A steel rod ($Y = 2.0 \times 10^{11}$ Nm$^{-2}$; and $\alpha = 10^{-5}\,{}^{\circ}\text{C}^{-1}$) of length 1 m and area of cross-section 1 cm$^2$ is heated from 0°C to 200°C, without being allowed to extend or bend. What is the tension produced in the rod?

A steel rod (Y = 2.0 × 10¹¹ Nm⁻²; and α = 10⁻⁵°C⁻¹) of length 1 m and area of cross-section 1 cm² is heated fr

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Y = 2.0 \times 10^{11}

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Concept understanding — Youngs Modulus

Young’s Modulus: The Stretchiness of a Solid

When you pull on a rubber band, it stretches easily. When you pull on a steel rod of the same size, it barely moves. Both are elastic — they return to their original shape when you let go — but they resist stretching very differently. Young’s modulus is the number that tells you exactly how much a material resists being stretched or compressed lengthwise.

The Intuition: Stiffness per Unit Size

Think of a spring. A stiff spring requires a large force to stretch it a little. A soft spring stretches a lot with a small force. Young’s modulus is like the “stiffness” of a material, but it’s cleverly designed to be independent of the object’s shape and size.

If you take a thick steel rod and a thin steel wire of the same length, the rod is harder to stretch. That’s because you’re pulling on more material. Young’s modulus removes this size effect — it tells you the stiffness of the material itself, not the particular piece you’re holding.

The Precise Definition

Young’s modulus ( $E$ or $Y$ ) is defined as the ratio of tensile stress to tensile strain, as long as the material obeys Hooke’s law (the deformation is reversible and proportional to the force).

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

Let’s break down the two parts.

Tensile Stress ( $\sigma$ ) is the force per unit area. If you pull with a force $F$ on a rod of cross-sectional area $A$ , the stress is:

\sigma = \frac{F}{A}

Stress has units of pressure — pascals (Pa) or $\text{N/m}^2$ . It tells you how “intense” the pulling is, regardless of the rod’s thickness.

Tensile Strain ( $\varepsilon$ ) is the fractional change in length. If the original length is $L_0$ and it stretches by $\Delta L$ , the strain is:

\varepsilon = \frac{\Delta L}{L_0}

Strain is a pure number — it has no units. A strain of 0.01 means the rod stretched by 1% of its original length.

Putting it together:

Y = \frac{F/A}{\Delta L / L_0} = \frac{F L_0}{A \Delta L}

What the Number Tells You

A high Young’s modulus means the material is very stiff — it takes a huge stress to produce even a tiny strain. Steel has $Y \approx 200 \times 10^9$ Pa. A low Young’s modulus means the material is easily stretched. Rubber has $Y \approx 0.01 \times 10^9$ Pa — about 20,000 times smaller than steel.

Note

Young’s modulus is only valid in the elastic region — where the material returns to its original shape after the force is removed. If you stretch too far (past the elastic limit), the material deforms permanently or breaks, and Young’s modulus no longer applies.

A Worked Example

A steel wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ is pulled by a force of 100 N. How much does it stretch? (Young’s modulus of steel = $2.0 \times 10^{11}$ Pa)

From $Y = \frac{F L_0}{A \Delta L}$ , rearrange:

$\Delta L = \frac{F L_0}{A Y} = \frac{100 \times 2.0}{(1.0 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{200}{2.0 \times 10^5} = 1.0 \times 10^{-3} \text{ m} = 1.0 \text{ mm}$ …

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When a material is heated, it tends to expand. If this expansion is prevented, an internal stress, known as thermal stress, is developed within the material. This stress is directly related to the material's Young's Modulus and its coefficient of linear expansion.

First, calculate the strain that would occur if the rod were allowed to expand freely due to the temperature change:

\text{Thermal Strain} = \alpha \Delta T

\text{Thermal Strain} = (10^{-5}\,{}^{\circ}\text{C}^{-1}) \times (200\,{}^{\circ}\text{C}) = 2 \times 10^{-3}

Since the rod is not allowed to extend, this thermal strain is effectively counteracted by an elastic strain, leading to a stress. The stress developed in the rod is given by Young's Modulus multiplied by this strain:

\text{Stress} (\sigma) = Y \times \text{Thermal Strain}

$$ \sigma = (2.0 \times 10^{11} \text{ Nm}^{-2}) \times (2 \times 10^{-3}) = 4.0 \times 10^8 \text{ Nm}^{-2} $$ …

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When a steel rod is heated but prevented from expanding, it develops internal compressive stress. This stress results in a force (often termed tension in such problems, referring to its magnitude) that can be calculated using Young's Modulus and the coefficient of thermal expansion. The tension produced in the rod is $\boxed{4.0 \times 10^4 \text{ N}}$ .

When a material is heated, its constituent particles vibrate with greater amplitude, causing the material to expand. This phenomenon is called thermal expansion. If this natural expansion is prevented, the material experiences internal forces and stresses.

Imagine the rod is first allowed to expand freely due to the temperature increase. It would become longer. Now, to bring it back to its original length (because it's "not allowed to extend"), we would have to apply a compressive force. This compressive force creates a stress within the rod. The magnitude of this stress and the resulting force are what we need to calculate.

The relationship between stress, strain, and Young's Modulus is fundamental here. Young's Modulus ( $Y$ ) is a measure of a material's stiffness, defined as the ratio of stress to strain in the elastic region:

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

Here's how we approach the problem:

Calculate the free thermal expansion: Determine how much the rod would expand if it were allowed to do so freely.
Determine the effective strain: Since the rod is prevented from extending, the amount it would have expanded is effectively the amount it is compressed back to its original length. This gives us the strain.
Calculate the stress: Use Young's Modulus to find the stress corresponding to this strain.
Calculate the tension (force): Multiply the stress by the cross-sectional area to find the total force.

Let's proceed with the calculations.

Calculate the change in temperature ( $\Delta T$ ):

The rod is heated from $0\,{}^{\circ}\text{C}$ to $200\,{}^{\circ}\text{C}$ .

$\Delta T = T_{\text{final}} - T_{\text{initial}} = 200\,{}^{\circ}\text{C} - 0\,{}^{\circ}\text{C} = 200\,{}^{\circ}\text{C}$ .
Calculate the free thermal expansion ( $\Delta L$ ):

If the rod were free to expand, its change in length would be given by the formula:

$\Delta L = L \alpha \Delta T$

where $L$ is the original length, $\alpha$ is the coefficient of linear thermal expansion, and $\Delta T$ is the change in temperature.

Given:

$L = 1 \text{ m}$

$\alpha = 10^{-5}\,{}^{\circ}\text{C}^{-1}$

$\Delta T = 200\,{}^{\circ}\text{C}$

Substituting these values:

$\Delta L = (1 \text{ m}) \times (10^{-5}\,{}^{\circ}\text{C}^{-1}) \times (200\,{}^{\circ}\text{C})$

$\Delta L = 200 \times 10^{-5} \text{ m} = 2 \times 10^{-3} \text{ m}$
Determine the thermal strain ( $\epsilon$ ):

Since the rod is not allowed to extend, it means its final length is the same as its initial length. The expansion of $2 \times 10^{-3}$ m is effectively prevented. This prevented expansion is equivalent to a compressive strain applied to the rod.

The strain is defined as the fractional change in length:

$\epsilon = \frac{\Delta L}{L}$

$\epsilon = \frac{2 \times 10^{-3} \text{ m}}{1 \text{ m}} = 2 \times 10^{-3}$ …

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