Structural loads — axial (thrust), bending (wind shear), dynamic (vibration, acoustics, shock): Worked examples

This is the drill page for the parent topic. We will not re-derive the formulas — the parent note earned every symbol. Here we hunt down every kind of situation the three load families can throw at you and solve one of each, out loud.

Symbols used everywhere below (all defined in the parent note — glance back if any is unfamiliar):

$m (x)$ = mass sitting above a cut at height $x$ (kg).
$g = 9.81 m/s^{2}$ = gravity. $n$ = axial load factor in "g's" (dimensionless).
$F_{axial} = m n g$ , and $σ = F / A$ (force spread over cross-section area $A$ ).
$M$ = bending moment (N·m). $y$ = the distance of a given fiber from the neutral axis (m), measured perpendicular to the bending axis — positive on the stretched side, negative on the squashed side. $c$ = the value of $y$ at the outer skin, i.e. the largest distance (m). $I$ = second moment of area (m⁴), $σ_{bend} = M c / I$ .
$θ$ = the angle around the ring (radians), running from $0$ to $2 π$ as you walk once around the circular cross-section; a fiber at angle $θ$ sits at height $y = R sin θ$ above the neutral axis. It is just a bookkeeping angle for adding up all the fibers.
$f_{n}$ = natural frequency (Hz), $Q = 1/ (2 ζ)$ = resonance amplification, $ζ$ = damping ratio.
$g_{RMS} = \frac{π}{2} f_{n} Q W$ = Miles' random-vibration response, $W$ = PSD (g²/Hz).
$k$ = stiffness constant (N/m): how much force it takes to deflect a part by one metre — a "spring rate". A structure that barely bends under load has a large $k$ . It sets the natural frequency through $f_{n} = \frac{1}{2 π} k / m$ .
$x_{static}$ = deflection the part would have under a steadily applied load; $x_{max}$ = the peak deflection when that load oscillates at the resonant frequency. Their ratio is exactly $Q$ : $x_{max} \approx Q x_{static}$ .

The scenario matrix

Every problem in this topic lands in exactly one of these cells. Our job is to cover them all.

Cell	Load family	Special feature being tested	Killed by
A1	Axial	ordinary compression at a cut	Example 1
A2	Axial	sign flip → tension (upper stage coasting, $n < 0$ / drag)	Example 2
A3	Axial	degenerate: $n = 0$ (free-fall / engine cutoff)	Example 3
B1	Bending	ordinary flexure stress $M c / I$	Example 4
B2	Combined	axial + bending add on same fiber at max-Q	Example 5
B3	Bending	limiting geometry: thin-ring $I = π R^{3} t$ , effect of doubling $R$	Example 6
D1	Dynamic	Miles' equation, then $3 σ$ design load	Example 7
D2	Dynamic	frequency-separation pass/fail decision	Example 8
D3	Dynamic	degenerate: $ζ \to 0$ (undamped), what happens to $Q$ and $x_{max}$	Example 9
W/X	Any	word problem + exam twist (opposite fiber sign)	Example 10

Axial cases

Worked example Example 1 — Cell A1: ordinary compression

A cut is made partway up a booster. Above the cut sits $m (x) = 1500 kg$ . The load-bearing wall has cross-section area $A = 0.015 m^{2}$ . At this instant the vehicle pulls $n = 5$ g's. Find the axial stress and say whether it is compression or tension.

Forecast: A few thousand kg × a few g's ≈ 70–80 kN, over ~1/60th of a m² → tens of MPa, compressive (thrust squeezes). Guess ~5 MPa? Let's see.

Internal force $F_{axial} = m n g = 1500 \times 5 \times 9.81 = 73, 575 N$ . Why this step? The structure at the cut must support the weight and accelerate everything above it; $g + a = n g$ bundles both into one term.
Stress $σ = F / A = 73, 575/0.015 = 4.905 \times 1 0^{6} Pa = 4.905 MPa$ . Why this step? Stress = force spread over the material actually carrying it.
Sign: the engine below pushes up on the mass above → the wall is squeezed → compressive (positive by our convention). Why this step? Direction matters: compression can buckle a thin shell, so we must name it.

