Sandwich structures — face sheets, core: Visual walkthrough

Before line one, three plain words we will lean on. Bending = the beam curving under load. Stiffness = how hard it is to make it curve (big stiffness → tiny bend). Mass = how heavy it is, which for a spacecraft is money burned at launch. Everything below is a fight to raise the first while lowering the third.

Step 1 — What "bending stiffness" even means

WHAT. When you push down on the middle of a beam, it sags by some amount $δ$ (the Greek letter delta, our symbol for "how far it drooped", measured in metres). A stiff beam gives a small $δ$ ; a floppy one gives a big $δ$ .

WHY. We need a single number that captures "resistance to bending" so we can compare designs fairly. That number is written $(E I)$ and it always shows up in the denominator of every sag formula, e.g. for a cantilever (fixed at one end, free at the other) pushed by a force $F$ at its tip:

$δ = \frac{F L ^{3}}{3 ( E I )} .$

$F$ — the push, in newtons.
$L$ — the beam's length, in metres.
$(E I)$ — our hero number. It is big when the beam is stiff. Because it sits underneath, a bigger $(E I)$ means a smaller $δ$ .

Where do the $L^{3}$ and the $\frac{1}{3}$ come from? Beam Bending Theory shows that the beam's local curvature at every point equals (bending moment) $/ (E I)$ . To get the tip droop you must add up that curvature twice along the length (once to turn curvature into slope, once to turn slope into deflection) — two integrations along $L$ , each contributing one power of $L$ beyond the moment's own single power of $L$ , giving $L^{3}$ in total. The clean fraction $\frac{1}{3}$ is simply the number those two integrations spit out for the triangular moment shape of a tip-loaded cantilever; a different load or support would give a different fraction (we meet $5/384$ and $1/8$ in Step 6). The takeaway you need: droop scales with $L^{3}$ and inversely with $(E I)$ , so raising $(E I)$ is the direct lever on stiffness.

The letters split into two jobs: $E$ (Young's modulus) is how stiff the material is — a property of the substance, like carbon fibre vs. jelly. $I$ (the second moment of area) is how the material is arranged in the cross-section — a property of the shape. Our whole game is to make $I$ enormous without buying more material. That geometry idea comes straight from Beam Bending Theory.

PICTURE. Same material, two shapes: laid flat it sags a lot; stood on edge it barely moves. Nothing changed but the arrangement — that is $I$ talking.

Figure — Sandwich structures — face sheets, core

Step 2 — Why the outer fibres matter most

WHAT. When a beam bends, its top surface gets squeezed shorter and its bottom surface gets stretched longer. Somewhere in the middle is a line that neither stretches nor squeezes — the neutral axis. We measure a distance $y$ from that line: $y = 0$ at the neutral axis, growing to $y_{m a x}$ at the top and bottom faces.

WHY. The bending stress $σ$ (Greek sigma, the force-per-area a fibre feels, in pascals) grows straight-line with $y$ :

$σ = \frac{M y}{I} .$

$M$ — the bending moment (how hard the load twists the cross-section), in newton-metres.
$y$ — distance from the neutral axis. The star of the show.
$I$ — the same arrangement number as before.

Read it plainly: stress is proportional to $y$ . A fibre sitting at the neutral axis ( $y = 0$ ) feels zero stress — it is dead weight. A fibre at the far edge ( $y = y_{m a x}$ ) feels the most. So material near the middle is loafing, and material far out is doing the heavy lifting.

PICTURE. A triangle of stress: zero on the centre line, longest arrows at the surfaces. The message glows: put your good material where the arrows are longest.

Step 3 — The trick: pull the material apart

WHAT. Take the material and split it into two thin sheets — the face sheets — and shove them as far from the neutral axis as you can. Between them put something almost weightless whose only job is to hold them apart: the core.

WHY. Because of Step 2, moving material outward increases how much bending it can resist. The amount of this reward is what the next step measures. The core is not there to resist bending; it is there to keep the faces from collapsing toward each other (and to carry shear — Step 6). This is the Composite Materials philosophy: let each part do the one thing it is best at.

