Production Engineering Test 2

Concept:

Extrusion ratio: \(r = \frac{{Initial\;area}}{{Final\;area}} = \frac{{{A_i}}}{{{A_f}}}\)

Area of circle: \(A = \frac{\pi }{4}{d^2}\)

Calculation:

r = 25, A_i = 300 cm²

\({A_f} = \frac{{{A_i}}}{r} = \frac{{300}}{{25}} = 12\;c{m^2}\)

\(\frac{\pi }{4}d_f^2 = 12 \Rightarrow {d_f} = 3.91\;cm\)

Concept:

True strain \(\Rightarrow {E_T} = \ln \frac{{{L_f}}}{{{L_i}}}\)

L_f = final length, L_i = initial length

Calculation:

\({\epsilon_T} = \ln \left( {\frac{{280}}{{220}}} \right)\)

∴ ϵ_T = 0.24

Concept:

Maximum possible reduction per pass (ΔH) = μ²R

μ = coefficient of friction, R = radius of roll.

Now,

\({\rm{\% \;increase}} = \frac{{{{\left( {{\rm{\Delta }}H} \right)}_n} - {{\left( {{\rm{\Delta }}H} \right)}_o}}}{{{{\left( {{\rm{\Delta }}H} \right)}_o}}} \times 100\)

Calculation:

(ΔH)_o = μ²R

(ΔH)_o = (1.6)² μ²R

% increase = [(1.6)² - 1] × 100

∴ % increase = 156 %

Concept:

Degree of drawing (D) is given by:

\(D = \frac{{{A_0} - {A_f}}}{{{A_0}}}\)

Calculation:

Given:

d₀ = 18 mm

Since degree of drawing (D) is given by:

\(D = \frac{{{A_0} - {A_f}}}{{{A_0}}}\)

\(0.36 = 1 - \frac{{{A_f}}}{{{A_0}}}\)

\(\frac{{{A_f}}}{{{A_o}}} = 0.64\)

\(\frac{{\frac{\pi }{4}d_f^2}}{{\frac{\pi }{4}d_0^2}} = 0.64\)

d_f = 0.8 × 18

= 14.4

% reduction in diameter \( = \frac{{{d_0} - {d_f}}}{{{d_0}}} \times 100\)

= \(\frac{{18 - 14.4}}{{18}} \times 100\) 

= 20%

Metal forming process may be classified on the basis of type of forces applied to the work piece as it is formed into direct shape.

• Direct compression processes, where compressive force is applied to the work surface and the metal flows at right angles to the direction of compression. e.g.-Forging, Rolling
• Indirect compression processes, where the applied forces are tensile (drawing), or compressive (extrusion), but the indirect compressive forces developed on the die walls are very effective in metal flow. e.g.-Extrusion, Wire Drawing
• Tension type forming, as in stretch forming, where the metal sheet is wrapped to the contour of a die under the application of tensile forces
• Bending type processes, as in sheet bending
• Shearing processes, where excessive shearing force is applied to rupture the metal (Punching, Blanking, Perforating, Parting etc.)

The correct match is:

Concept:

Power required in drawing = drawing force × velocity

Drawing stress, \({\sigma _d} = {\sigma _0}\left( {\frac{{1 + B}}{B}} \right)\left[ {1 - {{\left( {\frac{{{r_i}}}{{{r_0}}}} \right)}^{2B}}} \right]\) 

B = μ cot α

For maximum reduction, σ_d = σ₀

When back pressure is given,

\({\sigma _d} = {\sigma _0}\left[ {\frac{{1 + B}}{B}} \right]\left[ {1 - {{\left( {\frac{{{r_i}}}{{{r_0}}}} \right)}^{2B}}} \right] + {\sigma _B}{\left( {\frac{{{r_i}}}{{{r_0}}}} \right)^{2B}}\)

Calculation:

Given:

σ₀ = 310 MPa, α = 2.5°, d_i = 20 mm, μ = 0.2, d₀ = 14 mm, r = 110 m/min

B = μ cot α

B = 0.2 cot (2.5°)

∴ B = 4.58

Now,

\({\sigma _d} = 310\left[ {\frac{{5.58}}{{4.58}}} \right]\left[ {1 - {{\left( {\frac{{14}}{{20}}} \right)}^{9.16}}} \right] = 363.30\;MPa\)

\({\rm{Power}} = 363.30 \times \frac{\pi }{4} \times {\left( {\frac{{14}}{{1000}}} \right)^2} \times \frac{{110}}{{60}} \times {10^6}\)

∴ Power = 102.53 kW

Now,

For maximum reduction σ_d = σ₀

\(\frac{{4.58}}{{5.58}} = \left( {1 - {{\left( {\frac{{{d_i}}}{{{d_0}}}} \right)}^{9 - 16}}} \right) \Rightarrow \frac{{{d_i}}}{{{d_0}}} = 0.828\)

\({\rm{\% \;reduction\;}} = 1 - \frac{{{d_i}}}{{{d_0}}} = 0.17 \sim 17.11\% \)

Now,

Back pressure = 60 MPa.

