Exercises — Classical laminate theory — ABD matrix

6×6 ki pehli row dekho (parent note se). Woh row jo $N_x$ produce karti hai, woh kuch aise padhti hai: $N_x=A_{11}{ϵ}_x^0+A_{12}{ϵ}_y^0+A_{16}{\gamma }_{xy}^0+B_{11}{\kappa }_x+B_{12}{\kappa }_y+B_{16}{\kappa }_{xy}.$

• ${ϵ}_x^0\to N_x$ ka coefficient ==$A_{11}$== hai (top-left corner, extensional stiffness).
• ${\kappa }_x\to N_x$ ka coefficient ==$B_{11}$== hai (yeh top-right $[B]$ block mein rehta hai).

Kaisa dikhta hai: $[A]$ top-left 3×3 block hai, $[D]$ bottom-right hai, aur $[B]$ woh shared off-diagonal block hai jo do baar aata hai.

Har term $Q\times (\text{height ki ek power})$ hai:

• $A$: $Q\times (h_k-h_{k-1})$ → Pa·m${}^1$ = Pa·m.
• $B$: $Q\times (h_k^2-h_{k-1}^2)$ → Pa·m${}^2$ = Pa·m².
• $D$: $Q\times (h_k^3-h_{k-1}^3)$ → Pa·m${}^3$ = Pa·m³.

Kyun zaroori hai: agar kabhi $D$ compute karo aur Pa·m² mile, toh integrate galat kiya. Units sabse sasta error detector hai.

$A_{11}={\sum }_k{Q}_{11}^{(k)}(h_k-h_{k-1})$. Har ply ki thickness $t=0.5\times 10^{-3}$ m hai, toh: $A_{11}=t\left(Q_{11}^0+Q_{11}^{90}+Q_{11}^{90}+Q_{11}^0\right).$ Humne kya kiya: kyunki thickness equally cancel hoti hai, hum bas har ply ka $Q_{11}$ add karte hain aur $t$ se multiply karte hain. $A_{11}=0.5\times 10^{-3}(140+10+10+140)\times 10^9=0.5\times 10^{-3}\times 300\times 10^9$ $\boxed{A_{11}=1.50\times 10^8\text{Pa⋅m}}$

$D_{11}=\frac{1}{3}{\sum }_k{Q}_{11}^{(k)}(h_k^3-h_{k-1}^3)$. Metres mein kaam karo; $u=10^{-3}$ m maano.

Plies (bottom→top) $[0,90,90,0]$ with $Q_{11}=[140,10,10,140]$ GPa. Boundaries ke cubes ($u^3$ units mein): $(-1{)}^3=-1,(-0.5{)}^3=-0.125,0^3=0,0.5^3=0.125,1^3=1.$ Differences $h_k^3-h_{k-1}^3$ ($u^3$ mein): $[-0.125-(-1),0-(-0.125),0.125-0,1-0.125]=[0.875,0.125,0.125,0.875]$. $\sum Q_{11}(\Delta z^3)=(140\cdot 0.875+10\cdot 0.125+10\cdot 0.125+140\cdot 0.875)\times 10^9{u}^3$ $=(122.5+1.25+1.25+122.5)\times 10^9{u}^3=247.5\times 10^9{u}^3.$ $u^3=10^{-9}{\text{m}}^3$ ke saath: $\sum =247.5$ Pa·m³. Phir $D_{11}=\frac{1}{3}\times 247.5$. $\boxed{D_{11}=82.5{\text{Pa⋅m}}^3}$

Matlab kya hai: dono $0°$ plies extreme heights par hain ($|z|$ near $1$ mm). $z^3$ weighting un outer plies ko bending mein dominate karti hai — yehi wajah hai ki bending panel ke baahri surface par stiff plies rakhte hain.

$B_{11}=\frac{1}{2}{\sum }_k{Q}_{11}^{(k)}(h_k^2-h_{k-1}^2)$. $u^2=10^{-6}{\text{m}}^2$ maano. Squares: $(-0.5{)}^2=0.25,0^2=0,0.5^2=0.25$.

