The given system is a top gate system

$T_f=\frac{V_m}{A_g{v}_g}$

V_m = volume of mould, A_g = Area of gate, V_g = Velocity

h = 15 cm, g = 9.81 m/s² = 9.81 cm/s²

$V_g=\sqrt{2gh}$

$V_g=\sqrt{2\times 981\times 15}=171.6cm/s$

Volume of mould = (50 × 40 × 20) cm³ = 40000 cm³

A_g = 6 cm²

$\therefore T_f=\frac{V_m}{Agvg}=\frac{40000}{6\times 171.6}\Rightarrow T_f=38.85sec$

Concept:

$H_S=\frac{VI}{v}\left({\eta }_h\right)$

Where, H_S = Heat input per unit length, V = Arc voltage, I = Current supplied, η_h = Arc or thermal efficiency, v = welding speed

Calculation:

Given: Here, V = 40 V, I = 300 A, v = 9 mm/s, η_h = 0.75

$\therefore H_s=\frac{40\times 300}{9}\times 0.75=1000J/mm=1kJ/mm$

⇒ H_S = 1 kJ/mm

Points to remember:

• If heat input is asked in per unit length, use $H_S=\frac{VI}{v}\times {\eta }_h$
• If heat input is asked in per unit volume, use $H_S=\frac{VI}{A_b\times v}\times {\eta }_h$ where A_b is weld bead area

Explanation:

Primary Shear Zone (PSZ):

• The primary shear zone lies between the workpiece and metal chip.
• In the PSZ when shearing action is taking place, the atomic bond present between the atoms of the material is getting breaking.
• For breaking the atomic bond it needs to supply a certain amount of energy but during the breaking of the atomic bond, they release an equal amount of energy in the form of heat energy.
• Out of the heat generated the maximum (60 to 65%) amount of the heat is carried away by the chip.

Secondary Shear Zone (SSZ):

• It lies between the metal chip and cutting tool.
• In SSZ the energy supplied is converted into heat energy because of the presence of friction at the chip tool interface.
• About 30 to 35% of the energy supplied is converted into heat energy in the SSZ.
• Out of the heat generated the maximum amount of the heat is carried away by the chip, Only a small amount is transferred to the tool.
• This is because the thermal conductivity of the tool is less than the chip.

Tertiary Shear Zone(TSZ):

• It lies between workpiece and tool.
• In TSZ, the energy supplied is converted into heat energy is due to the presence of friction of the tool work interface.
• About 5 to 10% of the energy supplied is converted into heat energy in this zone.

Explanation:

Given:

Rake angle, α = 0°, Cutting force (F_c) = 510 N, Thrust force (F_t) = 320 N

Calculation:

Friction force (F) = F_c sin α + F_t cos α

F = F_c (0) + F_t cos 0°

∴ F = 320 N

Now,

Normal force (N) = F_c cos α - F_t sin α

Normal force (N) = (510) cos 0° - 0

∴ N = 510 N

Now,

Friction angle (ϕ) $={\tan }^{-1}\left(\frac{F}{N}\right)$

Friction angle (ϕ) $={\tan }^{-1}\left(\frac{320}{510}\right)$

ϕ = 32.1°

∴ ϕ = 0.56 radian

Concept:

Electro-Discharge Machining (EDM):

• Electrical Discharge Machining (EDM) is a manufacturing process whereby a desired shape is obtained by using electrical discharges (sparks). 
• Material is removed from the work-piece by a series of rapidly recurring current discharges between two electrodes, separated by a dielectric liquid which controls spark discharges and subject to an electric voltage.
• EDM has the lowest specific power requirement and can achieve sufficient accuracy.

