Classical laminate theory — ABD matrix: Worked examples

3.6.16 · D3 · Physics › Spacecraft Structures & Systems Engineering › Classical laminate theory — ABD matrix

Kuch bhi karne se pehle, vocabulary ka ek reminder — kyunki hum ise har example ki pehli line mein use karte hain.

Definition Teen sums, seedhe alfaazon mein

Har laminate $n$ plies ka stack hota hai. Ply $k$ , height $h_{k - 1}$ (bottom) aur $h_{k}$ (top) ke beech mein hoti hai, mid-plane se naapa hua (toh $z = 0$ middle hai, neeche negative, upar positive). Har ply ka ek $3 \times 3$ reduced stiffness matrix hota hai, jise hum poore block ke roop mein $[Q]_{k}$ likhte hain, aur iske individual entries ko $Q_{ij}^{(k)}$ kehte hain — toh $[Q]_{k}$ matrix hai aur $Q_{11}^{(k)}$ uska top-left number hai ( $x$ -direction resistance). Yeh pehle se laminate ke axes par rotate ho chuke hain (dekho Reduced Stiffness Matrix Q aur Transformation of Stiffness). $[A] = \sum_{k} [Q]_{k} (h_{k} - h_{k - 1}) [B] = \frac{1}{2} \sum_{k} [Q]_{k} (h_{k}^{2} - h_{k - 1}^{2}) [D] = \frac{1}{3} \sum_{k} [Q]_{k} (h_{k}^{3} - h_{k - 1}^{3})$ Entry-by-entry padhne par iska matlab hai $A_{ij} = \sum_{k} Q_{ij}^{(k)} (h_{k} - h_{k - 1})$ , aur isi tarah $B_{ij}$ , $D_{ij}$ ke liye. Hum neeche sirf ek entry (usually $ij = 11$ ) calculate karte hain.

$[A]$ thickness count karta hai — kitna material hai.
$[B]$ thickness × position count karta hai — material middle ke relative kahan baitha hai. Yeh lever-arm term hai: middle se door material, ek side par, plate ko tilt karta hai.
$[D]$ thickness × position² count karta hai — material middle se kitna door hai, squared mein. Door baitha material bending se zyaada resist karta hai (jaise ek deep diving board).

Figure 1 (neeche): teen sums alag kyun hain. Plot ek vertical coordinate $z$ dikhata hai (mid-plane $0$ par) jo teen alag weightings mein daala gaya hai. Flat black line $[A]$ weighting hai (use parwah nahi ki material kahan hai). Tilted dashed black line $[B]$ weighting hai (yeh $z$ ke saath linearly badhti hai, toh middle ke neeche negative aur upar positive — yeh wahi lever arm hai jo symmetric stacks mein cancel hota hai). Red bowl $[D]$ weighting hai ( $z^{2}$ ): middle se door material, kisi bhi side par, heavily weighted hai — isliye outer plies bending ko dominate karti hain.

Figure — Classical laminate theory — ABD matrix

Yeh akela picture hi poori kahaani hai — symmetry $[B]$ kyun khatam kar deti hai (tilted line top-to-bottom cancel hoti hai) lekin $[A]$ ya $[D]$ kabhi nahi (flat aur bowl kabhi cancel nahi hote).

The scenario matrix

Har ABD problem jo aap dekhenge woh in cells mein se kisi ek mein aata hai. Neeche diye examples ko unke cell ke saath label kiya gaya hai.

