Hooke's law and elastic constants E, G, nu

Hooke’s law states that within the proportional limit, stress is proportional to strain: and. The elastic constants,,, and are related by (RK Bansal / GATE SOM).

State Hooke’s law for uniaxial loading and name the constant of proportionality.

Within the proportional limit, stress is proportional to strain:. The constant is Young’s modulus (modulus of elasticity).

Write Hooke’s law in shear form.

where is the modulus of rigidity (shear modulus) and is the shear strain (in radians).

Why does Hooke’s law fail beyond the elastic limit?

Dislocations and permanent slip make the – relation nonlinear; unloading does not return to the origin. The material is no longer uniquely described by a single.

How is obtained experimentally from a tension test?

From the initial linear portion of the engineering stress–strain curve: slope (secant/tangent in the proportional range).

Relate,, and Poisson’s ratio.

. Measuring any two of determines the third for isotropic materials.

What is the difference between elastic limit and proportional limit?

Proportional limit: end of linearity. Elastic limit: maximum stress with no permanent set on unloading. They are close for many metals; proportional limit elastic limit.

Express strain energy density for uniaxial Hookean behaviour.

. This is the area under the linear – line.

If, what does Hooke’s law imply for lateral strain under uniaxial load?

Lateral strain is zero — no Poisson effect; when only acts.

True/False: Hooke’s law is valid up to ultimate tensile strength.

False — only up to the proportional/elastic limit; beyond that the relation is nonlinear.

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Study notes

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Qwestrum Engineering360 · Mechanical Engineering · Strength of Materials (SOM)

Hooke's Law & Elastic Constants

Hooke’s law states that within the proportional limit, stress is proportional to strain:

\sigma=E\varepsilon

and

\tau=G\gamma

. The elastic constants

E

G

K

, and

\nu

are related by

E=2G(1+\nu)=3K(1-2\nu)

(RK Bansal / GATE SOM).

Exam tip: keep units consistent — N with mm² gives MPa directly (1 MPa = 1 N/mm²).

Key formulas & points

Skim these first — then read the full notes below.

Uniaxial Hooke: $\sigma=E\varepsilon$ , $\delta=\dfrac{PL}{AE}$ .
Shear Hooke: $\tau=G\gamma$ ; $G$ = modulus of rigidity (shear modulus).
Bulk modulus: $K=-\dfrac{p}{\varepsilon_v}$ (hydrostatic pressure vs volumetric strain).
Poisson: $\nu=-\varepsilon_{\mathrm{lat}}/\varepsilon_{\mathrm{long}}$ .
Interrelations: $E=2G(1+\nu)$ , $E=3K(1-2\nu)$ , $E=\dfrac{9KG}{3K+G}$ .
Generalised Hooke (3-D isotropic): $\varepsilon_x=\dfrac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)]$ .
Only two independent elastic constants for isotropic linear elasticity.

Topic details

Definition and physical meaning

Hooke’s law (uniaxial): within the proportional limit, normal stress is linearly proportional to normal strain:

\sigma=E\varepsilon\qquad\Rightarrow\qquad E=\frac{\sigma}{\varepsilon}

E

is Young’s modulus (modulus of elasticity) — slope of the initial straight portion of the

\sigma

–

\varepsilon

curve.

Shear form:

\tau=G\gamma\qquad\Rightarrow\qquad G=\frac{\tau}{\gamma}

G

is the modulus of rigidity (shear modulus).

Bulk form (hydrostatic pressure

p=-\sigma_h

K=-\frac{p}{\Delta V/V}=-\frac{p}{\varepsilon_v}

Physical parameters

Symbol	Meaning	SI unit
$E$	Young’s modulus	$\mathrm{Pa}$ (often GPa)
$G$	Shear modulus	$\mathrm{Pa}$
$K$	Bulk modulus	$\mathrm{Pa}$
$\nu$	Poisson’s ratio	—
$\sigma$ , $\tau$	Normal / shear stress	$\mathrm{Pa}$
$\varepsilon$ , $\gamma$	Normal / shear strain	—
$\varepsilon_v$	Volumetric strain	—

Typical values (steel):

E\approx 200\,\mathrm{GPa}

G\approx 80\,\mathrm{GPa}

\nu\approx 0.3

K\approx 140

–

160\,\mathrm{GPa}

Fig 1.3 — Blue: ductile (mild steel). Red: brittle (cast iron). Ductile: large plastic strain. Brittle: fracture near σᵤ with minimal elongation. % elongation and cup-cone fracture distinguish ductile break in lab tests.

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Core assumptions (state these in exams)

1. Linear elastic behaviour — stress ∝ strain (proportional limit not exceeded).
2. Isotropic material — same

E

G

\nu

in all directions.
3. Homogeneous continuum.
4. Small strains — geometry linearised; superposition valid.
5. Isothermal (or constants measured at the working temperature).
6. No time-dependent effects (creep/relaxation neglected).

If the material is anisotropic (composites, wood), a stiffness matrix with more independent constants is required — elementary Hooke formulas do not apply directly.

