Problems
21.1 In an orthogonal cutting operation, the tool has a rake angle = 15°. The chip thickness before
the cut = 0.30 mm and the cut yields a deformed chip thickness = 0.65 mm. Calculate (a)
the shear plane angle and (b) the shear strain for the operation.
Solution
: (a)
r
=
t
o
/
t
c
= 0.30/0.65 = 0.4615
φ = tan
-1
(0.4615 cos 15/(1 - 0.4615 sin 15)) = tan
-1
(0.5062) =
26.85
°
(b) Shear strain γ = cot 26.85 + tan (26.85 - 15) = 1.975 + 0.210 =
2.185
21.2 In Problem 21.1, suppose the rake angle were changed to α = 0°. Assuming that the friction
angle remains the same, determine (a) the shear plane angle, (b) the chip thickness, and
(c) the shear strain for the operation.
Solution
: From Problem 21.1, α = 15° and φ = 26.85°. Using the Merchant Equation, Eq.
(21.16):
φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ
β = 2(45) +
α
- 2(φ) = 90 + 15 – 2(26.85) = 51.3°
Now, with α = 0 and β remaining the same at 51.3°, φ = 45 + 0/2 – 51.3/2 =
19.35
°
(b) Chip thickness at α = 0:
t
c
=
t
o
/tan φ = 0.30/tan 19.35 =
0.854 mm
(c) Shear strain γ = cot 19.35 + tan (19.35 - 0) = 2.848 + 0.351 =
3.199
21.5 The cutting force and thrust force in an orthogonal cutting operation are 1470 N and 1589 N,
respectively. The rake angle = 5°, the width of the cut = 5.0 mm, the chip thickness
before the cut = 0.6, and the chip thickness ratio = 0.38. Determine (a) the shear strength
of the work material and (b) the coefficient of friction in the operation.
Solution
: (a) φ = tan
-1
(0.38 cos 5/(1 - 0.38 sin 5)) = tan
-1
(0.3916) = 21.38°
F
s
= 1470 cos 21.38 – 1589 sin 21.38 = 789.3 N
A
s
= (0.6)(5.0)/sin 21.38 = 3.0/.3646 = 8.23 mm
2
S
= 789.3/8.23 = 95.9 N/mm
2
=
95.9 MPa
(b) φ = 45 + α/2 - β/2; rearranging, β = 2(45) + α - 2φ
β = 2(45) +
α
- 2(φ) = 90 + 5 – 2(21.38) = 52.24°
μ = tan 52.24 =
1.291
21.9 In an orthogonal cutting operation, the rake angle = -5°, chip thickness before the cut = 0.2
mm and width of cut = 4.0 mm. The chip ratio = 0.4. Determine (a) the chip thickness
after the cut, (b) shear angle, (c) friction angle, (d) coefficient of friction, and (e) shear
strain.
Solution
: (a)
r
=
t
o
/
t
c
,
t
c
=
t
o
/
r
= 0.2/.4 =
0.5 mm
(b) φ = tan
-1
(0.4 cos(–5)/(1 - 0.4 sin(–5))) = tan
-1
(0.3851) =
21.1
°
(c) β = 2(45) +
α
- 2(φ) = 90 + (-5) - 2(21.8) =
42.9
°
(d) μ = tan 42.9 =
0.93
(e) γ = cot 31.8 + tan(31.8 - 15) = 2.597 + 0.489 =
3.09
21.12 A carbon steel bar 7.64 inch in diameter has a tensile strength of 65,000 lb/in
2
and a shear
strength of 45,000 lb/in
2
. The diameter is reduced using a turning operation at a cutting
speed of 400 ft/min. The feed is 0.011 in/rev and the depth of cut is 0.120 in. The rake
angle on the tool in the direction of chip flow is 13°. The cutting conditions result in a
chip ratio of 0.52. Using the orthogonal model as an approximation of turning, determine