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Qwestrum Engineering360 · Civil Engineering · Strength of Materials

Elastic Constants

Remember the two linking identities E = 2G(1 + μ) and E = 3K(1 − 2μ); given any two elastic constants you can solve for the other two by simple substitution.
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.

  • Only two of E, G, K, μ are independent for isotropic materials
  • For steel: E ≈ 200 GPa, μ ≈ 0.3
  • Compliance is inverse of stiffness (1/E, 1/G)

Topic details

Introduction

Elastic constants questions are almost always short: given E and μ, find G and K, or given E and G, back-calculate μ. Because an isotropic material has only two independent constants, the whole topic reduces to the two relations that connect E, G, K and μ.

Scope in B.Tech and GATE syllabus

The common slip is a sign or bracket error in E = 3K(1 − 2μ); as μ approaches 0.5 the (1 − 2μ) term goes to zero, which is why an incompressible material (rubber) has a very large bulk modulus. Keeping this limiting behaviour in mind is a quick sanity check on your answer.

Why this topic matters in practice

Universities also ask for the physical meaning of each modulus, so pair the numerical with one line stating that E governs axial stiffness, G governs shear/twist and K governs volumetric response.

Key relations & formulas

E=σεE = \frac{\sigma}{\varepsilon}
(Young modulus)
G=τγG = \frac{\tau}{\gamma}
(modulus of rigidity)

Formulas (Indian textbook notation)

  • E=2G(1+μ)=3K(1−2μ)E = 2G(1 + \mu) = 3K(1 - 2\mu)
K=E/[3(1−2μ)]K = E / [3(1 - 2\mu)]
(bulk modulus)

Notation and sign conventions

Relation 1 —
E=σεE = \frac{\sigma}{\varepsilon}
E=σεE = \frac{\sigma}{\varepsilon}
(Young modulus)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
G=τγG = \frac{\tau}{\gamma}
G=τγG = \frac{\tau}{\gamma}
(modulus of rigidity)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
E=2GE = 2G

Formulas (Indian textbook notation)

  • E=2G(1+μ)=3K(1−2μ)E = 2G(1 + \mu) = 3K(1 - 2\mu)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 4 —
K=E/[3K = E / [3
K=E/[3(1−2μ)]K = E / [3(1 - 2\mu)]
(bulk modulus)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

Young’s modulus E is the slope of the linear part of the stress-strain curve and represents resistance to axial deformation. Modulus of rigidity G plays the same role for shear, relating shear stress to shear strain, and it governs the torsional stiffness of shafts.

Governing relations in practice

Bulk modulus K relates hydrostatic (all-round) pressure to volumetric strain and therefore describes how a material resists uniform compression. Poisson’s ratio μ is the coupling constant that ties lateral to longitudinal strain and appears in both bridging relations.

Design and analysis considerations

The identity E = 2G(1 + μ) comes from resolving a pure shear state into equivalent tension and compression at 45°, while E = 3K(1 − 2μ) follows from summing the three principal strains under hydrostatic stress. Together they mean that once two constants are known the material is fully characterised elastically.

Advanced theory and extensions

For structural steel the standard set is E ≈ 200 GPa, μ ≈ 0.3, giving G ≈ 77 GPa and K ≈ 167 GPa — memorising these lets you instantly check whether a computed constant is physically reasonable.

Assumptions and validity limits

State assumptions explicitly before using any relation for elastic constants — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Mechanics of Materials (Civil) viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Mechanics of Materials (Civil) papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to elastic constants.
4. Use equation 1:
E=σεE = \frac{\sigma}{\varepsilon}
.
5. Use equation 2:
G=τγG = \frac{\tau}{\gamma}
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Elastic Constants appears in beams, slabs, and columns. In Indian civil curricula this topic is tested because it connects theory to stress and deformation in civil structures.
GATE and semester exams often combine elastic constants with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use elastic constants?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Assuming all four constants are independent and trying to “look up” a fourth value instead of deriving it.
• Dropping a factor in E = 3K(1 − 2μ) and getting K smaller than E, which is impossible for μ = 0.3.
• Forgetting μ is dimensionless and inserting it in percent.
• Rounding G before using it downstream, propagating error into torsion problems.

Quick revision checklist

Before attempting elastic constants problems, confirm you can:
1. Only two of E, G, K, μ are independent for isotropic materials
2. For steel: E ≈ 200 GPa, μ ≈ 0.3
3. Compliance is inverse of stiffness (1/E, 1/G)
Revise the solved examples in Strength of Materials — RK Bansal and one previous-year GATE or university paper for this unit.

Worked examples

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

Finding G and K from E and Poisson’s ratio

Problem

For a steel specimen E = 200 GPa and μ = 0.3. Determine the modulus of rigidity G and the bulk modulus K.

Solution

From E = 2G(1 + μ): G = E / [2(1 + μ)] = 200 / (2 × 1.3) = 76.9 GPa. From E = 3K(1 − 2μ): K = E / [3(1 − 2μ)] = 200 / (3 × 0.4) = 166.7 GPa. Both results exceed the typical textbook values only marginally and satisfy K > E > G, confirming consistency for a normal metal with μ = 0.3.

Conceptual check — Elastic Constants

Problem

In a Mechanics of Materials (Civil) semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of elastic constants." What should a complete answer include?

Exams & GATE

Memorise E–G–μ relation — common 2-mark derivation in university papers.

📖 Standard books (India)

  • Strength of Materials — RK Bansal

    Read: Syllabus unit

    SOM — beams, torsion, columns, and deflection

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