Production Engineering Test 1

Concept:

\(\frac{{Time}}{{hole}} = \frac{L}{{f.N}}\)

Length of machining = (thickness) + (approach) + (overrun) + (compulsory approach)

Calculation:

Given:

Thickeness = 60 mm, Approach = 3 mm, Overrun = 3 mm, feed = 0.3 mm/rev, diameter = 20 mm, N = 600 rpm

Compulsory approach \(= \frac{{D/2}}{{\tan \left( {\beta /2\;} \right)}} = \frac{{10}}{{{\rm{tan}}\left( {59^\circ } \right)}} = 6.008\) 

⇒ L = 60 + 3 + 3 + 6.008  = 72.008 mm

\(\frac{{Time}}{{hole}} = \frac{L}{{f.N}} = \frac{{72.008}}{{0.3 \times 600}} = 0.40\;min\)

Concept:

\(\text{Heat}\text{ input }\!\!~\!\!\text{ }\left( H \right)=uA\ell =\eta VI\times \frac{\ell }{V}\)

u = specific energy, v = welding speed, V = Voltage, I = current, L = length of weld

Calculation:

\(u=\eta VI\times \frac{L}{v}\)

\(u=0.86\times 24\times 480\times \frac{1}{20}\times 60~J/mm\) 

Concept:

Mass of material at inlet = Mass of material at outlet

ṁ = ṁ

ρ V̇  = ρ V̇

A_iV_i = A₀V₀

Calculation:

Given:

A_i = 60, V_i = 0.02 m/s, A₀ = 2.2 m²

A_iV_i = A₀V₀

60 × 0.02 = 2.2 × V₀

∴ V₀ = 0.5454 m/s

Concept:

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

F = 96,500 (Faraday’s constant), ρ = density, I = current, Z = valency, A = atomic weight

Calculation:

\(I = \frac{{\rho ZF}}{A} \times MRR\)

\(I = \frac{{10\; \times \;2\; \times \;96500}}{{58}}\; \times \;\frac{3}{{60}}\)

∴ I = 1663.79 A

Concept:

Add all the basic size, upper tolerance limit, and lower tolerance limit individually.

Calculation:

Given:

\({\rm{Part\;}}1 = 25.32_{ - 0.01}^{ + 0.02},{\rm{\;Part\;}}2 = 18.91_{ - 0.03}^{ + 0.03},{\rm{\;Part\;}}3 = 62.17_{ - 0.01}^{ + 0.05}{\rm{\;and\;Part\;}}4 = 46.25_{ - 0.04}^{ + 0.04}\)

Assembled basic size = 25.32 + 18.91 + 62.17 + 46.25 = 152.65

Assembled upper tolerance limit = 0.02 + 0.03 + 0.05 + 0.04 ⇒ 0.14

Assembled lower tolerance limit = 0.01 + 0.03 + 0.01 + 0.04 ⇒ 0 .09

\(\therefore {\bf{Assembled}}\;{\bf{length}} = 1525.65_{ - 0.09}^{ + 0.14}\)

Concept:

In blanking operation, die is exact and clearance is provided on the punch

In punching operation, punch is exact and clearance is provided on the die

Calculation:

Given:

Thickness of the sheet (t) = 2.5 mm, % of clearance (c) = 2 %

Blanking operation:

Punch size \(= {\rm{}}40 - 2\left[ {\frac{2}{{100}}} \right]2.5 = 39.90\;mm\)

Die size = 40 mm

Punching operation:

Punch size = 40 mm

Die size \(= 40 + 2\left[ {\frac{2}{{100}}} \right]2.5 = 40.10\;mm\)

Note: Die size is always greater than the punch size

Concept:

True strain \(= ln\;\left[ {\frac{{{h_f}}}{{{h_i}}}} \right]\)

Calculation:

Given:

D_f = 650 mm, h_f = 100 mm, h_i = 180 mm

True strain \(= \;ln\;\left[ {\frac{{100}}{{180}}} \right]\)

∴ True strain = - 0.5877

Explanation:

Given:

T_c = 12 min, n = 0.3, c = 200

Calculation:

For maximum productivity,

\({\rm{Tool\;life\;}}\left( {{T_0}} \right) = {T_c}\left( {\frac{{1 - n}}{n}} \right)\)

\({T_0} = 12\left( {\frac{{1 - 0.3}}{{0.3}}} \right)\)

∴ T₀ = 28 min

Since, V₀T₀ⁿ = c

V₀ (28)^0.3 = 200

∴ V₀ = 73.6 m/min

Explanation:

Coordinate system:

Two types of coordinate systems are used to define and control the position of the tool in relation to the work-piece.

