Composite materials — fiber-matrix, ply properties, laminate theory: Visual walkthrough

3.6.15 · D2 · Physics › Spacecraft Structures & Systems Engineering › Composite materials — fiber-matrix, ply properties, laminate

Shuru karne se pehle, teen plain-word anchors jo hum baar baar use karenge:

Step 1 — Do blocks, glued, ek ki tarah dekhe

KYA. Hum poore composite ko ek chhote se representative cube mein shrink karte hain jiska side $1$ hai: fiber ka ek block, matrix ka ek block, ek doosre se weld kiya hua. Yeh sabse chhota chunk hai jo phir bhi material jaisa "dikhta" hai.

KYUN. Agar hum ek cube samajh lete hain, toh hum slab samajh lete hain — slab sirf laakhon copies hai. Side $1$ choose karna ek trick hai: iska matlab "volume fraction" aur "thickness fraction" same number ban jaate hain, toh $V_{f}$ literally fiber block ki width hai.

TASVEER. Figure mein fiber block (red) matrix block (black) ke paas baitha hai. Uski width $V_{f}$ hai; matrix ki width $V_{m}$ hai; saath milke woh unit cube bhar dete hain.

Step 2 — Fibers ke SAATH kheecho: parallel springs

KYA. Cube ko left-to-right kheecho, fiber axis ke saath (ise direction 1 kaho). Dono blocks same do end plates se gripped hain, toh unhe same amount stretch hone par majboor kiya jaata hai.

KYUN. Perfect glue (hum perfect adhesion assume karte hain) ka matlab fiber aur matrix ek doosre se slide nahi kar sakte. Same original length, same stretch ⇒ same strain. Yeh exactly wahi hai jo same do walls ke beech side by side lagaye hue do springs karte hain — wall kheecho, dono equally stretch hote hain.

$fiber strain ϵ_{f} = matrix strain ϵ_{m} = common strain ϵ$

TASVEER. Red fiber block aur black matrix block same moving plate ke beech pinned do parallel springs hain. Plate ek distance move karti hai; dono springs wahi same stretch feel karte hain.

Har block apni stiffness ki wajah se apna stress deta hai:

$σ_{f} = E_{f} ϵ σ_{m} = E_{m} ϵ$

$E_{f}$ = fiber modulus (bahut bada, ~ $230$ GPa) → fiber zyada push back karta hai.
$E_{m}$ = matrix modulus (chhota, ~ $3$ GPa) → matrix mushkil se resist karta hai.

Step 3 — Forces add karo, stresses nahi: $E_{1}$ ka janam

KYA. End plate ko kheechne wala total force fiber ki force plus matrix ki force hai. Force = stress × area, aur unit cube ke saath areas sirf $V_{f}$ aur $V_{m}$ hain.

KYUN. Parallel springs apni forces add karke load share karte hain. Hum stresses ko forces mein convert karte hain (har block jo area present karta hai usse multiply karo), unhe add karo, phir poore face pe "average stress" mein wapas convert karo.

$σ_{comp} = fiber ka share σ_{f} V_{f} + matrix ka share σ_{m} V_{m}$

$σ_{f} = E_{f} ϵ$ aur $σ_{m} = E_{m} ϵ$ substitute karo, phir dono sides ko common $ϵ$ se divide karo (woh division hi definition hai $E_{1} = σ_{comp} / ϵ$ ):

$E_{1} = E_{f} V_{f} + E_{m} V_{m}$

$E_{f} V_{f}$ = stiff fiber, kitna cube fill karta hai usse weighted.
$E_{m} V_{m}$ = soft matrix, apne share se weighted.
Result ek plain weighted average hai — Rule of Mixtures. Kyunki $E_{f}$ , $E_{m}$ se bahut bada hai, fibers dominate karte hain: woh almost saara kaam karte hain.

TASVEER. Bar chart dono contributions ko stacked dikhata hai; tall red slab ( $E_{f} V_{f}$ ) thin black slab ( $E_{m} V_{m}$ ) se kaafi oopar hai.

Step 4 — Ab fibers ke ACROSS kheecho: series springs

KYA. Pull ko $90°$ rotate karo — top-to-bottom kheecho, fibers ke across (direction 2). Ab load ko fiber block se guzarna hoga aur phir matrix block se, ek ke baad ek.

KYUN. Load path mein stacked blocks series springs hain. Rule ulta ho jaata hai:

Same stress dono se guzarta hai (same force har layer ki face se squeeze hoti hai) — jaise ek single force do springs ki line mein push ki jaaye.
Strains add up hote hain (total stretch = fiber ki stretch + matrix ki stretch).

$σ_{f} = σ_{m} = σ ϵ_{total} = ϵ_{f} V_{f} + ϵ_{m} V_{m}$

Yahan $ϵ_{f}$ aur $ϵ_{m}$ ko $V_{f}, V_{m}$ se weight kiya jaata hai kyunki fiber layer sirf $V_{f}$ tall hai, toh total stretch mein uska contribution uski strain hai jo uski height se scale hoti hai.

TASVEER. Ab do blocks end-to-end springs hain; ek force line seedhi dono se guzarti hai. Soft matrix (black) zyaadatar stretching karta hai kyunki woh weak link hai.

Step 5 — Inverse rule: $E_{2}$ ka janam

KYA. Har block ka strain uske apne stress se likho ( $ϵ_{f} = σ / E_{f}$ , $ϵ_{m} = σ / E_{m}$ ), sum mein plug karo, aur $σ$ se divide karo.

