Question 1 |
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 1 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 2 |
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 2 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}
Question 3 |
The machining process that involves ablation is
Abrasive Jet Machining | |
Chemical Machining | |
Electrochemical Machining | |
Laser Beam Machining |
Question 3 Explanation:
Laser beam machining (LBM) is a nonconventional machining process, which broadly refers to the process of material removal, accomplished through the interactions between the laser and target materials. The processes can include laser drilling, cutting, grooving, writing, scribing, ablation, welding, cladding, milling, and so on. LBM is a thermal process, and unlike conventional mechanical processes, LBM removes material without mechanical engagement. In general, the workpiece is heated to melting or boiling point and removed by melt ejection, vaporization, or ablation.
Question 4 |
In a grinding operation of a metal, specific energy consumption is 15 J/mm^3. If a grinding wheel with a diameter of 200 mm is rotating at 3000 rpm to obtain a material removal rate of 6000 mm^3/min, then the tangential force on the wheel is _______N (round off to two decimal places).
42.25 | |
62.45 | |
47.75 | |
28.24 |
Question 4 Explanation:
\begin{aligned} E_{s p} &=15 \mathrm{~J} / \mathrm{mm}^{3} \\ \mathrm{MRR} &=6000 \mathrm{~mm}^{3} / \mathrm{min} \\ N &=3000 \mathrm{rpm} \\ D &=200 \mathrm{~mm} \\ E_{s p} &=\frac{F_{C} \mathrm{~V}}{L d V} \\ 15 &=\frac{F \times \frac{m}{m i n}}{600 \frac{m m^{3}}{\min }} \\ V &=\frac{\pi D N}{1 m}=\frac{\pi(200)(3 m)}{1000}=600 \times 3.14\\ 15 &=\frac{F \times 600 \times 3 \pi y}{6000} \\ F &=\frac{150}{\pi}=47.746 \mathrm{~N} \simeq 47.75 \mathrm{~N} \end{aligned}
Question 5 |
An orthogonal cutting operation is performed using a single point cutting tool with a rake angle of 12^{\circ} on a lathe. During turning, the cutting force and the friction force are 1000 N and 600 N, respectively. If the chip thickness and the uncut chip thickness during turning are 1.5 mm and 0.75 mm, respectively, then the shear force is _______N (round off to two decimal places).
247.65 | |
685.77 | |
457.36 | |
825.66 |
Question 5 Explanation:
F_{c}=1000 \mathrm{~N}, F=600 \mathrm{~N}, t=0.75 \mathrm{~mm}, t_{c}=1.5 \mathrm{~mm}, \alpha=12^{\circ}
This is orthogonal turning operation
\begin{aligned} r &=\frac{t}{t_{C}}=\frac{0.75}{1.5}=0.5 \\ \tan \phi &=\frac{0.5 \cos 12^{\circ}}{1-0.5 \sin 12^{\circ}} \\ \Rightarrow\quad \phi &=28.63^{\circ} \end{aligned}
Method - 1: Using force relations
\begin{aligned} F &=F_{c} \sin \alpha+F_{t} \cos \alpha \\ 600 &=1000 \sin 12^{\circ}+F_{t} \cos 12^{\circ}\\ \text{or}\qquad F_{t}&=400.85 \mathrm{~N} \\ \therefore \qquad F_{s}&=F_{c} \cos \phi-F_{t} \sin \phi\\ &=1000 \cos 28.63^{\circ}-400.85 \sin 28.63^{\circ} \\ &=685.66 \mathrm{~N} \end{aligned}
Method - 2 : Using Merchant circle
\begin{aligned} F_{c} &=R \cos (\beta-\alpha) \\ F &=R \sin \beta \\ \text{or}\quad \frac{F}{F_{c}} &=\frac{R \sin \beta}{R \cos (\beta-\alpha)} \end{aligned}
\begin{aligned} \frac{600}{1000} &=\frac{\sin \beta}{\cos (\beta-12)} \\ 0.6 \cos \beta \cos 12^{\circ}+0.6 & \sin \beta \sin 12^{\circ}=\sin \beta \\ 0.586888 \cos \beta &=\left(1-0.6 \sin 12^{\circ}\right) \sin \beta=0.875253 \sin \beta \\ \tan \beta &=\frac{0.586888}{0.875253} \\ \Rightarrow \quad \quad \beta &=33.84^{\circ} \\ \therefore \quad F &=R \sin \beta \\ 600&=R \sin 33.84^{\circ}\\ \Rightarrow R&=1077.44 \mathrm{~N}\\ \therefore \quad F_{s}&=R \cos (\phi+\beta-\alpha)\\ &=1077.44 \cos \left(28.63^{\circ}+33.84^{\circ}-12^{\circ}\right)\\ &=685.77 \mathrm{~N} \end{aligned}
