Concept:

3-2-1 Principle :

• There are 12 degree of freedom of a work piece i.e., movement about negative and positive axis and clockwise and anti-clockwise rotation about the three axis.
• By providing three pins in the base, five degree of freedom will be arrested i.e. movement about negative Z-axis and the clockwise and anti-clockwise rotation about X and Y axis.
• These three pins also ensure that machining always takes place on single plane perpendicular to the base, along the length direction.
• By providing two pins at front side, this arrests movement of work piece along negative Y-direction in clockwise and anti-clock wise rotation about Z-axis.
• A sixth pin is provided on a plane perpendicular to the previous two planes, which also arrests movement about negative X-axis.
• By providing six pins, nine degree of freedom is arrested.
• Hence we can see that this principle helps us in locating the devices.

Concept:

\(BLU = \frac{{Lead\;screw\;pitch}}{{steps\;per\;revolution}}\)

Calculation:

Given:

Stepper motor = 300 steps per revolution, Lead screw pitch = 5 mm

Now,

\(BLU = \frac{5}{{300}}\)

∴ BLU = 0.0167 

Concept:

The power density is given by \(J = η \frac{{VI}}{A}\)

Calculation:

Given:

Volatge (V) = 45 kV, Current (I) = 60 mA, diameter (d) = 0.25 mm, heat transfer factor (η) = 0.87

\(J = 0.87 \times \frac{{45 \times 1000 \times 60 \times {{10}^{ - 3}} \times 4}}{{\pi \times {{0.25}^2}}} = 47853.435\;W/m{m^2} \)

∴ J = 47.853 kW/mm²

Explanation:

Important G codes

• G 00 – Rapid Transverse
• G 01 – Linear Interpolation
• G 02 – CW Circular Interpolation
• G 03 – CCW Circular Interpolation
• G 04 – Dwell
• G 97 – Spindle Speed

Important M codes

• M 00 – Program Stop
• M 01 – Optional Stop
• M 02 - End of Program
• M 03 – Spindle Start (CW)
• M 04 – Spindle Start (CCW)
• M 05 – Spindle Stop
• M 06 – Tool Change 
• M 08 – Coolant on
• M 09 – Coolant off
• M 10 – Clamp-on
• M 11 – Clamp off

Concept:

\(Metal\;removal\;rate,\;Q = \frac{{AI}}{{\rho ZF}} = c{m^3}/sec\)

A = Gram atomic weight of the metallic iron, I = Current (amp), ρ = Density of anode (g/cm³)

Z = Valence of the cation, F = Faraday = 96500 coulombs

Calculation:

Given:

\(Q = 1\frac{{c{m^3}}}{{min}} = \frac{1}{{60}}c{m^3}/sec\),

A = 56 gm, Z = 2, ρ = 7.8 gm/cm³, F = 1609 amp-min = 1609 × 60 amp-s (coulombs)

Now,

\(Q = \frac{{AI}}{{\rho ZF}} \)

\(∴ \frac{1}{{60}} = \frac{{56 \times I}}{{7.8 \times 2 \times 1609 \times 60}}\)

Concept:

800 pulses = 1 revolution of motor = 1 revolution of lead screw = 2 × pitch of lead screw

Calculation:

800 pulses = 2 × pitch of lead screw

∴ 800 pulses = 2 × 4 mm

∴ 800 pulses = 8 mm

Now,

Position accuracy of motor = distance travelled in 1 pulses

∴ positional accuracy \(= \frac{8}{{800}}\)

Concept:

The G00 code gives the cutter a rapid transverse motion.

If the X-axis and Y-axis have separate speed then the time taken to reach the final point separately in X and Y direction is given by

\({t_x} = \frac{{{\rm{\Delta }}x}}{{{V_x}}}\)

\({t_y} = \frac{{{\rm{\Delta }}y}}{{{V_y}}}\)

The time to reach the destination will be maximum of the t_x and t_y because the cutter will start the motion in X and Y direction simultaneously.

Calculation:

Given:

X₁ = 10 mm, X₂ = 160 mm and V_x = 10000 mm/min 

\({t_x} = \frac{{{\rm{\Delta }}x}}{{{V_x}}} = \frac{{160 - 10}}{{10000}} = 0.015\;minutes = 0.9\;s\)

In Y-direction,

Y₁ = 20 mm, X₂ = 120 mm and V_x = 5000 mm/min

\({t_y} = \frac{{{\rm{\Delta }}y}}{{{V_y}}} = \frac{{120 - 20}}{{5000}} = 0.02\;minutes = 1.2\;seconds\)

∴ The time required to reach the destination will be 1.2 seconds.

∴ The motor operating in X-direction will stop operating after 0.9 s and the motor in Y-direction will run for 1.2 s. 

