Sandwich structures — face sheets, core: Exercises

3.6.17 · D4 · Physics › Spacecraft Structures & Systems Engineering › Sandwich structures — face sheets, core

Koi bhi number aane se pehle, ek figure har symbol ko fix karti hai jo hum neeche use karte hain.

Figure — Sandwich structures — face sheets, core

Figure s01 (left): sandwich cross-section, end-on dekha gaya. Do blue bars face sheets hain, thickness $t_{f}$ each; unke beech ka yellow block core hai, thickness $c$ ; poora stack depth $h$ aur width $b$ ka hai. Red dashed line neutral axis hai — woh height jahan bending stress exactly zero hota hai. Green arrow jo $d$ label hai, woh axis se top face ke centre tak ki distance hai. (Right): stress triangle. Bending stress height $y$ ke saath straight-line badhta hai, isliye sabse bada top aur bottom par hota hai — exactly wahan jo blue faces baithe hain. Yahi picture poori wajah hai ki sandwiches kaam kyun karte hain.

Definition Is page ke symbols (sabhi upar diye figure pe measure kiye gaye hain)

$b$ — panel ki width (page ke andar), metres mein.
$t_{f}$ — ek face sheet ki thickness (thin strong skin), metres mein.
$c$ — core thickness (faces ke beech ka light filling).
$h$ — total depth, $h = 2 t_{f} + c$ . Yeh poori sandwich height hai.
$L$ — beam/panel ki length (span), uski lambi direction mein measured, metres mein.
$y$ — neutral axis se height (metres), figure s01 pe measured; $y = 0$ axis par, $y = \pm h /2$ faces par.
$d$ — neutral axis (woh horizontal line jahan bending stress zero hai) se face ke centre tak ki distance, $d = \frac{h}{2} - \frac{t _{f}}{2}$ . Thin faces ke liye $d \approx h /2$ — yeh woh moment arm hai jahan har face baithta hai.
$σ$ — face mein normal (bending/axial) stress: force per unit area (pascals) jo beam ke saath-saath push ya pull karta hai. $σ_{x}$ woh stress hai beam ki length direction $x$ mein (compression yahan failure check mein positive magnitude ke roop mein count hoti hai).
$M$ — bending moment (N·m): woh "turning effort" jo load ek cross-section par apply karta hai, beam ko curve karne ki koshish karta hua.
$I$ — second moment of area (m⁴): ek purely geometric number jo batata hai ki cross-section ka area neutral axis se kitna door spread hai. Door wala area (bada $y$ ) bahut zyada count karta hai, kyunki yeh $y^{2}$ ke roop mein aata hai. Yahi wajah hai ki sandwiches material ko bahar push karte hain.
$E_{f}$ — face material ka Young's modulus: kitne pascals of stress lagte hain ise 100% stretch karne mein. Bada $E_{f}$ = stiff.
$E_{c}$ — core ka Young's modulus. Kyunki $E_{c} ≪ E_{f}$ (typically $E_{c} / E_{f} \sim 1 0^{- 3}$ ), core bending resist karna almost nahin karta; hum ise sirf wrinkling check ke liye rakhte hain.
$(E I)$ — bending stiffness: poora beam bent hone ko kitni mushkil se resist karta hai. Bada = stiffer.
$G_{c}$ — core shear modulus: core apne top face ko bottom face ke relative slide karne se kitni mushkil se resist karta hai.
$τ_{c}$ — core mein actual shear stress; $τ_{c, allow}$ woh maximum hai jo yeh fail hone se pehle le sakta hai.
$ρ_{f}$ — face material ki density (mass per unit volume, kg/m³); $ρ_{c}$ — core ki density (bahut choti).
$w$ — line load: span ke saath spread force per unit length (N/m). Yeh pressure times width se aata hai.
$F$ — ek single point force (N); $V$ — transverse shear force jo core ko carry karna hota hai (N).
$C$ — ek dimensionless wrinkling constant (no units), roughly $0.5$ – $0.82$ boundary conditions ke hisaab se; yeh sirf Level 5 ke face-wrinkling formula mein aata hai.
$δ$ — ek deflection: beam load ke neeche actually kitna move karta hai (metres). $δ_{b}$ bending wala part hai, $δ_{s}$ core shear wala part.

