A sheet metal with a stock hardness of 250 HRC has to be sheared using a punch and a die having a clearance of 1 mm between them. If the stock hardness of the sheet metal increases to 400 HRC, the clearance between the punch and the die should be_________ mm.

The correct answer is 1.265.

Underlying Principle:
In sheet metal shearing operations, the clearance (C) between the punch and the die is proportional to the sheet thickness and the shear strength of the material. Since the shear strength of a metal is directly related to its hardness, the clearance is proportional to the square root of the hardness of the material. This relationship can be expressed as:
$C\propto \sqrt{HRC}$
where HRC represents the Rockwell C hardness of the sheet metal.

Therefore, we can establish a ratio between the clearances and the hardness values for the two different cases:
$\frac{C_2}{C_1}=\frac{\sqrt{HR{C}_2}}{\sqrt{HR{C}_1}}$

Given Data:
Initial hardness, HRC₁ = 250 HRC
Initial clearance, C₁ = 1 mm
Increased hardness, HRC₂ = 400 HRC

Step-by-step Calculation:
Rearranging the ratio formula to solve for the new clearance (C₂):
$C_2=C_1\times \sqrt{\frac{HR{C}_2}{HR{C}_1}}$

Substitute the given values into the equation:
$C_2=1\times \sqrt{\frac{400}{250}}$

Simplify the fraction inside the square root:
$\frac{400}{250}=1.6$

Calculate the square root of 1.6:
$C_2=1\times 1.26491\approx 1.265\text{ mm}$

Thus, the required clearance between the punch and the die when the hardness increases to 400 HRC is 1.265 mm.