• The accuracy and surface finish which are dependent on the over cut produced can be easily controlled by varying the frequency and current. The over cut is increased by increasing current and by decreasing frequency.
• For optimum metal removal and better surface finish, high frequency and maximum possible current is used.
• For roughing operation, low frequency and high current are used and for finishing application, high frequency and low current settings are used.
• In EDM a fluid is used to act as a dielectric and to help carry away debris.
• Quite often kerosene based oil is used as dielectric in EDM .
• The dielectric fluid is circulated through the tool at a pressure of 0.35 N/m² or less to free it from eroded metal particles, it is circulated through a filter and acts as a coolent,

Concept:

Taylor's tool life is given by - 

VTⁿ = C

where V = Cutting velocity (m/min), T = Tool life (min), n = Taylor's exponent and C = Taylor's constant.

Calculation:

Given:

V₁ = V m/min, T₁ = T min, n = 0.5, $V_2=\frac{V}{2}m/min$

Applying tool life equation:

$V_1{T}_1^{0.5}=V_2{T}_2^{0.5}$

$\Rightarrow (\frac{T_2}{T_1}{)}^{0.5}=\frac{V_1}{V_2}$

$\Rightarrow \frac{T_2}{T_1}=(2{)}^{\frac{1}{0.5}}$

∴ T₂ = 4T₁

% increase in tool life is -

$\mathrm{\%}increase=(\frac{T_2-T_1}{T_1})\times 100\mathrm{\%}$

$\mathrm{\%}increase=(\frac{4T_1-T_1}{T_1})\times 100\mathrm{\%}$

∴ % increase = 300 %.

Explanation:

Counter boring: Already existing holes in the components can be further machined by counter boring as shown in the figure. In counter boring operations, the hole is already enlarged with a flat bottom to provide proper seating for the bolt head or a nut, which is flushed from the outer surface.

Explanation:

Given:

d = 35 mm, h = 55 mm, t = 0.8 mm

Calculation:

$\mathrm{D}\mathrm{i}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{b}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{k}\left(D\right)=\sqrt{d^2+4dh}$

$D=\sqrt{{35}^2+4\left(35\right)\left(55\right)}=94.47mm$ 

Now,

First draw (45 % reduction)

$0.45=1-\left(\frac{d_1}{D}\right)$

$0.45=1-\left(\frac{d_1}{94.47}\right)$

∴ d₁ = 51.958 mm

Now,

Second draw (35 % reduction)

$0.35=1-\left(\frac{d_2}{x_1}\right)$

$\frac{d_2}{51.958}=0.65$

∴ d₂ = 33.77 mm

• Insufficient blank holder pressure causes wrinkles.
• Too much of blank holder pressure causes fracture.

Concept:

Maximum Uncut chip thickness in milling is given as, $t_{max}=\frac{2f_m}{NZ}\sqrt{\frac{d}{D}}$

where, fm =  table feed, N = rpm, Z = number of teeth, d =  depth of cut, D = diameter of cutter

Calculation:

Given:

D = 70 mm, d = 2 mm, fm = 20 mm/min, N = 40 rpm, Z = 30

Therefore, $t_{max}=\frac{2\times 20}{40\times 30}\sqrt{\frac{2}{70}}$

Explanation:

$\begin{array}{l}D=\sqrt{18\times 30}=23.24mm\\ i=0.45\sqrt[3]{D}+0.001D=1.3\mu m\end{array}$

Now,

Fundamental deviation of hole = 0

Fundamental deviation of shaft = - 5.5 (D)^0.41

Fundamental deviation of shaft = -19.97 μm

IT7 = 16i = 20.8 μm = 0.0208 mm

IT8 = 25i = 32.5 μm = 0.033 mm

Now,

For Hole = 30 (Basic size)

Maximum size = (30 + 0.02) mm = 30.02 mm

Now,

For shaft:

Maximum size = 30 – F.D. = 30 – 0.01997 = 29.98

Concept:

$r=\frac{t}{t_c}=\frac{V_c}{V}$

r = chip thickness ratio

t = chip thickness before cutting/(uncut chip thickness) (mm)

t_c = chip thickness after cutting (mm)

V = cutting speed (m/s)

V_c = chip velocity (m/s)

Calculation:

Given:

