Worked examples — Composite materials — fiber-matrix, ply properties, laminate theory

3.6.15 · D3 · Physics › Spacecraft Structures & Systems Engineering › Composite materials — fiber-matrix, ply properties, laminate

Yeh page ek drill deck hai. Parent note ne machinery build ki thi; yahan hum use har us tarah ke input ke against run karte hain jo yeh topic de sakta hai. Har worked example se pehle, tum ek forecast banate ho — ek guess. Pehle guess karna hi ideas ko dimaag mein chipaata hai.

Neeche ki sab cheez sirf wahi tools use karti hai jo parent ne earn ki hain: Rule of Mixtures ($E_1=E_f{V}_f+E_m{V}_m$), transverse aur shear ke liye uska inverse cousin, ply stiffness matrix $[Q]$, aur angle transformation $[\bar{Q}]$. Kuch naya smuggle nahi kiya gaya.

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The scenario matrix

Is topic ke har problem ko is table ke ek cell mein rehne wala samjho. Hamara kaam: inhe sab cover karna.

#	Case class	Kya tricky hai	Covered by
A	Longitudinal $E_1$ (parallel springs)	seedha, lekin baseline	Ex 1
B	Transverse $E_2$ / shear $G_{12}$ (series springs)	matrix dominate karta hai — counter-intuitive	Ex 2
C	Degenerate: $V_f=0$ aur $V_f=1$	limiting values — kya formula pure matrix / pure fiber pe reduce hota hai?	Ex 3
D	$[Q]$ banana $E_1,E_2,G_{12},{\nu }_{12}$ se	Poisson coupling, $1-{\nu }_{12}{\nu }_{21}$ denominator	Ex 4
E	Off-axis at $\theta =0^{\circ }$ aur $90^{\circ }$	check karo ki transform trivial angles pe jhooth na bole	Ex 5
F	Off-axis at $\theta =45^{\circ }$	extension–shear coupling aa jaati hai — yahi toh saara point hai	Ex 6
G	Real-world word problem: mass budget	composite vs aluminium ko specific stiffness se choose karna	Ex 7
H	Exam twist: ulta kaam karo	diya $E_1$, required $V_f$ nikalo	Ex 8

Hamare saath do golden rules chalta hain:

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Ex 1 — Case A: Longitudinal modulus (parallel springs)

Forecast: fibers matrix se ~77 guna stiffer hain aur 60% volume occupy karti hain. Guess karo: $E_1$ $230$ ke kareeb aayega ya $3$ ke? (Fiber ke kareeb — stiff phase jeet jaata hai jab strains share hoti hain.)

1. Matrix fraction nikalo. $V_m=1-V_f=1-0.60=0.40$. Yeh step kyun? Volume fractions ka sum 1 hona chahiye — cube ka har hissa ya fiber hai ya matrix, kuch aur nahi.
2. Stiffnesses ko volume se weight karke add karo (parallel springs — figure ke left mein dono arrows dekho, dono same amount stretch ho rahe hain): $E_1=E_f{V}_f+E_m{V}_m=230(0.60)+3(0.40)=138+1.2=139.2\text{ GPa}.$ Yeh step kyun? Bonded fiber aur matrix equally stretch hote hain, toh har ek stress contribute karta hai $\propto$ apni khud ki stiffness ke. Total stress ÷ shared strain = moduli ka weighted sum.

Verify: $E_1$ ko $E_m$ aur $E_f$ ke beech rehna chahiye ($3<139.2<230$ ✓) aur fiber ki taraf lean karna chahiye kyunki wahi sum dominate karta hai. $139.2/230=0.605$, matlab pure fiber ka 60.5% — sensible, $V_f$ se match karta hai. Units: throughout GPa ✓.

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Ex 2 — Case B: Transverse & shear (series springs)

Forecast: ab load soft matrix ko series mein cross karta hai. Guess karo: $E_2$ $139$ GPa ke kareeb aata hai ya ek chhote number pe? (Chhota — series mein sabse kamzor link control karta hai.)

