Welding professionals in Kenya frequently collect and analyze data to monitor quality, improve processes, and ensure safety. Understanding how to present data clearly is essential for interpreting measurements such as weld dimensions, material properties, or defect frequencies. This chapter focuses on methods for presenting continuous data and graphical tools like histograms and line graphs that facilitate decision-making in welding operations.

4.1 Data presentation

Data presentation is fundamental in welding to communicate measurements and trends effectively. Welding data often involves continuous variables such as temperature, tensile strength, or bead width, which require appropriate graphical representation. This section covers continuous variables, histograms, and line graphs, providing tools to visualize welding data for analysis and reporting.

4.1.1 Continuous variables

Continuous variables represent measurements that can take any value within a range, crucial in welding for precise parameter control. Examples include weld bead length, current intensity, and gas flow rate. These variables are often recorded to monitor process stability and product quality.

The main characteristic of continuous variables is that they are measurable quantities that can be subdivided infinitely, unlike discrete variables which count separate items.

$$ \text{Continuous Variable} = x \in [a, b], \quad a,b \in \mathbb{R} $$

Worked Examples

Example 1:
A welder measures the bead width on a steel plate 5 times, obtaining values (in mm): 6.2, 6.5, 6.3, 6.4, 6.6. Calculate the average bead width.

Given: Measurements \(x_1=6.2, x_2=6.5, x_3=6.3, x_4=6.4, x_5=6.6\) mm$$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$$$$\bar{x} = \frac{6.2 + 6.5 + 6.3 + 6.4 + 6.6}{5}$$$$\bar{x} = \frac{31.0}{5}$$ Answer: 6.2 mm

Example 2:
The welding current is recorded continuously between 120 A and 140 A during a procedure. Express the current as a continuous variable interval.

Given: Range \(120 \text{ A} \leq I \leq 140 \text{ A}\)$$I \in [120, 140] \text{ A}$$ Answer: Welding current varies continuously from 120 A to 140 A

Example 3:
Calculate the range of weld temperatures measured as 1350°C, 1365°C, 1348°C, 1355°C, and 1370°C.

Given: Temperatures \(T = \{1350, 1365, 1348, 1355, 1370\}\) °C$$\text{Range} = \max(T) - \min(T)$$$$= 1370 - 1348$$$$= 22 \text{ °C}$$ Answer: Temperature range is 22 °C

4.1.2 Histogram

Histograms are graphical representations that organize continuous data into intervals or bins, showing frequency distribution. In welding, histograms help visualize the spread of measurements such as weld thickness or defect sizes, aiding quality control and process improvement.

The construction of histograms involves grouping data into classes and plotting the frequency of each class as bars.

$$ \text{Frequency} = f_i, \quad i=1,2,...,k $$

Worked Examples

Example 1:
A welder records bead widths (mm) from 20 welds: 5.9, 6.1, 6.0, 6.4, 6.2, 6.3, 6.0, 6.1, 6.5, 6.2, 6.3, 6.4, 6.1, 6.0, 6.3, 6.2, 6.1, 6.4, 6.3, 6.2. Group the data into intervals 5.9–6.0, 6.1–6.2, 6.3–6.4, 6.5–6.6 and find the frequency for each.

Given: Data set as above
Class intervals:

• 5.9–6.0
• 6.1–6.2
• 6.3–6.4
• 6.5–6.6

Count frequencies:

• 5.9–6.0: 5.9, 6.0, 6.0, 6.0 → 4
• 6.1–6.2: 6.1, 6.2, 6.1, 6.2, 6.1, 6.2 → 6
• 6.3–6.4: 6.3, 6.4, 6.3, 6.4, 6.3 → 5
• 6.5–6.6: 6.5 → 1

Answer: Frequencies are 4, 6, 5, 1 respectively

Example 2:
Calculate the relative frequency for each class interval from Example 1 if total data points are 20.

