Question 1 |

In an orthogonal machining operation, the cutting
and thrust forces are equal in magnitude. The uncut
chip thickness is 0.5 mm and the shear angle is
15^{\circ} . The orthogonal rake angle of the tool is 0^{\circ} and
the width of cut is 2 mm. The workpiece material
is perfectly plastic and its yield shear strength is
500 MPa. The cutting force is _________ N (round
off to the nearest integer).

5247 | |

1245 | |

2732 | |

3214 |

Question 1 Explanation:

Given, F_C=F_T

Uncut chip thickness (t_1)=0.5mm

Shear angle (\phi )=15^{\circ}

Orthogonal rake angle \alpha =0^{\circ}

Width of cut w=2mm

Shear strength (\tau _S) =500MPa

Cutting force f_c= ____N

\begin{aligned} F_S&=F_C \cos \phi -F_T \sin \phi \\ &=F_C \cos \phi -F_C \sin \phi \\ \\ &= F_C(\cos \phi - \sin \phi)\\ F_C&=\frac{F_S}{\cos \phi - \sin \phi} \\ &= \frac{\tau _S\cdot w\cdot t_1}{\sin \phi (\cos \phi - \sin \phi)}\\ &=\frac{500 \times 2 \times 0.5}{\sin 15^{\circ}(\cos 15^{\circ}-\sin 15^{\circ})}\\ &=2732 N \end{aligned}

Uncut chip thickness (t_1)=0.5mm

Shear angle (\phi )=15^{\circ}

Orthogonal rake angle \alpha =0^{\circ}

Width of cut w=2mm

Shear strength (\tau _S) =500MPa

Cutting force f_c= ____N

\begin{aligned} F_S&=F_C \cos \phi -F_T \sin \phi \\ &=F_C \cos \phi -F_C \sin \phi \\ \\ &= F_C(\cos \phi - \sin \phi)\\ F_C&=\frac{F_S}{\cos \phi - \sin \phi} \\ &= \frac{\tau _S\cdot w\cdot t_1}{\sin \phi (\cos \phi - \sin \phi)}\\ &=\frac{500 \times 2 \times 0.5}{\sin 15^{\circ}(\cos 15^{\circ}-\sin 15^{\circ})}\\ &=2732 N \end{aligned}

Question 2 |

A straight-teeth horizontal slab milling cutter is
shown in the figure. It has 4 teeth and diameter (D)
of 200 mm. The rotational speed of the cutter is
100 rpm and the linear feed given to the workpiece
is 1000 mm/minute. The width of the workpiece
(w) is 100 mm, and the entire width is milled in
a single pass of the cutter. The cutting force/tooth
is given by F = K t_cw, where specific cutting force
K = 10 N/mm^2
, \; w is the width of cut, and t_c
is the
uncut chip thickness.

The depth of cut (d) is D/2, and hence the assumption of \frac{d}{D} \lt \lt 1 is invalid. The maximum cutting force required is __________ kN (round off to one decimal place).

The depth of cut (d) is D/2, and hence the assumption of \frac{d}{D} \lt \lt 1 is invalid. The maximum cutting force required is __________ kN (round off to one decimal place).

1.8 | |

2.5 | |

3.2 | |

2.9 |

Question 2 Explanation:

Given data, No. of teeth (n)=4

Diameter of cutter (D) = 200 mm

Rotational speed (N) = 100 rpm

Linear feed to work piece = 1000 mm/min

Width of work piece (w) = 100 mm

Cutting force/tooth = F = K t_c w

Specific cutting force = K = 10 N/mm^2

Depth of cut (d)= D/2

Feed(f) = 1000/100 = 10 mm/rev

Uncut chip thickness =t_c

Maximum uncut chip thickness (t_c)_{max}

\begin{aligned} &=\frac{2f}{n}\sqrt{d/D(1-d/D)}\\ &=\frac{2 \times 10}{4}\sqrt{\frac{D/2}{D}\left ( 1- \frac{D/2}{D}\right )}\\ &=5\sqrt{1/2(1-1/2)}=2.5mm\\ &\text{Maximum force }\\ (F)_{max}&=K(t_c)_{max}\cdot w\\ &=10 \times 2.5 \times 100\\ &=2500N=2.5kN \end{aligned}

