6.1.1 Normal Stress, Shear Stress, Normal Strain and Shear Strain

Normal stress acts perpendicular to a section. For axial force P on cross-sectional area A, the average normal stress is sigma equal to P divided by A. Tensile stress is taken as positive. Shear stress acts parallel to a section. For transverse shear force V on area A, the average shear stress is tau equal to V divided by A. Complementary shear stresses on adjacent faces are equal in magnitude for moment equilibrium of a small element.

Normal strain is elongation per unit original length. For extension delta of a gauge length L, epsilon equals delta divided by L, dimensionless. Shear strain gamma is the change of an initially right angle, measured in radians, positive when the right angle decreases under the sign convention used here. Tensile test landmarks are proportional limit, elastic limit, yield strength sigma y, ultimate tensile strength sigma u, breaking stress, percentage elongation, and reduction of area. The initial linear portion defines elastic response.

6.1.2 Hooke Law, Poisson Ratio and Volumetric Strain

Below the proportional limit, normal stress is proportional to normal strain. Hooke law in uniaxial form is sigma equals E times epsilon, where E is Young modulus. In shear, tau equals G times gamma, where G is shear modulus. Lateral contraction accompanies axial extension. Poisson ratio nu is the negative ratio of lateral strain to axial strain in uniaxial tension, a positive number for ordinary metals, about 0.3 for steel. Bulk modulus K relates hydrostatic pressure to volumetric contraction.

For small strains, volumetric strain epsilon v equals the sum of the three normal strains epsilon x plus epsilon y plus epsilon z. Under uniaxial tension with epsilon x equal to epsilon and epsilon y equal to epsilon z equal to minus nu times epsilon, the volumetric strain is epsilon times one minus two nu. Positive nu less than one half gives dilatation in tension. Incompressibility in the elastic sense corresponds to nu approaching one half, with zero volumetric strain.

6.1.3 Derivation of the E, G, K and Nu Interrelations

An isotropic linear elastic solid has only two independent elastic constants. The standard set E, G, K, nu is therefore linked by two independent relations, with a third relation dependent on them. The derivations use superposition of uniaxial Hooke responses.

First relation, E in terms of G and nu. Consider a small square element in pure shear tau. Its principal stresses are plus tau and minus tau at 45 degrees. Principal strains follow from generalized Hooke law. The major principal strain is tau divided by E times one plus nu. The same deformation viewed as shear distortion gives principal strain difference related to engineering shear strain gamma equal to tau divided by G. Geometry of Mohr strain circle for pure shear gives maximum normal strain difference equal to gamma divided by two magnified by orientation, leading to gamma equal to two times one plus nu times tau divided by E. Comparison with tau divided by G yields G equal to E divided by two times one plus nu. Rearranged, E equals two G times one plus nu.

Second relation, E in terms of K and nu. Consider a small cube under equal triaxial tension sigma in all three directions. Each normal strain is sigma divided by E times one minus two nu by superposition of one direct strain sigma over E and two Poisson contractions minus nu sigma over E. Volumetric strain is three times that value, equal to three sigma times one minus two nu divided by E. The same state viewed as hydrostatic tension with mean stress sigma gives volumetric strain equal to three sigma divided by three K, that is sigma divided by K with sign care for pressure versus tension. Equating the two expressions gives E equal to three K times one minus two nu.

Third relation, E in terms of K and G. Eliminate nu between the first two relations. From the first, nu equals E divided by two G minus one. Substitute into the second and solve for E. The algebra gives E equal to nine K G divided by three K plus G. Equivalent forms express G and K in terms of E and nu and are used interchangeably in data tables.

Physical reading: shear stiffness G controls shape change at constant volume, bulk stiffness K controls volume change at constant shape, Young modulus E controls extension under lateral freedom, and nu couples the two modes. Increasing nu at fixed E lowers G and raises K in the isotropic model.

6.2.1 Uniform Bar Derived from Elongation of a Slice

For a prismatic bar under axial force P, constant cross section A, modulus E, the internal normal stress is uniform and equal to P divided by A. A slice of thickness dx extends by P dx divided by A E from Hooke law applied locally. Integration from zero to L gives total elongation delta equal to P L divided by A E. Extension is positive in tension. Shortening in compression uses the same magnitude with compressive force. Stiffness k equals P divided by delta, equal to A E divided by L. Flexibility is the reciprocal.

Quick example — bar elongation: Steel bar with $P = 50000$ N, $L = 2000$ mm, $A = 500$ mm squared, $E = 200000$ MPa.

Stress is $\sigma = P/A = 100$ MPa, well within elastic range for steel.

Elongation is $\delta = PL/(AE) = 100$ MPa times $L/E$, giving $\delta = 1.0$ mm.

Trap: mixing metres and millimetres breaks the N-mm-MPa set — keep $L$ in mm throughout.

6.2.2 Stepped Bars in Series and Parallel Bars with Compatibility

A stepped bar is a series chain. Each segment i carries its own internal force P i, length L i, area A i, modulus E i. Equilibrium gives each P i from free-body cuts. Total elongation is the sum over segments of P i L i divided by A i E i. Elongations add because displacements accumulate along the chain. Tensile and compressive segments contribute with opposite signs when a consistent sign convention is adopted.

Parallel bars between rigid end plates share the same end displacement. Compatibility states that elongations of all branches are equal. Equilibrium states that the sum of branch forces equals the external load. Branch force is stiffness times common displacement, so load divides in proportion to A E divided by L of each branch. The stiffer branch attracts more load. Thermal terms are added in the same framework when temperature changes, as developed in 6.3.