Each system has its own applications and both systems may be used independently or may be mixed within a CNC part programming.

Absolute system:

A (2.5,1.2) → B (12.5,24.2)

Both A and B are written with respect to origin (0,0) as the datum.

Incremental system:

A (2.5,1.2)

Here co-ordinate of B (x₂,y₂) will be taken with respect to A (x₁,y₁).

∴ B (x₂ - x₁,y₂ – y₁)

⇒ B (12.5 – 2.5,24.2 – 1.2)

⇒ B (10,23)

Concept:

The roll strip contact length is given by:

\(L = \sqrt {R\Delta h}\) 

And the roll separating force is \(F = {\sigma _f} \times wL\)

Calculation:

∆h = 10 – 6 = 4 mm

\(L = \sqrt {R\Delta h} = \sqrt {100 \times 4} = 20\;mm\)

The roll separating force is:

\(F = {\sigma _f} \times wL = 500 \times 20 \times 200 = 2000000\;N = 2000\;kN\)

Concept:

Heat input per unit length can be obtained from the following equation

\(\frac{Q}{\ell }=\frac{V\cdot I}{velocity~}× \eta\)

V = voltage

I = current

η = Heat transfer efficiency

Calculation:

Given, for the First coupon: V = 15 V, I = 300 A, velocity = 30 mm/min 

\({{\left( \frac{Q}{\ell } \right)}_{I}}=\frac{15× 300}{30\left( \frac{mm}{min} \right)}× {{\eta }_{1}}~Joules\)

For second coupon: V = 60 kV = 60 × 10³ V, I = 200 mA = 200 × 10^-3 A, velocity = 25 mm/s

\({{\left( \frac{Q}{\ell } \right)}_{II}}=\frac{60× {{10}^{3}}× 200× {{10}^{-3}}}{\frac{25}{60}\left( \frac{mm}{min} \right)}× {{\eta }_{2}}~Joules\)

It is given that \({{\eta }_{2}}=\frac{{{\eta }_{1}}}{2}\)

Now the ratio:

\(\frac{{{\left( \frac{Q}{\ell } \right)}_{I}}}{{{\left( \frac{Q}{\ell } \right)}_{II}}}=\frac{\frac{15× 300}{30}× {{\eta }_{1}}}{\frac{60× {{10}^{3}}× 200× {{10}^{-3}}}{\frac{25}{60}}× {{\eta }_{2}}}\)

\(\frac{{{\left( \frac{Q}{\ell } \right)}_{I}}}{{{\left( \frac{Q}{\ell } \right)}_{II}}}=\frac{15× 300}{30}× \frac{25}{60× 200× 60}× \frac{1}{2}\)  (Given η₂ = η₁/2)   

\(\frac{{{\left( \frac{Q}{\ell } \right)}_{I}}}{{{\left( \frac{Q}{\ell } \right)}_{II}}}=\frac{75}{8}\)

Explanation:

Given:

V₁ = 90 m/min, T₁ = 75 min, V₂ = 70 m/min, T₂ = 140 min

Calculation:

Taylor’s tool life equation is VTⁿ = C

Thus,

V₁T₁ⁿ = c

90 (75)ⁿ = c      …1)

Also, V₂T₂ⁿ = c

70 (140)ⁿ = c     …2)

Now,

From 1) & 2)

90 (75)ⁿ = 70 (140)ⁿ

n = 0.402 → n ≃ 0.4

Putting in 1)-

90 (75)^0.4 = c

Concept:

For a top gating system, the poring time or filling time is given by

\(Filling\;time = \frac{{Volume}}{{Flow\;rate}} = \frac{{lbh}}{{{A_C}{V_{max}}}}\)

Where lbh = volume of mold;

A_C = choke area = min (A_S(sprue c-s area)_, A_r (runner c-s area), A_g (gate c-s area));

V_max = √2gh;

h = height of the sprue;

The gating ratio is given by

GR = A_s : A_r : A_g

Calculation:

Given:

l = 80 cm; b = 50 cm; h = 20 cm; h = 15 cm;

Time = t = 20 sec; GR = 2 : 1 : 2 ⇒ A_s: A_r: A_g = 2 : 1 : 2;

From the ratio, minimum area is for A_r that is Runner (Option 2 is right)

Without gating ratio, he would consider sprue cross-section area,

\(30 = \frac{{80\; \times \;50\; \times\; 20}}{{{A_s}\; \times \;\sqrt {2\; \times \;981\; \times \;15} }} \Rightarrow {A_s} = 15.5\;c{m^2}\)

\( \Rightarrow \frac{{\pi d_s^2}}{4} = 15.5 \Rightarrow {d_s} = 4.44\;cm\;\) (Option 1)

