Worked examples — Classical laminate theory — ABD matrix

Before anything, one reminder of the vocabulary we lean on, because we use it in line one of every example.

Figure 1 (below): why the three sums differ. The plot shows one vertical coordinate $z$ (mid-plane at $0$) fed into three different weightings. The flat black line is the $[A]$ weighting (it does not care where the material sits). The tilted dashed black line is the $[B]$ weighting (it grows linearly with $z$, so it is negative below the middle and positive above — that is the lever arm that cancels for symmetric stacks). The red bowl is the $[D]$ weighting ($z^2$): material far from the middle, on either side, is weighted heavily — this is why outer plies dominate bending.

That single picture is the whole story of why symmetry kills $[B]$ (the tilted line cancels top-to-bottom) but never $[A]$ or $[D]$ (flat and bowl never cancel).

The scenario matrix

Every ABD problem you will ever see falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	Case class	What's special	Example
C1	Symmetric stack, $[B]=0$	above/below plies mirror → coupling vanishes	Ex 1
C2	Asymmetric stack, $[B]\ne 0$	pulling → bending; sign of $B_{11}$ matters	Ex 2
C3	Sign flip of the load	negative $N_x$ → mirror-image response	Ex 3
C4	Single ply (degenerate $n=1$)	check formulas reduce to plate theory	Ex 4
C5	Zero-load / zero-thickness limit	drop a ply's thickness $\to 0$; sanity of sums	Ex 5
C6	Invert ABD for strains (symmetric shortcut)	$[A{]}^{-1}$ block only	Ex 6
C7	Full coupled inversion ($[B]\ne 0$)	must invert the whole $6\times 6$; curvature appears	Ex 7
C8	Real-world word problem	a satellite panel; pick a stack	Ex 8
C9	Exam twist — balanced but antisymmetric	$A_{16}=0$ yet $B_{16}\ne 0$	Ex 9

Shared material for all examples. Carbon/epoxy: $E_1=140$ GPa, $E_2=10$ GPa, $G_{12}=5$ GPa, ${\nu }_{12}=0.3$. From these, the reduced stiffness of a ply along its own fibres is $Q_{11}=\frac{E_1}{1-{\nu }_{12}{\nu }_{21}},Q_{22}=\frac{E_2}{1-{\nu }_{12}{\nu }_{21}},Q_{12}=\frac{{\nu }_{12}{E}_2}{1-{\nu }_{12}{\nu }_{21}},Q_{66}=G_{12},$ where ${\nu }_{21}={\nu }_{12}{E}_2/E_1$. Why these formulas? They are the plane-stress inverse of the compliance matrix — the parent note's $[Q]$ for a ply whose fibres run along $x$. Plugging numbers: ${\nu }_{21}=0.3\cdot 10/140=0.02143$, so $1-{\nu }_{12}{\nu }_{21}=1-0.3\cdot 0.02143=0.99357$. Then

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

For a 90° ply, fibres run along $y$, so the roles of $1$ and $2$ swap in the $11$ position: $Q_{11}^{(90°)}=Q_{22}^{(0°)}=10.065$ GPa. This single fact — the stiff number moves depending on fibre direction — drives every coupling result below.

Recall Why does 90° swap the numbers?

A 0° ply resists $x$-stretch with its fibres ($140$ GPa). A 90° ply's fibres point along $y$, so $x$-stretch only pulls the weak matrix ($10$ GPa). ::: The $11$ entry of $[Q]$ therefore reads $140.9$ for 0° and $10.06$ for 90°.

---

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

Ply heights (mm), mid-plane at 0: $h_0=-1,h_1=-0.5,h_2=0,h_3=0.5,h_4=1$. Fibre pattern: ply 1 = 0°, ply 2 = 90°, ply 3 = 90°, ply 4 = 0°. $Q_{11}$ per ply: $140.905,10.065,10.065,140.905$ GPa.

Step 1 — $A_{11}$. Why: $[A]$ counts thickness only, so just add each ply's $Q_{11}$ times its 0.5 mm. $A_{11}=(140.905+10.065+10.065+140.905)\times 0.5=150.970\text{GPa⋅mm}.$ Apply the $[A]$ factor $10^6$: $A_{11}=150.970\times 10^6=1.50970\times 10^8$ N/m.