We name every symbol we will ever use now, so nothing later is undefined:

$b$ — width of the panel (into the page), in metres.
$t_{f}$ — thickness of one face sheet, in metres. Thin, e.g. $0.5 mm$ .
$c$ — thickness of the core, in metres. Fat, e.g. $19 mm$ .
$h = 2 t_{f} + c$ — total depth of the sandwich, face-outside to face-outside, in metres.
$E_{f}$ — Young's modulus (material stiffness) of the face material, in pascals.
$E_{c}$ — Young's modulus of the core material, in pascals. Tiny, because honeycomb is mostly air.
$G_{c}$ — shear modulus of the core (how hard it resists sliding, a stiffness property), in pascals. Meets us properly in Step 6.
$τ_{c, allow}$ — the core's shear strength (the stress at which it ruptures, a different property from $G_{c}$ ), in pascals. Also Step 6.
$ρ$ — density (mass packed into each cubic metre of a material), in kilograms per cubic metre ( $kg/m^{3}$ ). We use $ρ_{f}$ for the face material and $ρ_{c}$ for the core; low $ρ$ is exactly what makes a material "light".

PICTURE. The flat slab of Step 2 splits and its two halves migrate to the surfaces, a light honeycomb core inflating between them. Same material, dramatically taller shape.

Step 4 — Measuring one face's contribution (the parallel-axis theorem)

WHAT. We compute the arrangement number $I$ for a single face sheet sitting off the neutral axis. The tool for "how much does a shape resist bending when it is not centred?" is the parallel-axis theorem — that is exactly the question it answers, which is why we reach for it and not anything else.

WHY this tool. A shape's $I$ comes in two parts: (1) its own stiffness about its own centre, plus (2) a bonus for how far its centre is shoved off the beam's neutral axis. First name the face's cross-sectional area, since it appears in both the theorem and later in the mass count:

$A_{f} = b t_{f} (width \times face thickness, in m^{2}) .$

The theorem then states:

$I_{face} = own shape \frac{b t _{f}^{3}}{12} + distance bonus A_{f} d^{2} .$

$\frac{b t _{f}^{3}}{12}$ — the face's stiffness about its own centreline. A thin sheet is floppy on its own, so this is tiny.
$A_{f} = b t_{f}$ — the cross-sectional area of the face, just named above.
$d$ — distance from the face's centre to the beam's neutral axis. Squared, because the reward grows with distance squared (echoing the $σ \propto y$ of Step 2).

The face's centre sits at $d = \frac{c}{2} + \frac{t _{f}}{2} = \frac{h}{2} - \frac{t _{f}}{2}$ . For thin faces ( $t_{f} ≪ h$ ) we round this to $d \approx \frac{h}{2}$ . State that approximation out loud every time — it is only true when the faces are thin compared to the depth.

PICTURE. One face highlighted, an arrow of length $d$ from the neutral axis to its centre, with a little " $d^{2}$ " reward badge that dwarfs the "own shape" badge.

Step 5 — Why we throw away two terms (and when we may not)

WHAT. Add both faces and the core, then discard the small pieces to reach the clean formula.

WHY term 1 dies. Compare the two parts of $I_{face}$ : the own-shape term carries $t_{f}^{3}$ , the distance term carries $A_{f} (h /2)^{2} = t_{f} h^{2} /4 \cdot b$ . Their ratio (per unit width) is

$\frac{t _{f}^{3} /12}{t _{f} h ^{2} /4} = \frac{4}{12} \cdot \frac{t _{f}^{2}}{h ^{2}} = \frac{1}{3} (\frac{t _{f}}{h})^{2} .$

With $t_{f} = 0.5 mm$ and $h = 20 mm$ this is $\frac{1}{3} (0.5/20)^{2} \approx 2 \times 1 0^{- 4}$ — about $0.02%$ . Negligible.

WHY the core dies too. The core has its own bending stiffness $E_{c} I_{core} = E_{c} \frac{b c ^{3}}{12}$ , but its material stiffness $E_{c}$ is roughly a thousand times smaller than the face material $E_{f}$ (honeycomb is mostly empty air). So $E_{c} I_{core} ≪ E_{f} I_{faces}$ and it drops out. Add the two faces:

$I_{faces} \approx 2 A_{f} (\frac{h}{2})^{2} = \frac{b t _{f} h ^{2}}{2} .$

The " $2$ " — one for each face.
$(\frac{h}{2})^{2}$ — each face is half the depth away, squared.

