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)
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)
5000 | |
6000 | |
8000 | |
9000 |
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?
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?
1.84 \times 10^4 | |
5.20 \times 10^4 | |
0.71 \times 10^4 | |
1.30 \times 10^4 |
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}
\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?
Configuration P1 arrests 6 degrees of freedom, while Configurations P2 and P4 are
over-constrained and Configuration P3 is under-constrained.
| |
Configuration P2 arrests 6 degrees of freedom, while Configurations P1 and P3 are
over-constrained and Configuration P4 is under-constrained. | |
Configuration P3 arrests 6 degrees of freedom, while Configurations P2 and P4 are
over-constrained and Configuration P1 is under-constrained. | |
Configuration P4 arrests 6 degrees of freedom, while Configurations P1 and P3 are over-constrained and Configuration P2 is under-constrained. |
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.
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).
5247 | |
1245 | |
2732 | |
3214 |
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}
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).
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 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}
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.