1. Softnesses (reciprocals) add karo, kyunki series mein dono stress share karte hain aur unki strains add hoti hain: $\frac{1}{E_2}=\frac{V_f}{E_f}+\frac{V_m}{E_m}=\frac{0.60}{230}+\frac{0.40}{3}=0.002609+0.13333=\mathrm{0.13594.}$ Yeh step kyun? Series mein (figure ka right side), wahi stress pehle fiber phir matrix se guzarti hai; total stretch har layer ke stretch ka sum hai → compliances add hote hain.
2. Invert karo modulus recover karne ke liye: $E_2=\frac{1}{0.13594}=7.36\text{ GPa}.$
3. Shear mein bhi same series rule laagoo hota hai (shear bhi soft matrix se flow karta hai): $\frac{1}{G_{12}}=\frac{0.60}{9}+\frac{0.40}{1.1}=0.06667+0.36364=0.43030\Rightarrow G_{12}=2.32\text{ GPa}.$

Verify: $E_2=7.36$ GPa barely matrix modulus $3$ GPa se double hai — fibers transversely almost kuch help nahi karte. Anisotropy ratio $E_1/E_2=139.2/7.36\approx 18.9$, parent ke "$\approx 19$" se match karta hai. Units consistent ✓. Yahi wajah hai ki parent warn karta hai ki composites ko kabhi isotropic treat mat karo.

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Ex 3 — Case C: Degenerate limits $V_f=0$ aur $V_f=1$

Forecast: ek formula jis par trust kar sako, $V_f=0$ pe pure matrix properties dena chahiye aur $V_f=1$ pe pure fiber. Padhne se pehle charon numbers guess karo.

1. $V_f=0$ (sab matrix), longitudinal: $E_1=230(0)+3(1)=3$ GPa. Yeh step kyun? Koi fiber nahi bacha — composite hai hi matrix, toh $E_1$ ko $E_m$ ke barabar hona chahiye.
2. $V_f=0$, transverse: $1/E_2=0/230+1/3=1/3\Rightarrow E_2=3$ GPa. Yeh step kyun? Same reasoning — dono formulas $E_m$ pe collapse ho jaate hain. Achha hai, woh is endpoint pe agree karte hain.
3. $V_f=1$ (sab fiber), longitudinal: $E_1=230(1)+3(0)=230$ GPa.
4. $V_f=1$, transverse: $1/E_2=1/230+0/3=1/230\Rightarrow E_2=230$ GPa. Yeh step kyun? Pure fiber ka in idealised formulas mein koi direction preference nahi hota, toh $E_1=E_2=E_f$.

Verify: dono extremes pe $E_1=E_2$, exactly jaisa ek single material demand karta hai (anisotropy sirf mixing se aati hai). Endpoints $E_m$ aur $E_f$ hain — $0\le V_f\le 1$ mein kahin bhi koi blow-up nahi, koi negative value nahi, koi division by zero nahi ✓.

Recall Practice mein

$V_f$ 1 tak kyun nahi pahunch sakta? Real fibers round hote hain; packed cylinders matrix ke liye gaps chhodte hain, toh $V_f$ near $\approx 0.65$–$0.75$ pe top out karta hai. Formula $V_f=1$ pe behave karta hai, lekin manufacturing ise forbid karti hai. ::: Round packing + har fiber ko wet karne ki zaroorat $V_f$ ko around 0.65–0.75 cap karti hai.

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Ex 4 — Case D: Ply stiffness matrix $[Q]$ banana

Forecast: denominator $1-{\nu }_{12}{\nu }_{21}$ sirf slightly less than 1 hai, toh guess karo: $Q_{11}$ $140$ GPa ke kareeb hoga ya bahut alag? (Kareeb — Poisson correction chhoti hai.)

1. Minor Poisson ratio symmetry se. ${\nu }_{21}={\nu }_{12}{E}_2/E_1=0.30\times 10/140=0.02143$. Yeh step kyun? Stiffness matrices symmetric honi chahiye (energy load order pe depend nahi kar sakti); woh ${\nu }_{21}{E}_1={\nu }_{12}{E}_2$ force karta hai.
2. Shared denominator. $D=1-{\nu }_{12}{\nu }_{21}=1-0.30(0.02143)=1-0.006429=0.99357$. Yeh step kyun? Compliance matrix invert karne pe yeh determinant term produce hoti hai; yeh Poisson coupling ko "un-mix" karti hai taaki stress↔strain consistent rahe.
3. Entries fill karo. $Q_{11}=\frac{E_1}{D}=\frac{140}{0.99357}=140.91\text{ GPa},Q_{22}=\frac{E_2}{D}=\frac{10}{0.99357}=10.065\text{ GPa},$ $Q_{12}=\frac{{\nu }_{12}{E}_2}{D}=\frac{0.30(10)}{0.99357}=3.019\text{ GPa},Q_{66}=G_{12}=5\text{ GPa}.$

Verify: $Q_{11}\gtrsim E_1$ (140.91 vs 140) — Poisson coupling slightly stiffen karta hai, jaisa predict kiya. Symmetry check: $Q_{12}$ ya toh ${\nu }_{12}{E}_2/D$ se compute ho ya ${\nu }_{21}{E}_1/D=0.02143(140)/0.99357=3.019$ GPa se — identical ✓. Shear akela baithta hai $Q_{66}=G_{12}$ pe (material axes mein normal stress se koi coupling nahi) ✓.