6.2.3 Tapered Circular Bar Derived by Integration

Consider a solid circular bar tapering linearly from diameter D at one end to diameter d at the other end over length L. Let x be measured from the large end. Diameter at station x is D minus D minus d times x divided by L. Area A of x is pi divided by four times diameter squared. Local extension is P dx divided by A of x times E. Total elongation is the integral from zero to L of P dx divided by E A of x.

The integral is elementary after substitution u equal to diameter as a linear function of x. The result is delta equal to four P L divided by pi E D d. When D equals d, the expression reduces to P L divided by A E for the uniform bar, since A equals pi D squared divided by four. Narrowing the small end raises elongation strongly because the integrand weights the small-area region most. The formula is valid for a truncated cone with nonzero small diameter. A sharp cone to a point is excluded because the integral diverges at zero area, reflecting singular stress at the tip in the idealized model.

The same integral method gives elongation for any prescribed area variation, including exponential horns and stepped approximations, by direct substitution of A of x.

6.2.4 Hanging Bar Under Self-Weight Derived by Integration

A vertical prismatic bar of length L, cross-sectional area A, mass density rho, hung from the top, carries its own weight. Let x be measured from the free bottom end upward. Internal tensile force at height x equals weight below that station, equal to rho g A x. Local stress is rho g x, zero at the free end and maximum rho g L at the support. Maximum stress therefore equals weight density times length, independent of area for a prismatic bar.

Local extension of slice dx is internal force dx divided by A E, equal to rho g x dx divided by E. Integration from zero to L gives delta equal to rho g L squared divided by two E. Since total weight W equals rho g A L, the same result is W L divided by two A E, exactly half the elongation of the same bar under an end load equal to W. The factor one half reflects the triangular distribution of internal force along the bar.

6.3.1 Free Thermal Expansion and Fully Constrained Stress

Free thermal expansion of an unconstrained bar is delta T equal to alpha times T rise times L, where alpha is the linear expansion coefficient. No stress develops under free expansion. If expansion is fully blocked between rigid walls, a compressive mechanical deformation of equal magnitude must cancel the thermal expansion. Mechanical shortening under compressive force P is P L divided by A E. Equating magnitudes gives P divided by A equal to E alpha T rise. Fully constrained thermal stress is therefore sigma equal to E alpha T rise, compressive for temperature rise and tensile for temperature drop. Yielding or buckling then depends on slenderness and support detail in real hardware.

6.3.2 The Three-Step Compatibility and Equilibrium Method

Composite and thermally loaded systems are solved by a uniform three-step method. First, release the constraints and compute the free change of each part, thermal plus load-induced. Second, impose compatibility of final geometry, meaning parts joined to common plates end at the same position, or a stated gap just closes. Third, impose equilibrium of forces on the freed bodies, meaning internal forces balance with any external load.

The method separates kinematics from statics. Compatibility provides deformation relations among unknown forces. Equilibrium provides force balance. Constitutive relations delta equal to P L divided by A E and delta T equal to alpha T L close the system. The same structure handles bolt plus sleeve assemblies, concrete plus steel columns, shrink fits expressed as interference, and temperature plus preload combinations.

6.3.3 Parallel Composite System Under Temperature Change

Consider steel and copper bars in parallel between rigid plates, no external load, temperature rise T. Free expansions alpha c T L and alpha s T L differ. The common final length lies between the two free lengths. The bar with larger alpha is compressed back from its free position, the bar with smaller alpha is stretched forward. Compatibility equates the two final lengths. Equilibrium with no external load states that the compressive force in one bar equals the tensile force in the other.

With copper as the higher-alpha material, the compatibility statement is alpha c T L minus P c L divided by A c E c equals alpha s T L plus P s L divided by A s E s, with P c equal to P s from equilibrium. Solution gives the common force magnitude, then sigma equals P divided by A in each bar. Signs follow the convention that compression reduces length. When an external load is also present, it is added to the equilibrium statement while compatibility is unchanged in form.

6.4.1 Strain Energy in an Axial Bar Derived from Load Deflection Work

Load a linear elastic bar gradually from zero to P. At intermediate load p with extension x, incremental work is p dx. Since x equals p L divided by A E along the linear path, total work is the integral from zero to P of p L dp divided by A E, equal to P squared L divided by two A E. This stored work is the strain energy U. Equivalent forms are one half P delta and sigma squared divided by two E times volume V, where V equals A L. Strain energy density u equals sigma squared divided by two E for uniaxial stress. The triangular area under the load deflection line is the geometric reading of the factor one half.

Generalization to bending, torsion, and shear uses the same work idea with the appropriate load deflection pair, namely moment with rotation and torque with twist, integrated over the member length.

6.4.2 Resilience, Proof Resilience and Modulus of Resilience

Resilience is strain energy stored at a stated stress. Proof resilience is U at first yield, equal to sigma y squared times V divided by two E. Modulus of resilience is proof resilience per unit volume, equal to sigma y squared divided by two E. A high yield strength raises energy capacity quadratically, while a low modulus raises it inversely. This ordering explains why a high-strength low-modulus alloy can absorb more elastic energy per unit volume than a mild steel of equal volume, even when the steel is stiffer. The result applies to elastic storage only and does not measure fracture toughness, which includes plastic work.

6.4.3 Sudden Load Giving a Factor of Two Derived by Energy Balance

Apply a weight W instantaneously at zero drop height to the end of an elastic bar and allow the system to settle to maximum momentary extension without damping. Loss of potential of the weight is W times delta max. Stored strain energy at delta max is one half k delta max squared, where k equals A E divided by L. Equating loss to storage in the idealized undamped model gives W delta max equal to one half k delta max squared, so delta max equals two W divided by k. Static deflection under W is W divided by k. Maximum sudden deflection and stress are therefore twice the static values. The factor two is the zero-height limit of the general impact expression.

