Classical laminate theory — ABD matrix: Visual walkthrough

3.6.16 · D2 · Physics › Spacecraft Structures & Systems Engineering › Classical laminate theory — ABD matrix

Step 0 — Woh ek picture jis par sab kuch tikaa hai

KYA. Koi bhi equation se pehle, ruler fix karo. Hamara laminate얇e flat layers ka ek stack hai jo ek doosre se glued hain, aur hum layers ki total number ko $n$ bolte hain (to plies ko bottom se top ki taraf $k = 1, 2, \dots, n$ se number kiya gaya hai). Hum ek vertical arrow kheenchte hain — isse $z$ bolte hain — jo stack ke through upar ki taraf point karta hai. Iska zero exactly geometric middle mein rakha jaata hai, yaani mid-plane.

YAHAN KYU. Neeche har quantity usi mid-plane ke relative measured ki jaati hai: ek layer ki height, ek fibre centre se kitna upar ya neeche hai, ek force ka lever arm. $z = 0$ ko middle mein rakhna koi decoration nahi hai — yeh woh choice hai jo symmetry ko cheezein baad mein cancel karwati hai. Yeh galat karo to poora $[B]$ wala story toot jaata hai.

PICTURE. Figure dekho. Total thickness $h$ hai, isliye top face $z = + h /2$ par hai aur bottom $z = - h /2$ par. Har layer $k$ ek lower edge $h_{k - 1}$ se upper edge $h_{k}$ tak failti hai. Chaar coloured bands chaar plies hain (to yahan $n = 4$ ); dashed white line woh mid-plane hai jahan $z = 0$ hai.

Figure — Classical laminate theory — ABD matrix

Step 1 — Geometry freeze karo: "plane sections plane rehti hain"

KYA. Unloaded plate ke through ek seedhi vertical line kheencho — material ka ek column. Ab plate ko bend aur stretch karo. Classical laminate theory ki kinematic assumption kehti hai ki woh line seedhi rehti hai aur mid-surface ke perpendicular rehti hai; woh tilt aur slide ho sakti hai, lekin kabhi curl ya warp nahi hoti.

YEH KYU AUR FULL 3D KYU NAHI. Ek real solid apni thickness ke through wildly complicated tarike se deform ho sakta hai. Usse solve karne ke liye har ply mein 3D elasticity chahiye — haath se impossible aur computer par expensive. "Plane stays plane" ka rule (Kirchhoff–Love assumption) ek deliberate simplification hai jo accurate hai jab tak plate얇i hai (thickness ≪ width). Yeh through-thickness behaviour ko sirf do cheezein mein collapse karta hai: middle kaise move karta hai, aur plate kaise tilt hoti hai.

PICTURE. Figure same vertical fibre ko loading se pehle (grey) aur baad mein (blue) dikhata hai. Do cheezein huyeen: middle ek uniform amount se sideways slide kiya — yeh membrane strain hai — aur line tilt ho gayi, zyada tilt ka matlab zyada curvature. Height $z$ par total strain ek constant part aur ek part ka sum hai jo $z$ ke saath straight-line grow karta hai.

Formula Strain thickness ke through straight-line vary karta hai

$height z par strain ϵ (z) = middle par strain ϵ^{0} + z tilt per unit height κ$

$ϵ^{0} = [ϵ_{x}^{0}, ϵ_{y}^{0}, γ_{x y}^{0}]^{T}$ ::: teen mid-plane strains — constant offset, har height par same. Yahan $ϵ_{x}^{0}$ aur $ϵ_{y}^{0}$ , $x$ aur $y$ ke along direct stretches (fractional length change) hain, jabki $γ_{x y}^{0}$ engineering shear strain hai — $x$ aur $y$ fibres ke beech right angle ka total change, radians mein (tensor shear se double).
$κ = [κ_{x}, κ_{y}, κ_{x y}]^{T}$ ::: teen curvatures, units $1/ m$ ; $z$ se multiply karke yeh bolte hain "jitna upar ho, utna bending strain add ya subtract karta hai." Yahan $κ_{x}$ aur $κ_{y}$ bending curvatures hain (plate $y$ aur $x$ axes ke baare mein kitni sharply cup karti hai), jabki $κ_{x y}$ twist curvature hai — plate propeller jaisi kitni warp karti hai, ek corner upar aur diagonal corner neeche.
$z$ ::: woh lever jo curvature ko extra strain mein badalta hai. $z = 0$ par (mid-plane) yeh vanish ho jaata hai, isliye $ϵ (0) = ϵ^{0}$ — exactly yahi wajah hai ki humne zero middle mein rakha.

