Sandwich structures — face sheets, core: Worked examples

3.6.17 · D3 · Physics › Spacecraft Structures & Systems Engineering › Sandwich structures — face sheets, core

Yeh page parent topic ki practice arena hai. Parent ne formulae build kiye; yahan hum unhe har possible case mein exercise karte hain — thin faces, thick faces, dead core, dominant shear, degenerate zero-core, aur kuch exam-style traps. Har example ko steps padhne se pehle solve karo: "Forecast" line tumhara cue hai guess karne ka.

Kisi bhi formula se pehle, yeh hai symbols ki cast — har ek real, measurable cheez hai:

Definition Symbols, plain words aur pictures mein

$b$ — beam ki width (metres), panel ke across, page ke andar.
$t_{f}$ — face-sheet thickness (metres): do thin skins mein se har ek kitni moti hai.
$h$ — total depth (metres): top surface se bottom surface tak, $h = 2 t_{f} + c$ jahan $c$ core thickness hai.
$L$ — span (metres): beam jis do supports ke beech bridge karta hai unke beech ki free distance.
$E_{f}$ — face Young's modulus (Pa): skin material ki stiffness (stress needed per unit stretch).
$E_{c}, G_{c}$ — core ki stretch stiffness aur shear stiffness (Pa): $G_{c}$ hai stress ÷ shear angle, yaani core ko sideways slide karna kitna mushkil hai.
$w$ — line load (N/m): span ke per metre downward force, jaise panel pressure $p$ times width, $w = p b$ .
$δ$ — deflection (m): beam mid-span par kitna jhukta hai.
$(E I)$ — bending stiffness (N·m²): bending ke against resistance, modulus $\times$ second moment of area $I$ se banta hai.

Recall Teen formulae jo hum stress-test karenge (sab symbols upar hain)

Bending stiffness (thin faces): $(E I) = \frac{E _{f} b t _{f} h ^{2}}{2}$ — $b, t_{f}, h, E_{f}$ 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 aur sab, Cell C mein).

The scenario matrix

Jo variables decide karte hain kaun si physics dominate karti hai woh hain: face-to-depth ratio $t_{f} / h$ , span-to-depth ratio $L / h$ , core stiffness $E_{c} / E_{f}$ , aur failure mode jo hum test kar rahe hain. Yeh hai us topic ke case classes ka full grid jo tumhare saamne aata hai, aur woh example jo har cell ko hit karta hai.

Cell	Case class	Kya extreme hai	Covered by
A	Thin faces, dono terms	$t_{f} ≪ h$ , bending AND shear dono honestly compute karo	Ex 1
B	Thick faces (approx. fails)	$t_{f} / h$ chhota nahi → exact $I$ rakho	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 karte hain	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	pehle kya tootega	Ex 6
G	Real-world word problem	mass-budget optimisation	Ex 7
H	Exam twist: unit-slip + limiting sanity	decimal trap pakdo	Ex 8

Matrix ko top-to-bottom padho: jaise $t_{f} / h$ badhta hai thin-face shortcut khatam hota hai (A→B), jaise $L / h$ badhta hai shear jaag jaata hai (A→C), aur jaise core degenerate hota hai puri idea solid ya hollow plate mein collapse ho jaati hai (D, E). Yeh sab hum dikhate hain.

Figure — Sandwich structures — face sheets, core

Cell A — Thin faces, dono terms honestly compute kiye

Forecast: Computing se pehle guess karo — kya bending deflection sub-millimetre, kuch millimetres, ya centimetres hogi? (Teen boxes mein se ek chuno.)

$(E I)$ compute karo. Yeh step kyun? Deflection ke liye stiffness chahiye; $t_{f} / h = 0.5/20 = 0.025 ≪ 1$ hone se thin-face formula legal hai. $(E I) = \frac{E _{f} b t _{f} h ^{2}}{2} = \frac{150 \times 1 0 ^{9} \cdot 0.8 \cdot 0.0005 \cdot 0.0 2 ^{2}}{2} = 1.2 \times 1 0^{4} Nm^{2}$
Deflection mein plug karo. Yeh step kyun? Load uniform hai aur beam simply supported hai, isliye $\frac{5 w L ^{4}}{384 E I}$ shape apply hoti hai. $δ_{b} = \frac{5 w L ^{4}}{384 ( E I )} = \frac{5 \cdot 160 \cdot 1. 5 ^{4}}{384 \cdot 1.2 \times 1 0 ^{4}} = \frac{4050}{4.608 \times 1 0 ^{6}} \approx 8.79 \times 1 0^{- 4} m$

Verify: $δ_{b} \approx 0.88 mm$ — sub-millimetre box, 5 mm target se comfortably neeche. Units: $\frac{N/m \cdot m ^{4}}{Nm ^{2}} = m$ . ✓ Agar tumne "sub-millimetre" guess kiya tha to bilkul sahi tha.