Cell	Case class	Kya Khaas Hai	Example
C1	Symmetric stack, $[B] = 0$	upar/neeche plies mirror → coupling gayab	Ex 1
C2	Asymmetric stack, $[B] \neq = 0$	kheenchna → bending; $B_{11}$ ka sign matter karta hai	Ex 2
C3	Sign flip of the load	negative $N_{x}$ → mirror-image response	Ex 3
C4	Single ply (degenerate $n = 1$ )	check karein formulas plate theory tak reduce hon	Ex 4
C5	Zero-load / zero-thickness limit	ek ply ki thickness $\to 0$ karein; sums ki sanity	Ex 5
C6	Invert ABD for strains (symmetric shortcut)	$[A]^{- 1}$ block only	Ex 6
C7	Full coupled inversion ( $[B] \neq = 0$ )	poora $6 \times 6$ invert karna hoga; curvature aata hai	Ex 7
C8	Real-world word problem	ek satellite panel; stack chunein	Ex 8
C9	Exam twist — balanced but antisymmetric	$A_{16} = 0$ yet $B_{16} \neq = 0$	Ex 9

Saare examples ke liye shared material. Carbon/epoxy: $E_{1} = 140$ GPa, $E_{2} = 10$ GPa, $G_{12} = 5$ GPa, $ν_{12} = 0.3$ . Inse, apne fibres ke saath ek ply ka reduced stiffness hai: $Q_{11} = \frac{E _{1}}{1 - ν _{12} ν _{21}}, Q_{22} = \frac{E _{2}}{1 - ν _{12} ν _{21}}, Q_{12} = \frac{ν _{12} E _{2}}{1 - ν _{12} ν _{21}}, Q_{66} = G_{12},$ jahan $ν_{21} = ν_{12} E_{2} / E_{1}$ . Yeh formulas kyun? Yeh compliance matrix ka plane-stress inverse hain — parent note ka $[Q]$ ek ply ke liye jiske fibres $x$ ke along chalte hain. Numbers daalne par: $ν_{21} = 0.3 \cdot 10/140 = 0.02143$ , toh $1 - ν_{12} ν_{21} = 1 - 0.3 \cdot 0.02143 = 0.99357$ . Phir

$Q_{11} = \frac{140}{0.99357} = 140.905 GPa, Q_{22} = \frac{10}{0.99357} = 10.065 GPa .$

90° ply ke liye, fibres $y$ ke along chalte hain, toh $11$ position mein $1$ aur $2$ ki roles swap ho jaati hain: $Q_{11}^{(90°)} = Q_{22}^{(0°)} = 10.065$ GPa. Yeh ek simple fact — stiff number fibre direction par depend karke move karta hai — neeche har coupling result drive karta hai.

Recall 90° numbers kyun swap karta hai?

Ek 0° ply $x$ -stretch ko apne fibres se resist karta hai ( $140$ GPa). Ek 90° ply ke fibres $y$ ki taraf point karte hain, toh $x$ -stretch sirf weak matrix ko kheenchta hai ( $10$ GPa). ::: $[Q]$ ki $11$ entry isliye 0° ke liye $140.9$ aur 90° ke liye $10.06$ padhti hai.

Example 1 — Symmetric [0/90]ₛ (Cell C1: symmetric, B = 0)

Ply heights (mm), mid-plane $0$ par: $h_{0} = - 1, h_{1} = - 0.5, h_{2} = 0, h_{3} = 0.5, h_{4} = 1$ . Fibre pattern: ply 1 = 0°, ply 2 = 90°, ply 3 = 90°, ply 4 = 0°. $Q_{11}$ har ply ke liye: $140.905, 10.065, 10.065, 140.905$ GPa.

Step 1 — $A_{11}$ . Kyun: $[A]$ sirf thickness count karta hai, toh bas har ply ka $Q_{11}$ uski 0.5 mm se multiply karke add karo. $A_{11} = (140.905 + 10.065 + 10.065 + 140.905) \times 0.5 = 150.970 GPa\cdotmm .$ $[A]$ factor $1 0^{6}$ apply karo: $A_{11} = 150.970 \times 1 0^{6} = 1.50970 \times 1 0^{8}$ N/m.