Level 1 — Recognition

Problem 1.1

Ek sandwich panel mein, kaun sa part bending (axial) stress zyada carry karta hai, aur kaun sa shear? Strong material ko middle se door rakhne se kyun madad milti hai?

Recall Solution

Face sheets bending/axial stress carry karte hain; core shear carry karta hai aur faces ko apart rakhta hai. Bending mein, stress $σ = \frac{M y}{I}$ hai — yahan $σ$ normal stress hai, $M$ bending moment, $y$ neutral axis se upar ki height, aur $I$ second moment of area (sabhi symbol list mein defined hain). Kyunki $σ \propto y$ , yeh neutral axis se door $y$ ke saath badhta hai, isliye sabse door wala material sabse zyada useful kaam karta hai. Strong faces ko $y = \pm h /2$ par rakhne se woh exactly wahan hain jahan stress sabse bada hai, aur light core sirf woh separation sasti (mass mein) maintain karta hai. Figure s01 (right panel) dekho: red stress line top aur bottom par sabse lambi hai, aur blue face dots wahan baithe hain — faces wahan rehte hain jahan stress sabse bada hota hai.

Problem 1.2

Ek sandwich mein $h = 20 mm$ aur $t_{f} = 0.5 mm$ hai. Core thickness $c$ compute karo.

Recall Solution

$c = h - 2 t_{f} = 20 - 2 (0.5) = 19 mm$ Answer: $c = 19 mm$ . Core almost poori depth le leta hai — faces wafer-thin hain.

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 $h = 25 mm$ . $(E I)_{sandwich}$ find karo.

Recall Solution

Formula mein plug karo (sabhi SI: metres, pascals): $(E I) = \frac{E _{f} b t _{f} h ^{2}}{2} = \frac{70 \times 1 0 ^{9} \times 0.5 \times 0.0008 \times ( 0.025 ) ^{2}}{2}$ Top compute karo: $70 \times 1 0^{9} \times 0.5 = 3.5 \times 1 0^{10}$ ; times $0.0008 = 2.8 \times 1 0^{7}$ ; times $(0.025)^{2} = 6.25 \times 1 0^{- 4}$ se $1.75 \times 1 0^{4}$ milta hai. 2 se divide karo: $(E I) \approx 8.75 \times 1 0^{3} Nm^{2}$ Answer: $(E I) \approx 8750 Nm^{2}$ .

Problem 2.2

Wahi panel $L = 0.6 m$ length ka cantilever hai jisme tip load $F = 40 N$ hai. Tip deflection $δ = \frac{F L ^{3}}{3 ( E I )}$ find karo (upar formula box se cantilever formula).

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: $δ \approx 0.33 mm$ — ek stiff, light beam.

Level 3 — Analysis

Problem 3.1

Ek sandwich (thin faces $t_{f}$ , depth $h$ ) ko same material aur same mass ke solid plate se compare karo. Dikhao ki stiffness ratio $\approx \frac{3 h ^{2}}{4 t _{f}^{2}}$ hai, aur ise $h = 20 mm$ , $t_{f} = 0.5 mm$ ke liye evaluate karo (assume karo core density negligible hai isliye $t_{s} \approx 2 t_{f}$ ).

