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

3.6.4 · D3 · Physics › Spacecraft Structures & Systems Engineering › Hooke's law in 3D — generalized stress-strain (tensor)

Yeh page Hooke's law in 3D — generalized stress-strain (tensor) ki exercise ground hai. Hum isotropic law lete hain aur use har tarah ki loading se guzarte hain jo ek spacecraft part feel kar sakta hai — ek pull, chaaron taraf se squeeze, pure twist, do directions ek saath, tricky sign cases, aur woh degenerate limits jahan formula almost toot jaata hai.

Pehli line se pehle, yahan sirf teen tools hain jo hum use karenge, plain words mein:

Sign convention ka matlab, hamesha: positive stress = pull (tension), negative stress = push (compression); positive strain = us direction mein lamba hua, negative strain = chhota hua.

Scenario matrix

Neeche har problem is grid ka ek cell hai. Columns hain kaunsi tarah ki loading; rows hain kya cheez usse tricky banati hai. Hum har cell tick karenge.

Cell	Loading type	Woh twist jo dikhana zaroori hai
A	Single pull (uniaxial, $σ > 0$ )	1D recover karo + sideways shrink dekho
B	Single push (uniaxial, $σ < 0$ )	Negative sign cleanly flow karta hai
C	All-round squeeze (hydrostatic)	Volume-only, $K$ define karta hai
D	Pure twist (shear only)	Volume change nahi, sirf $G$ enter karta hai
E	Do directions ek saath (biaxial)	Donon ke beech Poisson coupling
F	Mixed tension + compression	Signs bracket ke andar partly cancel hote hain
G	Degenerate limit $ν \to \frac{1}{2}$	Incompressible — $K \to \infty$
H	Degenerate limit $ν \to 0$	No coupling — axes independent ho jaate hain
I	Real-world word problem	Spacecraft tank, numbers with units
J	Exam twist	" $y$ mein zero strain" (ek constraint, stress nahi)

Geometric cells (A/B, C, D) ke saath figures bhi hain.

Worked examples

Worked example Cell A — Single pull (uniaxial tension)

Ek aluminium strut ( $E = 70 GPa$ , $ν = 0.33$ ) ko $σ_{xx} = + 140 MPa$ se khicha ja raha hai; baaki faces par kuch nahi.

Forecast: aage padhne se pehle $ε_{y y}$ ka sign guess karo. Jab tum ise kheencho toh bar sideways mein moti hoti hai ya patli?

Pull ki direction mein: $ε_{xx} = \frac{1}{E} [σ_{xx} - ν (0 + 0)] = \frac{140}{70000} = 2.0 \times 1 0^{- 3}$ . Yeh step kyun? Sirf ek stress hone par, Poisson bracket khaali ho jaata hai aur 3D seedha plain $σ = E ε$ ban jaata hai.
Sideways: $ε_{y y} = ε_{z z} = \frac{1}{E} [0 - ν σ_{xx}] = \frac{- 0.33 \times 140}{70000} = - 6.6 \times 1 0^{- 4}$ . Yeh step kyun? $- ν σ_{xx}$ term hi Poisson effect hai — $x$ mein kheenchne par $y$ aur $z$ ki width chheen li jaati hai. Negative = patla. Dekho
— laal bar stretch ho raha hai, mint arrows dikhate hain ki woh pinch ho raha hai.

Verify: $ε_{xx} / ε_{y y} = 2.0 \times 1 0^{- 3} / (- 6.6 \times 1 0^{- 4}) = - 1/0.33 = - 1/ ν$ . Yahi Poisson's ratio ka matlab hai: sideways strain, axial strain ka $- ν$ times hota hai. Units: strain dimensionless hai (MPa/MPa). ✔

Worked example Cell B — Single push (uniaxial compression)

Same strut, ab dhaka diya: $σ_{xx} = - 140 MPa$ .

