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

3.6.15 · D4 · Physics › Spacecraft Structures & Systems Engineering › Composite materials — fiber-matrix, ply properties, laminate

Yeh aapka self-testing sheet hai parent topic on composites ke liye (fiber–matrix, ply properties, laminate theory). Har problem ko solution kholne se pehle attempt karo. Problems seedhi "kya formula yaad hai?" se shuru hokar "kya aap ek layup design karke usse defend kar sakte ho?" tak jaati hain.

Poore note mein hum wohi symbols reuse karte hain jo parent note ne define kiye the. Ek quick reminder taaki koi bhi symbol use hone se pehle unfamiliar na lage:

Recall Symbol dictionary (koi bhi symbol unfamiliar lage to open karo)

$E_{f}$ ::: Young's modulus (stiffness) of the fiber alone — yeh stretching ko kitna resist karta hai, GPa mein. $E_{m}$ ::: Young's modulus of the matrix (glue/resin) alone. $V_{f}, V_{m}$ ::: Volume fractions: material ka kitna hissa fiber hai aur kitna matrix. Dono milake 1 hote hain. $E_{1}$ ::: Composite stiffness fibers ke saath measure ki gayi (strong direction). $E_{2}$ ::: Composite stiffness fibers ke across measure ki gayi (weak direction). $G_{12}$ ::: In-plane shear modulus — material ke "scissoring" ka resistance. $ν_{12}$ ::: Major Poisson's ratio — fibers ke along khicho, toh kitna sideways patla hota hai. $[Q]$ ::: Ply stiffness matrix material axes mein (1 = fiber, 2 = transverse). $[\overset{ˉ}{Q}]$ ::: Wohi ply ki stiffness laminate x–y axes mein rotate ki gayi jab fibers angle $θ$ par hoon. $θ$ ::: Fiber direction aur laminate x-axis ke beech ka angle.

Level 1 — Recognition

L1.1 — Rule ka naam batao

Problem. Tum ek unidirectional composite ko uske fibers ke saath khinchte ho. Kaun sa averaging rule $E_{1}$ deta hai, aur yeh direct (inverse nahi) sum kyun hai?

Recall Solution

Rule of Mixtures: $E_{1} = E_{f} V_{f} + E_{m} V_{m}$ . Fibers aur matrix ko do springs ki tarah socho jo side by side (parallel) khade hain — poore cube ko khiincho toh dono ek jaisi amount stretch hote hain (equal strain). Parallel springs apni stiffnesses directly add karte hain, isliye stiffness contributions directly weighted sum mein add hoti hain. Isliye yeh ek direct weighted sum hai.

L1.2 — Direction ko formula se match karo

Problem. Batao kaun sa formula (direct ya inverse Rule of Mixtures) kahan lagta hai: (a) $E_{1}$ , (b) $E_{2}$ , (c) $G_{12}$ .

Recall Solution

(a) $E_{1}$ → direct (parallel springs, equal strain). (b) $E_{2}$ → inverse: $\frac{1}{E _{2}} = \frac{V _{f}}{E _{f}} + \frac{V _{m}}{E _{m}}$ (series springs, equal stress). (c) $G_{12}$ → inverse: $\frac{1}{G _{12}} = \frac{V _{f}}{G _{f}} + \frac{V _{m}}{G _{m}}$ (yeh bhi series-jaisa hai: load soft matrix se hokar guzarna padta hai).

Figure — Composite materials — fiber-matrix, ply properties, laminate theory

L1.3 — Anisotropy padho

Problem. Ek ply mein $E_{1} = 139.2$ GPa aur $E_{2} = 7.4$ GPa hai. Anisotropy ratio $E_{1} / E_{2}$ compute karo aur ek sentence mein batao ki yeh physically kya matlab rakhta hai.

Recall Solution

$E_{1} / E_{2} = 139.2/7.4 \approx 18.8$ . Material fibers ke saath roughly 19× zyada stiff hai fibers ke across se — yeh strongly anisotropic hai, isliye orientation bahut zyada matter karta hai.

