Sandwich structures — face sheets, core: Worked examples

This page is the practice arena for the parent topic. The parent built the formulae; here we exercise them across every case they can face — thin faces, thick faces, dead core, dominant shear, degenerate zero-core, and a couple of exam-style traps. Work each example before reading the steps: the "Forecast" line is your cue to guess.

Before any formula, here is the cast of symbols — every one is a real, measurable thing:

Definition The symbols, in plain words and pictures

$b$ — the width of the beam (metres), measured across the panel, into the page.
$t_{f}$ — the face-sheet thickness (metres): how thick each of the two thin skins is.
$h$ — the total depth (metres): top surface to bottom surface, $h = 2 t_{f} + c$ where $c$ is core thickness.
$L$ — the span (metres): the free distance between the two supports the beam bridges.
$E_{f}$ — the face Young's modulus (Pa): stiffness of the skin material (stress needed per unit stretch).
$E_{c}, G_{c}$ — the core's stretch stiffness and shear stiffness (Pa): $G_{c}$ is stress ÷ shear angle, i.e. how hard it is to slide the core sideways.
$w$ — the line load (N/m): downward force per metre of span, e.g. panel pressure $p$ times width, $w = p b$ .
$δ$ — the deflection (m): how far the beam droops at mid-span.
$(E I)$ — the bending stiffness (N·m²): resistance to bending, built from modulus $\times$ second moment of area $I$ .

Recall The three formulae we will stress-test (all symbols above)

Bending stiffness (thin faces): $(E I) = \frac{E _{f} b t _{f} h ^{2}}{2}$ — with $b, t_{f}, h, E_{f}$ as defined above. Simply-supported beam deflection under uniform line load $w$ : $δ_{b} = \frac{5 w L ^{4}}{384 ( E I )}$ . Shear deflection add-on: $δ_{s} = \frac{w L ^{2}}{8 G _{c} b h}$ (derived, coefficient and all, in Cell C).

The scenario matrix

The variables that decide which physics dominates are: the face-to-depth ratio $t_{f} / h$ , the span-to-depth ratio $L / h$ , the core stiffness $E_{c} / E_{f}$ , and the failure mode we test against. Here is the full grid of case classes this topic throws at you, and the example that hits each cell.

Cell	Case class	What is extreme	Covered by
A	Thin faces, both terms	$t_{f} ≪ h$ , compute bending AND shear honestly	Ex 1
B	Thick faces (approx. fails)	$t_{f} / h$ not small → keep exact $I$	Ex 2
C	Long span, shear-dominated	$L / h$ large → $δ_{s}$ matters	Ex 3
D	Degenerate: zero core thickness	$h \to 2 t_{f}$ , faces touch	Ex 4
E	Degenerate: dead/soft core	$E_{c} / E_{f} \to 0$ vs. finite	Ex 5
F	Failure mode: core shear vs. face wrinkling	which breaks first	Ex 6
G	Real-world word problem	mass-budget optimisation	Ex 7
H	Exam twist: unit-slip + limiting sanity	catch the decimal trap	Ex 8

Read the matrix top-to-bottom: as $t_{f} / h$ grows the thin-face shortcut dies (A→B), as $L / h$ grows shear wakes up (A→C), and as the core degenerates the whole idea collapses back into a solid or hollow plate (D, E). We show all of it.

Figure — Sandwich structures — face sheets, core

Cell A — Thin faces, both terms computed honestly

Forecast: Guess before computing — will the bending deflection be sub-millimetre, a few millimetres, or centimetres? (Pick one of those three boxes.)

Compute $(E I)$ . Why this step? Deflection needs stiffness; with $t_{f} / h = 0.5/20 = 0.025 ≪ 1$ the thin-face formula is legal. $(E I) = \frac{E _{f} b t _{f} h ^{2}}{2} = \frac{150 \times 1 0 ^{9} \cdot 0.8 \cdot 0.0005 \cdot 0.0 2 ^{2}}{2} = 1.2 \times 1 0^{4} Nm^{2}$
Plug into deflection. Why this step? The load is uniform and the beam simply supported, so the $\frac{5 w L ^{4}}{384 E I}$ shape applies. $δ_{b} = \frac{5 w L ^{4}}{384 ( E I )} = \frac{5 \cdot 160 \cdot 1. 5 ^{4}}{384 \cdot 1.2 \times 1 0 ^{4}} = \frac{4050}{4.608 \times 1 0 ^{6}} \approx 8.79 \times 1 0^{- 4} m$

Verify: $δ_{b} \approx 0.88 mm$ — the sub-millimetre box, comfortably under a 5 mm target. Units: $\frac{N/m \cdot m ^{4}}{Nm ^{2}} = m$ . ✓ If you guessed "sub-millimetre" you nailed it.