Forecast: sideways faces ab kis taraf move karengi — andar ya bahar?

Along: $ε_{xx} = \frac{- 140}{70000} = - 2.0 \times 1 0^{- 3}$ (yeh chhota hota hai — negative, jaise expected). Yeh step kyun? Same formula; stress ka minus sign seedha chala jaata hai. Compression ke liye kuch special nahi hota — linear law symmetric hai.
Sideways: $ε_{y y} = \frac{1}{E} [0 - ν (- 140)] = + 6.6 \times 1 0^{- 4}$ (positive → yeh bahar bulge karta hai). Yeh step kyun? Ab Poisson term hai $- ν \times (negative) = +$ . Lengthwise squeeze karne par sides moti ho jaati hain — bilkul foam-block wali picture.

Verify: har number Cell-A ka answer hai apna sign flip karke. Linear elasticity ka matlab hai "push = negative pull", aur ratio $- ν$ preserve rehta hai: $6.6 \times 1 0^{- 4} / (- 2.0 \times 1 0^{- 3}) = - 0.33 = - ν$ . ✔

Worked example Cell C — All-round squeeze (hydrostatic)

Strut ko pressure chamber ke bottom mein daal diya: $σ_{xx} = σ_{y y} = σ_{z z} = - p$ , $p = 140 MPa$ . Koi shear nahi.

Forecast: saathi chehron par equal push ke saath, kya shape badlegi, ya sirf size?

Ek normal strain: $ε_{xx} = \frac{1}{E} [- p - ν (- p - p)] = \frac{- p ( 1 - 2 ν )}{E}$ . Yeh step kyun? Har face apna $- p$ feel karta hai, phir doosre do ki Poisson bulge se thoda push-back milta hai, jisse factor $(1 - 2 ν)$ banta hai.
Symmetry se $ε_{y y} = ε_{z z} = ε_{xx}$ , toh koi bhi face alag nahi — shape unchanged, sirf volume. Dekho
: cube uniformly shrink karta hai, cube hi rehta hai.
Volume change: $ε_{k k} = 3 ε_{xx} = \frac{- 3 p ( 1 - 2 ν )}{E}$ . Yeh step kyun? Strain ka trace = fractional volume change; yahan teen equal terms ka sum hai.
Numbers ( $ν = 0.33$ ): $ε_{k k} = \frac{- 3 \times 140 \times ( 1 - 0.66 )}{70000} = \frac{- 3 \times 140 \times 0.34}{70000} = - 2.04 \times 1 0^{- 3}$ .

Verify: bulk modulus hai $K = - p / ε_{k k} = \frac{140}{2.04 \times 1 0 ^{- 3}} = 68.6 GPa$ , jo equal hona chahiye $\frac{E}{3 ( 1 - 2 ν )} = \frac{70000}{3 \times 0.34} = 68.6 GPa$ . ✔

Worked example Cell D — Pure twist (shear only)

Ek spacecraft drive shaft $τ_{x y} = τ_{0} = 50 MPa$ carry karta hai; saare normal stresses zero hain. Same material.

Forecast: kya yeh loading shaft ka volume bilkul bhi badlegi?

Shear strain: $γ_{x y} = \frac{τ _{0}}{G}$ , jahan $G = \frac{E}{2 ( 1 + ν )} = \frac{70000}{2 \times 1.33} = 26.3 GPa$ . Yeh step kyun? Compliance matrix ki shear rows, normal rows se decoupled hain — ek shear stress sirf shear strain hi produce kar sakta hai.
Number: $γ_{x y} = \frac{50}{26315} = 1.90 \times 1 0^{- 3} rad$ .
Volume? Saare $ε_{xx} = ε_{y y} = ε_{z z} = 0$ , toh $ε_{k k} = 0$ : pure shape-change, zero volume change. Dekho
— pastel square ek rhombus mein rack ho jaata hai lekin same area maintain karta hai.

