Sandwich structures — face sheets, core: Exercises

Before any numbers, one figure fixes every symbol we use below.

Figure — Sandwich structures — face sheets, core

Figure s01 (left): the sandwich cross-section, seen end-on. The two blue bars are the face sheets, thickness $t_{f}$ each; the yellow block between them is the core, thickness $c$ ; the whole stack has depth $h$ and width $b$ . The red dashed line is the neutral axis — the height where bending stress is exactly zero. The green arrow labelled $d$ is the distance from that axis up to the centre of the top face. (Right): the stress triangle. Bending stress grows straight-line with height $y$ , so it is largest at the very top and bottom — which is exactly where the blue faces sit. That picture is the whole reason sandwiches work.

Definition The symbols on this page (all measured on the figure above)

$b$ — width of the panel (into the page), in metres.
$t_{f}$ — thickness of one face sheet (the thin strong skin), in metres.
$c$ — core thickness (the light filling between the faces).
$h$ — total depth, $h = 2 t_{f} + c$ . This is the full sandwich height.
$L$ — length (span) of the beam/panel, measured along its long direction, in metres.
$y$ — height above the neutral axis (metres), measured on figure s01; $y = 0$ at the axis, $y = \pm h /2$ at the faces.
$d$ — distance from the neutral axis (the horizontal line where bending stress is zero) to the centre of a face, $d = \frac{h}{2} - \frac{t _{f}}{2}$ . For thin faces $d \approx h /2$ — this is the moment arm each face sits at.
$σ$ — normal (bending/axial) stress in a face: force per unit area (pascals) pushing or pulling along the beam. $σ_{x}$ is that stress in the beam's length direction $x$ (compression counted here as a positive magnitude in the failure check).
$M$ — bending moment (N·m): the "turning effort" the load applies at a cross-section, trying to curve the beam.
$I$ — second moment of area (m⁴): a purely geometric number saying how the cross-section's area is spread away from the neutral axis. Area far out (big $y$ ) counts much more, because it enters as $y^{2}$ . This is exactly why sandwiches push material outward.
$E_{f}$ — Young's modulus of the face material: how many pascals of stress it takes to stretch it by 100%. Big $E_{f}$ = stiff.
$E_{c}$ — Young's modulus of the core. Because $E_{c} ≪ E_{f}$ (typically $E_{c} / E_{f} \sim 1 0^{- 3}$ ), the core barely resists bending; we keep it only for the wrinkling check.
$(E I)$ — bending stiffness: how hard the whole beam resists being bent. Bigger = stiffer.
$G_{c}$ — core shear modulus: how hard the core resists sliding its top face relative to its bottom face.
$τ_{c}$ — the actual shear stress carried by the core; $τ_{c, allow}$ is the most it can take before failing.
$ρ_{f}$ — density of the face material (mass per unit volume, kg/m³); $ρ_{c}$ — density of the core (much smaller).
$w$ — line load: force per unit length spread along the span (N/m). It comes from a pressure times the width.
$F$ — a single point force (N); $V$ — the transverse shear force the core must carry (N).
$C$ — a dimensionless wrinkling constant (no units), roughly $0.5$ – $0.82$ depending on boundary conditions; it appears only in the face-wrinkling formula in Level 5.
$δ$ — a deflection: how far the beam actually moves under load (metres). $δ_{b}$ is the part from bending, $δ_{s}$ the part from core shear.

Level 1 — Recognition

Problem 1.1

In a sandwich panel, which part carries most of the bending (axial) stress, and which part carries the shear? Why does putting strong material far from the middle help?

Recall Solution

The face sheets carry the bending/axial stress; the core carries the shear and holds the faces apart. In bending, stress is $σ = \frac{M y}{I}$ — here $σ$ is the normal stress, $M$ the bending moment, $y$ the height above the neutral axis, and $I$ the second moment of area (all defined in the symbol list). Since $σ \propto y$ , it grows with distance $y$ from the neutral axis, so the material farthest out does the most useful work. Placing the strong faces at $y = \pm h /2$ puts them exactly where stress is largest, and the light core just maintains that separation cheaply (in mass). Look at figure s01 (right panel): the red stress line is longest at the top and bottom, and the blue face dots sit right there — the faces live where the stress is biggest.

Problem 1.2

A sandwich has $h = 20 mm$ and $t_{f} = 0.5 mm$ . Compute the core thickness $c$ .

Recall Solution

$c = h - 2 t_{f} = 20 - 2 (0.5) = 19 mm$ Answer: $c = 19 mm$ . The core takes up almost the whole depth — the faces are wafer-thin.

Level 2 — Application

Problem 2.1

Face sheets: $E_{f} = 70 GPa$ , $t_{f} = 0.8 mm$ . Panel width $b = 0.5 m$ , total depth $h = 25 mm$ . Find $(E I)_{sandwich}$ .