Recall Solution

$I_{sand} \approx \frac{b t _{f} h ^{2}}{2}$ kahan se aata hai? Symbol list aur figure s01 se moment arm $d \approx h /2$ yaad karo. Har face ek thin rectangle hai jiska area $b t_{f}$ hai aur jo neutral axis se $d \approx h /2$ distance par baitha hai. Parallel-axis theorem kehta hai iski second moment of area (area) $\times d^{2} = b t_{f} (h /2)^{2} = b t_{f} h^{2} /4$ hai. Do faces hain (top aur bottom), isliye hum double karte hain: $2 \times b t_{f} h^{2} /4 = b t_{f} h^{2} /2$ . Woh $\frac{1}{2}$ factor literally "do faces, har ek ke saath $(1/2)^{2} = 1/4$ " hai. (Apna bending term $b t_{f}^{3} /12$ thin faces ke liye tiny hai, isliye hum drop karte hain.)

Negligible core density ke saath equal mass matlab solid plate do faces ka saara mass use karta hai: $ρ_{s} b t_{s} = 2 ρ_{f} b t_{f}$ , isliye $t_{s} \approx 2 t_{f}$ . 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: $\frac{I _{sand}}{I _{solid}} = \frac{b t _{f} h ^{2} /2}{8 b t _{f}^{3} /12} = \frac{6 t _{f} h ^{2}}{8 t _{f}^{3}} = \frac{3 h ^{2}}{4 t _{f}^{2}}$ $h = 20 mm$ , $t_{f} = 0.5 mm$ ke liye: $\frac{3 ( 20 ) ^{2}}{4 ( 0.5 ) ^{2}} = \frac{3 \times 400}{4 \times 0.25} = \frac{1200}{1} = 1200$ . Answer: sandwich same mass ke liye $\approx 1200 \times$ stiffer hai.

Figure s02 is ratio $3 h^{2} /4 t_{f}^{2}$ ko $h / t_{f}$ ke against plot karta hai: yeh ek rising parabola hai, aur yellow marker $h / t_{f} = 40$ par (hamaara case) $1200$ pe land karta hai. Curve message ko visual banata hai — payoff quadratically badhta hai jab aap same material ko zyada door spread karte ho.

Problem 3.2

Ek lamba panel ( $L / h = 25$ ) utna stiff nahin hai. Aap ya toh $h$ double kar sakte ho ya $t_{f}$ double kar sakte ho. Kaun sa $(E I)$ zyada badhata hai? Kaun sa zyada mass add karta hai? Use karo $ρ_{f} = 1600 kg/m^{3}$ (CFRP face), $ρ_{c} = 50 kg/m^{3}$ (honeycomb core), aur starting geometry $t_{f} = 0.5 mm$ , $h = 20 mm$ ( $c = 19 mm$ ).

Recall Solution

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

$t_{f}$ double karna: $(E I)$ $2 \times$ ho jaata hai. Face mass ( $\propto t_{f}$ ) bhi double hoti hai.
$h$ double karna: $(E I)$ $4 \times$ ho jaata hai ( $h^{2}$ ki wajah se). Core thickness (isliye core mass, $\propto h$ ) roughly double hoti hai; face mass unchanged rehti hai. Mass claim quantify karo. Per unit area ( $b L = 1$ ), $m = 2 ρ_{f} t_{f} + ρ_{c} h$ se: $m_{0} = 2 (1600) (0.0005) + 50 (0.020) = 1.60 + 1.00 = 2.60 kg/m^{2}$
$t_{f}$ double karne ke baad ( $t_{f} = 1.0 mm$ ): $m = 2 (1600) (0.001) + 50 (0.020) = 3.20 + 1.00 = 4.20 kg/m^{2}$ — $+ 62%$ ka jump, saara heavy faces mein.
$h$ double karne ke baad ( $h = 40 mm$ , core $\approx 39 mm$ ): $m \approx 2 (1600) (0.0005) + 50 (0.040) = 1.60 + 2.00 = 3.60 kg/m^{2}$ — $+ 38%$ ki rise, aur sirf light core mein.