Derivation summary — elongation of a prismatic bar

Consider a uniform bar of length

L

, area

A

, under axial load

P

\sigma=\frac{P}{A},\qquad\varepsilon=\frac{\delta}{L}

Hooke:

\sigma=E\varepsilon

⇒

\frac{P}{A}=E\frac{\delta}{L}\qquad\Rightarrow\qquad\delta=\frac{PL}{AE}

Strain energy (elastic, gradually applied load):

U=\frac{1}{2}P\delta=\frac{P^2 L}{2AE}=\frac{\sigma^2}{2E}\cdot(AL)

Resilience (strain energy density)

u=\sigma^2/(2E)

; proof resilience uses

\sigma_y

Relations among E, G, K, and ν

For an isotropic linear elastic solid, only two constants are independent. Standard identities:

Formulas (Indian textbook notation)

$E=2G(1+\nu)$
$E=3K(1-2\nu)$
$E=\frac{9KG}{3K+G}$
$\nu=\frac{3K-2G}{2(3K+G)}$
$G=\frac{E}{2(1+\nu)},\qquad K=\frac{E}{3(1-2\nu)}$

Derivation sketch of

E=2G(1+\nu)

: pure shear is equivalent to equal tension and compression at

45^\circ

. Relating shear strain

\gamma

to principal strains and using

\tau=G\gamma

with

\sigma=E\varepsilon

yields the identity.

Exam tip: Given any two of

\{E,G,K,\nu\}

, compute the rest. Check

\nu<0.5

so that

K>0

Generalised Hooke’s law (3-D)

For principal (or Cartesian) stresses on an isotropic body:

\varepsilon_x=\frac{1}{E}\big[\sigma_x-\nu(\sigma_y+\sigma_z)\big]

\varepsilon_y=\frac{1}{E}\big[\sigma_y-\nu(\sigma_z+\sigma_x)\big]

\varepsilon_z=\frac{1}{E}\big[\sigma_z-\nu(\sigma_x+\sigma_y)\big]

Shear strains decouple:

\gamma_{xy}=\frac{\tau_{xy}}{G},\quad\gamma_{yz}=\frac{\tau_{yz}}{G},\quad\gamma_{zx}=\frac{\tau_{zx}}{G}

Volumetric strain:

\varepsilon_v=\varepsilon_x+\varepsilon_y+\varepsilon_z=\frac{1-2\nu}{E}(\sigma_x+\sigma_y+\sigma_z)=\frac{3(1-2\nu)}{E}\sigma_{\mathrm{mean}}

with

\sigma_{\mathrm{mean}}=(\sigma_x+\sigma_y+\sigma_z)/3

Step-by-step problem approach

1. Confirm loading is within the elastic range (or problem states elastic constants apply).
2. Identify known constants; pick the relation that solves for the unknown directly.
3. For

\delta

: use

\delta=PL/(AE)

with consistent units.
4. For 3-D stress: write generalised Hooke component-wise; sum for

\varepsilon_v

if needed.
5. Convert GPa ↔ MPa ↔ Pa carefully (

1\,\mathrm{GPa}=10^3\,\mathrm{MPa}=10^9\,\mathrm{Pa}

).
6. Sanity-check:

G\approx E/2.6

for

\nu\approx 0.3

;

K

should exceed

\frac{E}{3}

Common mistakes in exams

• Using

E=2G(1-\nu)

instead of

E=2G(1+\nu)

.
• Forgetting that

\nu

is dimensionless while mixing

E

in GPa with

\sigma

in MPa.
• Applying

\delta=PL/AE

beyond the elastic limit.
• Using

K=E/[3(1+\nu)]

(wrong sign/factor) instead of

3(1-2\nu)

.
• Treating

E

and

G

as independent for isotropic materials when a third constant is also freely chosen (over-constrained).
• Confusing modulus of rigidity

G

with bulk modulus

K

Calculator

Axial deformation

\delta = \frac{P L}{A E}

Load PNLength LmmArea Amm²Modulus EN/mm²

Result

0.9524mm

δ = PL/(AE) = (80000×1000) / (400×2.100e+5) = 0.9524 mm

Worked examples

Try the problem first — open the solution when you are ready to check.

Find G and K from E and ν

Problem

For steel,

E=210\,\mathrm{GPa}

and

\nu=0.3

. Find

G

and

K

Solution

Formulas (Indian textbook notation)

$G=\frac{E}{2(1+\nu)}=\frac{210}{2(1.3)}=\frac{210}{2.6}=80.77\,\mathrm{GPa}$

Formulas (Indian textbook notation)

$K=\frac{E}{3(1-2\nu)}=\frac{210}{3(1-0.6)}=\frac{210}{3\times 0.4}=\frac{210}{1.2}=175\,\mathrm{GPa}$

Check:

E=2G(1+\nu)=2\times 80.77\times 1.3\approx 210\,\mathrm{GPa}

Elongation using Hooke’s law

Problem

A brass bar (

E=100\,\mathrm{GPa}

) has

A=500\,\mathrm{mm^2}

L=2\,\mathrm{m}

, and axial load

P=50\,\mathrm{kN}

. Find

\delta

and strain energy

U

Volumetric strain under hydrostatic pressure

Problem

A solid copper sphere has

K=140\,\mathrm{GPa}

. It is subjected to hydrostatic pressure

p=70\,\mathrm{MPa}

. Find volumetric strain and percentage change in volume.

