Sandwich structures — face sheets, core: Visual walkthrough

3.6.17 · D2 · Physics › Spacecraft Structures & Systems Engineering › Sandwich structures — face sheets, core

Pehli line se pehle, teen seedhe words jinpar hum tike rahenge. Bending = beam load ke neeche curve karna. Stiffness = usse curve karna kitna mushkil hai (badi stiffness → chhota bend). Mass = yeh kitna bhaari hai, jo ek spacecraft ke liye launch ke waqt jala hua paisa hai. Neeche sab kuch pehli ko badhate hue teesri ko ghatane ki ladaai hai.

Step 1 — "Bending stiffness" ka matlab kya hai

KYA. Jab tum beam ke beech mein neeche dabaate ho, toh yeh kuch matra $δ$ (Greek letter delta, hamaara symbol "kitna jhuka", metres mein maapa gaya) se jhuk jaata hai. Ek stiff beam chhota $δ$ deta hai; ek floppy wala bada $δ$ deta hai.

KYUN. Humein ek aisa akela number chahiye jo "resistance to bending" capture kare taaki hum designs ko fairly compare kar sakein. Woh number $(E I)$ likha jaata hai aur yeh hamesha har sag formula ke denominator mein dikhta hai, jaise ki ek cantilever (ek end pe fixed, doosre pe free) ke liye jo apni tip pe force $F$ se push ho raha ho:

$δ = \frac{F L ^{3}}{3 ( E I )} .$

$F$ — push, newtons mein.
$L$ — beam ki length, metres mein.
$(E I)$ — hamaara hero number. Yeh bada hota hai jab beam stiff hoti hai. Kyunki yeh neeche baith ta hai, bada $(E I)$ matlab chhota $δ$ .

$L^{3}$ aur $\frac{1}{3}$ kahan se aate hain? Beam Bending Theory dikhata hai ki beam ki local curvature har point pe (bending moment) $/ (E I)$ ke barabar hoti hai. Tip droop paane ke liye tumhe length ke saath woh curvature do baar add karni padti hai (ek baar curvature ko slope mein badalne ke liye, ek baar slope ko deflection mein badalne ke liye) — $L$ ke saath do integrations, jismein se har ek moment ke apne ek power of $L$ se aage ek power of $L$ contribute karta hai, total mein $L^{3}$ deta hai. Saaf fraction $\frac{1}{3}$ simply woh number hai jo woh do integrations tip-loaded cantilever ke triangular moment shape ke liye nikalte hain; alag load ya support ek alag fraction deta (Step 6 mein hum $5/384$ aur $1/8$ milte hain). Jo takeaway tumhe chahiye: droop $L^{3}$ ke saath scale karta hai aur $(E I)$ ke inverse mein, isliye $(E I)$ badhana stiffness par seedha lever hai.

Letters do kaam mein bante hain: $E$ (Young's modulus) hai material kitna stiff hai — substance ki ek property, jaise carbon fibre vs. jelly. $I$ (second moment of area) hai cross-section mein material kaise arrange hai — shape ki ek property. Hamaara poora game $I$ ko enormous banana hai bina zyada material kharide. Woh geometry idea seedha Beam Bending Theory se aata hai.

PICTURE. Same material, do shapes: flat rakhne pe zyada jhukti hai; edge pe khadi karne pe muskil se hilti hai. Kuch nahi badla siwaaye arrangement ke — yeh $I$ bol raha hai.

Figure — Sandwich structures — face sheets, core

Step 2 — Outer fibres sabse zyada kyun matter karti hain

KYA. Jab beam bends karti hai, uski top surface squeeze hokar chhoti ho jaati hai aur bottom surface stretch hokar lambi. Beech mein kahin ek line hai jo na stretch hoti hai na squeeze — neutral axis. Hum us line se ek distance $y$ measure karte hain: $y = 0$ neutral axis pe, top aur bottom faces pe $y_{m a x}$ tak badhta hua.

