A strip with cross-sectional area 150 mm - 4-5 mm is being rolled with 20

Bite angle: nbsp; θ = tan^ 1 ( √(RΔ h)R Δ h2 ) From question, we have 20 150 × 4.5 nbsp;× 0.8 = 150 nbsp;× h h = 3.6 mm Δ h = 4.5 3.6 = 0.9 mm Give ...

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Question

A strip with cross-sectional area 150 mm $\times$ 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rollers. The angle subtended by the deformation zone at the roll centre is (in radians)

0.01

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0.06

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Solution

The correct option is D 0.06
Bite angle:
$θ = t a n^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\sqrt{R Δ h}}{R - \frac{Δ h}{2}} ⎞ ⎟ ⎟ ⎟ ⎠$

From question, we have 20% reduction in cross-sectional area

150 $\times$ 4.5 $\times$ 0.8 = 150 $\times$ h

$h = 3.6$ mm

$Δ h = 4.5 - 3.6 = 0.9$ mm

Given, $R = \frac{450}{2} = 225$ mm

$θ = t a n^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\sqrt{225 \times 0.9}}{225 - \frac{0.9}{2}} ⎞ ⎟ ⎟ ⎟ ⎠$

$= {3.62}^{\circ} = 0.06$ rad

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A strip with cross-sectional area 150 mm × 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rollers. The angle subtended by the deformation zone at the roll centre is (in radians)

A strip with cross-sectional area 150 mm $\times$ 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rollers. The angle subtended by the deformation zone at the roll centre is (in radians)