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: Track the three dimensions through the draw.
Call length, width, thickness $L, W, T$. Stretching changes $L$ and $T$ directly; $W$ follows from constant volume.

Step 2: Get the width change from no volume loss.
$L$ becomes $1.30L_0$ and $T$ becomes $0.85T_0$, so $W_1/W_0 = 1/(1.30 \times 0.85) = 1/1.105 = 0.905$.
This means the width shrinks by about 9.5 percent while the sheet stretches in length.

Step 3: Convert both changes to true (logarithmic) strain.
$\varepsilon_w = \ln(0.905) = -0.0998$ and $\varepsilon_t = \ln(0.85) = -0.1625$.

Step 4: Take the ratio that defines normal anisotropy.
$R = \varepsilon_w / \varepsilon_t = 0.614$, which rounds to 0.61.

Final Answer:
The sheet shows a normal anisotropy close to 0.61. \[ \boxed{R \approx 0.61} \]