14. In a tensile test on a metal specimen, the following extensions were recorded: Load (Mg) & 0 & 5 & 10 & 15 & 20 Extension (mm) & 0 & 0.013 & 0.025 & 0.037 & 0,051 From the values plot a graph and determine the extension under a load of 12 mathrm(Mg)

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14. In a tensile test on a metal specimen, the following extensions were recorded: Load (Mg) & 0 & 5 & 10 & 15 & 20 Extension (mm) & 0 & 0.013 & 0.025 & 0.037 & 0,051 From the values plot a graph and determine the extension under a load of 12 mathrm(Mg)

Answer 1

The extension of the metal under a load of \(12 Mg\) is approximately \(0.042 mm\).

Answer 2

## Step 1: Firstly, understanding the concept of a tensile test is important: a tensile test is an engineering or materials science test from which one can infer several material properties. It measures the resistance of a material to a static or slowly applied force and measures characteristics such as elasticity, tensile strength, elongation, hardness and elastic limit. ## Step 2: We got the values of load and its corresponding extension. We have to determine the extension under a load of 12 Mg. Laying out the observed data points on a plain X,Y graph will help. ## Step 3: It is reasonable to use linear approximation to determine the extension under \(12 Mg\). Here we can see the behavior is approximately linear in the given load range. ## Step 4: Construct a linear approximation from the values. For this, we create a linear relationship \(y = mx + c\). Here \(m = (y_2 - y_1) / (x_2 - x_1)\) is the gradient and y-axis intercept, \(c\), can be extrapolated. We choose two data points near the value 12 from the data provided so the gradient \(m\) calculation has meaning (10 and 15 Mg, with extension 0.025 mm and 0.037 mm respectively). ### \[ m = \frac{0.037-0.025}{15 - 10} = \frac{0.012}{5} = 0.0024 \frac{mm}{Mg} \] This gradient \(m\) is the rate which extension changes relative to load. To find \(c\) (intercept with y-axis), we substitute values for known point (10, .025) into \(y = mx + c\) and solve. ### \(c = y - mx = 0.025 - 0.0024 \times 10 = 0.013 mm \) Those, the corresponding linear relation we have derived is \( Y = (0.0024)X + 0.013 \) where X = load and Y = Extension ## Step 5: Apply \( X = 12 mg \) in derived relation to get Y = Extension corresponding to it. ### \( Y = (0.0024 \times 12) + 0.013= 0.0288 + 0.013 \approx 0.042 mm \)

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14. In a tensile test on a metal specimen, the following extensions were recorded: Load (Mg) & 0 & 5 & 10 & 15 & 20 Extension (mm) & 0 & 0.013 & 0.025 & 0.037 & 0,051 From the values plot a graph and determine the extension under a load of 12 mathrm(Mg)

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