Mechanics of MaterialsStudy Guide

Force alone tells you nothing about whether a part will survive: a 50 kN pull is nothing to a bridge chord and fatal to a coat hanger. What matters is the force spread over the material actually resisting it, and that intensity is stress. Claude-Louis Navier formalised it in his 1826 Leçons, the book that turned the craft of building into the discipline of strength of materials. Work an example: a 50 kN load on a rod of 500 mm² cross-section gives σ = 50 000 N ÷ 0.0005 m² = 100 000 000 Pa = 100 MPa, comfortably inside structural steel's 250 MPa yield.

The classic trap is using the gross area when a hole has been drilled through it. A 1/2-inch bolt hole through a 3-inch by 1/4-inch bar removes a sixth of the section, and the stress at that net section is a sixth higher: this is why steel design codes make you check gross yielding and net-section rupture separately. Threaded rod is the same story: a 1/2-13 rod has 0.196 in² of shank but only 0.1419 in² of tensile stress area, and the threads are where it breaks.

Strain is stretch per unit of original length, so it carries no units at all: a 2 mm elongation over a 4 m member is ε = 0.002 ÷ 4 = 0.0005, or 500 microstrain. Engineers quote microstrain (µε, parts per million) because real elastic strains are tiny: steel yields at roughly 1200 µε, barely a millimetre per metre. Measuring numbers that small was impossible until 1938, when Edward Simmons at Caltech and Arthur Ruge at MIT independently glued fine wire to a specimen and watched its electrical resistance change: the bonded foil strain gauge, still the backbone of every load cell and truck scale today.

The trap is the denominator: strain is always referred to the original length, not the stretched one. Divide by the final length and you have computed "true strain", which differs from engineering strain by less than a tenth of a percent in the elastic range but diverges wildly once a tensile specimen starts necking. Keep both lengths in the same unit and the ratio takes care of itself.

This is Hooke's law promoted from a particular spring to a whole material: stress and strain rise together in fixed proportion, and the constant of proportionality is the stiffness of the substance itself. Thomas Young presented it in his 1807 lecture series, though Leonhard Euler had the idea in 1727 and Giordano Riccati measured it in 1782, and Young's own prose was so impenetrable that his publisher's reviewers complained they could not follow it. Steel comes in at E = 200 GPa (29 000 ksi), aluminium at 69 GPa (10 000 ksi), concrete near 30 GPa, wood along the grain around 11 GPa. Apply 200 MPa to steel and you get ε = 200 ÷ 200 000 = 0.001, one millimetre per metre.

The trap that surprises everyone new to the trade: all structural steels have the same E. A992 grade-50 steel is no stiffer than plain A36: it is only stronger. Upgrading grade buys you more capacity before yield, never less deflection, so a bouncy floor is never fixed by a better alloy, only by a deeper section. And E only describes the straight part of the curve; past the proportional limit the material yields and this formula quietly stops being true.

Chain σ = P/A, ε = δ/L and E = σ/ε together and everything collapses into one line: δ = PL/AE. The product AE is the axial stiffness of the member, and dividing it by L gives the spring constant of a rod: a structural member really is just a very stiff spring. A 3 m steel hanger rod of 1000 mm² carrying 100 kN stretches δ = (100 000 × 3) ÷ (0.001 × 200 × 10⁹) = 0.0015 m, one and a half millimetres.

Two practical consequences. First, length is linear, so the elevator ropes at the bottom of a 400 m shaft stretch dramatically more than the same ropes near the top: the reason high-rise lifts need compensating ropes and re-levelling. Second, in a run of pipe supported by rods of different lengths, the long rods stretch more under the same load and quietly shed their share onto the short ones; in a redundant hanger array the load distributes by stiffness, not by fairness. If several segments are in series with different A, E or P, compute δ for each and add them.

Where normal stress pulls a section apart, shear stress slides it sideways: the scissors action that cuts a bolt or tears a weld. The formula assumes the stress is spread evenly over the sheared area, which is a fiction (in a round bar under transverse shear the true peak is about a third higher at the centre), but it is the fiction every bolt code is calibrated to. A bolt with 200 mm² of shank carrying 20 kN across a splice sees τ = 20 000 ÷ 0.0002 = 100 MPa.

The detail that gets missed is counting shear planes. A bolt in a lap joint has one plane; the same bolt in a double-shear butt splice has two, and doubling the area halves the stress, so use $A = n \times A_{\mathrm{bolt}}$. The consequences of getting connections wrong are not academic: the Hyatt Regency walkway collapse in Kansas City in 1981 killed 114 people because a shop-drawing change doubled the load on a single rod-and-washer connection, and it tore straight through the box beam. The joint, not the beam, is almost always the weakest link.

Shear strain γ is not a stretch but a change of angle: push the top of a block sideways by Δx while its height stays h, and γ = Δx/h, the small angle in radians by which the originally square corner has been racked out of true. The shear modulus is the stress needed per unit of that racking. Charles-Augustin de Coulomb measured it first, in the 1784 torsion-balance experiments that also gave him the law of electrostatics: he twisted fine wires and timed their oscillations. Steel runs G ≈ 77 GPa (11 600 ksi), aluminium ≈ 26 GPa. A shear stress of 60 MPa producing γ = 0.0008 rad implies G = 60 ÷ 0.0008 = 75 000 MPa = 75 GPa.