6.4.4 Impact Factor Derived from Drop Height and Static Deflection

Drop a weight W from height h above the stop attached to an elastic bar. At maximum momentary extension delta max, the weight has fallen through h plus delta max. Energy balance without rebound loss gives W times h plus delta max equal to one half k delta max squared. This is a quadratic in delta max. Divide by static deflection delta st equal to W divided by k. The normalized equation is delta max squared divided by two minus delta st delta max minus delta st h equal to zero. Positive root gives delta max equal to delta st times one plus square root of one plus two h divided by delta st. The bracket is the impact factor. Maximum stress follows by linearity, sigma max equal to sigma st times the same factor. When h equals zero, the factor reduces to two, recovering the sudden-load result. When h is much larger than delta st, the factor approaches square root of two h divided by delta st. Consistent length units for h and delta st are essential, commonly mm throughout under the N mm MPa discipline.

Damping, mass of the bar, and stress-wave effects are neglected in this elementary model. The model is conservative for design screening and shows the strong penalty of small static deflection, since a stiff stop amplifies drop energy into high force.

6.5.1 Transformation Equations Derived on an Inclined Wedge

Consider a small element with stresses sigma x on the x face, sigma y on the y face, and shear stress tau xy positive on the positive x face in the positive y direction, with complementary shear on the y face for moment balance. Cut the element by an oblique plane whose outward normal makes angle theta anticlockwise from the x axis. The oblique face has area dA. The x face has area dA cos theta and the y face has area dA sin theta by projection.

Resolve forces normal and tangential to the oblique plane and divide by dA. Normal equilibrium gives sigma theta equal to sigma x cos squared theta plus sigma y sin squared theta plus two tau xy sin theta cos theta. Tangential equilibrium gives tau theta equal to sigma y minus sigma x times sin theta cos theta plus tau xy times cos squared theta minus sin squared theta. Double-angle identities reduce these to the standard transformation equations: sigma theta equals average plus half-difference times cos two theta plus tau xy times sin two theta, and tau theta equals minus half-difference times sin two theta plus tau xy times cos two theta, where average is sigma x plus sigma y divided by two and half-difference is sigma x minus sigma y divided by two.

6.5.2 Principal Planes and Principal Stresses

Principal planes carry zero shear stress. Setting tau theta to zero gives tan two theta p equal to two tau xy divided by sigma x minus sigma y. Two roots separated by 90 degrees in physical angle, 180 degrees on the circle, define the two orthogonal principal planes. Substitution of theta p into sigma theta gives the principal values sigma one and sigma two equal to average plus or minus R, where R is the square root of half-difference squared plus tau xy squared. The sum of principal stresses equals sigma x plus sigma y, an invariant under rotation, and the determinant sigma x sigma y minus tau xy squared is likewise invariant.

6.5.3 Maximum In-Plane Shear and Absolute Maximum Shear

Maximum in-plane shear magnitude equals R and acts on planes at 45 degrees to the principal planes in physical space. Its value is half the difference of the in-plane principals. The associated normal stress on those planes equals the average C. In three dimensions, absolute maximum shear is the largest of the three pairwise half-differences among sigma one, sigma two, sigma three. For plane stress, the out-of-plane principal is zero. When the two in-plane principals have opposite signs, the in-plane maximum is also the absolute maximum. When they share the same sign, the absolute maximum is half the largest absolute principal, involving the out-of-plane zero, and exceeds the in-plane value. This distinction matters for Tresca checks.

6.5.4 Mohr Circle Construction Proved Step by Step

Rearrange the transformation equations by subtracting C from sigma theta. Then sigma theta minus C equals half-difference times cos two theta plus tau xy times sin two theta, and tau theta equals minus half-difference times sin two theta plus tau xy times cos two theta. Squaring and adding eliminates theta and gives sigma theta minus C squared plus tau theta squared equal to half-difference squared plus tau xy squared, which is R squared. The locus as theta varies is therefore a circle in sigma tau space with centre C on the sigma axis and radius R.

Construction proceeds in five linked steps. First, mark centre C equal to sigma x plus sigma y divided by two on the sigma axis. Second, compute radius R as defined above. Third, plot point X at sigma x, tau xy and point Y at sigma y, minus tau xy. The segment XY is a diameter passing through C because the two points are diametrically opposite by construction. Fourth, read sigma one equal to C plus R, sigma two equal to C minus R, and maximum in-plane shear equal to R. Fifth, measure angle two theta p from X toward the sigma axis on the circle. The physical plane rotates by theta p, half the circle angle, in the same rotational sense. Consistency of the sign convention for tau xy and the plotting of Y at minus tau xy is essential and is maintained throughout this chapter.

Quick example — Mohr 100/0 case: State $\sigma_x = 100$ MPa, $\sigma_y = 0$ MPa, $\tau_{xy} = 0$ MPa.

Centre is $C = (\sigma_x + \sigma_y)/2 = 50$ MPa, with radius $R = 50$ MPa.

Principals read $C + R = 100$ MPa and $C - R = 0$ MPa, circle through origin and $100$ MPa mark.

Trap: out-of-plane shear is still $50$ MPa, so Tresca sees more than the in-plane picture shows.

6.6.1 Rankine Maximum Principal Stress Theory

Rankine theory states that failure occurs when the largest absolute principal stress reaches the ultimate strength measured in uniaxial tension or compression. Equivalent stress is the maximum of absolute values of sigma one, sigma two, sigma three. The criterion is sensitive to hydrostatic tension and matches the cleavage-dominated fracture of brittle solids such as cast iron and concrete. It is unsafe for ductile metals because ductile yielding is governed by shear and distortion rather than by one principal value alone.

