Visual walkthrough — Structural loads — axial (thrust), bending (wind shear), dynamic (vibration, acoustics, shock)

Before any letters, meet the object we are cutting apart.

We treat the rocket as a hollow tube standing on its engine. Two things attack it at the same instant near max-Q: the engine shoves it up the long axis (red), and a sideways gust pushes the middle (orange). Our goal is the stress on the one worst spot — the windward fiber (the outer skin line facing into the wind, marked in blue).

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Step 1 — What "stress" even means (force spread over skin)

WHAT we are doing: giving ourselves a single number to compare against "how much can the metal take." WHY a ratio and not just force: a big rocket and a small strut can carry very different forces yet fail at the same stress, because failure is a property of the material, not the size.

PICTURE: the same red force $F$ arrows squeezed through a small area (crowded, high stress) vs a large area (spread out, low stress). Every term in every later formula will end up being some force ÷ some area — this is the only unit we ever care about.

• $\sigma$ ::: stress, force per unit area, in pascals (Pa)
• Why compressive stress is what breaks tubes ::: a thin can fails by crumpling (buckling) under compression, so the compressive side is the dangerous side

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Step 2 — Cut the rocket: where does axial force come from?

WHAT we do: slice the vehicle horizontally at height $x$ and look only at the piece above the cut.

WHY we cut: the internal force at a spot is invisible until we expose it. Cutting turns a hidden internal push into an external force we can balance with Newton's law.

The chunk above the cut has mass $m(x)$ (all the fuel + structure sitting above that height). Three forces act on it, drawn as arrows:

• Gravity $m(x)g$ pulling down (green).
• The structure below pushing up on it with force $F_{\text{axial}}(x)$ (red) — this is the internal force we want.
• The whole chunk is accelerating up at $a$.

Newton's second law ($\text{net force}=\text{mass}\times \text{acceleration}$), taking up as positive:

$\underset{\text{skin pushes up}}{\underbrace{F_{\text{axial}}(x)}}-\underset{\text{weight down}}{\underbrace{m(x)g}}=\underset{\text{mass}\times \text{accel}}{\underbrace{m(x)a}}$

Rearranging (move the weight to the right):

$\boxed{F_{\text{axial}}(x)=m(x)(g+a)}$

• $m(x)$ ::: total mass located above the cut at height $x$
• $g+a$ ::: gravity plus rocket acceleration, the total downward "pull" the structure must fight

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Step 3 — Package $g+a$ into the load factor $n$

WHAT we do: rename $g+a$ as $ng$ where $n=\frac{g+a}{g}$.

WHY engineers do this: they want one dimensionless number that says "the structure feels $n$ times its own weight." Saying "the payload sees 6 g" is instantly meaningful; saying "$a=49.05{\text{m/s}}^2$" is not.

$F_{\text{axial}}(x)=m(x)ng,{\sigma }_{\text{axial}}=\frac{F_{\text{axial}}}{A}=\frac{m(x)ng}{A}$

• $n$ ::: axial load factor in g's; $n=1$ means sitting still on the pad, $n=6$ means 6× weight
• $A$ ::: cross-sectional area of the load-bearing skin (the metal ring you'd see if you sawed the tube)

This connects directly to Rocket Equation & Thrust — the thrust that sets $a$ is the same thrust that squeezes the skin.

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Step 4 — The gust: why a straight tube starts to bend

WHAT we do: add the sideways wind and watch the tube curve.

WHY it bends: the rocket is long and slender, clamped at the bottom by its own downward momentum — exactly like a broomstick you grab at one end and shove in the middle. Engineers call this a cantilever beam: fixed one end, free to curve along its length.

The gust pushes sideways (orange). The tube curves. Now look closely at the bent shape:

• The outer side (facing away from the curve centre) gets stretched longer → tension.
• The inner side gets squashed shorter → compression.
• Somewhere in the middle a line is neither stretched nor squashed — the neutral axis (dashed grey).

