The cutting length of the broach for the key way cutting in a $150 m m$ long cast iron gear hub. given the width of the key way is $9 m m$ depth of key way is $4.5 m m$ of finishing teeth $= 4$ and no.of semi finishing teeth $= 8$

Rise per tooth Pitch

Roughing teeth $0.1 m m$ $22 m m$

Semi finishing teeth $0.0125 m m$ $20 m m$

Finishing teeth $0$ $20 m m$

968 m m

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160 m m

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1208 m m

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None of these

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Solution

The correct option is C $1208 m m$
$d - 4.5 m m$
$d_{f i n i s h i n g} = 0$
$d_{s} = 0.0125 \times 8 = 0.1$
$d_{r} = d_{t o t a l} - (d_{f} + d_{s})$
$4.5 - 01 = 4.4$
$η_{r} = \frac{d_{r}}{h_{r}} = \frac{4.4}{0.1} = 44$ teeth
Cutting length $=$ effective length
$= L_{r} + L_{s} + L_{r}$
$= 44 \times 22 + 8 \times 20 + 4 \times 20$
$= 1208 m m$

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