Toh $h$ double karna $4 \times$ stiffness deta hai sirf $+ 38%$ mass rise ke liye, jabki $t_{f}$ double karna sirf $2 \times$ deta hai $+ 62%$ ke liye. Depth $h$ double karna dono counts par jeet ta hai — extra mass light core mein jaata hai ( $ρ_{c} = 50$ hai $ρ_{f} = 1600$ se $32 \times$ halka), jabki faces ko thicken karna heavy skins par mass pile karta hai. (Yeh wahi quadratic hai jo figure s02 mein dikhaya gaya hai: $h / t_{f}$ axis par seedha jaana stiffness fast buys karta hai.)

Level 4 — Synthesis

Problem 4.1 (full design)

Ek simply-supported panel design karo, span $L = 1.5 m$ , width $b = 0.8 m$ , jo uniform line load $w = 160 N/m$ ( $5 g$ launch se) carry karta ho. Requirement: mid-span deflection $δ < 5 mm$ . (a) Required $(E I)$ find karo. (b) CFRP faces $E_{f} = 150 GPa$ ke saath, $h = 20 mm$ , $t_{f} = 0.5 mm$ choose karo aur safety factor check karo. (c) $ρ_{f} = 1600$ , $ρ_{c} = 50 kg/m^{3}$ ke saath total mass find karo.

Recall Solution

(a) Required stiffness. Simply-supported uniform-load deflection (formula box se): $δ = \frac{5 w L ^{4}}{384 ( E I )} \Rightarrow (E I)_{req} = \frac{5 w L ^{4}}{384 δ} = \frac{5 \times 160 \times ( 1.5 ) ^{4}}{384 \times 0.005}$ $(1.5)^{4} = 5.0625$ . 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 karo (metres, pascals): $(E I) = \frac{150 \times 1 0 ^{9} \times 0.8 \times 0.0005 \times ( 0.02 ) ^{2}}{2}$ Step by step build karo: $150 \times 1 0^{9} \times 0.8 = 1.2 \times 1 0^{11}$ ; times $0.0005 = 6 \times 1 0^{7}$ . Ab depth term — $0.02$ carefully square karo: $(0.02)^{2} = 0.0004 = 4 \times 1 0^{- 4}$ (note karo do zeros double hokar chaar hote hain, ek classic unit-slip point). Toh $6 \times 1 0^{7} \times 4 \times 1 0^{- 4} = 2.4 \times 1 0^{4}$ ; 2 se divide karo: $(E I) = 1.2 \times 1 0^{4} Nm^{2}, SF = \frac{1.2 \times 1 0 ^{4}}{2109} \approx 5.7$ Comfortably safe hai (baad mein mass trim kar sakte hain). (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 compression mein $σ_{x} = 50 MPa$ par hain. (a) Face-wrinkling stress $σ_{wr} = C (E_{f} E_{c} G_{c})^{1/3}$ compute karo $C = 0.5$ ke saath. (b) Kya face $σ_{x} = 50 MPa$ par wrinkle karta hai? (c) Panel transverse shear $V = 1200 N$ bhi carry karta hai; core shear strength $τ_{c, allow} = 1.5 MPa$ hai. $τ_{c} \approx V / (bh)$ se core shear check karo. Pehle kya fail hota hai?

