On a certain mill the rolling load for a 30 percent reduction is 7.6 kN/mm of width. What is the rolling load when a back tension of 150 MPa and a front tension of 120 MPa are applied? Assume that S 500 MPa.
The Correct Answer and Explanation is :
To solve this problem, we can use the relationship between rolling load, tension, and reduction in width. The formula for the rolling load is:
[
P = \sigma \cdot A
]
Where:
- ( P ) is the rolling load
- ( \sigma ) is the average flow stress, which is the average of the back and front tension in this case
- ( A ) is the contact area (width of the mill and reduction in thickness)
Given:
- Rolling load for 30% reduction is ( 7.6 \, \text{kN/mm of width} )
- Back tension = 150 MPa
- Front tension = 120 MPa
- Material yield stress ( S ) = 500 MPa
Step 1: Determine the average flow stress
The flow stress is the average of the back and front tension:
[
\sigma = \frac{\text{Back Tension} + \text{Front Tension}}{2} = \frac{150 \, \text{MPa} + 120 \, \text{MPa}}{2} = 135 \, \text{MPa}
]
Step 2: Find the relationship between rolling load and tension
The rolling load is proportional to the tension and width of the mill. Since we know the rolling load for a 30% reduction is 7.6 kN/mm of width, we can calculate the rolling load for the given tension values by scaling accordingly. The stress ( \sigma ) affects the rolling load in proportion to its value relative to the yield stress ( S ).
Step 3: Calculate the new rolling load
Since the back and front tension are now 135 MPa on average, which is less than the material’s yield stress (500 MPa), the rolling load will adjust proportionally. The proportional factor is the ratio of the average flow stress to the material’s yield stress:
[
\text{Proportional Factor} = \frac{\sigma}{S} = \frac{135}{500} = 0.27
]
Now, we calculate the new rolling load by multiplying the given rolling load by this proportional factor:
[
P_{\text{new}} = 7.6 \, \text{kN/mm} \times 0.27 = 2.05 \, \text{kN/mm of width}
]
Thus, the rolling load when a back tension of 150 MPa and a front tension of 120 MPa are applied is approximately 2.05 kN/mm of width.