The trap is treating G as an independent property to look up separately: for an isotropic material it is locked to E and Poisson's ratio by E = 2G(1 + ν), so with ν ≈ 0.3 you always get G ≈ 0.385 E. Enter γ as a plain ratio (radians), not degrees: a shear strain of 0.001 is a milliradian, about 0.057°, and mixing the two inflates G by a factor of 57.

Stretch a bar and it gets thinner; squeeze it and it bulges. Poisson's ratio is the bookkeeping for that sideways response: this calculator uses magnitudes, so a positive ν means the usual behaviour (the formal definition carries a minus sign, ν = −ε_lat/ε_ax). Pull steel to an axial strain of 0.0003 and each transverse dimension shrinks by 0.00009, giving ν = 0.30. Siméon Denis Poisson derived a universal value of exactly 1/4 in 1829 from a molecular model of matter; the model was wrong and the measurements soon proved it, but the ratio kept his name.

Thermodynamics caps isotropic materials at ν = 0.5, the incompressible limit that rubber nearly reaches, which is why a rubber block confined in a steel cavity behaves like a hydraulic fluid and will burst its container rather than squash. At the other end, cork sits near zero, which is exactly why a cork pushes into a bottle neck without fattening while a rubber bung fights you. Engineered auxetic foams even manage negative values, getting fatter when stretched. The practical trap: ν only applies in the elastic range; once a metal yields, plastic flow conserves volume and the effective ratio jumps to 0.5.

An isotropic material (one with no grain direction, which covers most metals, glass and concrete well enough) has only two independent elastic constants. Pick any two of E, G, ν and K, and the rest follow. This particular identity falls out of resolving a pure shear into equal tension and compression on planes at 45°, and it is the fastest sanity check in the business: steel's E = 200 GPa with ν = 0.30 demands G = 200 ÷ (2 × 1.3) = 76.9 GPa, which is what handbooks list (77 GPa).

Whether elasticity needed two constants or one was a genuine nineteenth-century brawl. Navier, Poisson and Cauchy's early "rari-constant" molecular theory insisted a single constant sufficed and forced ν = 1/4 for everything; Green and Stokes argued for two on energy grounds. Careful measurement settled it: real materials refuse to sit at ν = 0.25. The trap: this relation applies only to isotropic materials. Wood, fibre composites and rolled sheet with strong texture have direction-dependent moduli, and plugging one axis's E into this formula will give you a shear modulus that is off by a factor of ten.

A factor of safety is the ratio between the stress that breaks the material and the stress you allow it to see in service. Structural steel with a 400 MPa ultimate strength worked at 160 MPa carries FS = 400 ÷ 160 = 2.5. American steel practice long used FS = 1.67 on yield for tension members, which is where the familiar 0.6F_y allowable comes from: 58 ksi ÷ 1.67 ≈ 34.7 ksi for A36. Elevator suspension ropes are held to 10 or 12, pressure-vessel plate to about 3.5, aircraft structure to a lean 1.5, because on an aircraft, every extra kilogram of margin is a kilogram not carried.

Old engineers called it the "factor of ignorance", and the name is honest: it covers material scatter, corrosion, fabrication tolerance, overload and the sheer possibility that the analysis missed a load path. Two traps. First, always state what the numerator is: a factor of 2 on ultimate is a far thinner margin than a factor of 2 on yield, since steel's ultimate is well above its yield. Second, a factor of safety is not a licence to overload: it is consumed by the things you did not model, and a rope rated 5:1 that is jerked, shock-loaded and worn has already spent most of it.

A free member heated by ΔT would stretch by strain αΔT and feel nothing. Clamp both ends and it cannot move, so the material must carry exactly the stress needed to undo that strain: σ = EαΔT. Notice what is missing: length and area do not appear. A 3 m pipe spool and a 300 m pipeline reach the same thermal stress for the same temperature swing, which is the single most counter-intuitive result in the subject. Steel with E = 200 GPa and α = 12 × 10⁻⁶/°C heated 50 °C develops 200 × 10⁹ × 12 × 10⁻⁶ × 50 = 120 MPa, roughly half of A36's yield, from a temperature change any black pipe sees between a winter install and a summer steam-up.

Railways learned this the hard way: continuously welded rail is laid pre-tensioned at a "stress-free temperature" precisely so that summer heat does not build enough compression to throw the track sideways into a sun kink. The same physics puts expansion loops, bellows and slide guides into every hot-water and steam distribution system, and cracks the anchor lugs of any pipe rigidly clamped at both ends. The trap in the arithmetic is unit pairing: α in 1/°F must go with ΔT in Fahrenheit degrees, α in 1/K with kelvin or Celsius degrees. The two scales differ by 9/5 and mixing them is a 44% error.

Torsion is bending's rotational twin, and J plays the role I plays in bending: it measures how far the section's area lies from the shaft's centre rather than from a line through it. By the perpendicular-axis theorem $J = I_x + I_y$, and for a circle those two are equal, so J = 2I = πd⁴/32 exactly. A 50 mm shaft has J = π × 0.05⁴ ÷ 32 = 6.136 × 10⁻⁷ m⁴.

