# Machining and Machine Tool Operation

 Question 1
A solid part (see figure) of polymer material is to be fabricated by additive manufacturing (AM) in square-shaped layers starting from the bottom of the part working upwards. The nozzle diameter of the AM machine is $a/10$ mm and the nozzle follows a linear serpentine path parallel to the sides of the square layers with a feed rate of $a/5$ mm/min.
Ignore any tool path motions other than those involved in adding material, and any other delays between layers or the serpentine scan lines.
The time taken to fabricate this part is _______ minutes. (Answer in integer) A 5000 B 6000 C 8000 D 9000
GATE ME 2023   Manufacturing Engineering
Question 1 Explanation:
\begin{aligned} V&=(3a)^2 \times 1.5a+ (2a)^2 \times a+(a)^2 \times 0.5a\\ &= 18a^3\\ T&=\frac{V}{\left ( \frac{a}{10} \right )^2 \times \left ( \frac{a}{5} \right )} \\ &=\frac{18a^3}{a^3} \times 100 \times 5=9000 min \end{aligned}
 Question 2
In an ideal orthogonal cutting experiment (see figure), the cutting speed $V$ is 1 m/s, the rake angle of the tool $\alpha =5^{\circ}$, and the shear angle, $\phi$, is known to be $45^{\circ}$.
Applying the ideal orthogonal cutting model, consider two shear planes $PQ$ and $RS$ close to each other. As they approach the thin shear zone (shown as a thick line in the figure), plane $RS$ gets sheared with respect to $PQ$ (point $R1$ shears to $R2$, and $S1$ shears to $S2$).
Assuming that the perpendicular distance between $PQ$ and $RS$ is $\delta =25\mu m$, what is the value of shear strain rate (in $s^{-1}$) that the material undergoes at the shear zone? A $1.84 \times 10^4$ B $5.20 \times 10^4$ C $0.71 \times 10^4$ D $1.30 \times 10^4$
GATE ME 2023   Manufacturing Engineering
Question 2 Explanation:
Data given $\alpha=5^{\circ}$
\begin{aligned} \phi & =45^{\circ} \\ \Delta H & =25 \mu \mathrm{m} \\ \mathrm{V} & =1 \mathrm{~m} / \mathrm{sec} \\ \mathrm{V}_{\mathrm{s}} & =\frac{\mathrm{V} \cos \alpha}{\cos (\phi-\alpha)}=\frac{1 \times \cos 5^{\circ}}{\cos (45-5)} \\ & =\frac{\cos 5}{\cos 40}=1.3 \mathrm{~m} / \mathrm{sec} \end{aligned}
Shear strain rate $=\frac{\mathrm{V}_{\mathrm{s}}}{\delta \mathrm{H}}=\frac{1.3}{25 \times 10^{-6}}=5.20 \times 10^{4} \mathrm{s}^{-1}$

 Question 3
A cuboidal part has to be accurately positioned first, arresting six degrees of freedom and then clamped in a fixture, to be used for machining. Locating pins in the form of cylinders with hemi-spherical tips are to be placed on the fixture for positioning. Four different configurations of locating pins are proposed as shown. Which one of the options given is correct? A Configuration P1 arrests 6 degrees of freedom, while Configurations P2 and P4 are over-constrained and Configuration P3 is under-constrained. B Configuration P2 arrests 6 degrees of freedom, while Configurations P1 and P3 are over-constrained and Configuration P4 is under-constrained. C Configuration P3 arrests 6 degrees of freedom, while Configurations P2 and P4 are over-constrained and Configuration P1 is under-constrained. D Configuration P4 arrests 6 degrees of freedom, while Configurations P1 and P3 are over-constrained and Configuration P2 is under-constrained.
GATE ME 2023   Manufacturing Engineering
Question 3 Explanation:
3-2-1 principle of location
The 3-2-1 principle of location (six point location principle) is used to constrain the movement of workpiece along the three axes XX, YY and ZZ.
This is achieved by providing six locating points, 3-pins in base plate, 2-pins in vertical plane and 1-pin in a plane which is perpendicular to first two planes.
 Question 4
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 4 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 5
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 5 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}

There are 5 questions to complete.

### 4 thoughts on “Machining and Machine Tool Operation”

1. Question1 forget the pie in denominator

2. Q16 . Doubt in second method since soln of 2.shear angle-frictn angle+rake angle not equal to 90 deg.

3. 4. 