This is orthogonal turning operation
\begin{aligned} r &=\frac{t}{t_{C}}=\frac{0.75}{1.5}=0.5 \\ \tan \phi &=\frac{0.5 \cos 12^{\circ}}{1-0.5 \sin 12^{\circ}} \\ \Rightarrow\quad \phi &=28.63^{\circ} \end{aligned}
Method - 1: Using force relations
\begin{aligned} F &=F_{c} \sin \alpha+F_{t} \cos \alpha \\ 600 &=1000 \sin 12^{\circ}+F_{t} \cos 12^{\circ}\\ \text{or}\qquad F_{t}&=400.85 \mathrm{~N} \\ \therefore \qquad F_{s}&=F_{c} \cos \phi-F_{t} \sin \phi\\ &=1000 \cos 28.63^{\circ}-400.85 \sin 28.63^{\circ} \\ &=685.66 \mathrm{~N} \end{aligned}
Method - 2 : Using Merchant circle
\begin{aligned} F_{c} &=R \cos (\beta-\alpha) \\ F &=R \sin \beta \\ \text{or}\quad \frac{F}{F_{c}} &=\frac{R \sin \beta}{R \cos (\beta-\alpha)} \end{aligned}
\begin{aligned} \frac{600}{1000} &=\frac{\sin \beta}{\cos (\beta-12)} \\ 0.6 \cos \beta \cos 12^{\circ}+0.6 & \sin \beta \sin 12^{\circ}=\sin \beta \\ 0.586888 \cos \beta &=\left(1-0.6 \sin 12^{\circ}\right) \sin \beta=0.875253 \sin \beta \\ \tan \beta &=\frac{0.586888}{0.875253} \\ \Rightarrow \quad \quad \beta &=33.84^{\circ} \\ \therefore \quad F &=R \sin \beta \\ 600&=R \sin 33.84^{\circ}\\ \Rightarrow R&=1077.44 \mathrm{~N}\\ \therefore \quad F_{s}&=R \cos (\phi+\beta-\alpha)\\ &=1077.44 \cos \left(28.63^{\circ}+33.84^{\circ}-12^{\circ}\right)\\ &=685.77 \mathrm{~N} \end{aligned}
Question 6 |
The correct sequence of machining operations to be performed to finish a large diameter through hole is
drilling, boring, reaming | |
boring, drilling, reaming | |
drilling, reaming, boring | |
boring, reaming, drilling |
Question 6 Explanation:
Drilling: to produce a hole, which then may be followed by boring it to improve its dimensional accuracy and surface finish.
Boring: to enlarge a hole or cylindrical cavity made by a previous process or to produce circular internal grooves.
Reaming: is an operation used to (a) make an existing hole dimensionally more accurate than can br achived by drilling alone and (b) improve its surface finish. The most accurate holes in workpieces generally are produced by the following sequence of operation.
Centering -> Drilling -> Boring -> Reaming.
Boring: to enlarge a hole or cylindrical cavity made by a previous process or to produce circular internal grooves.
Reaming: is an operation used to (a) make an existing hole dimensionally more accurate than can br achived by drilling alone and (b) improve its surface finish. The most accurate holes in workpieces generally are produced by the following sequence of operation.
Centering -> Drilling -> Boring -> Reaming.
Question 7 |
In a machining operation, if a cutting tool traces the workpiece such that the directrix is perpendicular to the plane of the generatrix as shown in figure, the surface generated is
plane | |
cylindrical | |
spherical | |
a surface of revolution |
Question 7 Explanation:
Question 8 |
Bars of 250 mm length and 25 mm diameter are to be turned on a lathe with a feed
of 0.2 mm/rev. Each regrinding of the tool costs Rs. 20. The time required for each tool
change is 1 min. Tool life equation is given as VT^{ 0.2} = 24 (where cutting speed V is
in m/min and tool life T is in min). The optimum tool cost per piece for maximum
production rate is Rs. ________ (round off to 2 decimal places).
