Key_HW_2-1 - Production Engineering HW#2 Fall 07 2 3 4 5.A...

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Production Engineering HW #2 Fall 07 2. 3. 4. 5.A rolling operation is to reduce a 8 foot wide sheet of 7 gage (0.1793") steel to 12 gage (0.1046") in a single pass. 12 GA 0.1046” 7 GA 0.1793” 8 ft = 96 in
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a) What is the minimum roll diameter if you assume the coefficient of friction ( μ ) is 0.15? Δ = h 1 – h 0 = 0.1046 - 0.1793 = 0.0747; R min = Δ / μ 2 = 0.0747 / (.15) 2 = 3.32 in => D min = 6.64 in b) Double that roll diameter to allow for some margin of safety in all (b - e) the following. R’=6.64” c) If the resistance to deformation (1.2 Y) is 50,000 psi, what is the force on a roll in short tons? The force on a roll is expressed as: 1.2 1.2(50000)(96) (6.64)(0.0747) 1.417 6 709 F Yw R e lb T = Δ = = = F=709 T d) What is the torque on the roll? The torque on a roll is expressed as: 1.2 / 2 (50000)(96)(6.64)(0.0747)/ 2 1,428,500 T YwR lbs in = Δ = = e) If the roll turns at 500 rpm, what is the velocity in fpm (feet/min) of the sheet? (Note: since Δ << R, assume v = R ω ) Roll rpm = 500 rpm v = R ω = (6.64)(500 rev/min)(2 π rad/rev)(1ft/12 in) v = 1738 ft/min f) What power (HP) is required to drive both rolls?
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