Level 2 — Application

L2.1 — Longitudinal modulus

Problem. Glass–epoxy: $E_{f} = 72$ GPa, $E_{m} = 3.5$ GPa, $V_{f} = 0.55$ . $E_{1}$ nikalo.

Recall Solution

$V_{m} = 1 - 0.55 = 0.45$ . $E_{1} = 72 (0.55) + 3.5 (0.45) = 39.6 + 1.575 = 41.175 GPa .$ Toh $E_{1} \approx 41.2$ GPa.

L2.2 — Transverse modulus

Problem. Wahi glass–epoxy ( $E_{f} = 72$ , $E_{m} = 3.5$ , $V_{f} = 0.55$ ). $E_{2}$ nikalo.

Recall Solution

$\frac{1}{E _{2}} = \frac{0.55}{72} + \frac{0.45}{3.5} = 0.007639 + 0.128571 = 0.136210.$ $E_{2} = \frac{1}{0.136210} = 7.342 GPa .$ Ghaur karo: 55% stiff fiber hone ke bawajood, transverse stiffness soft matrix ki taraf khich jaati hai, kyunki load series mein weak resin se hokar guzarna padta hai.

L2.3 — Minor Poisson's ratio

Problem. Ek ply mein $ν_{12} = 0.28$ , $E_{1} = 130$ GPa, $E_{2} = 9$ GPa hai. $ν_{21}$ nikalo.

Recall Solution

Compliance matrix ki symmetry demand karti hai $ν_{21} = ν_{12} E_{2} / E_{1}$ . $ν_{21} = 0.28 \times \frac{9}{130} = 0.28 \times 0.069231 = 0.019385 \approx 0.0194.$ Minor ratio bahut chhota hai — fibers ke across khiinchne par stiff fiber direction barely contract hoti hai.

Level 3 — Analysis

L3.1 — Ply stiffness matrix $[Q]$ banao

Problem. Ek ply ke liye $E_{1} = 140$ GPa, $E_{2} = 10$ GPa, $G_{12} = 5$ GPa, $ν_{12} = 0.30$ diya hai, $Q_{11}, Q_{22}, Q_{12}, Q_{66}$ compute karo.

Recall Solution

Step 1 — minor Poisson's ratio. $ν_{21} = 0.30 \times \frac{10}{140} = 0.021429.$ Step 2 — coupling denominator. (Yeh term har $Q$ mein aata hai; yeh is baat ka account karta hai ki ek direction mein stretch karne se doosri direction mein stress induce hoti hai, isliye stiffness thodi boost hoti hai.) $1 - ν_{12} ν_{21} = 1 - 0.30 (0.021429) = 1 - 0.006429 = 0.993571.$ Step 3 — entries. $Q_{11} = \frac{140}{0.993571} = 140.906 GPa, Q_{22} = \frac{10}{0.993571} = 10.065 GPa,$ $Q_{12} = \frac{0.30 \times 10}{0.993571} = 3.0194 GPa, Q_{66} = G_{12} = 5 GPa .$ $[Q] = 140.9 3.02 0 3.02 10.06 0 005 GPa .$ $Q_{11}$ barely $E_{1}$ se upar hai — Poisson boost sirf 0.6% hai. Lekin $Q_{12} \neq = 0$ batata hai ki direction 1 mein khinchne se direction 2 mein bhi stress generate hoti hai.

L3.2 — Degenerate case: $θ = 0°$ par kya hota hai?

Problem. Dikhao ki $θ = 0°$ par transformation matrix $[T]$ identity mein reduce ho jaati hai, aur explain karo ki $[\overset{ˉ}{Q}]$ ke liye iska kya matlab hai.

Recall Solution

$θ = 0$ par: $cos θ = 1$ , $sin θ = 0$ , isliye $cos^{2} θ = 1$ , $sin^{2} θ = 0$ , $sin θ cos θ = 0$ . $[T] = 100010001 = [I] .$ Phir $[\overset{ˉ}{Q}] = [T]^{- 1} [Q] [T]^{- T} = [I] [Q] [I] = [Q]$ . Matlab: jab fibers already laminate x-axis ke saath aligned hain, toh rotate karne ke liye kuch nahi hai — material axes aur laminate axes coincide karte hain, isliye $[\overset{ˉ}{Q}] = [Q]$ . Yeh sanity check hai jo har transformation ko pass karna chahiye.

