Exercises — Classical laminate theory — ABD matrix

Look at the first row of the 6×6 (from the parent note). The row that produces $N_x$ reads $N_x=A_{11}{ϵ}_x^0+A_{12}{ϵ}_y^0+A_{16}{\gamma }_{xy}^0+B_{11}{\kappa }_x+B_{12}{\kappa }_y+B_{16}{\kappa }_{xy}.$

• ${ϵ}_x^0\to N_x$ is the coefficient ==$A_{11}$== (top-left corner, extensional stiffness).
• ${\kappa }_x\to N_x$ is the coefficient ==$B_{11}$== (it lives in the top-right $[B]$ block).

What it looks like: $[A]$ is the top-left 3×3 block, $[D]$ the bottom-right, and $[B]$ is the shared off-diagonal block that appears twice.

Each term is $Q\cdot (\text{a power of height})$:

• $A$: $Q\times (h_k-h_{k-1})$ → 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.

Why it matters: if you ever compute $D$ and get Pa·m², you integrated wrong. Units are the cheapest error detector.

$A_{11}={\sum }_k{Q}_{11}^{(k)}(h_k-h_{k-1})$. Every ply has thickness $t=0.5\times 10^{-3}$ m, so: $A_{11}=t\left(Q_{11}^0+Q_{11}^{90}+Q_{11}^{90}+Q_{11}^0\right).$ What we did: because thickness cancels equally, we just add the $Q_{11}$ of every ply and multiply by $t$. $A_{11}=0.5\times 10^{-3}(140+10+10+140)\times 10^9=0.5\times 10^{-3}\times 300\times 10^9$ $\boxed{A_{11}=1.50\times 10^8\text{Pa⋅m}}$

$D_{11}=\frac{1}{3}{\sum }_k{Q}_{11}^{(k)}(h_k^3-h_{k-1}^3)$. Work in metres; let $u=10^{-3}$ m.

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

What it means: the two $0°$ plies sit at the extreme heights ($|z|$ near $1$ mm). The $z^3$ weighting makes those outer plies dominate bending — exactly why you put stiff plies on the outside of a bending panel.

$B_{11}=\frac{1}{2}{\sum }_k{Q}_{11}^{(k)}(h_k^2-h_{k-1}^2)$. Let $u^2=10^{-6}{\text{m}}^2$. 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 meaning: $B_{11}<0$. The stiff $0°$ ply sits below the mid-plane ($z<0$), the compliant $90°$ ply above. Pulling in $x$ ($N_x>0$) stretches both, but the stiff bottom resists more, so the plate curls — a nonzero ${\kappa }_x$ appears even though no moment was applied. This is the tension–bending coupling that makes asymmetric laminates warp after cure. See Symmetric and Balanced Laminates for why we usually design it away.

The argument: $B_{11}=\frac{1}{2}\sum Q_{11}^{(k)}(h_k^2-h_{k-1}^2)$. Symmetry means for every ply at heights $[h_{k-1},h_k]$ there is a mirror ply at $[-h_k,-h_{k-1}]$ with the same $Q$. Its contribution is $(-h_{k-1}{)}^2-(-h_k{)}^2=h_{k-1}^2-h_k^2$ — the exact negative of its partner. They cancel in pairs.

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: $140(-0.75)+140(+0.75)=0$. ✔ The $90°$ inner pair cancels identically. Hence $B_{11}=0$.

• $A_{16}=0$: no in-plane shear–extension coupling. A pure $N_x$ produces no ${\gamma }_{xy}^0$ shear strain — the stack is balanced (every $+\theta$ ply matched by a $-\theta$ ply).
• $B_{16}\ne 0$: there is bend–twist / shear-curvature coupling. A shear-in-plane load will still cause a twisting curvature ${\kappa }_{xy}$ because the $+\theta$ and $-\theta$ plies are at different heights (balanced ≠ symmetric).

Takeaway: balanced kills $A_{16},A_{26}$; symmetric kills all of $[B]$. You need both to fully decouple.

