Grips must transmit axial force without slip. Misalignment adds bending moment and makes opposite sides experience different axial strains; an instrument on just one side may then report a biased modulus. A changing extension offset, hysteresis between loading and unloading, or load-dependent fitted slope is evidence to investigate seating, slip or material response. A straight line over a short interval alone does not identify which is responsible.

A three-point flexural test loads a specimen through one central contact while two separated supports carry the reactions. In a small-deflection simply supported beam model, an incremental central force and the resulting midspan displacement determine the midspan compliance of a simply supported beam. With a known second moment of area, that response yields a flexural modulus.^[ref]

The apparatus must distinguish specimen deflection from machine travel. The manufacturer's fixture description uses adjustable lower anvils and a central loading member; a direct midspan plunger can transmit deflection to an extensometer. A load cell supplies the force record.^[ref] The crosshead displacement can also contain deformation of the frame and contacts.

For an original readout example, stipulate span $L=0.120\mathrm{m}$, width $b=12\mathrm{mm}$ and thickness $h=4\mathrm{mm}$, so $I=6.4\times 10^{-11}{\mathrm{m}}^4$. Suppose a linear loading interval gives direct midspan compliance $C_s=0.020mm/N$, while machine travel gives $C_m=0.025mm/N$. Applying the same beam relation to the two readouts givesThe apparent modulus is twenty percent lower. The numbers are chosen to isolate the displacement-channel effect; they do not characterize a particular fixture. Inferring a material Young's modulus further assumes the beam's linear, homogeneous bending model is adequate. Contact indentation, support motion and transverse shear strain can otherwise alter the measured compliance. The parent simply supported beam supplies the deflection derivation and support-comparison figure.

Unilateral contact permits bodies to separate while preventing them from passing through each other. In an ideal rigid, nonadhesive model, the supporting normal contact force can push but cannot pull. It is therefore different from a permanently attached support: an upward pull may lift a body away instead of producing a downward support reaction.

Let $d$ be the signed normal gap, positive when the bodies are separated, and $N$ the outward compressive normal contact force. The nonpenetration constraint and absence of adhesion give the unilateral contact complementarity conditionsAn open gap requires $N=0$, while $N>0$ requires a closed gap. The pair $d=N=0$ is also allowed: touching does not by itself imply a loaded contact.^[ref] Tangential friction requires an additional relation such as Coulomb friction; it is not specified by the normal-gap conditions.

The equation of motion is still needed to determine the reaction. Consider a block of mass $m$ on a fixed horizontal floor, initially at rest at $d=0$, with an upward applied force $u$ through its center of mass. With upward positive, Newton's second law isIf the contact remains closed, $\ddot{d}=0$ would require $N=mg-u$. That is feasible for $u\le mg$. For $u>mg$, a negative normal reaction would violate unilateral contact, so the contact opens with $N=0$ and $\ddot{d}=(u-mg)/m>0$. At a closed contact with zero normal velocity, the appropriate acceleration-level contact complementarity iswhich yieldsThese acceleration conditions are being applied at a closed, initially stationary contact; they are not a rule forbidding downward acceleration during separated free flight.

For an original numerical example, let $m=2\mathrm{kg}$ and $g=9.81m/{\mathrm{s}}^2$. Pulling upward with $u=5\mathrm{N}$ leaves a normal reaction of $14.62\mathrm{N}$. At $u=25\mathrm{N}$, the reaction is zero and the initial upward acceleration is $2.69m/{\mathrm{s}}^2$. At $u=19.62\mathrm{N}$, both the reaction and the initial acceleration vanish, while touching remains possible.

A simple apparatus test places the block on a calibrated load cell and applies the upward pull through a spring balance, keeping the pull vertical through the center of mass to avoid an extra overturning torque. The support reading should decrease with the pull while contact persists, then vanish at lift-off. Recording the gap as well distinguishes an unloaded touch from a genuinely open contact. Real support elastic deformation and adhesion must be accounted for before treating that apparatus as the ideal rigid, nonadhesive model.

A penalty approximation of unilateral contact replaces perfect rigidity by a chosen stiff response, for exampleHere a small negative gap represents an allowed modeled overlap relative to undeformed reference surfaces. It is not an exact solution of $d\ge 0$. For the $14.62\mathrm{N}$ supported load above and chosen penalty stiffness $k=10^{5}N/m$, equilibrium requires $d=-0.1462\mathrm{mm}$. Increasing $k$ reduces that overlap at the same load. The penalty method and the ideal constraint are different approximations of the contact response.

