Question 1 |

A cylindrical billet of 100 mm diameter and 100
mm length is extruded by a direct extrusion process
to produce a bar of L-section. The cross sectional
dimensions of this L-section bar are shown in the
figure. The total extrusion pressure (p) in MPa for
the above process is related to extrusion ratio (r) as

p=K_s\sigma _m\left [ 0.8+1.5 \ln (r)+\frac{2l}{d_0} \right ]

where \sigma _m is the mean flow strength of the billet material in MPa, l is the portion of the billet length remaining to be extruded in mm, d_0 is the initial diameter of the billet in mm, and K is the die shape factor.

If the mean flow strength of the billet material is 50 MPa and the die shape factor is 1.05, then the maximum force required at the start of extrusion is ________ kN (round off to one decimal place).

p=K_s\sigma _m\left [ 0.8+1.5 \ln (r)+\frac{2l}{d_0} \right ]

where \sigma _m is the mean flow strength of the billet material in MPa, l is the portion of the billet length remaining to be extruded in mm, d_0 is the initial diameter of the billet in mm, and K is the die shape factor.

If the mean flow strength of the billet material is 50 MPa and the die shape factor is 1.05, then the maximum force required at the start of extrusion is ________ kN (round off to one decimal place).

865.3 | |

2429.3 | |

2145.6 | |

1254.5 |

Question 1 Explanation:

Original length of the billet (L_0)=100mm

Original diameter of billet (d_0)=100mm

Mean flow stress of billet material (\sigma _m)=50MPa

Die Shape factor K_S=1.05

Original Cross Sectional Area of Billet

A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}100^2=7853.981

Cross sectional area of extruded product

A_f=(10 \times 50)+ (10 \times 50)=1000 mm^2

Extrusion ratio r=\frac{A_0}{A_f}=\frac{7853.9816}{1000}=7.854

Extrusion pressure will be maximum at the start of extrusion process where L_0=100mm

\begin{aligned} P_{max}&= K_S\sigma _m\left [ 0.8+1.5 \ln (r)+\frac{2L_0}{d_0} \right ]\\ &=1.05 \times 50 \left [ 0.8+1.5 \ln (7.854)+\frac{2 \times 100}{100} \right ]\\ &=309.305 MPa \end{aligned}

Maximum Extrusion force

\begin{aligned} F_{max}&= P_{max} \times A_0\\ &=309.305 \times \frac{\pi}{4}(100)^2\\ &=2429265.97N\\ &=2429.3kN \end{aligned}

Original diameter of billet (d_0)=100mm

Mean flow stress of billet material (\sigma _m)=50MPa

Die Shape factor K_S=1.05

Original Cross Sectional Area of Billet

A_0=\frac{\pi}{4}d_0^2=\frac{\pi}{4}100^2=7853.981

Cross sectional area of extruded product

A_f=(10 \times 50)+ (10 \times 50)=1000 mm^2

Extrusion ratio r=\frac{A_0}{A_f}=\frac{7853.9816}{1000}=7.854

Extrusion pressure will be maximum at the start of extrusion process where L_0=100mm

\begin{aligned} P_{max}&= K_S\sigma _m\left [ 0.8+1.5 \ln (r)+\frac{2L_0}{d_0} \right ]\\ &=1.05 \times 50 \left [ 0.8+1.5 \ln (7.854)+\frac{2 \times 100}{100} \right ]\\ &=309.305 MPa \end{aligned}

Maximum Extrusion force

\begin{aligned} F_{max}&= P_{max} \times A_0\\ &=309.305 \times \frac{\pi}{4}(100)^2\\ &=2429265.97N\\ &=2429.3kN \end{aligned}

Question 2 |

A 4 mm thick aluminum sheet of width w=100mm
is rolled in a two-roll mill of roll diameter 200 mm
each. The workpiece is lubricated with a mineral
oil, which gives a coefficient of friction, \mu =0.01 .
The flow stress ( \sigma ) of the material in MPa is \sigma=204+414\varepsilon , where \varepsilon is the true strain. Assuming
rolling to be a plane strain deformation process, the
roll separation force (F) for maximum permissible
draft (thickness reduction) is _________ kN (round
off to the nearest integer).

Use : F=1.15 \bar{\sigma }\left ( 1+\frac{\mu L}{2\bar{h}} \right )wL

where \bar{\sigma } is average flow stress, L is roll-workpiece contact length, and \bar{h} is the average sheet thickness.

Use : F=1.15 \bar{\sigma }\left ( 1+\frac{\mu L}{2\bar{h}} \right )wL

where \bar{\sigma } is average flow stress, L is roll-workpiece contact length, and \bar{h} is the average sheet thickness.

