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

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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.

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The thickness of the plate at neutral point is

✞︎ 7.27 mm.

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

$\boxed{ \red{\bf \:h = hf + D(1 - cos θ )}} \color{red} \ast$

Where ,

$\boxed{\bfθ = angle \: of \: contact}$

$\boxed{\bf D = diameter \: of \: the \: rolls}$

${\boxed{\bf h_f = final \: thickness \: of \: the \: strip}}$

$\diamond \: \sf \color{red}Given$

$\footnotesize \bf \: Original \: Thickness (h_o) = 10 mm$

$\footnotesize \bf \: final \: Thickness (h_f) = 7mm$

$\footnotesize\bf Diameter \: of \: the \: rolls = 1000 \: mm$

$\footnotesize\bf Angle \: of \: neutral \: point = 0.3 × bite \: angle \: ( \color{red}{α}).$

We know that ,

$: \implies\bf \cos( \alpha ) = 1 - \dfrac{h_o - h_f}{D}$

$: \implies1 - \dfrac{3}{100} = 0.997$

Therefore

$\leadsto \boxed{\bf \alpha = 0.775 \: rad }$

Now, the angle of neutral point θ

= 0.3 × 4.932 = 1.479 °

∴The thickness of the plate at neutral point

$\red{ \bf h = h_f+ D(1 - cos θ )}$

H = 7 + 1000(1 - cos 1.4796)

= 7.27 mm [ correct upto 2 dp ] .

Conclusion

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

Thankyou

Answer 2

7.27 mm is the required answer