Step 2 — $B_{11}$ . Kyun: $[B]$ $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ se weight karta hai. Kyunki stack ek mirror hai (ply 1 ↔ ply 4, ply 2 ↔ ply 3), har pair equal-and-opposite $z$ -lever arms contribute karta hai. $B_{11} = \frac{1}{2} [140.905 (0.25 - 1) + 10.065 (0 - 0.25) + 10.065 (0.25 - 0) + 140.905 (1 - 0.25)] GPa\cdotmm^{2} .$ Pehla aur aakhri cancel ho jaate hain; beech ke do cancel ho jaate hain. $B_{11} = 0$ .

Step 3 — $D_{11}$ . Kyun: $[D]$ $\frac{1}{3} (h_{k}^{3} - h_{k - 1}^{3})$ se weight karta hai; $z^{3}$ outer plies ke liye sabse bada hota hai. $D_{11} = \frac{1}{3} [140.905 (- 0. 5^{3} - (- 1)^{3}) + 10.065 (0 - (- 0.5)^{3}) + 10.065 (0. 5^{3} - 0) + 140.905 (1^{3} - 0. 5^{3})] GPa\cdotmm^{3} .$ Height brackets compute karo: $(- 0.125 + 1) = 0.875$ , $(0.125)$ , $(0.125)$ , $(0.875)$ . $D_{11} = \frac{1}{3} [140.905 (0.875) + 10.065 (0.125) + 10.065 (0.125) + 140.905 (0.875)] .$ Inner sum hai $2 (140.905 \cdot 0.875) + 2 (10.065 \cdot 0.125) = 2 (123.292) + 2 (1.258) = 246.584 + 2.516 = 249.100$ . $D_{11} = \frac{1}{3} (249.100) = 83.03 GPa\cdotmm^{3} .$ $[D]$ factor apply karo (jo $1$ hai): $D_{11} = 83.03$ N·m.

Verify: Units — $A$ N/m mein ✓, $B = 0$ jaise symmetry demand karti hai ✓, $D$ N·m mein ✓. Sanity: outer 0° plies $140.905 \times 0.875 = 123.29$ contribute karti hain total $249.10$ bracket mein se har ek — yeh bending dominate karti hain, forecast se match karta hai. Cell C1 confirmed: symmetric ⇒ $B = 0$ . Dekho Symmetric and Balanced Laminates.

Example 2 — Asymmetric [0/90] (Cell C2: B ≠ 0)

Step 1 — heights. $h_{0} = - 0.5, h_{1} = 0, h_{2} = 0.5$ mm.

Step 2 — $B_{11}$ raw GPa·mm² number ke roop mein. Kyun: sirf $[B]$ hi top/bottom imbalance sense karta hai. $Q$ ko GPa mein rakho, heights mm mein, aur koi bhi unit factor lagate se pehle arithmetic karo. $B_{11} = \frac{1}{2} [Q_{11}^{0°} (0^{2} - (- 0.5)^{2}) + Q_{11}^{90°} (0. 5^{2} - 0^{2})]$ $= \frac{1}{2} [140.905 (- 0.25) + 10.065 (0.25)] = \frac{1}{2} (- 35.226 + 2.516) = \frac{1}{2} (- 32.710) = - 16.355 GPa\cdotmm^{2} .$

Step 3 — ek conversion. Kyun: ab, aur sirf ab, single $[B]$ factor $1 0^{3}$ apply karo (bookkeeping box se: GPa·mm² $\to$ N). $B_{11} = - 16.355 \times 1 0^{3} = - 16, 355 N \approx - 1.636 \times 1 0^{4} N .$

$B_{11} = - 1.636 \times 1 0^{4} N$

Step 4 — sign ka matlab. Kyun: sign aapko induced curvature ki direction batata hai. $B_{11} < 0$ hai kyunki negative- $z$ (bottom) ply stiff wali hai, jo negative lever-arm contribution deti hai. Physically, $N_{x} > 0$ se kheenchne par invert karne ke baad positive curvature $κ_{x}$ milti hai — plate is tarah bow karti hai ki stiff side tension relief mein jaaye.