The same fourth-power leverage applies: go from a 50 mm to a 60 mm shaft and torsional stiffness rises by (60/50)⁴ = 2.07, more than double for a 20% size step. Hollow shafts win even harder here than in bending, since torsional shear is zero at the centre and maximum at the surface: the core is dead weight, which is why propeller shafts and torque tubes are bored out. Two warnings: J is returned in m⁴ as a plain number (1 in⁴ = 4.162314 × 10⁻⁷ m⁴), and this simple formula holds only for circular sections. Saint-Venant showed in 1855 that non-circular bars warp out of plane when twisted, and a square bar's torsional constant is about 0.844 (a/2)⁴ × 2.25, not its polar moment at all.

Twist a round shaft and every cross-section rotates rigidly relative to its neighbour, so the shear strain, and therefore the shear stress, grows linearly from zero at the centre to a maximum at the outer surface. That is the whole content of τ = Tr/J. Coulomb established the linear torque-twist behaviour experimentally in 1784 while building his torsion balance, and the modern form followed from Navier's generation. A 1 kN·m torque on a 50 mm shaft (J = 6.136 × 10⁻⁷ m⁴, r = 0.025 m) produces τ = 1000 × 0.025 ÷ 6.136 × 10⁻⁷ = 40.7 MPa at the surface.

Because only the surface matters, hollow shafts are the efficient answer, and any surface defect is disproportionately dangerous: a machining groove, a keyway or a sharp fillet at a shoulder is exactly where the stress is already highest, which is why fatigue failures of drive shafts nearly always start at a step or a keyway corner. Enter J in m⁴ and T in newton-metres; also remember that a shaft carrying torque plus bending must be checked on the combined stress, not each separately, and that ductile shafts fail on a plane perpendicular to the axis while brittle ones (cast iron, chalk) break on a 45° helix, following the principal tension.

This is the torsional counterpart of δ = PL/AE: the product JG is the torsional rigidity, and dividing it by L gives the shaft's torsional spring rate. A 2 m length of 50 mm steel shaft (J = 6.136 × 10⁻⁷ m⁴, G = 80 GPa) under 1 kN·m twists φ = 1000 × 2 ÷ (6.136 × 10⁻⁷ × 80 × 10⁹) = 0.0407 rad = 2.33°. The formula returns radians internally; this calculator displays degrees, so watch which one you are reading.

Long shafts are usually governed by twist rather than by stress. The old machine-design rule of thumb is to hold a power transmission shaft to about one degree of twist per twenty diameters of length, because more than that and the driven end lags the driver enough to upset timing, chatter a cutting tool, or set up torsional vibration. Torsional resonance is a real killer in reciprocating engine and compressor drivelines: the crankshaft, flywheel and driven mass form a torsional oscillator whose stiffness is exactly JG/L, and running near its natural frequency has snapped many crankshafts. Enter J in m⁴ as a plain number, and note this holds for circular sections only.

Power is the rate of doing work, and for a rotating shaft that is torque times angular speed:

$$P = T\omega \qquad \Longrightarrow \qquad T = \frac{P}{\omega}$$

with $\omega$ in radians per second. The radian is not decorative: it is what makes the equation dimensionally clean, since $\omega$ is really an angle per unit time and the radian is the angle for which arc length equals radius. From rev/min it is $\omega = 2\pi n/60$.

The 5252

Every North American shop has the rule $T = 5252\,\text{HP}/\text{RPM}$ written on a wall somewhere, and it is exactly this equation with the unit conversions folded in. One horsepower is 33,000 ft·lbf per minute, so $T = 33000\,\text{HP}/(2\pi\,\text{RPM})$, and $33000/2\pi = 5252.11$. Nothing else is going on. A pleasant consequence: at 5,252 rpm a motor's torque in ft·lbf and its power in horsepower are numerically equal, which is why every dynamometer plot of torque and power crosses at that speed.

The design lesson: slow shafts are fat shafts

At constant power, torque is inversely proportional to speed. Put a 10:1 reduction after a motor and the output shaft carries ten times the torque. That is why the low-speed end of any gearbox is always the heavy end, and it is worth doing the arithmetic once to feel the scale of it. Ten kilowatts at 3,000 rpm is about 32 N·m: a 20 mm shaft handles it comfortably. The same 10 kW at 30 rpm is about 3,180 N·m, a hundred times the torque, and since shaft diameter grows as the cube root of torque, the shaft needs to be roughly 4.6 times thicker. Low-speed machinery is heavy not because it is old-fashioned but because torque is what steel has to resist and power is not.

It also explains the direction of most drivetrains. Electric motors and engines are cheap and light at high speed and expensive and heavy at low speed, so almost every machine puts a fast, small prime mover behind a reduction, and pays for the reduction rather than for a slow motor.

Two cautions

Nameplate power is output power at rated speed and rated load. A motor started under load, or stalled, or accelerating a large inertia, produces torque that has nothing to do with this equation: starting torque can be several times full-load torque, and the shaft and coupling have to survive it. That is what a service factor is for.

And this gives the torque, not the stress. What the shaft actually feels also includes bending from the pulleys, gears and couplings hung on it, and because the shaft rotates, that bending stress fully reverses every turn, which puts it into fatigue. See the shaft sizing page.

This is $P = Fv$ rewritten for a rotating shaft: power equals torque times angular velocity, $P = \tau\omega$. Torque alone tells you how hard something is being twisted and says nothing about how fast work is being done; multiply by the rotation rate and you have the rate of energy delivery. It is the single most useful equation in drivetrain work, because torque is what a shaft has to be built to survive and power is what it actually delivers.