Recall Solution

Plug into the formula (all SI: metres, pascals): $(E I) = \frac{E _{f} b t _{f} h ^{2}}{2} = \frac{70 \times 1 0 ^{9} \times 0.5 \times 0.0008 \times ( 0.025 ) ^{2}}{2}$ Compute the top: $70 \times 1 0^{9} \times 0.5 = 3.5 \times 1 0^{10}$ ; times $0.0008 = 2.8 \times 1 0^{7}$ ; times $(0.025)^{2} = 6.25 \times 1 0^{- 4}$ gives $1.75 \times 1 0^{4}$ . Divide by 2: $(E I) \approx 8.75 \times 1 0^{3} Nm^{2}$ Answer: $(E I) \approx 8750 Nm^{2}$ .

Problem 2.2

The same panel is a cantilever of length $L = 0.6 m$ with a tip load $F = 40 N$ . Find the tip deflection $δ = \frac{F L ^{3}}{3 ( E I )}$ (the cantilever formula from the formula box above).

Recall Solution

$δ = \frac{40 \times ( 0.6 ) ^{3}}{3 \times 8750} = \frac{40 \times 0.216}{26250} = \frac{8.64}{26250} \approx 3.29 \times 1 0^{- 4} m$ Answer: $δ \approx 0.33 mm$ — a stiff, light beam.

Level 3 — Analysis

Problem 3.1

Compare a sandwich (thin faces $t_{f}$ , depth $h$ ) against a solid plate of the same material and same mass. Show the stiffness ratio is $\approx \frac{3 h ^{2}}{4 t _{f}^{2}}$ , and evaluate it for $h = 20 mm$ , $t_{f} = 0.5 mm$ (assume core density negligible so $t_{s} \approx 2 t_{f}$ ).

Recall Solution

Where does $I_{sand} \approx \frac{b t _{f} h ^{2}}{2}$ come from? Recall the moment arm $d \approx h /2$ from the symbol list and figure s01. Each face is a thin rectangle of area $b t_{f}$ sitting at distance $d \approx h /2$ from the neutral axis. The parallel-axis theorem says its second moment of area is (area) $\times d^{2} = b t_{f} (h /2)^{2} = b t_{f} h^{2} /4$ . There are two faces (top and bottom), so we double it: $2 \times b t_{f} h^{2} /4 = b t_{f} h^{2} /2$ . That factor of $\frac{1}{2}$ is literally "two faces, each with a $(1/2)^{2} = 1/4$ ." (Its own bending term $b t_{f}^{3} /12$ is tiny for thin faces, so we drop it.)

Equal mass with negligible core density means the solid plate uses all the mass of the two faces: $ρ_{s} b t_{s} = 2 ρ_{f} b t_{f}$ , so $t_{s} \approx 2 t_{f}$ . Second moments of area: $I_{sand} \approx \frac{b t _{f} h ^{2}}{2}, I_{solid} = \frac{b t _{s}^{3}}{12} = \frac{b ( 2 t _{f} ) ^{3}}{12} = \frac{8 b t _{f}^{3}}{12}$ Ratio: $\frac{I _{sand}}{I _{solid}} = \frac{b t _{f} h ^{2} /2}{8 b t _{f}^{3} /12} = \frac{6 t _{f} h ^{2}}{8 t _{f}^{3}} = \frac{3 h ^{2}}{4 t _{f}^{2}}$ For $h = 20 mm$ , $t_{f} = 0.5 mm$ : $\frac{3 ( 20 ) ^{2}}{4 ( 0.5 ) ^{2}} = \frac{3 \times 400}{4 \times 0.25} = \frac{1200}{1} = 1200$ . Answer: the sandwich is $\approx 1200 \times$ stiffer for the same mass.

Figure s02 plots this ratio $3 h^{2} /4 t_{f}^{2}$ against $h / t_{f}$ : it is a rising parabola, and the yellow marker at $h / t_{f} = 40$ (our case) lands on $1200$ . The curve makes the message visual — the payoff climbs quadratically as you spread the same material further apart.

Problem 3.2

A long panel ( $L / h = 25$ ) is not stiff enough. You may either double $h$ or double $t_{f}$ . Which raises $(E I)$ more? Which adds more mass? Use $ρ_{f} = 1600 kg/m^{3}$ (CFRP face), $ρ_{c} = 50 kg/m^{3}$ (honeycomb core), and a starting geometry $t_{f} = 0.5 mm$ , $h = 20 mm$ ( $c = 19 mm$ ).

Recall Solution

$(E I) \propto t_{f} h^{2}$ .

