Young's Modulus

Young’s Modulus is the ratio of stress to strain within the elastic limit of a material. It is named after the British physicist Thomas Young and is also called the Modulus of Elasticity.

It tells us how stiff a material is — that is, how much it resists stretching or compressing when a force is applied.

When a material is stretched by a force, it experiences:

• Stress = Force per unit area
• Strain = Change in length / Original length

Within the elastic limit, the material follows a linear stress–strain relationship (Hooke’s Law region). In this region, the material returns to its original shape once the force is removed.

Y =\frac{Stress}{Strain}= \frac{σ} {ϵ}

Y= \frac{\frac{F}{A}}{\frac{\Delta L}{L}} =\frac {F.L}{A. \Delta L}

where ,
Y is Young’s Modulus of the material
σ is the stress applied to the material
ϵ is the strain corresponding to the applied stress

Units and Dimensions

• The SI unit for Young's Modulus is the Pascal (Pa), which is equivalent to Newton per square meter (N/m²).
• The dimensional formula for Young’s Modulus is [ML⁻¹T⁻²], representing mass (M), length (L), and time (T) in terms of their respective powers.
• The values of Young's Modulus are often expressed in megapascals (MPa), gigapascals (GPa), Newtons per square millimeter (N/mm²), or kilonewtons per square millimeter (kN/mm²).

Derivation of Young's Modulus

This derivation below illustrates how Young's modulus (Y) is calculated by relating stress (σ) and strain (ϵ) :

We know that,

Y =\frac{ σ}{ϵ}...(1)

Also,

σ =\frac{ F}{A} = \frac{\Delta L}{L_0}

Putting these values in eq(1)

Y =\frac{ σ}{ϵ}

   =\frac{ F}{A} . \frac{\Delta L}{L_0}

Y = \frac{FL_0} { AΔL}

Where,

• Y is Young’s modulus
• σ is Stress applied
• ε is Strain related to the applied stress
• F is Force exerted by the object
• A is Actual cross-sectional area
• ▵L is change in the length
• L₀ is actual length

Factors Affecting Young's Modulus

Young's Modulus, which measures a material's stiffness, can be influenced by several factors:

1. Material Type: Different materials, such as metals and rubber, have varying Young’s Modulus values due to differences in atomic structure and bonding.
2. Temperature: As temperature increases, the stiffness of most materials decreases, lowering Young’s Modulus.
3. Impurities and Alloying: Impurities can reduce stiffness, while alloying elements may either enhance or decrease it, depending on the material.
4. Strain Rate: The speed at which stress is applied can affect the material's Young’s Modulus, with faster strain rates often showing higher stiffness.
5. Microstructure and Grain Size: Smaller grains or more uniform structures generally lead to higher stiffness and a higher Young's Modulus.
6. Material Density: Denser materials tend to have stronger atomic bonds, resulting in higher stiffness and Young’s Modulus.
7. Phase and State of Material: Solid materials, particularly crystalline ones, have a higher Young's Modulus compared to liquids or gases.
8. Stress State and Loading Conditions: The type of stress (tensile, compressive, or shear) can affect the observed Young’s Modulus.

How to Calculate Young’s Modulus?

Young's Modulus can be determined experimentally by plotting a stress-strain curve. The slope of the initial, linear portion of this curve represents the Young's Modulus. This portion corresponds to the material's elastic region, where the material returns to its original shape once the stress is removed.

As stress continues to increase, the material eventually reaches its elastic limit, beyond which it no longer returns to its original shape after stress is removed. At this point, the material begins to experience permanent deformation.

If the stress is increased further, the material reaches the plastic limit, where it starts to deform plastically without the need for further increase in stress. Beyond this point, the material undergoes irreversible deformation.

Derivation of Young's Modulus for a Bar

Let a force F be applied on the bar along its length (L), i.e., normal to the surface of the bar as shown in the figure. If △L is the change in length of the bar, then Tensile stress (σ = \frac {F}{A}), where A is the area of the cross-section of the bar,r, and the Longitudinal strain (ϵ = \frac{ \Delta L} {L}).

Therefore, Young's Modulus for this case is given by:

Y = \frac {\frac{ F}{A} }{\frac{\Delta L}{L}}

= \frac{F .L }{(A .▵L}     

If the extension is produced by the load of mass m, then Force, F is mg, where m is the mass and g is the gravitational acceleration. 

And the area of the cross-section of the bar, A is πr² 

Therefore, the above expression can be written as:

Y =\frac{m g L }{πr^2 \Delta L}

Young's Modulus of Some Materials

The table below shows the Young's Modulus values for various materials, indicating their stiffness and resistance to deformation under stress.

Related Articles,

• Elasticity and Plasticity
• Modulus of Elasticity: Definition, Formula, Unit
• Modulus of Rigidity: Shear Modulus

Solved Examples

Problem 1: A steel wire of length 2 m and radius 1 mm is stretched by a force of 200 N. Find the extension. (Y_steel=2×10¹¹ N/m²)

Solution: Given: L = 2m

r = 1 mm = 10⁻³m

A = πr² = π (10⁻³)² = π × 10⁻⁶

Using:

\Delta L = \frac{F L}{A Y}

\Delta L = \frac{200 \times 2}{\pi \times 10^{-6} \times 2 \times 10^{11}}

\Delta L = \frac{200 \times 2}{\pi \times 10^{-6} \times 2 \times 10^{11}}

\Delta L \approx 6.37 \times 10^{-4}

\boxed{\Delta L \approx 0.64 \text{ mm}}

Problem 2: Find the and extension of a vertical wire of length L, density ρ, Young’s modulus Y.

Solution: Consider small element at distance x.

Force on it = weight of wire below it:

F=ρAg(L−x)

Small extension: dL = \frac{F dx}{AY}

dL = \frac{\rho A g (L-x)}{AY} dx

Cancel A: dL = \frac{\rho g (L-x)}{Y} dx

Integrating from 0 to L:

\Delta L = \frac{\rho g}{Y} \int_0^L (L-x) dx

= \frac{\rho g}{Y} \left[ Lx - \frac{x^2}{2} \right]_0^L

= \frac{\rho g}{Y} \left( L^2 - \frac{L^2}{2} \right)

\boxed{\Delta L = \frac{\rho g L^2}{2Y}}

Unsolved Problems

Problem 1: Two wires of the with same length but radii in the the ratio 1:2 are joined in parallel. Find effective Young’s modulus.

Problem 2: A rod of density 8000 kg/m³, length 2 m. Find the extension due to its own weight. Y=2×10¹¹

Problem 3: A wire fixed between two rigid walls is heated by 100°C. Find thermal stress.
(Given: α=1×10⁻⁵ & Y=2×10¹¹)