Yeh equation $z$ mein ek straight line hai: intercept $ϵ^{0}$ , slope $κ$ .

Step 2 — Har ply apne stress se respond karti hai

KYA. Strain ek shape change hai; stress push-back force per area hai. Layer $k$ ke andar yeh us layer ki stiffness matrix $[Q]_{k}$ se linked hain: strain multiply karo aur stress milta hai.

$[Q]$ PAR SUBSCRIPT $k$ KYU HAI. Do plies bilkul same carbon/epoxy se kaati ja sakti hain aur phir bhi alag tarah se push back karti hain, kyunki unki fibres alag directions mein point karti hain. $x$ ke along fibres wali ply $x$ ke along stiff hoti hai; ply ko $90°$ rotate karo aur woh ab $y$ ke along stiff ho jaati hai. Woh rotation fibre angle $θ_{k}$ ke through $[Q]_{k}$ mein bake in hoti hai — $[Q]$ kahaan se aata hai uske liye Reduced Stiffness Matrix Q dekho aur angle use kaise rotate karta hai uske liye Transformation of Stiffness dekho.

PICTURE. Step 1 ka strain profile stack ke through ek smooth straight line hai (left). Isse alag-alag $[Q]_{k}$ ke through ply-by-ply feed karna ek stress profile deta hai jo har glue line par jump karta hai (right): same strain, lekin ek stiffer ply zyada stress ke saath react karti hai. Woh jumps hi poori wajah hain ki ek laminate interesting hota hai.

Step 3 — Stresses ko add karo force paane ke liye: $[A]$ aur $[B]$ matrices

KYA. Practice mein hum har height par stress measure nahi kar sakte; hum total in-plane force per unit width $N$ measure kar sakte hain. Iske teen components hain $N = [N_{x}, N_{y}, N_{x y}]^{T}$ — $N_{x}$ aur $N_{y}$ $x$ aur $y$ ke along direct pulls hain, aur $N_{x y}$ in-plane shear force hai, har ek edge width ke metre per measured (units N/m, yaani force divided by woh width jis par edge chalti hai). $N$ paane ke liye, poori thickness par stress add up karo (integrate karo). Kyunki $[Q]$ plies ke beech jump karta hai, smooth integral ply integrals ka sum ban jaata hai.

INTEGRATE KYU KARO — YEH TOOL KYU. Force ek stress hai jo area par spread hoti hai. "Thickness par spread" precisely wahi hai jo integral $\int σ d z$ compute karta hai: yeh har얇e slice ka contribution accumulate karta hai. Koi sasta tool nahi hai — ek plain sum yeh ignore karta ki stress ek ply ke andar change hota hai, aur woh change hota hai kyunki $z κ$ isse tilt karta hai.

PICTURE. Figure stress-vs- $z$ curve ke neeche ke area ko shade karta hai. Woh shaded area hi $N$ hai. Isse flat piece ( $ϵ^{0}$ se driven) aur sloped piece ( $κ$ se driven) mein split karo: flat piece ko $\int d z = t_{k}$ chahiye, sloped piece ko $\int z d z$ chahiye.

N = \int_{- h /2}^{h /2} σ d z = k = 1 \sum n [Q]_{k} = t_{k} \int_{h_{k - 1}}^{h_{k}} d z ϵ^{0} + k = 1 \sum n [Q]_{k} = \frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2}) \int_{h_{k - 1}}^{h_{k}} z d z κ

jahan sum saari $n$ plies par run karta hai.

$[B]$ KYU VANISH HO SAKTA HAI. Weight $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ mid-plane ke neeche waali ply ke liye negative hota hai aur uski mirror ply ke liye upar positive. Agar stack $z = 0$ ke baare mein mirror image hai (ek symmetric laminate), to har pair cancel ho jaata hai aur $[B] = 0$ . Dekho Symmetric and Balanced Laminates.