Verify: Units: $kg \cdot (m/s^{2}) / m^{2} = N/m^{2} = Pa$ ✓. Ballpark ~5 MPa matched the forecast ✓.

Worked example Example 2 — Cell A2: the sign flip (tension)

After burnout the stage coasts and aerodynamic drag now decelerates it. Effective acceleration is downward-along-axis, so the upper mass tends to keep going and pulls on the structure. Model this as $n = - 0.5$ (net 0.5 g of "negative" load along the axis). Same $m = 1500 kg$ , $A = 0.015 m^{2}$ . Find the stress and its sign.

Forecast: Same numbers but a tenth the magnitude and the opposite sign — so a small tensile stress, well under 1 MPa.

Force $F_{axial} = m n g = 1500 \times (- 0.5) \times 9.81 = - 7357.5 N$ . Why this step? $n$ can be negative when the axial inertial load reverses (drag deceleration, coast). The formula still holds; the sign carries the physics.
Stress $σ = F / A = - 7357.5/0.015 = - 4.905 \times 1 0^{5} Pa = - 0.4905 MPa$ . Why this step? Magnitude via the same division; the negative sign is the whole point here.
Interpret sign: negative $σ$ = tension (by our master convention) — the wall is being stretched. Why this step? Thin shells shrug off tension but crumple under compression, so knowing you are in tension is reassuring.

Verify: Magnitude is exactly $1/10$ of Example 1 (since $∣ - 0.5∣/5 = 1/10$ ) → $4.905/10 = 0.4905$ MPa ✓. Sign flipped as promised ✓.

Worked example Example 3 — Cell A3: degenerate,

$n = 0$ (free fall) The engine cuts off and, ignoring drag, the whole stage is in free fall: it accelerates downward at exactly $g$ , so $a = - g$ and $g + a = 0$ , i.e. $n = 0$ . Same $m$ , $A$ . What axial stress?

Forecast: In free fall everything falls together — nothing pushes on anything. Guess: zero stress.

Force $F_{axial} = m n g = 1500 \times 0 \times 9.81 = 0 N$ . Why this step? This is the degenerate check: when $n = 0$ , gravity and acceleration exactly cancel in the free body of the upper mass.
Stress $σ = 0/0.015 = 0 Pa$ . Why this step? Zero force → zero stress. The structure carries no axial load in free fall.

Verify: Physically, an object in free fall is "weightless" — a scale reads zero. Our formula gives $0$ ✓. This is the boundary between compression ( $n > 0$ ) and tension ( $n < 0$ ).

Bending & combined cases

Worked example Example 4 — Cell B1: ordinary flexure

A thin cylinder shell: radius $R = 1.2 m$ , wall $t = 4 mm = 0.004 m$ . A gust makes a bending moment $M = 6 \times 1 0^{5} N\cdotm$ . Find the peak bending stress. (Thin-ring second moment: $I = π R^{3} t$ ; outer fiber $c = R$ .)

Forecast: $I$ scales like $R^{3} t \approx π (1.7) (0.004) \approx 0.02$ m⁴, and $M c / I \approx 6 \times 1 0^{5} \times 1.2/0.02 \approx 3 \times 1 0^{7}$ → tens of MPa.

Reading the figure (alt text): On the left is the round shell cross-section seen end-on. The horizontal dashed line through the centre is the neutral axis — the one layer of material that neither stretches nor squashes when the tube bends. The magenta arrow points to the top outer fiber, which is stretched (tension); the orange arrow points to the bottom outer fiber, which is compressed. Both are a distance $c = R$ from the neutral axis, the farthest any material gets — that is why the extreme stress lives there. The right panel plots stress $σ = M y / I$ against the distance $y$ of a fiber from the neutral axis (recall $y$ is defined in the symbol list): it is a straight line through the origin, magenta (tension) above the axis and orange (compression) below. The key insight the picture carries: stress is zero at the centre and grows in a straight line to its maximum $\pm M c / I$ at the skin — so only the outermost fiber ever needs checking.