KYUN. Bending stress $σ$ (Greek sigma, woh force-per-area jo ek fibre feel karta hai, pascals mein) $y$ ke saath seedha-line mein badhta hai:

$σ = \frac{M y}{I} .$

$M$ — bending moment (load cross-section ko kitna mushkil se twist karta hai), newton-metres mein.
$y$ — neutral axis se distance. Show ka star.
$I$ — pehle wala wahi arrangement number.

Seedha padho: stress $y$ ke proportional hai. Neutral axis pe baith ne wali fibre ( $y = 0$ ) zero stress feel karti hai — yeh dead weight hai. Dur edge pe wali fibre ( $y = y_{m a x}$ ) sabse zyada feel karti hai. Isliye beech ke paas ka material aaram kar raha hai, aur bahut baahir wala hard work kar raha hai.

PICTURE. Stress ka ek triangle: centre line pe zero, surfaces pe sabse lambe arrows. Message chamakta hai: apna achha material wahan rakhho jahan arrows sabse lambe hain.

Step 3 — Trick: material ko alag kar do

KYA. Material lo aur use do patli sheets mein split karo — face sheets — aur unhe neutral axis se jitna ho sake utna door dhakelo. Unke beech kuch almost weightless rakhho jiska kaam sirf unhe alag rakhna hai: core.

KYUN. Step 2 ki wajah se, material ko baahir move karna badhata hai ki yeh bending ko kitna resist kar sakti hai. Is reward ki miqdar ko agla step measure karta hai. Core wahan nahi hai bending resist karne ke liye; yeh wahan hai taaki faces ek doosre ki taraf collapse na ho jayein (aur shear carry karne ke liye — Step 6). Yeh Composite Materials ki philosophy hai: har part ko woh ek kaam karne do jo woh sabse achhe se kare.

Har symbol jo hum kabhi bhi use karenge ab naam lete hain, taaki baad mein kuch undefined na rahe:

$b$ — panel ki width (page ke andar), metres mein.
$t_{f}$ — ek face sheet ki thickness, metres mein. Patli, jaise $0.5 mm$ .
$c$ — core ki thickness, metres mein. Moti, jaise $19 mm$ .
$h = 2 t_{f} + c$ — sandwich ki total depth, face-outside se face-outside tak, metres mein.
$E_{f}$ — face material ka Young's modulus (material stiffness), pascals mein.
$E_{c}$ — core material ka Young's modulus, pascals mein. Chhota, kyunki honeycomb mostly air hai.
$G_{c}$ — core ka shear modulus (yeh sliding ko kitna resist karta hai, ek stiffness property), pascals mein. Step 6 mein theek se milta hai.
$τ_{c, allow}$ — core ki shear strength (woh stress jis par yeh toot ta hai, $G_{c}$ se alag property), pascals mein. Yeh bhi Step 6.
$ρ$ — density (har cubic metre material mein kitna mass packed hai), kilograms per cubic metre ( $kg/m^{3}$ ) mein. Hum face material ke liye $ρ_{f}$ aur core ke liye $ρ_{c}$ use karte hain; kam $ρ$ exactly wahi hai jo ek material ko "light" banata hai.

PICTURE. Step 2 ka flat slab split hota hai aur uske do hisse surfaces ki taraf migrate karte hain, unke beech ek light honeycomb core phoolti hai. Same material, dramatically lamba shape.

Step 4 — Ek face ka contribution measure karna (parallel-axis theorem)

KYA. Hum ek single face sheet ke liye arrangement number $I$ calculate karte hain jo neutral axis se door baith i hui hai. "Koi shape bending ko kitna resist karti hai jab woh centred nahi hoti?" ka tool hai parallel-axis theorem — exactly yahi sawaal hai jiska woh jawab deta hai, isliye hum isi ko use karte hain aur kuch aur nahi.

YEH TOOL KYUN. Ek shape ka $I$ do parts mein aata hai: (1) apne apne centre ke baare mein khud ki stiffness, plus (2) ek bonus ki beam ke neutral axis se uska centre kitna door hai. Pehle face ke cross-sectional area ko naam do, kyunki yeh theorem aur baad mein mass count dono mein aata hai:

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

Theorem tab kehta hai:

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

$\frac{b t _{f}^{3}}{12}$ — apne khud ke centreline ke baare mein face ki stiffness. Ek patli sheet akele floppy hoti hai, isliye yeh chhota hai.
$A_{f} = b t_{f}$ — face ka cross-sectional area, abhi upar naam diya gaya.
$d$ — face ke centre se beam ke neutral axis tak ki distance. Squared, kyunki reward distance squared ke saath badhta hai (Step 2 ke $σ \propto y$ ki echo).