An electric motor producing 200 N·m at 3000 rpm: convert first, $\omega = 3000 \times 0.10472 = 314$ rad/s, so $P = 200 \times 314 = 62\,800$ W, about 63 kW or 84 hp. Run the same motor at 1500 rpm at the same torque and it delivers half the power, having done nothing different at the shaft except turn more slowly.

This is what a dyno chart is plotting, and it explains the shape everyone recognises. An engine's torque curve peaks somewhere in the mid range and falls away, yet its power keeps climbing past that point, because $\omega$ is still rising faster than $\tau$ is falling. Power finally peaks where the two rates of change balance. It is also why gearing works: a gearbox trades $\tau$ against $\omega$ at constant power, so a low gear multiplies torque and divides speed, and the product, the useful output, is unchanged apart from losses.

Units are the whole difficulty here, and North American practice hides one conversion inside a magic number. The shop formula $\text{hp} = \tau_{\text{lb·ft}} \times \text{rpm}/5252$ is exactly this equation with the pound-foot, the revolution and the horsepower folded into a single constant. That is why every horsepower and pound-foot curve ever plotted on shared axes crosses at 5252 rpm, not a property of engines, a property of the unit system. In SI the equation needs no constant at all, but it does need $\omega$ in rad/s: feed rpm in directly and the power comes out 9.55 times too high, which is the difference between an 84 hp motor and an 800 hp one. Two further cautions. Torque and power do not peak at the same speed, so a machine specified by its peak torque and a machine specified by its peak power are being described at different operating points, and quoting one at the other's rpm is meaningless. And this is power at the shaft; a motor's electrical input is larger by whatever its efficiency costs, and the output at the far end of a gearbox is smaller again.

The torsion formula for a round shaft says the shear stress at radius $r$ is $\tau = Tr/J$. For a solid round section the polar second moment of area is $J = \pi d^4/32$, and the stress is worst at the surface, where $r = d/2$. Substitute both and the $d$s collapse:

$$\tau_{max} = \frac{T(d/2)}{\pi d^4/32} = \frac{16T}{\pi d^3} \qquad \Longrightarrow \qquad d = \sqrt[3]{\frac{16T}{\pi\tau}}$$

Torsional shear varies linearly from zero on the axis to maximum at the surface, which tells you immediately that the core of a solid shaft is nearly idle. That is the argument for hollow shafts: removing the inner half of the diameter takes away only about 6% of the torsional strength while removing 25% of the mass, and it is why driveshafts and aircraft transmission shafts are tubes.

The cube root is generous

Because diameter enters cubed, torque capacity is extremely sensitive to size, and size is correspondingly insensitive to torque. Eight times the torque needs only twice the diameter. A 10% increase in diameter buys a 33% increase in torque capacity. This is why shafts rarely look as heavy as the loads they carry would suggest, and why "go up one size" is such an effective piece of shop advice.

What this equation leaves out, which is most of it

Almost no real shaft carries steady torsion alone. A shaft with a pulley, a gear or a sprocket on it is also a beam, carrying bending from the belt pull or the tooth loads. And because the shaft turns, a point on its surface moves from tension to compression and back on every revolution: the bending stress is fully reversed, at the shaft speed, and the shaft is in high-cycle fatigue rather than static loading. A shaft at 1,750 rpm accumulates a million cycles in under ten hours.

The classical way to combine them is an equivalent torque. On the maximum-shear-stress (Tresca) theory, $T_e = \sqrt{M^2 + T^2}$, and the same equation is then used with $T_e$ in place of $T$. Modern practice follows a fatigue method instead: the ASME shaft equation, or a Soderberg/Goodman construction with the mean and alternating components treated separately, and with stress-concentration and surface-finish factors applied. Either way, the number from this page is a floor, not an answer.

Two more subtractions

A keyway removes material and, more importantly, concentrates stress at its corners, with a stress concentration factor around 2 to 3 depending on the fillet radius. The common allowance is to size a keyed shaft 5–10% larger than the plain calculation, and to insist on a properly radiused keyway rather than a sharp-cornered one. Every shoulder, snap-ring groove and cross-hole does something similar.

And stiffness can govern instead of strength. A shaft that is strong enough may still wind up too far under load, which upsets gear mesh alignment and timing. The traditional limit is about one degree of twist per 20 diameters of length, from $\phi = TL/(GJ)$. Lateral stiffness matters too: the shaft's first bending critical speed must sit well clear of the running speed, usually by 20% or more.

Choosing the allowable

An allowable shear stress is a design decision, not a material property. It comes from the material's yield or endurance strength divided by a factor of safety, reduced by whatever the fatigue analysis demands, and it is what codes and standards exist to pin down for a given application. This page will use whatever number you give it, and the honesty of the answer is entirely the honesty of that number.

A key is a small rectangular bar sitting half in a slot in the shaft and half in a slot in the hub, and its only job is to stop the two from turning relative to one another. The torque arrives at the shaft surface as a tangential force, $F = 2T/d$, and the key resists it across a shear plane whose area is its width times its length. So:

$$\tau = \frac{F}{wL} = \frac{2T}{d\,w\,L}$$

The shear plane is the one lying flush with the shaft surface, splitting the key into its shaft half and its hub half. That is why the width appears and not the height: the key is being sliced along its length, not crushed flat.