This is the seed of everything: bending is just one side longer, the other shorter. To turn that picture into stress we need one more measurement.

• neutral axis ::: the line through the tube that keeps its original length while bending; strain is zero there
• Why one side stretches and the other compresses ::: bending curves the tube, and the outside of a curve has farther to travel than the inside

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Step 5 — From "stretch" to stress: build $\sigma =\frac{Ey}{R}$

WHAT we do: measure how much a fibre at distance $y$ from the neutral axis stretches, then convert stretch → stress with Hooke's law.

WHY we need $R$: to say how much a fibre stretches we need to know how sharply the tube is curved. We describe the curve by the radius of curvature $R$ — the radius of the circle that hugs the bend (small $R$ = tight bend, big $R$ = gentle bend).

Look at a short arc of the bent tube. A fibre at distance $y$ from the neutral axis lies on a circle of radius $R+y$; the neutral fibre lies on radius $R$. Over the same angle, arc length is (radius × angle), so the extra length of the outer fibre, divided by its original length, is:

$\epsilon =\frac{\text{extra length}}{\text{original length}}=\frac{y}{R}$

• $\epsilon$ ::: strain, the fractional stretch (a pure number, no units)
• $y$ ::: distance of the fibre from the neutral axis — zero at centre, biggest at the skin
• $R$ ::: radius of curvature; the same for all fibres in one slice

Now Hooke's law — the material's own rule that stress is proportional to strain, with stiffness $E$ (Young's modulus):

$\sigma =E\epsilon =\frac{Ey}{R}$

• $E$ ::: Young's modulus, how hard the metal resists stretching (Pa)
• Where is bending stress largest ::: at the outer skin, the fibre with the greatest $y$

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Step 6 — Trade $R$ for the moment $M$: the birth of $I$

WHAT we do: get rid of the awkward $R$ (we never actually measure curvature) and replace it with the bending moment $M$ — the turning effect the gust applies, which we can compute.

WHY: $R$ is an effect, not a cause. The cause is the gust's twisting action $M$. We connect them by adding up every fibre's little contribution.

Each fibre carries stress $\sigma =Ey/R$ over its tiny area $dA$, at lever arm $y$ from the neutral axis. Its contribution to the internal turning moment is $(\text{force})\times (\text{lever arm})=\sigma dA\cdot y$. Summing (integrating) over the whole cross-section:

$M=\int y\sigma dA=\int y\cdot \frac{Ey}{R}dA=\frac{E}{R}\underset{\text{call this }I}{\underbrace{\int y^{2}dA}}$

We name that integral the second moment of area $I\equiv \int y^{2}dA$. It measures how far the material is spread from the centre — material far out (big $y^2$) resists bending far more.

PICTURE: two cross-sections of equal material. The one with metal pushed to the rim has a huge $I$; the compact one has a small $I$. This is why rockets are hollow tubes, not solid rods — the same metal, moved outward, buys enormous bending resistance. See Beam Bending & Second Moment of Area.

From $M=\frac{E}{R}I$ we read off $\frac{E}{R}=\frac{M}{I}$. Substitute back into $\sigma =\frac{Ey}{R}$:

${\sigma }_{\text{bend}}=\frac{My}{I},\text{max at the skin }y=c:\boxed{{\sigma }_{\text{bend}}=\frac{Mc}{I}}$

• $M$ ::: bending moment, the gust's total turning effect (N·m)
• $I$ ::: second moment of area — the shape's resistance to bending (m⁴)
• $c$ ::: distance from the neutral axis to the outermost fibre (for a tube, $c=R$, the radius)

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Step 7 — All four cases on the ring: where do the stresses ADD?

WHAT we do: walk around the tube's cross-section and check the sign of stress at every point.

WHY this matters most: the parent's boxed result adds the two terms — but that is only true on one specific fibre. Get the sign wrong and you design the wrong side.