Face ka centre $d = \frac{c}{2} + \frac{t _{f}}{2} = \frac{h}{2} - \frac{t _{f}}{2}$ pe baith ta hai. Patli faces ke liye ( $t_{f} ≪ h$ ) hum ise $d \approx \frac{h}{2}$ round kar dete hain. Woh approximation har baar oonchi awaaz mein bolte raho — yeh tabhi true hai jab faces depth ki tulna mein patli hon.

PICTURE. Ek face highlighted, neutral axis se uske centre tak lambaai $d$ ka ek arrow, ek chhota sa " $d^{2}$ " reward badge ke saath jo "own shape" badge ko daba deta hai.

Step 5 — Do terms kyun hatate hain (aur kab nahi hata sakte)

KYA. Dono faces aur core jodo, phir clean formula tak pahunchne ke liye chhote pieces hata do.

TERM 1 KYUN MARTA HAI. $I_{face}$ ke do parts compare karo: own-shape term mein $t_{f}^{3}$ hai, distance term mein $A_{f} (h /2)^{2} = t_{f} h^{2} /4 \cdot b$ hai. Unka ratio (per unit width) hai

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

$t_{f} = 0.5 mm$ aur $h = 20 mm$ ke saath yeh $\frac{1}{3} (0.5/20)^{2} \approx 2 \times 1 0^{- 4}$ hai — lagbhag $0.02%$ . Negligible.

CORE KYUN MARTA HAI. Core ki apni bending stiffness $E_{c} I_{core} = E_{c} \frac{b c ^{3}}{12}$ hai, lekin uski material stiffness $E_{c}$ face material $E_{f}$ se roughly ek hazaar guna chhoti hai (honeycomb mostly empty air hai). Isliye $E_{c} I_{core} ≪ E_{f} I_{faces}$ aur yeh nikal jaata hai. Do faces jodo:

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

" $2$ " — ek har face ke liye.
$(\frac{h}{2})^{2}$ — har face half depth door baith ti hai, squared.

Face material stiffness $E_{f}$ se multiply karo aur hum pahunch gaye:

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

PICTURE. Poora expression dono doomed terms grey aur shrink hote hue, survivor chamakta hua.

Step 6 — Woh case jo stiffness formula chhupa leta hai: core mein shear

KYA. Boxed $(E I)$ sirf story ka bending wala aadha hissa batata hai. Ek lamba, slender panel ke liye woh kaafi hai. Lekin panels aksar stubby hoti hain, aur tab ek doosra sag mechanism aata hai: poora core shears karta hai jaise taash ki deck slide kar rahi ho.

YEH KYUN MATTER KARTA HAI. Ek common, well-defined case consider karo: ek simply-supported beam (har end pe support pe tikaa hua) jo span $L$ ka hai, uniform line load $w$ carry kar raha hai — yeh woh force hai jo beam ke length ke har metre pe push hota hai, newtons per metre mein (width $b$ ke panel pe pressure $p$ ke liye, $w = p b$ ). Us case ke liye total deflection bending plus shear hai:

