Classical laminate theory — ABD matrix: Question bank

3.6.16 · D5 · Physics › Spacecraft Structures & Systems Engineering › Classical laminate theory — ABD matrix

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Ek laminate patli plies ka stack hota hai. Hum ise ek vertical coordinate $z$ se slice karte hain, jo mid-plane (woh surface jahan $z = 0$ hai) se measure hota hai, positive direction "upar" ki taraf. Figure 1 dekhein: har ply $k$ do heights ke beech ek slab occupy karta hai.

Figure — Classical laminate theory — ABD matrix

Recall Reduced stiffness

$[Q]_{k}$ ek breath mein $[Q]_{k}$ ply $k$ ki $3 \times 3$ transformed reduced stiffness hai — woh law $σ_{k} = [Q]_{k} ϵ$ jo us ply ke andar kaam karta hai. Ye aapko ply ki own-axis stiffness (jo $E_{1}, E_{2}, G_{12}, ν_{12}$ se banti hai) leke aur use fiber angle $θ_{k}$ se rotate karke milti hai (dekho Reduced Stiffness Matrix Q aur Transformation of Stiffness). Iski form: $[Q]_{k} = Q_{11} Q_{12} Q_{16} Q_{12} Q_{22} Q_{26} Q_{16} Q_{26} Q_{66}_{k}$ . Same material lekin alag angle wale do plies ke alag $[Q]_{k}$ hote hain.

Figure 2 dikhata hai ki teen sub-matrices kaise aati hain $[Q]_{k}$ ko $z$ ki alag powers ke against integrate karke — ye usi geometric root se aata hai jo questions probe karte hain.

Figure 3 dikhata hai ki kyun mirror symmetry $[B] = 0$ force karti hai: mirror-paired plies ke signed $z$ -lever contributions cancel ho jaate hain.

True ya false — justify karo

[B] = 0

kisi bhi laminate mein guaranteed hota hai agar uske top aur bottom pe same plies hon.

False — aapko mid-plane ke baare mein mirror symmetry chahiye (same material, angle, AUR same distance

∣ z ∣

, jaisa Fig. 3 mein hai). [0/90/0/90] jaisa stack matching plies rakhta hai lekin mirror-placed nahi hain, isliye

[B] \neq = 0

Ek symmetric laminate phir bhi bend kar sakta hai jab aap ise pull karo.

False — symmetric laminate ke liye Eq. (4) mein

[B] = 0

hota hai, isliye pure in-plane force

N

sirf mid-plane strain

ϵ^{0}

produce karta hai aur curvature

κ

zero rehti hai. Bend-stretch coupling exactly wohi cheez hai jo

[B]

carry karta hai, aur woh khatam ho gayi.

[A]

us order pe depend karta hai jisme aap plies stack karte hain.

False — Eq. (1) har ply ko sirf uski thickness

t_{k}

se weight karta hai, position se nahi. Same plies ko reshuffle karne se

[A]

identical rehta hai; sirf

[B]

aur

[D]

order feel karte hain.

[D]

stacking order pe depend karta hai.

True — Eq. (3) har ply ko

\frac{1}{3} (h_{k}^{3} - h_{k - 1}^{3})

se weight karta hai, jo mid-plane se door hone ke saath fast badhta hai, isliye surface pe plies dominate karte hain. Ek stiff ply ko bahar move karne se

[D]

badhta hai.

Ek balanced laminate mein hamesha

[B] = 0

hota hai.

False — "balanced" (

+ θ

aur

- θ

plies ki equal numbers)

A_{16}

aur

A_{26}

ko khatam karta hai, yaani

[A]

mein shear-extension coupling. Yeh mid-plane symmetry ke baare mein kuch nahi kehta, isliye

[B]

phir bhi non-zero ho sakta hai.

[A B D]

matrix hamesha symmetric hoti hai.

True — kyunki

[B]

Eq. (4) ke force row aur moment row dono mein aata hai, poori

6 \times 6

symmetric hai (

A B B D

). Ye ek single strain energy ke consequence hai jo dono ko govern karta hai, isliye mixed partials equal hote hain.

Agar do plies ka same fiber material hai toh unka

[Q]_{k}

same hoga.