Look at the ring, wind coming from the left. Axial thrust squeezes the whole ring uniformly (every point feels $-F/A$, compression). Bending adds a swirl: compression on one side, tension on the other. Walk around:

Fibre location	Axial part	Bending part	Total
Windward (into wind)	$-\frac{F}{A}$	$-\frac{Mc}{I}$	most compressive — they ADD
Leeward (downwind)	$-\frac{F}{A}$	$+\frac{Mc}{I}$	they cancel — safest
Neutral axis (sides)	$-\frac{F}{A}$	$0$	only axial

So the worst fibre is windward, and there — and only there — the magnitudes add:

$\boxed{{\sigma }_{\text{total}}=\frac{F_{\text{axial}}}{A}+\frac{Mc}{I}}$

Degenerate cases (never leave a scenario unshown):

• No gust ($M=0$): the bending term vanishes, ${\sigma }_{\text{total}}=F/A$ — pure Step 3 axial.
• Engine off / coasting ($n=0\Rightarrow F=0$): only bending, ${\sigma }_{\text{total}}=Mc/I$.
• Solid rod vs hollow tube: same material, but the tube's larger $I$ makes $Mc/I$ far smaller — this is why the airframe is hollow.

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Step 8 — Plug in the numbers: the sizing verdict

WHAT we do: add our two worked halves at the windward fibre.

${\sigma }_{\text{total}}=\underset{\text{axial, Step 3}}{\underbrace{5.89\text{MPa}}}+\underset{\text{bending, Step 6}}{\underbrace{50.9\text{MPa}}}=56.8\text{MPa (compressive)}$

WHY it matters: this single peak number is what you compare to the metal's allowable stress, after multiplying by a factor of safety. If the aluminium yields near $270\text{MPa}$, we have healthy margin — but bending (50.9 MPa) dominates axial (5.89 MPa), so the design battle at max-Q is against the wind, not the thrust.

Recall Why this is the max-Q worst case

Dynamic pressure $q=\frac{1}{2}\rho v^2$ peaks partway up the atmosphere. Side force ∝ $q$, so the bending moment $M$ peaks there too — while thrust is still hammering. Both terms are large at the same instant, on the same windward fibre. That is the sizing point. (See Max-Q and Dynamic Pressure.)

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The one-picture summary

Left: cut → Newton → $F=mng$ → divide by $A$ gives the uniform axial squeeze. Right: gust → cantilever bend → fibre strain $y/R$ → Hooke → sum over area gives $Mc/I$. Bottom: the two arrive on the windward fibre of the ring and add: ${\sigma }_{\text{total}}=\frac{F_{\text{axial}}}{A}+\frac{Mc}{I}.$

Recall Feynman retelling — the whole walkthrough in plain words

A rocket is a hollow tube. Slice it and look at the part above the cut: gravity pulls it down, the skin below pushes it up, and the whole thing is accelerating up — Newton's law says the skin must push with $m(g+a)$, which we rename $mng$. Spread that force over the ring of metal and you get the axial stress $F/A$: a uniform squeeze, biggest at the base where the most mass sits above. Now the wind slaps the middle. The tube is long and skinny, clamped at the bottom, so it bends like a broomstick — outside of the bend stretches, inside squashes, the centreline stays put. How much a fibre stretches is just how far it sits from the centre divided by how tight the curve is ($y/R$), and Hooke turns stretch into stress ($Ey/R$). We don't know the curve $R$, but we know the wind's turning effect $M$; adding up every fibre's contribution births a shape number $I$ (bigger when metal sits far out — that's why tubes beat rods) and gives $\sigma =Mc/I$, biggest at the skin. Finally, walk around the ring: thrust squeezes it all evenly, bending compresses the windward side and stretches the leeward side. On the windward fibre both are compression, so they add — and that one number, $F/A+Mc/I$, is what the metal must survive at max-Q.