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

$w$ — uniform line load, newtons per metre. ( $5/384$ aur $1/8$ numbers is simply-supported uniform-load case ke liye standard coefficients hain — yeh Step 1 mein describe ki gayi curvature ki same double-integration se nikalte hain, bas uniform load aur end supports ke liye tip force ki jagah; cantilever ya point load alag coefficients use karta — formula ko hamesha boundary conditions se match karo.)
$(E I)$ — hamaara boxed bending stiffness.
$G_{c}$ — core ka shear modulus (core sliding ko kitna resist karta hai), ek stiffness property.
$b c$ — shear-carrying area. Sirf thickness $c$ ka core shear carry karta hai, poori depth $h$ nahi, kyunki patli faces shear mein almost kuch contribute nahi karti.
$κ$ — shear-correction factor. Beam theory maanta hai ki shear stress section par evenly spread hai, lekin actually yeh neutral axis pe peak pe bulge karta hai aur surfaces pe zero ho jaata hai. $κ$ woh akela number hai jo us jhooth ko theek karta hai taaki simple formula abhi bhi sahi stored shear energy de; rectangular cross-section ke liye actual parabolic shear distribution solve karne par $κ = 5/6 \approx 0.83$ milta hai. Isliye hamare rectangular panel ke liye $κ \approx 5/6$ aata hai — alag cross-section shape alag $κ$ carry karta.

Bending term $L^{4}$ ke saath scale karta hai; shear term sirf $L^{2}$ ke saath. Isliye chhote, deep panels ( $L / h < 20$ ) ke liye shear term dominate kar sakta hai — isse miss karo aur stiffness bahut zyada predict karte ho.

Alag se, core fail ho sakta hai agar shear stress core ki shear strength $τ_{c, allow}$ se zyada ho jaaye. Transverse shear force $V$ (newtons mein) core area $b c$ mein carry hoti hai, isliye:

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

Dono properties ko confuse mat karo: $G_{c}$ extra sag control karta hai, $τ_{c, allow}$ rupture control karta hai. Yahi shear channel kyun hai ki honeycomb, apne high $G_{c} / ρ$ (shear stiffness per unit density — bahut kam mass ke liye bahut zyada sliding-resistance) ke saath, structural panels ke liye foam se behtar hai.

PICTURE. Left, ek lamba slender panel smoothly bending kar raha hai. Right, ek chhota deep panel jiska core saaf dikhta shear kar raha hai — parallel arrows faces ko depth $c$ mein sideways slide kar rahe hain.

Step 7 — Do degenerate limits, sanity-checked

KYA. Geometry ko uski extremes pe push karo aur formula ko kisi aisi cheez se check karo jis par hum pehle se trust karte hain.

KYUN. Ek trustworthy formula ko edges pe sensibly behave karna chahiye — bina ek bhi test rig ke galtiyan pakadne ka yahi tarika hai. Lekin hum ise sirf wahan apply karte hain jahan yeh valid ho.

Core collapse karo, $h \to 2 t_{f}$ (ek solid double-thick plate). Yahan boxed formula ki khud ki assumption $t_{f} ≪ h$ toot gayi hai — faces ab poori depth hain — isliye humein boxed result nahi quote karna chahiye. Exact expression ki taraf wapas jaate hain: $2 t_{f}$ thickness ki solid plate ki honest value $I_{solid} = \frac{b ( 2 t _{f} ) ^{3}}{12} = \frac{2}{3} b t_{f}^{3}$ hai, yaani $(E I)_{exact} = \frac{2}{3} E_{f} b t_{f}^{3}$ . (Boxed formula naively plug in karne par $2 E_{f} b t_{f}^{3}$ milta, teen guna bada — exactly woh warning ki thin-face approximation yahan illegal hai.) Sabak: sandwich magic tabhi hai jab faces sach mein door hon. Jab woh touch karti hain, tumhare paas ordinary plate hai aur ordinary formula use karna padega. Achha: koi free lunch nahi, aur formula honestly announce karta hai kahan yeh apply karna band ho jaata hai.
Weightless spacer jo badhta hai, $h$ bada, mass fixed. Kyunki $(E I) \propto h^{2}$ , depth double karne se stiffness (almost) koi mass cost ke bina chaar guna ho jaati hai — jab tak Step 6 ka shear term ya face wrinkling nahi aa jaata aur limit nahi kar deta ki tum kitna patla/lamba dare jaate ho. Formula depth ko reward karta hai; physics eventually isko tax karta hai.

PICTURE. Ek dial $h$ ko chhote se bade tak sweep karta hai, stiffness curve $h^{2}$ ke roop mein badhti hai, ek red "wrinkling / shear limit" wall ke saath jahan reward free hona band ho jaata hai — aur chhote- $h$ end pe ek marker "thin-face formula invalid, use exact plate" flag kiya hua.

