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

3.6.4 · D5 · Physics › Spacecraft Structures & Systems Engineering › Hooke's law in 3D — generalized stress-strain (tensor)

Notation reminder taaki yahan kuch bhi unearned na lage:

$σ_{ij}$ = stress (force per area; pehla index = face normal, doosra = force direction).
$ε_{ij}$ = tensorial strain (fractional deformation, symmetric $3 \times 3$ ).
Sign convention (is page par har jagah use hota hai): tension positive, compression negative. Isliye ek pressure $p$ jo andar push karta hai woh $σ_{xx} = - p$ ke roop mein aata hai (hydrostatic items dekho).
$γ_{ij}$ = engineering shear strain, defined hai $γ_{ij} = 2 ε_{ij}$ for $i \neq = j$ . Yeh do originally perpendicular fibres ke beech angle change ka total measure karta hai, jabki tensor shear $ε_{ij}$ us angle change ko dono directions mein symmetrically split karta hai — isliye 2 ka factor aata hai. Neeche diya figure dekho.
$C_{ij k l}$ = stiffness tensor (4-index map $σ_{ij} = C_{ij k l} ε_{k l}$ ).
$S_{ij k l}$ = compliance tensor, iska inverse: $ε_{ij} = S_{ij k l} σ_{k l}$ . Voigt matrix form mein yeh $6 \times 6$ matrix $[S]$ hai jiske rows 1–3 normal rows hain aur rows 4–6 shear rows hain.
$E$ = Young's modulus, $ν$ = Poisson's ratio, $G = μ$ = shear modulus, $λ$ = Lamé's first parameter, $K$ = bulk modulus.
$ε_{k k} = ε_{11} + ε_{22} + ε_{33}$ = volumetric strain (trace).

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

Baayein taraf ki picture dikhati hai ki strain kya preserve karta hai: ek rotation block ko rigidly slide karta hai (angles unchanged, koi stretch nahi) isliye woh koi energy store nahi karta aur discard ho jaata hai; ek shear block ko tilt karta hai (corner angle actually change hota hai) isliye woh strain hai. Right panel Poisson contraction dikhata hai — $x$ mein kheeencho, block $y$ mein patla ho jaata hai.

Yeh figure $6 \times 6$ Voigt compliance matrix $[S]$ ko lay out karta hai: top-left $3 \times 3$ block normal stresses ko normal strains se couple karta hai (wahin $- ν$ Poisson terms rehte hain), jabki bottom-right block diagonal hai — har shear stress sirf apne shear strain ko feed karta hai. Yeh structural separation exactly wahi hai jise neeche ke questions mein "shear rows are decoupled from the normals" kaha gaya hai.

True or false — justify

A single stiffness number like

E

is enough to describe any isotropic 3D material.

False — isotropic solids ko bhi do constants chahiye (

E, ν

λ, μ

), kyunki ek direction mein kheenchne se baaki do bhi Poisson effect ke zariye change hote hain, jise

E

akela encode nahi kar sakta.

The stress tensor has 9 independent components.

False — moment balance

σ_{ij} = σ_{j i}

force karta hai, isliye sirf 6 independent hain; off-diagonal shears equal pairs mein aate hain.

A rigid-body rotation of a loaded part increases its stored elastic energy.

False — displacement gradient ka antisymmetric part pure rotation hai aur

ε

banate waqt discard ho jaata hai, isliye rotation zero strain aur zero elastic energy contribute karta hai.

For a fully anisotropic solid,

C_{ij k l}

has 81 truly independent numbers.

False — do minor symmetries aur major symmetry (energy ke scalar hone se)

81 \to 36 \to 21

kar deti hain; sirf 81 positions hain, 81 freedoms nahi.

The major symmetry

C_{ij k l} = C_{k l ij}

holds for every linear-elastic material regardless of energy considerations.

False — iske liye ek conservative material chahiye jahan

σ_{ij} = \partial W / \partial ε_{ij}

; tab mixed partials commute karte hain. Ek hypothetical non-conservative "linear" material ise violate kar sakta hai.

In pure shear (

τ_{x y} = τ_{0}

only), the material changes volume.

