Hooke's law in 3D — generalized stress-strain (tensor): Exercises

3.6.4 · D4 · Physics › Spacecraft Structures & Systems Engineering › Hooke's law in 3D — generalized stress-strain (tensor)

Kuch bhi calculate karne se pehle, har word aur symbol ko fix kar lete hain taaki koi bhi use hone se pehle clearly define ho.

Definition Woh symbols jo tumhe baar baar milenge

$σ_{ij}$ = stress: ek tiny cube ke face par force per unit area. Pehla index = kaun sa face (uska outward normal), doosra = force kis direction mein push kar raha hai. Units: pascals (Pa = N/m²).
Jab dono indices equal hon ( $σ_{xx}$ ) to yeh normal (push/pull) stress hai; jab woh alag hon to yeh shear hai. Shear stresses itne common hain ki inhe apna alag letter milta hai: $τ_{x y} \equiv σ_{x y}$ (sirf ek rename hai — $x$ -face par $y$ direction mein point karne wala force).
$ε_{ij}$ = strain: fractional deformation, ek pure number (metre-per-metre).
$γ_{x y}$ = engineering shear strain: define hai $γ_{x y} = 2 ε_{x y}$ ke roop mein. Yeh ek right corner ke total angle change ko measure karta hai; $2$ ka factor isliye hai kyunki dono $x$ - aur $y$ -sides tilt karte hain, aur engineers dono tilts ko ek number mein add kar dete hain.
$δ_{ij}$ = Kronecker delta: ek bookkeeping switch jo $1$ hota hai jab $i = j$ aur $0$ jab $i \neq = j$ . Yeh "identity" hai — yeh ek formula ko sirf matching indices (diagonal / normal directions) ke liye ON karta hai aur off-diagonal (shear) waalon ke liye OFF.
$E$ = Young's modulus: kitna stress lagta hai ek given fraction se stretch karne ke liye (Pa). Bada $E$ = stiff.
$ν$ (nu) = Poisson's ratio: isko ek direction mein lambe kheencho, yeh number batata hai ki yeh sideways kitna thin hoga. Ordinary solids ke liye $0$ aur $\frac{1}{2}$ ke beech ek pure ratio.
$λ, μ$ = Lamé parameters: woh do numbers jo ek isotropic material ko fully describe karte hain. $μ \equiv G$ shear modulus hai (twist ke khilaf resistance); $λ$ ek volume change ko normal stress se link karta hai. Yeh sirf $E, ν$ ki repackaging hain (formulas neeche hain). Repeated index (jaise $σ_{k k}$ ) ka matlab hai jodo $σ_{11} + σ_{22} + σ_{33}$ — the trace.

Poori notes mein, jab kisi material ka naam aaye, yeh values use karo (spacecraft ke liye typical aluminium alloy): $E = 70 GPa, ν = 0.33, G = \frac{E}{2 ( 1 + ν )} \approx 26.3 GPa .$

L1 — Recognition

1.1 Symbol $σ_{23}$ mein, cube ka kaun sa face loaded hai aur force kis direction mein point karta hai? Kya $σ_{23}$ equal hai $σ_{32}$ ke, aur kyun?

1.2 Ek material mein $ε_{11} = ε_{22} = ε_{33} = 0.001$ hai aur saare shear strains zero hain. Volumetric strain $ε_{k k}$ likho aur words mein batao iska matlab kya hai.

1.3 Upar reproduce ki gayi compliance matrix $[S]$ use karte hue, kaun si single entry tumhe $x$ mein stress aur $y$ mein uske kaaran hone wale strain ke beech Poisson coupling batati hai? Uski value $E, ν$ ke terms mein batao.

Recall Solutions — L1

1.1 Pehla index $= 2$ : woh face jiska outward normal $2$ -axis (the $y$ -face) ke along point karta hai. Doosra index $= 3$ : us face par force $3$ -axis ( $z$ ) ke along point karta hai. Toh $σ_{23}$ ek shear hai $y$ -face par jo $z$ mein push kar raha hai. Haan, $σ_{23} = σ_{32}$ : agar yeh alag hote to tiny cube ek net twisting moment feel karta aur infinitely fast spin up ho jaata — moment balance ise forbid karta hai, isliye stress tensor symmetric hota hai.

1.2 $ε_{k k} = ε_{11} + ε_{22} + ε_{33} = 0.001 + 0.001 + 0.001 = 0.003$ . Yeh volume mein fractional change hai: is material ka ek cube volume mein $0.3%$ swell karta hai.

1.3 Top-left $3 \times 3$ block mein off-diagonal entry: $S_{12} = - \frac{ν}{E}$ . Toh $ε_{xx}$ ko $- \frac{ν}{E} σ_{y y}$ milta hai — Poisson pull-in.