Figure 2 (neeche): ek picture mein coupling. Red rectangle stiff 0° ply hai jo mid-plane (dotted line) ke neeche baitha hai; black rectangle weak 90° ply uske upar hai. Horizontal black arrows applied tension $N_{x}$ hain. Red dashed arc consequence dikhata hai: kyunki strong material sab ek side par hai, pure kheenchna plate ko bow kar deta hai — woh curved arc hi $B_{11} \neq = 0$ ko visible banata hai.

Verify: Units: GPa·mm² $\times 1 0^{3}$ = Pa·m² = N ✓. Symmetry cross-check: agar dono plies 0° hoti, $140.905 (- 0.25) + 140.905 (0.25) = 0$ ⇒ $B_{11} = 0$ , symmetric case recover hota. Hamara nonzero value sirf aur sirf $140.905 \neq = 10.065$ se aata hai. Cell C2 confirmed.

Example 3 — Load ka Sign Flip (Cell C3)

Forecast: Equations linear hain. Guess karo: kya curvature sirf sign flip karegi, ya kuch aur subtle hoga?

Step 1 — linearity. Kyun: ABD relation $N, M = [A B D] (ϵ^{0}, κ)$ ek linear map hai, toh uska inverse bhi linear hai. Load ko $- 1$ se scale karne par har response $- 1$ se scale hota hai.

Step 2 — apply karo. Maano $κ_{x}^{(+)}$ woh curvature hai jo $N_{x} = + 1000$ N/m se milti hai (uska value Ex 7 mein calculate hoga). Linearity se, $N_{x} = - 1000$ N/m se $κ_{x} = - κ_{x}^{(+)}$ milti hai: same magnitude, ulta bow. Jo plate tension mein upar dome karti thi woh compression mein neeche dome karegi.

Verify: Koi naye numbers calculate nahi karne — yeh ek structural check hai ki hamare machine mein koi hidden asymmetry nahi hai (koi thermal/prestress term nahi; woh Thermal and Hygroscopic Effects mein hain). Agar $N_{x} = 0$ par residual curvature rehti, toh hum ek term miss kar dete. Zero-load state mein zero curvature hai ⇒ pure linearity ✓. Cell C3 confirmed.

Example 4 — Single Ply (Cell C4: degenerate n = 1)

Forecast: Ek single centred ply apne middle ke baare mein symmetric hai. Guess karo: $B_{11} = 0$ , aur $D_{11} = A_{11} t^{2} /12$ (classic $b t^{3} /12$ shape).

Step 1 — $A_{11}$ . Kyun: sirf ek term hai. Raw: $140.905 \times (0.5 - (- 0.5)) = 140.905 \times 1 = 140.905$ GPa·mm. $[A]$ factor $1 0^{6}$ apply karo: $A_{11} = 140.905 \times 1 0^{6} = 1.40905 \times 1 0^{8} N/m .$

Step 2 — $B_{11}$ . Raw: $\frac{1}{2} \cdot 140.905 (0. 5^{2} - (- 0.5)^{2}) = \frac{1}{2} \cdot 140.905 (0.25 - 0.25) = 0$ GPa·mm². Toh $B_{11} = 0$ (centred ⇒ no coupling).

Step 3 — $D_{11}$ . Kyun: $\frac{1}{3} (0. 5^{3} - (- 0.5)^{3}) = \frac{1}{3} (0.125 + 0.125) = \frac{0.25}{3} = 0.083333 mm^{3}$ . Raw: $D_{11} = 140.905 \times 0.083333 = 11.742 GPa\cdotmm^{3} .$ $[D]$ factor apply karo (jo $1$ hai, kyunki GPa·mm³ = N·m directly): $D_{11} = 11.742$ N·m. Koi hidden power-of-ten juggling nahi — GPa·mm³ hi N·m hai.