Because $[B]=0$, the ABD is block-diagonal and ${\mathbf{ϵ}}^0=[A{]}^{-1}\mathbf{N}$ (curvatures stay zero — no moments). We only need the top-left $2\times 2$ (shear decouples since $A_{16}=A_{26}=0$): $\left[\begin{array}{c}{ϵ}_x^0\\ {ϵ}_y^0\end{array}\right]={\left[\begin{array}{cc}1.5& 0.2\\ 0.2& 1.5\end{array}\right]}^{-1}\times 10^{-8}\left[\begin{array}{c}1000\\ 0\end{array}\right].$ Determinant $=1.5^2-0.2^2=2.25-0.04=2.21$ (in units of $10^{16}$). Inverse of the $2\times 2$ (pull out $10^8$): $[A{]}_{2\times 2}^{-1}=\frac{1}{2.21\times 10^{16}}\left[\begin{array}{cc}1.5& -0.2\\ -0.2& 1.5\end{array}\right]\times 10^8.$ ${ϵ}_x^0=\frac{1.5\times 10^8}{2.21\times 10^{16}}\times 1000=\frac{1500}{2.21\times 10^8}\approx 6.787\times 10^{-6}.$ ${ϵ}_y^0=\frac{-0.2\times 10^8}{2.21\times 10^{16}}\times 1000=\frac{-200}{2.21\times 10^8}\approx -9.05\times 10^{-7}.$ $\boxed{{ϵ}_x^0\approx 6.79\times 10^{-6},{ϵ}_y^0\approx -9.05\times 10^{-7}}$

What it means: pulling in $x$ stretches it in $x$ (positive) and, through the $A_{12}$ coupling (a Poisson-like effect), it contracts slightly in $y$ (negative). No curvature — clean membrane response, exactly what symmetric design buys you.

$D_{11}$ rewards putting high $Q_{11}$ plies at large $|z|$ (the $z^3$ weight). The $0°$ ply has $Q_{11}=140$, the $90°$ ply has $Q_{11}=10$.

• $[0/90/90/0{]}_s$: $0°$ plies are outer (large $|z|$) → this is the L2 stack, $D_{11}=82.5$ Pa·m³.
• $[90/0/0/90{]}_s$: $0°$ plies are inner → weaker. Compute: Using the $\Delta z^3$ list $[0.875,0.125,0.125,0.875]u^3$ 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: $[0/90/90/0{]}_s$ wins with $82.5$ vs $17.5$ Pa·m³ — a factor of $\approx 4.71$. Neither reaches $100$, so to meet the target you must add more $0°$ plies to the outside or thicken the laminate. Placement matters as much as material.

Two independent requirements:

1. Balanced (every $+45$ paired with a $-45$) → kills $A_{16},A_{26}$.
2. Symmetric (mirror about mid-plane) → kills all $[B]$, hence no warp.

A two-ply $[+45/-45]$ is balanced but not symmetric ⇒ $[B]\ne 0$ ⇒ warps. The thinnest stack satisfying both is the four-ply $\boxed{[+45/-45/-45/+45]}$ read bottom→top. Check: angles $\{+45,-45,-45,+45\}$ contain equal counts of $\pm 45$ ⇒ balanced. The sequence reversed about the centre is itself ⇒ symmetric ⇒ $[B]=0$. This is the standard $[\pm 45{]}_s$ notation. See Symmetric and Balanced Laminates. (Thermal residual stress that drives the warp is treated in Thermal and Hygroscopic Effects.)

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

What it means: a pure tension of $1000$ N/m makes this asymmetric plate curve at radius $\approx 1/0.0794\approx 12.6$ m. No moment was applied — the curvature is entirely the $B_{11}$ coupling at work. A symmetric plate ($B_{11}=0$) would give ${\kappa }_x=0$ exactly. This is the quantitative face of the L3 trap.

Which matrix relates moments to curvatures? ::: $[D]$, the bending stiffness (Pa·m³). What makes $[B]=0$? ::: Symmetry of the stacking sequence about the mid-plane. What makes $A_{16}=A_{26}=0$? ::: A balanced layup ($+\theta$ paired with $-\theta$). Can $N_x$ alone cause curvature? ::: Yes, whenever $[B]\ne 0$ (asymmetric laminate). Which plies dominate $D_{11}$? ::: The outermost stiff plies, because of the $z^3$ weighting.