Finally, impact requires a rule for the velocity change. The position-level inequalities alone do not decide whether an approaching body sticks or rebounds. A coefficient of restitution or another specified impact law determines that behavior, and the impulsive contact force is summarized by its impulse. Tedrake's particular time-stepping formulation identifies its block collisions as perfectly inelastic; that is a feature of the stated model, not a consequence of the three gap/reaction inequalities alone.^[ref]

The vibroscope area estimator converts the fundamental frequency of a tensioned material fibre into a cross-sectional area through a mechanical model. For a uniform flexible strand of length $L$, positive tension $T$, linear mass density $\mu$ and mass density $\rho$, the fixed-end string relation givesThe second equation uses $\mu =\rho A$: the frequency first identifies mass per length under the stated tension, and a chosen mass density then converts it into area. Circular geometry is needed only to infer a diameter from that area.^[ref]

The experimental study by Bo Madsen, Rajnish Kumar and Hans Lilholt used a Favimat+ fibre testing machine. Acoustic waves excited the strand, and a frequency sweep plus a polynomial fit located the resonance peak. A load cell measured tension, while grip separation supplied $L$. The reported frequency range was 500–4000 Hz, with an estimated resolution of 0.1 Hz; the stated force and length readout resolutions were $1\mu \mathrm{N}$ and $0.1\mu \mathrm{m}$. Those readout resolutions do not themselves specify the estimator's full measurement uncertainty.^[ref]

The area and mass density must refer to the same loaded or reference configuration. The companion theory derives volume and area corrections for recovering an untensioned reference area from the tensioned measurement.^[ref]

The vibroscope area uncertainty follows locally by logarithmic differentiation:Treat the small input errors as independent random variables, and let each $u$ denote the corresponding standard uncertainty. First-order uncertainty propagation then givesCorrelated inputs require their covariances as well. Four independent 1% relative standard uncertainties give $\sqrt{10}\%\approx 3.16\%$ for the area. A systematic 1% overestimate of frequency alone instead multiplies the area estimate by $(1.01{)}^{-2}\approx 0.9803$: it produces about a 1.97% underestimate.

Bending resistance can also increase the observed frequency. The companion theory paper models fully clamped supports and derives the corresponding frequency equation and deformation corrections.^[ref] An independently derived, simply supported example isolates the bending-stiffness bias in the vibroscope area estimator. Let $B=EI$ be the filament bending modulus and impose zero deflection and zero bending moment at each endpoint. Substituting $y=\sin (n\pi x/L)\cos (2\pi f_{n}t)$ into the tensioned Euler-Bernoulli beam equation $\mu y_{tt}+B{y}_{xxxx}-T{y}_{xx}=0$ givesWith $r=b/a={\pi }^{2}B/(T{L}^2)$, the uncorrected flexible-string estimate satisfies $\hat{A}/A=1/(1+r)$. At $r=0.10$, it is 9.09% too small. The same model yields two-mode separation of tension and bending stiffness:For $T>0$ and $B\ge 0$, the predicted mode ratio is $f_2/f_1=2\sqrt{(1+4r)/(1+r)}$, ranging from 2 toward 4. These identities belong to the simply supported model; clamped grips require their own boundary conditions. Neither a single resonance peak nor an assumed diameter establishes the appropriate support, density or bending correction.

W. F. Finney, published in full as William F. Finney, collaborated with M. D. Morris on using Raman spectroscopy to study bone under mechanical loading. Their experiment connected a change in a Raman spectrum to deformation of a living material's mineral component, rather than merely identifying its chemical ingredients.^[ref]

The apparatus held hydrated mouse femora in a mechanical tester on a Raman spectroscopy microprobe stage. A load cell monitored the applied loading, which stayed below yielding, while the experiment followed shifts of a phosphate vibrational band. This joined a mechanical measurement to a molecular-scale probe of the loaded bone.^[ref]

For an original illustration of the mechanical scale, a specimen segment of initial length $10\mathrm{mm}$ that extends by $0.02\mathrm{mm}$ has engineering strain $\epsilon =0.02/10=0.002$, or $0.2\%$. A simultaneous Raman shift measurement could track how a molecular vibration changes during that extension. The numerical extension is illustrative; it is not a reported value from Finney's experiment.