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In an orthogonal cutting, the following observations were made. Rake Angle: 10, Cutting Speed: 50 m/min, Chip Thickness 0.4 mm

, 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.

1 Answer
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Given

γ = 10°

Vc=50m/min

t2= 0.4mm

t1= 0.148mm

b= 2mm

Fc=1500N

Ft=1000N

Soln:

(i) Chip reduction coefficient:

Rc=t1t2=0.1480.4=0.37

(ii) Shear Angle:

tan=rc.cosγ1r(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

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