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 ...

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Analysis of Orthogonal Machining

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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

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0.46

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0.847

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1.16

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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

$ϕ + β - α = 45^{\circ}$

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

$β = {40.26}^{\circ}$

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

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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