False —

[Q]_{k}

laminate axes mein transformed reduced stiffness hai, isliye ply ka angle

θ_{k}

ise change karta hai. Same material ke 0° aur 90° ply ke alag

[Q]_{k}

hote hain (dekho Transformation of Stiffness).

Mid-plane se door ek patla ply add karna

[D]

ko zyada change karta hai baniswat usse near add karne ke.

True — Eq. (3) mein

z^{3}

weighting ka matlab hai ki large

∣ z ∣

pe ek ply bending stiffness mein bahut contribute karta hai, jabki

z \approx 0

wala barely

[D]

move karta hai (uska

h_{k}^{3} - h_{k - 1}^{3}

tiny hota hai).

Error dhundho

"Laminate symmetric hai, isliye

[A] = 0

Claim confuse kar raha hai ki kaun si matrix vanish hoti hai. Symmetry deta hai

[B] = 0

, na ki

[A] = 0

; Eq. (1) se

[A]

positive-definite

[Q]_{k} t_{k}

terms ka sum hai aur real plies ke liye kabhi zero nahi hota.

[B]

ki units stiffness jaisi hain, same as

[A]

Wrong units.

[A]

Pa·m hai,

[B]

Pa·m

^{2}

hai (yeh curvature

κ

ko 1/m mein multiply karta hai force/width

N

dene ke liye). Aap

[A]

aur

[B]

entries ko directly add ya compare nahi kar sakte.

"Strains paane ke liye, loads ko

[A B D]

matrix se multiply kar lo."

Eq. (4) kehta hai

[N M] = [A B D] [ϵ^{0} κ]

, isliye loads se strains tak jaane ke liye aapko invert karna hoga:

[A B D]^{- 1}

se multiply karo.

"Symmetric laminate ke liye

N_{x}

only ke under, mujhe phir bhi full

6 \times 6

inverse chahiye."

Jab

[B] = 0

hota hai Eq. (4) block-diagonal ho jaata hai, isliye

ϵ^{0} = [A]^{- 1} N

aur

κ = [D]^{- 1} M

decouple ho jaate hain.

M = 0

ke saath aap sirf

3 \times 3

[A]

invert karte ho.

\int_{h_{k - 1}}^{h_{k}} z d z

har ply ke liye vanish hota hai, isliye

[B]

hamesha zero hota hai."

Woh integral sirf us ply ke liye vanish hota hai jo mid-plane ko symmetrically straddle karta hai; ek ply ke liye jo entirely ek side pe hai,

\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2}) \neq = 0

. Eq. (2) mein

[B]

plies ke upar ek sum hai, aur yah tab vanish hota hai jab woh signed contributions cancel ho jaate hain.

"Curvature

κ

aur strain

ϵ^{0}

ke same units hain, isliye

[A]

aur

[D]

interchangeable hain."

Nahi —

ϵ^{0}

dimensionless hai jabki

κ

ki units 1/m hain. Ye difference exactly wohi reason hai kyun

[A]

(Pa·m) aur

[D]

(Pa·m

^{3}

) length ke do powers se differ karte hain.

"[0/90] two-ply laminate symmetric hai kyunki isme ek 0 aur ek 90 hai."

Yeh antisymmetric hai, symmetric nahi — 0° bottom ply ka mirror ek 0° top ply hoga, lekin top 90° hai. Isliye

[B] \neq = 0

aur pull karne se curvature aata hai.

Why questions

Same integral

\frac{1}{2} (h_{k}^{2} - h_{k - 1}^{2})

dono force aur moment equations mein kyun aata hai?

N

mein yeh

z κ

integrate karne se aata hai (curvature ka force pe effect);

M

mein

z ϵ^{0}

integrate karne se (stretch ka lever-arm moment). Eq. (2) ka shared integral exactly wohi reason hai kyun ek matrix

[B]

dono directions couple karta hai.

Kirchhoff-Love assumption (

ϵ (z) = ϵ^{0} + z κ

)

z

mein linear kyun hai?

"Plane sections plane aur perpendicular rehte hain" ka matlab hai thickness ke through ek straight fiber loading ke baad straight rehta hai, isliye uski strain height

z

ke saath linearly vary karti hai — koi bulging nahi, koi shear warping nahi.

[D]

beam quantity

E I

jaisa kyun behave karta hai?

