# Manufacturing Engineering Questions & Answers – Cutting Forces Power

This set of Manufacturing Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Cutting Forces and Power”.

1. A single-point cutting tool with 120C rake angle is used to machine a steel work-piece. The depth of cut, i.e., uncut thickness is 0.81 mm. The chip thickness under orthogonal machining condition is 1.8 mm. The shear angle is approximately
a) 220C
b) 260C
c) 560C
d) 760C

Explanation: Given : α= 12c, t= 0.81mm, tc= 1.8 mm
Shear angle, tanφ = [rcosα/1 – rsinα] ……….(i)

Chip thickness ratio, r = t/tc
= 0.81/1.8
= 0.45

From equation (i), tanφ = [0.45cos120/1 – 0.45sin120]

φ = tan-1(0.486) = 25.910 – 260.

2. In a single point turning tool, the side rake angle and orthogonal rake angle are equal. ϕ is the principal cutting edge angle and its range is 00<ϕ<900. The chip flows in the orthogonal plane. The value of ϕ is closest to
a) 00
b) 450
c) 600
d) 900

Explanation: Interconversion between ASA (American Standards Association) system and ORS (Orthogonal Rake System)
tanαs = sinφtanα − cosφtani
where αs = Side rake angle
α = orthogonal rake angle
φ = principle cutting edge angle = 00<ϕ<900
i = inclination angle (i = 0 for ORS)
αs = α (Given)
tanαs =sinφtanα − cosφtan(00)
tanαs = sinφtanα
tanαs/tanα = sinφ

φ =sin-1(1)=900.

3. In an orthogonal machining operation :
Uncut thickness = 0.5 mm
Cutting speed = 20 m/min
Rake angel = 150
Width of cut = 5 mm Chip thickness = 0.7 mm
Thrust force = 200 N Cutting force = 1200 N
The values of shear angle and shear strain, respectively, are
a) 30.30 and 1.98
b) 30.30 and 4.23
c) 40.20 and 2.97
d) 40.20 and 1.65

Explanation: Given : t= 0.5mm, V = 20 m/min, α= 150, w= 5mm, tc = 0.7 mm,
Ft = 200 N, Fc = 1200 N
We know, from the merchant’s theory
Chip thickness ratio, r = t/tc = 0.5/0.7 = 0.714

For shear angle, tanφ = [rcosα/1 – rsinα]

Substitute the values, we get
tanφ = [0.714cos150/1 – 0.714sin150] = 0.689/0.815 = 0.845

φ = tan-1(0.845) = 40.20
Shear strain, s = cotφ + tan(φ − α)
s = cot(40.20) + tan(40.20 − 150)
= cot 40.20 + tan 25.2 = 1.183 + 0.470 = 1.65.

4. Which of the following parameters govern the value of the shear angle in continuous chip formation?
a) true feed
b) chip thickness
c) rake angle of the cutting tool
d) all of the mentioned

Explanation: To find a shear angle in continuous chip feed, rake angle, chip thickness, cutting ratio are required.

5. In determining the various forces on the chip, Merchant assumed that the
a) cutting edge of the tool is sharp and it does not make any flank contact with the workpiece
b) only continuous chip without built up edge is produced
c) cutting velocity remains constant
d) all of the mentioned

Explanation: None

6. Cutting forces can be measured using a
a) transducer
b) dynamometer
d) all of the mentioned

Explanation: Cutting forces can be measured using a force transducer (typically with quartz piezoelectric sensors), a dynamometer, or a load cell (with resistance-wire strain gages placed on octagonal rings) mounted on the cutting-tool holder.

7. Transducers have a much __________ natural frequency and stiffness than dynamometers.
a) higher
b) lower
c) equal
d) none of the mentioned

Explanation: Transducers have a much higher natural frequency and stiffness than dynamometers, which are prone to excessive deflection and vibration.

8. In metal machining, the zone where the heat is generated due to friction between the moving chip and the tool face is called
a) friction zone
b) work tool contact zone
c) shear zone
d) none of the mentioned

Explanation: In metal machining, the zone where the maximum heat is generated due to the plastic deformation of metal, is called shear zone while the zone where the heat is generated due to friction between the moving chip and the tool face, is called friction zone.

9. In metal machining, the zone where the maximum heat is generated due to the plastic deformation of metal is called:
a) friction zone
b) work tool contact zone
c) shear zone
d) none of the mentioned

Explanation: In metal machining, the zone where the maximum heat is generated due to the plastic deformation of metal, is called shear zone while the zone where the heat is generated due to friction between the moving chip and the tool face, is called friction zone.

10. The ratio of the cutting force to the cross-sectional area being cut is called:
a) specific cutting force
b) thrust force
c) frictional force
d) none of the mentioned

Explanation: The ratio of the cutting force to the cross-sectional area being cut (i.e., the product of the width of cut and depth of cut) is referred to as the specific cutting force. The thrust force, Ft, acts in a direction normal to the cutting force.

Sanfoundry Global Education & Learning Series – Manufacturing Engineering.

To practice all areas of Manufacturing Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers. 