Question

A solid cylinder of diameter 100 mm and height 50 mm is forged between two frictionless flat dies to a height of 25 mm. The percentage change in diameter is

• A. 0
• B. 2.07
• C. 20.7
• D. 41.4

Hint:

Since height is halved ($50 \text{ mm}$ to $25 \text{ mm}$), the area must double to keep volume constant.
Thus, $D_2^2 = 2 D_1^2 \implies D_2 = \sqrt{2} D_1 \approx 1.414 D_1$. This immediately gives a $41.4\%$ increase.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This is a metal forming problem where a solid cylinder undergoes plastic deformation during upset forging.
We need to calculate the percentage change in the cylinder's diameter, assuming volume is conserved during plastic deformation.

Step 2: Key Formula or Approach:
During plastic deformation, the volume of the material remains constant:
\[ V_1 = V_2 \]
where:
\[ V_1 = \frac{\pi}{4} D_1^2 h_1 \quad \text{and} \quad V_2 = \frac{\pi}{4} D_2^2 h_2 \]

Step 3: Detailed Explanation:

• The initial diameter is $D_1 = 100 \text{ mm}$, and the initial height is $h_1 = 50 \text{ mm}$.

• The final height after forging is $h_2 = 25 \text{ mm}$.

• Equating the initial and final volumes:
\[ \frac{\pi}{4} D_1^2 h_1 = \frac{\pi}{4} D_2^2 h_2 \]
\[ D_1^2 h_1 = D_2^2 h_2 \]

• Substitute the known values:
\[ 100^2 \times 50 = D_2^2 \times 25 \]
\[ D_2^2 = 10000 \times \frac{50}{25} = 20000 \]
\[ D_2 = \sqrt{20000} = 100 \sqrt{2} \approx 141.42 \text{ mm} \]

• Now, calculate the percentage change in diameter:
\[ \% \text{ Change} = \frac{D_2 - D_1}{D_1} \times 100 \]
\[ \% \text{ Change} = \frac{141.42 - 100}{100} \times 100 = 41.42\% \approx 41.4\% \]

Step 4: Final Answer:
The percentage change in diameter is approximately $41.4\%$.