V = 2 m/s

Depth of cut = 0.5 mm

In orthogonal cutting,

t = d = 0.5 mm

t_c = 0.6 mm

$\frac{t}{t_c}=\frac{V_c}{V}$

$\frac{0.5}{0.6}=\frac{V_c}{2}\Rightarrow V_c=1.66m/s$

Concept:

Maximum reduction per pass:

Δh = μ²R

μ is coefficient of friction, R is radius of roll

Calculation:

Given: μ₂ = 1.08 μ₁

Δh₁ = μ₁²R

Δh₂ = μ₂²R = (1.08 μ₁)²R = (1.08)²μ₁²R

% change in maximum possible reduction:

$\frac{\mathrm{\Delta }{\mathrm{h}}_2-\mathrm{\Delta }{h}_1}{\mathrm{\Delta }{h}_1}\times 100=\frac{{1.08}^2-1}{1}\times 100=16.64\mathrm{\%}$

Note: Take care whether the answer is asked in % or not. If it is not asked in percentage then the answer will be 0.1664

Concept:

Since, volume remains constant.

$\frac{\pi }{4}{d}_i^2{h}_i=\frac{\pi }{4}{d}_0^2{h}_0$

Forging force (F) $=A_f{\sigma }_y\left(1+\frac{2\mu r_0}{3h_0}\right)$

A_f = area of cross-section

Calculation:

Given:

d_i = 200 mm, h_i = 160 mm, h₀ = 80 mm

$d_0=d_i\sqrt{\frac{h_i}{h_0}}$

∴ d₀ = 282.84 mm

Now,

σ_y = 1100 MPa, μ = 0.30

$F=\frac{\pi }{4}\times {\left(282.84\right)}^2\times 1100\times \left[1+\frac{2\times 0.3\times 141.42}{3\times 80}\right)$

∴ F = 93.55 MN  

Explanation:

Given:

Linear V-I characteristic,

$\frac{V}{V_0}+\frac{I}{I_s}=1$

Where,

V₀ = 100 V

I_s = 500 A  (Given)

$\frac{V}{100}+\frac{I}{500}=1$      ---(1)

Now,

Since, L = 0.4 cm

V = 25 + 50(0.4) = 25 + 20

∴ V = 45 volts

Now,

Putting into equation (1) –

$\frac{45}{100}+\frac{I}{500}=1$

∴ I = 275 A

Now,

Heat input = V.I = 45 × 275

Heat input = 12375 W

∴ Heat input = 12.375 kW

AC = AP + PC = 35 + 5 = 40 mm

Now, AB = 40 – BC = 40 – 20 = 20 mm

TU = O₂U – O₁S = 10 – 7.5 = 2.5

ST = O₁O₂ = O₁A + AB + BO₂

= 7.5 + 20 + 10 = 37.5

In Δ STU:

tanθ = TU/ST= 2.5/37.5

Area of cross section = 25 × 25 = 625 mm²

Gap (H) = 0.25 mm

Voltage (V) = 12 V

δ = 3 Ω cm.

Valency of iron (Z) = 2

Atomic weight A = 55.85

Density ${\delta }_a=7860\frac{kg}{m^3}$

Gap resistance R is given by

$R=\frac{3\times 10\times 0.25}{625}$

= 0.012 Ω

$I=\frac{V}{R}=\frac{12}{0.012}=1000A$

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

$=\frac{55.85\times 1000}{2\times 96540}$

= 289.3 × 10^-3 g/s = 289.3 × 10^-6 kg/s

= 0.03677 × 10^-6 m³/s

Feed rate of electrode

$=\frac{MRR}{Surfacearea}=\frac{0.03677\times {10}^{-6}\times 60}{625\times {10}^{-3}}$

Explanation:

Given:

r_s = r_c = 7 cm, t_c = 60 sec

V_c = V_s

$\pi r^{2}h=\frac{4}{3}\pi r^3$

∴ h = 9.33 cm

Now,

Surface area of cylinder (A_c) = 2πrh + 2πr²

Surface area of cylinder (A_c) = 2πr (h + r)