Verify: Beam-shape identity $D_{11} = ? A_{11} t^{2} /12$ : $A_{11} t^{2} /12 = 1.40905 \times 1 0^{8} \times (1 0^{- 3})^{2} /12 = 1.40905 \times 1 0^{8} \times 1 0^{- 6} /12 = 11.742$ N·m ✓. Exact match — hamara tensorial $[D]$ textbook ke $I = b t^{3} /12$ per unit width tak reduce hota hai. Cell C4 confirmed.

Example 5 — Zero-Thickness Limit (Cell C5)

Forecast: Jaise 90° ply gayab hoti hai, $A, B, D$ mein uska contribution sab $\to 0$ hona chahiye, sirf bottom 0° ply bachni chahiye jo $- 0.5$ se $0$ tak span karti hai. Limiting $A_{11}$ guess karo.

Step 1 — parametrise karo. Maano ply 2, $0$ se $δ$ tak span kare. Uska $A$ -contribution $Q_{11}^{90°} \cdot δ$ hai, $B$ -contribution $\frac{1}{2} Q_{11}^{90°} (δ^{2} - 0)$ , $D$ -contribution $\frac{1}{3} Q_{11}^{90°} δ^{3}$ (sab raw GPa·mm-something). Kyun: hum exact formulas padhte hain $h_{1} = 0, h_{2} = δ$ ke saath.

Step 2 — $δ \to 0$ lo. Teeno $\to 0$ (orders $δ, δ^{2}, δ^{3}$ ). $B$ aur $D$ terms zyaada tezi se vanish hote hain — ek patla off-axis surface ply bending ko barely affect karta hai, yeh ek well-known design fact hai.

Step 3 — jo bachta hai woh 0° ply hai, $- 0.5$ se $0$ . Har raw value compute karo, phir ek-ek factor.

$A_{11} = 140.905 \times (0 - (- 0.5)) = 140.905 \times 0.5 = 70.4525$ GPa·mm $\to \times 1 0^{6} = 7.0452 \times 1 0^{7}$ N/m.
$B_{11} = \frac{1}{2} \cdot 140.905 (0^{2} - (- 0.5)^{2}) = \frac{1}{2} \cdot 140.905 (- 0.25) = - 17.613$ GPa·mm² $\to \times 1 0^{3} = - 1.7613 \times 1 0^{4}$ N.
$D_{11} = \frac{1}{3} \cdot 140.905 (0^{3} - (- 0.5)^{3}) = \frac{1}{3} \cdot 140.905 (0.125) = 5.871$ GPa·mm³ $\to \times 1 = 5.871$ N·m.

Verify: Residual single ply centred nahi hai (yeh $- 0.5$ se $0$ span karti hai), toh $B_{11} \neq = 0$ — yeh ek achha trap hai notice karne ka. Uska magnitude $\approx - 1.76 \times 1 0^{4}$ N, Ex 2 ke $- 1.64 \times 1 0^{4}$ N ke same order mein hai (dono single-sided 0.5 mm lever arms hain), jo sahi scale hai. Ex 4 se compare karo jahan ply centred thi aur $B_{11} = 0$ . Position, sirf presence nahi, coupling set karti hai. Cell C5 confirmed — limits continuous hain.

Example 6 — ABD Invert Karo, Symmetric Shortcut (Cell C6)

Forecast: $[B] = 0$ ke saath top aur bottom blocks split ho jaate hain. Guess karo: $κ_{x} = 0$ exactly (koi moment apply nahi), aur $ϵ_{x}^{0}$ tiny hoga (composites stiff hote hain).

Step 1 — partitioned ABD explicitly likhna. Kyun: novices ko shortcut par trust karne se pehle block structure dekhna zaroori hai. Parent note se poora $6 \times 6$ relation hai: $[N M] = [[A] [B] [B] [D]] [ϵ^{0} κ],$ jahan $[A], [B], [D]$ har ek $3 \times 3$ blocks hain, $N, M, ϵ^{0}, κ$ har ek $3 \times 1$ .