6.6.2 Tresca Maximum Shear Stress Theory

Tresca theory states that yielding begins when the maximum shear stress reaches the shear yield measured in uniaxial tension. Since shear yield in tension is sigma y divided by two, the condition is sigma max minus sigma min equal to sigma y. Equivalent stress is sigma max minus sigma min. In plane stress with ordered principals including the out-of-plane zero when needed, the expression is evaluated over all three values. The Tresca locus in the sigma one sigma two plane is a hexagon inscribed in the von Mises ellipse. Tresca predicts yielding at or before von Mises everywhere, lower by up to about 15.5 percent, and is therefore the conservative ductile choice. For pure shear tau, Tresca predicts yielding at tau equal to sigma y divided by two.

6.6.3 von Mises Distortion Energy Theory

von Mises theory states that yielding begins when the distortion energy density reaches the distortion energy density at yield in uniaxial tension. Hydrostatic energy, which changes volume only, is subtracted from total energy, leaving the part that distorts shape. Equivalent stress in full form is one over root two times the square root of sigma one minus sigma two squared plus sigma two minus sigma three squared plus sigma three minus sigma one squared. In plane stress with sigma three equal to zero, this reduces to the square root of sigma one squared minus sigma one sigma two plus sigma two squared. The locus is an ellipse through the uniaxial yield points. Test data for ductile steels fall closest to this ellipse. For pure shear, von Mises predicts yielding at tau equal to sigma y divided by root three, about 15.5 percent above the Tresca value.

6.6.4 Selection Logic for Ductile Response and Brittle Response

Ductile metals yield by slip, a shear-driven mechanism, so a shear-based or distortion-based criterion is physically appropriate. Use von Mises for best-estimate ductile checks such as shafts under bending plus torsion, and use Tresca when a safe-sided ductile result is required, as in pressure-vessel screening and examination defaults keyed to maximum shear stress theory. Brittle solids fracture by separation on planes of maximum tension with little prior slip, so Rankine maximum principal stress theory is appropriate. Never apply Rankine to ductile design. The design inequality is equivalent stress less than or equal to allowable stress, where allowable equals strength divided by factor of safety, with yield strength for ductile criteria and ultimate strength for brittle criteria.

6.7.1 Differential Relations Derived on a Beam Slice

Consider a short beam slice of length dx carrying distributed transverse load w positive upward per unit length. Let shear V act on the faces and bending moment M act as sagging-positive couples. Vertical equilibrium gives V plus w dx minus V plus dV equal to zero, so dV divided by dx equals w with sign following the chosen upward-positive convention. Many texts take w positive downward, giving dV divided by dx equal to minus w. Either form is correct once the sign of w is fixed. Moment equilibrium about the right face, neglecting second-order terms, gives dM divided by dx equal to V. Differentiation once more gives second derivative of M equal to w in the upward-positive convention. Consequences: shear diagram slope equals load intensity, moment diagram slope equals shear, moment maximum or minimum occurs where shear crosses zero, point load causes a jump in shear and a kink in moment, and a concentrated couple causes a jump in moment.

6.7.2 Sign Conventions and Diagram Reading Rules

Shear is positive when the left segment is pushed up relative to the right segment. Moment is positive when the beam sags concave upward, compression on the top fibres. Shear force diagram starts from the left reaction, moves linearly under uniform load, jumps at point loads, and returns to close at the right reaction for overall equilibrium. Bending moment diagram starts from the end-moment value, rises or falls with slope equal to local shear, curves parabolically under uniform load, peaks where shear is zero, and shows a kink under a point load. Point of contraflexure is the station where moment crosses zero and curvature changes sign. It does not occur in simple cantilever or simply supported beams under the elementary loadings below, and appears in overhanging and indeterminate beams.

6.7.3 Standard Beam Cases Explained

Cantilever with end load P has constant shear magnitude P along the span and triangular moment rising linearly to P L at the wall, hogging in the fixed-end sense. Cantilever with full uniform load w has linear shear from zero at the free end to w L at the wall and parabolic moment reaching w L squared divided by two at the wall. Simply supported beam with central load P has opposite rectangular shear blocks plus P over two and minus P over two, with triangular moment peaking at P L over four at midspan. Simply supported beam with full uniform load w has linear shear from plus w L over two to minus w L over two, zero at midspan, with parabolic moment peaking at w L squared over eight at midspan. Simply supported beam with eccentric load P at distance a from the left and b from the right has reactions P b over L and P a over L, rectangular shear blocks of those magnitudes, and peak moment P a b over L under the load. These five patterns are the building blocks for superposition and for rapid checking of computer output.

6.8.1 Flexure Formula Derived from Kinematics and Equilibrium

Classical bending rests on the hypothesis that plane cross sections remain plane and perpendicular to the deformed neutral surface for a long prismatic beam in pure bending. A fibre at distance y from the neutral axis lies on an arc of radius rho minus or plus y, where rho is the radius of curvature of the neutral surface. Its strain is minus y divided by rho, linear through the depth, zero at the neutral axis. Hooke law gives sigma equal to minus E y divided by rho.

Axial equilibrium over the section requires the integral of sigma dA to vanish, which places the neutral axis through the centroid for a homogeneous section. Moment equilibrium requires the integral of minus sigma y dA to equal applied moment M. Substitution gives M equal to E divided by rho times the integral of y squared dA. The integral is the second moment of area I about the neutral axis. Curvature is therefore M divided by E I. Elimination of rho gives the flexure formula sigma equal to M y divided by I. Maximum stress at outer fibre y max is M divided by section modulus Z, where Z equals I divided by y max. Compression on the sagging top side and tension on the bottom side follow the sagging-positive convention.