L2 — Application

2.1 Ek bar ko uniaxially $σ_{xx} = 140$ MPa se load kiya gaya hai, baki sab zero hai ( $E = 70$ GPa, $ν = 0.33$ ). $ε_{xx}$ , $ε_{y y}$ , $ε_{z z}$ nikalo.

2.2 Ek block equal triaxial tension $σ_{xx} = σ_{y y} = σ_{z z} = 100$ MPa ke under hai, koi shear nahi. $ε_{xx}$ aur volumetric strain $ε_{k k}$ nikalo.

2.3 Pure shear $τ_{x y} = 60$ MPa akela act kar raha hai (yaad karo $τ_{x y} \equiv σ_{x y}$ , aur engineering shear strain $γ_{x y} = 2 ε_{x y} = τ_{x y} / G$ hai). $γ_{x y}$ nikalo. Confirm karo ki koi normal strains appear nahi hote.

Recall Solutions — L2

2.1 Sirf $σ_{xx}$ nonzero hai, isliye $σ_{k k} = σ_{xx}$ . $ε_{xx} = \frac{1}{E} [σ_{xx} - ν (0 + 0)] = \frac{140 \times 1 0 ^{6}}{70 \times 1 0 ^{9}} = 2.0 \times 1 0^{- 3} .$ WHAT humne kiya: dono zero cross-stresses drop kar diye. WHY: uniaxial ka matlab hai $y, z$ stresses vanish ho jaate hain. $ε_{y y} = ε_{z z} = \frac{1}{E} [0 - ν σ_{xx}] = - 0.33 \times 2.0 \times 1 0^{- 3} = - 6.6 \times 1 0^{- 4} .$ Bar $x$ ke along $0.2%$ stretch hoti hai aur har transverse direction mein $0.066%$ thin hoti hai — Poisson effect.

2.2 Teeno normal stresses $σ_{0} = 100$ MPa ke equal hain, $σ_{k k} = 3 σ_{0}$ .

=\frac{100\times10^6\,(1-0.66)}{70\times10^9}=4.857\times10^{-4}.$$ $$\varepsilon_{kk}=3\varepsilon_{xx}=1.457\times10^{-3}.$$ Note karo yeh positive hai (tension volume swell karti hai) aur chhota hai kyunki $(1-2\nu)$ chhota hai. **2.3** Shear decoupled hai, isliye $\gamma_{xy}=\tau_{xy}/G$ use karo with $G=\dfrac{E}{2(1+\nu)}=\dfrac{70\times10^9}{2(1.33)}=26.316\times10^9$ Pa: $$\gamma_{xy}=\frac{60\times10^6}{26.316\times10^9}=2.28\times10^{-3}.$$ Normal strains sirf normal stresses se aate hain compliance matrix mein; yahan saare normal stresses zero hain, isliye $\varepsilon_{xx}=\varepsilon_{yy}=\varepsilon_{zz}=0$. Pure distortion, koi volume change nahi.

L3 — Analysis

3.1 (Plane stress) Ek thin flat panel mein $σ_{z z} = 0$ hai (kuch bhi uski thickness se push nahi karta). $σ_{xx} = 200$ MPa, $σ_{y y} = 80$ MPa, $τ_{x y} = 0$ diya gaya hai, $ε_{z z}$ nikalo. Explain karo kyun yeh nonzero hai jabki $σ_{z z} = 0$ hai.

3.2 (Plane strain) Ek lamba tunnel-jaisa body $z$ ke along fully constrained hai isliye $ε_{z z} = 0$ hai. $σ_{xx} = 200$ MPa, $σ_{y y} = 80$ MPa diya gaya hai, woh induced $σ_{z z}$ nikalo jo constraint supply karta hoga.

3.3 Upar wale plane-strain body ke liye, apna nikala hua $σ_{z z}$ use karte hue, $ε_{xx}$ compute karo.

Neeche ki figure dono cases contrast karti hai. Left (plane stress): ek thin panel in-plane forces (black arrows) se kheechi jaa rahi hai jabki uske dono flat faces free hain; red arrows dikhate hain ki yeh thickness ke through thin ho rahi hai — $σ_{z z} = 0$ lekin $ε_{z z} \neq = 0$ hai. Right (plane strain): wahi in-plane pull, lekin top aur bottom faces clamped hain (red hatching) isliye thickness nahi badal sakti — $ε_{z z} = 0$ hai, aur clamps ko ek hidden $σ_{z z} \neq = 0$ supply karna padta hai. Accent colour mark karta hai, har panel mein, woh cheez jo zero nahi hai.