Step 2 — $[B] = 0$ set karo aur split padho. Kyun: agar off-diagonal block zero hai, toh do block-rows decouple ho jaati hain: $N = [A] ϵ^{0} \Rightarrow ϵ^{0} = [A]^{- 1} N, M = [D] κ \Rightarrow κ = [D]^{- 1} M .$ Strains sirf forces par depend karti hain, curvatures sirf moments par — koi cross-talk nahi.

Step 3 — moments. $M = 0 \Rightarrow κ = [D]^{- 1} 0 = 0$ . Toh $κ_{x} = 0$ .

Step 4 — strain (uniaxial estimate). $x$ -direction ko dominant term maante hue, $ϵ_{x}^{0} \approx N_{x} / A_{11} = 1000/1.5097 \times 1 0^{8} = 6.624 \times 1 0^{- 6}$ (dimensionless strain), yaani $6.6$ microstrain.

Verify: Units: (N/m)/(N/m) = dimensionless ✓. Magnitude sanity: 2 mm stiff plate par 1000 N/m se micro-strain milna chahiye — stiffer laminate ⇒ smaller strain, aur $6.6 μ ε$ exactly usi regime mein hai ✓. Kyunki $[B] = 0$ , koi curvature leak nahi hoti ✓. Cell C6 confirmed ( $A_{12}$ ke saath full-matrix cross-coupling FE tools mein handle hoti hai — Finite Element Modeling of Composites).

Example 7 — Full Coupled Inversion (Cell C7: B ≠ 0)

Forecast: Kyunki $B_{11} \neq = 0$ hai, pure tension zaroor curvature produce karegi. Guess karo: is baar $κ_{x} \neq = 0$ hoga.

Step 1 — $2 \times 2$ . Kyun: coupled pair isolate karo. $[N_{x} M_{x}] = [A_{11} B_{11} B_{11} D_{11}] [ϵ_{x}^{0} κ_{x}] = [7.5485 \times 1 0^{7} - 1.6355 \times 1 0^{4} - 1.6355 \times 1 0^{4} 6.290] [ϵ_{x}^{0} κ_{x}] .$

Step 2 — determinant. Kyun: Cramer's rule ko iska zaroort hai. $Δ = A_{11} D_{11} - B_{11}^{2} = (7.5485 \times 1 0^{7}) (6.290) - (- 1.6355 \times 1 0^{4})^{2} = 4.7480 \times 1 0^{8} - 2.675 \times 1 0^{8} = 2.0730 \times 1 0^{8} .$ (Yahan $B^{2}$ term negligible nahi hai — do-ply laminate ke liye coupling strong hai.)

Step 3 — solve karo $M_{x} = 0$ ke saath. $ϵ_{x}^{0} = \frac{D _{11} N _{x} - B _{11} \cdot 0}{Δ} = \frac{6.290 \cdot 1000}{2.0730 \times 1 0 ^{8}} = 3.034 \times 1 0^{- 5},$ $κ_{x} = \frac{A _{11} \cdot 0 - B _{11} N _{x}}{Δ} = \frac{- ( - 1.6355 \times 1 0 ^{4} ) ( 1000 )}{2.0730 \times 1 0 ^{8}} = 7.890 \times 1 0^{- 2} m^{- 1} .$

Verify: $κ_{x} \neq = 0$ — pure tension ne plate bend kiya, exactly Ex 2 ki physics. Sign: $κ_{x} > 0$ , plate positive $z$ (weak 90° side) ki taraf bow karti hai, stiff side ko relief deti hai — forecast se consistent. Units: $[N] \cdot [N/m] / [N^{2}] = 1/ m$ ✓ (curvature). Cell C7 confirmed — jab $[B] \neq = 0$ ho toh inversion full hona chahiye, kabhi block-split nahi.

Example 8 — Real-World Word Problem (Cell C8)

Forecast: Ex 6 mein $1000$ N/m par $6.6 μ ε$ mila tha. Load double karo — strain guess karo aur $20 μ ε$ limit se compare karo.

