Modulus of Elasticity (original) (raw)

Last Updated : 18 May, 2026

Modulus of elasticity is a fundamental material property that quantifies a material’s stiffness or resistance to elastic deformation when subjected to stress. It is defined as the ratio of stress (σ) to the corresponding strain (ε) within the elastic limit of the material, expressed as:

E = \frac{Stress (σ)}{Strain (ε)}

This modulus indicates how much a material will deform under a given load and helps predict its behavior under mechanical forces. A higher modulus of elasticity means the material is stiffer and deforms less, while a lower modulus indicates it is more flexible.

**Examples: Steel is a common example of a material with a high modulus of elasticity, widely used in construction due to its strength and durability. It resists deformation and maintains its shape under stress, allowing it to bear heavy loads without bending or breaking. Steel is relatively brittle, so safety precautions are necessary during use.

Some examples of elastic modulus values for common materials:

Calculate Modulus of Elasticity

Modulus of Elasticity, in terms of the stress-strain curve, is the slope of the stress-strain curve in the region of elastic behavior, where stress is linearly proportional to strain.

stress_and_strain_curve

Modulus of elasticity is the slope of the stress-strain curve and represents the material's stiffness in the elastic region, which extends from 0 up to the yield strength. The yield strength is the stress required to produce a small amount of plastic deformation. Beyond this elastic region the material enters the plastic deformation region, where permanent changes in shape occur. The ultimate strength is the maximum stress a material can withstand before failure. Once this point is exceeded, the material can no longer support the load and breaks.

Determining the Modulus of Elasticity of Wire

Determining the Young's modulus (also known as the modulus of elasticity) of a material typically involves performing a tensile test on a wire or specimen.

Formula

The formula of Elastic Modulus is given as the ratio of stress and strain

\boxed {E = \fracσ ε}

where

We can also define the modulus of elasticity by

E=\frac{\text Stress} {\text Strain} = \frac{\frac F A}{\frac {\Delta l} {l} }

\boxed {E = \frac{F \times L}{A \times \Delta L}}

where

The dimensional formula for modulus of elasticity can be written as **[M 1 L -1 T -2 ]

Modulus of Elasticity vs Modulus of Rigidity

Modulus of Elasticity Modulus of Rigidity
Calculate the deformations of an object when the deforming force is applied parallel to the surface. Calculate the deformation of an object when the deforming force is tangential to the surface.
Denoted by E denoted by G
E = Stress/Strain G = Shear Stress/Shear Strain
stretching or compressing type of deformation sliding or twisting type of deformation
E = 3K(1 - 2v) G = E/(2 + 2v)

Elastic Constants

Elastic constants are used to determine the deformation produced by stress. They are used to obtain the relationship between stress and strain. There are basically 4 types of elastic constants that we use, namely:

frcergrdfttfrgt

A. Young’s Modulus (E)

It is the ratio of stress to strain within the elastic limit of a material, and it measures the material’s ability to resist deformation under tensile or compressive forces.

In formula form:

Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L}

Where:

**B. Bulk modulus (K)

It is a measure of a material's resistance to uniform compression. In other words, it tells us how difficult it is to compress a substance.

A material's volume tends to decrease when exposed to external pressure. Bulk modulus relates the applied pressure to the fractional change in volume.

B = \frac{\text{Applied Pressure}}{\text{Fractional Change in Volume}}

= \frac{\Delta P}{\frac{\Delta V}{V}}

Where:

The negative sign is often used to indicate that volume decreases under pressure.

B = -V \frac{\Delta P}{\Delta V}

C. Shear Modulus (G)

Shear modulus, also called the modulus of rigidity, measures a material's resistance to shear deformation. It describes how a material changes shape when a force is applied parallel to its surface, without changing its volume. It tells us how stiff a material is against forces that try to make it slide.

G = \frac{\text{Shear Stress}}{\text{Shear Strain}}

= \frac{F/A}{\Delta x / h}

= \frac{F \cdot h}{A \cdot \Delta x}

Relation with Young's Modulus and Poisson's Ratio

G = \frac{Y}{2(1 + \sigma)}

D. Poisson’s ratio (**µ)

When a material is stretched or compressed along one axis, it tends to contract or expand along the perpendicular axes. Poisson's ratio (σ) is defined as the ratio of the lateral (transverse) strain to the longitudinal (axial) strain.