False — compliance matrix

[S]

ke shear rows normal rows se decoupled hain, isliye

ε_{k k} = 0

; shear pure shape change hai, koi dilation nahi.

Hydrostatic pressure produces shear strain.

False — equal normal stresses

σ_{xx} = σ_{y y} = σ_{z z} = - p

(compression negative) sirf equal normal strains aur zero off-diagonal terms dete hain, isliye koi shear nahi, sirf uniform volume change.

If you pull a rod in

x

, its length in

y

genuinely gets shorter.

True —

ν > 0

ke saath,

ε_{y y} = - ν σ_{0} / E < 0

, yeh Poisson contraction hai; rod physically patli ho jaati hai.

The two Lamé parameters

λ

and

μ

are both always positive.

λ

ke liye generally False — stability ke liye

μ = G > 0

hamesha hona chahiye, lekin

λ = \frac{E ν}{( 1 + ν ) ( 1 - 2 ν )}

tab ==negative ho jaata hai jab

ν < 0

== (auxetic materials), jo physically allowed hai.

Spot the error

"Since

τ = G γ

looks just like

σ = E ε

, the compliance entry for shear must be

1/ G

with no factor of 2."

Tensor shear

ε_{x y} = \frac{1}{2} γ_{x y}

hai; compliance entry

2 (1 + ν) / E = 1/ G

tab hi close hoti hai jab Voigt vector mein engineering shear

γ_{x y}

(tensor

ε_{x y}

nahi) rakho.

ε_{k k}

means multiply the three normal strains:

ε_{11} ε_{22} ε_{33}

Nahi — repeated index ek sum hai (Einstein convention), isliye

ε_{k k} = ε_{11} + ε_{22} + ε_{33}

, yeh trace aur fractional volume change hai.

"In

σ_{ij} = λ ε_{k k} δ_{ij} + 2 μ ε_{ij}

, the

λ ε_{k k} δ_{ij}

term contributes to shear stress

σ_{x y}

Kronecker delta

δ_{ij}

i \neq = j

ke liye value zero hota hai, isliye woh term sirf diagonal par rehta hai; shear stresses purely

2 μ ε_{ij}

se aati hain.

"To relate two

3 \times 3

tensors linearly I just need another

3 \times 3

matrix."

Ek matrix vector ko vector mein map karta hai; ek 2-index object ko doosre mein linearly map karne ke liye 4 indices wala object chahiye, isliye

C_{ij k l}

"For a thin pressure-vessel wall, axial strain uses only the axial stress:

ε_{z} = σ_{z} / E

Tumne hoop stress se Poisson pull-in drop kar diya:

ε_{z} = \frac{1}{E} [σ_{z} - ν σ_{θ}]

, jo bolted flanges size karte waqt matter karta hai.

"Because

C_{ij k l} = C_{j ik l}

, we also automatically have

C_{ij k l} = C_{ik j l}

(swap middle indices)."

Nahi — minor symmetries sirf pehle pair mein (

ij

) ya doosre pair mein (

k l

) swap allow karti hain; pairs ke across swap karna koi symmetry nahi hai aur yeh galat hoga.

"Plane strain just means the same thing as plane stress — pick whichever is easier."

Ye dono opposite constraints hain: plane stress ==out-of-plane stress

σ_{z z} = 0

set karta hai (thin plate), plane strain out-of-plane strain

ε_{z z} = 0

== set karta hai (long body); ek ko doosre ki jagah use karna galat stiffness deta hai.

Why questions

Why do only Kronecker-delta combinations appear in the isotropic stiffness tensor?

Isotropy ka matlab hai

C

ko kisi bhi rotation ke baad identical dikhna chahiye; sirf wahi 4th-rank tensors rotation-invariant hain jo

δ_{ij} δ_{k l}

δ_{ik} δ_{j l}

δ_{i l} δ_{j k}

se bane hain.

Why do the last two delta terms merge into a single coefficient

μ

Minor symmetry (

ε_{k l} = ε_{l k}

) force karti hai ki

δ_{ik} δ_{j l}

aur

δ_{i l} δ_{j k}

hamesha saath mein act karein, isliye woh ek constant share karte hain.

Why do we symmetrize the displacement gradient when defining strain?