Step 2 — $B_{11}$. Why: $[B]$ weights by $\frac{1}{2}(h_k^2-h_{k-1}^2)$. Because the stack is a mirror (ply 1 ↔ ply 4, ply 2 ↔ ply 3), each pair contributes equal-and-opposite $z$-lever arms. $B_{11}=\frac{1}{2}[140.905(0.25-1)+10.065(0-0.25)+10.065(0.25-0)+140.905(1-0.25)]{\text{GPa⋅mm}}^2.$ The first and last cancel; the middle two cancel. $B_{11}=0$.

Step 3 — $D_{11}$. Why: $[D]$ weights by $\frac{1}{3}(h_k^3-h_{k-1}^3)$; $z^3$ is largest for the outer plies. $D_{11}=\frac{1}{3}[140.905(-0.5^3-(-1{)}^3)+10.065(0-(-0.5{)}^3)+10.065(0.5^3-0)+140.905(1^3-0.5^3)]{\text{GPa⋅mm}}^3.$ Compute the height brackets: $(-0.125+1)=0.875$, $(0.125)$, $(0.125)$, $(0.875)$. $D_{11}=\frac{1}{3}[140.905(0.875)+10.065(0.125)+10.065(0.125)+140.905(0.875)].$ The inner sum is $2(140.905\cdot 0.875)+2(10.065\cdot 0.125)=2(123.292)+2(1.258)=246.584+2.516=249.100$. $D_{11}=\frac{1}{3}(249.100)=83.03{\text{GPa⋅mm}}^3.$ Apply the $[D]$ factor (which is $1$): $D_{11}=83.03$ N·m.

Verify: Units — $A$ in N/m ✓, $B=0$ as symmetry demands ✓, $D$ in N·m ✓. Sanity: the outer 0° plies supply $140.905\times 0.875=123.29$ of the $249.10$ bracket total each — they dominate bending, matching the forecast. Cell C1 confirmed: symmetric ⇒ $B=0$. See 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 — $B_{11}$ as a raw GPa·mm² number. Why: only $[B]$ senses the top/bottom imbalance. Keep $Q$ in GPa, heights in mm, and do the arithmetic before any unit factor. $B_{11}=\frac{1}{2}[Q_{11}^{0°}(0^2-(-0.5{)}^2)+Q_{11}^{90°}(0.5^2-0^2)]$ $=\frac{1}{2}[140.905(-0.25)+10.065(0.25)]=\frac{1}{2}(-35.226+2.516)=\frac{1}{2}(-32.710)=-16.355{\text{GPa⋅mm}}^2.$

Step 3 — one conversion. Why: now, and only now, apply the single $[B]$ factor $10^3$ (from the bookkeeping box: GPa·mm² $\to$ N). $B_{11}=-16.355\times 10^3=-16,355\text{N}\approx -1.636\times 10^4\text{N}.$

$\boxed{B_{11}=-1.636\times 10^4\text{N}}$

Step 4 — meaning of the sign. Why: the sign tells you the direction of induced curvature. $B_{11}<0$ because the negative-$z$ (bottom) ply is the stiff one, giving a negative lever-arm contribution. Physically, pulling with $N_x>0$ produces a positive curvature ${\kappa }_x$ once you invert — the plate bows so the stiff side goes into tension relief.

Figure 2 (below): the coupling in one picture. The red rectangle is the stiff 0° ply sitting below the mid-plane (dotted line); the black rectangle is the weak 90° ply above it. The horizontal black arrows are the applied tension $N_x$. The red dashed arc shows the consequence: because the strong material is all on one side, pure pulling makes the plate bow — that curved arc is $B_{11}\ne 0$ made visible.

Verify: Units: GPa·mm²$\times 10^3$ = Pa·m² = N ✓. Symmetry cross-check: if both plies were 0°, $140.905(-0.25)+140.905(0.25)=0$ ⇒ $B_{11}=0$, recovering the symmetric case. Our nonzero value comes entirely from $140.905\ne 10.065$. Cell C2 confirmed.

---

Example 3 — Sign flip of the load (Cell C3)

Forecast: The equations are linear. Guess: does the curvature just flip sign, or something subtler?