Cell B — Thick faces (the shortcut breaks)

Forecast: The shortcut drops the face's own self-inertia $\frac{b t _{f}^{3}}{12}$ and pretends the face centroid sits at $h /2$ . Guess: does the shortcut over- or under-estimate stiffness?

Exact face centroid distance. Why this step? The face centre is at $d = \frac{h}{2} - \frac{t _{f}}{2}$ , not $\frac{h}{2}$ . With $t_{f} / h = 0.2$ the difference is 10%. $d = \frac{h - t _{f}}{2} = \frac{0.020 - 0.004}{2} = 0.008 m$
Exact $I$ per unit modulus (two faces, each with self-inertia + parallel axis): $I_{exact} = 2 [\frac{b t _{f}^{3}}{12} + b t_{f} d^{2}] = 2 [\frac{0.8 \cdot 0.00 4 ^{3}}{12} + 0.8 \cdot 0.004 \cdot 0.00 8^{2}]$ $= 2 [4.267 \times 1 0^{- 9} + 2.048 \times 1 0^{- 7}] = 4.181 \times 1 0^{- 7} m^{4}$
Shortcut $I$ . Why this step? To measure the sin we replace $d \to h /2$ and drop self-inertia: $I_{thin} = \frac{b t _{f} h ^{2}}{2} = \frac{0.8 \cdot 0.004 \cdot 0.0 2 ^{2}}{2} = 6.4 \times 1 0^{- 7} m^{4}$

Verify: ratio $I_{thin} / I_{exact} = 6.4/4.181 \approx 1.53$ . The shortcut over-predicts stiffness by 53% — dangerously unconservative for fat faces. Forecast confirmed: it over-estimates.

Cell C — Long span, shear wakes up

First we must earn the shear-deflection formula, coefficient and all.

Deriving the coefficient (where the "8" comes from). Why bother? So the constant is earned, not memorised.

Shear strain from shear force. The shear angle is $γ = \frac{V}{G _{c} A _{s}}$ , where $V$ is the transverse shear force at a section. Why? By definition $G_{c} = shear stress / γ$ and shear stress $= V / A_{s}$ ; rearrange.
Extra slope = shear angle. Each slice tilts by $γ$ , so the shear contribution to beam slope is $\frac{d δ _{s}}{d x} = γ = \frac{V ( x )}{G _{c} A _{s}}$ . Why integrate $V$ ? Slope accumulates into deflection along the span.
Uniform load gives $V (x)$ . For a simply-supported beam with line load $w$ , $V (x) = w (\frac{L}{2} - x)$ (zero at mid-span, $\pm w L /2$ at supports). Integrate the slope twice with $δ_{s} = 0$ at both supports: $δ_{s}^{m a x} = \frac{1}{G _{c} A _{s}} \int \int V d x d x_{mid} = \frac{w L ^{2}}{8 G _{c} A _{s}} .$ The 8 is purely the geometry of integrating the triangular $V (x)$ of a uniformly loaded simple beam — a different load (point load) would give a different constant.

Where does the $L / h > 20$ rule come from? Why a dimensionless threshold? Form the ratio of the two deflections; everything but the geometry cancels. $\frac{δ _{s}}{δ _{b}} = \frac{w L ^{2} / ( 8 G _{c} bh )}{5 w L ^{4} / ( 384 E I )} = \frac{384 ( E I )}{40 G _{c} bh L ^{2}} .$ Substituting $(E I) = E_{f} b t_{f} h^{2} /2$ gives $\frac{δ _{s}}{δ _{b}} \propto \frac{E _{f} t _{f} h}{G _{c} L ^{2}}$ — it falls like $(h / L)^{2}$ . So as the beam gets long ( $L / h$ big) shear shrinks relative to bending... but only if $G_{c}$ is large. The rough " $L / h > 20$ " is not universal: it is the point where, for typical honeycomb ratios $E_{f} / G_{c} \sim 200$ – $500$ , $δ_{s} / δ_{b}$ climbs past a few percent. The honest rule is "compute the dimensionless ratio above," which we do next.

Forecast: Guess whether $δ_{s}$ is a small correction (<20%) or comparable to $δ_{b}$ .