• Ply 1 ($0°$): $h_1^2-h_0^2=0-0.25=-0.25u^2$, $Q_{11}=140$ GPa.
• Ply 2 ($90°$): $h_2^2-h_1^2=0.25-0=+0.25u^2$, $Q_{11}=10$ GPa. $B_{11}=\frac{1}{2}(140(-0.25)+10(0.25))\times 10^9{u}^2=\frac{1}{2}(-35+2.5)\times 10^9\times 10^{-6}$ $=\frac{1}{2}(-32.5)\times 10^3=\boxed{-1.625\times 10^4{\text{Pa⋅m}}^2}$

Sign ka matlab: $B_{11}<0$. Stiff $0°$ ply mid-plane ke neeche hai ($z<0$), compliant $90°$ ply upar. $x$ mein pull karne par ($N_x>0$) dono stretch hote hain, lekin stiff bottom zyaada resist karta hai, toh plate curl karti hai — ek nonzero ${\kappa }_x$ aata hai bina koi moment apply kiye. Yeh woh tension–bending coupling hai jo asymmetric laminates ko cure ke baad warp karti hai. Dekho Symmetric and Balanced Laminates ki hum ise zyaadatar design mein kyun khatam kar dete hain.

Argument: $B_{11}=\frac{1}{2}\sum Q_{11}^{(k)}(h_k^2-h_{k-1}^2)$. Symmetry ka matlab hai ki $[h_{k-1},h_k]$ heights par har ply ke liye ek mirror ply $[-h_k,-h_{k-1}]$ par hai jisme same $Q$ hai. Uska contribution $(-h_{k-1}{)}^2-(-h_k{)}^2=h_{k-1}^2-h_k^2$ hai — apne partner ka exact negative. Yeh pairs mein cancel ho jaate hain.

Numeric check (outer $0°$ pair):

• Bottom $0°$: $h_0=-1,h_1=-0.5$ mm → $h_1^2-h_0^2=0.25-1=-0.75u^2$.
• Top $0°$: $h_3=0.5,h_4=1$ mm → $h_4^2-h_3^2=1-0.25=+0.75u^2$.
• Sum: $140(-0.75)+140(+0.75)=0$. ✔ Inner $90°$ pair identically cancel karta hai. Isliye $B_{11}=0$.

• $A_{16}=0$: koi in-plane shear–extension coupling nahi. Ek pure $N_x$ koi ${\gamma }_{xy}^0$ shear strain produce nahi karta — stack balanced hai (har $+\theta$ ply ke saath ek $-\theta$ ply matched hai).
• $B_{16}\ne 0$: hai bend–twist / shear-curvature coupling. Ek in-plane shear load phir bhi ek twisting curvature ${\kappa }_{xy}$ cause karega kyunki $+\theta$ aur $-\theta$ plies alag heights par hain (balanced ≠ symmetric).

Takeaway: balanced $A_{16},A_{26}$ ko khatam karta hai; symmetric poore $[B]$ ko khatam karta hai. Fully decouple karne ke liye dono chahiye.

Kyunki $[B]=0$ hai, ABD block-diagonal hai aur ${\mathbf{ϵ}}^0=[A{]}^{-1}\mathbf{N}$ (curvatures zero rahenge — koi moments nahi). Hume sirf top-left $2\times 2$ chahiye (shear decouple ho jaata hai kyunki $A_{16}=A_{26}=0$): $\left[\begin{array}{c}{ϵ}_x^0\\ {ϵ}_y^0\end{array}\right]={\left[\begin{array}{cc}1.5& 0.2\\ 0.2& 1.5\end{array}\right]}^{-1}\times 10^{-8}\left[\begin{array}{c}1000\\ 0\end{array}\right].$ Determinant $=1.5^2-0.2^2=2.25-0.04=2.21$ ($10^{16}$ ki units mein). $2\times 2$ ka inverse ($10^8$ bahar nikalo): $[A{]}_{2\times 2}^{-1}=\frac{1}{2.21\times 10^{16}}\left[\begin{array}{cc}1.5& -0.2\\ -0.2& 1.5\end{array}\right]\times 10^8.$ ${ϵ}_x^0=\frac{1.5\times 10^8}{2.21\times 10^{16}}\times 1000=\frac{1500}{2.21\times 10^8}\approx 6.787\times 10^{-6}.$ ${ϵ}_y^0=\frac{-0.2\times 10^8}{2.21\times 10^{16}}\times 1000=\frac{-200}{2.21\times 10^8}\approx -9.05\times 10^{-7}.$ $\boxed{{ϵ}_x^0\approx 6.79\times 10^{-6},{ϵ}_y^0\approx -9.05\times 10^{-7}}$

Matlab kya hai: $x$ mein pull karne se $x$ mein stretch hota hai (positive) aur, $A_{12}$ coupling ke through (ek Poisson-jaisa effect), $y$ mein thodi si contraction hoti hai (negative). Koi curvature nahi — clean membrane response, exactly wahi jo symmetric design deta hai.