Multiply by the face material stiffness $E_{f}$ and we have arrived:

$(E I)_{sandwich} = \frac{E _{f} b t _{f} h ^{2}}{2} (t_{f} ≪ h) .$

PICTURE. The full expression with the two doomed terms greyed out and shrinking, the survivor glowing.

Step 6 — The case the stiffness formula hides: shear in the core

WHAT. The boxed $(E I)$ only tells the bending half of the story. For a long, slender panel that is enough. But panels are often stubby, and then a second sag mechanism appears: the whole core shears like a deck of cards sliding.

WHY it matters. Consider one common, well-defined case: a simply-supported beam (resting on a support at each end) of span $L$ , carrying a uniform line load $w$ — that is the force pushed onto the beam per metre of length, in newtons per metre (for a pressure $p$ over a panel of width $b$ , $w = p b$ ). For that case the total deflection is bending plus shear:

$δ_{total} = bending (simply-supported UDL) \frac{5 w L ^{4}}{384 ( E I )} + core shear \frac{w L ^{2}}{8 κ G _{c} b c} .$

$w$ — the uniform line load, newtons per metre. (The numbers $5/384$ and $1/8$ are the standard coefficients for this simply-supported uniform-load case — they come out of the same double-integration of curvature described in Step 1, just for a uniform load and end supports rather than a tip force; a cantilever or a point load would use different coefficients — always match the formula to the boundary conditions.)
$(E I)$ — our boxed bending stiffness.
$G_{c}$ — the core's shear modulus (how hard the core resists sliding), a stiffness property.
$b c$ — the shear-carrying area. Only the core of thickness $c$ carries the shear, not the full depth $h$ , because the thin faces contribute almost nothing to shear.
$κ$ — the shear-correction factor. Beam theory pretends the shear stress is spread evenly over the section, but in reality it bulges to a peak at the neutral axis and falls to zero at the surfaces. $κ$ is the single number that repairs that lie so the simple formula still gives the right stored shear energy; solving the actual parabolic shear distribution for a rectangular cross-section yields $κ = 5/6 \approx 0.83$ . That is why $κ \approx 5/6$ appears for our rectangular panel — a different cross-section shape would carry a different $κ$ .

The bending term scales with $L^{4}$ ; the shear term only with $L^{2}$ . So for short, deep panels ( $L / h < 20$ ) the shear term can dominate — miss it and you over-predict stiffness badly.

Separately, the core can fail if the shear stress exceeds the core's shear strength $τ_{c, allow}$ . The transverse shear force $V$ (in newtons) is carried across the core area $b c$ , so:

$τ_{c} \approx \frac{V}{b c} \leq τ_{c, allow} .$

Do not confuse the two properties: $G_{c}$ controls extra sag, $τ_{c, allow}$ controls rupture. This shear channel is why honeycomb, with its high $G_{c} / ρ$ (shear stiffness per unit density — lots of sliding-resistance for very little mass), beats foam for structural panels.

PICTURE. Left, a long slender panel bending smoothly. Right, a short deep panel whose core visibly shears — parallel arrows sliding the faces sideways across the depth $c$ .

Step 7 — Two degenerate limits, sanity-checked

WHAT. Push the geometry to its extremes and check the formula against something we already trust.

WHY. A trustworthy formula must behave sensibly at the edges — this is how we catch mistakes without a single test rig. But we must apply it only where it is valid.