The second check, which is often the one that governs

A key can also fail in crushing (bearing) on its side faces, where the hub pushes against it. The bearing area is the key length times only the half of the height standing proud of the shaft, so for a key sunk half its depth:

$$\sigma_{crush} = \frac{4T}{d\,h\,L}$$

Compare the two. For a square key, $h = w$, so the crushing stress is exactly twice the shear stress. Whether that matters depends on the material: a ductile steel's allowable compressive stress is roughly twice its allowable shear, so for a square key in steel the two checks come out close together and either can govern. For a flat (rectangular) key, where $h < w$, crushing is more likely to be the limit. Check both, always.

The key is meant to break

This is the design intent and it is worth stating plainly: the key is normally specified in a material softer than both the shaft and the hub, so that an overload shears the key. A sheared key is a part you drive out with a brass punch and replace in ten minutes. A twisted shaft or a split hub is a rebuild. Specifying a key harder than the shaft (which happens when someone substitutes whatever bar stock is on the shelf) inverts the intended failure order and turns a cheap failure into an expensive one.

Length has diminishing returns

Doubling a key's length does not double its capacity in practice. The shaft winds up elastically along the key's length, so the end where the torque enters carries far more than its share and the far end carries almost nothing. Beyond roughly 1.5 shaft diameters of key length the extra material contributes very little. When one key is genuinely not enough, the answers are two keys at 180°, a splined hub (which is effectively many small keys cut integrally, and spreads the load properly), or a keyless friction fit: a taper lock, a shrink disc, or a polygon connection.

Standard proportions, and why they exist

Key sizes are standardised against shaft diameter: the width is conventionally about a quarter of the shaft diameter and the height a quarter or a sixth, with the exact series given in the public dimensional standards (ANSI B17.1 and ISO/R 773). Those proportions are not arbitrary: they are chosen so that a key of the standard size, in the standard material, is a little weaker than the shaft it sits in. If your calculation says a standard key is badly overstressed for the torque you are passing, the real conclusion is usually that the shaft is undersized too.

The second moment of area measures how far a section's material sits from the axis it bends about, and because the contribution goes as distance squared, it rewards depth enormously. For a rectangle, I = bh³/12 about the axis through the centroid perpendicular to h. A 50 mm × 100 mm bar bending about its strong axis has I = 0.05 × 0.1³ ÷ 12 = 4.167 × 10⁻⁶ m⁴; turn it flat and h and b swap, giving 1.042 × 10⁻⁶ m⁴, a quarter as stiff. Double the depth alone and stiffness goes up eightfold: the whole reason joists stand on edge and I-beams put their steel in the flanges.

This engine has no m⁴ unit type, so I is entered and returned as a plain number in m⁴: 1 in⁴ = 4.162314 × 10⁻⁷ m⁴ and 1 mm⁴ = 10⁻¹² m⁴. The trap is the cube: get b and h the wrong way round and you are not off by a little, you are off by (h/b)⁴. For a 2 × 8 that is a factor of 16. And this formula is only valid about the centroidal axis; if the rectangle is a flange offset from the neutral axis of a built-up section, you must add the parallel-axis term Ad².

A circle is the same in every direction, so a solid round bar has the same I about every axis through its centre: πd⁴/64. A 100 mm bar gives I = π × 0.1⁴ ÷ 64 = 4.909 × 10⁻⁶ m⁴. The fourth power is brutal in both directions: a 2-inch shaft is sixteen times stiffer in bending than a 1-inch shaft, and a 10% undersize bar has lost a third of its stiffness.

That fourth power is also the argument for hollow sections. Subtracting the inner circle gives I = π(d⁴ − dᵢ⁴)/64, and because the removed core sits close to the neutral axis it was contributing almost nothing: bore a 100 mm bar out to 50 mm and you lose 6% of the stiffness while shedding 25% of the weight. It is why scaffold tube, bicycle frames and drill pipe are all round and hollow. Remember the value here is in m⁴, convert with 1 in⁴ = 4.162314 × 10⁻⁷ m⁴, and do not confuse it with the polar moment J = πd⁴/32, which is exactly twice as large and belongs to torsion, not bending.

Before you can compute anything about a built-up section (its second moment of area, its section modulus, the stress at the top fibre) you have to know where its neutral axis is. For a symmetric shape you know by inspection. For a T, an angle, a channel, a flitch beam or a plate welded to a rolled section, you have to find it, and the way you find it is an area-weighted average: $\bar{y} = \frac{\sum A_i y_i}{\sum A_i}$.

Work a T-section: a 200 × 20 mm flange sitting on a 20 × 180 mm stem, measured from the bottom of the stem. The flange has $A_1 = 4000$ mm² with its own centroid at $y_1 = 190$ mm; the stem has $A_2 = 3600$ mm² at $y_2 = 90$ mm. So $\bar{y} = (4000 \times 190 + 3600 \times 90)/7600 = 1\,084\,000/7600 = 142.6$ mm from the bottom. The neutral axis sits well up in the stem, nowhere near mid-depth, which is exactly why a T-section is so much stronger one way up than the other.