Diameter of cutter (D) = 200 mm

Rotational speed (N) = 100 rpm

Linear feed to work piece = 1000 mm/min

Width of work piece (w) = 100 mm

Cutting force/tooth = F = K t_c w

Specific cutting force = K = 10 N/mm^2

Depth of cut (d)= D/2

Feed(f) = 1000/100 = 10 mm/rev

Uncut chip thickness =t_c

Maximum uncut chip thickness (t_c)_{max}

\begin{aligned} &=\frac{2f}{n}\sqrt{d/D(1-d/D)}\\ &=\frac{2 \times 10}{4}\sqrt{\frac{D/2}{D}\left ( 1- \frac{D/2}{D}\right )}\\ &=5\sqrt{1/2(1-1/2)}=2.5mm\\ &\text{Maximum force }\\ (F)_{max}&=K(t_c)_{max}\cdot w\\ &=10 \times 2.5 \times 100\\ &=2500N=2.5kN \end{aligned}

Question 3 |

Which of these processes involve(s) melting in
metallic workpieces?

**MSQ**Electrochemical machining | |

Electric discharge machining | |

Laser beam machining | |

Electron beam machining |

Question 3 Explanation:

Non traditional thermal energy process includes electrical discharge machining (EDM), electron beam machining (EBM), and laser beam machining (LBM). EDM removes metal by a series of discrete electrical discharge (sparks) that cause localized temperatures high enough to melt or vaporize the metal in the immediate vicinity of the discharge. EBM adopts a high velocity stream of electrons focused on the workpiece surface to remove material by melting and vaporization. LBM employs the light energy from a laser to remove material by vaporization and ablation.

Question 4 |

A 1 mm thick cylindrical tube, 100 mm in diameter,
is orthogonally turned such that the entire wall
thickness of the tube is cut in a single pass. The
axial feed of the tool is 1 m/minute and the specific
cutting energy (u) of the tube material is 6 J/mm^3.
Neglect contribution of feed force towards power.
The power required to carry out this operation is
_________ kW (round off to one decimal place).

31.4 | |

25.6 | |

87.5 | |

65.5 |

Question 4 Explanation:

Given data, cylindrical tube thickness (t) = 1 mm

Diameter (D) = 100 mm

Orthogonal cutting such that entire wall thickness of tube is cut in single pass.

Therefore, Tube thickness = depth of cut = d = 1 mm

Axial feed of the tool = 1 m/min = f.N

Specific cutting Energy (U) = 6 \; J/mm^3

Power _________ kW

Specific cutting energy(U) = 6 \; J/mm^3=\frac{\text{Power}}{\text{MRR}}

\begin{aligned} Power &= 6 \times MRR \\ &= 6 \times f.d.v\\ &= 6 \times f.d. \pi.D.N\\ &= 6 \times d \times \pi \times D \times f.N\\ &= 6 \times 1 \times \pi \times 100 \times 1 \times \frac{1000}{60}\\ &= 31415.9\;J/sec \; or \; Watt\\ &=31.4 \;kW \end{aligned}

Diameter (D) = 100 mm

Orthogonal cutting such that entire wall thickness of tube is cut in single pass.

Therefore, Tube thickness = depth of cut = d = 1 mm

Axial feed of the tool = 1 m/min = f.N

Specific cutting Energy (U) = 6 \; J/mm^3

Power _________ kW

Specific cutting energy(U) = 6 \; J/mm^3=\frac{\text{Power}}{\text{MRR}}

\begin{aligned} Power &= 6 \times MRR \\ &= 6 \times f.d.v\\ &= 6 \times f.d. \pi.D.N\\ &= 6 \times d \times \pi \times D \times f.N\\ &= 6 \times 1 \times \pi \times 100 \times 1 \times \frac{1000}{60}\\ &= 31415.9\;J/sec \; or \; Watt\\ &=31.4 \;kW \end{aligned}

Question 5 |

Under orthogonal cutting condition, a turning
operation is carried out on a metallic workpiece at a
cutting speed of 4 m/s. The orthogonal rake angle of
the cutting tool is 5^{\circ} . The uncut chip thickness and
width of cut are 0.2 mm and 3 mm, respectively. In
this turning operation, the resulting friction angle
and shear angle are 45^{\circ} and 25^{\circ} , respectively. If
the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa,
then the cutting force is ______________ N (round
off to one decimal place).

