Manufacturing Engineering

Original length of the billet (L_0)=100mm
Original diameter of billet (d_0)=100mm
Mean flow stress of billet material (\sigma _m)=50MPa
Die Shape factor K_S=1.05
Original Cross Sectional Area of Billet
A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}100^2=7853.981
Cross sectional area of extruded product
A_f=(10 \times 50)+ (10 \times 50)=1000 mm^2
Extrusion ratio r=\frac{A_0}{A_f}=\frac{7853.9816}{1000}=7.854
Extrusion pressure will be maximum at the start of extrusion process where L_0=100mm
\begin{aligned} P_{max}&= K_S\sigma _m\left [ 0.8+1.5 \ln (r)+\frac{2L_0}{d_0} \right ]\\ &=1.05 \times 50 \left [ 0.8+1.5 \ln (7.854)+\frac{2 \times 100}{100} \right ]\\ &=309.305 MPa \end{aligned}
Maximum Extrusion force
\begin{aligned} F_{max}&= P_{max} \times A_0\\ &=309.305 \times \frac{\pi}{4}(100)^2\\ &=2429265.97N\\ &=2429.3kN \end{aligned}

Given, OCV = 100 V and SCC = 1000 A
Voltage Arc length characteristic V_{arc}=10+5l
We have,
\begin{aligned} V_P&=OCV-\frac{OCV}{SCC}I\\ V_P&=100-\frac{100}{1000}I\\ _P&=100-\frac{I}{10} \end{aligned}
For stable arc
\begin{aligned} V_{arc}&=V_P\\ 10+5l&=100-\frac{I}{10}\\ I&=900-50l\\ \text{Arc Power} (P)&=V \times I\\ P&=(10+5l)(900-50l)\\ P&=9000+4000l-250l^2 \end{aligned}
For Maximum Arc Power
\begin{aligned} \frac{dP}{dl}&=0\\ \frac{d}{dl}(9000+400l-250l^2)&=0\\ l=4000/500&=8mm \end{aligned}
Hence, for Maximum arc power, arc length is 8 mm
Voltage at l=8mm
\begin{aligned} V&=10+5l\\ V&=10+( 5 \times 8)\\ V&=50 \;Volt \end{aligned}
Current at l=8mm
\begin{aligned} I&=900-50l\\ I&=900-( 50 \times 8)\\ I&=500 \;A \end{aligned}
Maximum Arc Power
\begin{aligned} P_{max}&= V \times I\\ P_{max}&=50 \times 500\\ &=25000VA\\ &=25 kVA \end{aligned}

Pitch (p) = 4 mm
Thread angle 2\alpha =60^{\circ}
Semi Thread Angle \alpha =30^{\circ}
Diameter of best size wire
\begin{aligned} d_w&=\frac{p}{2}\sec \alpha \\ d_w&=\frac{4}{2} \sec 30^{\circ}\\ d_w&=2.039 mm=2.31mm \end{aligned}

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.

Riser should take more time for solidification than casting \left ( \frac{V}{A} \right )_r\geq\left ( \frac{V}{A} \right )_c
For top riser bottom area is not cooling surface and it is given that top cross section is also insulated. So only lateral area is cooling area.
\frac{\frac{\pi}{4}D^2H}{\pi DH}\geq \frac{a^3}{6a^2}\Rightarrow \frac{D}{4}\geq \frac{a}{6}\Rightarrow D\geq \frac{2a}{3}
Therefore, Radius R\geq \frac{a}{3} .
only answer is available \frac{a}{2}.

Chain and Sprocket mechanism is not used in CNC Machines.
Linear Motors Recently linear motors are being increasingly considered for use in high performance CNC machine tools. The linear motor consists of a series of magnets attached to the machine base and a set of electrical coils potted around a steel laminate core attached to the moving slide.
The fact that there are no mechanical parts in contact means that there is no wear periodic maintenance required. Linear motors are not limited in travel like ball screws. Larger bare required to achieve high velocity, with a longer travel to prevent undue vibration.
This larger ball screw results in a higher inertia. This means a larger motor with more torque is required (introduction inertia) and the responsiveness and bandwidth of the system is reduced, resulting in poor servo performance.
Machines built with linear motors and all-digital drive systems can produce parts with higher accuracy and tighter tolerances at higher feeds and speeds. Also, they reduce significantly the non-machining time with high acceleration and deceleration rates.

Fused-deposition modeling consists of a computercontrolled extruder, through which a polymer filament is deposited to produce a part slice by slice.

Laminated-object manufacturing uses a laser beam or vinyl cutter to first cut the slices on paper or plastic sheets (laminations); then it applies an adhesive layer, if necessary and finally stacks the sheets to produce the part.

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.

(1) Allowance = Lower limit of hole - upper limit of shaft
Allowance = 25.00 - 24.96 = 0.04 mm

(2) Maximum clearance C_{max} = Upper limit of hole - lower limit of shaft
C_{max} = 25.02 - 24.93 = 0.09 mm

(3) Maximum material limit for hole = minimum size of hole = 25.00