Surface area of cylinder (A_c) = 2π (7) (9.33 + 7)

∴ A_c = 718.23 cm²

Now,

Surface area of sphere (A_s) = 4πr²

Surface area of sphere (A_s) = 4π (49)

∴ A_s = 615.75 cm²

Since, solidification time,

$t\propto {\left(\frac{V}{A}\right)}^2$

And, V_s = V_c

Thus,

$\frac{t_s}{t_c}={\left(\frac{A_c}{A_s}\right)}^2$

$\frac{t_s}{60}={\left(\frac{718.23}{615.75}\right)}^2$

∴ t_s = 81.63 seconds.

Concept:

G – codes are used in CNC programming for machining of different profiles like turning, boring, facing, step turning, threading, chamfer, radius profile, etc.

The trajectory of the point is in the clockwise direction and is circular.

For clockwise circular interpolation, the code used is G02.

The X and Y coordinates of the final point is to be mentioned in the absolute programming block

So X120 and Y60 will be mentioned.

G00 – Point to point positioning in rapid travel – Moves the tool to the given position at a rapid rate

G01 – Linear interpolation – Take straight cut at cutting feed rate

G02 – Circular Interpolation – clockwise (CW) – Moves the tool along the curved path from the curved start point to the endpoint to in clockwise direction with respect to the arc center

G03 – Circular Interpolation – counter-clockwise (CCW)

Concept:

Shear Power = Fs.Vs

Calculation:

Given:

Uncut-chip thickness (t) = 0.40 mm, Chip thickness (t_c) = 1.6 mm, Chip thickness ratio (r) = t/t_c = 0.25, rake angle = 12°  

Now,

$Tan\varphi =\frac{rcos\alpha }{1-rsin\alpha }$

∴ ϕ = 14.4638°

∵ Shear energy = F_s.V_s

Now,

F_s = F_c cos ϕ - F_t sin ϕ

Fs= 80 cos 14.4638° – 20 sin 14.4638°

Fs= 72.4691 N

Now,

V_s= 156.6484 m/min

V_s= 2.61080 m/s

Now,

Shear power = F_s.V_s

​Shear power= 72.4691 × 2.61080

Shear power = 189.202 N-m/s

Explanation:

• Weld porosity this is due to atmospheric gases are trapped inside the liquid metal during solidification of metal.
• Slag inclusion caused by trapping of compound by oxides, fluxes and electrode coating materials in weld zone.
• Weld crack caused due to non-uniform cooling and internal stresses generated in weld Beed. If stress is more than the strength of material cracks will be formed.
• Weld spatter caused due to high welding current and low welding seed and arc blow
• Incomplete fusion caused due to insufficient heat and too fast/quick travel of torch or electrode.

Explanation:

Given:

V₁ = 50 m/min, T₁ = 50 min, V₂ = 80 m/min, T₂ = 15 min, T_c = 10 min, C_t = Rs. 80, C_m = Rs. 5/min

Calculation:

$V_1{T}_1^n=V_2{T}_2^n$

50(50)ⁿ = 80(15)ⁿ

∴ n = 0.39

For maximum productivity,

$T_{opt}=T_c\left(\frac{1-n}{n}\right)$

$T_{opt}=10\left(\frac{1-0.39}{0.39}\right)=15.64min$

Now,

For maximum profit,

$T_{opt}=\left(T_c+\frac{C_t}{C_m}\right)\left(\frac{1-n}{n}\right)$

$T_{opt}=\left(10+\frac{80}{5}\right)\left(\frac{1-0.39}{0.39}\right)$

∴ T_opt = 40.66 minutes.