Step 1 — linearity. Why: the ABD relation $\mathbf{N},\mathbf{M}=[ABD]({\mathbf{ϵ}}^0,\mathbf{\kappa })$ is a linear map, so its inverse is linear too. Scaling the load by $-1$ scales every response by $-1$.

Step 2 — apply. Let ${\kappa }_x^{(+)}$ denote the curvature produced by $N_x=+1000$ N/m (its value is computed in Ex 7). By linearity, $N_x=-1000$ N/m produces ${\kappa }_x=-{\kappa }_x^{(+)}$: same magnitude, opposite bow. A plate that domed upward under tension domes downward under compression.

Verify: No new numbers to compute — this is a structural check that our machine has no hidden asymmetry (no thermal/prestress term; those live in Thermal and Hygroscopic Effects). If a residual curvature remained at $N_x=0$, we'd have missed a term. The zero-load state has zero curvature ⇒ pure linearity ✓. Cell C3 confirmed.

---

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

Forecast: A single centred ply is symmetric about its own middle. Guess: $B_{11}=0$, and $D_{11}=A_{11}{t}^2/12$ (the classic $b{t}^3/12$ shape).

Step 1 — $A_{11}$. Why: only one term. Raw: $140.905\times (0.5-(-0.5))=140.905\times 1=140.905$ GPa·mm. Apply the $[A]$ factor $10^6$: $A_{11}=140.905\times 10^6=1.40905\times 10^8\text{N/m}.$

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. So $B_{11}=0$ (centred ⇒ no coupling).

Step 3 — $D_{11}$. Why: $\frac{1}{3}(0.5^3-(-0.5{)}^3)=\frac{1}{3}(0.125+0.125)=\frac{0.25}{3}=0.083333{\text{mm}}^3$. Raw: $D_{11}=140.905\times 0.083333=11.742{\text{GPa⋅mm}}^3.$ Apply the $[D]$ factor (which is $1$, since GPa·mm³ = N·m directly): $D_{11}=11.742$ N·m. No hidden power-of-ten juggling — GPa·mm³ is N·m.

Verify: The beam-shape identity $D_{11}\stackrel{?}{=}{A}_{11}{t}^2/12$: $A_{11}{t}^2/12=1.40905\times 10^8\times (10^{-3}{)}^2/12=1.40905\times 10^8\times 10^{-6}/12=11.742$ N·m ✓. Exact match — our tensorial $[D]$ reduces to the textbook $I=b{t}^3/12$ per unit width. Cell C4 confirmed.

---

Example 5 — Zero-thickness limit (Cell C5)

Forecast: As the 90° ply vanishes, its contribution to $A,B,D$ should all $\to 0$, leaving just the bottom 0° ply spanning $-0.5$ to $0$. Guess the limiting $A_{11}$.

Step 1 — parametrise. Let ply 2 span $0$ to $\delta$. Its $A$-contribution is $Q_{11}^{90°}\cdot \delta$, its $B$-contribution $\frac{1}{2}{Q}_{11}^{90°}({\delta }^2-0)$, its $D$-contribution $\frac{1}{3}{Q}_{11}^{90°}{\delta }^3$ (all raw GPa·mm-something). Why: we read the exact formulas with $h_1=0,h_2=\delta$.

Step 2 — take $\delta \to 0$. All three $\to 0$ (orders $\delta ,{\delta }^2,{\delta }^3$). The $B$ and $D$ terms vanish faster — a thin off-axis surface ply barely affects bending, a well-known design fact.

Step 3 — what remains is the 0° ply, $-0.5$ to $0$. Compute each raw value, then one factor apiece.

• $A_{11}=140.905\times (0-(-0.5))=140.905\times 0.5=70.4525$ GPa·mm $\to \times 10^6=7.0452\times 10^7$ 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 $\to \times 10^3=-1.7613\times 10^4$ 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: The residual single ply is not centred (it spans $-0.5$ to $0$), so $B_{11}\ne 0$ — a good trap to notice. Its magnitude $\approx -1.76\times 10^4$ N is the same order as Ex 2's $-1.64\times 10^4$ N (both are single-sided 0.5 mm lever arms), which is the right scale. Contrast Ex 4 where the ply was centred and $B_{11}=0$. Position, not just presence, sets coupling. Cell C5 confirmed — limits are continuous.

