10. 10
Orthogonal cutting
1- Cutting tool travel in the
direction perpendicular to
the cutting edge.
2-The cutting edge clear
either end of work piece.
3- Chip flows in the
direction perpendicular to
the cutting edge.
4-Two mutually
perpendicular cutting
forces act on the work
piece.
Oblique cutting
1-cutting edge travels,
making an angle with the
normal of cutting edge.
2-The cutting edge may or
may not clear either end of
work piece.
3- chip flows, making an
angle with normal of
cutting edge.
4-Three mutually
perpendicular forces are
involved.
12. Cutting forces acting during oblique
turning operation
Feed force, Ff - along z-axis
Radial force, Fr - along x-axis, radial direction
Cutting force, Fc- along y-axis perpendicular to x-z plane
Merchant circle diagram is drawn for the
forces acting in orthogonal plane, i.e. the
plane containing cutting force Fc and Ft (
resultant of Ff and Fr)
12
13. Normal rake angle (αn)
Where αb and αs = back and side rake angles
Ψ =side cutting edge angle
Shear angle (φn)
13
nc
nc
n
sinr1
cosr
tan
16. 16
cos
d
wcut.of.Width
cosftthickness.chip.Uncut 1
cc1 fdvvwtmrrrate.removal.Metal
Where,
Cutting velocity = vc = πDN
D = Average diameter of the
work
N = RPM
ψ = side cutting edge
angle
d, f = depth of cut, feed/
revolution
18. Machining time (tc)
(for a single cut)
fN
t
L
c
18
Where,
L = Length of the work along the axis
F = feed/ revolution
N = rpm of the work
19. 19Problem
The following data are available from a turning operation:
* s stands for (second digit of serial number)
Determine the (a) shear angle (b) friction coefficient (c) shear stress and shear strain
on shear plane(d) chip velocity and shear velocity and (e) energies uf, us and ut.
Work material Aluminium Steel
Tool signature 9,s*,6,7,10,15,9 mm 9,s*,6,7,10,15,9
mm
Depth of cut, mm 2.5 2.5
Cutting speed, m/s 2 2
Chip thickness, mm 0.23 0.58
Cutting force, Fc, N 430 890
Thrust force, Ft, N 280 800