L3.3 — Off-axis coupling at $θ = 45°$

Problem. Poora matrix compute kiye bina explain karo ki $[\overset{ˉ}{Q}]$ mein $θ = 45°$ par kaun sa naya coupling term aata hai jo $[Q]$ mein zero tha, aur yeh kaunsa physical behaviour cause karta hai. Phir $45°$ par $sin θ cos θ$ aur $cos^{2} θ - sin^{2} θ$ compute karo aur justify karo.

Recall Solution

$45°$ par: $sin θ cos θ = \frac{1}{2}$ (iska maximum) aur $cos^{2} θ - sin^{2} θ = 0$ . Terms $\overset{ˉ}{Q}_{16}$ aur $\overset{ˉ}{Q}_{26}$ — extension–shear coupling terms — non-zero ho jaate hain (yeh material-axis $[Q]$ mein exactly 0 the). Physical matlab: laminate ko x ke saath seedha khiincho, aur yeh shear karta hai (square parallelogram mein distort ho jaata hai). Isliye ek single off-axis ply simple tension ke neeche "twist" karti hai — ek aisa behaviour jo metals mein kabhi nahi dikhta.

Level 4 — Synthesis

L4.1 — Specific stiffness trade study

Problem. Specific stiffness $E / ρ$ (units GPa per g/cm³) compare karo:

Carbon–epoxy: $E_{1} = 139.2$ GPa, $ρ = 1.60$ g/cm³
Aluminium 7075: $E = 71.7$ GPa, $ρ = 2.81$ g/cm³ Kaun jeetega, aur kitne factor se? Phir ek direction batao jahan composite haarta hai.

Recall Solution

Composite: $139.2/1.60 = 87.0$ GPa·cm³/g. Aluminium: $71.7/2.81 = 25.5$ GPa·cm³/g. Ratio: $87.0/25.5 = 3.41$ → composite apne fibers ke saath ~3.4× zyada efficient hai. Jahan yeh haarta hai: fibers ke across, $E_{2} \approx 7.4$ GPa, specific stiffness $7.4/1.60 = 4.6$ deta hai — aluminium ke isotropic 25.5 se kaafi neeche. Isliye ek naive single-direction layup transverse direction mein aluminium se bhi bura hai. Yahi reason hai multi-directional laminates aur sandwich cores ka, naa ki raw unidirectional plies ka. Seedha mass budgeting mein feed hota hai.

L4.2 — Quasi-isotropic layup design karna

Problem. Tumhe L3.1 ki ply use karke ek aisa panel banana hai jo sab in-plane directions mein (approximately) equally stiff ho. Ek standard ply-angle set batao jo yeh achieve kare aur explain karo kyun woh angles.

Recall Solution

Ek quasi-isotropic stack: $[0°/ + 45° / - 45° /90°]_{s}$ ( $_{s}$ matlab symmetry ke liye mirror karo). Kyun yeh angles: in-plane directions ke 180° mein stiffness evenly average karne ke liye, tum equal ply angles evenly spaced rakho. Chaar angles 45° apart spaced ( $0, 45, 90, 135 \equiv - 45$ ) half-circle uniformly tile karte hain, isliye averaged in-plane modulus (nearly) direction-independent ho jaata hai. $\pm 45$ pair balanced hai, L3.3 ka extension–shear coupling cancel karta hai; mirror symmetry ( $_{s}$ ) bending–stretching coupling cancel karta hai. Finite-element analysis se practice mein verified.

Level 5 — Mastery

L5.1 — Fiber volume fraction back-solve karo

Problem. Ek carbon–epoxy ply measure ki gayi toh $E_{1} = 120$ GPa nikla. $E_{f} = 230$ GPa aur $E_{m} = 3$ GPa ke saath, manufacturing mein kaunsa fiber volume fraction $V_{f}$ achieve hua?