Collapse the core, $h \to 2 t_{f}$ (a solid double-thick plate). Here the boxed formula's own assumption $t_{f} ≪ h$ has broken — the faces now are the whole depth — so we must not quote the boxed result. Going back to the exact expression: a solid plate of thickness $2 t_{f}$ has the honest value $I_{solid} = \frac{b ( 2 t _{f} ) ^{3}}{12} = \frac{2}{3} b t_{f}^{3}$ , i.e. $(E I)_{exact} = \frac{2}{3} E_{f} b t_{f}^{3}$ . (The boxed formula naively plugged in would give $2 E_{f} b t_{f}^{3}$ , three times too big — precisely the warning that the thin-face approximation is illegal here.) The lesson: the sandwich magic exists only while the faces are genuinely far apart. When they touch, you have an ordinary plate and must use the ordinary formula. Good: no free lunch, and the formula honestly announces where it stops applying.
Weightless spacer that grows, $h$ large, mass fixed. Since $(E I) \propto h^{2}$ , doubling depth quadruples stiffness at (almost) no mass cost — until Step 6's shear term or face wrinkling steps in and caps how thin/tall you dare go. The formula rewards depth; physics eventually taxes it.

PICTURE. A dial sweeping $h$ from small to large, the stiffness curve rising as $h^{2}$ , with a red "wrinkling / shear limit" wall where the reward stops being free — and a marker at the small- $h$ end flagged "thin-face formula invalid, use exact plate".

The one-picture summary

Everything above in a single frame: the flat slab (weak), the same mass pulled into faces + core (strong), the $σ \propto y$ triangle explaining why, and the boxed $(E I) = E_{f} b t_{f} h^{2} /2$ with its two survivors and two casualties labelled.

Recall Feynman retelling — say it back in plain words

When a beam bends, the fibres near the middle barely feel anything and the fibres at the surface feel the most — stress grows straight-line with distance from the centre line. So keeping strong material in the middle is a waste. The clever move is to slice the material into two thin sheets and shove them as far apart as possible, with a nearly weightless spacer holding them there. How much do we gain? The parallel-axis theorem says each sheet's contribution to the stiffness grows with the square of how far it sits from the centre line, so tall beats thick every time — that squared distance is where all the reward lives. We then throw away two tiny terms — each sheet's own floppy stiffness (it goes like thickness cubed, which is minuscule for a thin sheet) and the airy core's own bending (its material stiffness is a thousand times softer than the faces) — and what's left is $(E I) = E_{f} b t_{f} h^{2} /2$ : face material stiffness times width times face thickness times depth-squared, all over two. To use it you drop that stiffness into any sag formula: droop scales as length-cubed over $(E I)$ , so a big $(E I)$ means a tiny droop. Three warnings ride along. First, the formula only works when the faces are genuinely thin next to the depth — collapse the core so the faces touch and it overpredicts threefold, so there you must use the ordinary solid-plate formula instead. Second, it only counts bending sag, so short stubby panels also sag by shearing the core sideways like a sliding deck of cards; only the core of thickness $c$ carries that shear, softened by a $κ \approx 5/6$ correction that repairs the "uniform shear" fib. Third, if you make the panel too tall-and-thin the faces buckle or wrinkle — so the $h^{2}$ reward is free only up to a limit, beyond which physics collects its tax. That is the whole story: put strong material far out, let a light core hold it there, and count only what actually carries the load.

Recall Quick self-test

Why is material at the neutral axis wasted? ::: Because bending stress $σ = M y / I$ is proportional to $y$ , so at $y = 0$ the stress is zero — that material carries no bending load. Which term in $I_{face}$ dominates and why? ::: The parallel-axis term $A_{f} d^{2}$ , because it scales with $t_{f} h^{2}$ while the own-shape term scales with $t_{f}^{3}$ ; their ratio is $\frac{1}{3} (t_{f} / h)^{2} ≪ 1$ . What area carries the core shear stress, and what does $G_{c}$ do versus $τ_{c, allow}$ ? ::: The core area $b c$ carries it; $G_{c}$ (shear modulus, stiffness) sets extra shear deflection, $τ_{c, allow}$ (shear strength) sets when the core ruptures. When does the boxed $(E I)$ formula break down? ::: When faces are not thin ( $t_{f} \neq ≪ h$ ) or the core is stiff ( $E_{c} \neq ≪ E_{f}$ ); at $h \to 2 t_{f}$ it overpredicts threefold, so use the exact plate formula there.

Related vault topics: Beam Bending Theory, Composite Materials, Buckling and Instability, Adhesive Bonding, Vibration and Modal Analysis, Finite Element Analysis.