Dono bending resistance measure karte hain, aur dono ko unki strength material stiffness times square-ish moment of area se milti hai — yahan Eq. (3) mein

z^{3}

integration

\int z^{2} d A

ka tensor cousin hai jo

I

banata hai.

Hum ek continuous integral karne ki jagah ply-by-ply integrate kyun kar sakte hain?

Har ply ka

[Q]_{k}

apni thickness mein constant hota hai, isliye through-thickness integral saaf per-ply pieces ke sum mein split ho jaata hai (Fig. 2). Discontinuous stiffness summation form force karti hai.

Ek asymmetric laminate sirf thanda hone pe kyun warp ho jaata hai, chahe koi load apply na ho?

Uneven ply placement

[B] \neq = 0

deta hai, isliye thermal contraction strains (ek stretching effect)

[B]

ke through curvature mein feed hote hain — wahi coupling jo Thermal and Hygroscopic Effects mein cover hoti hai.

Engineers spacecraft panels ke liye symmetric layups kyun prefer karte hain?

[B] = 0

stretch aur bend ko decouple karta hai, isliye orbit mein thermal cycling koi unwanted warping produce nahi karta aur panel flat rehta hai — mirror mounts aur antennas ke liye critical hai.

Edge cases

Ek single isotropic ply mid-plane pe centered:

[B]

kya hai?

Zero —

z = 0

ke baare mein symmetric ek ply mein

\frac{1}{2} (h_{1}^{2} - h_{0}^{2})

hota hai

h_{1} = - h_{0}

ke saath, isliye Eq. (2) mein term vanish ho jaata hai. Ek akela centered ply trivially symmetric hai.

[A]

[B]

[D]

ka kya hota hai agar aap uniformly har ply ki thickness double kar do (total thickness 2 se scale ho)?

[A]

2 se scale hoga (linear in

t

, Eq. 1),

[B]

4 se (

h^{2}

, Eq. 2), aur

[D]

8 se (

h^{3}

, Eq. 3). Thick laminates bending stiffness sabse fast gain karte hain.

Ek laminate jisme saari plies same angle pe hain (e.g. [0/0/0]): kya yeh coupling sense mein "laminate" hai?

Angle differences se effectively koi coupling nahi — har

[Q]_{k}

identical hai, isliye yeh ek single thick orthotropic plate jaisa behave karta hai aur

[B] = 0

hota hai (yeh apne center ke baare mein symmetric hai).

Ek laminate ka

[B]

kya hoga jo geometry mein symmetric hai lekin dono halves alag materials ki hain jo mirror images mein rakhi hain?

Phir bhi

[B] = 0

— mirror-image condition ke liye matching

[Q]_{k}

AUR matching

∣ z ∣

chahiye; agar dono halves exactly mirror karte hain (same material mirrored positions pe), symmetry hold karti hai aur

[B]

vanish ho jaata hai.

Zero applied moment lekin pure

N_{x}

ke under ek non-symmetric laminate: kya curvature zero hai?

Nahi —

[B] \neq = 0

ke saath,

M = 0

κ = 0

force nahi karta; coupled Eq. (4) solve karne se generally non-zero

κ_{x}

milta hai. Load use bend karta hai chahe aapne koi moment apply nahi kiya.

[A B D]

matrix kab ill-conditioned hoti hai (accurately invert karna mushkil)?

Jab entries bahut alag magnitudes span karti hain —

[A] \sim

Pa·m aur

[D] \sim

Pa·m

^{3}

thickness squared se differ karte hain, isliye unhe ek

6 \times 6

mein mix karne se large condition numbers produce ho sakte hain, especially thin laminates ke liye.

Kya ek laminate jo First Ply Failure karta hai suddenly apna

[A B D]

matrix change kar leta hai?

Linear ABD intact ply stiffnesses se compute hota hai; First Ply Failure predict karne ke liye ABD ply stresses paane ke liye use hota hai, lekin ABD itself tabhi update hota hai jab aap ek progressive-damage model mein failed ply ka

[Q]_{k}

degrade karte hain.

Kya classical laminate theory (aur is tarah

[A B D]

) soft core wala thick sandwich handle kar sakta hai?

Zyada nahi — CLT assume karta hai ki plies thin hain aur transverse shear nahi hai; ek soft-core sandwich core ke through shear karta hai, isliye aapko Sandwich Panel Theory ya shear-deformable / FE approach chahiye.