Question

A 10 mm thick plate is rolled to 7 mm thickness in a rolling mill using 1000 mm diameter rigid rolls. The neutral point is located at an angle of 0.3 times the bite angle from the exit. The thickness (in mm, up to two decimal places) of the plate at the neutral point is

Answer 1

Answer:

Given:

Original thickness ho = 10 mm, final thickness hf = 7 mm, the diameter of rolls D = 1000 mm and angle of neutral point = 0.3 × bite angle (α).

We know,

⇒ α = 0.775 rad or 4.932°

Now, the angle of neutral point θ = 0.3 × 4.932 = 1.479°

∴ The thickness of the plate at neutral point h = hf + D(1 - cos θ )

⇒ h = 7 + 1000(1 - cos 1.4796) = 7.27 mm

∴ The thickness of the plate at neutral point is 7.27 mm.

Explanation:

hope it was useful.

Answer 2

Answer:

The thickness of the plate in rolling operation can be given as follow:

h = hf + D(1 - cos θ ), hf = final thickness of the strip, D = diameter of the rolls and θ = angle of contact.

Calculation:

Given:

Original thickness ho = 10 mm, final thickness hf = 7 mm, the diameter of rolls D = 1000 mm and angle of neutral point = 0.3 × bite angle (α).

We know,

⇒ α = 0.775 rad or 4.932°

Now, the angle of neutral point θ = 0.3 × 4.932 = 1.479°

∴ The thickness of the plate at neutral point h = hf + D(1 - cos θ )

⇒ h = 7 + 1000(1 - cos 1.4796) = 7.27 mm

∴ The thickness of the plate at neutral point is 7.27 mm.