A cylindrical cup of $48.5 m m$ diameter and $52 m m$ height has a corner radius of $1.5 m m$ . Cold rolled steel sheet of thickness $1.5 m m$ is used to produce the cup. Assume trim allowance as $2 m m$ per $25 m m$ of cup diameter. What is the blank size in $m m$ ?

102.0

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115.4

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120.5

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128.5

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Solution

The correct option is B $115.4$
Here, we have
$48.5$ (cup dia) $> 20 r = 20 \times 1.5 = 30 m m$
So, Blank size $= \sqrt{d^{2} + 4 d h}$
$= \sqrt{(48.5)^{2} + 4 \times 48.5 \times 52} = 111.53 m m$
$Trim allowance = \frac{2}{25} \times 48.5 = 3.88 m m$
$Total blank size = 111.53 + 3.88 = 115.41 m m$

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A cylindrical cup of 48.5mm diameter and 52mm height has a corner radius of 1.5mm. Cold rolled steel sheet of thickness 1.5mm is used to produce the cup. Assume trim allowance as 2mm per 25mm of cup diameter. What is the blank size in mm?

The correct option is B 115.4Here, we have 48.5 (cup dia) >20r=20×1.5=30mm So, Blank size =√d2+4dh =√(48.5)2+4×48.5×52=111.53mm Trim allowance=225×48.5=3.88mm Total blank size=111.53+3.88=115.41mm

A cylindrical cup of $48.5 m m$ diameter and $52 m m$ height has a corner radius of $1.5 m m$ . Cold rolled steel sheet of thickness $1.5 m m$ is used to produce the cup. Assume trim allowance as $2 m m$ per $25 m m$ of cup diameter. What is the blank size in $m m$ ?