Question

A strip of 200 mm wide and 10 mm thick is reduced to 8 mm in one pass. Roll diameter = 100 mm, average flow stress = 200 MPa. Rolling load is:

• A. 0.4 MN
• B. 0.8 MN
• C. 1.2 MN
• D. 1.6 MN

Hint:

Be careful with the roll diameter and roll radius.
Always convert the given diameter to radius ($R = D/2$) before substituting into the contact length formula $L = \sqrt{R \Delta h}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This manufacturing engineering question asks for the rolling load (separating force) required during a flat rolling process.

Step 2: Key Formula or Approach:
The rolling load ($F$) can be estimated as the product of the contact area and the average flow stress ($\sigma_m$):
\[ F = L \cdot w \cdot \sigma_m \]
where:
$w$ is the width of the strip,
$L$ is the projected contact length between the rolls and the sheet, calculated as:
\[ L = \sqrt{R \cdot \Delta h} \]
and $\Delta h = h_0 - h_f$ is the draft, with $R$ being the roll radius.

Step 3: Detailed Explanation:

• Identify the given parameters:
- Width, $w = 200 \text{ mm}$
- Initial thickness, $h_0 = 10 \text{ mm}$
- Final thickness, $h_f = 8 \text{ mm}$
- Roll diameter, $D = 100 \text{ mm} \implies$ Roll radius, $R = 50 \text{ mm}$
- Average flow stress, $\sigma_m = 200 \text{ MPa} = 200 \text{ N/mm}^2$

• Calculate the draft ($\Delta h$):
\[ \Delta h = h_0 - h_f = 10 - 8 = 2 \text{ mm} \]

• Calculate the projected contact length ($L$):
\[ L = \sqrt{R \cdot \Delta h} = \sqrt{50 \times 2} = \sqrt{100} = 10 \text{ mm} \]

• Calculate the rolling load ($F$):
\[ F = L \cdot w \cdot \sigma_m \]
\[ F = 10 \text{ mm} \times 200 \text{ mm} \times 200 \text{ N/mm}^2 \]
\[ F = 400,000 \text{ N} = 0.4 \text{ MN} \]

Step 4: Final Answer:
The required rolling load is $0.4 \text{ MN}$.