Shear deflection. Why this step? The core sliding adds droop the bending formula missed; use the just-derived $δ_{s}$ with $A_{s} \approx bh$ . $δ_{s} = \frac{160 \cdot 1. 5 ^{2}}{8 \cdot 300 \times 1 0 ^{6} \cdot 0.8 \cdot 0.02} = \frac{360}{3.84 \times 1 0 ^{7}} \approx 9.375 \times 1 0^{- 6} m$
Total. Why this step? Deflections from independent mechanisms superpose. $δ = δ_{b} + δ_{s} = 8.79 \times 1 0^{- 4} + 9.375 \times 1 0^{- 6} \approx 8.88 \times 1 0^{- 4} m$

Verify: $δ_{s} / δ_{b} = 9.375 \times 1 0^{- 6} /8.79 \times 1 0^{- 4} \approx 1.1%$ . Units of $δ_{s}$ : $\frac{( N/m ) m ^{2}}{Pa \cdot m ^{2}} = \frac{N \cdot m}{N} = m$ ✓.

Cell D — Degenerate: zero core thickness

Forecast: With no core, the two faces sit back-to-back — this is just a solid plate of thickness $2 t_{f}$ . Predict what $(E I)$ the thin-face formula gives and whether it matches the honest solid-plate value.

Thin-face formula at $h = 2 t_{f}$ . Why this step? Test the boundary of validity. $(E I)_{thin} = \frac{E _{f} b t _{f} h ^{2}}{2} = \frac{E _{f} b t _{f} ( 2 t _{f} ) ^{2}}{2} = 2 E_{f} b t_{f}^{3}$
Honest solid plate of thickness $2 t_{f}$ : $(E I)_{solid} = E_{f} \cdot \frac{b ( 2 t _{f} ) ^{3}}{12} = \frac{8 E _{f} b t _{f}^{3}}{12} = \frac{2}{3} E_{f} b t_{f}^{3}$

Verify: ratio $= \frac{2 E _{f} b t _{f}^{3}}{\frac{2}{3} E _{f} b t _{f}^{3}} = 3$ . The shortcut is 3× too stiff at zero core — exactly the failure we saw in Ex 2 taken to the limit ( $t_{f} / h = 0.5$ ). This is the same disease: the thin-face formula is only trustworthy far from this degenerate corner.

Cell E — Degenerate: dead vs. soft core

Forecast: The parent claimed $E_{c} / E_{f} \sim 1 0^{- 3}$ makes core bending negligible. Guess the percentage contribution of the core.

Face stiffness. Why this step? Baseline the dominant term. $E_{f} I_{faces} = E_{f} \cdot \frac{b t _{f} h ^{2}}{2} = 70 \times 1 0^{9} \cdot \frac{0.8 \cdot 0.0008 \cdot 0.025 6 ^{2}}{2} \approx 1.468 \times 1 0^{4} Nm^{2}$
Core stiffness (honeycomb). Why this step? Core is a solid rectangle of depth $c$ about the neutral axis. $E_{c} I_{core} = 0.1 \times 1 0^{9} \cdot \frac{b c ^{3}}{12} = 0.1 \times 1 0^{9} \cdot \frac{0.8 \cdot 0.02 4 ^{3}}{12} \approx 92.2 Nm^{2}$

Verify: contribution $= 92.2/ (1.468 \times 1 0^{4} + 92.2) \approx 0.62%$ . Below 1% — parent's neglect justified. For the dead core ( $E_{c} = 0$ ) the contribution is exactly $0%$ and $(E I)$ is unchanged to within 0.62%. Both degenerate limits land within experimental scatter of each other. ✓

Cell F — Failure mode: core shear vs. face wrinkling

Where the cube root comes from. Why derive it? So $σ_{wr} = C (E_{f} E_{c} G_{c})^{1/3}$ is understood, not memorised.

Forecast: Guess which margin is tighter — wrinkling or shear.

Wrinkling stress. Why this step? Compare applied $σ_{x}$ to the buckling threshold. $σ_{wr} = C (E_{f} E_{c} G_{c})^{1/3} = 0.5 (70 \times 1 0^{9} \cdot 0.1 \times 1 0^{9} \cdot 300 \times 1 0^{6})^{1/3}$ $= 0.5 (2.1 \times 1 0^{27})^{1/3} = 0.5 \cdot 1.281 \times 1 0^{9} \approx 6.40 \times 1 0^{8} Pa = 640 MPa$ Margin against wrinkling: $640/50 = 12.8$ .
Core shear stress. Why this step? Compare $τ_{c}$ to core strength. $τ_{c} = \frac{V}{bh} = \frac{4000}{0.8 \cdot 0.025} = 2.0 \times 1 0^{5} Pa = 0.2 MPa$ Margin against shear: $2.0/0.2 = 10$ .