If gating ratio considered,

A_c = A_r = 15.5 cm²;

A_s : A_r : A_g = 2 : 1 : 2 ⇒ A_s = 31 cm²

\(\therefore \frac{{\pi d_s^2}}{4} = 31 \Rightarrow {d_s} = 6.28\;cm\)   (Option 4)

Concept:

\(r = \frac{t}{{{t_c}}} = \frac{{uncut\;chip\;thickness}}{{chip\;thickness\;after\;cut}}\)

\(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }}\)

\(\frac{V}{{\cos \left( {\phi - \alpha } \right)}} = \frac{{{V_s}}}{{\cos \alpha }}\)

V = cutting speed, V_s = shear speed

\({\rm{Shear\;strain\;rate}} = \varepsilon = \frac{{{V_s}}}{{{t_s}}}\)

Calculation:

Given:

α = 14°, V = 3 m/s

\(r = \frac{{0.5}}{{0.7}}\)

\(\phi = {\tan ^{ - 1}}\left( {\frac{{r\cos \alpha }}{{1 - r\sin \alpha }}} \right)\)

∴ ϕ = 39.957°

Now,

\({V_s} = \frac{{V\cos \alpha }}{{\cos \left( {\phi - \alpha } \right)}} = \frac{{3\cos 14}}{{\cos \left( {39.957 - 14} \right)}}\)

∴ V_s = 3.24 m/s

Now,

\(\varepsilon = \frac{{{V_s}}}{{{t_s}}} = \frac{{3.24}}{{30 \times {{10}^{ - 6}}}}\)

\(\therefore \epsilon = 1.08 \times {10^5}{s^{ - 1}}\)

Concept:

Energy required = F_max × Punch travel

Calculation:

Given:

Blank diameter = 60 mm, Shear strength (τ_s) = 260 N/mm², Thickness of sheet = 2 mm

Penetration (p) = 30 %

Now,

F_max = πDt × τ_s

D = punch diameter

Punch diameter = Black – 2C

∴ Punch diameter = 60 – 2C

Now,

\({\rm{Clearance\;}}\left( C \right) = 0.0032\;t\sqrt {{\tau _s}} \)

C = 0.1031 mm

Punch diameter = 60 – 2 × 0.1031

∴ Punch diameter = 59.79360 mm

Now,

F_max = πDt × τ_s = 97.680 kN

Energy required = F_max × Punch travel

Energy required = 97.680 × Pt = 97.680 × 0.3 × 2

Concept:

At coulomb friction condition,

α = β & ΔH = μ²R

Where α = bite angle; β = friction angle;

Roller separating force is given by

F_avg = P_avg × projected area = P_avg × b × L

Where b = width of sheet; L = deformation zone length;

\({P_{avg}} = \frac{2}{{\sqrt 3 }}{\sigma _y}\left( {1 + \frac{{\mu L}}{{4H}}} \right)\)

Where σ_y = yield stress; H = average of H_o, H_f

Now,

Power required is given by

P = 2 T ω; T = F_avg × λ × L

Where λ = arm factor

Calculation:

Given:

H_o = 50 mm; H_f = 30 mm; b = 150 mm; R = 400 mm;

σ_y = 200 MPa; N = 10 rpm ⇒ ω = 1.04 rad/s; λ = 0.3;

Δ H = H_o – H_f = 50 – 30 = 20 mm;

Now,

Coulomb friction condition ⇒

20 = μ² × 400 ⇒ μ = 0.22 (Option 1)

β = tan^-1 μ = tan^-1 0.22 = 12.6°

⇒ α = β = 12.6° (Option 2)

Now,

Deformation zone length is given by

L = √R ΔH = √400 × 20 = 89.5 mm

H = avg (50, 30) = 40 mm

\({P_{avg}} = \frac{2}{{\sqrt 3 }}200\left( {1 + \frac{{0.22\; \times \;89.5}}{{4\; \times \;40}}} \right) = 259.36\;MPa\)

F_avg = 259.36 × 150 × 89.5 = 3481910 = 3481 kN (Option 3 is wrong)

Now,

Power required is given by

P = 2 × F_avg × λ × L × ω = 2 × 3481910 × 0.3 × 0.0895 × 1.04 = 194457

∴ P = 194.4 kW (Option 4)

Explanation:

Given:

 α = 0°

\(\tan ϕ = \frac{{r\cos α }}{{1 - r\sin α }}\)

Chip thickness ratio \(\left( r \right) = \frac{t}{{{t_c}}} = \frac{{0.25}}{{0.75}}\)

\(= \frac{1}{3}\)