Given: Frequencies \(f = \{4, 6, 5, 1\}\), total \(N=20\)$$\text{Relative frequency} = \frac{f_i}{N}$$$$\text{For 5.9–6.0} = \frac{4}{20} = 0.2$$$$\text{For 6.1–6.2} = \frac{6}{20} = 0.3$$$$\text{For 6.3–6.4} = \frac{5}{20} = 0.25$$$$\text{For 6.5–6.6} = \frac{1}{20} = 0.05$$ Answer: Relative frequencies are 0.2, 0.3, 0.25, 0.05

Example 3:
Determine the midpoint of each class interval in Example 1 for histogram plotting.

Given: Intervals 5.9–6.0, 6.1–6.2, 6.3–6.4, 6.5–6.6
Midpoint formula:$$\text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$$ Calculate:$$\frac{5.9 + 6.0}{2} = 5.95$$$$\frac{6.1 + 6.2}{2} = 6.15$$$$\frac{6.3 + 6.4}{2} = 6.35$$$$\frac{6.5 + 6.6}{2} = 6.55$$ Answer: Midpoints are 5.95, 6.15, 6.35, 6.55 mm

4.1.3 Line

Line graphs display data points connected by straight lines, used to show trends or changes over time. Welders use line graphs to track parameters such as temperature variation during welding or changes in tensile strength across different weld runs.

Line graphs emphasize the relationship between two continuous variables, often time and a measured quantity.

$$ y = f(x) $$

Worked Examples

Example 1:
Plot temperature readings (°C) taken every 10 minutes during welding: 1300, 1320, 1340, 1330, 1350. Find the average rate of temperature increase per minute between the first and last reading.

Given:
Initial temperature \(T_1 = 1300^\circ C\) at \(t_1 = 0\) min
Final temperature \(T_5 = 1350^\circ C\) at \(t_5 = 40\) min
Rate of increase:$$\text{Rate} = \frac{T_5 - T_1}{t_5 - t_1}$$$$= \frac{1350 - 1300}{40 - 0}$$$$= \frac{50}{40}$$$$= 1.25^\circ C/\text{min}$$ Answer: Temperature increases at 1.25 °C per minute

Example 2:
The tensile strength (MPa) of five weld samples is recorded as 420, 430, 425, 435, 440. Calculate the percentage increase from the first to the last sample.

Given: Initial \(S_1=420\) MPa, Final \(S_5=440\) MPa$$\%\text{ Increase} = \frac{S_5 - S_1}{S_1} \times 100$$$$= \frac{440 - 420}{420} \times 100$$$$= \frac{20}{420} \times 100$$$$= 4.76\%$$ Answer: The tensile strength increased by 4.76%

Example 3:
A welder records the gas flow rate (L/min) every 5 minutes: 12, 13, 13.5, 14, 15. Determine the average flow rate over the 20-minute period.

Given: Flow rates \(Q = \{12, 13, 13.5, 14, 15\}\) L/min$$\bar{Q} = \frac{1}{5} \sum_{i=1}^5 Q_i$$$$= \frac{12 + 13 + 13.5 + 14 + 15}{5}$$$$= \frac{67.5}{5}$$$$= 13.5 \text{ L/min}$$ Answer: Average gas flow rate is 13.5 L/min

Chapter Summary

This chapter covers the presentation of data, distinguishing between continuous variables using histograms and line graphs, and discrete variables with bar and pie graphs. It further explores grouped data presentation through histograms, bar charts, cumulative frequency tables, and ogives, as well as ungrouped data displayed by line graphs and cumulative frequency. Measures of central tendency including mean, mode, and median are calculated for both grouped and ungrouped data sets. The chapter also addresses measures of dispersion, focusing on standard deviation and variance for grouped and ungrouped data. Probability concepts are introduced, considering scenarios with and without replacement, followed by probability distribution functions such as the binomial and Poisson distributions. Finally, the chapter examines the normal distribution, providing a foundation for understanding statistical variability and probabilistic models in civil engineering contexts.