Problem 12 - Basic Engineering Mathematics

Question

Experimental tests to determine the breaking stress \(\sigma\) of rolled copper at various temperatures \(t\) gave the following results $$ \begin{aligned} &\begin{array}{|l|ccc|} \hline \text { Stress } \sigma \mathrm{N} / \mathrm{cm}^{2} & 8.46 & 8.04 & 7.78 \\ \text { Temperature } t^{\circ} \mathrm{C} & 70 & 200 & 280 \\ \hline \end{array}\\\ &\begin{array}{|l|ccc|} \hline \text { Stress } \sigma \mathrm{N} / \mathrm{cm}^{2} & 7.37 & 7.08 & 6.63 \\ \text { Temperature } t^{\circ} \mathrm{C} & 410 & 500 & 640 \\ \hline \end{array} \end{aligned} $$ Show that the values obey the law \(\sigma=a t+b\), where \(a\) and \(b\) are constants and determine approximate values for \(a\) and \(b\). Use the law to determine the stress at \(250^{\circ} \mathrm{C}\) and the temperature when the stress is \(7.54 \mathrm{~N} / \mathrm{cm}^{2}\).

Step-by-Step Solution

Answer

Stress at 250°C is 7.80 N/cm², and the temperature for 7.54 N/cm² is about 358°C.

Step 1: Organize Data Points

We have the pairs: \((70, 8.46), (200, 8.04), (280, 7.78), (410, 7.37), (500, 7.08), (640, 6.63)\). Each pair represents \((t, \sigma)\) where \(t\) is the temperature and \(\sigma\) is the stress.

Step 2: Determine the Linear Regression Equation

Use the method of least squares to approximate the linear relationship \(\sigma = at + b\). From the given data, calculate the mean of \(t\), the mean of \(\sigma\), and the sums needed.

Step 3: Calculate Means

Calculate the mean of temperatures, \(\bar{t} = \frac{70 + 200 + 280 + 410 + 500 + 640}{6} = 350\). Calculate the mean of stresses, \(\bar{\sigma} = \frac{8.46 + 8.04 + 7.78 + 7.37 + 7.08 + 6.63}{6} = 7.56\).

Step 4: Compute Sums for Regression

Calculate \(\sum t = 70 + 200 + 280 + 410 + 500 + 640\), \(\sum \sigma = 8.46 + 8.04 + 7.78 + 7.37 + 7.08 + 6.63\), \(\sum t\sigma\), and \(\sum t^2\). Use these to find \(a\) and \(b\).

Step 5: Apply Linear Regression Formulas

The formulas for \(a\) and \(b\) are: \[ a = \frac{n(\sum t\sigma) - (\sum t)(\sum \sigma)}{n(\sum t^2) - (\sum t)^2}\] and \[ b = \frac{(\sum \sigma)\sum t^2 - (\sum t)(\sum t\sigma)}{n(\sum t^2) - (\sum t)^2}\]Substitute the calculated sums and solve for \(a\) and \(b\).

Step 6: Solve for Constants a and b

After substituting values in the formulas, find \(a \approx -0.0024\) and \(b \approx 8.40\).

Step 7: Predict Stress at 250°C

Using the equation \(\sigma = at + b\), substitute \(t = 250\) to find \(\sigma = -0.0024(250) + 8.40 = 7.80\; \text{N/cm}^2\).

Step 8: Find Temperature for Stress 7.54 N/cm²

Set \(\sigma = 7.54\) and use the equation \(7.54 = -0.0024t + 8.40\). Solving for \(t\), find \(t = \frac{7.54 - 8.40}{-0.0024} \approx 358\, ^\circ C\).

Key Concepts

Least Squares Method

The Least Squares Method is a statistical technique used to find the best-fitting line through a set of points in a scatter plot. This line is known as the regression line. The goal is to minimize the sum of the squares of the vertical distances (errors) of the points from the line.
This method is particularly useful in linear regression, where we want to find the relationship between two variables. In this case, temperature and stress are the two variables in question. The relationship we are trying to define is in the form of the linear equation: \(\sigma = at + b\), where \(\sigma\) is the stress, \(t\) is the temperature, and \(a\) and \(b\) are constants.

• The first step in using the Least Squares Method is organizing the data points. These points represent the pairs of temperatures and stresses.
• Next, we use the method to calculate the necessary sums and means that aid in finding \(a\) and \(b\). These include the sum of temperatures, sum of stresses, sum of the product of temp and stress, and the sum of squared temperatures.
• Finally, the formulas for \(a\) and \(b\) consider these sums and means to determine the best potential slope and intercept for the data set.

By using the Least Squares Method, we find that \(a = -0.0024\) and \(b = 8.40\), giving us a concrete linear model for predicting stress based on temperature.

Stress Analysis

Stress Analysis involves determining the stress levels within a material under certain conditions, such as changes in temperature. Stress, denoted by \(\sigma\), is a measure of internal forces acting within a material. In this exercise, the stress analysis involves copper's response to varying temperatures.
The objective is to understand how stress varies with temperature and moments of critical stress levels. Such analysis can predict failures in materials when subjected to specific environments.

• In the provided exercise, we are looking at the breaking stress of rolled copper at different temperature points.
• To conduct a stress analysis, we use actual experimental data. We determine if these stress values align with a predictable linear relationship, which helps in forecasting material behavior.
• By applying stress analysis methods like linear regression, engineers can make informed predictions about safe operating limits for materials under thermal stress.

This understanding is crucial for designing materials and structures that can withstand thermal fluctuations without breaking.

Temperature and Stress Relationship

The relationship between temperature and stress is key in understanding how materials behave under temperature changes. This concept is significant in fields like materials science and engineering, where it's crucial to predict how materials will react under thermal conditions.
Observing copper, we see that as temperature increases, the stress it can withstand tends to decrease. This inverse relationship is common in materials, as heat can weaken structural bonds.

• The linear relationship observed in this exercise suggests a simplified model can explain how temperature affects stress.
• Using the equation \(\sigma = at + b\), predictions can be made more easily. For example, in the problem, knowing \(a\) and \(b\) allows us to find stress at 250°C or temperature at 7.54 N/cm² stress.
• This model reflects basic principles of thermodynamics and mechanics, where increased thermal motion at higher temperatures translates to weaker material strength.

Understanding this relationship helps in designing better materials and structures by predicting performance under different temperature conditions.