Cell B — Thick faces (shortcut toot jaata hai)

Forecast: Shortcut face ki apni self-inertia $\frac{b t _{f}^{3}}{12}$ drop kar deta hai aur pretend karta hai ki face centroid $h /2$ par hai. Guess karo: shortcut stiffness ko over- ya under-estimate karta hai?

Exact face centroid distance. Yeh step kyun? Face centre $d = \frac{h}{2} - \frac{t _{f}}{2}$ par hai, $\frac{h}{2}$ par nahi. $t_{f} / h = 0.2$ hone se difference 10% hai. $d = \frac{h - t _{f}}{2} = \frac{0.020 - 0.004}{2} = 0.008 m$
Exact $I$ per unit modulus (do faces, har ek mein 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 $I$ . Yeh step kyun? Galti measure karne ke liye $d \to h /2$ replace karo aur self-inertia drop karo: $I_{thin} = \frac{b t _{f} h ^{2}}{2} = \frac{0.8 \cdot 0.004 \cdot 0.0 2 ^{2}}{2} = 6.4 \times 1 0^{- 7} m^{4}$

Verify: ratio $I_{thin} / I_{exact} = 6.4/4.181 \approx 1.53$ . Shortcut stiffness ko 53% over-predict karta hai — fat faces ke liye dangerously unconservative hai. Forecast confirmed: yeh over-estimate karta hai.

Cell C — Long span, shear jaag jaata hai

Pehle hume shear-deflection formula, coefficient samet, earn karna hoga.

Coefficient derive karna (yeh "8" kahan se aaya). Kyun bother karna? Taaki constant earned ho, memorised nahi.

Shear force se shear strain. Shear angle hai $γ = \frac{V}{G _{c} A _{s}}$ , jahan $V$ section par transverse shear force hai. Kyun? Definition se $G_{c} = shear stress / γ$ aur shear stress $= V / A_{s}$ ; rearrange karo.
Extra slope = shear angle. Har slice $γ$ se tilt karta hai, isliye slope mein shear contribution hai $\frac{d δ _{s}}{d x} = γ = \frac{V ( x )}{G _{c} A _{s}}$ . $V$ integrate kyun karna? Slope span ke along deflection mein accumulate hota hai.
Uniform load $V (x)$ deta hai. Simply-supported beam ke liye line load $w$ ke saath, $V (x) = w (\frac{L}{2} - x)$ (mid-span par zero, supports par $\pm w L /2$ ). Slope ko dono supports par $δ_{s} = 0$ ke saath twice integrate karo: $δ_{s}^{m a x} = \frac{1}{G _{c} A _{s}} \int \int V d x d x_{mid} = \frac{w L ^{2}}{8 G _{c} A _{s}} .$ 8 purely uniformly loaded simple beam ke triangular $V (x)$ ko integrate karne ki geometry hai — ek alag load (point load) ek alag constant deta.

$L / h > 20$ rule kahan se aaya? Dimensionless threshold kyun? Do deflections ka ratio banao; geometry ke alawa sab cancel ho jaata hai. $\frac{δ _{s}}{δ _{b}} = \frac{w L ^{2} / ( 8 G _{c} bh )}{5 w L ^{4} / ( 384 E I )} = \frac{384 ( E I )}{40 G _{c} bh L ^{2}} .$ $(E I) = E_{f} b t_{f} h^{2} /2$ substitute karne se $\frac{δ _{s}}{δ _{b}} \propto \frac{E _{f} t _{f} h}{G _{c} L ^{2}}$ milta hai — yeh $(h / L)^{2}$ ki tarah girta hai. Isliye jaise beam lamba hota hai ( $L / h$ bada) shear bending ke relative shrink karta hai... lekin sirf tab jab $G_{c}$ bada ho. Rough " $L / h > 20$ " universal nahi hai: yeh woh point hai jahan, typical honeycomb ratios $E_{f} / G_{c} \sim 200$ – $500$ ke liye, $δ_{s} / δ_{b}$ kuch percent se upar chad jaata hai. Honest rule hai "upar wala dimensionless ratio compute karo," jo hum aage karte hain.

Forecast: Guess karo ki $δ_{s}$ ek chhoti correction hai (<20%) ya $δ_{b}$ ke comparable.