6.8.2 Second Moment and Section Modulus Table with Reasoning

Bending uses I about the horizontal centroidal axis parallel to the neutral axis, never the polar moment. Parallel-axis theorem shifts tabulated centroidal values to any offset axis. Deeper sections gain I with the cube of depth, which is the structural reason for I-beams placing material in flanges far from the axis.

Section geometry	Second moment I about neutral axis	Section modulus Z equal to I over y max
Solid rectangle width b, depth d	b d cubed over 12	b d squared over 6
Solid circle diameter d	pi d to the fourth over 64	pi d cubed over 32
Hollow circle outer D, inner d	pi times D to the fourth minus d to the fourth over 64	pi times D to the fourth minus d to the fourth over 32 D
Hollow rectangle outer B by D, inner b by d	B D cubed minus b d cubed over 12	B D cubed minus b d cubed over 6 D
Triangle base b, height h, centroidal axis parallel to base	b h cubed over 36 about centroidal axis, b h cubed over 12 about apex-parallel axis through centroid in alternate tabulation, b h cubed over 4 about base itself in elementary statement	Centroidal value divided by distance to extreme fibre
Built-up I-section, total depth H	Flange Steiner terms plus web term, approximately two times flange area times H over two squared plus web thickness times web height cubed over 12	I divided by H over two

The triangle row is included to show the strong axis-shift effect. The I-section row shows that flange area times the square of half-depth dominates, which is the reason an I-section carries far more moment per unit mass than a solid rectangle of equal area.

6.8.3 Transverse Shear Stress Formula Derived from Horizontal Equilibrium

Transverse shear force produces horizontal shear flow between adjacent layers. Isolate a top slice of the beam cut at level y. Bending stresses on the two ends differ because moment changes by dM over length dx. Net axial imbalance on the slice must be balanced by horizontal shear on the cut plane. Summation gives shear flow q equal to V Q divided by I, where Q is the first moment of the area above the cut about the neutral axis. Dividing by width b at that level gives tau equal to V Q divided by I b. Shear stress is complementary, so the same magnitude acts vertically on the cross section. The distribution is parabolic in character for solid sections, zero at free top and bottom surfaces where Q vanishes, and maximum generally at the neutral axis where Q is largest.

6.8.4 Rectangular Maximum 1.5 Times Average and Circular Maximum 4 over 3 Times Average

For a solid rectangle b by d, Q at distance y from the neutral axis is b over two times d squared over four minus y squared. Substitution gives tau equal to six V divided by b d cubed times d squared over four minus y squared. At y equal to zero, tau max equals three V divided by two A, which is 1.5 times average shear V divided by A. At top and bottom, tau is zero.

For a solid circle, integration of Q over circular chords gives a similar parabolic-like profile with maximum at the neutral axis equal to four V divided by three A, which is 4 over 3 times the average. The derivation uses chord width varying with y and the centroid of the circular segment above the cut. Thin-walled and wide-flange sections concentrate shear in the web, with flange shear low, a direct consequence of Q divided by b at each level.

6.9.1 Moment Curvature Relation and the Integration Idea

For small slopes, curvature is approximated by second derivative of transverse deflection v with respect to x. The moment curvature relation is E I d squared v divided by dx squared equal to M of x, with sign following the sagging-positive and upward-positive conventions adopted here. Integration once gives slope, integration again gives deflection, with two constants fixed by supports. The method is exact within Euler Bernoulli assumptions of slender beams, linear elasticity, and negligible shear deformation. Shear deformation is added separately for deep beams and is outside the elementary table.

6.9.2 Boundary Condition Logic at Supports and Symmetry Points

A simple pin support allows rotation and enforces zero deflection. A roller support likewise enforces zero deflection with free rotation. A fixed wall enforces zero deflection and zero slope. A free end enforces zero moment and zero shear, which translate to zero second and third derivatives in the integration scheme. Symmetry at midspan of a symmetric load enforces zero slope. An internal hinge enforces zero moment. Each span and each load region needs its own M of x expression, with continuity of deflection and slope at junctions. Counting conditions against integration constants is the systematic check that the solution is determinate.

6.9.3 Eight-Case Deflection Table with Behavioural Notes

E is Young modulus, I is bending I from 6.8, x is measured from the stated origin, downward deflection is reported as magnitude. Fixed-fixed is also termed clamped-clamped.

Number	Beam and loading	Maximum deflection magnitude	Location and logic
1	Cantilever, end load P	P L cubed over 3 E I	Free end, slope P L squared over 2 E I at the same end, stiffness 3 E I over L cubed
2	Cantilever, full UDL w	w L to the fourth over 8 E I	Free end, fixed wall carries w L force and w L squared over two moment
3	Cantilever, end moment M	M L squared over 2 E I	Free end, uniform curvature M over E I along the span
4	Simply supported, central load P	P L cubed over 48 E I	Midspan by symmetry, end slope P L squared over 16 E I
5	Simply supported, full UDL w	5 w L to the fourth over 384 E I	Midspan by symmetry, end slope w L cubed over 24 E I
6	Simply supported, equal end moments M in pure bending	M L squared over 8 E I	Midspan, circular arc under uniform moment
7	Fixed-fixed, central load P	P L cubed over 192 E I	Midspan, four times stiffer than the simply supported counterpart in case 4 because end fixity halves effective span in curvature terms
8	Fixed-fixed, full UDL w	w L to the fourth over 384 E I	Midspan, five times stiffer than case 5 in the tabulated ratio

Stiffness k equal to P over delta follows directly for load cases, for instance 48 E I over L cubed for case 4. Superposition adds deflections of elementary loadings when the system stays linear. Consistent N mm MPa units give delta in mm directly.