Three things go wrong with this calculation, and none of them announces itself. The datum has to be the same for every part: mixing "from the bottom" for one piece and "from the top" for another produces a perfectly plausible wrong answer. The $y_i$ are the parts' own centroids, not their edges. And more than two parts is handled by applying this twice: combine any two, treat the result as a single part of area $A_1 + A_2$ at $\bar{y}$, then bring in the third. Cut-outs enter as negative areas in the general method; here, subtract them by hand and keep the datum straight. Once $\bar{y}$ is right, the parallel axis theorem measures every transfer distance from it, and an error here propagates squared into everything downstream.

A section's second moment of area is only ever quoted about one particular axis, and the moment you build a shape out of parts, most of those parts are nowhere near the axis the whole thing bends about. The transfer term fixes that in one line: move an area A a distance d away from its own centroid and its contribution grows by Ad². Take a 200 × 20 mm flange plate sitting 240 mm above the neutral axis of a girder. Its own $I_c = bh^3/12 = 133\,333$ mm⁴; the transfer term is $4000 \times 240^2 = 230{,}400{,}000$ mm⁴. The plate's own stiffness is 0.06% of what it actually delivers.

That ratio is the whole argument for the I-beam. Material only earns its keep by being far from the neutral axis, and the transfer term is quadratic while the local term is not. It also explains a fact that surprises people: you can nearly always ignore $I_c$ for a thin flange and lose almost nothing, but you can never ignore it for the web, whose centroid is right on the axis and whose entire contribution is the local term.

Two traps. The theorem only works from the part's own centroidal axis outward, never between two arbitrary parallel axes, so if you already transferred once you must come back to the centroid before transferring again. And d is measured to the centroid of the composite section, which you have to locate first by taking area moments. Getting the composite centroid wrong is the most common failure in a built-up section calculation, and it is silent: the arithmetic all works, the answer is just wrong.

The fastest way to the strong-axis $I$ of a doubly symmetric I-shape is not to add flanges and web, it is to subtract. Take the full bounding rectangle $BH^3/12$ and remove the two rectangular voids that flank the web, each of width $(B-t_w)/2$ and depth $h = H - 2t_f$. A 200 × 400 welded girder with 20 mm flanges and a 10 mm web gives $(200 \times 400^3 - 190 \times 360^3)/12 = 327{,}946{,}667$ mm⁴. Check it the long way (web $10 \times 360^3/12 = 38.88$ million, plus two flanges each contributing $133{,}333 + 4000 \times 190^2 = 144.53$ million) and you land on the same number. The box-minus-voids route is one expression instead of five and it never asks you to find a centroid.

What the arithmetic shows is where the stiffness lives. In that girder the two flanges supply 88% of $I$ and the web supplies 12%, even though the web is nearly half the steel. The web is not there for bending at all; it is there to carry shear and to hold the flanges apart. Change the depth from 400 to 500 mm with the same plates and $I$ rises by roughly 60% for no extra weight, which is the reason plate girders get deep and thin rather than short and fat.

The limits worth knowing: this is the strong axis only, it assumes sharp corners rather than the rolled fillets that make a real W-shape a percent or two stiffer than the formula, and it says nothing about whether the section can actually reach that stiffness. A deep thin web buckles, and an unbraced compression flange rolls sideways long before the fibres reach yield. Compactness and lateral bracing are separate checks that $I$ cannot see.

Bending stress is Mc/I, and for a given shape the c and the I never change, so designers precombine them into one number, S = I/c, and the stress check becomes the one-liner σ = M/S. A 50 mm × 100 mm rectangle has I = 4.167 × 10⁻⁶ m⁴ and c = 0.05 m, so S = 8.333 × 10⁻⁵ m³. Every steel manual on every job site is really a table of S values: pick the lightest beam whose S beats M/σ_allow and the sizing is done.

Two cautions. This engine has no m³ unit type, so S is a plain number in m³ (1 in³ = 1.6387 × 10⁻⁵ m³, 1 mm³ = 10⁻⁹ m³). And a section that is not symmetric about its bending axis (a channel, a tee, an unequal-flange girder) has two different values of c and therefore two section moduli, one for the tension face and one for the compression face. Use the smaller S, which belongs to the fibre furthest from the neutral axis, or you will check the wrong face. Note also that S here is the elastic modulus; plastic design uses Z, which for a rectangle is 1.5 times larger.

Drop a single load P at the middle of a beam that simply rests on two supports and each support takes P/2. Walk out to midspan and the moment there is (P/2)(L/2) = PL/4: the largest anywhere on the beam, and it falls off linearly to zero at each support. A 10 kN hoist hung at the centre of a 6 m span produces M = 10 000 × 6 ÷ 4 = 15 000 N·m = 15 kN·m.

The lesson buried in the L is that span is expensive: doubling the span doubles the moment for the same load, and the beam you need grows much faster than that once deflection is added to the picture. The trap is the word simply supported. This result assumes the ends are free to rotate: a beam of the same span with fixed ends carries only PL/8 at midspan, and a cantilever of length L with the load at its tip carries PL, four times as much. Check the end conditions before reaching for the coefficient, and remember to add the beam's own weight, which contributes its own wL²/8.

wL²/8 is the most-used number in structural engineering. Spread a load evenly along a simply supported beam (its own weight, a floor, snow on a roof, a run of water-filled pipe) and the moment diagram is a parabola peaking at midspan with the value wL²/8. A 5 kN/m load on a 4 m span gives M = 5000 × 16 ÷ 8 = 10 000 N·m. The square on L is the important part: stretch the span by 50% and the moment goes up by 125%.

