Elastic Modulus & Energy

Calculate stress, deformation and elastic energy via Hooke''s Law (σ = Eε, τ = Gγ, U = W²l/2AE).

Material Preset:

Cross-sectional Area:

Inputs

W

Applied Force

l

Length

d

Diameter

E

Young's Modulus

GPa

Formula Interpretation

Normal Stress — Formula ①

\sigma = E\varepsilon = \dfrac{W \cdot l}{A \cdot \lambda} \quad \text{(MPa)}

$E$ is Young's modulus (GPa); $\varepsilon$ is axial strain; $W$ is force (N); $l$ is length (mm); $A$ is area (mm²); $\lambda$ is deformation (mm).

Shear Stress — Formula ②

\tau = G\gamma = \dfrac{W_s \cdot l}{A \cdot \lambda_s} = \dfrac{W_s}{A\varphi} \quad \text{(MPa)}

$G$ is shear modulus (GPa); $\gamma$ is shear strain; $W_s$ is shear force (N); $\varphi$ is shear angle (rad).

Elastic Energy — Formula ③

U = \dfrac{\sigma^2}{2E}\,Al = \dfrac{1}{2}W\lambda = \dfrac{W^2 l}{2AE} \quad \text{(N·m)}

$U$ is elastic energy stored (N·m); $\sigma$ is normal stress (MPa); $E$ is Young's modulus (GPa); $W$ is force (N); $\lambda$ is deformation (mm).

Knowledge Points

Young''s Modulus E

Young''s modulus E (longitudinal elastic modulus) is the ratio of normal stress to axial strain: E = σ/ε. It is a material constant that quantifies stiffness under axial loading. Higher E means greater resistance to deformation.

Shear Modulus G

Shear modulus G (transverse elastic modulus) is the ratio of shear stress to shear strain: G = τ/γ. It quantifies resistance to shear deformation. For isotropic materials, G = E / [2(1+ν)].

Elastic Strain Energy

When a body deforms elastically under load, work is stored as elastic strain energy U = ½Wλ. This energy is fully recoverable upon unloading within the elastic limit. Energy density is u = σ²/(2E) per unit volume.

Worked Example

A tensile force of $W = 80\,\text{kN}$ is applied to a mild steel rod with diameter $d = 25\,\text{mm}$ and length $l = 1\,\text{m}$ . Given Young's modulus $E = 200\,\text{GPa}$ , find the normal stress, deformation, and elastic energy stored in the rod.

Step 1 — Compute cross-sectional area

A = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4} \times 25^2 = \dfrac{625\pi}{4} \approx 490.87\,\text{mm}^2

Step 2 — Compute normal stress σ

\sigma = \dfrac{W}{A} = \dfrac{80{,}000}{490.87} \approx 163.0\,\text{MPa}

Step 3 — Compute deformation λ = Wl/(AE)

\lambda = \dfrac{W \cdot l}{A \cdot E} = \dfrac{80{,}000 \times 1000}{490.87 \times 200{,}000} \approx 0.815\,\text{mm}

Step 4 — Compute elastic energy U = W²l/(2AE)

U = \dfrac{W^2 l}{2AE} = \dfrac{(80 \times 10^3)^2 \times 1000}{2 \times 490.87 \times 200{,}000} \approx 32.6\,\text{N·m}

The elastic energy stored is $U \approx 32.6\,\text{N·m}$ .

Extended Knowledge

Material	E (GPa)	G (GPa)	ν
Low Carbon Steel	206	82	0.29
High Carbon Steel	200	78	0.28
Cast Iron	157	61	0.26
Copper	123	46	0.34
Brass	100	37	0.35
Titanium	103	—	—
Aluminum	73	26	0.34
Hard Aluminum	72	27	0.34
Glass	71	29	0.35
Concrete	20	—	0.2

•The relationship G = E/[2(1+ν)] links the three elastic constants for isotropic materials. Knowing any two determines the third.
•Elastic energy per unit volume (strain energy density) u = σ²/(2E) is used in fracture mechanics to determine the energy release rate at crack tips.
•In spring design, the elastic energy storage capacity determines how much energy a spring can absorb or release — critical for shock absorbers and valve springs.
•Material selection based on specific stiffness (E/ρ, where ρ is density) is important in aerospace where minimum weight with maximum stiffness is required.
•Beyond the elastic limit, energy is dissipated as heat and permanent deformation occurs. The elastic–plastic boundary defines the material''s yield strength.

Material

E (GPa)

G (GPa)

Low Carbon Steel

206

0.29

High Carbon Steel

200

0.28

Cast Iron

157

0.26

Copper

123

0.34

Brass

100

0.35

Titanium

103

—

Aluminum

0.34

Hard Aluminum

0.34

Glass

0.35

Concrete

—

0.2