1245.3 | |

1547.4 | |

3652.4 | |

2573.4 |

Question 5 Explanation:

Cutting speed (V) = 4m/sec

Orthogonal rake angle \alpha =5^{\circ}

Uncut chip thickness ((t_1)=.0.2mm

Width of cut (w) = 3 mm

Friction angle \beta =45^{\circ}

Shear angle (\phi ) =25^{\circ}

Shear strength (\tau _0 ) =1000MPa

Cutting force (F_c)=____N

\begin{aligned} F_S &= R \cos (\phi +\beta -\alpha )\\ R&=\frac{\tau _0 \cdot A_S}{\cos (\phi +\beta -\alpha )} \\ R&=\frac{\tau _0 \cdot wt_1}{\sin \phi \cos (\phi +\beta -\alpha )} \\ F_c&= R \cdot \cos (\beta -\alpha )\\ &=\frac{\tau _0 \cdot wt_1\cdot \cos (\beta -\alpha )}{\sin \phi \cos (\phi +\beta -\alpha )} \\ &= \frac{1000 \times 3 \times 0.2 \times \cos (45-5)}{\sin 25\cdot \cos (25+45-5)}\\ &=2573.4N \end{aligned}

Orthogonal rake angle \alpha =5^{\circ}

Uncut chip thickness ((t_1)=.0.2mm

Width of cut (w) = 3 mm

Friction angle \beta =45^{\circ}

Shear angle (\phi ) =25^{\circ}

Shear strength (\tau _0 ) =1000MPa

Cutting force (F_c)=____N

\begin{aligned} F_S &= R \cos (\phi +\beta -\alpha )\\ R&=\frac{\tau _0 \cdot A_S}{\cos (\phi +\beta -\alpha )} \\ R&=\frac{\tau _0 \cdot wt_1}{\sin \phi \cos (\phi +\beta -\alpha )} \\ F_c&= R \cdot \cos (\beta -\alpha )\\ &=\frac{\tau _0 \cdot wt_1\cdot \cos (\beta -\alpha )}{\sin \phi \cos (\phi +\beta -\alpha )} \\ &= \frac{1000 \times 3 \times 0.2 \times \cos (45-5)}{\sin 25\cdot \cos (25+45-5)}\\ &=2573.4N \end{aligned}

Question 6 |

Electrochemical machining operations are
performed with tungsten as the tool, and copper and
aluminum as two different workpiece materials.
Properties of copper and aluminum are given in the
table below.

\begin{array}{|c|c|c|c|} \hline \text{Material} & \text{Atomic mass (amu)} & \text{Valency} & \text{Density }(g/cm^3)\\ \hline Copper &63 &2 & 9\\ Aluminum &27 & 3 & 2.7 \\ \hline \end{array}

Ignore overpotentials, and assume that current efficiency is 100% for both the workpiece materials. Under identical conditions, if the material removal rate (MRR) of copper is 100 mg/s, the MRR of aluminum will be ________________ mg/s (roundoff to two decimal places).

\begin{array}{|c|c|c|c|} \hline \text{Material} & \text{Atomic mass (amu)} & \text{Valency} & \text{Density }(g/cm^3)\\ \hline Copper &63 &2 & 9\\ Aluminum &27 & 3 & 2.7 \\ \hline \end{array}

Ignore overpotentials, and assume that current efficiency is 100% for both the workpiece materials. Under identical conditions, if the material removal rate (MRR) of copper is 100 mg/s, the MRR of aluminum will be ________________ mg/s (roundoff to two decimal places).