Concept:

Energy released per spark

\(E = \frac{1}{2}CV_d^2\)

For maximum Power ⇒   V_d = 0.72 V_s

\({\rm{Cycle\;time\;}}\left( {{{\rm{t}}_{\rm{C}}}} \right){\rm{\;}} = RC\ln \left( {\frac{{Vs}}{{Vs - Vd}}} \right)\)

\({\rm{Average\;power\;input\;}}\left( W \right) = \frac{E}{{{t_C}}}\)

Calculation:

V_d = 0.72 V_s

V_d = 0.72 × 250

∴ V_d = 180 V

Now,

\(E = \frac{1}{2}CV_d^2\)

\(E = \frac{1}{2} \times 10 \times {10^{ - 6}} \times {180^2} = 0.162\;J\)

Now,

\({t_C} = 45 \times 10 \times {10^{ - 6}}\ln \left( {\frac{{250}}{{250 - 180}}} \right)\)

∴ t_c = 5.728 × 10^-4 sec

Now,

\(W = \frac{{0.162}}{{5.728\; \times \;{{10}^{ - 4}}}} = 282.80\;W\)

MRR = θ = 27 (0.2828)^1.70

∴ MRR = 3.15 mm³/min

Now,

Material to be removed = 12 × 8 × 6 mm³

\({\rm{Time\;required\;}} = \frac{{12\; \times \;8\; \times \;6}}{{3.15}}min\)

∴ Time required = 3.043 hour

Concept:

Material removal rate (MRR) \(= \frac{{AI}}{{\rho ZF}}\)

A = atomic weight, I = current, ρ = density, F = faraday constant, Z = valency

\(I = \frac{{{\rm{\Delta }}V}}{R} = \frac{{Voltage}}{{Resistance}}\)

R = ρ_sL/ A

Electrode feed rate = s × S

Electrode feed rate = specific MRR × current density

\(s = \frac{e}{{F\rho }},\;s = \frac{{{\rm{\Delta }}V}}{{{\rho _s}L}},\;e = \frac{A}{Z}\)

ρ_s = specific resistance

Calculation:

A = 56, Z = 3, ρ = 7700 kg/m³ = 7.70 gm/cc, ρ_s = 4 Ω cm, ΔV = 14 V, A_r = 3 × 3 cm², L = 0.3 mm = 0.03 cm

e = A/Z = 56/3

\(I = \frac{{{\rm{\Delta }}V}}{R} = \frac{{14}}{{4 \times \frac{{0.03}}{{3\; \times \;3}}}} = 1050\;A\)

\({\rm{MRR\;}} = \frac{{56\; \times \;1050}}{{96500\; \times \;3\; \times\; 7.70}}\)

∴ MRR = 0.02638 cm³/s

Now,

s = e/FP

\(s = \frac{{\frac{{56}}{3}}}{{96500\; \times \;7.70}}\)

Now,

\(S = \frac{{\Delta V}}{{{P_s}L}}\)

\(s = \frac{{14}}{{4\; \times \;0.03}}\)

Now,

\({\rm{Feed\;rate\;}} = \left( {\frac{{\frac{{56}}{3}}}{{96500\; \times \;7.70}}} \right) \times \frac{{14}}{{4 \times 0.03}}\)

∴ Feed rate = 0.00293 cm/s

(i) Concept:

When a fast moving electron impinges on a material surface, it penetrates through a layer undisturbed. Then it starts colliding with the molecules and ultimately is brought to rest. The layer through which the electron penetrates undisturbed is called the transparent layer. When electron begins colliding with the lattice atoms, it starts giving up its kinetic energy. The total distance to which electron can penetrate (δ) depends upon kinetic energy. The distance δ is given by

\(\delta =2.6\times {{10}^{-17}}\frac{{{V}^{2}}}{\rho }\)

Here, δ = total range in mm, V = Accelerating voltage in volts

ρ = Density of work material in kg/mm³

(ii) Calculation:

Given data, V = 150 kV = 150 × 10³ volts

\(\rho =7500\frac{kg}{{{m}^{3}}}=\frac{7500}{{{10}^{9}}}=75\times {{10}^{-7}}~kg/m{{m}^{3}}\)

\(\delta =2.6\times {{10}^{-17}}\frac{{{\left( 150\times {{10}^{3}} \right)}^{2}}}{75\times {{10}^{-7}}}\)

\(=\frac{2.6\times {{10}^{-17}}\times {{\left( 150 \right)}^{2}}\times {{10}^{6}}\times {{10}^{7}}}{75}\)

\(=\frac{2.6\times 150\times 150\times {{10}^{-4}}}{75}\)

= 0.078 mm = 78 μm

δ = 78 μm