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

Milling:

• Milling is a process of producing flat and complex shapes with the use of a multi-point (or multi-tooth) cutting tool. The axis of rotation of the cutting tool is perpendicular to the direction of feed, either parallel or perpendicular to the machined surface.

Milling machines are superior to other machines in terms of accuracy and surface finish.

There are two basic types of milling operations:

Down milling: 

• The cutter rotation is in the same direction as the motion of the workpiece being fed.
• The cut starts with the full chip thickness 
• The cutting force is maximum at the beginning and minimum at the end of the cut.
• In down milling, the cutting force is directed on to the work table, which allows thinner parts to be machined without susceptibility to breakage, but it requires the optimum holding device as the cutting tool forces the work-piece in the opposite direction of the tool motion. Therefore the gap between the lead screw and the half nut increases which requires backlash eliminator. 
• A better surface finish is obtained.

Up milling:

• The workpiece is moving towards the cutter, opposing the cutter direction of rotation.
• The cutting force is minimum during the beginning of the cut and maximum at the end of the cut.

Explanation:-

The surface roughness on a drawing is represented by triangles.

Surface texture or roughness representation

The basic symbol consists of two legs of unequal length inclined at approximately 60° to the line representing the considered surface.

The symbol must be represented by a thin line.

The value of roughness is added to the symbols.

1. Roughness ‘a’ obtained by any production process.

2. Roughness ‘a’ obtained by machining.

3. Roughness ‘a’ obtained without removal of material.

If it is necessary to impose maximum and minimum limits of surface roughness, both values are indicated. a1 = Maximum limit; a2 = Minimum limit

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,

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

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

Rake angle (γ):

• It is the angle of inclination of the cutting tool rake surface from the reference plane, i.e. the plane perpendicular to the velocity vector.
• It is an important parameter in various cutting and machining process.

There are three types of rake angle i.e. positive, negative and zero.

Positive rake angle:

• It is used when there is requirement of less cutting force and used for low strength materials like mild steel etc.

Negative rake angle:

• It is used for hard metals which requires high cutting force and machined at very high speed.

Zero rake angle:

• It is used for gear cutting or thread manufacturing. Brass and cast iron are machined with zero rake.

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

Honing

• Honing is an abrasion process that involves the use of abrasive stones or tools to remove material from the surface of a workpiece, typically to improve its surface finish or to achieve precise dimensions. This process is widely used in manufacturing and engineering industries for finishing cylindrical surfaces, such as engine cylinders, hydraulic cylinders, and other precision components.
• In the honing process, a tool called a honing stone, which is embedded with abrasive particles, is rotated or oscillated against the surface of the workpiece. The motion of the honing tool can be linear, rotary, or a combination of both. The abrasive particles on the honing stone cut away material from the surface of the workpiece, resulting in a highly smooth and dimensionally accurate surface.
• Honing typically involves the use of a lubricant or coolant to reduce friction, dissipate heat, and carry away debris generated during the process. The combination of abrasive action and lubrication ensures that the surface finish is consistent and free from defects such as scratches or tool marks.

Applications of Honing:

• Improving the surface finish and dimensional accuracy of engine cylinders, ensuring proper sealing and efficient operation of pistons.
• Finishing hydraulic cylinders to achieve precise dimensions and smooth surfaces, which are essential for proper functioning and minimizing wear.
• Removing small amounts of material from bores or cylindrical surfaces to achieve tight tolerances.
• Creating cross-hatch patterns on surfaces to improve lubrication retention, especially in applications involving sliding motion.

Advantages of Honing:

• Produces extremely smooth and accurate surfaces.
• Can achieve very tight tolerances, making it suitable for precision applications.
• Improves the wear resistance and service life of components.
• Enhances the performance of engine cylinders by creating a cross-hatch pattern that retains lubrication.

Correct Answer is: Honing

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 

Fillet Gauge

Definition: A fillet gauge is a tool specifically designed to measure the radius of curvature of convex and concave surfaces. This tool is used across a wide range of applications, such as engineering, manufacturing, and quality control, to ensure that the curvature of surfaces conforms to specified dimensions. Fillet gauges are available in various configurations, with each blade typically representing a specific radius or curvature. These gauges are particularly useful for checking radii ranging from small values (e.g., 1 mm) to larger values (e.g., 25 mm).

Working Principle: A fillet gauge consists of multiple blades, each shaped to a specific radius or curvature. To use the tool, the operator selects the appropriate blade and places it against the curved surface to be measured. If the curvature of the surface matches the curvature of the blade, the fit will be precise, confirming the radius of the surface. This tool is easy to use, requires no additional setup, and provides quick and accurate measurements.

Advantages:

• Simplicity: Fillet gauges are straightforward to use, requiring no complex setup or calibration.
• Cost-Effective: These gauges are relatively inexpensive, making them accessible to a wide range of industries.
• Accuracy: Fillet gauges provide precise measurements of small radii, which are critical in manufacturing and quality control.
• Portability: The compact design of fillet gauges makes them easy to carry and use in various settings.

Disadvantages:

• Limited Range: Fillet gauges are typically designed for specific ranges of radii, so multiple sets may be required for broader applications.
• Subjectivity: The accuracy of the measurement can depend on the user's ability to visually align the gauge with the surface.

Applications:

• Machining: Ensuring that machined parts have the correct radius of curvature.
• Welding: Checking the radius of weld fillets for compliance with design specifications.
• Inspection: Verifying the dimensions of parts during quality control processes.
• Design Engineering: Ensuring that components meet ergonomic or aesthetic requirements involving curved surfaces.