Recall Solution

Rule of Mixtures se shuru karo, $V_{m} = 1 - V_{f}$ ke saath: $E_{1} = E_{f} V_{f} + E_{m} (1 - V_{f}) = E_{m} + (E_{f} - E_{m}) V_{f} .$ $V_{f}$ ke liye solve karo: $V_{f} = \frac{E _{1} - E _{m}}{E _{f} - E _{m}} = \frac{120 - 3}{230 - 3} = \frac{117}{227} = 0.5154.$ Toh $V_{f} \approx 0.52$ — thoda low fraction, jo under-consolidation ya excess resin ka hint deta hai, material selection / QA ko flag karne layak.

L5.2 — $E_{2}$ ka full limiting-behaviour sweep

Problem. Transverse modulus $\frac{1}{E _{2}} = \frac{V _{f}}{E _{f}} + \frac{V _{m}}{E _{m}}$ ke liye, $E_{f} = 230$ , $E_{m} = 3$ GPa ke saath, do limiting cases $V_{f} \to 0$ aur $V_{f} \to 1$ evaluate karo, aur explain karo ki practical range ke zyaadatar hisse mein $E_{2}$ matrix se kyun dominated hoti hai.

Recall Solution

Case $V_{f} \to 0$ (sab matrix): $\frac{1}{E _{2}} = \frac{0}{230} + \frac{1}{3} \Rightarrow E_{2} = 3$ GPa. Pure matrix — sahi. Case $V_{f} \to 1$ (sab fiber): $\frac{1}{E _{2}} = \frac{1}{230} + 0 \Rightarrow E_{2} = 230$ GPa. Pure fiber — sahi. Beech mein matrix kyun dominate karta hai: typical $V_{f} = 0.6$ par, $\frac{1}{E _{2}} = \frac{0.6}{230} + \frac{0.4}{3} = 0.002609 + 0.133333 = 0.135942, E_{2} = 7.36 GPa .$ Fiber term ( $0.0026$ ) matrix term ( $0.133$ ) se ~50× chhota hai. Ek series path mein, sabse soft element compliance control karta hai (jaise spring chain mein sabse choda gap), isliye $E_{2}$ matrix value ke paas rehta hai jab tak $V_{f}$ nearly 1 na ho.

L5.3 — Strain energy se consistency check

Problem. Confirm karo ki $Q_{12}$ ke do expressions, yaani $\frac{ν _{12} E _{2}}{1 - ν _{12} ν _{21}}$ aur $\frac{ν _{21} E _{1}}{1 - ν _{12} ν _{21}}$ , equal hain — $E_{1} = 140$ , $E_{2} = 10$ GPa, $ν_{12} = 0.30$ use karke. Explain karo ki yeh equal kyun hone chahiye.

Recall Solution

$ν_{21} = 0.30 \times 10/140 = 0.0214286$ ; denominator $= 0.993571$ (L3.1 se). Pehla form: $\frac{0.30 \times 10}{0.993571} = \frac{3.0}{0.993571} = 3.01942$ GPa. Doosra form: $\frac{0.0214286 \times 140}{0.993571} = \frac{3.0}{0.993571} = 3.01942$ GPa. Match karte hain. Kyun hone chahiye: reciprocity relation $ν_{12} E_{2} = ν_{21} E_{1}$ (L2.3 se) dono numerators ko identical bana deti hai. Yeh coincidence nahi hai — yeh enforce karta hai ki ply ki stiffness matrix symmetric hai, jo zaroori hai taaki material strain energy consistently store kare (loading cycle se koi free energy na mile).

Self-test cloze recap

Direct Rule of Mixtures isliye lagta hai kyunki fibers ke saath, fiber aur matrix parallel springs ki tarah equal strain par kaam karte hain. Transverse modulus $E_{2}$ inverse Rule of Mixtures use karta hai kyunki phases series mein equal stress par kaam karte hain. $θ = 45°$ par naye non-zero coupling terms $\overset{ˉ}{Q}_{16}, \overset{ˉ}{Q}_{26}$ panel ko ek saath extend aur shear karna cause karte hain. Reciprocity $ν_{12} E_{2} = ν_{21} E_{1}$ guarantee karta hai ki stiffness matrix symmetric hai.