Machining and Machine Tool Operation

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}

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}

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.

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}

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}

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}

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.

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}

\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}

\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}