Answer:
Roll force, F = 5.6 MN
Explanation:
Sheet width, b = 0.5 m
Roll contact length, [tex]l_{p} = 40 mm[/tex]
Strength coefficient, [tex]\sigma_{0} = 200 MPa[/tex]
Thickness, h = 11 mm
Since the sheet of metal is cold worked by 40%, the reduction in thickness will be:
Δh = 40% * 11 = 0.4 * 11 = 4.4 mm
Strain, e = (Δh)/h
e = 4.4/11 = 0.4
The roll force is calculated by the formula:
[tex]F = \sigma_{f} l_{p} b[/tex]
[tex]\sigma_{f} = \sigma_{0} (e+1)\\\sigma_{f} = 200 (0.4+1)\\\sigma_{f} = 200 *1.4\\\sigma_{f} = 280 MPa[/tex]
Substituting the value of [tex]\sigma_{f}[/tex], [tex]l_{p}[/tex], and b into the formula for the roll force:
[tex]F = \sigma_{f} l_{p} b\\F = 280 * 0.04 * 0.5\\F = 5.6 MN[/tex]