Enter w as force per unit length: this calculator borrows the N/m unit family, which also offers kN/m, lbf/ft and lbf/in. Two traps. First, converting an area load to a line load requires the tributary width: 2.4 kPa of floor load on joists at 400 mm centres is 2.4 × 0.4 = 0.96 kN/m per joist, and forgetting the tributary width is the most common error in the whole calculation. Second, this is the simply supported case; a fixed-fixed beam peaks at wL²/12 over the supports and only wL²/24 at midspan, and a propped cantilever gives wL²/8 at the fixed end. Continuous multi-span beams are somewhere in between, which is why they use less steel.

Bend a beam and the top fibres shorten while the bottom fibres lengthen; somewhere between runs a neutral axis that does neither, and the stress grows linearly with distance c from it, peaking at the outermost fibre. Galileo posed the problem in Two New Sciences in 1638, sketching a cantilever built into a wall, but he assumed the beam pivoted about its bottom face, with the entire section in tension, and so overestimated a rectangular beam's strength by a factor of three. Antoine Parent in 1713 and Coulomb in 1773 put the neutral axis where it belongs, at the centroid, and the formula settled into the shape used ever since.

A 10 kN·m moment on a 50 × 100 mm rectangle (I = 4.167 × 10⁻⁶ m⁴, c = 0.05 m) gives σ = 10 000 × 0.05 ÷ 4.167 × 10⁻⁶ = 120 MPa. Two traps. First, I must be in m⁴ and M in N·m here: mixing in⁴ with N·m produces a number wrong by seven orders of magnitude. Second, c is measured from the neutral axis, which for a symmetric section is the mid-depth (half the total depth, not the full depth) and for an unsymmetric one must be located by finding the centroid first. The formula also assumes pure bending of a straight, prismatic, elastic beam; sharp notches, holes and welds concentrate stress well above what it predicts.

This is how beams are actually selected in the field. Work out the maximum moment, divide by the allowable stress, and you have the section modulus you must buy: S ≥ M/σ_allow. A 20 000 ft·lbf moment at an allowable 24 ksi needs S = 240 000 in·lbf ÷ 24 000 psi = 10 in³: flip to the table, find the lightest shape with S above 10 in³, done. In SI the same check on a 10 kN·m moment through a section with S = 8.333 × 10⁻⁵ m³ gives σ = 10 000 ÷ 8.333 × 10⁻⁵ = 120 MPa.

Enter S as a plain number in m³ (1 in³ = 1.6387 × 10⁻⁵ m³) and M in newton-metres; the answer comes back in pascals. The trap is that passing the stress check is not the same as passing the design. A beam sized purely on S may still deflect visibly, may buckle sideways if its compression flange is unbraced, and may crush its web where it lands on a bearing plate. Stress, deflection, lateral-torsional buckling and bearing are four separate checks, and in long shallow spans deflection usually governs first.

Bending stress varies linearly from the neutral axis and peaks at the extreme fibres. Shear stress does the exact opposite: it is zero at the top and bottom faces (there is nothing outside to shear against) and maximum on the neutral axis, the one place where the bending stress is nil. For a solid rectangle the algebra collapses to a memorable shortcut: $\tau_{max} = 1.5\,V/A$. A 50 × 150 mm section carrying 30 kN gives $1.5 \times 30{,}000/0.0075 = 6$ MPa, and running the full $VQ/Ib$ with $Q = 140{,}625$ mm³, $I = 14.06 \times 10^6$ mm⁴ and $b = 50$ mm returns the same 6 MPa. Two routes, one answer.

Dmitrii Zhuravskii derived this in the 1840s after timber bridges on the Saint Petersburg–Moscow railway kept splitting along their length rather than snapping across, and the split is the signature: a wooden beam overloaded in shear fails horizontally near mid-depth, right where this formula says the stress is worst. The same logic sizes the web of a steel beam, where the shortcut is different again: for a wide flange, nearly all the shear rides in the web, and designers approximate $\tau \approx V/(d\,t_w)$ and get within a few percent.

The trap is $b$: it is the width at the cut you are examining, not the overall width. Step from a flange into a web on a tee or an I-shape and $b$ drops by an order of magnitude while $Q$ barely changes, so the stress jumps discontinuously at the fillet. That step is real in the formula and smoothed in the actual part, which is why the theory overpredicts slightly at re-entrant corners and why the underlying assumption, shear uniform across the width, quietly fails in a wide, shallow section.

Shear flow is the answer to a practical question: if I make a beam out of two pieces, how hard do they try to slide past each other? Bending stress varies along the length, so the force pulling on the top piece at one cross-section differs from the force at the next, and the difference has to cross the glue line. $q = VQ/I$ gives that difference per unit length, in newtons per metre or pounds per inch. With $V = 60$ kN, $Q = 1.2 \times 10^6$ mm³ and $I = 4 \times 10^8$ mm⁴, $q = 180$ kN/m, so a pair of nails good for 3 kN together need spacing $6000/180{,}000 = 33$ mm.