Figure — Hooke's law in 3D — generalized stress-strain (tensor)

Recall Solutions — L3

3.1 Plane stress: $σ_{z z} = 0$ , isliye $σ_{k k} = σ_{xx} + σ_{y y} = 280$ MPa.

=\frac{1}{70\times10^9}[0-0.33(280\times10^6)]=-1.32\times10^{-3}.$$ *WHY nonzero:* koi *force* thickness ke through push nahi kar raha, lekin in-plane pulls panel ko Poisson effect ke through thinner squeeze kar dete hain. Zero stress $\ne$ zero strain. **3.2** Plane strain $\varepsilon_{zz}=0$ force karta hai. $z$-strain equation ko zero set karo: $$0=\frac1E[\sigma_{zz}-\nu(\sigma_{xx}+\sigma_{yy})]\ \Rightarrow\ \sigma_{zz}=\nu(\sigma_{xx}+\sigma_{yy})=0.33(280\times10^6)=92.4\ \text{MPa}.$$ *WHY:* natural Poisson thinning ko rokne ke liye, walls ko $z$ ke along exactly is stress se *push back* karna padega. Constraint sirf geometry se ek real stress manufacture kar deta hai. **3.3** Ab $\sigma_{kk}=\sigma_{xx}+\sigma_{yy}+\sigma_{zz}=200+80+92.4=372.4$ MPa hai. $\varepsilon_{xx}$ par Poisson pull-in $\sigma_{yy}+\sigma_{zz}=80+92.4=172.4$ MPa se aata hai: $$\varepsilon_{xx}=\frac1E[\sigma_{xx}-\nu(\sigma_{yy}+\sigma_{zz})] =\frac{[200-0.33(172.4)]\times10^6}{70\times10^9}=2.04\times10^{-3}.$$ Compare karo: induced $\sigma_{zz}$ ek free panel ke versus extra Poisson pull-in add karta hai, lekin $\sigma_{xx}$ phir bhi dominate karta hai, isliye $\varepsilon_{xx}$ positive rehta hai (body $x$ ke along stretch hoti hai).

L4 — Synthesis

4.1 (Pressure-vessel wall) Ek thin cylindrical propellant tank mein hoop stress $σ_{θ} = \frac{p r}{t}$ hai, axial stress $σ_{z} = \frac{p r}{2 t}$ hai, radial $\approx 0$ hai (plane stress). $p = 2$ MPa, $r = 0.5$ m, $t = 4$ mm, $E = 70$ GPa, $ν = 0.33$ ke saath, $σ_{θ}$ , $σ_{z}$ , phir axial strain $ε_{z}$ aur hoop strain $ε_{θ}$ nikalo. Dekho Thin-walled pressure vessels.

4.2 Usi tank ke liye, von Mises equivalent stress $σ_{v} = σ_{θ}^{2} - σ_{θ} σ_{z} + σ_{z}^{2}$ (plane-stress form) compute karo aur $270$ MPa ki yield strength se compare karo — kya wall yield karti hai? Dekho Von Mises yield criterion.

4.3 Pure hydrostatic pressure $σ_{xx} = σ_{y y} = σ_{z z} = - p$ load karke, isotropic law se purely bulk modulus $K$ derive karo, phir number ke liye $E = 70$ GPa, $ν = 0.33$ plug karo.

Recall Solutions — L4

4.1 Pehle stresses:

\sigma_z=\tfrac12\sigma_\theta=125\ \text{MPa}.$$ Axial strain (radial stress $=0$, isliye sirf $\sigma_\theta$ Poisson partner hai): $$\varepsilon_z=\frac1E[\sigma_z-\nu\sigma_\theta] =\frac{[125-0.33(250)]\times10^6}{70\times10^9}=6.07\times10^{-4}.$$ Hoop strain (uska Poisson partner $\sigma_z$ hai): $$\varepsilon_\theta=\frac1E[\sigma_\theta-\nu\sigma_z] =\frac{[250-0.33(125)]\times10^6}{70\times10^9}=2.98\times10^{-3}.$$ *WHY coupling:* thin wall mein hoop aur axial directions perpendicular hain, isliye har ek *doosre* ka Poisson partner hai. Bada hoop stress wall ko axially thin karne ki koshish karta hai ($\varepsilon_z$ ko $-\nu\sigma_\theta$ term ke through kam karta hai), jabki chhota axial stress hoop stretch ko sirf mildly reduce karta hai ($-\nu\sigma_z$ term). Isliye $\varepsilon_\theta$ end mein $\varepsilon_z$ se bahut bada hota hai — aur isliye bolted flanges sizing mein is coupled thinning ka account karna zaroori hai. **4.2** $\sigma_v=\sqrt{250^2-250\cdot125+125^2}=\sqrt{62500-31250+15625}=\sqrt{46875}=216.5$ MPa. Kyunki $216.5 < 270$ MPa, wall **yield nahi karti** — lekin margin sirf $20\%$ ke aaspaas hai, flight hardware ke liye tight hai. **4.3** $\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=-p$ ke saath: har normal strain $\varepsilon_{xx}=\frac1E[-p-\nu(-2p)]=\frac{-p(1-2\nu)}{E}$ hai. Phir $\varepsilon_{kk}=3\varepsilon_{xx}=\frac{-3p(1-2\nu)}{E}$. Bulk modulus $K\equiv -p/\varepsilon_{kk}$ hai: $$K=\frac{E}{3(1-2\nu)}=\frac{70\times10^9}{3(1-0.66)}=\frac{70\times10^9}{1.02}=68.6\ \text{GPa}.$$

