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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.

Bending stress is the longitudinal normal stress associated with a bending moment in a beam. In a homogeneous slender elastic beam with small deformation, plane sections remaining plane give longitudinal strain ${\epsilon }_{xx}=-\kappa y$ relative to the neutral axis. Applying Hooke's law gives ${\sigma }_{xx}=-E\kappa y$.^[ref]

Choose the moment convention $M=-{\int }_{A}y{\sigma }_{xx}dA$. ThenHere $I$ is the second moment of area, and the neutral axis passes through the section centroid when axial resultant is zero and $E$ is uniform. The sign tells which side is in tension; the magnitude grows linearly with distance from that axis.

For an original rectangular section with width $b=12\mathrm{mm}$ and thickness $h=6\mathrm{mm}$, $I=b{h}^3/12=216\mathrm{m}{\mathrm{m}}^4$. A moment $M=1.2\mathrm{N}\mathrm{m}$ produces maximum surface magnitudeThe moment is $1200\mathrm{N}\mathrm{mm}$ when using millimetre section dimensions. At fixed $M$ and $b$, doubling $h$ reduces this nominal surface stress fourfold, not merely twofold, because $I$ grows as $h^3$.

These are beam-model stresses away from local load contacts and discontinuities. Added axial force contributes $N/A$ to the longitudinal stress; a notch can produce a stress concentration beyond the nominal distribution. For inhomogeneous sections the neutral-axis position must be obtained from the actual stiffness distribution rather than assumed from geometry alone.

Rotating shaft bending fatigue occurs when a shaft rotates while carrying a bending moment whose direction is approximately fixed in space. A material point on the outer surface moves alternately through the tensile and compressive sides of the bent section. A constant external bending load can therefore produce cyclic local stress.

For an ideal circular shaft in elastic pure bending, with diameter $d$, second moment of area $I=\pi d^4/64$ and steady moment magnitude $M$, the nominal surface stress amplitude isA rotating surface point has $\sigma (t)={\sigma }_a\cos (\omega t+\varphi )$ in this model. The mean is zero, ${\sigma }_{\max }={\sigma }_a$, ${\sigma }_{\min }=-{\sigma }_a$ and stress ratio $R={\sigma }_{\min }/{\sigma }_{\max }=-1$. One shaft revolution produces one full stress cycle at that point. A bending moment rotating with the shaft would require a different analysis; shaft rotation alone does not guarantee this cycle.

For an original example, $M=20\mathrm{N}\mathrm{m}$ and $d=20\mathrm{mm}$ give ${\sigma }_a\approx 25.5\mathrm{MPa}$. At $1800$ revolutions per minute, one hour supplies $108000$ cycles. These values specify a nominal loading example, not a predicted life or a safe design stress. Stress concentration, surface finish, material condition and other loads affect the local response and mechanical fatigue behavior.

A motivating apparatus is four-point rotating-bending testing. A NIST/NASA benchmark used machined and polished titanium-alloy specimens, a $100\mathrm{Hz}$ test frequency and fully reversed loading with $R=-1$. Comparing differently treated specimens under a controlled arrangement helps connect material state to fatigue response.^[ref] A lifetime claim needs appropriate experimental fatigue data and treatment of scatter; the elastic stress formula does not supply one.

A stress concentration is a localized increase of stress caused by geometry or another disturbance of the load path. Holes, notches and abrupt section changes can make the largest local normal stress much greater than a stress calculated from the total force divided by a reference cross-sectional area. A stress concentration factor $K_t={\sigma }_{\max }/{\sigma }_{\mathrm{ref}}$ is meaningful only after both the stress component and its reference have been specified.

The Kirsch equations give an instructive exact result for a circular hole of radius $a$ in an infinite, isotropic linear elastic elastic plate under remote tensile stress ${\sigma }_{\infty }$ in one direction, using a plane stress model. Measure the polar angle $\theta$ from the loading direction. At the hole boundary,At $\theta =\pi /2,3\pi /2$, the tangential normal stress reaches $3{\sigma }_{\infty }$, giving $K_t=3$ relative to remote stress. At $\theta =0,\pi$, it is $-{\sigma }_{\infty }$. A traction-free hole therefore does not require every component of its stress tensor to vanish: it requires zero traction on the radial boundary.^[ref]

For a finite plate, distinguish gross and net reference areas. A plate of width $W$, thickness $t$ and central hole diameter $d$ transmits force $P$ through gross area $A_g=tW$ and minimum net area $A_n=t(W-d)$. Defining ${\sigma }_g=P/A_g$ and ${\sigma }_n=P/A_n$ gives the exact conversionThis is a change of normalization, not a solution for the finite plate's stress field. For an original numerical example, take $W=40\mathrm{mm}$, $t=2\mathrm{mm}$, $d=10\mathrm{mm}$ and $P=800\mathrm{N}$. The reference stresses are $10$ and $13.333\mathrm{MPa}$. If the local maximum is stipulated as $32\mathrm{MPa}$, the same physical maximum gives $K_{tg}=3.2$ and $K_{tn}=2.4$. Neither number changes the local stress by choosing a different denominator.

The distinction appears in real historical measurements. In 1939, C. Dumont loaded a perforated aluminum-alloy plate in out-of-plane bending and measured strain and deflection at two loads. The reported peak tangential stress was 1.59 times the computed net-section stress, but 1.85 times the computed stress for an unperforated plate under the same bending forces. These are bending results for that geometry, rather than the infinite-plate tensile factor three.^[ref]

Modern demonstrations can combine a metal plate instrumented with strain gauges and a transparent polymer viewed in a polariscope. The latter uses stress-induced birefringence; its fringes respond to differences of principal stresses. The MIT laboratory compares local and remote strain measurements with qualitative photoelasticity observations around holes and a slot.^[ref]

The elastic stress concentration factor is dimensionless. It differs from the stress intensity factor used to describe a sharp crack's singular field. Local yielding or a crack changes the mechanical problem; multiplying a remote stress by the elastic $K_t$ does not, by itself, establish a failure load.