In turning trail using orthogonal cutting- a chip length of 84mm was obtained for uncut chip length of 200 mm- The cutting conditions were-Vc - 30 m-min- uncut chip thickness is 0-5 mm- rake angle is 20- width of cut is 2 mm- cutting force is 900 N- and cutting tool is HSS- The coefficient of friction- if the machining is taking place under Lee and Shaffer shear angle conditions- is

L_2 =84 mm, L_1 =200 mm V_c =30 m/min, t_1 =0.5 mm, b=2 nbsp; mm, α=20^∘ F_c=900 nbsp; N r=84200=0.42 ϕ=tan^ 1(rcosα1 rsinα)=24.74^∘ Lee and shaff ...

You visited us 9 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Production

Analysis of Orthogonal Machining

In turning tr...

Question

In turning trail using orthogonal cutting, a chip length of 84mm was obtained for uncut chip length of 200 mm. The cutting conditions were:
$V_{c} = 30$ m/min, uncut chip thickness is 0.5 mm, rake angle is $20^{\circ}$ , width of cut is 2 mm, cutting force is 900 N, and cutting tool is HSS.

The coefficient of friction, if the machining is taking place under Lee and Shaffer shear angle conditions, is

0.42

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.46

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.847

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.847
$L_{2} = 84 mm, L_{1} = 200 mm$

$V_{c} = 30 m/min, t_{1} = 0.5 mm,$

$b = 2 mm, α = 20^{\circ}$

$F_{c} = 900 N$

$r = \frac{84}{200} = 0.42$

$ϕ = {tan}^{- 1} (\frac{r cos α}{1 - r sin α}) = {24.74}^{\circ}$

Lee and shaffer

$ϕ + β - α = 45^{\circ}$

${24.74}^{\circ} + β - 20^{\circ} = 45^{\circ}$

$β = {40.26}^{\circ}$

$μ = tan β = tan ({40.26}^{\circ}) = 0.847$

Suggest Corrections

Similar questions

Q. In turning trail using orthogonal cutting, a chip length of 84mm was obtained for uncut chip length of 200 mm. The cutting conditions were:

V_{c} = 30

m/min, uncut chip thickness is 0.5 mm, rake angle is

20^{\circ}

, width of cut is 2 mm, cutting force is 900 N, and cutting tool is HSS.

The shear stress induced in the layer of work material is

Q. In a turning process using orthogonal tool geometry, a chip length of 100 mm is obtained for an uncut chip length of 250 mm. The cutting conditions are cutting speed = 30 m/min, rake angle = 20°.
The shear plane angle is degrees (round off to one decimal place).

Q. In an orthogonal machining with a single point cutting tool of rake angle

10^{o}

, the uncut chip thickness and the chip thickness are

0.125 m m

and

0.22 m m

respectivley. Using Merchant's first solution for the condition of minimum cutting force, the coefficient of friction at the chip-tool interface is
(round off to two decimal places).

Common Data:
In an orthogonal machining operation:
Uncut thickness $= 0.5 m m$
Cutting speed $= 20 m / m i n$
Rake angle $= 15^{o}$
Width of cut $= 5 m m$
Chip thickness $= 07 m m$
Thrust force $= 200 N$
Cutting force $= 1200 N$
Assume Merchant's theory.

The coefficient of friction at the tool-chip interface is

Q. 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 shear angle (in degrees) is

Join BYJU'S Learning Program

The correct option is C 0.847L2=84 mm, L1=200 mm Vc=30 m/min, t1=0.5 mm, b=2 mm, α=20∘ Fc=900 N r=84200=0.42 ϕ=tan−1(rcosα1−rsinα)=24.74∘ Lee and shaffer ϕ+β−α=45∘ 24.74∘+β−20∘=45∘ β=40.26∘ μ=tanβ=tan(40.26∘)=0.847