---

Example 6 — Invert ABD, symmetric shortcut (Cell C6)

Forecast: With $[B]=0$ the top and bottom blocks split. Guess: ${\kappa }_x=0$ exactly (no moment applied), and ${ϵ}_x^0$ is tiny (composites are stiff).

Step 1 — write the partitioned ABD explicitly. Why: novices need to see the block structure before trusting the shortcut. The full $6\times 6$ relation from the parent note is $\left[\begin{array}{c}\mathbf{N}\\ \mathbf{M}\end{array}\right]=\left[\begin{array}{cc}[A]& [B]\\ [B]& [D]\end{array}\right]\left[\begin{array}{c}{\mathbf{ϵ}}^0\\ \mathbf{\kappa }\end{array}\right],$ where $[A],[B],[D]$ are each $3\times 3$ blocks, $\mathbf{N},\mathbf{M},{\mathbf{ϵ}}^0,\mathbf{\kappa }$ each $3\times 1$.

Step 2 — set $[B]=0$ and read off the split. Why: if the off-diagonal block is zero, the two block-rows decouple: $\mathbf{N}=[A]{\mathbf{ϵ}}^0\Rightarrow {\mathbf{ϵ}}^0=[A{]}^{-1}\mathbf{N},\mathbf{M}=[D]\mathbf{\kappa }\Rightarrow \mathbf{\kappa }=[D{]}^{-1}\mathbf{M}.$ The strains depend only on forces, the curvatures only on moments — no cross-talk.

Step 3 — moments. $\mathbf{M}=0\Rightarrow \mathbf{\kappa }=[D{]}^{-1}0=0$. So ${\kappa }_x=0$.

Step 4 — strain (uniaxial estimate). Treating the $x$-direction as the dominant term, ${ϵ}_x^0\approx N_x/A_{11}=1000/1.5097\times 10^8=6.624\times 10^{-6}$ (dimensionless strain), i.e. $6.6$ microstrain.

Verify: Units: (N/m)/(N/m) = dimensionless ✓. Magnitude sanity: 1000 N/m over a 2 mm stiff plate should give micro-strain — a stiffer laminate ⇒ smaller strain, and $6.6\mu \epsilon$ is exactly that regime ✓. Because $[B]=0$, no curvature leaks in ✓. Cell C6 confirmed (full-matrix cross-coupling with $A_{12}$ handled in FE tools — Finite Element Modeling of Composites).

---

Example 7 — Full coupled inversion (Cell C7: B ≠ 0)

Forecast: Because $B_{11}\ne 0$, pure tension must produce curvature. Guess: ${\kappa }_x\ne 0$ this time.

Step 1 — the $2\times 2$. Why: isolate the coupled pair. $\left[\begin{array}{c}{N}_x\\ M_x\end{array}\right]=\left[\begin{array}{cc}{A}_{11}& B_{11}\\ B_{11}& D_{11}\end{array}\right]\left[\begin{array}{c}{ϵ}_x^0\\ {\kappa }_x\end{array}\right]=\left[\begin{array}{cc}7.5485\times 10^7& -1.6355\times 10^4\\ -1.6355\times 10^4& 6.290\end{array}\right]\left[\begin{array}{c}{ϵ}_x^0\\ {\kappa }_x\end{array}\right].$

Step 2 — determinant. Why: Cramer's rule needs it. $\Delta =A_{11}{D}_{11}-B_{11}^2=(7.5485\times 10^7)(6.290)-(-1.6355\times 10^4{)}^2=4.7480\times 10^8-2.675\times 10^8=2.0730\times 10^8.$ (Here the $B^2$ term is not negligible — coupling is strong for a two-ply laminate.)