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

1
State Hooke’s law for uniaxial loading and name the constant of proportionality.
Model answer
Within the proportional limit, stress is proportional to strain: $\sigma=E\varepsilon$ . The constant $E$ is Young’s modulus (modulus of elasticity).
2
Write Hooke’s law in shear form.
Model answer
$\tau=G\gamma$ where $G$ is the modulus of rigidity (shear modulus) and $\gamma$ is the shear strain (in radians).
3
What is the generalized Hooke’s law for a 3-D isotropic linear elastic solid?
Model answer
$\varepsilon_x=\dfrac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)]$ and cyclic for $\varepsilon_y,\varepsilon_z$ ; shear: $\gamma_{xy}=\tau_{xy}/G$ etc.
4
Why does Hooke’s law fail beyond the elastic limit?
Model answer
Dislocations and permanent slip make the $\sigma$ – $\varepsilon$ relation nonlinear; unloading does not return to the origin. The material is no longer uniquely described by a single $E$ .
5
How is $E$ obtained experimentally from a tension test?
Model answer
From the initial linear portion of the engineering stress–strain curve: slope $E=\Delta\sigma/\Delta\varepsilon$ (secant/tangent in the proportional range).
6
Relate $E$ , $G$ , and Poisson’s ratio $\nu$ .
Model answer
$E=2G(1+\nu)$ . Measuring any two of $E,G,\nu$ determines the third for isotropic materials.
7
What is the difference between elastic limit and proportional limit?
Model answer
Proportional limit: end of linearity ( $\sigma\propto\varepsilon$ ). Elastic limit: maximum stress with no permanent set on unloading. They are close for many metals; proportional limit $\le$ elastic limit.
8
Express strain energy density for uniaxial Hookean behaviour.
Model answer
$u=\dfrac{1}{2}\sigma\varepsilon=\sigma^2/(2E)=E\varepsilon^2/2$ . This is the area under the linear $\sigma$ – $\varepsilon$ line.
9
State Hooke’s law for plane stress ( $\sigma_z=0$ ).
Model answer
$\varepsilon_x=(\sigma_x-\nu\sigma_y)/E$ , $\varepsilon_y=(\sigma_y-\nu\sigma_x)/E$ , $\varepsilon_z=-\nu(\sigma_x+\sigma_y)/E$ , $\gamma_{xy}=\tau_{xy}/G$ .
10
State Hooke’s law for plane strain ( $\varepsilon_z=0$ ).
Model answer
$\sigma_z=\nu(\sigma_x+\sigma_y)$ appears. Effective relations use $E'=E/(1-\nu^2)$ and $\nu'=\nu/(1-\nu)$ in 2-D form.
11
Can Hooke’s law be written in terms of bulk modulus?
Model answer
Yes for hydrostatic loading: $p=-K(\Delta V/V)$ . Also $E=3K(1-2\nu)$ links $E$ , $K$ , and $\nu$ .
12
What does a nonlinear but elastic material imply for Hooke’s law?
Model answer
Stress and strain remain uniquely related and recoverable, but not linear — classical Hooke’s law ( $\sigma=E\varepsilon$ ) does not hold; a nonlinear constitutive law is needed.
13
Why is $E$ nearly the same in tension and compression for mild steel in the elastic range?
Model answer
Atomic bond stiffness is essentially the same for small tension/compression; the initial slope of $\sigma$ – $\varepsilon$ is nearly identical until yielding differs.
14
How does temperature typically affect $E$ for metals?
Model answer
Young’s modulus decreases with rising temperature as interatomic forces soften. Design at elevated temperature must use $E(T)$ from data sheets.
15
Write the compliance form: strain in terms of stress for isotropic material (summary).
Model answer
Normal strains depend on all three normal stresses via $E$ and $\nu$ ; each shear strain depends only on its conjugate shear stress via $G$ . No coupling between shear and normal for isotropic linear elasticity.

Exams & GATE

1
Textbook: RK Bansal Ch. 2–3.
2
Memorise $E=2G(1+\nu)$ and $E=3K(1-2\nu)$ — extremely frequent GATE numericals.
3
Units: $E$ , $G$ , $K$ in GPa or MPa; $\nu$ is dimensionless.
4
State “within elastic/proportional limit” in every derivation.

📖 Standard books (India)

Strength of Materials — RK Bansal
Read: Ch. 2–3
SOM — beams, torsion, columns, and deflection

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📚Key formulas & points

Topic details

Definition and physical meaning

Core assumptions (state these in exams)

Derivation summary — elongation of a prismatic bar

Relations among E, G, K, and ν

Generalised Hooke’s law (3-D)

Step-by-step problem approach

Common mistakes in exams

Axial deformation

Worked examples

Find G and K from E and ν

Elongation using Hooke’s law

Volumetric strain under hydrostatic pressure

Practice questions

📝Exams & GATE

📖 Standard books (India)

Explore related topics

See real mechanical engineering careers

Key formulas & points

Exams & GATE