KYUN. Hum $E_{2} = σ / ϵ_{total}$ chahte hain. Kyunki strains add hote hain aur stress shared hai, hume reciprocals milte hain jo add hote hain — series elements ka yahi khaas sign hai.

$ϵ_{total} = σ (\frac{V _{f}}{E _{f}} + \frac{V _{m}}{E _{m}}) ⟹ \frac{1}{E _{2}} = \frac{V _{f}}{E _{f}} + \frac{V _{m}}{E _{m}}$

$\frac{V _{f}}{E _{f}}$ = fiber ki floppiness (compliance) ka share — chhota, kyunki $E_{f}$ bahut bada hai.
$\frac{V _{m}}{E _{m}}$ = matrix ki floppiness ka share — bada, kyunki $E_{m}$ chhota hai. Yeh term sum ko dominate karta hai.

Sabse weak link (matrix) series chain ko control karta hai. Toh $E_{2}$ matrix-jaisi values pe aa jaata hai.

TASVEER. Ek "floppiness" bar chart — ab black matrix bar tall wali hai. Material ki crosswise stiffness soft phase ki qaidi hai.

Step 6 — Edge cases: extremes par kya hota hai?

KYA. Do formulas ko un inputs pe test karo jahan answer obvious hai, yeh prove karne ke liye ki woh trustworthy hain.

KYUN. Jo formula extremes par nahi toota, woh formula beech mein trust kiya ja sakta hai. Hum $V_{f}$ ko $0$ (poora matrix) se $1$ (poora fiber) tak sweep karte hain.

$V_{f} = 0$ (pure matrix): $E_{1} = E_{m}$ aur $E_{2} = E_{m}$ . Dono matrix dete hain — sahi, koi fiber nahi.
$V_{f} = 1$ (pure fiber): $E_{1} = E_{f}$ aur $E_{2} = E_{f}$ . Dono fiber dete hain — sahi.
Beech mein: $E_{1}$ ek straight line pe $E_{m}$ se $E_{f}$ tak jaata hai (weighted average). $E_{2}$ bottom ko hug karta hai, matrix value se sirf $V_{f} \to 1$ ke paas uthta hai — kyunki jab tak fibers load ke across ek unbroken column nahi banate, soft matrix tab bhi har load path ko gate karta hai.

$E_{1} (0) = E_{2} (0) = E_{m}, E_{1} (1) = E_{2} (1) = E_{f}$

TASVEER. Dono curves $V_{f}$ ke against plot kiye gaye hain. Woh dono ends par milte hain (jaisa hona chahiye) aur beech mein bahut zyada spread ho jaate hain. Red $E_{1}$ line seedhi aur oopar hai; black $E_{2}$ curve floor ke saath rengti hai.

Ek-tasveer summary

Poora derivation ek fork in the road hai: aap kis direction mein kheench rahe ho? Along kheecho → parallel springs → forces add → arithmetic average $E_{1}$ (fiber jeet jaata hai). Across kheecho → series springs → stretches add → reciprocal average $E_{2}$ (matrix haar jaata hai).

Recall Feynman retelling — ek story ki tarah bolo

Socho ek stiff steel wire ko rubber block mein glue kar rahe ho. Ise wire ke saath end-to-end kheecho: wire aur rubber same do haathon ke beech clamp hain, toh dono same thodi si amount stretch hote hain — lekin wire bahut zyada fight back karta hai aur rubber shrug karta hai. Unke pushbacks add up hote hain, har ek ko kitna room leta hai usse count kiya jaata hai. Woh sum, per unit stretch, $E_{1}$ hai, aur yeh basically wire hai — ek plain average jise stiff phase jeet jaata hai. Ab sideways, wire ke across kheecho: force ko wire layer se aur phir rubber layer se guzarna padta hai, ek ke peechhe ek. Same force dono se squeeze hoti hai, lekin rubber bahut zyada stretch karta hai aur wire almost nahi — total give floppy rubber se dominate hota hai. Floppinesses add karna, stiffnesses ki jagah, $E_{2}$ ke liye reciprocal formula deta hai, aur soft rubber ise neeche kheench leta hai. Ends check karo: koi wire nahi → sab rubber hai; sab wire → sab wire hai; dono formulas wahaan agree karte hain. Toh ek material, do bilkul alag stiffnesses, sab kuch sirf yeh poochne se ki main kis direction mein kheench raha hoon?

Cloze checkpoints:

Fibers ke along, fiber aur matrix ek hi same share karte hain

strain (woh equally stretch hote hain — parallel springs).

Fibers ke across, fiber aur matrix ek hi same share karte hain

stress (load dono se guzarta hai — series springs).

Longitudinal modulus ek weighted average hai

stiffnesses

E

ka (Rule of Mixtures).

Transverse modulus ek weighted average hai

compliances

1/ E

ka (inverse Rule of Mixtures).

Jab

V_{f} \to 1

, toh

E_{1}

aur

E_{2}

dono

E_{f}

approach karte hain (sab fiber, koi matrix nahi bachi).

Yeh aage kahan jaata hai: in plies ko angles par stack karna poora laminate banata hai; fiber vs metal choose karna material selection hai; ek soft-core version sandwich structures hai; yeh moduli finite element models aur launch load sizing ko feed karte hain, jabki low expansion thermal design ke liye matter karta hai aur har gram mass budget mein count karta hai.