38.25 | |

31.5 | |

24.86 | |

28.57 |

Question 6 Explanation:

Copper At. Wt = 63, Valency = 2

e=\frac{\text{At.\; Wt}}{\text{Valency}}=\frac{63}{2}=31.5

Aluminium At. Wt = 27, Valency = 3

e=\frac{27}{3}=9

Material Removal Rate in g/s is given by

MRR=\frac{e.I}{F}

e = Chemical equivalent

I = Current

F = Faraday constant

At Constant current & Faraday's constant

\begin{aligned} MRR\propto e\\ \frac{MRR_{Cu}}{MRR_{Al}}&=\frac{e_{Cu}}{e_{Al}}\\ \frac{100}{MRR_{Al}}&=\frac{31.5}{9}\\ MRR_{Al}&=28.57 mg/s \end{aligned}

e=\frac{\text{At.\; Wt}}{\text{Valency}}=\frac{63}{2}=31.5

Aluminium At. Wt = 27, Valency = 3

e=\frac{27}{3}=9

Material Removal Rate in g/s is given by

MRR=\frac{e.I}{F}

e = Chemical equivalent

I = Current

F = Faraday constant

At Constant current & Faraday's constant

\begin{aligned} MRR\propto e\\ \frac{MRR_{Cu}}{MRR_{Al}}&=\frac{e_{Cu}}{e_{Al}}\\ \frac{100}{MRR_{Al}}&=\frac{31.5}{9}\\ MRR_{Al}&=28.57 mg/s \end{aligned}

Question 7 |

Which of the following additive manufacturing technique(s) can use a wire as a feedstock material?

Stereolithography | |

Fused deposition modeling | |

Selective laser sintering | |

Directed energy deposition processes |

Question 7 Explanation:

Fused-deposition modeling consists of a computercontrolled extruder, through which a polymer
filament is deposited to produce a part slice by
slice.

Selective laser sintering uses a high-powered laser beam to sinter powders or coatings on the powders in a desired pattern. Selective laser sintering has been applied to polymers, sand, ceramics, and metals.

Stereolithography involves a computer-controlled laser-focusing system, that cures a liquid thermosetting polymer containing a photosensitive curing agent.

Directed energy deposition (DED) processes enable the creation of parts by melting material as it is being deposited. Although this basic approach can work for polymers, ceramics, and metal matrix composites, it is predominantly used for metal powders. Thus, this technology is often referred to as "metal deposition" technology.

Selective laser sintering uses a high-powered laser beam to sinter powders or coatings on the powders in a desired pattern. Selective laser sintering has been applied to polymers, sand, ceramics, and metals.

Stereolithography involves a computer-controlled laser-focusing system, that cures a liquid thermosetting polymer containing a photosensitive curing agent.

Directed energy deposition (DED) processes enable the creation of parts by melting material as it is being deposited. Although this basic approach can work for polymers, ceramics, and metal matrix composites, it is predominantly used for metal powders. Thus, this technology is often referred to as "metal deposition" technology.

Question 8 |

A CNC worktable is driven in a linear direction by
a lead screw connected directly to a stepper motor.
The pitch of the lead screw is 5 mm. The stepper
motor completes one full revolution upon receiving
600 pulses. If the worktable speed is 5 m/minute
and there is no missed pulse, then the pulse rate
being received by the stepper motor is

20 KHz | |

10 kHz | |

3 kHz | |

15 kHz |

Question 8 Explanation:

No. of steps required for one full revolution of
stepper motor shaft or lead screws n_S= 600

Pitch (p) = 5 mm

Linear table speed V_{table}= 5 m/min = 5000 mm/min

RPM of lead Screw (N_S)= \frac{V_{table}}{p}=1000 rpm

We have equation of frequency of pulse generator

\begin{aligned} f_p&= N_s \times n_S\\ f_p&= 1000 \times 600=600,000 pulses/min\\ f_p&=\frac{600000}{60}pulses/sec\\ f_p&=10000 pulses/sec \text{ or }Hz\\ f_p&=10kHz \end{aligned}