Step 4 — Torques add karo moment paane ke liye: $[D]$ matrix

KYA. Mid-plane se door ka stress sirf push nahi karta, woh plate ko twist bhi karta hai — uska ek lever arm $z$ hota hai. Total bending moment per unit width $M = [M_{x}, M_{y}, M_{x y}]^{T}$ (units N, yaani moment per metre of width) woh stress hai apne lever arm $z$ se multiply hokar, thickness ke through integrate kiya hua.

$z$ SE KYU MULTIPLY KARO. Torque = force × pivot se distance. Yahan pivot mid-plane hai aur distance $z$ hai. Isliye har stress slice moment mein $z σ d z$ contribute karta hai. Yeh ek extra factor $z$ hi force aur moment ke beech poora farq hai — aur yahi wajah hai ki outer plies bending ko dominate karti hain.

PICTURE. Figure same stress profile ko $z$ se weight karta hai: mid-plane ke paas $z \approx 0$ isliye contributions tiny hain; outer faces par $z$ sabse bada hota hai isliye woh plies dominate karti hain. Signed shaded area $M$ hai. Strain term $z κ$ ko phir se $z$ se multiply karne par $z^{2}$ milta hai — hamesha positive — isliye $[D]$ hamesha positive-definite hota hai.

M = \int_{- h /2}^{h /2} z σ d z = k = 1 \sum n [Q]_{k} \frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2}) \int_{h_{k - 1}}^{h_{k}} z d z ϵ^{0} + k = 1 \sum n [Q]_{k} \frac{1}{3} (h_{k}^{3} - h_{k - 1}^{3}) \int_{h_{k - 1}}^{h_{k}} z^{2} d z κ

Step 5 — Chaar blocks ko ek saath snap karo

KYA. Force equation ko moment equation ke upar stack karo. Load ke do $3$ -vectors, response ke do $3$ -vectors, chaar $3 \times 3$ blocks — ek single $6 \times 6$ machine.

ASSEMBLE KYU KARO. Alag-alag $[A], [B], [D]$ sub-questions ka jawab dete hain. Saath mein woh asli sawal ka jawab dete hain: forces aur moments ka koi bhi combination diya, plate kya karti hai? Aur block layout physics ko ek nazar mein visible karta hai — off-diagonal $[B]$ blocks exactly stretch–bend coupling hain.

PICTURE. Figure $6 \times 6$ ko colour karta hai: blue $[A]$ top-left, green $[D]$ bottom-right, red $[B]$ dono off-diagonals par. Load vector right mein, response far right mein.

Units check (har block integrate kiye gaye $z$ 's ki power ke hisaab se length ki apni power kamaata hai): $[A]$ hai $Pa\cdotm$ , $[B]$ hai $Pa\cdotm^{2}$ , $[D]$ hai $Pa\cdotm^{3}$ .

Step 6 — Edge cases jinpar tum kabhi mat fisal jaana

KYA / KYU / PICTURE — teeno cases ek figure mein. Real derivations corner cases par mar jaate hain. Yeh teen hain jo hamesha aate hain.

Symmetric stack ( $[B] = 0$ ). $z = 0$ ke baare mein mirror-image plies har $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ pair ko cancel kar deti hain. $6 \times 6$ block-diagonal ban jaata hai: $ϵ^{0} = [A]^{- 1} N$ aur $κ = [D]^{- 1} M$ decouple ho jaate hain. Kheencho, woh sirf stretch hoga; koi surprise bending nahi. Yahi wajah hai ki aerospace laminates usually symmetric hote hain.
Antisymmetric $[0/90]$ ( $[B] \neq = 0$ ). Do plies lo har ek $0.5 mm$ thick, stiff $0°$ ply centre ke neeche (from $z = - 0.5 mm$ to $0$ ) aur compliant $90°$ ply upar ( $0$ to $+ 0.5 mm$ ). Coupling term hai $B_{11} = \frac{1}{2} [Q_{11}^{(0°)} (0^{2} - (- 0.5 \times 1 0^{- 3})^{2}) + Q_{11}^{(90°)} ((0.5 \times 1 0^{- 3})^{2} - 0^{2})] .$ Har squared height metres mein hai, isliye $(0.5 \times 1 0^{- 3} m)^{2} = 0.25 \times 1 0^{- 6} m^{2}$ — yahi woh jagah hai jahan se tiny factor $1 0^{- 6}$ aata hai: yeh simply $mm^{2}$ hai $m^{2}$ mein likhaa hua. Bottom ply $- 0.25 \times 1 0^{- 6}$ contribute karta hai aur top ply $+ 0.25 \times 1 0^{- 6}$ ; kyunki stiff $0°$ ply negative slot mein hai aur soft $90°$ ply positive mein, aur $Q_{11}^{(0°)} \neq = Q_{11}^{(90°)}$ , do cancel nahi hote. $x$ mein kheeencho: stiff bottom soft top se zyada resist karta hai → plate curve ho jaati hai.
Single ply / zero thickness (degenerate). Ek layer $h_{0} = - t /2, h_{1} = + t /2$ ke saath: tab $[B] = \frac{1}{2} Q ((t /2)^{2} - (t /2)^{2}) = 0$ automatically (ek single ply apna khud ka mirror hai), $[A] = Qt$ , $[D] = Q t^{3} /12$ — familiar plate result recover karta hai. Aur agar koi $t_{k} \to 0$ , woh ply simply teeno sums se drop out ho jaati hai. Kisi bhi cheez se division nahi hoti, isliye kuch bhi blow up nahi hota.