Ek-picture summary

Upar sab kuch ek single frame mein: flat slab (weak), same mass faces + core mein khicha hua (strong), $σ \propto y$ triangle explain karta kyun, aur boxed $(E I) = E_{f} b t_{f} h^{2} /2$ apne do survivors aur do casualties ke saath labelled.

Recall Feynman retelling — plain words mein wapas bolo

Jab beam bends karti hai, beech ke paas ki fibres almost kuch feel nahi karti aur surface ki fibres sabse zyada feel karti hain — stress centre line se distance ke saath seedha-line mein badhta hai. Isliye strong material ko beech mein rakhna waste hai. Clever move yeh hai ki material ko do patli sheets mein cut karo aur unhe jitna ho sake utna door karo, ek almost weightless spacer unhe wahan rakhe. Kitna gain milta hai? Parallel-axis theorem kehta hai ki har sheet ka stiffness mein contribution centre line se kitni door baith ti hai uske square ke saath badhta hai, isliye tall beats thick har baar — woh squared distance wahan hai jahan saara reward rehta hai. Hum phir do chhote terms hata dete hain — har sheet ki khud ki floppy stiffness (yeh thickness cubed jaisi hai, jo patli sheet ke liye minuscule hai) aur hawa bhari core ki khud ki bending (uski material stiffness faces se ek hazaar guna softer hai) — aur jo bachta hai woh hai $(E I) = E_{f} b t_{f} h^{2} /2$ : face material stiffness times width times face thickness times depth-squared, sab do se divide. Isse use karne ke liye woh stiffness kisi bhi sag formula mein daal do: droop length-cubed over $(E I)$ ke saath scale karta hai, isliye bada $(E I)$ matlab chhota droop. Teen warnings saath aate hain. Pehla, formula tabhi kaam karta hai jab faces sach mein depth ki tulna mein patli hon — core collapse karo taaki faces touch karein aur yeh threefold overpredicts karta hai, isliye wahan ordinary solid-plate formula use karo. Doosra, yeh sirf bending sag count karta hai, isliye chhote stubby panels bhi core ko sideways shear karke sag karte hain jaise sliding deck of cards; sirf thickness $c$ ka core woh shear carry karta hai, ek $κ \approx 5/6$ correction se soften kiya hua jo "uniform shear" jhooth ko theek karta hai. Teesra, agar panel bahut tall-and-thin banao toh faces buckle ya wrinkle karti hain — isliye $h^{2}$ reward tab tak free hai jab tak ek limit nahi aata, jiske baad physics apna tax collect karta hai. Yeh poori kahani hai: strong material ko door rakhho, ek light core use unhe wahan rakhe, aur sirf wahi count karo jo actually load carry karta hai.

Recall Quick self-test

Neutral axis pe material waste kyun hai? ::: Kyunki bending stress $σ = M y / I$ , $y$ ke proportional hai, isliye $y = 0$ pe stress zero hai — woh material koi bending load carry nahi karta. $I_{face}$ mein kaun sa term dominate karta hai aur kyun? ::: Parallel-axis term $A_{f} d^{2}$ , kyunki yeh $t_{f} h^{2}$ ke saath scale karta hai jabki own-shape term $t_{f}^{3}$ ke saath; unka ratio $\frac{1}{3} (t_{f} / h)^{2} ≪ 1$ hai. Core shear stress kaun sa area carry karta hai, aur $G_{c}$ kya karta hai versus $τ_{c, allow}$ ? ::: Core area $b c$ carry karta hai; $G_{c}$ (shear modulus, stiffness) extra shear deflection set karta hai, $τ_{c, allow}$ (shear strength) set karta hai ki core kab rupture karta hai. Boxed $(E I)$ formula kab break down karta hai? ::: Jab faces patli nahi hain ( $t_{f} \neq ≪ h$ ) ya core stiff hai ( $E_{c} \neq ≪ E_{f}$ ); $h \to 2 t_{f}$ pe yeh threefold overpredicts karta hai, isliye wahan exact plate formula use karo.

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