a rectangular slab of 2024-T6 alluminium alloy is to be machined by a cutting tool of...

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.

Solutions

Expert Solution

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 F_c =

Where Fs = shear force

= rake angle = 6.5

Coefficient of friction =

Angle == 17.74⁰.

Shear force = Shear yield strength * shear area.

Shear area = A_s =

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 = A_s = 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 F_c =

Cutting force = Fc = 264.6 cos(17.74-6.5) / cos(39.4 + 17.74 - 6.5) = 409 N

samet mamat answered 1 year ago

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