Verify: $γ$ ko $E, ν$ se rewrite karo: $γ_{x y} = \frac{2 ( 1 + ν )}{E} τ_{0} = \frac{2 \times 1.33 \times 50}{70000} = 1.90 \times 1 0^{- 3}$ . Same value — compliance shear entry $2 (1 + ν) / E$ sach mein $1/ G$ hai. ✔

Worked example Cell E — Do directions ek saath (biaxial)

Ek flat panel in-plane khicha ja raha hai: $σ_{xx} = + 120 MPa$ , $σ_{y y} = + 60 MPa$ , $σ_{z z} = 0$ (plane stress). Same material.

Forecast: kya $ε_{z z}$ (thickness strain) zero hogi, positive hogi, ya negative?

$x$ -strain: $ε_{xx} = \frac{1}{E} [120 - ν (60 + 0)] = \frac{120 - 0.33 \times 60}{70000} = \frac{100.2}{70000} = 1.431 \times 1 0^{- 3}$ . Yeh step kyun? $y$ -pull, Poisson ke zariye $x$ -stretch ka thoda sa hissa cheen leta hai.
$y$ -strain: $ε_{y y} = \frac{1}{E} [60 - ν (120 + 0)] = \frac{60 - 39.6}{70000} = \frac{20.4}{70000} = 2.914 \times 1 0^{- 4}$ .
Thickness strain: $ε_{z z} = \frac{1}{E} [0 - ν (120 + 60)] = \frac{- 0.33 \times 180}{70000} = - 8.486 \times 1 0^{- 4}$ . Yeh step kyun? Flat faces par kuch bhi push nahi karta, phir bhi dono in-plane pulls thickness ko squeeze karte hain — isliye thin sheets biaxially stretch hone par patli ho jaati hain.

Verify: $ε_{z z}$ equal hona chahiye $- \frac{ν}{E} (σ_{xx} + σ_{y y}) = - \frac{0.33}{70000} \times 180 = - 8.486 \times 1 0^{- 4}$ . ✔ Aur yeh negative hai, jaise forecast kiya tha.

Worked example Cell F — Mixed tension + compression

Ek bracket $σ_{xx} = + 90 MPa$ (pull) dekhta hai lekin $σ_{y y} = - 90 MPa$ (push), $σ_{z z} = 0$ .

Forecast: $x$ -strain mein, kya Poisson term direct pull mein add hoga ya subtract hoga?

$x$ -strain: $ε_{xx} = \frac{1}{E} [90 - ν (- 90 + 0)] = \frac{90 + 0.33 \times 90}{70000} = \frac{119.7}{70000} = 1.710 \times 1 0^{- 3}$ . Yeh step kyun? Yahan neighbour stress negative hai, toh $- ν σ_{y y} = + 29.7$ stretch mein add ho jaata hai: $x$ kheenchte hue $y$ squeeze karne par $x$ aur bhi zyada elongate hota hai. Signs bahut zyada matter karte hain.
$y$ -strain: $ε_{y y} = \frac{1}{E} [- 90 - ν (90 + 0)] = \frac{- 90 - 29.7}{70000} = \frac{- 119.7}{70000} = - 1.710 \times 1 0^{- 3}$ .
Volume: $ε_{z z} = \frac{1}{E} [0 - ν (90 - 90)] = 0$ , toh $ε_{k k} = ε_{xx} + ε_{y y} + 0 = 0$ . Yeh step kyun? Equal-and-opposite normal stresses ek pure shear state disguise mein hai — koi volume change nahi, sirf distortion.

Verify: $ε_{xx} = - ε_{y y}$ aur $ε_{k k} = 0$ , yeh shape-only deformation ki pehchaan hai. Check karo $ε_{xx} = \frac{90 ( 1 + ν )}{E} = \frac{90 \times 1.33}{70000} = 1.710 \times 1 0^{- 3}$ . ✔

Worked example Cell G — Degenerate limit

$ν \to \frac{1}{2}$ (incompressible) Cell-C ka hydrostatic result lo aur $ν \to \frac{1}{2}$ karo (rubber-like seal material), $p = 140 MPa$ .