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Ex 5 — Case E: Trivial angles $\theta =0^{\circ },90^{\circ }$ pe transform karo

Forecast: $0^{\circ }$ pe fibers pehle se hi $x$ ke saath aligned hain, toh kuch nahi badalna chahiye. $90^{\circ }$ pe ply quarter turn ho gayi — stiff direction ab $y$ ke along point karti hai, toh $x$-stiffness transverse value padni chahiye. Guess karo: ${\bar{Q}}_{11}(0^{\circ })=?$ aur ${\bar{Q}}_{11}(90^{\circ })=?$

Hum standard result use karte hain (jo $[\bar{Q}]=[T{]}^{-1}[Q][T{]}^{-T}$ ke equivalent hai) jahan $c=\cos \theta$, $s=\sin \theta$: ${\bar{Q}}_{11}=Q_{11}{c}^4+2(Q_{12}+2Q_{66})s^2{c}^2+Q_{22}{s}^4.$

1. $\theta =0^{\circ }$: $c=1,s=0$. Sirf pehla term bachta hai: ${\bar{Q}}_{11}=Q_{11}(1)=140.91$ GPa. Yeh step kyun? Koi rotation nahi → material axes = laminate axes → matrix unchanged. Agar badal jaata, toh haara transform broken hota.
2. $\theta =90^{\circ }$: $c=0,s=1$. Sirf aakhri term bachta hai: ${\bar{Q}}_{11}=Q_{22}(1)=10.065$ GPa. Yeh step kyun? Quarter turn directions 1 aur 2 ke roles swap karta hai, toh $x$-stiffness transverse stiffness $Q_{22}$ ban jaati hai — ply ab $x$ ke along soft hai.

Verify: ${\bar{Q}}_{11}(0^{\circ })=Q_{11}$ aur ${\bar{Q}}_{11}(90^{\circ })=Q_{22}$ — exactly $[Q]$ ke do endpoints. Transform dono trivial tests pass karta hai, toh hum ise messy angles pe trust kar sakte hain ✓.

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Ex 6 — Case F: $\theta =45^{\circ }$ pe transform — coupling appear hoti hai

Forecast: $45^{\circ }$ pe $x$ ke along pull karna dono fiber aur transverse directions pe tug karta hai aur ply ko twist karta hai. Toh predict karo ${\bar{Q}}_{11}$ $Q_{11}$ aur $Q_{22}$ ke beech land kare, aur ${\bar{Q}}_{16}$ nonzero ho jaaye (material axes mein woh zero tha).

$\theta =45^{\circ }$ pe: $c=s=\frac{1}{\sqrt{2}}$, toh $c^2=s^2=\frac{1}{2}$, $c^4=s^4=\frac{1}{4}$, $s^2{c}^2=\frac{1}{4}$, aur $c^2-s^2=0$.

1. ${\bar{Q}}_{11}$ compute karo: ${\bar{Q}}_{11}=Q_{11}\frac{1}{4}+2(Q_{12}+2Q_{66})\frac{1}{4}+Q_{22}\frac{1}{4}.$ $=\frac{1}{4}[140.91+2(3.019+2\cdot 5)+10.065]=\frac{1}{4}[140.91+26.038+10.065]=\frac{1}{4}(177.01)=44.25\text{ GPa}.$ Yeh step kyun? $Q_{11},Q_{22}$ aur shear-laden middle term ko equally weight karna (har ek $\times \frac{1}{4}$) stiff, soft, aur shear contributions ko mix karta hai — ply $x$ ke along na poori tarah stiff hai na poori tarah soft.
2. Coupling term compute karo ${\bar{Q}}_{16}=(Q_{11}-Q_{12}-2Q_{66})c^{3}s+(Q_{12}-Q_{22}+2Q_{66})c{s}^3$. $45^{\circ }$ pe, $c^{3}s=c{s}^3=\frac{1}{4}$: ${\bar{Q}}_{16}=\frac{1}{4}[(Q_{11}-Q_{12}-2Q_{66})+(Q_{12}-Q_{22}+2Q_{66})]=\frac{1}{4}(Q_{11}-Q_{22})=\frac{1}{4}(140.91-10.065)=32.71\text{ GPa}.$ Yeh step kyun? $\pm 2Q_{66}$ aur $\pm Q_{12}$ pieces cancel ho jaate hain, $\frac{1}{4}(Q_{11}-Q_{22})$ bacha rehta hai — coupling purely isi baat se driven hai ki dono direct stiffnesses kitni unequal hain.