6.10.1 Torsion Formula Derived from Shear Strain and Equilibrium

A circular shaft under torque T twists with uniform rate theta per unit length along the axis for a prismatic segment. A generator at radius r rotates through arc r times twist per unit length, so shear strain gamma equals r times twist rate. Hooke law gives tau equal to G r times twist rate, linear with radius, zero at the centre. Moment equilibrium over the section requires T equal to the integral of tau r dA, equal to G times twist rate times the integral of r squared dA. The integral is the polar second moment J. Hence twist rate equals T divided by G J and tau equals T r divided by J. Maximum shear at outer radius R is T R divided by J. Total angle over length L is T L divided by G J.

For a solid shaft of diameter d, J equals pi d to the fourth over 32, giving tau max equal to 16 T divided by pi d cubed and twist equal to 32 T L divided by G pi d to the fourth. For a hollow shaft with outer D and inner d, J equals pi times D to the fourth minus d to the fourth over 32, with tau max evaluated at D over two and twist using the same J.

Quick example — shaft stress via 16T over pi d cubed: Solid shaft with $T = 1000000$ N mm, $d = 50$ mm.

Section term is $\pi d^3 = \pi \times 125000 \approx 392699$ mm cubed.

Shear is $\tau_{max} = 16T/(\pi d^3) \approx 40.7$ MPa.

Trap: the $16T$ form is solid-shaft only — hollow shafts need $J$ from $D^4 - d^4$.

6.10.2 Hollow Shaft Advantage Proved for Equal Weight

Equal weight means equal cross-sectional area and equal length, hence equal material volume. Let a solid shaft of diameter d s have the same area as a hollow shaft with outer D and inner d and ratio k equal to d divided by D. Area equality gives d s squared equal to D squared times one minus k squared. Torque capacity at fixed allowable shear is proportional to J divided by outer radius. Ratio of hollow capacity to solid capacity is one plus k squared divided by the square root of one minus k squared, which exceeds unity for all nonzero k and grows as the wall thins within thin-wall validity. Twist per unit torque is inversely proportional to J, and J ratio hollow to solid is one plus k squared divided by one minus k squared, likewise above unity. A hollow shaft therefore carries more torque and twists less per unit mass. This is the reason drive and pump shafts use hollow sections when weight is critical, subject to local buckling and manufacturing limits for very thin walls.

6.10.3 Power Transmission Relation

Power transmitted at steady rotation is torque times angular velocity. With N in revolutions per minute and T in N m, angular velocity is two pi N divided by 60 radians per second, so P in watts equals two pi N T divided by 60. In N mm MPa work, convert torque to N m before applying the power formula or carry the factor of 1000 explicitly. Higher speed transmits the same power at lower torque and lower shaft stress, which is the reason high-speed shafts are smaller for equal power.

6.10.4 Combined Bending and Torsion in Shaft Design

Outer-fibre material of a shaft under bending moment M and torque T carries normal stress sigma equal to 32 M divided by pi d cubed and shear stress tau equal to 16 T divided by pi d cubed for a solid shaft. Principal stresses follow from the plane-stress formulas in 6.5. Tresca equivalent torque is T e equal to the square root of M squared plus T squared, with equivalent moment M e equal to M plus T e divided by two. von Mises equivalent moment uses the square root of M squared plus 0.75 T squared. Ductile shaft sizing uses one of these equivalents against allowable stress, with Tresca as the safe-sided route and von Mises as the best-estimate route per 6.6.

6.11.1 Stiffness and Deflection from Torsion plus Direct Shear

A close-coiled helical spring with mean coil diameter D, wire diameter d, active turns n, under axial load P, loads the wire mainly in torsion with a small direct shear. Torque on the wire section is P D over two. Angle of twist of the wire over developed length pi D n is torque times length divided by G times polar moment of the wire. Axial deflection follows from twist times lever arm D over two. The result is delta equal to eight P D cubed n divided by G d to the fourth. Stiffness k equal to P over delta is G d to the fourth divided by eight D cubed n. More turns or larger coil diameter soften the spring strongly, while thicker wire stiffens it with the fourth power. Stored energy is P squared divided by two k, mirroring the axial-bar form.

6.11.2 Wahl Correction Meaning Curvature plus Direct Shear

Nominal shear stress from torsion alone is eight P D divided by pi d cubed. Direct transverse shear adds P divided by wire area, and curvature concentrates stress on the inner side of the coil. Wahl factor K combines both effects into a single multiplier on the nominal stress, giving tau equal to eight P D K divided by pi d cubed. Spring index C equals D divided by d. Wahl expression is K equal to four C minus one divided by four C minus four plus 0.615 divided by C. As C grows large, the coil straightens toward a straight torsion bar and K approaches unity. Low C means a tight coil with strong curvature correction. Design keeps C in a moderate range to limit stress concentration while avoiding buckling and surging in long springs.

6.11.3 Series and Parallel Combinations and Energy Storage

Springs in series carry the same force. Total deflection is the sum of individual deflections, so reciprocal stiffness adds: one over k eq equals the sum of one over k i. The series set is softer than its softest member. Springs in parallel share the same deflection. Total force is the sum of individual forces, so stiffness adds: k eq equals the sum of k i. The parallel set is stiffer than its stiffest member. Energy stored at load P is P squared divided by two k eq, or one half k eq delta squared. Damping and surge are neglected in the static stiffness model and are treated separately in vibration analysis.