Doubling $t_{f}$ : $(E I)$ becomes $2 \times$ . Face mass ( $\propto t_{f}$ ) also doubles.
Doubling $h$ : $(E I)$ becomes $4 \times$ (the $h^{2}$ ). Core thickness (hence core mass, $\propto h$ ) roughly doubles; face mass is unchanged. Quantify the mass claim. Per unit area ( $b L = 1$ ), from $m = 2 ρ_{f} t_{f} + ρ_{c} h$ : $m_{0} = 2 (1600) (0.0005) + 50 (0.020) = 1.60 + 1.00 = 2.60 kg/m^{2}$
After doubling $t_{f}$ ( $t_{f} = 1.0 mm$ ): $m = 2 (1600) (0.001) + 50 (0.020) = 3.20 + 1.00 = 4.20 kg/m^{2}$ — a $+ 62%$ jump, all in the heavy faces.
After doubling $h$ ( $h = 40 mm$ , core $\approx 39 mm$ ): $m \approx 2 (1600) (0.0005) + 50 (0.040) = 1.60 + 2.00 = 3.60 kg/m^{2}$ — a $+ 38%$ rise, and only in the light core.

So doubling $h$ gives $4 \times$ the stiffness for a $+ 38%$ mass rise, while doubling $t_{f}$ gives only $2 \times$ for $+ 62%$ . Doubling the depth $h$ wins on both counts — the extra mass sits in the light core ( $ρ_{c} = 50$ is $32 \times$ lighter than $ρ_{f} = 1600$ ), whereas thickening the faces piles mass onto the heavy skins. (This is the same quadratic shown in figure s02: stepping right on the $h / t_{f}$ axis buys stiffness fast.)

Level 4 — Synthesis

Problem 4.1 (full design)

Design a simply-supported panel, span $L = 1.5 m$ , width $b = 0.8 m$ , carrying a uniform line load $w = 160 N/m$ (from a $5 g$ launch). Requirement: mid-span deflection $δ < 5 mm$ . (a) Find the required $(E I)$ . (b) With CFRP faces $E_{f} = 150 GPa$ , choose $h = 20 mm$ , $t_{f} = 0.5 mm$ and check the safety factor. (c) Find total mass with $ρ_{f} = 1600$ , $ρ_{c} = 50 kg/m^{3}$ .

Recall Solution

(a) Required stiffness. Simply-supported uniform-load deflection (from the formula box): $δ = \frac{5 w L ^{4}}{384 ( E I )} \Rightarrow (E I)_{req} = \frac{5 w L ^{4}}{384 δ} = \frac{5 \times 160 \times ( 1.5 ) ^{4}}{384 \times 0.005}$ $(1.5)^{4} = 5.0625$ . Top $= 5 \times 160 \times 5.0625 = 4050$ . Bottom $= 384 \times 0.005 = 1.92$ . $(E I)_{req} = \frac{4050}{1.92} \approx 2109 Nm^{2}$ (b) Provided stiffness. Substitute (metres, pascals): $(E I) = \frac{150 \times 1 0 ^{9} \times 0.8 \times 0.0005 \times ( 0.02 ) ^{2}}{2}$ Build it step by step: $150 \times 1 0^{9} \times 0.8 = 1.2 \times 1 0^{11}$ ; times $0.0005 = 6 \times 1 0^{7}$ . Now the depth term — square $0.02$ carefully: $(0.02)^{2} = 0.0004 = 4 \times 1 0^{- 4}$ (note the two zeros doubling to four, a classic unit-slip point). So $6 \times 1 0^{7} \times 4 \times 1 0^{- 4} = 2.4 \times 1 0^{4}$ ; divide by 2: $(E I) = 1.2 \times 1 0^{4} Nm^{2}, SF = \frac{1.2 \times 1 0 ^{4}}{2109} \approx 5.7$ Comfortably safe (we could trim mass later). (c) Mass. $m = 2 ρ_{f} b L t_{f} + ρ_{c} b L h$ $= 2 (1600) (0.8) (1.5) (0.0005) + (50) (0.8) (1.5) (0.02) = 1.92 + 1.20 = 3.12 kg$ Answers: $(E I)_{req} \approx 2109 Nm^{2}$ ; provided $1.2 \times 1 0^{4} Nm^{2}$ (SF $\approx 5.7$ ); mass $\approx 3.12 kg$ .

Level 5 — Mastery

Problem 5.1 (competing failure modes)

Aluminium-faced panel: $E_{f} = 70 GPa$ , $t_{f} = 0.8 mm$ , $h = 25 mm$ , honeycomb $E_{c} = 0.1 GPa$ , $G_{c} = 300 MPa$ , width $b = 0.8 m$ . Faces are in compression at $σ_{x} = 50 MPa$ . (a) Compute the face-wrinkling stress $σ_{wr} = C (E_{f} E_{c} G_{c})^{1/3}$ with $C = 0.5$ . (b) Does the face wrinkle at $σ_{x} = 50 MPa$ ? (c) The panel also carries transverse shear $V = 1200 N$ ; core shear strength is $τ_{c, allow} = 1.5 MPa$ . Check core shear with $τ_{c} \approx V / (bh)$ . What fails first?