Forecast: jaise $ν$ $0.5$ ki taraf badhta hai, volume change ka kya hoga?

Volume change: $ε_{k k} = \frac{- 3 p ( 1 - 2 ν )}{E}$ . Jaise $ν \to \frac{1}{2}$ , factor $(1 - 2 ν) \to 0$ , toh $ε_{k k} \to 0$ . Yeh step kyun? Material kitna bhi squeeze karo, volume change karne se mana kar deta hai — sirf shape redistribute karta hai.
Bulk modulus: $K = \frac{E}{3 ( 1 - 2 ν )} \to \infty$ jaise $ν \to \frac{1}{2}$ . Yeh step kyun? Volume change ke liye infinite resistance = incompressible. Physically $ν$ kabhi $\frac{1}{2}$ se zyada nahi ho sakta, warna $K$ negative ho jaata (material squeeze hone par expand karta — unstable).

Verify (numeric): $ν = 0.499$ par, $K = \frac{70000}{3 ( 1 - 0.998 )} = \frac{70000}{0.006} = 1.167 \times 1 0^{7} MPa$ — bahut bada, infinity ki taraf ja raha hai. ✔

Worked example Cell H — Degenerate limit

$ν \to 0$ (uncoupled) Cell E jaisi hi biaxial load ( $σ_{xx} = 120$ , $σ_{y y} = 60$ , $σ_{z z} = 0$ ), lekin cork-jaisi material ke saath jismein $ν = 0$ hai.

Forecast: jab koi Poisson coupling nahi, toh $ε_{z z}$ kya hogi?

Har strain standalone ban jaata hai: $ε_{xx} = \frac{120}{70000} = 1.714 \times 1 0^{- 3}$ , $ε_{y y} = \frac{60}{70000} = 8.571 \times 1 0^{- 4}$ . Yeh step kyun? $ν = 0$ ke saath har Poisson bracket term gayab ho jaata hai, toh har direction apne independent 1D spring ki tarah behave karta hai.
Thickness: $ε_{z z} = \frac{1}{E} [0 - 0] = 0$ . Yeh step kyun? Flat faces par koi stress nahi aur koi coupling nahi → bilkul koi thickness change nahi, Cell E ki tarah nahi jahan woh shrink hui thi.

Verify: yeh diagonal-only compliance matrix hai; strains exactly $σ / E$ ke barabar hain. $ε_{xx} = 120/70000 = 1.714 \times 1 0^{- 3}$ . ✔

Worked example Cell I — Real-world word problem (spacecraft tank)

Ek thin-walled cylindrical propellant tank (thin-walled vessel): radius $r = 1.0 m$ , wall $t = 5 mm$ , internal pressure $p = 2.0 MPa$ . Aluminium, $E = 70 GPa$ , $ν = 0.33$ . Axial strain $ε_{z}$ nikalo (yeh check karne ke liye ki bolted flanges leak toh nahi karengi).

Forecast: hoop stress, axial stress se double hoti hai. Kya Poisson coupling axial strain ko akele axial stress se zyada karega ya kam?

Hoop stress: $σ_{θ} = \frac{p r}{t} = \frac{2.0 \times 1.0}{0.005} = 400 MPa$ . Yeh step kyun? Standard thin-wall formula; hoop direction sabse zyada load carry karta hai.
Axial stress: $σ_{z} = \frac{p r}{2 t} = \frac{2.0 \times 1.0}{2 \times 0.005} = 200 MPa$ . Radial $\approx 0$ (plane stress).
Axial strain: $ε_{z} = \frac{1}{E} [σ_{z} - ν (σ_{θ} + 0)] = \frac{200 - 0.33 \times 400}{70000} = \frac{68}{70000} = 9.714 \times 1 0^{- 4}$ . Yeh step kyun? Badi hoop stress wall ko sideways kheenchti hai aur (Poisson ke zariye) tank ko axially chhota karti hai, axial strain ko almost aadha kar deti hai.

