Civil engineering projects require precise interpretation of data and accurate graphical representation to analyze trends, relationships, and behaviors of materials and structures. The ability to apply linear graphs effectively is essential for civil engineers in Kenya, especially when dealing with site measurements, material testing, or structural analysis. This chapter focuses on identifying relevant information from data sets and selecting appropriate scales to construct meaningful linear graphs that support engineering decisions.

6.1 Identifying Information

Civil engineers often receive large volumes of data from site surveys, laboratory tests, or project monitoring. Extracting the correct information for graphing is critical to visualize relationships and make informed decisions. This section develops skills to identify relevant data points and understand problem contexts for effective linear graph construction.

6.1.1 Understanding Given Data Sets and Problems

Linear graphs represent relationships between two variables, usually dependent and independent variables. Understanding the data set involves recognizing what each variable represents, units of measurement, and the problem’s context.

The general equation of a straight line is:

$$ y = mx + c $$

where \(y\) is the dependent variable, \(x\) is the independent variable, \(m\) is the slope, and \(c\) is the y-intercept.

Worked Examples

Example 1: A civil engineer records the settlement \(y\) (in mm) of a foundation over time \(x\) (in days). The data set is \(y = 0.5x + 2\). Find the settlement after 10 days.

Given: \(y = 0.5x + 2\), \(x = 10\) days

$$ y = 0.5 \times 10 + 2 $$

$$ y = 5 + 2 $$

Answer: 7 \, \text{mm}

Example 2: The stress \(y\) (in MPa) in a beam is related to strain \(x\) (in microstrain) by \(y = 200x + 5\). Calculate the stress when strain is 0.002.

Given: \(y = 200x + 5\), \(x = 0.002\)

$$ y = 200 \times 0.002 + 5 $$

$$ y = 0.4 + 5 $$

Answer: 5.4 \, \text{MPa}

Example 3: A contractor measures water flow \(y\) (litres per second) through a pipe at different pressures \(x\) (kPa). The relationship is \(y = -0.3x + 15\). Find flow when pressure is 20 kPa.

Given: \(y = -0.3x + 15\), \(x = 20\)

$$ y = -0.3 \times 20 + 15 $$

$$ y = -6 + 15 $$

Answer: 9 \, \text{L/s}

Example 4: During a soil compaction test, the dry density \(y\) (g/cm³) varies linearly with moisture content \(x\) (%) as \(y = -0.02x + 1.8\). Find the dry density at 12% moisture.

Given: \(y = -0.02x + 1.8\), \(x = 12\)

$$ y = -0.02 \times 12 + 1.8 $$

$$ y = -0.24 + 1.8 $$

Answer: 1.56 \, \text{g/cm}^3

Example 5: The deflection \(y\) (mm) of a beam under load \(x\) (kN) is given by \(y = 0.1x + 0.5\). Find deflection at 50 kN.

Given: \(y = 0.1x + 0.5\), \(x = 50\)

$$ y = 0.1 \times 50 + 0.5 $$

$$ y = 5 + 0.5 $$

Answer: 5.5 \, \text{mm}

6.1.2 Extracting Relevant Information for Graphing

To graph data correctly, it is necessary to extract the independent and dependent variables, units, and range of values from the problem. This allows plotting points and interpreting trends.

The linear graph plotting formula remains:

$$ y = mx + c $$

where the slope \(m\) and intercept \(c\) may need calculation from raw data.

Worked Examples

Example 1: From the data points: (2, 5), (4, 9), (6, 13), find the slope \(m\) and intercept \(c\) of the line.

Given points: \(x_1=2, y_1=5; x_2=4, y_2=9\)

$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2 $$

Using point (2,5) to find \(c\):

$$ y = mx + c \Rightarrow 5 = 2 \times 2 + c $$

$$ 5 = 4 + c $$

$$ c = 1 $$

Answer: \(y = 2x + 1\)

Example 2: A site engineer records temperature \(y\) (°C) as 20, 25, 30 for times \(x\) (hours) 1, 2, 3 respectively. Find the equation of the line.

Given points: (1,20), (3,30)

$$ m = \frac{30 - 20}{3 - 1} = \frac{10}{2} = 5 $$

Using (1,20):

$$ 20 = 5 \times 1 + c $$

$$ c = 15 $$

Answer: \(y = 5x + 15\)

Example 3: For points (0, 4) and (5, 14), calculate the linear equation.

Given: \(x_1=0, y_1=4; x_2=5, y_2=14\)

$$ m = \frac{14 - 4}{5 - 0} = \frac{10}{5} = 2 $$

Using (0,4):

$$ 4 = 2 \times 0 + c $$

$$ c = 4 $$

Answer: \(y = 2x + 4\)

Example 4: A soil test shows moisture content \(x\) (%) and unit weight \(y\) (kN/m³) as (5, 17), (10, 15), (15, 13). Find slope and intercept using first two points.

Given: (5,17), (10,15)

$$ m = \frac{15 - 17}{10 - 5} = \frac{-2}{5} = -0.4 $$

Using (5,17):

$$ 17 = -0.4 \times 5 + c $$

$$ 17 = -2 + c $$

$$ c = 19 $$

Answer: \(y = -0.4x + 19\)

Example 5: For points (3, 12) and (7, 20), determine the line equation.

Given: (3,12), (7,20)

$$ m = \frac{20 - 12}{7 - 3} = \frac{8}{4} = 2 $$

Using (3,12):

$$ 12 = 2 \times 3 + c $$

$$ 12 = 6 + c $$

$$ c = 6 $$

Answer: \(y = 2x + 6\)

Chapter Summary

This chapter focused on applying linear graphs by first identifying and understanding the given data sets and problems, emphasizing the importance of extracting relevant information for accurate graphing. It then covered how to choose appropriate scales for the graph axes and explained how different scale selections can affect the interpretation of the graph. Proper labeling of graph axes was highlighted as essential for clear data representation and understanding. The chapter detailed techniques for plotting linear graphs, including accurately placing points and connecting them to form meaningful lines. Finally, it addressed how to analyze the plotted graphs by identifying trends and patterns, enabling drawing valid conclusions based on the visualized data. Throughout, the chapter reinforced the critical steps necessary to construct and interpret linear graphs effectively in practical contexts.