Question

For plane strain fracture toughness (\(K_{Ic}\)) testing of a Maraging steel specimen, find the minimum thickness needed for a valid \(K_{Ic}\) measurement (answer as an integer, in mm).
Given: \(K_{Ic} = 90\ \text{MPa}\sqrt{\text{m}}\) and yield stress \(\sigma_y = 900\ \text{MPa}\).

Hint:

Use the ASTM E399 validity condition \(B \geq 2.5(K_{Ic}/\sigma_y)^2\) to size the specimen.

Answer 1

Correct Answer: 25

Step 1: Recall the ASTM E399 validity condition.
A plane strain fracture toughness test is valid only when the crack front sits inside enough material for plane strain conditions to hold, so the plastic zone stays small next to the specimen size. The standard needs the specimen thickness \(B\) to satisfy
\[ B \geq 2.5\left(\frac{K_{Ic}}{\sigma_y}\right)^2 \]
Here \(K_{Ic}\) is the plane strain fracture toughness and \(\sigma_y\) is the yield stress of the material.

Step 2: Write down the given values.
We are given
\[ K_{Ic} = 90\ \text{MPa}\sqrt{\text{m}}, \qquad \sigma_y = 900\ \text{MPa} \]

Step 3: Find the ratio \(K_{Ic}/\sigma_y\).
\[ \frac{K_{Ic}}{\sigma_y} = \frac{90}{900} = 0.1\ \sqrt{\text{m}} \]
The units work out to \(\sqrt{\text{m}}\) because \(K_{Ic}\) carries units of \(\text{MPa}\sqrt{\text{m}}\) and \(\sigma_y\) carries units of \(\text{MPa}\); dividing cancels the stress unit and leaves \(\sqrt{\text{m}}\).

Step 4: Square the ratio.
\[ \left(\frac{K_{Ic}}{\sigma_y}\right)^2 = (0.1)^2 = 0.01\ \text{m} \]
Squaring \(\sqrt{\text{m}}\) turns the unit into plain metres, which is why this formula gives a length directly.

Step 5: Apply the factor of 2.5 and get the minimum thickness.
\[ B_{min} = 2.5 \times 0.01 = 0.025\ \text{m} \]
Converting to millimetres,
\[ B_{min} = 0.025 \times 1000 = 25\ \text{mm} \]

Final Answer:
The minimum specimen thickness needed for a valid \(K_{Ic}\) test is 25 mm.
\[ \boxed{B_{min} = 25\ \text{mm}} \]