Verify: wrinkling margin 12.8 vs. shear margin 10. Core shear is the tighter constraint — it will fail first if loads scale up. Units: $σ_{wr}$ from $(Pa^{3})^{1/3} = Pa$ ✓; $τ_{c}$ from $N / m^{2} = Pa$ ✓. Forecast: whoever guessed "shear" wins.

Cell G — Real-world word problem: mass-budget optimisation

Forecast: $(E I) \propto h^{2}$ , so shrinking $h$ from 20→14 mm cuts stiffness by $(14/20)^{2} = 0.49$ . Predict whether SF stays above 2, and roughly how much mass drops.

New $(E I)$ . Why this step? Confirm we still clear the requirement. $(E I) = \frac{150 \times 1 0 ^{9} \cdot 0.8 \cdot 0.0005 \cdot 0.01 4 ^{2}}{2} = 5.88 \times 1 0^{3} Nm^{2}$
Safety factor. Why this step? Divide by requirement. $SF = \frac{5.88 \times 1 0 ^{3}}{2637} \approx 2.23 \geq 2 ✓$
New mass. Why this step? Only the core term shrinks ( $h$ : 20→14 mm); faces unchanged. $m = 2 ρ_{f} b L t_{f} + ρ_{c} b L h = 1.92 + 50 \cdot 0.8 \cdot 1.5 \cdot 0.014 = 1.92 + 0.84 = 2.76 kg$

Verify: SF $2.23 \geq 2$ ✓, mass fell from $3.12 \to 2.76 kg$ — a $0.36 kg$ (11.5%) saving, all from the core. Forecast confirmed: stiffness roughly halved ( $0.49 \times$ ) yet stayed above threshold because we started 4.5× over-designed.

Cell H — Exam twist: unit-slip and a limiting sanity check

Forecast: The face-mass term is $2 ρ_{f} b L t_{f}$ , linear in $t_{f}$ . Predict the corrupted face mass if $t_{f}$ is wrongly 10× too big.

Reproduce the wrong number. Why this step? See where 20.4 comes from. $m_{face,wrong} = 2 \cdot 1600 \cdot 0.8 \cdot 1.5 \cdot 0.005 = 19.2 kg, m_{wrong} = 19.2 + 1.2 = 20.4 kg$
Correct face mass with $t_{f} = 0.0005$ m. Why this step? Restore the true thickness. $m_{face} = 2 \cdot 1600 \cdot 0.8 \cdot 1.5 \cdot 0.0005 = 1.92 kg, m = 1.92 + 1.2 = 3.12 kg$
Limiting sanity check. Why this step? Turn intuition into a guard rail. A $0.5 mm$ CFRP skin over $0.8 \times 1.5 m$ has face volume $2 \cdot 0.0005 \cdot 1.2 = 1.2 \times 1 0^{- 3} m^{3}$ ; at $1600 kg/m^{3}$ that is $1.92 kg$ . Compare to a solid CFRP slab filling the whole $20 mm$ gap: $1600 \cdot 0.8 \cdot 1.5 \cdot 0.02 = 38.4 kg$ . Two thin 0.5 mm skins cannot be half that slab's mass — 19.2 kg is physically absurd for skins that are $1/20$ of the depth.

Verify: wrong 20.4 kg vs. correct 3.12 kg — factor $20.4/3.12 \approx 6.5$ off (the 10× face-term error diluted by the correct 1.2 kg core term). Correct total 3.12 kg matches Ex 1's parent value ✓.

Recall Self-test

Why does the thin-face $(E I)$ formula over-predict at $t_{f} / h = 0.5$ ? ::: The neglected self-inertia $\frac{b t _{f}^{3}}{12}$ and the $d \approx h /2$ approximation both break; at the zero-core limit it is 3× too stiff (Ex 4). In Ex 3, what fraction was shear deflection and why so small? ::: About 1%, because aluminium honeycomb's $G_{c} = 300$ MPa is high; a soft foam would make it ~30%. Where does the "8" in $δ_{s} = w L^{2} / (8 G_{c} A_{s})$ come from? ::: From integrating the triangular shear force $V (x) = w (L /2 - x)$ of a uniformly loaded simple beam — pure geometry of the load. In Ex 6, which mode governed and by what margin? ::: Core shear, margin 10 vs. wrinkling margin 12.8.

How every cell got covered

Related tools if you want to go further: Beam Bending Theory for the deflection formulae, Buckling and Instability for face wrinkling, Composite Materials for CFRP face properties, Adhesive Bonding for the shear-transfer assumption, and Finite Element Analysis to check these hand calcs.