Verify: ${\bar{Q}}_{11}=44.25$ GPa, $Q_{22}=10.06$ aur $Q_{11}=140.9$ ke beech hai ✓. ${\bar{Q}}_{16}=32.71\ne 0$ — $x$ ke along pull karna ab shear produce karta hai (extension–shear coupling jo parent ne flag kiya tha). Yahi wajah hai ki single off-axis plies avoid ki jaati hain aur balanced $\pm 45^{\circ }$ pairs use kiye jaate hain (unke ${\bar{Q}}_{16}$ terms cancel ho jaate hain). Figure mein tilted fibers dekho — pull ab kisi bhi weave direction ke saath align nahi karti.

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Ex 7 — Case G: Real-world word problem (mass budget)

Forecast: composite stiffer bhi hai aur lighter bhi — guess karo specific-stiffness advantage "kaafi × hai", 100× nahi.

1. Har ek ki Specific stiffness (stiffness per unit density — fair figure of merit jab mass budgets rule karte hain): ${\left(\frac{E}{\rho }\right)}_{\text{comp}}=\frac{139.2}{1600}\times 10^6=87.0\times 10^3,{\left(\frac{E}{\rho }\right)}_{\text{Al}}=\frac{70}{2700}\times 10^6=25.9\times 10^3{\text{(SI: Pa⋅m}}^3\text{/kg).}$ Yeh step kyun? Strength-limited ya stiffness-limited parts ke liye, spec meet karne wale mass $\rho /E$ ke saath scale karta hai — toh $E/\rho$ maximize karna mass minimize karta hai.
2. Ratio: $87.0/25.9=3.36$. Yeh step kyun? Divide karne se head-to-head advantage milta hai, panel geometry cancel ho jaati hai jo dono designs share karte hain.

Verify: parent ke "3.3× better" specific stiffness se match karta hai ✓. Interpretation: same stiffness hold karne wala ek composite bench roughly $1/3.36\approx 30\%$ weight karta hai aluminium wale ka — exactly yahi wajah hai ki launch-load-driven (quasi-static loads) structures composite jaate hain. Units pure ratio mein cancel ho jaate hain ✓.

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Ex 8 — Case H: Exam twist — $V_f$ ke liye ulta kaam karo

Forecast: $150$, $V_f=0.6$ pe mila $139.2$ se comfortably upar hai, toh guess karo $V_f$ badhna chahiye — lekin kya yeh $\approx 0.70$ ceiling ke neeche hai?

1. Rule of Mixtures ko ek unknown ke saath likhho. $V_m=1-V_f$ ke saath: $E_1=E_f{V}_f+E_m(1-V_f)=E_m+(E_f-E_m)V_f.$ Yeh step kyun? Single $V_f$ mein collapse karna design goal ko ek linear equation bana deta hai jise hum invert kar sakte hain.
2. $V_f$ ke liye solve karo: $V_f=\frac{E_1-E_m}{E_f-E_m}=\frac{150-3}{230-3}=\frac{147}{227}=\mathrm{0.6476.}$ Yeh step kyun? Seedha algebra — wahi "endpoints ke beech" line, doosri taraf se padhi.

Verify: wapas plug karo: $E_1=3+227(0.6476)=3+147=150$ GPa ✓. Aur $V_f=0.648$, Ex 3 ke practical $\approx 0.65$–$0.70$ ceiling ke neeche hai, toh yeh manufacturable hai — barely. Agar target $E_1=200$ GPa hota, toh $V_f=(200-3)/227=0.868>0.70$ chahiye hota: impossible is fiber se, jo tumhe ek stiffer fiber ya sandwich construction ki taraf push karta.

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Recall Kaun si rule kaun si property ke liye?

$E_1$ direct Rule of Mixtures use karta hai (parallel). ::: $E_1=E_f{V}_f+E_m{V}_m$. $E_2$, $G_{12}$ inverse (series) rule use karte hain. ::: $1/E_2=V_f/E_f+V_m/E_m$. $V_f\to 0$ pe dono moduli approach karte hain ::: $E_m$ (pure matrix). $45^{\circ }$ pe off-axis coupling term ${\bar{Q}}_{16}$ equal hoti hai ::: $\frac{1}{4}(Q_{11}-Q_{22})$.

Yeh worked cases directly laminate stacking mein aur FE models mein feed hote hain, jahan har ply ka $[\bar{Q}]$ ek element property ban jaata hai; usi algebra ke thermal versions thermal-expansion design drive karte hain.