Shear deflection. Yeh step kyun? Core sliding ne jo droop miss ki thi bending formula ne, use add karo; just-derived $δ_{s}$ use karo $A_{s} \approx bh$ ke saath. $δ_{s} = \frac{160 \cdot 1. 5 ^{2}}{8 \cdot 300 \times 1 0 ^{6} \cdot 0.8 \cdot 0.02} = \frac{360}{3.84 \times 1 0 ^{7}} \approx 9.375 \times 1 0^{- 6} m$
Total. Yeh step kyun? Independent mechanisms ke deflections superpose hote hain. $δ = δ_{b} + δ_{s} = 8.79 \times 1 0^{- 4} + 9.375 \times 1 0^{- 6} \approx 8.88 \times 1 0^{- 4} m$

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: Koi core nahi to do faces back-to-back baith jaate hain — yeh sirf ek solid plate of thickness $2 t_{f}$ hai. Predict karo ki thin-face formula kya $(E I)$ deta hai aur kya yeh honest solid-plate value se match karta hai.

Thin-face formula $h = 2 t_{f}$ par. Yeh step kyun? Validity ki boundary test karo. $(E I)_{thin} = \frac{E _{f} b t _{f} h ^{2}}{2} = \frac{E _{f} b t _{f} ( 2 t _{f} ) ^{2}}{2} = 2 E_{f} b t_{f}^{3}$
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 $= \frac{2 E _{f} b t _{f}^{3}}{\frac{2}{3} E _{f} b t _{f}^{3}} = 3$ . Shortcut 3× zyada stiff hai zero core par — exactly wahi failure jo Ex 2 mein limit tak le jaane par dikhi ( $t_{f} / h = 0.5$ ). Yeh same bimari hai: thin-face formula sirf is degenerate corner se door trustworthy hai.

Cell E — Degenerate: dead vs. soft core

Forecast: Parent ne claim kiya $E_{c} / E_{f} \sim 1 0^{- 3}$ core bending negligible banata hai. Core ka percentage contribution guess karo.

Face stiffness. Yeh step kyun? Dominant term ko baseline karo. $E_{f} I_{faces} = E_{f} \cdot \frac{b t _{f} h ^{2}}{2} = 70 \times 1 0^{9} \cdot \frac{0.8 \cdot 0.0008 \cdot 0.025 6 ^{2}}{2} \approx 1.468 \times 1 0^{4} Nm^{2}$
Core stiffness (honeycomb). Yeh step kyun? Core depth $c$ ka neutral axis ke baare mein solid rectangle hai. $E_{c} I_{core} = 0.1 \times 1 0^{9} \cdot \frac{b c ^{3}}{12} = 0.1 \times 1 0^{9} \cdot \frac{0.8 \cdot 0.02 4 ^{3}}{12} \approx 92.2 Nm^{2}$

Verify: contribution $= 92.2/ (1.468 \times 1 0^{4} + 92.2) \approx 0.62%$ . 1% se neeche — parent ka neglect justified. Dead core ( $E_{c} = 0$ ) ke liye contribution exactly $0%$ hai aur $(E I)$ 0.62% ke andar unchanged hai. Dono degenerate limits ek doosre ke experimental scatter ke andar land karte hain. ✓

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

Cube root kahan se aaya. Derive kyun karna? Taaki $σ_{wr} = C (E_{f} E_{c} G_{c})^{1/3}$ samjha jaaye, memorise nahi kiya jaaye.

Forecast: Guess karo kaunsa margin tighter hai — wrinkling ya shear.

Wrinkling stress. Yeh step kyun? Applied $σ_{x}$ ko buckling threshold se compare karo. $σ_{wr} = C (E_{f} E_{c} G_{c})^{1/3} = 0.5 (70 \times 1 0^{9} \cdot 0.1 \times 1 0^{9} \cdot 300 \times 1 0^{6})^{1/3}$ $= 0.5 (2.1 \times 1 0^{27})^{1/3} = 0.5 \cdot 1.281 \times 1 0^{9} \approx 6.40 \times 1 0^{8} Pa = 640 MPa$ Wrinkling ke against margin: $640/50 = 12.8$ .
Core shear stress. Yeh step kyun? $τ_{c}$ ko core strength se compare karo. $τ_{c} = \frac{V}{bh} = \frac{4000}{0.8 \cdot 0.025} = 2.0 \times 1 0^{5} Pa = 0.2 MPa$ Shear ke against margin: $2.0/0.2 = 10$ .

