In: Mechanical Engineering
a rectangular slab of 2024-T6 alluminium alloy is to be machined by a cutting tool of a rake angle of 6.5 degrees. the friction coefficiet between the tool and the slabmaterial was found to be 0.32 . if the shear yield stress of the alluminium alloy is 210 MN/m^2 , calculate the force in the direction of the tool feed (Fx) for a depth cut (t) of 0.8 mm and a unit tool width of 1mm.
Rake angle = 6.5
friction coefficient between tool and slab material = 0.32
Shear yield stress = 210 MN/m2.
Depth of cut =t = 0.8 mm.
Tool width =w=1 mm.
Cutting force Fc =
Where Fs = shear force
= rake angle = 6.5
Coefficient of friction =
Angle == 17.740.
Shear force = Shear yield strength * shear area.
Shear area = As =
where = shear plane angle.
From merchant relation, the shear plane angle is determined as
Shear plane angle = = 45+(6.5/2) - (17.74/2) = 39.40.
Shear area = As = t*w / sin 39.4 = 1.26 * 10-6 m2.
Shear force = Yield shear stress * shear area = 210*106 * 1.26*10-6 = 264.6 N.
Cutting force Fc =
Cutting force = Fc = 264.6 cos(17.74-6.5) / cos(39.4 + 17.74 - 6.5) = 409 N