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 \(\sigma_y = 900\ \text{MPa}\).
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Use the ASTM E399 validity condition \(B \geq 2.5(K_{Ic}/\sigma_y)^2\) to size the specimen.
Step 1: Note what the ASTM E399 thickness rule is checking.
For a $K_{Ic}$ test to be trusted as a true plane strain value, the specimen must be thick enough that the crack tip plastic zone is small next to the thickness. The rule used to fix this minimum thickness is
\[ B \geq 2.5\left(\frac{K_{Ic}}{\sigma_y}\right)^2 \]
Step 2: Put the given data in consistent units first.
Take $K_{Ic}=90$ MPa$\sqrt{m}$ and $\sigma_y=900$ MPa. Since both are already in MPa, no unit conversion is needed; the ratio itself will carry the units of $\sqrt{m}$.
Step 3: Simplify the ratio before squaring.
\[ \frac{K_{Ic}}{\sigma_y}=\frac{90}{900}=\frac{1}{10} \]
So the ratio is exactly $0.1\ \sqrt{m}$, a clean number because 900 is ten times 90.
Step 4: Square and scale in one step.
\[ B_{min}=2.5\times(0.1)^2=2.5\times0.01=0.025\ \text{m} \]
Step 5: Change metres to millimetres at the end.
One metre is $1000$ mm, so
\[ B_{min}=0.025\times1000=25\ \text{mm} \]
This matches the earlier working. The minimum valid thickness comes out to a round number because the given $K_{Ic}$ to $\sigma_y$ ratio was already a clean $1/10$.
\[ \boxed{B_{min}=25\ \text{mm}} \]