Verify: wrinkling margin 12.8 vs. shear margin 10. Core shear tighter constraint hai — yeh pehle fail karega agar loads scale up hoti hain. Units: $σ_{wr}$ se $(Pa^{3})^{1/3} = Pa$ ✓; $τ_{c}$ se $N / m^{2} = Pa$ ✓. Forecast: jisne "shear" guess kiya tha woh jeeta.

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

Forecast: $(E I) \propto h^{2}$ , isliye $h$ 20→14 mm se shrink karna stiffness $(14/20)^{2} = 0.49$ se cut karta hai. Predict karo ki SF 2 se upar rahega ya nahi, aur roughly kitna mass drop hoga.

New $(E I)$ . Yeh step kyun? Confirm karo ki hum requirement clear karte hain. $(E I) = \frac{150 \times 1 0 ^{9} \cdot 0.8 \cdot 0.0005 \cdot 0.01 4 ^{2}}{2} = 5.88 \times 1 0^{3} Nm^{2}$
Safety factor. Yeh step kyun? Requirement se divide karo. $SF = \frac{5.88 \times 1 0 ^{3}}{2637} \approx 2.23 \geq 2 ✓$
New mass. Yeh step kyun? Sirf core term shrink hota hai ( $h$ : 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 $3.12 \to 2.76 kg$ se gira — $0.36 kg$ (11.5%) saving, sab core se. Forecast confirmed: stiffness roughly half hui ( $0.49 \times$ ) phir bhi threshold se upar rahi kyunki hum 4.5× over-designed the.

Cell H — Exam twist: unit-slip aur ek limiting sanity check

Forecast: Face-mass term $2 ρ_{f} b L t_{f}$ hai, $t_{f}$ mein linear. Predict karo ki corrupted face mass kya hoga agar $t_{f}$ wrongly 10× bada ho.

Wrong number reproduce karo. Yeh step kyun? Dekho 20.4 kahan se aata hai. $m_{face,wrong} = 2 \cdot 1600 \cdot 0.8 \cdot 1.5 \cdot 0.005 = 19.2 kg, m_{wrong} = 19.2 + 1.2 = 20.4 kg$
Correct face mass $t_{f} = 0.0005$ m ke saath. Yeh step kyun? Sahi thickness restore karo. $m_{face} = 2 \cdot 1600 \cdot 0.8 \cdot 1.5 \cdot 0.0005 = 1.92 kg, m = 1.92 + 1.2 = 3.12 kg$
Limiting sanity check. Yeh step kyun? Intuition ko guard rail banao. Ek $0.5 mm$ CFRP skin $0.8 \times 1.5 m$ par ka face volume $2 \cdot 0.0005 \cdot 1.2 = 1.2 \times 1 0^{- 3} m^{3}$ hai; $1600 kg/m^{3}$ par woh $1.92 kg$ hai. Compare karo ek solid CFRP slab se jo poora $20 mm$ gap fill kare: $1600 \cdot 0.8 \cdot 1.5 \cdot 0.02 = 38.4 kg$ . Do thin 0.5 mm skins us slab ki mass ki aadhi nahi ho sakti — 19.2 kg un skins ke liye physically absurd hai jo depth ka $1/20$ hain.

Verify: wrong 20.4 kg vs. correct 3.12 kg — factor $20.4/3.12 \approx 6.5$ off (10× face-term error sahi 1.2 kg core term se dilute hua). Correct total 3.12 kg Ex 1 ke parent value se match karta hai ✓.

Recall Self-test

Thin-face $(E I)$ formula $t_{f} / h = 0.5$ par over-predict kyun karta hai? ::: Neglected self-inertia $\frac{b t _{f}^{3}}{12}$ aur $d \approx h /2$ approximation dono break hoti hain; zero-core limit par yeh 3× too stiff hai (Ex 4). Ex 3 mein shear deflection kaunsa fraction tha aur itna chhota kyun? ::: Lagbhag 1%, kyunki aluminium honeycomb ka $G_{c} = 300$ MPa high hai; soft foam ise ~30% bana deta. $δ_{s} = w L^{2} / (8 G_{c} A_{s})$ mein "8" kahan se aata hai? ::: Uniformly loaded simple beam ke triangular shear force $V (x) = w (L /2 - x)$ ko integrate karne se — pure load ki geometry. Ex 6 mein kaunsa mode govern kiya aur kitne margin se? ::: Core shear, margin 10 vs. wrinkling margin 12.8.

Har cell kaise cover hua

Aage jaane ke liye related tools: Beam Bending Theory deflection formulae ke liye, Buckling and Instability face wrinkling ke liye, Composite Materials CFRP face properties ke liye, Adhesive Bonding shear-transfer assumption ke liye, aur Finite Element Analysis in hand calcs check karne ke liye.