solved anna university problems on cutting forces

SOLVED ANNA UNIVERSITY PROBLEMS ON CUTTING FORCES

AU Problem 1.1

The following data pertain to an orthogonal cutting test. Determine the forces on the tool face.

Rake angle = 15°

Uncut chip thickness = 0.38

Chip thickness ratio = 0.5 mm

Width of cut = 3 mm

Yield stress of metal in shear = 280 N/mm2.

Coefficient of tool-chip friction = 0.7.

Similar to Problem 1.5 on Page 1.39.

 [Ans:- Shear force, Fs = 1113.96 N, Resultant force, F = 1502.53 N, Tangential force or cutting force, Fz = 1411.0 N, Radial force (or) Feed force, Fx = 513.89 N and Normal force or feed force, N = 1230.78 N]

AU Problem 1.2

During machining C20 steel with a carbide cutting tool having a tool geometry given by 0-5-6-6-8-75-1 mm ORS, the following forces have been recorded by a two dimensional dynamometer:

Cutting force = 1300 N

Feed force = 800 N

Determine the following:

Radial component of force, frictional force and normal force and kinetic coefficient of friction μ

Given data:

Rake angle, α = 5°

Side cutting angle, γ = 75°

Fz = 1300 N

Fx = 800 N

Solution:

Normal force, N = Fz cos α - Fx sin α

AU Problem 1.3

In orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness before the cut = 0.3 mm and the cut yields a deformed chip thickness = Calculate the (1) shear plane angle and (2) shear strain for the operation.

Given data:

α = 15°

t1 = 0.3 mm

t2 = 0.65 mm

Solution:

AU Problem 1.4

In an orthogonal cutting test with a tool of rake angle 10o, the following observations were made:

Chip thickness ratio = 0.3

Horizontal component of cutting force = 1290 N

Vertical component of cutting force = 1650 N

From Merchant's theory, calculate the various components of the cutting forces and the coefficient of friction at the chip tool interface.

Given data:

α = 10°

Fz = 1290 N

Fx = 1650 N

r = 0.3

Solution:

Coefficient of friction at chip-tool interface

AU Problem 1.5

In an orthogonal cutting test with a tool of rake angle 8o, the following observations were made:

Chip thickness ratio = 0.2

Horizontal component of the cutting force = 1190 N

Vertical component of the cutting force = 1450 N

From Merchant's theory, calculate the various components of the cutting forces and the coefficient of friction at the chip tool interface. 

Similar to AU Problem 1.4 on Page 1.54.

 [Ans:- P = 1601.51 N, N = 976.62N, Fs = 876.45 N, Fn = 1658.44 N and F = 1875.79 N]

AU Problem 1.6

A specimen of 100 mm length along the stroke of shaper is machined with a tool with 15o rake angle. The uncut chip thickness is 1.5 mm. If a chip length of 40 mm is obtained during one stroke of machining, find the shear plane angle and thickness of cut-chip.

Similar to Problem 1.2 on Page 1.37.

[Ans:- Shear angle, β = 23.3° and chip thickness, t2 = 3.75 mm]

AU Problem 1.7

The following data was obtained from an orthogonal cutting test. Rake angle = 20o, depth of cut 6 mm, feed rate = 0.25 mm/rev, cutting speed = 0.6 m/s, chip length before cutting - 29.4 mm, vertical cutting force = 1050 N, horizontal cutting force = 630 N, chip length after cutting = 12.9 mm. Using Merchant's analysis, calculate the (i) magnitude of resultant force, (ii) shear plane angle, (iii) friction force and friction angle and (iv) various energies consumed.

Given data:

α = 20°

d = 6 mm

t1 = 0.25 mm/rev.

V = 0.6 m/s

l1 = 29.4 mm 

l2 = 12.9 mm 

Fx = 1050 N

Fz = 630 N

Solution:

From Merchant's diagram, normal force can be calculated by

Coefficient of friction at chip-tool interface is calculated by