The variable that trips everyone is $Q$, the first moment of the area beyond the cut: take only the part of the section on one side of the joint, multiply its area by the distance from its centroid to the neutral axis. Not the whole section. Not the area on the axis side. And $Q$ is largest at the neutral axis, so a joint placed there is the hardest working one in the beam. That is exactly why a plywood box beam gets its web glued continuously rather than with a few fasteners, and why the flange-to-web weld of a plate girder is sized on shear flow rather than on anything the bending calculation produces.

Something the formula quietly reveals: shear flow does not depend on how strongly the pieces are pressed together, only on the shear force and the geometry. Two boards stacked loose and two boards glued into a solid section carry the same total load very differently: the loose pair each bend about their own axis, and the glued pair are four times stiffer for a doubled depth. The connection is what buys the composite action, and $q$ is the price.

Strength decides whether a beam breaks; stiffness decides whether anyone will walk on it happily. A 20 kN load at the centre of a 4 m steel span with I = 4.167 × 10⁻⁶ m⁴ deflects δ = 20 000 × 4³ ÷ (48 × 200 × 10⁹ × 4.167 × 10⁻⁶) = 0.032 m: 32 mm, or L/125, far past any serviceability limit even if the stresses are fine. Codes usually cap live-load deflection at L/360 for plastered ceilings (the number dates from the era when plaster cracked visibly at about that curvature) and L/240 overall.

The L³ is the headline: doubling the span multiplies the sag by eight, and the only real cures are more depth (I goes as h³) or a shorter span. Note that E and I always appear together as the product EI, the flexural rigidity, which is why "use stronger steel" never fixes a bouncy floor, since all structural steels share E = 200 GPa. Remember to enter I as a plain number in m⁴, and to superimpose cases: a beam carrying both its own uniform weight and a point load deflects by the sum of 5wL⁴/384EI and PL³/48EI.

The odd-looking 5/384 comes straight from integrating the parabolic moment diagram of a uniformly loaded simply supported beam twice; it works out to 0.013021. A 10 kN/m load on a 5 m span with E = 200 GPa and I = 10⁻⁴ m⁴ sags δ = 5 × 10 000 × 5⁴ ÷ (384 × 200 × 10⁹ × 10⁻⁴) = 0.00407 m, about 4 mm, or L/1230: a comfortably stiff floor.

Now the fourth power. Going from a 4 m to a 5 m span with everything else unchanged multiplies the deflection by (5/4)⁴ = 2.44, and it is this exponent, not stress, that governs most residential floor framing: the joists are almost always chosen for bounce, not for strength. A useful shortcut is that the uniform-load deflection is 5/8 of the deflection a single mid-span point load of the same total magnitude would cause, so wL total spread out sags less than the same weight concentrated at the middle. Enter w in force per unit length (N/m, kN/m, lbf/ft) and I as a plain number in m⁴; forgetting the tributary width when turning a floor pressure into w is the classic mistake here.

The cantilever is the shape Galileo drew in 1638 and the one every diving board, balcony and pipe-support bracket still copies. With the load at the tip, δ = PL³/3EI: sixteen times the sag a simply supported beam of the same span, load and section would show, because the fixed end must resist the full moment PL rather than sharing it with a second support. A 2 kN load on the end of a 2 m steel cantilever with I = 4.167 × 10⁻⁶ m⁴ droops δ = 2000 × 8 ÷ (3 × 200 × 10⁹ × 4.167 × 10⁻⁶) = 0.0064 m, 6.4 mm.

Everything hinges on the word fixed. The formula assumes the built-in end has zero rotation, and in real life that is the hardest thing to achieve: a bracket bolted to a flexible column, a beam pocketed into masonry, or a weld that is not full-strength will all rotate, and the tip deflection then comes out well above prediction. Measure the true rotation, or design conservatively. If instead the load is spread uniformly along the cantilever, the coefficient becomes wL⁴/8EI; and enter I as a plain number in m⁴, since this engine has no m⁴ unit.

Spread a load along a cantilever instead of hanging it at the tip and the coefficient becomes $wL^4/8EI$. A 2 m steel cantilever with $I = 10^7$ mm⁴ carrying 5 kN/m droops $5000 \times 16/(8 \times 2.0\times10^{11} \times 10^{-5}) = 5$ mm. Compare it against the same total load, 10 kN, concentrated at the free end, which gives $PL^3/3EI = 13.3$ mm. The distributed case is exactly three-eighths of that, because most of the load sits closer to the fixed end where it has less leverage.

The fourth power on length is the part that governs design. Add a third to the projection of a balcony, from 1.5 m to 2 m, and the deflection multiplies by $(2/1.5)^4 = 3.16$ for the same slab and the same load per square metre. Cantilevers are the members where serviceability nearly always beats strength: the usual limit is $L/180$ on the projection, which for a 2 m balcony means 11 mm, and there is no visual reference point out there to hide the sag against, so people notice a droop on a cantilever that they would never see mid-span.

Everything here also assumes the fixed end is genuinely fixed. It rarely is. A bracket bolted to a flexible column, a joist pocketed into masonry, or a slab cantilevering off a beam that itself twists will all rotate at the support, and that rotation adds $\theta L$ directly to the tip deflection: often more than the flexural term this formula computes. Also remember to superimpose: a balcony carries its own uniform weight and a rail load at the tip, and the two deflections simply add.