Recall Solution

Here $σ_{x}$ is the compressive normal stress in the face (length direction $x$ ), $E_{c}$ is the core's Young's modulus, and $C$ is the dimensionless wrinkling constant (0.5–0.82) — all defined in the symbol list. We use $C = 0.5$ (a conservative choice). (a) Wrinkling stress. Put all moduli in pascals: $E_{f} = 70 \times 1 0^{9}$ , $E_{c} = 0.1 \times 1 0^{9} = 1 0^{8}$ , $G_{c} = 300 \times 1 0^{6} = 3 \times 1 0^{8}$ . Product $= 70 \times 1 0^{9} \times 1 0^{8} \times 3 \times 1 0^{8} = 2.1 \times 1 0^{27}$ . Cube root: $(2.1 \times 1 0^{27})^{1/3}$ . Since $1 0^{27}$ cube-roots to $1 0^{9}$ and $2. 1^{1/3} \approx 1.28$ , we get $\approx 1.28 \times 1 0^{9}$ . $σ_{wr} = 0.5 \times 1.28 \times 1 0^{9} \approx 6.4 \times 1 0^{8} Pa = 640 MPa$ (b) Wrinkling check. Applied $σ_{x} = 50 MPa ≪ 640 MPa$ , so no wrinkling — huge margin ( $640/50 \approx 12.8 \times$ ). (c) Core shear. $τ_{c} \approx \frac{V}{bh} = \frac{1200}{0.8 \times 0.025} = \frac{1200}{0.02} = 6.0 \times 1 0^{4} Pa = 0.06 MPa$ Compare to $τ_{c, allow} = 1.5 MPa$ : margin $= 1.5/0.06 = 25 \times$ . Core does not fail either. Which is closer to failing? Wrinkling margin $\approx 12.8$ , core-shear margin $\approx 25$ . Both are safe, but the smaller margin is the nearer limit, so face wrinkling governs — it is the mode you would hit first if the load grew. Figure s03 draws these two margins as bars against the yellow "failure line" at 1: both bars tower over it, but the red wrinkling bar (12.8) is shorter than the green core-shear bar (25), showing at a glance which mode you would hit first if the load grew. Answers: $σ_{wr} \approx 640 MPa$ ; $τ_{c} \approx 0.06 MPa$ ; both safe, wrinkling is the governing (nearer) mode.

Problem 5.2 (shear deflection matters)

A slender panel has bending deflection $δ_{b} = 4 mm$ under load. The extra shear deflection is $δ_{s} = \frac{V L}{G _{c} bh}$ with $V = 500 N$ , $L = 1.6 m$ , $G_{c} = 50 MPa$ (soft foam core), $b = 0.6 m$ , $h = 0.03 m$ . Compute $δ_{s}$ and the total. Why can't we ignore it here?

Recall Solution

Why can we simply add $δ_{b}$ and $δ_{s}$ ? Under the small-deflection, linear-elastic assumptions stated at the top of this page, the bending deformation and the core-shear deformation are independent responses to the same load — they superpose. So the beam's total sag is the bending sag plus the shear sag, added directly. $δ_{s} = \frac{V L}{G _{c} bh} = \frac{500 \times 1.6}{50 \times 1 0 ^{6} \times 0.6 \times 0.03} = \frac{800}{9.0 \times 1 0 ^{5}} \approx 8.89 \times 1 0^{- 4} m \approx 0.89 mm$ Total $δ = δ_{b} + δ_{s} = 4 + 0.89 = 4.89 mm$ . Why it matters: the soft core ( $G_{c}$ only $50 MPa$ ) lets the faces slide relative to each other, adding $\sim 0.9 mm$ — over 20% of the bending deflection. For long, soft-cored panels ( $L / h$ large, small $G_{c}$ ) shear deflection is not a rounding error; ignoring it would predict $4 mm$ and quietly bust a $5 mm$ budget.

Recall Self-check summary

Which single geometric change quadruples bending stiffness at almost no mass cost? ::: Doubling the total depth $h$ (since $(E I) \propto h^{2}$ while the added mass sits in the light core). For which panels must you add shear deflection? ::: Long panels ( $L / h > 20$ ) or those with a soft, low- $G_{c}$ core. What two assumptions let us add bending and shear deflections directly? ::: Small deflections and linear elasticity, which together allow superposition.