\({\sigma _d} = 363.30 + 60 \times {\left( {\frac{{14}}{{20}}} \right)^{9.16}}\)

∴ σ_d = 365.60 MPa  

Concept:

When corner radius is given

\(D = \sqrt {{d^2} + 4dh} - \frac{r}{2}\)

r = radius, h = depth, d = diameter.

Now,

d_n = d_n-1 (1 - DRR_n)

d₁ = D (1 – DRR₁)

Calculation:

Given:

h = 180 mm, r = 8 mm, d = 120 mm

\(D = \sqrt {{d^2} + 4\;dh} - \frac{r}{2}\)

∴ D = 313.49 mm

Now,

d₁ = D (1 – DRR₁)

d₁ = 313.49 × 0.6 = 188.09 mm > 120 mm

Now,

d₂ = d₁ (1 – DRR₂)

d₂ = 188.09 × 0.8

d₂ = 150.47 mm > 120 mm

Now,

d₃ = 150.47 × 0.8

d₃ = 120.38 mm

d₄ = 120.38 × 0.8 = 96.30 mm < 120.38 mm

∴ 4 draws are required.

Concept:

As the sheet passes through the rolls due to conservation of mass at inlet and outlet, the velocity at the outlet is more than the velocity of rolls and that at inlet is less than the velocity of rolls. Hence, the velocity of sheet varies and at the neutral point no slip occurs since velocity of sheet is equal to the velocity of roll at that point.

w_ih_iv_i = w_fh_fv_f

\(\frac{{{v_f}}}{{{v_i}}} = \frac{{{h_i}}}{{{h_n}}}\)

\({\rm{Forward\;slip\;}} = \frac{{{\rm{Velocity\;at\;exit\;}} - {\rm{\;neutral\;point\;velocity}}}}{{{\rm{neutral\;point\;velocity}}}}{\rm{\;}} \times 100{\rm{\;}}\)

\({\rm{Backward\;slip\;}} = \frac{{{\rm{Neutral\;point\;velocity\;}} - {\rm{\;velocity\;at\;inlet}}}}{{{\rm{\;Neutral\;point\;velocity}}}}{\rm{\;}} \times 100\)

Calculation:

In a single pass rolling operation a sheet of dimensions 120 mm × 20 mm is reduced to 14 mm thickness. Rolls are of diameter 300 mm and rotate at 180 rpm. If the sheet thickness at the neutral plane is 16 mm, calculate the sum of forward and backward slip in % respectively.

Given: h_i = 20 mm, h_f = 14 mm, h_n = 16 mm

\(Forward\;slip\; = \;\frac{{{v_f} - \;{v_n}}}{{{v_n}\;}} = \frac{{{v_f}}}{{{v_n}}} - 1 = \frac{{{h_n}}}{{{h_f}\;}} - 1 = \frac{{16}}{{14}} - 1 = \frac{2}{{14}}\;\)

\(Forward\;slip = \frac{2}{{14}} \times 100 = 14.28\%\)

\(Backward\;slip\; = \frac{{{v_n} - \;{v_i}}}{{{v_n}}} = 1 - \frac{{{v_i}}}{{{v_n}}} = 1 - \;\frac{{{h_n}}}{{{h_i}\;}} = 1 - \frac{{16}}{{20}} = \frac{4}{{20}}\)

\(Backward\;slip = \frac{4}{{20\;}}\; \times 100 = 20\%\)

Sum of forward and backward slip = 14.28 + 20 = 34.28 %

Concept:

Forging force (F) in open die forging can be determined by

\(F = {\sigma _f}.\frac{\pi }{4}d_f^2\left[ {1 + \frac{{\mu {d_f}}}{{3{h_f}}}} \right]\) 

Where, σ_f = flow stress of the material (MPa)

d_f = final diameter (mm)

μ = coefficient of friction

h_f = final height (mm)

Calculation:

Given data

d_i = 150 mm, h_i = 100 mm, d_f = ?, h_f = 50 mm, μ = 0.2

Volume of component is conserved.

\(\frac{\pi }{4}d_i^2{h_i} = \frac{\pi }{4}d_f^2{h_f}\)

(150)². (100)² = d_f­².50 ⇒ d_f = 150√2 mm

Forging force is calculated by

\(F = 1000 \times \frac{\pi }{4} \times {\left( {150} \right)^2} \times 2\left( {1 + \frac{{\left( {0.2} \right)\left( {150} \right)\sqrt 2 }}{{3 \times 50}}} \right)\)