L5 — Mastery

5.1 Strain-from-stress law $ε_{ij} = \frac{1 + ν}{E} σ_{ij} - \frac{ν}{E} σ_{k k} δ_{ij}$ se shuru karke, shear relation $γ_{x y} = τ_{x y} / G$ derive karo aur isse $G = \frac{E}{2 ( 1 + ν )}$ first principles se prove karo.

5.2 (Energy) Elastic strain-energy density $W = \frac{1}{2} σ_{ij} ε_{ij}$ hai (dekho Elastic strain energy density). 2.1 ki uniaxial bar ke liye ( $σ_{xx} = 140$ MPa, baki zero), dikhao $W = \frac{1}{2} σ_{xx} ε_{xx}$ aur ise J/m³ mein evaluate karo.

5.3 (Incompressibility limit) Algebraically dikhao ki jab $ν \to \frac{1}{2}$ , bulk modulus $K \to \infty$ hota hai, aur physically explain karo kyun $ν > \frac{1}{2}$ forbidden hai. Phir compute karo ki $K$ kitna change hota hai agar ek rubber-like material mein $ν = 0.499$ vs $ν = 0.4999$ ho ( $E = 10$ MPa lo).

Recall Solutions — L5

5.1 Off-diagonal component $i \neq = j$ ke liye, Kronecker delta $δ_{ij} = 0$ hai, isliye poora trace term drop ho jaata hai: $ε_{x y} = \frac{1 + ν}{E} σ_{x y} .$ Engineering shear $γ_{x y} = 2 ε_{x y} = \frac{2 ( 1 + ν )}{E} τ_{x y}$ hai. Definition se $γ_{x y} = τ_{x y} / G$ hai, isliye coefficients match karo: $\frac{1}{G} = \frac{2 ( 1 + ν )}{E} \Rightarrow G = \frac{E}{2 ( 1 + ν )} .$ WHAT/WHY: delta term hi shear ko volumetric part se link karta hai; ise kill karna ( $i \neq = j$ ) pure shear isolate karta hai, aur factor-of-2 seedha $γ = 2 ε$ convention se aata hai.

5.2 Sirf $σ_{xx}$ nonzero hai, isliye sum $W = \frac{1}{2} σ_{ij} ε_{ij}$ mein sirf ek surviving term hai (baki saare $σ_{ij} = 0$ ):

=\tfrac12\,\frac{(140\times10^6)^2}{70\times10^9}=1.4\times10^{5}\ \text{J/m}^3.$$ ($\sigma_{yy}\varepsilon_{yy}$ jaise cross terms vanish ho jaate hain kyunki $\sigma_{yy}=0$ hai, chahe $\varepsilon_{yy}\ne0$ ho — energy ko *dono* factors chahiye.) **5.3** $K=\dfrac{E}{3(1-2\nu)}$. Jab $\nu\to\tfrac12$, denominator $(1-2\nu)\to0^+$, isliye $K\to+\infty$: volume change karne ke liye infinite pressure chahiye — material **incompressible** hai. Agar $\nu>\tfrac12$ to $(1-2\nu)<0$ aur $K<0$ hoga, matlab squeeze karne se yeh *expand* ho jaayega: ek spontaneous instability, passive solid ke liye physically impossible. Isliye $\nu\le\tfrac12$. $E=10$ MPa ke saath numbers: $$K(0.499)=\frac{10^7}{3(1-0.998)}=\frac{10^7}{0.006}=1.667\times10^{9}\ \text{Pa},$$ $$K(0.4999)=\frac{10^7}{3(1-0.9998)}=\frac{10^7}{0.0006}=1.667\times10^{10}\ \text{Pa}.$$ $\nu$ ka $\tfrac12$ ki taraf tenfold approach $K$ ko das guna bada kar deta hai — blow-up steep hai.

Recall Jaane se pehle ek one-line self-check

Kaun sa convention pitfall tumhe sabse zyada laga — face-vs-force indices, $γ$ -vs- $ε$ shear, ya plane-stress-vs-plane-strain? ::: Teeno "tumne ek symmetric pair ka galat member fix kiya" hain — pehle physics padho, phir decide karo kaun sa member zero hai.