, Uncut chip thickness 0.148 mm, Depth of Cut 2 mm, Cutting Force 1500N, Thrust force: 1000N.

Calculate

(i) Chip reduction coefficient,

(ii) Shear Angle,

(iii) Shear Force,

(iv) Force Normal to the shear plane,

(v) Frictional Force

(vi) Normal to frictional force

(vii) Shear stress

(viii) Shear strain

(ix) coefficient of friction

(x) Resultant Force.

Given

γ = $10°$

$V_{c}$=50m/min

$t_{2}$= 0.4mm

$t_{1}$= 0.148mm

b= 2mm

$F_{c}$=1500N

$F_{t}$=1000N

Soln:

(i) Chip reduction coefficient:

$Rc=\frac{t_{1}}{t_{2}} =\frac{0.148}{0.4}$=0.37

(ii) Shear Angle:

$tan ∅=\frac{r_c .cosγ}{1-r_(c ).sinγ}$

∴tan ∅=$\frac{0.37 .cos10}{1-0.37.sin10}$

∴tan ∅=0.38

$∴∅=21.27°$

(iii) Shear Force:

$F_{s}=F_{c} cos∅ - F_{t} sin∅$

$∴ F_{s}= 1500 \times 0.93 – 1000 \times 0.36$

∴ $F_{s} = 1032.23 N$

(iv) Force Normal to the shear plane:

$F_{N}=F_{t} cos∅ + F_{c} sin∅$

∴ $F_{N} = 1000 \times 0.93 + 1500 \times 0.36$

∴ $F_{N}$ = 1470N

(v) Frictional Force:

F=$F_{t} cosγ + F_{c} sinγ$

∴ F= 984.8 + 260.47

∴ F= 1245.27N

(vi) Normal to frictional force

N=$F_{c} cosγ – F_{t} sinγ$

∴ N= 1477.2 – 173.65

∴ N= 1300.55N

(vii) Shear stress :

$f_{s}=\frac{F_{s}}{A_{S}} =\frac{F_{s}}{\frac{t_{1}.b}{sin∅}}$

∴$f_s=\frac{1032.23}{\frac{0.148 ×2}{sin⁡(21.27)}} = 1265.05 N⁄mm{^2}$

(viii) Shear strain :

Shear strain=$cot∅ + tan (∅-γ)$

=2.56 + 0.199

= 2.759

(ix) coefficient of friction :

$μ=\frac{F}{N}=\frac{1245.27}{1300.55}$=0.95

(x) Resultant Force.:

R=$\sqrt{F^2+N^2}= 1800.59N$