Step 3 — solve with $M_x=0$. ${ϵ}_x^0=\frac{D_{11}{N}_x-B_{11}\cdot 0}{\Delta }=\frac{6.290\cdot 1000}{2.0730\times 10^8}=3.034\times 10^{-5},$ ${\kappa }_x=\frac{A_{11}\cdot 0-B_{11}{N}_x}{\Delta }=\frac{-(-1.6355\times 10^4)(1000)}{2.0730\times 10^8}=7.890\times 10^{-2}{\text{m}}^{-1}.$

Verify: ${\kappa }_x\ne 0$ — pure tension bent the plate, exactly the Ex 2 physics. Sign: ${\kappa }_x>0$, the plate bows toward positive $z$ (the weak 90° side), relieving the stiff side — consistent with the forecast. Units: $[\text{N}]\cdot [\text{N/m}]/[{\text{N}}^2]=1/\text{m}$ ✓ (curvature). Cell C7 confirmed — inversion must be full, never block-split, when $[B]\ne 0$.

---

Example 8 — Real-world word problem (Cell C8)

Forecast: Ex 6 gave $6.6\mu \epsilon$ at $1000$ N/m. Double the load — guess the strain and compare to the $20\mu \epsilon$ limit.

Step 1 — strain. Why: symmetric ⇒ decoupled (Ex 6 Step 2), ${ϵ}_x^0=N_x/A_{11}$. ${ϵ}_x^0=\frac{2000}{1.5097\times 10^8}=1.3248\times 10^{-5}=13.25\mu \epsilon .$

Step 2 — compare. $13.25<20$. Pass, with $34\%$ margin.

Step 3 — check no curvature. Why: a bonded sensor also fails from bending. Symmetric ⇒ $[B]=0$ ⇒ ${\kappa }_x=0$. Safe on both counts. (A sandwich construction would raise $[D]$ further — see Sandwich Panel Theory; first-ply margins in First Ply Failure.)

Verify: Linearity check — doubling $1000\to 2000$ N/m doubled $6.624\mu \epsilon \to 13.25\mu \epsilon$ ✓. Margin $=(20-13.25)/20=33.8\%$ ✓. Cell C8 confirmed.

---

Example 9 — Exam twist: balanced but antisymmetric (Cell C9)

Forecast: "Balanced" means equal $+\theta$ and $-\theta$ plies. Guess which of $A_{16}$, $B_{16}$ survives.

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 — $A_{16}$. Why: $[A]$ ignores position, so opposite signs cancel. Raw: $A_{16}=(+32.4)(0.5)+(-32.4)(0.5)=0\text{GPa⋅mm}\Rightarrow A_{16}=0.$ This is the balanced property: no extension–shear coupling.

Step 3 — $B_{16}$. Why: $[B]$ weights by $\frac{1}{2}(h_k^2-h_{k-1}^2)$: bottom ply $\frac{1}{2}(0-0.25)=-0.125$, top ply $\frac{1}{2}(0.25-0)=+0.125$ (mm²). Raw: $B_{16}=(+32.4)(-0.125)+(-32.4)(+0.125)=-4.05-4.05=-8.10{\text{GPa⋅mm}}^2.$ Apply the single $[B]$ factor $10^3$: $B_{16}=-8.10\times 10^3=-8.10\times 10^3$ N $=-8100$ N $\ne 0$. Both terms had the same sign — the sign flip of $Q_{16}$ times the sign flip of the lever arm gives a reinforcing (non-cancelling) result.

Verify: The twist resolved: balanced ($A_{16}=0$) does not imply no coupling — antisymmetry leaves $B_{16}\ne 0$, so twisting a $\pm 45°$ antisymmetric plate under load is real. Numeric (raw): $32.4\cdot 0.125\cdot 2=8.10$ GPa·mm² ⇒ $-8100$ N ✓. A symmetric $[\pm 45{]}_s$ would flip the top-pair lever arms and give $B_{16}=0$ too. Cell C9 confirmed — this is the classic exam gotcha. See Symmetric and Balanced Laminates.

---

Recall Which cell forces a full inversion, and why?

Cell C7 (asymmetric, $[B]\ne 0$). ::: Because $[B]$ links the force and moment blocks, so strains and curvatures cannot be found independently.

Recall Balanced vs symmetric — what does each kill?

Balanced kills $A_{16},A_{26}$ (extension–shear coupling). Symmetric kills all of $[B]$. ::: They are independent — Ex 9 is balanced yet has $B_{16}\ne 0$.