# Machining and Machine Tool Operation

 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).
 A 5247 B 1245 C 2732 D 3214
GATE ME 2022 SET-2   Manufacturing Engineering
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}
 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).

 A 1.8 B 2.5 C 3.2 D 2.9
GATE ME 2022 SET-2   Manufacturing Engineering
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}
 Question 3
Which of these processes involve(s) melting in metallic workpieces?

MSQ
 A Electrochemical machining B Electric discharge machining C Laser beam machining D Electron beam machining
GATE ME 2022 SET-2   Manufacturing Engineering
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).
 A 31.4 B 25.6 C 87.5 D 65.5
GATE ME 2022 SET-1   Manufacturing Engineering
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}
 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).
 A 1245.3 B 1547.4 C 3652.4 D 2573.4
GATE ME 2022 SET-1   Manufacturing Engineering
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}
 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).
 A 38.25 B 31.5 C 24.86 D 28.57
GATE ME 2022 SET-1   Manufacturing Engineering
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}
 Question 7
Which of the following additive manufacturing technique(s) can use a wire as a feedstock material?
 A Stereolithography B Fused deposition modeling C Selective laser sintering D Directed energy deposition processes
GATE ME 2022 SET-1   Manufacturing Engineering
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.
 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
 A 20 KHz B 10 kHz C 3 kHz D 15 kHz
GATE ME 2022 SET-1   Manufacturing Engineering
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}
 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).
 A 458.25 B 392.23 C 254.36 D 214.36
GATE ME 2021 SET-2   Manufacturing Engineering
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).
 A 24.8 B 32.6 C 38.4 D 48.2
GATE ME 2021 SET-2   Manufacturing Engineering
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}

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

There are 10 questions to complete.