Step 1 — strain. Kyun: symmetric ⇒ decoupled (Ex 6 Step 2), $ϵ_{x}^{0} = N_{x} / A_{11}$ . $ϵ_{x}^{0} = \frac{2000}{1.5097 \times 1 0 ^{8}} = 1.3248 \times 1 0^{- 5} = 13.25 μ ε .$

Step 2 — compare karo. $13.25 < 20$ . Pass, $34%$ margin ke saath.

Step 3 — check karo ki koi curvature nahi. Kyun: bonded sensor bending se bhi fail hota hai. Symmetric ⇒ $[B] = 0$ ⇒ $κ_{x} = 0$ . Dono counts par safe. (Ek sandwich construction $[D]$ aur badhayega — dekho Sandwich Panel Theory; first-ply margins First Ply Failure mein.)

Verify: Linearity check — $1000 \to 2000$ N/m double karne par $6.624 μ ε \to 13.25 μ ε$ double hua ✓. Margin $= (20 - 13.25) /20 = 33.8%$ ✓. Cell C8 confirmed.

Example 9 — Exam Twist: Balanced Lekin Antisymmetric (Cell C9)

Forecast: "Balanced" matlab equal $+ θ$ aur $- θ$ plies. Guess karo in dono mein se kaun bachega — $A_{16}$ ya $B_{16}$ .

Step 1 — heights. $h_{0} = - 0.5, h_{1} = 0, h_{2} = 0.5$ mm. $Q_{16}$ : $+ 32.4$ (bottom), $- 32.4$ (top) GPa.

Step 2 — $A_{16}$ . Kyun: $[A]$ position ignore karta hai, toh opposite signs cancel ho jaate hain. Raw: $A_{16} = (+ 32.4) (0.5) + (- 32.4) (0.5) = 0 GPa\cdotmm \Rightarrow A_{16} = 0.$ Yeh balanced property hai: koi extension–shear coupling nahi.

Step 3 — $B_{16}$ . Kyun: $[B]$ $\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})$ se weight karta hai: bottom ply $\frac{1}{2} (0 - 0.25) = - 0.125$ , top ply $\frac{1}{2} (0.25 - 0) = + 0.125$ (mm²). Raw: $B_{16} = (+ 32.4) (- 0.125) + (- 32.4) (+ 0.125) = - 4.05 - 4.05 = - 8.10 GPa\cdotmm^{2} .$ Single $[B]$ factor $1 0^{3}$ apply karo: $B_{16} = - 8.10 \times 1 0^{3} = - 8100$ N $\neq = 0$ . Dono terms ka same sign tha — $Q_{16}$ ka sign flip aur lever arm ka sign flip milkar ek reinforcing (non-cancelling) result deta hai.

Verify: Twist resolve hua: balanced ( $A_{16} = 0$ ) ka matlab no coupling nahi hota — antisymmetry $B_{16} \neq = 0$ chodh deti hai, toh load mein $\pm 45°$ antisymmetric plate ko twist karna real hai. Numeric (raw): $32.4 \cdot 0.125 \cdot 2 = 8.10$ GPa·mm² ⇒ $- 8100$ N ✓. Ek symmetric $[\pm 45]_{s}$ top-pair lever arms flip kar deta aur $B_{16} = 0$ bhi deta. Cell C9 confirmed — yeh classic exam gotcha hai. Dekho Symmetric and Balanced Laminates.

Recall Kaun sa cell full inversion force karta hai, aur kyun?

Cell C7 (asymmetric, $[B] \neq = 0$ ). ::: Kyunki $[B]$ force aur moment blocks ko link karta hai, toh strains aur curvatures independently nahi nikaali ja sakti.

Recall Balanced vs symmetric — har ek kya khatam karta hai?

Balanced $A_{16}, A_{26}$ (extension–shear coupling) khatam karta hai. Symmetric poora $[B]$ khatam karta hai. ::: Yeh independent hain — Ex 9 balanced hai phir bhi $B_{16} \neq = 0$ hai.