Verify: tidy form use karo $ε_{z} = \frac{p r}{2 tE} (1 - 2 ν) = \frac{2 \times 1 \times ( 1 - 0.66 )}{2 \times 0.005 \times 70000} = \frac{0.68}{700} = 9.714 \times 1 0^{- 4}$ . Step 3 se match karta hai. ✔

Worked example Cell J — Exam twist (ek constraint, stress nahi)

Ek block ko $σ_{xx} = + 150 MPa$ se khicha ja raha hai lekin $y$ -direction mein rigidly pakda gaya hai taaki $ε_{y y} = 0$ rahe ( $y$ mein contract nahi kar sakta). $z$ -face free hai ( $σ_{z z} = 0$ ). Same material. Reaction stress $σ_{y y}$ aur resulting $ε_{xx}$ nikalo.

Forecast: block ko $y$ mein shrinkne se rokne ke liye, kya wall push (compress) karegi ya pull (tension) karegi usse?

$y$ -equation par constraint impose karo: $ε_{y y} = \frac{1}{E} [σ_{y y} - ν (σ_{xx} + σ_{z z})] = 0$ . Yeh step kyun? Hum $σ_{y y}$ nahi jaante; bajaaye uske hum jaante hain ki wahan strain zero hai — wall jo bhi stress chahiye supply karta hai. Yahi exam ka trap hai: fixed strain ek unknown stress ban jaata hai.
$σ_{y y}$ solve karo: $σ_{y y} = ν (σ_{xx} + 0) = 0.33 \times 150 = + 49.5 MPa$ (tension — wall pull back karta hai usse pakdne ke liye). Yeh step kyun? Wall ke bina block $y$ mein contract ho jaata; wall tension apply karta hai us contraction ko exactly cancel karne ke liye.
Ab $x$ -strain: $ε_{xx} = \frac{1}{E} [σ_{xx} - ν (σ_{y y} + σ_{z z})] = \frac{150 - 0.33 \times 49.5}{70000} = \frac{133.665}{70000} = 1.909 \times 1 0^{- 3}$ . Yeh step kyun? Ab nonzero $σ_{y y}$ Poisson ke zariye feedback karta hai aur free tension ke comparison mein $x$ -response ko thoda stiffen kar deta hai.

Verify: $σ_{y y} = 49.5$ ko $y$ -equation mein wapas plug karo: $\frac{1}{E} [49.5 - 0.33 (150 + 0)] = \frac{1}{E} [49.5 - 49.5] = 0$ . Constraint satisfy hua. ✔

Recall Main kaun se cell mein hoon?

Sirf ek stress diya, baaki zero ::: uniaxial (Cell A/B) — $σ = E ε$ tak collapse hota hai sides par Poisson ke saath. Saare faces par equal normal stresses ::: hydrostatic (Cell C) — volume only, $K$ define karta hai. Sirf ek shear stress diya ::: pure shear (Cell D) — $γ = τ / G$ use karo, koi volume change nahi. Ek strain kisi value par fixed hai (jaise $ε_{y y} = 0$ ) stress ki jagah ::: constraint problem (Cell J) — pehle reaction stress solve karo. $ν \to \frac{1}{2}$ ::: incompressible, $K \to \infty$ . $ν \to 0$ ::: fully uncoupled, har axis ek independent 1D spring.

Related tools jab yeh cases master kar lo: Elastic strain energy density (har case mein store ki gayi energy), Von Mises yield criterion (yeh stresses yielding kab cause karte hain), aur Finite Element Method — stiffness matrices (computer ek saath millions aisa kaise solve karta hai).