Second moment $I = π R^{3} t = π (1.2)^{3} (0.004) = π \times 1.728 \times 0.004 = 0.021714 m^{4}$ . Why this step? For a thin ring all material sits ~distance $R$ from the neutral axis. To add up $\int y^{2} d A$ we walk once around the ring using the angle $θ$ (defined in the symbol list, running $0$ to $2 π$ ): a fiber at angle $θ$ has height $y = R sin θ$ , and the arithmetic collapses to $π R^{3} t$ . $I$ measures how far material is spread from the bending axis — the further, the stiffer.
Outer fiber $c = R = 1.2 m$ — the most-stretched (and most-compressed) fiber is the outer skin (in the figure: the fiber the magenta/orange arrows point to, at the top and bottom of the ring). Why this step? Stress grows linearly with distance $y$ from the neutral axis (right panel of the figure), so it is maximal at $y = c$ .
Peak stress $σ_{bend} = M c / I = 6 \times 1 0^{5} \times 1.2/0.021714 = 3.316 \times 1 0^{7} Pa = 33.16 MPa$ . Why this step? Flexure formula: bending moment divided by section's spread-of-material, times lever arm.

Verify: Units $N\cdotm \cdot m / m^{4} = N/m^{2}$ ✓. ~33 MPa sits in the forecast band ✓.

Worked example Example 5 — Cell B2: axial + bending on the same fiber (max-Q)

Same shell as Example 4 ( $I = 0.021714 m^{4}$ , $c = 1.2$ m, so bending $= 33.16$ MPa). At max-Q the stage also carries axial compression: mass above $m = 3000 kg$ , $n = 4$ , and the wall area is $A = 2 π R t = 2 π (1.2) (0.004) = 0.030159 m^{2}$ . Find the total compressive stress on the windward fiber.

Forecast: Axial part is a few MPa; bending is ~33 MPa; they add → total in the high-30s MPa.

Axial force $F = m n g = 3000 \times 4 \times 9.81 = 117, 720 N$ . Why this step? Axial load at the cut, load-factor form.
Axial stress $σ_{ax} = F / A = 117, 720/0.030159 = 3.904 \times 1 0^{6} Pa = 3.904 MPa$ (compressive, positive). Why this step? Thin-ring wall area is circumference × thickness $= 2 π R t$ .
Add on the compressed fiber $σ_{total} = σ_{ax} + σ_{bend} = 3.904 + 33.16 = 37.06 MPa$ (both positive → both compression → add). Why this step? On the windward/leeward compressed fiber both effects push the same direction, so they add. This is the worst-case sizing point — never analyze them apart at max-Q.

Verify: Total > either piece alone ✓. The windward fiber (compression + compression) is worse than the leeward fiber (compression − tension = $3.904 - 33.16 = - 29.3$ MPa, i.e. net tension), confirming we picked the critical side ✓.

Worked example Example 6 — Cell B3: limiting geometry (double the radius)

Take Example 4's shell ( $R = 1.2$ m, $t = 4$ mm, $M = 6 \times 1 0^{5}$ N·m, $σ = 33.16$ MPa). A designer doubles the radius to $R^{'} = 2.4 m$ but keeps $t$ and $M$ the same. What is the new bending stress? What is the ratio to the old?

Forecast: $I \propto R^{3}$ and $c \propto R$ , so $σ = M c / I \propto R / R^{3} = 1/ R^{2}$ . Doubling $R$ should quarter the stress → about $33/4 \approx 8$ MPa.

New second moment $I^{'} = π R^{'3} t = π (2.4)^{3} (0.004) = π \times 13.824 \times 0.004 = 0.173712 m^{4}$ . Why this step? $I^{'} = 2^{3} I = 8 I$ — pushing material twice as far out multiplies stiffness by $8$ .
New stress $σ^{'} = M R^{'} / I^{'} = 6 \times 1 0^{5} \times 2.4/0.173712 = 8.289 \times 1 0^{6} Pa = 8.289 MPa$ . Why this step? $c$ doubled (only $\times 2$ ) but $I$ grew $\times 8$ , so net $\times 2/8 = \times 1/4$ .
Ratio $σ^{'} / σ = 8.289/33.16 = 0.25$ . Why this step? Confirms the $1/ R^{2}$ scaling law — the single most useful lever a structures engineer has: a wider tube is dramatically stiffer in bending.

Verify: Predicted $1/ 2^{2} = 1/4$ ✓; $33.16 \times 0.25 = 8.29$ MPa matches ✓.