\(\tan ϕ = \frac{{\left( {\frac{1}{3}} \right)\cos 0^\circ }}{{1 - \left( {\frac{1}{3}} \right)\sin 0^\circ }}\)

\(\tan ϕ = \frac{1}{3}\)

ϕ = 18.435°

Now,

since rake angle α = 0 hence F_c = F_n

∴ F_c = 950 N

Now,

Fs = F_ccosϕ - F_tsinϕ  

∴ Fs = 750.96 N

Shear plane area \(\left( {{A_S}} \right) = \frac{{bt}}{{\sin ϕ }}\)

Where, b: width of cut, t = uncut chip thickness

\({A_s} = \frac{{2.5 \times 0.25}}{{\sin 18.435^\circ }}\)

∴ A_s = 1.9764 mm²

Now,

\(Shear\;stress\;\left( {{\tau _s}} \right) = \frac{{{F_s}}}{{{A_s}}} = \frac{{750.96}}{{1.9764}}\)

Explanation:

The relation between ORS and ASA tool designation is

tan α₀ = sin λ ⋅ tan α_s + cos λ ⋅ tan α_b     …1)

where, α₀ → orthogonal rake angle

λ → principal cutting edge angle

α_s → side rake angle

α_b → back rake angle

ASA tool designation is

α_b – α_s – γ_e – γ_s – ψ_e – ψ_s – r

Now,

Comparing, we get

α_b = 8°, α_s = 11°, ψ_s = 25°

Now,

Principal cutting edge angle

λ = (90 – ψ_s)

Thus, λ = 90 – 25 = 65°

Now,

Putting in equation 1)

tan α₀ = (sin 65 × tan 11) + (cos 65 × tan 8)

tan α_s = 0.2355

∴ α₀ = 13.25°

Explanation:

Given:

I = 36000 A, t = 0.01 sec, R = 110 μ Ω

Nugget diameter = D = 6 mm, Nugget thickness = h = 2.5 mm, Heat required per mm³ = 12 J/mm³.

Now,

\({\rm{Volume\;of\;nugget\;}} = \left[ {\frac{\pi }{4}{D^2} \times h} \right]\)

\({\rm{Volume\;of\;nugget\;}} = \left[ {\frac{\pi }{4} \times {6^2} \times 2.5} \right]\)

 Volume of nugget = 70.685 mm³

Now,

Heat required for producing the weld bead = volume of weld bead × H.R/mm³

Heat required for producing the weld bead = 70.685 × 12

∴ Heat required for producing the weld bead = 848.23 J

Now,

Heat generated = [I²Rt]

Heat generated = [36000² × 110 × 10^-6 × 0.01]

∴ Heat generated = 1425.6 J

Now,

The amount of heat distributed to the surroundings = heat generated – heat required

The amount of heat distributed to the surroundings = 1425.6 – 848.23

∴ The amount of heat distributed to the surroundings = 577.37 J

Concept:

Drawing force (F_d) = Drawing stress × Final area

Drawing force (F_d) = σ_d × A_f

\({\rm{Drawing\;stress\;}}\left( {{\sigma _d}} \right) = {\sigma _0}\left[ {\frac{{1 + B}}{B}} \right]\left[ {1 - {{\left[ {\frac{{{D_f}}}{{{D_0}}}} \right]}^{2B}}} \right]\)

Where

B = μ cot α, α = semi-die angle, σ₀ = stress of work material

Now,

Power required in drawing = Drawing force × Drawing speed

P = F_d × V_d

Calculation:

Given:

D_i = 15 mm, D_f = 11 mm, V_d = 100 m/min, α = 7°, μ = 0.25, σ₀ = 450 MPa   

\({\sigma _d} = {\sigma _0}\left[ {\frac{{1 + B}}{B}} \right]\left[ {1 - {{\left[ {\frac{{{D_f}}}{{{D_i}}}} \right]}^{2B}}} \right]\)

\({\sigma _d} = 450\left[ {\frac{{1 + 0.25\cot 7^\circ }}{{0.25\cot 7^\circ }}} \right]\left[ {1 - {{\left[ {\frac{{11}}{{15}}} \right]}^{2 \times 0.25\cot 7^\circ }}} \right]\)

∴ σ_d = 481.235 MPa

Now,

\({\rm{Drawing\;force\;}} = 481.235 \times \frac{\pi }{4} \times {\left( {11} \right)^2}\)

∴ Drawing force = 45.733 kN

Now,

Power required = Drawing force × drawing speed

Power required = 45.733 × 100

\({\rm{Power\;required\;}} = \frac{{4573.3}}{{60}}\)

∴ Power required = 76.221 kW