Antisymmetric part rotation hai (koi stretch nahi, koi energy store nahi); sirf symmetric part lengths change karta hai, isliye strain sirf wahi rakhta hai.

Why does energy conservation give the major symmetry?

Stress stored energy ka gradient hai

σ_{ij} = \partial W / \partial ε_{ij}

, isliye

C_{ij k l} = \partial^{2} W / \partial ε_{ij} \partial ε_{k l}

; mixed partials commute karte hain, jo

C_{ij k l} = C_{k l ij}

force karta hai.

Why can't

ν

exceed

\frac{1}{2}

for a stable ordinary material?

Tab

K = \frac{E}{3 ( 1 - 2 ν )}

negative ho jaata, matlab material compression mein expand karta — ek self-amplifying instability.

Why does the shear modulus

G

, not

E

, govern the twist of a spacecraft drive shaft?

Twisting ek pure shear deformation hai; shear compliance

G = E / [2 (1 + ν)]

par depend karta hai, jo normal-stress response se independent hai.

Why do composite panels need up to 21 constants while aluminium needs 2?

Composites mein direction-dependent stiffness hoti hai (koi rotational symmetry nahi), isliye unka

C_{ij k l}

saare 21 anisotropic constants rakhta hai; isotropy ise 2 par collapse kar deta hai.

Why is the distinction between

ε_{ij}

and

γ_{ij}

worth policing so carefully?

Kyunki

γ_{ij} = 2 ε_{ij}

; inhe silently mix karna tumhara shear stress double ya half kar deta hai, jo kisi bhi von Mises check ko corrupt karta hai jo normal aur shear terms combine karta hai.

Edge cases

What happens to

K

ν \to \frac{1}{2}

K = \frac{E}{3 ( 1 - 2 ν )} \to \infty

: material incompressible ho jaata hai (rubber ki tarah), kisi bhi volume change ko resist karta hai jabki freely shear karta rehta hai.

What does

ν = 0

physically mean?

x

mein kheenchne se koi lateral contraction nahi hoti (

ε_{y y} = ε_{z z} = 0

); cork iska approximation hai, tab useful jab aap nahi chahte ki ek stretched part apne neighbours ko pinch kare.

Is a negative

ν

physically allowed, and what is such a material called?

Haan — stability ke liye

ν > - 1

tak; ye auxetic materials hain jo stretch karne par sideways mote ho jaate hain (kuch foams aur lattices).

What does the uniaxial load (

σ_{xx} = σ_{0}

, rest zero) reduce the 3D law to?

Cross terms vanish ho jaate hain, 1D Hooke

ε_{xx} = σ_{0} / E

recover hota hai, jabki lateral

ε_{y y} = ε_{z z} = - ν σ_{0} / E

bhi predict hota hai.

For plane stress in a thin vessel wall, which stress is set to zero and why?

==Out-of-plane (through-thickness) stress

σ_{z z} \approx 0

==, kyunki ek thin wall ke paas apni choti thickness ke across push back karne ke liye kuch nahi hota.

In plane strain, which quantity is zero and when does it apply?

==Out-of-plane strain

ε_{z z} = 0

==, jo long bodies (dam, buried pipe, long weld) par apply hota hai jinke ends axially move nahi kar sakte; note karo ki yeh ek non-zero constraint stress

σ_{z z} = ν (σ_{xx} + σ_{y y})

force karta hai.

Why does plane strain generate an out-of-plane stress even though the strain there is zero?

ε_{z z} = 0

ke saath material

z

mein Poisson-contract karna chahta hai lekin roka jaata hai, isliye constraint ==

σ_{z z} = ν (σ_{xx} + σ_{y y})

== ke roop mein push back karta hai.

At the incompressible limit, what happens to Lamé's

λ

λ = \frac{E ν}{( 1 + ν ) ( 1 - 2 ν )} \to \infty

jab

ν \to \frac{1}{2}

, isliye standard displacement finite elements incompressibility ke paas "lock" ho jaate hain aur special formulations chahiye hote hain.

Recall Ek-line self-test

Upar ke saare answers cover karo aur sirf "Spot the error" section dobara run karo — ye woh traps hain jo exam ya design review mein sabse zyada bite karte hain.