6.12.1 Hoop Stress Derived by Splitting the Cylinder

Consider a thin cylindrical shell with inner diameter d, wall thickness t, internal pressure p, length L, closed ends, with t over d at or below 0.05. Cut the cylinder by a longitudinal diametral plane. Pressure acts on the projected rectangular area d L, giving separating force p d L. Wall tension on the two cut edges gives resisting force two times sigma h t L. Equilibrium gives sigma h equal to p d divided by two t. This is the hoop or circumferential stress, tensile, acting around the circumference. Thin-wall uniformity through the thickness is assumed, valid only when the wall is thin enough that the gradient across t is negligible.

6.12.2 Longitudinal Stress Derived by Capping the Ends

Cut the cylinder transversely. Pressure acts on the circular bore area pi d squared over four, giving axial separating force. Wall tension acts on the annular metal area approximately pi d t for thin walls, giving resisting force sigma l pi d t. Equilibrium gives sigma l equal to p d divided by four t. Longitudinal stress is therefore exactly half the hoop stress for closed ends. A thin vessel reaches yield or bursts along the longitudinal seam first because the hoop direction is the more stressed direction. Open-ended pipes and cylinders with free pistons develop negligible longitudinal membrane stress from pressure alone.

Quick example — hoop versus longitudinal: Thin cylinder with $p = 2$ MPa, $d = 500$ mm, $t = 10$ mm.

Hoop stress is $\sigma_h = pd/(2t) = 50$ MPa, and longitudinal stress is $\sigma_l = pd/(4t) = 25$ MPa.

Hoop controls at twice longitudinal, so the longitudinal seam carries the higher demand.

Trap: thin formulas apply only for $t/d \le 0.05$, met here with $t/d = 0.02$.

6.12.3 Spherical Vessels and Biaxial Strain Consequences

A thin sphere under internal pressure has equal membrane stress in all tangent directions by symmetry. Cutting by any diametral plane and balancing pressure on the circular area against wall tension on the circumference gives sigma equal to p d divided by four t. Diameter change and volume change follow from biaxial Hooke law with sigma one equal to sigma two in the tangent plane. The sphere is the most efficient pressure shape per unit mass because the maximum stress is half the cylindrical hoop value for equal diameter, thickness, and pressure.

6.12.4 Thick Cylinders and the Lame Statement

When thickness ratio t over d exceeds 0.05, the thin formulas underpredict the peak stress at the bore and must not be used. Lame analysis for a thick cylinder with inner radius R i and outer radius R o under internal pressure p i and external pressure p o states that radial stress sigma r and hoop stress sigma h vary through the wall as sigma r equal to A minus B over r squared and sigma h equal to A plus B over r squared, where r is the local radius and A and B are constants fixed by sigma r equal to minus p i at r equal to R i and sigma r equal to minus p o at r equal to R o. Compressive radial stress is negative in the elasticity sign convention. Hoop stress is maximum at the inner surface. Maximum shear and von Mises equivalent are then evaluated at the bore from the local triaxial set including the longitudinal stress, which for closed ends follows from axial equilibrium over the annulus. The thin formulas emerge as the limiting case when the wall becomes thin relative to the radius.

6.13.1 Euler Critical Load Derived from the Bent Equilibrium Equation

A long straight column under concentric axial load P can remain straight or take a slightly bent shape. Bending moment at deflection v is P v. The moment curvature equation is E I d squared v divided by dx squared plus P v equal to zero for the buckled mode. General solution is sinusoidal with wave number root P over E I. Pinned ends enforce zero deflection at both ends, giving the lowest nontrivial mode as a half sine wave. The eigenvalue condition gives P cr equal to pi squared E I divided by L squared for pinned ends. Higher modes are integer multiples and are not reached before the fundamental mode in ideal loading. The derivation assumes linear elasticity, small deflections, straightness, concentric load, and slenderness sufficient that yielding does not precede bifurcation.

6.13.2 Four End Conditions Explained Through Effective Length

End restraint changes the buckled wavelength. Effective length L e is the half-wave length of the mode, equal to actual length times an effective factor. Both ends pinned allow rotation and give L e equal to L, the reference case. Both ends fixed prevent rotation and inflection points form at quarter points, giving L e equal to L over two and critical load four times the pinned value. One end fixed and one end free behaves as a quarter wave with the free end as an antinode, giving L e equal to two L and critical load one quarter of the pinned value. One end fixed and one end pinned gives L e about 0.7 L, more precisely L over root two in the elementary Euler statement, with critical load about twice the pinned value. Fixity raises capacity strongly because critical load varies with the inverse square of effective length. Real supports supply partial restraint between these ideals, and design codes account for that spread.

Quick example — Euler end effect times four: Same column with $E = 200000$ MPa, $I = 100000$ mm to fourth, $L = 2000$ mm.

Pinned load is $P_{cr} = \pi^2EI/L^2 \approx 49.3$ kN.

Fixed-fixed halves effective length, so $P_{cr} = \pi^2EI/(L/2)^2 \approx 197.4$ kN, four times the pinned value.

Trap: real end fixity is partial, so the ideal factor four is an upper bound, never a bonus to assume freely.

6.13.3 Slenderness Ratio and the Validity Limit of Euler Theory

Slenderness lambda equals L e divided by radius of gyration k, where k equals the square root of I divided by A. Minimum k among principal axes governs because buckling selects the weakest plane. Euler critical stress is pi squared E divided by lambda squared. Validity requires this stress to lie below yield, giving lambda above critical slenderness lambda c equal to the square root of two pi squared E divided by sigma y, about 100 for structural steel. Short columns with low lambda crush at sigma y A before bifurcation. Intermediate columns yield after partial bending magnification and follow empirical transition curves. Cross-section efficiency for buckling favours large k per unit area, which is the reason hollow circular and wide-flange struts outperform solid bars of equal mass.

