Production Engineering Test 2

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

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

⇒ 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:

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:

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:

\(V_1T_1^{0.5}=V_2T_2^{0.5}\)

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

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

∴ T₂ = 4T₁

% increase in tool life is -

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

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

∴ % increase = 300 %.

Explanation:

Given:

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

Calculation:

\({\rm{Diameter\;of\;blank\;}}\;\left( D \right) = \sqrt {{d^2} + 4dh} \)

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

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:

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

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

\(\frac{t}{{{t_c}}} = \frac{{{V_c}}}{V}\)

\(\frac{{0.5}}{{0.6}} = \frac{{{V_c}}}{2} \Rightarrow {V_c} = 1.66\;m/s\)

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:

\(\frac{{{\rm{\Delta }}{{\rm{h}}_2} - {\rm{\Delta }}{h_1}}}{{{\rm{\Delta }}{h_1}}} \times 100 = \frac{{{{1.08}^2} - 1\;}}{1} \times 100 = 16.64\% \)

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

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

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

\(= \frac{{MRR}}{{Surface\;area}} = \frac{{0.03677 \times {{10}^{ - 6}} \times 60}}{{625 \times {{10}^{ - 3}}}}\)

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:

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,

\(Tan\phi = \frac{{rcos\alpha }}{{1 - rsin\alpha }}\)

∴ 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,

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

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

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

\(\frac{V}{{{V}_{o}}}+\frac{I}{{{I}_{o}}}=1\) 

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

\(\frac{V}{80}=1-\frac{I}{1000}\) 

\(V=80\left[ 1-\frac{I}{1000} \right]=80-\frac{80}{1000}I\) …1)

V = 20 + 40l …2)

Equating 1) & 2)

\(80-\frac{80}{1000}I=20+40l\) 

\(\frac{80}{1000}I=80-20-40l\) 

\(\frac{8I}{100}=60-40l\) 

\(I=\frac{100}{8}\left[ 60-40l \right]\) 

P = VI

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

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

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

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

⇒ 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,

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

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

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)

\(x = \left[ {\begin{array}{*{20}{c}}0&0\\1&0\\1&1\\0&1\end{array}} \right]\)

\(T = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}0&0\\2&3\\8&4\\6&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&0\\1&0\\1&1\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}}0&0\\2&3\\8&4\\6&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&0\\a&b\\{a + c}&{b + d}\\c&d\end{array}} \right]\;\)

Giving:

\(a = 2\;,\;b = 3\;,c = 6\;and\;d = 1\)

Thus:

\(T = \left[ {\begin{array}{*{20}{c}}2&3\\6&1\end{array}} \right]\)