Question

A steel rod of length \(1.0000 \mathrm{~m}\) and cross-sectional area \(5.00 \cdot 10^{-4} \mathrm{~m}^{2}\) is placed snugly against two immobile end points. The rod is initially placed when the temperature is \(0^{\circ} \mathrm{C}\). Find the stress in the rod when the temperature rises to \(40.0^{\circ} \mathrm{C}\).

Short answer

Answer: To find the stress in the steel rod after the temperature change, follow the steps below: 1. Calculate the length change due to temperature change using the linear expansion formula: ∆L = L₀α∆T. The coefficient of linear expansion for steel (α) is 12 x 10^-6 K^-1, and the temperature difference (∆T) is 40.0°C, resulting in a change in length (∆L). 2. Calculate the force required for the length change using Hooke's Law: F = YA(∆L/L₀). The Young's Modulus (Y) for steel is 2 x 10^11 N/m², and the given cross-sectional area (A) is 5.00 x 10^-4 m². 3. Calculate the stress in the rod using the stress formula: σ = F/A. Plug in the force obtained in Step 2 and the given cross-sectional area to get the final stress value. After calculating the stress, you will have the final solution for the problem.

Step-by-step solution

Calculate the length change due to temperature change

To find the change in length, first, we need to know the initial and final temperatures. The initial temperature is \(T_1 = 0^{\circ} C\), and the final temperature is \(T_2 = 40.0^{\circ} C\). The temperature difference is \(\Delta T = T_2 - T_1 = 40.0^{\circ} C\). We also need the coefficient of linear expansion for steel, which is \(\alpha = 12 \times 10^{-6} \text{K}^{-1}\). Now we can use the linear expansion formula to find the change in length of the rod: $$\Delta L = L_0 \alpha \Delta T$$ Plug in the given values: $$\Delta L = (1.0000 \text{ meters}) \times (12 \times 10^{-6} \,\text{K}^{-1}) \times (40.0 \text{ K})$$

Calculate the force required for the length change

Now that we have the change in length, we can use Hooke's Law to find the force required to change the length by that amount. We need the Young's Modulus (Y) for steel, which is \(2 \times 10^{11} \text{N}/\text{m}^2\), and the initial cross-sectional area of the rod, which is given as \(A = 5.00 \times 10^{-4} \text{ m}^2\). Hooke's Law is given by \(F = YA \frac{\Delta L}{L_0}\): $$F = (2 \times 10^{11} \text{ N/m}^2) (5.00 \times 10^{-4} \text{ m}^2) \frac{\Delta L}{1.0000 \text{ meters}}$$

Calculate the stress in the rod

To find the stress in the rod, we need to use the stress formula, \(\sigma = \frac{F}{A}\). Simply plug in the force obtained in Step 2, and the given cross-sectional area of the rod: $$\sigma = \frac{F}{5.00 \times 10^{-4} \text{ m}^2}$$ Once you have calculated the stress, you will have the final solution for the problem.

Key concepts

Linear Expansion

When materials are heated or cooled, their dimensions often change. This change in dimension is known as thermal expansion. Linear expansion refers specifically to the change in length. The equation that describes this phenomenon is given by: \[ \Delta L = L_0 \alpha \Delta T, \] where \( \Delta L \) is the change in length, \( L_0 \) is the original length, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.

• **Original Length \( L_0 \):** The initial length of the object before any change in temperature.
• **Coefficient of Linear Expansion \( \alpha \):** This is a material-specific constant that indicates how much a material expands per degree change in temperature.
• **Temperature Change \( \Delta T \):** The difference between the final and initial temperatures.

In the example given, a steel rod at \(0^{\circ} \mathrm{C}\) increases to \(40.0^{\circ} \mathrm{C}\). By applying these values in the linear expansion formula with the known \( \alpha \) for steel, one can calculate \( \Delta L \), which will help in further calculations related to thermal stress.

Young's Modulus

Young's Modulus, often denoted as \( Y \), is a measure of the stiffness of a material. It is the ratio of stress (force per unit area) to strain (proportional deformation in length). The formula is given by: \[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L_0}, \] where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L_0 \) is the original length.

• **Stress \( \sigma \):** This is the force applied per unit area, expressed in N/m².
• **Strain:** A dimensionless measure of deformation representing the relative change in length, \( \Delta L / L_0 \).
• **Material Stiffness:** Higher Young's Modulus indicates a stiffer material, which means it will deform less when a force is applied.

Young's Modulus helps us compute the force \( F \) in the linear expansion scenario. For steel, it is about \(2 \times 10^{11} \text{ N/m}^2\). By using this information, the required force to achieve the given \( \Delta L \) from thermal expansion can be calculated.

Hooke's Law

Hooke's Law describes the behavior of materials that return to their original shape after the applied stress is removed, as long as the material's elastic limit is not exceeded. The formula is stated as: \[ F = YA \frac{\Delta L}{L_0}, \] where \( F \) is the restoring force, \( Y \) is Young's Modulus, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L_0 \) is the original length.

• The equation assumes that the material behaves elastically, meaning it will revert back once the stress is removed.
• **Elastic Limit:** The maximum extent a material can be stretched without undergoing permanent deformation.
• This law provides a simple linear relationship between stress and strain for small deformations.

In our specific exercise, Hooke's Law is utilized to link the thermal expansion to the force applied on the rod due to constraints by the immovable endpoints. With the calculated force, the stress can subsequently be derived using the formula for stress, \( \sigma = F / A \). This allows determination of the thermal stress experienced by the steel rod as its temperature rises.