Aage yeh ABD First Ply Failure feed karta hai, Thermal and Hygroscopic Effects mein temperature terms leta hai, Sandwich Panel Theory mein cores tak extend hota hai, aur Finite Element Modeling of Composites mein ek element ban jaata hai.

Ek-picture summary

Poori derivation ek single flow mein: ek straight-line strain ( $ϵ^{0} + z κ$ ) har ply ke $[Q]_{k}$ mein enter hoti hai, jumping stress produce karti hai; $\int d z$ aur $\int z d z$ integrate karo force ( $[A], [B]$ ) collect karne ke liye, $\int z d z$ aur $\int z^{2} d z$ integrate karo moment ( $[B], [D]$ ) collect karne ke liye, phir chaar blocks ko ek single $6 \times 6$ ABD matrix mein stack karo jo loads ${N, M}$ ko deformations ${ϵ^{0}, κ}$ se map karta hai. Woh ek box laminate ka complete stiffness passport hai.

Recall Feynman retelling — apne plain words mein wapas bolo

Imagine karo stiff sheets ka ek sandwich jo ek doosre se glued hain, aur ek seedha pin bottom se top tak uske through draw karo. Jab tum sandwich ko kheeencho aur modo, pin seedha rehta hai lekin slide aur tilt hota hai — sliding stretch hai, tilting bend hai, aur pin par kahin bhi strain hai "middle slide plus height times tilt." Har sheet apni stiffness se push back karti hai kyunki uski fibres apni taraf point karti hain, isliye push-back (stress) mein kinks hote hain jahan sheets milti hain. Total sideways force paane ke liye, woh saari push-back height par add up karo. Total twisting moment paane ke liye, isse phir add karo lekin har slice ko weight karo by wo kitni dur middle se hai, kyunki door ke pushes zyada twist karte hain. Woh do sums karne se numbers ki teen tables milti hain: stretching ke liye $[A]$ , bending ke liye $[D]$ , aur $[B]$ jo donon ko couple karta hai — $[B]$ zero hota hai jab sandwich apne middle ke baare mein perfect mirror hota hai, aur nonzero jab woh lopsided hota hai, yahi exactly wajah hai ki ek lopsided panel sirf kheeenche jaane par warp kyun karta hai. Teeno tables ko ek $6 \times 6$ box mein glue karo aur tumhare paas laminate ka complete instruction manual hai: loads do, woh deformations wapas karta hai.

Recall

$z$ ka zero kahan jaata hai, aur kyun? ::: Mid-plane par, taaki symmetric stacks $[B] = 0$ dein aur $ϵ (0) = ϵ^{0}$ . Kaun sa integral $[A]$ , $[B]$ , $[D]$ banata hai? ::: Thickness ke through $[Q]$ ka $\int d z$ , $\int z d z$ , $\int z^{2} d z$ respectively. $[B]$ $6 \times 6$ mein do baar kyun appear hota hai? ::: Force integral aur moment integral dono same $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ weight produce karte hain — ek coupling matrix, do appearances. $[B] \neq = 0$ ka physical meaning? ::: Kheenchne se curving hoti hai (aur bending se stretching); stack apne mid-plane ke baare mein mirror-symmetric nahi hai. $γ_{x y}$ physically kya hai? ::: Engineering shear strain — $x$ aur $y$ fibres ke beech right angle ka total change, radians mein.