Given data,

V_o = 80 volts, I_o = 1000 amp

$\frac{V}{V_o}+\frac{I}{I_o}=1$ 

$\Rightarrow \frac{V}{80}+\frac{I}{1000}=1$ 

$\frac{V}{80}=1-\frac{I}{1000}$ 

$V=80[1-\frac{I}{1000}]=80-\frac{80}{1000}I$ …1)

V = 20 + 40l …2)

Equating 1) & 2)

$80-\frac{80}{1000}I=20+40l$ 

$\frac{80}{1000}I=80-20-40l$ 

$\frac{8I}{100}=60-40l$ 

$I=\frac{100}{8}[60-40l]$ 

P = VI

$P=(20+40l).\frac{100}{8}(60-40l)$ 

$P=\frac{100}{8}\left[(20+40l)(60-40l)\right]$ 

For arc power (P) to be maximum, dP/dl = 0

$\Rightarrow \left(\frac{100}{8}\right)[(20+40l)(-40)+\left(40\right)(60-40l)]=0$ 

⇒ 20 + 40l = 60 – 40l

80l = 40

L = 0.5 cm

⇒ Optimum arc length (l_opt) = 0.5 cm

Arc power corresponding to optimum arc length is,

$P=\frac{100}{8}\left[\{20+\left(40\right)\left(0.5\right)\}\{60-\left(40\right)\left(0.5\right)\}\right]$ 

$=\frac{100}{8}\left[\left(40\right)\left(40\right)\right]$ 

In top gating system:

$t_f=\frac{Mouldvolume\left(V_m\right)}{Moltenmetalflowrate\left(A_c{V}_c\right)}$

As gating ratio is given:

A_choke : A_runner : A_ingate = 1 : 1.5 : 2

Given A_ingate = 400 sq-mm.

Thus, $A_{choke}=\frac{400}{2}=200sq-mm$

Choke area will be considered for molten metal flow rate because choke regulates the rate of pouring of molten metal.

$\begin{array}{l}{V}_c=\sqrt{2g{h}_t}\\ =\sqrt{2\times 9810\times 250}\end{array}$

= 2214.7234 mm/s

$\begin{array}{l}{t}_f=\frac{\frac{\pi }{4}{d}^{2}h}{200\times 2214.7234}\\ =\frac{\frac{\pi }{4}\times {150}^2\times 200}{200\times 2214.7234}=7.979sec\end{array}$

Concept:

The final position of the square can be related to the initial position as:

x' = xT

Where, T is the transformation matrix and x is the original position

Calculation:

Let coordinates of a unit square is at (0,0), (1, 0), (1, 1), (0, 1)

$x=\left[\begin{array}{cc}0& 0\\ 1& 0\\ 1& 1\\ 0& 1\end{array}\right]$

$T=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\left[\begin{array}{cc}0& 0\\ 2& 3\\ 8& 4\\ 6& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 1& 0\\ 1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\left[\begin{array}{cc}0& 0\\ 2& 3\\ 8& 4\\ 6& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ a& b\\ a+c& b+d\\ c& d\end{array}\right]$

Giving:

$a=2,b=3,c=6andd=1$

Thus:

$T=\left[\begin{array}{cc}2& 3\\ 6& 1\end{array}\right]$

Rake angle (α) = 45°

From Merchant’s analysis, Merchant constant

C = 2ϕ + β - α

Where,

ϕ = Shear angle

β = Friction angle

α = Rake angle

Now, for finding ϕ i.e., shear angle

$\begin{array}{l}r=\frac{t}{t_c}=\frac{0.15}{0.3}=0.5\\ \varphi ={\tan }^{-1}\left(\frac{r\cos \alpha }{1-r\sin \alpha }\right)={\tan }^{-1}\left(\frac{0.5\cos 45}{1-0.5\sin 45}\right)={28.67}^{\circ }\end{array}$

From Merchant circle diagram as shown above, $\tan \left(\beta -\alpha \right)=\frac{F_T}{F_C}$

Where,

F_T = Thrust force

F_C = Cutting force

$\begin{array}{l}\therefore \tan \left(\beta -\alpha \right)=\frac{120}{300}\\ \therefore \beta -\alpha ={\tan }^{-1}\left(\frac{120}{300}\right)={21.80}^{\circ }\end{array}$

∴ Merchant constant (C) = 2ϕ + (β - α)

= 2 × 28.67 + 21.80 = 79.14°