Two separate slab milling operations, $1$ and $2$ , are performed with identical milling cutters. The depth of cut in operation $2$ is twice that in operation $1$ . The other cutting parameters are identical. The ratio of maximum uncut chip thickness in operations $1$ and $2$ is

Open in App

Solution

$t_{m a x} = \frac{2 f}{N Z} \sqrt{\frac{d}{D}}$
Formula for maximum uncut thickness
$∴ \frac{t_{1}}{t_{2}} = \sqrt{\frac{d_{1}}{d_{2}}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = 0..707$
$∵, D_{1} = D_{2}$

Suggest Corrections

Similar questions

Q. During a conventional milling operation, if depth of cut gets doubled where as the cutter diameter reduces to half, then what will be the percentage change in maximum uncut chip thickness ?

Q. Common Data for Q 6.124 and Q.6.125
Data for plain milling operation are given below:
Length of work piece = 200 mm,
Cutter speed = 100 rpm,
Feed = 200 mm/min, Depth of cut = 2 mm,
Tota clearance (entry and exit) = 5 mm

Mean un-deformed chip thickness (in microns)

Q. Feed rate in slab milling operation is equal to

Common Data:
An orthogonal operation is carried out at $20 m / m i n$ cutting speed, using a cutting tool of rake angle $15$ degree. The chip thickness is $0.4 m m$ and uncut chip thickness is $0.2 m m$

The chip velocity (in $m / m i n$ ) is

Q. In orthogonal cutting operation, the ratio of the chip thickness to the depth of cut is 1.5. If the cutting speed is 10 m/s, then the chip flow velocity will be

Join BYJU'S Learning Program

Two separate slab milling operations,1 and 2, are performed with identical milling cutters. The depth of cut in operation 2 is twice that in operation 1. The other cutting parameters are identical. The ratio of maximum uncut chip thickness in operations 1 and 2 is

tmax=2fNZ√dD Formula for maximum uncut thickness ∴t1t2=√d1d2=√12=1√2=0..707 ∵,D1=D2

Two separate slab milling operations, $1$ and $2$ , are performed with identical milling cutters. The depth of cut in operation $2$ is twice that in operation $1$ . The other cutting parameters are identical. The ratio of maximum uncut chip thickness in operations $1$ and $2$ is

$t_{m a x} = \frac{2 f}{N Z} \sqrt{\frac{d}{D}}$
Formula for maximum uncut thickness
$∴ \frac{t_{1}}{t_{2}} = \sqrt{\frac{d_{1}}{d_{2}}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = 0..707$
$∵, D_{1} = D_{2}$