\sigma = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}

= \frac{\Delta d / d}{\Delta L / L}

Relation Between Elastic Constants

The relationship between these constants can be known by the given expression:

\frac{1}{K} - \frac{3}{G} = \frac{9}{E}

Where,

Applications

Modulus of elasticity is used in day-to-day life in many ways. It is used in engineering as well as medical science.

Solved Problems

**Question 1. A wire is 2 m long and has a cross-sectional area of 10⁻⁶ m. 2 A load of 980 N is suspended. Calculate the stress, the strain, and the energy stored in the wire

**Solution: Given: Y = 12 × 1010 N m-2

Stress = F / A = 980 / 10-6 = 98 × 107 N m-2

Strain = Stress / Y = 98 × 107 / 12 × 1010 = 8.17 × 10-3

Energy = 1/2 × (stress × strain) × volume = 1/2 × (98 × 107 × 8.17 × 10-3) × 2 × 10-6 = 8 Joules.

**Example 2. A metallic cube with a side of 0.20 m undergoes a shearing force of 1000 N. The top surface is displaced by 0.40 cm with respect to the bottom. Find the elasticity of the metal's shear modulus.

**Solution: Here, L = 0.20 m, F = 1000 N, x = 0.40 cm = 0.004 m and

Area A = L2= 0.04 m2

Therefore, Shear modulus = (1000/ 0.04) × (0.20 / 0.004) = 1.25 × 106 N m-2.

**Example 3. What must be the elongation of a wire 5 m long so that the strain is 1% of 0.1? If the wire has a cross-section of 1 mm 2 and is stretched by 10 kg, what is the stress?

**Solution: L = 5m, Strain = 1% of 0.1 = 1 × 10-2 × 0.1 = 1 × 10-3, Area of cross-section = 1 mm² = 1 × 10-6 m²

F = 10 kg= 10 × 9.8 N

Extension = Stress × L = 5mm

Stress = Force / Area = (10 × 9.8) / (1 × 10-6) = 9.8 × 107 N/m².

**Question 4: A steel wire 3 m long and cross-sectional area 2×10⁻⁶ m² is subjected to a load of 600 N; find the stress, strain, and extension if Young’s modulus is 2 × 10¹¹ N/m².

**Solution: Given

L = 3 m

A = 2 × 10-6 m²

F = 600 N

Y = 2 × 1011 N/m²

Stress

\text{Stress} = \frac{F}{A}

= \frac{600}{2 \times 10^{-6}}

= 3 \times 10^8 \,\text{N/m²} \\[1mm]

Strain

\text{Strain} = \frac{\text{Stress}}{Y}

= \frac{3 \times 10^8}{2 \times 10^{11}}

= 1.5 \times 10^{-3} \\[1mm]

Extension

\Delta L = \text{Strain} \times L = 1.5 \times 10^{-3} \times 3

= 0.0045\,\text{m} = 4.5\,\text{mm} \\[1mm]

Unsolved Problems

**Question 1: A rod 10 m long has a cross-sectional area of 1.5 × 10⁻⁴ m. It is subjected to a 10-kg load. If Young’s modulus of the material is 4 × 1010 N/m2, calculate the elongation produced in the wire.

**Question 2: A metallic cube of side 100 cm is subjected to a uniform force acting normal to the whole surface of the cube. The pressure is 106 pascal. If the volume changes by 1.25 x 10⁻³ m3, calculate the bulk modulus of the material.

**Question 3: A metal cube of side 0.45 m is subjected to a shearing force of 8000 N. The top surface is displaced by 0.30 cm with respect to the bottom. Calculate the shear modulus of metal.

**Question 4: A mild steel wire of radius 0.5 mm and length 3 m is stretched by a force of 49 N. Calculate a) longitudinal stress, b) longitudinal strain, and c) elongation produced in the body if Y for steel is 2.1 × 1011 N/m2.

**Question 5: A steel wire of length 4 m and cross-sectional area 2 × 10⁻⁶ m² is subjected to a tensile force of 1200 N; find the stress, strain, and extension if Young’s modulus is 2 × 10¹¹ N/m².