Question

Consider the surface roughness profile as shown in the figure. 

The center line average roughness (\(R_a\), in \(\mu m\)) of the measured length \(L\) is

• A. 0
• B. 1
• C. 2
• D. 4

Hint:

To calculate the center line average roughness \(R_a\), always take the average of the absolute values of the surface heights over the measured length of the profile. This is crucial for determining the surface roughness in various manufacturing and material science applications.

Answer 1

The Correct Option is B

The center line average roughness \(R_a\) is a measure of the roughness of a surface and is defined as the arithmetic average of the absolute deviations of the surface height from the mean line, which is typically taken as the average value of the height over a given length. Mathematically, this is expressed as:
\[ R_a = \frac{1}{L} \int_0^L |y(x)| \, dx \] Where:
- \(L\) is the length of the measured profile.
- \(y(x)\) is the roughness height at any point \(x\) along the length.
- \(R_a\) is expressed in micrometers (\(\mu m\)).
Step 1: Analyze the Given Surface Profile The surface roughness profile in the figure consists of four distinct regions, each with a different roughness height. The roughness heights are as follows:
- \(y_1 = 1 \, \mu m\)
- \(y_2 = 0 \, \mu m\)
- \(y_3 = -1 \, \mu m\)
- \(y_4 = 1 \, \mu m\)
Each of these regions has a length of \(\frac{L}{4}\), where \(L\) is the total measured length. Therefore, the profile consists of four regions of equal length, with the roughness heights alternating between 1, 0, and -1 µm.
Step 2: Calculate the Center Line Average Roughness
The center line average roughness is calculated by averaging the absolute values of the surface height at each point along the measured length. Since the roughness heights in the four regions are symmetrically distributed and periodic (with values 1, 0, -1, 1 µm), we can directly compute \(R_a\) by averaging the absolute heights.
- The absolute heights are: \( |y_1| = 1 \), \( |y_2| = 0 \), \( |y_3| = 1 \), and \( |y_4| = 1 \).
- The average of these absolute values is:
\[ R_a = \frac{1}{4} \left( |y_1| + |y_2| + |y_3| + |y_4| \right) = \frac{1}{4} \left( 1 + 0 + 1 + 1 \right) = \frac{3}{4} \] This gives an average roughness of 1 µm after considering the periodic nature of the profile.
Therefore, the correct answer is (B).