Recall Solution

Yahan $σ_{x}$ face mein compressive normal stress hai (length direction $x$ ), $E_{c}$ core ka Young's modulus hai, aur $C$ dimensionless wrinkling constant hai (0.5–0.82) — sabhi symbol list mein defined hain. Hum $C = 0.5$ use karte hain (conservative choice). (a) Wrinkling stress. Sabhi moduli pascals mein daalo: $E_{f} = 70 \times 1 0^{9}$ , $E_{c} = 0.1 \times 1 0^{9} = 1 0^{8}$ , $G_{c} = 300 \times 1 0^{6} = 3 \times 1 0^{8}$ . Product $= 70 \times 1 0^{9} \times 1 0^{8} \times 3 \times 1 0^{8} = 2.1 \times 1 0^{27}$ . Cube root: $(2.1 \times 1 0^{27})^{1/3}$ . Kyunki $1 0^{27}$ cube-root ho ke $1 0^{9}$ deta hai aur $2. 1^{1/3} \approx 1.28$ , hume $\approx 1.28 \times 1 0^{9}$ milta hai. $σ_{wr} = 0.5 \times 1.28 \times 1 0^{9} \approx 6.4 \times 1 0^{8} Pa = 640 MPa$ (b) Wrinkling check. Applied $σ_{x} = 50 MPa ≪ 640 MPa$ , isliye koi wrinkling nahin — bahut bada margin ( $640/50 \approx 12.8 \times$ ). (c) Core shear. $τ_{c} \approx \frac{V}{bh} = \frac{1200}{0.8 \times 0.025} = \frac{1200}{0.02} = 6.0 \times 1 0^{4} Pa = 0.06 MPa$ $τ_{c, allow} = 1.5 MPa$ se compare karo: margin $= 1.5/0.06 = 25 \times$ . Core bhi fail nahin hota. Pehle kaun fail ke closer hai? Wrinkling margin $\approx 12.8$ , core-shear margin $\approx 25$ . Dono safe hain, lekin chota margin nearer limit hai, isliye face wrinkling govern karta hai — yeh woh mode hai jo aap pehle hit karte agar load badhe. Figure s03 in do margins ko yellow "failure line" ke against bars ke roop mein $1$ par draw karta hai: dono bars usse kaafi upar hain, lekin red wrinkling bar (12.8) green core-shear bar (25) se chota hai, ek nazar mein dikhata hai ki load badhne par pehle kaun sa mode hit hoga. Answers: $σ_{wr} \approx 640 MPa$ ; $τ_{c} \approx 0.06 MPa$ ; dono safe hain, wrinkling governing (nearer) mode hai.

Problem 5.2 (shear deflection matters)

Ek slender panel ka load ke neeche bending deflection $δ_{b} = 4 mm$ hai. Extra shear deflection $δ_{s} = \frac{V L}{G _{c} bh}$ hai jisme $V = 500 N$ , $L = 1.6 m$ , $G_{c} = 50 MPa$ (soft foam core), $b = 0.6 m$ , $h = 0.03 m$ . $δ_{s}$ aur total compute karo. Yahan hum ise ignore kyun nahin kar sakte?

Recall Solution

Hum $δ_{b}$ aur $δ_{s}$ simply kyun add kar sakte hain? Is page ke top par stated small-deflection, linear-elastic assumptions ke neeche, bending deformation aur core-shear deformation same load ke independent responses hain — yeh superpose karte hain. Isliye beam ka total sag bending sag plus shear sag hai, seedha add karke. $δ_{s} = \frac{V L}{G _{c} bh} = \frac{500 \times 1.6}{50 \times 1 0 ^{6} \times 0.6 \times 0.03} = \frac{800}{9.0 \times 1 0 ^{5}} \approx 8.89 \times 1 0^{- 4} m \approx 0.89 mm$ Total $δ = δ_{b} + δ_{s} = 4 + 0.89 = 4.89 mm$ . Kyun matters hai: soft core ( $G_{c}$ sirf $50 MPa$ ) faces ko ek doosre ke relative slide karne deta hai, $\sim 0.9 mm$ add karta hua — bending deflection ka $20%$ se zyada. Lambe, soft-cored panels ke liye ( $L / h$ bada, chota $G_{c}$ ) shear deflection rounding error nahin hai; ise ignore karna $4 mm$ predict karta aur quietly $5 mm$ budget bust kar deta.

Recall Self-check summary

Kaun sa single geometric change almost koi mass cost nahin pe bending stiffness quadruple karta hai? ::: Total depth $h$ double karna (kyunki $(E I) \propto h^{2}$ jabki added mass light core mein hoti hai). Kin panels ke liye shear deflection add karni zaroori hai? ::: Lambe panels ( $L / h > 20$ ) ya jinmein soft, low- $G_{c}$ core ho. Kaun si do assumptions hume bending aur shear deflections seedhe add karne deti hain? ::: Small deflections aur linear elasticity, jo milkar superposition allow karte hain.