Dynamic cases

Worked example Example 7 — Cell D1: Miles' equation + 3σ

A component has natural frequency $f_{n} = 120 Hz$ , damping $ζ = 0.05$ , and rides a random-vibration PSD of $W = 0.05 g^{2} / Hz$ at its resonance. Find the RMS response $g_{RMS}$ , then the $3 σ$ design load.

Forecast: $Q = 1/ (2 \times 0.05) = 10$ . Under the root: $\frac{π}{2} \times 120 \times 10 \times 0.05 \approx 94$ , root ≈ 9.7 g. Times 3 ≈ 29 g.

Amplification $Q = 1/ (2 ζ) = 1/ (2 \times 0.05) = 10$ . Why this step? Resonance amplifies by $Q$ ; we need it before Miles' equation.
RMS response $g_{RMS} = \frac{π}{2} f_{n} Q W = \frac{π}{2} \times 120 \times 10 \times 0.05 = 94.25 = 9.708 g$ . Why this step? Miles' equation collapses the whole resonance-peak integral of random vibration into one closed form; the $π /2$ is the area under the SDOF peak.
Design load $3 σ = 3 \times 9.708 = 29.12 g$ . Why this step? Random peaks are Gaussian; $3 σ$ covers 99.7% of them, the standard qualification level.

Verify: $94.25 = 9.708$ and $\times 3 = 29.12$ ✓. Units: PSD is g²/Hz, times Hz → g², root → g ✓.

Worked example Example 8 — Cell D2: frequency-separation decision

A launch vehicle demands the payload's first lateral natural frequency stay above 15 Hz and first axial above 30 Hz. Analysis predicts the payload has $f_{lat} = 12 Hz$ and $f_{ax} = 42 Hz$ . Does it pass? If not, does making the structure stiffer (raising the stiffness constant $k$ , the force-per-metre-of-deflection defined at the top of this page) help, and by how much on $f$ if $k$ quadruples while the mass $m$ is unchanged?

Forecast: Lateral 12 < 15 → fail. Axial 42 > 30 → pass. Stiffening: $f_{n} \propto k$ , so $\times 4$ stiffness → $\times 2$ frequency → 12 becomes 24 Hz → passes.

Axial check: $42 Hz > 30 Hz$ → pass. Why this step? Frequency separation keeps the payload's resonance out of the LV's strong low-frequency drive band, so we compare each predicted mode to its own minimum.
Lateral check: $12 Hz < 15 Hz$ → fail. This mode could be shaken hard by the launcher. Why this step? Being below the minimum is the dangerous direction — the excitation energy lives down there, so a mode under the floor risks resonance.
Fix by stiffening: $f_{n} = \frac{1}{2 π} k / m$ , so with $m$ fixed $f_{n} \propto k$ . Quadrupling $k$ gives $f^{'} = 4 f = 2 \times 12 = 24 Hz > 15$ → now passes. Why this step? Stiffening helps only because it moves $f_{n}$ away from the band — not because stiffness ( $k$ ) is intrinsically "safer" (see the parent note's mistake box). If the band were higher, the same stiffening could push you into it.

Verify: $4 = 2$ , $12 \times 2 = 24 \geq 15$ ✓. Axial $42 \geq 30$ ✓, lateral $12 < 15$ (fail) ✓.

Worked example Example 9 — Cell D3: degenerate,

$ζ \to 0$ (undamped) A perfectly undamped idealization has $ζ = 0$ . Find $Q$ , and use $x_{max} \approx Q x_{static}$ (both symbols defined at the top of the page: $x_{static}$ is the deflection under a steady load, $x_{max}$ the peak deflection at resonance) to say what happens to the resonant deflection. Then take a realistic tiny damping $ζ = 0.01$ , recompute, and put numbers on a $x_{static} = 0.1 mm$ part.

Forecast: $Q = 1/ (2 ζ)$ blows up as $ζ \to 0$ → infinite amplification (physically impossible — the model breaks). At $ζ = 0.01$ , $Q = 50$ , an enormous but finite factor, turning $0.1$ mm into ~ $5$ mm.

