Question

In a deep drawing operation of a rectangular sheet metal, \(30\%\) stretching in length results in \(15\%\) reduction in thickness. Assuming volume constancy, the normal anisotropy of the sheet metal is ________ (rounded off to 2 decimal places).

Hint:

Use volume constancy to get the width change first, then take the ratio of true width strain to true thickness strain.

Answer 1

Correct Answer: 0.61

Step 1: Set up the volume constancy condition.
For a sheet, volume stays fixed during forming, so \(L_0 W_0 T_0 = L_1 W_1 T_1\).

Step 2: Put in the given length and thickness changes.
Length grows by 30%, so \(L_1 = 1.30 L_0\); thickness drops by 15%, so \(T_1 = 0.85 T_0\).
From volume constancy, \(W_1 = \dfrac{W_0}{1.30 \times 0.85} = \dfrac{W_0}{1.105} = 0.9050\,W_0\).

Step 3: Find the true strains in width and thickness.
Width strain: \(\varepsilon_w = \ln(0.9050) = -0.0998\).
Thickness strain: \(\varepsilon_t = \ln(0.85) = -0.1625\).

Step 4: Compute the normal anisotropy.
Normal anisotropy is the ratio of width strain to thickness strain: \(R = \dfrac{\varepsilon_w}{\varepsilon_t} = \dfrac{-0.0998}{-0.1625} = 0.614\).

Final Answer:
The normal anisotropy of the sheet metal is close to 0.61. \[ \boxed{R \approx 0.61} \]