12.56 | |
28.42 | |
26.98 | |
35.62 |
Question 8 Explanation:
\begin{aligned} \text{Optimum tool life } \left(T_{o}\right)&=\frac{T_{c}(1-n)}{n} \\ &=\frac{1(1-0.2)}{0.2}=4 \mathrm{min} \\ V_{o} T_{o}^{n} &=C \\ \text { or } V_{o}&=\frac{C}{T_{o}^{n}}\\ &=\frac{24}{4^{0.2}}=18.19 \mathrm{m} / \mathrm{min} \\ V &=\pi D \mathrm{N} \\ \Rightarrow \quad 18.19 &=\pi \times 0.025 \times N \end{aligned}
\Rightarrow N=231.6 \mathrm{rpm}
Maching time per piece
\left(T_{m}\right)=\frac{L}{f N}=\frac{250}{0.2 \times 231.6}=5.397 \mathrm{min}
Number of tool needed per piece work
=\frac{5.397}{4} \text { piece }
\therefore The optimum tool cost per piece
=\frac{5.397}{4} \times 20=26.985
\Rightarrow N=231.6 \mathrm{rpm}
Maching time per piece
\left(T_{m}\right)=\frac{L}{f N}=\frac{250}{0.2 \times 231.6}=5.397 \mathrm{min}
Number of tool needed per piece work
=\frac{5.397}{4} \text { piece }
\therefore The optimum tool cost per piece
=\frac{5.397}{4} \times 20=26.985
Question 9 |
A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0^{\circ}- 9^{\circ} - 7^{\circ} - 8^{\circ} - 15^{\circ} - 30^{\circ} - 0.05 inch (Coordinate system, ASA) resulting in a cutting force
F_{c1}. If the tool geometry is changed to 0^{\circ}- 9^{\circ} - 7^{\circ} - 8^{\circ} - 15^{\circ} - 0^{\circ} - 0.05 inch (Coordinate
system. ASA) and all other parameters remain unchanged, the cutting force changes
to F_{c2}. Specific cutting energy (in J/mm^3) is U_c = U_0 (t_1)^{-0.4}, where U_0 is the specific
energy coefficient, and t_1 is the uncut thickness in mm. The value of percentage change
in cutting force F_{c2}, i.e. \left ( \frac{F_{c2}-F_{c1}}{F_{c1}} \right ) \times 100, is _______ (round off to one decimal place).
-2.5 | |
-5.59 | |
5.59 | |
2.5 |
Question 9 Explanation:
\begin{aligned} C_{s 1}&=30^{\circ}\\ \therefore \quad \lambda_{1}&=90-30=60^{\circ} \\ C_{s 2}&=0^{\circ}\\ \therefore \lambda_{2}&=90-0=90^{\circ} \end{aligned}
We know that specific energy consumption
\begin{aligned} U_{C}&=\frac{F_{C}}{1000 \mathrm{fd}}=U_{0}\left(t_{1}\right)^{-0.4}(\text {given }) \\ \therefore \quad F_{C}&=U_{0}(f \sin \lambda)^{-0.4} \times 1000 \mathrm{fd}\\ \therefore \quad F_{c} &\propto(\sin \lambda)^{-0.4}\\ \therefore \quad &=\frac{F_{c 2}-F_{c 1}}{F_{c 1}} \times 100=\left[\left(\frac{\sin \lambda_{2}}{\sin \lambda_{1}}\right)^{-0.4}-1\right] \times 100 \\ &=\left[\left(\frac{\sin 90^{\circ}}{\sin 60^{\circ}}\right)^{-0.4}-1\right] \times 100=-5.59 \end{aligned}
We know that specific energy consumption
\begin{aligned} U_{C}&=\frac{F_{C}}{1000 \mathrm{fd}}=U_{0}\left(t_{1}\right)^{-0.4}(\text {given }) \\ \therefore \quad F_{C}&=U_{0}(f \sin \lambda)^{-0.4} \times 1000 \mathrm{fd}\\ \therefore \quad F_{c} &\propto(\sin \lambda)^{-0.4}\\ \therefore \quad &=\frac{F_{c 2}-F_{c 1}}{F_{c 1}} \times 100=\left[\left(\frac{\sin \lambda_{2}}{\sin \lambda_{1}}\right)^{-0.4}-1\right] \times 100 \\ &=\left[\left(\frac{\sin 90^{\circ}}{\sin 60^{\circ}}\right)^{-0.4}-1\right] \times 100=-5.59 \end{aligned}
Question 10 |
The process, that uses a tapered horn to amplify and focus the mechanical energy for machining of glass, is
electrochemical machining | |
electrical discharge machining | |
ultrasonic machining | |
abrasive jet machining |
Question 10 Explanation:
In Ultrasonic machining, the function of horn (also
called concentrator, it is a tapered metal bar) is to
amplify and focus vibration of the transducer to an
adequate intensity for driving the tool to fulfll the
cutting operation.
There are 10 questions to complete.