6.13.4 Rankine Formula for Intermediate Columns and the Secant Idea

Rankine combination bridges crushing and Euler buckling with reciprocal addition: one over P R equals one over P E plus one over P C, where P E is the Euler load and P C equals sigma y A. For very short columns, P E is large and P R tends to P C. For very long columns, P C is large relative to P E and P R tends to P E. Intermediate lengths interpolate smoothly and match test trends for design use.

Eccentric loading adds bending from the start, so bifurcation is replaced by magnified bending stress. The secant formula expresses maximum stress as direct stress P over A multiplied by one plus eccentricity times c over k squared times secant of L e over two k times the square root of P over E A, where c is the outer-fibre distance. The secant term grows rapidly as P approaches the Euler load, showing that small eccentricity strongly reduces capacity in slender struts. Code design for eccentrically loaded columns uses this magnification route with allowable-stress or limit-state margins.

• Linear elasticity is governed by E, G, K, and nu with only two independent constants for isotropic response. The interrelations E equals 2 G times 1 plus nu, E equals 3 K times 1 minus 2 nu, and E equals 9 K G over 3 K plus G follow from superposition of uniaxial and hydrostatic states. Volumetric strain is the sum of normal strains.
• Axial elongation follows from integration of P dx over A E. Uniform bars give P L over A E, series bars sum segment contributions, parallel bars share displacement with load split by A E over L, tapered circular bars give 4 P L over pi E D d, and self-weight gives rho g L squared over 2 E with peak stress rho g L at the support.
• Thermal and composite response uses the three-step method of free change, compatibility of final geometry, and equilibrium of forces. Fully blocked expansion gives sigma equal to E alpha T. Parallel dissimilar bars divide the mismatch into compression in the high-alpha branch and tension in the low-alpha branch.
• Strain energy in linear response is one half load times deflection, equal to sigma squared over 2 E times volume in uniaxial form. Proof resilience is sigma y squared V over 2 E. Sudden axial load doubles static response. Drop from height h multiplies static response by 1 plus the square root of 1 plus 2 h over delta st, derived from W times h plus delta max equated to stored energy.
• Plane-stress transformation follows from wedge equilibrium and double-angle reduction. Principal stresses are average plus or minus R. Maximum in-plane shear is R at 45 degrees to the principal planes. Absolute maximum shear includes the out-of-plane zero and exceeds the in-plane value when the in-plane principals share the same sign. Mohr construction is the circle identity sigma theta minus C squared plus tau squared equal to R squared, with X at sigma x comma tau xy and Y at sigma y comma minus tau xy.
• Ductile yielding is checked with von Mises distortion energy as best estimate and Tresca maximum shear as the safe-sided alternative inscribed as a hexagon in the von Mises ellipse. Brittle fracture is checked with Rankine maximum principal stress. Pure shear yields at sigma y over 2 by Tresca and sigma y over root three by von Mises.
• Beam internal resultants obey dV over dx equal to load intensity and dM over dx equal to shear, giving linear shear and parabolic moment under uniform load, jumps at point loads, kinks in moment under point loads, and peak moment where shear vanishes. The five standard cantilever and simply supported cases give wall moment P L or w L squared over 2 and midspan moment P L over 4 or w L squared over 8 as applicable.
• Flexure stress is M y over I with maximum M over Z, derived from linear strain through the depth plus axial and moment equilibrium. Transverse shear is V Q over I b, giving 1.5 times average at the neutral axis for a solid rectangle and 4 over 3 times average for a solid circle. I and Z for rectangle, circle, hollow sections, triangle with axis-shift care, and built-up I-sections follow the tabulated forms.
• Deflection follows from E I second derivative equal to M of x with support logic of zero deflection at pins and rollers, zero deflection and slope at walls, zero moment and shear at free ends, and zero slope at symmetry. The eight-case table spans cantilever, simply supported, and fixed-fixed loadings with midspan or free-end maxima and direct stiffness readings such as 3 E I over L cubed for the end-loaded cantilever.
• Torsion gives tau equal to T r over J with twist T L over G J, derived from linear shear strain with radius plus moment equilibrium defining J. Solid-shaft maxima use 16 T over pi d cubed. Equal-weight hollow shafts carry more torque and twist less, proved by area-constrained ratios exceeding unity. Power in watts is 2 pi N T over 60 with T in N m. Combined bending plus torsion uses Tresca root M squared plus T squared or von Mises root M squared plus 0.75 T squared.
• Close-coiled springs give deflection 8 P D cubed n over G d to the fourth and stiffness G d to the fourth over 8 D cubed n. Wahl factor K on 8 P D over pi d cubed captures direct shear plus curvature, approaching unity at large spring index C. Series springs add reciprocals of stiffness, parallel springs add stiffnesses, with energy P squared over 2 k.
• Thin cylinders with t over d at or below 0.05 give hoop p d over 2 t by diametral equilibrium and longitudinal p d over 4 t by cap equilibrium, so hoop controls. Thin spheres give p d over 4 t in all tangent directions. Thicker walls use Lame radial form A minus B over r squared and hoop form A plus B over r squared with constants from bore and outer pressure conditions, checked at the inner surface.
• Euler buckling gives P cr equal to pi squared E I over L e squared from the bent equilibrium equation, with L e equal to L pinned-pinned, L over 2 fixed-fixed, 2 L fixed-free, and about 0.7 L fixed-pinned. Validity needs slenderness above root 2 pi squared E over sigma y. Rankine reciprocal addition bridges crushing and Euler loads, and the secant expression adds eccentricity magnification diverging near the Euler load.