$D_{11}$ high $Q_{11}$ plies ko large $|z|$ par rakhne par reward karta hai ($z^3$ weight ki wajah se). $0°$ ply mein $Q_{11}=140$ hai, $90°$ ply mein $Q_{11}=10$ hai.

• $[0/90/90/0{]}_s$: $0°$ plies outer hain (large $|z|$) → yeh L2 stack hai, $D_{11}=82.5$ Pa·m³.
• $[90/0/0/90{]}_s$: $0°$ plies inner hain → kamzor. Compute karo: $\Delta z^3$ list $[0.875,0.125,0.125,0.875]u^3$ use karke with $Q_{11}=[10,140,140,10]$: $\sum =(10\cdot 0.875+140\cdot 0.125+140\cdot 0.125+10\cdot 0.875)\times 10^9{u}^3=(8.75+17.5+17.5+8.75)\times 10^9{u}^3=52.5{\text{Pa⋅m}}^3.$ $D_{11}=\frac{1}{3}\times 52.5=17.5{\text{Pa⋅m}}^3.$ Verdict: $[0/90/90/0{]}_s$ jeet ta hai $82.5$ vs $17.5$ Pa·m³ ke saath — $\approx 4.71$ ka factor. Koi bhi $100$ tak nahi pahuncha, toh target meet karne ke liye baahri taraf zyaada $0°$ plies add karni hongi ya laminate ko mota karna hoga. Placement utna hi matter karta hai jitna material.

Do independent requirements:

1. Balanced (har $+45$ ek $-45$ ke saath paired) → $A_{16},A_{26}$ khatam karta hai.
2. Symmetric (mid-plane ke baare mein mirror) → poore $[B]$ ko khatam karta hai, isliye koi warp nahi.

Ek two-ply $[+45/-45]$ balanced hai lekin symmetric nahi ⇒ $[B]\ne 0$ ⇒ warp karta hai. Dono ko satisfy karne wala sabse patla stack four-ply hai: $\boxed{[+45/-45/-45/+45]}$ bottom→top padha gaya. Check karo: angles $\{+45,-45,-45,+45\}$ mein $\pm 45$ ki equal counts hain ⇒ balanced. Centre ke baare mein reversed sequence khud hi hai ⇒ symmetric ⇒ $[B]=0$. Yeh standard $[\pm 45{]}_s$ notation hai. Dekho Symmetric and Balanced Laminates. (Thermal residual stress jo warp drive karta hai, usse Thermal and Hygroscopic Effects mein treat kiya gaya hai.)

$2\times 2$ invert karo. Determinant: $\Delta =A_{11}{D}_{11}-B_{11}^2=(7.5\times 10^7)(6.25)-(-1.625\times 10^4{)}^2.$ $A_{11}{D}_{11}=4.6875\times 10^8,B_{11}^2=2.640625\times 10^8,\Delta =2.046875\times 10^8.$ Inverse deta hai: ${\kappa }_x=\frac{-B_{11}{N}_x+A_{11}{M}_x}{\Delta }=\frac{-(-1.625\times 10^4)(1000)+0}{2.046875\times 10^8}.$ ${\kappa }_x=\frac{1.625\times 10^7}{2.046875\times 10^8}\approx 0.0794{\text{m}}^{-1}.$ $\boxed{{\kappa }_x\approx 7.94\times 10^{-2}{\text{m}}^{-1}}$

Matlab kya hai: ek pure tension of $1000$ N/m is asymmetric plate ko $\approx 1/0.0794\approx 12.6$ m ke radius par curve kara deta hai. Koi moment apply nahi kiya tha — curvature poori tarah $B_{11}$ coupling ka kaam hai. Ek symmetric plate ($B_{11}=0$) bilkul ${\kappa }_x=0$ deta. Yeh L3 trap ka quantitative chehra hai.

Kaun sa matrix moments ko curvatures se relate karta hai? ::: $[D]$, the bending stiffness (Pa·m³). $[B]=0$ kya banata hai? ::: Stacking sequence ki symmetry mid-plane ke baare mein. $A_{16}=A_{26}=0$ kya banata hai? ::: Ek balanced layup ($+\theta$ aur $-\theta$ paired). Kya $N_x$ akela curvature cause kar sakta hai? ::: Haan, jab bhi $[B]\ne 0$ ho (asymmetric laminate). Kaun si plies $D_{11}$ dominate karti hain? ::: Sabse baahri stiff plies, $z^3$ weighting ki wajah se.