Pitch (p) = 5 mm

Linear table speed V_{table}= 5 m/min = 5000 mm/min

RPM of lead Screw (N_S)= \frac{V_{table}}{p}=1000 rpm

We have equation of frequency of pulse generator

\begin{aligned} f_p&= N_s \times n_S\\ f_p&= 1000 \times 600=600,000 pulses/min\\ f_p&=\frac{600000}{60}pulses/sec\\ f_p&=10000 pulses/sec \text{ or }Hz\\ f_p&=10kHz \end{aligned}

Question 9 |

In a pure orthogonal turning by a zero rake angle single point carbide cutting tool, the shear force has been computed to be 400 N. The cutting velocity, V_c=100 \; m/min, depth of cut, t=2.2 \; mm, feed, s_0=0.1 \; mm/revolution and chip velocity, V_f=20 \; m/min, the shear strength, \tau _s of the material will be _______MPa (round off to two decimal places).

458.25 | |

392.23 | |

254.36 | |

214.36 |

Question 9 Explanation:

\begin{aligned} \alpha &=0^{\circ} \\ F_{s} &=400 \mathrm{~N} \\ \text { Cutting velocity }(\mathrm{V}) &=100 \mathrm{~m} / \min \left(V_{c}\right) \\ d &=2.0 \mathrm{~mm}(t) \\ f &=0.1 \mathrm{~mm} / \mathrm{rev}\left(S_{o}\right) \\ \text { Chip velocity }\left(V_{c}\right) &=20 \mathrm{~m} / \min \quad\left(V_{f}\right) \\ r &=\frac{t}{t_{c}}=\frac{l_{c}}{l}=\frac{V_{c}}{V} \quad\left(V_{c}=\text { chip velocity; } V=\right.\text { cutting velocity) }\\ &=\frac{20}{100}=0.2\\ \tan \phi &=\frac{r \cos \alpha}{1-r \sin \alpha}=\frac{0.2 \cos 0^{\circ}}{1-0.2 \sin 0^{\circ}}=0.2 \\ \phi &=11.31^{\circ} \\ & \frac{F_{S}}{A_{S}}=\frac{F_{S}}{\left(\frac{b t}{\sin \phi}\right)}=\frac{F_{S} \sin \phi}{f d} \\ t &=f \sin 90^{\circ} \\ b &=d \sin 90^{\circ} \\ b t &=f d \\ &=\frac{400 \sin 11.31^{\circ}}{0.1 \times 2.0} \mathrm{~N} / \mathrm{mm}^{2} \\ &=392.23 \mathrm{MPa} \end{aligned}

Question 10 |

A surface grinding operation has been performed on a Cast Iron plate having dimensions 300 mm (length) x 10 mm (width) x 50 mm (height). The grinding was performed using an alumina wheel having a wheel diameter of 150 mm and wheel width of 12 mm. The grinding velocity used is 40 m/s, table speed is 5 m/min, depth of cut per pass is 50 \mu m and the number of grinding passes is 20. The average tangential and average normal force for each pass is found to be 40 N and 60 N respectively. The value of the specific grinding energy under the aforesaid grinding conditions is __________J/mm^3 (round off to one decimal place).

24.8 | |

32.6 | |

38.4 | |

48.2 |

Question 10 Explanation:

\begin{aligned} \text { Power } &=F_{c} \cdot V \\ &=40 \mathrm{~N} \times 40 \mathrm{~m} / \mathrm{s}=1600 \mathrm{~W} \\ \mathrm{MRR} &=10 \times 0.050 \times \frac{5000}{60} \mathrm{~mm}^{3} / \mathrm{s} \\ &=41.667 \mathrm{~mm}^{3} / \mathrm{s} \end{aligned}

e=\frac{\text { Power }}{\text { MRR }}=\frac{1600 \mathrm{~J} / \mathrm{s}}{41.667 \mathrm{~mm}^{3} / \mathrm{s}}=38.40 \mathrm{~J} / \mathrm{mm}^{3}

e=\frac{\text { Power }}{\text { MRR }}=\frac{1600 \mathrm{~J} / \mathrm{s}}{41.667 \mathrm{~mm}^{3} / \mathrm{s}}=38.40 \mathrm{~J} / \mathrm{mm}^{3}

There are 10 questions to complete.