Undamped limit $Q = 1/ (2 \times 0) = \infty$ . Why this step? The limiting case exposes that zero damping is the worst case — amplification is unbounded. Real structures always have some $ζ$ , which is what saves them.
Deflection in that limit $x_{max} \approx Q x_{static} \to \infty$ . Why this step? With $Q$ infinite, the peak deflection $x_{max}$ (the amplitude the part actually swings to at resonance) also runs away — the physical warning that a truly undamped resonance would tear the part apart. This is why $ζ > 0$ always matters.
Realistic $ζ = 0.01 \Rightarrow Q = 1/ (2 \times 0.01) = 50$ . Why this step? Even 1% damping caps amplification at $50 \times$ — still huge, showing why low $ζ$ (not high) is the danger.
Put numbers on it $x_{max} \approx Q x_{static} = 50 \times 0.1 mm = 5 mm$ . Why this step? Turns the abstract $Q$ into a physical deflection you can picture: a hairline static sag becomes a violent $5$ mm swing at resonance.

Verify: As $ζ \to 0^{+}$ , $1/ (2 ζ) \to + \infty$ ✓, so $x_{max} \to \infty$ ✓. $ζ = 0.01 \Rightarrow Q = 50$ ✓, and $50 \times 0.1 mm = 5 mm$ ✓.

Word problem + exam twist

Worked example Example 10 — Cell W/X: real mission, opposite-fiber twist

Word problem: A rocket at max-Q has, on its critical cross-section, an axial compressive stress of $6 MPa$ (positive, by our convention) and a bending stress of magnitude $40 MPa$ . The material yields in tension at $35 MPa$ and in compression at $250 MPa$ (buckling aside). Exam twist: which fiber fails first — the windward (compressed) one or the leeward (stretched) one?

Forecast: Windward: $6 + 40 = 46$ MPa compression, far below $250$ → safe. Leeward: axial compression $6$ minus bending tension $40$ = net $34$ MPa tension, just under the $35$ MPa tensile limit → the leeward fiber is the near-failure one. Twist: the tension side, not the big-number compression side, is critical.

Windward (compressed) fiber: $σ = + 6 + 40 = + 46 MPa$ (both compressive, positive, add). Why this step? On the windward side bending compresses the same fiber thrust already compresses; positive-plus-positive.
Compare to compressive limit: $∣ + 46 ∣ = 46 < 250$ → margin huge → not critical. Why this step? Big number ≠ danger; you must compare to the relevant allowable (the compressive one here).
Leeward (stretched) fiber: $σ = + 6 - 40 = - 34 MPa$ — negative, so net tension (the bending tension overpowers the axial compression). Why this step? The bending tension on the leeward side fights the axial compression; whichever wins sets the sign, and here tension wins ( $- 34$ ).
Compare to tensile limit: $∣ - 34∣ = 34 < 35$ → survives but with only $35/34 - 1 = 2.9%$ margin → this is the critical fiber. Why this step? The tensile allowable ( $35$ ) is far lower than compressive ( $250$ ), so a smaller stress there is more dangerous — the exam-twist punchline.

Verify: Windward $∣ + 6 + 40∣ = 46$ MPa, margin vs 250 large ✓. Leeward $+ 6 - 40 = - 34$ MPa (net tension), $∣ - 34∣ = 34 < 35$ (passes) but margin $35/34 - 1 = 0.0294 \approx 2.9%$ ✓.

Recall Matrix self-check

Which cell has $n < 0$ ? ::: A2 — tension, coast/drag deceleration (Example 2). Which cell has $n = 0$ ? ::: A3 — free fall, zero axial stress (Example 3). Doubling tube radius changes bending stress by what factor? ::: $1/4$ (stress $\propto 1/ R^{2}$ ) — Example 6. Which fiber is critical when tensile allowable ≪ compressive allowable? ::: the leeward (net-tension) fiber — Example 10. As $ζ \to 0$ , what do $Q$ and $x_{max}$ do? ::: both $\to \infty$ ; low damping is the danger — Example 9.

Related: Factor of Safety & Margins of Safety · Beam Bending & Second Moment of Area · Random Vibration & PSD · Modal Analysis & Natural Frequencies · Max-Q and Dynamic Pressure · Rocket Equation & Thrust · Pyrotechnic Separation Systems.