0.12 | |

0.35 | |

0.55 | |

0.85 |

Question 2 Explanation:

Initial thickness h_1=4mm

Width w= 100 mm

Roll Diameter D= 200 mm

Roll Radius R= 100 mm

Coefficient of Friction \mu =0.1

Flow stress is given by \sigma =207+414\varepsilon

We have equation for maximum possible reduction is

\begin{aligned} \Delta h&=\mu ^2R\\ \Delta h&=(0.1)^2 \times 100=1mm=h_1-h_2\\ h_2&=h_1-\Delta h=4-1=3mm \end{aligned}

Now True strain \varepsilon = \ln\left ( \frac{h_1}{h_2} \right )= \ln \left ( \frac{4}{3} \right )=0.2876

Stress when no rolling \varepsilon =0, \sigma _1=207 MPa

Stress when material is rolled to maximum possible extent \varepsilon =0.2876, \sigma _2=207+414(0.2876)=326.06 MPa

Since Strain hardening characteristics is linear the average flow stress \bar{\sigma }=\frac{\sigma _1+\sigma _2}{2}=\frac{207+326.06}{2}=266.53MPa

Projected length of contact L=\sqrt{R\cdot \Delta h}=\sqrt{100 \times 1}=10mm

Rolls separation force,

\begin{aligned} F&=1.15\bar{\sigma }\left ( 1+\frac{\mu L}{2\bar{h}} \right )wL\\ &=1.15 \times 266.53\left ( 1+\frac{0.1 \times 10}{2 \times \frac{7}{2}} \right ) \times 100 \times 10\\ F&=350.3 N\\ F&=0.350kN \end{aligned}

Width w= 100 mm

Roll Diameter D= 200 mm

Roll Radius R= 100 mm

Coefficient of Friction \mu =0.1

Flow stress is given by \sigma =207+414\varepsilon

We have equation for maximum possible reduction is

\begin{aligned} \Delta h&=\mu ^2R\\ \Delta h&=(0.1)^2 \times 100=1mm=h_1-h_2\\ h_2&=h_1-\Delta h=4-1=3mm \end{aligned}

Now True strain \varepsilon = \ln\left ( \frac{h_1}{h_2} \right )= \ln \left ( \frac{4}{3} \right )=0.2876

Stress when no rolling \varepsilon =0, \sigma _1=207 MPa

Stress when material is rolled to maximum possible extent \varepsilon =0.2876, \sigma _2=207+414(0.2876)=326.06 MPa

Since Strain hardening characteristics is linear the average flow stress \bar{\sigma }=\frac{\sigma _1+\sigma _2}{2}=\frac{207+326.06}{2}=266.53MPa

Projected length of contact L=\sqrt{R\cdot \Delta h}=\sqrt{100 \times 1}=10mm

Rolls separation force,

\begin{aligned} F&=1.15\bar{\sigma }\left ( 1+\frac{\mu L}{2\bar{h}} \right )wL\\ &=1.15 \times 266.53\left ( 1+\frac{0.1 \times 10}{2 \times \frac{7}{2}} \right ) \times 100 \times 10\\ F&=350.3 N\\ F&=0.350kN \end{aligned}

Question 3 |

The thickness, width and length of a metal slab are 50 mm, 250 mm and 3600 mm, respectively. A rolling operation on this slab reduces the thickness by 10% and increases the width by 3%. The length of the rolled slab is ________mm (round off to one decimal place).

3254.2 | |

2453.6 | |

4521.3 | |

3883.5 |

Question 3 Explanation:

\begin{array}{l} h_{1}=50 \mathrm{~mm} ; \quad h_{2}=0.9 h_{1} ; \\ b_{1}=250 \mathrm{~mm} ; \quad b_{2}=1.03 b_{1} ; \\ L_{1}=3600 \mathrm{~mm} ; \quad L_{2}=? \end{array}

Volume remains const. in theory of plasticity

\begin{aligned} h_{1} b_{1} L_{1} &=h_{2} b_{2} L_{2} \\ &=0.9 h_{1} \times 1.03 b_{1} \times L_{2} \\ L_{2} &=\frac{3600}{0.9 \times 1.03}=3883.5 \mathrm{~mm}=3883.5 \mathrm{~mm} \end{aligned}

Volume remains const. in theory of plasticity

\begin{aligned} h_{1} b_{1} L_{1} &=h_{2} b_{2} L_{2} \\ &=0.9 h_{1} \times 1.03 b_{1} \times L_{2} \\ L_{2} &=\frac{3600}{0.9 \times 1.03}=3883.5 \mathrm{~mm}=3883.5 \mathrm{~mm} \end{aligned}

Question 4 |

The size distribution of the powder particles used in Powder Metallurgy process can be determined by

Laser scattering | |

Laser reflection | |

Laser absorption | |

Laser penetration |

Question 4 Explanation:

Particle Size, Shape, and Distribution:

Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:

1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.

2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.

3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.

4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.

5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.

Particle size is generally controlled by screening, that is, by passing the metal powder through screens (sieves) of various mesh sizes. Several other methods also are available for particle-size analysis:

1. Sedimentation, which involves measuring the rate at which particles settle in a fluid.

2. Microscopic analysis, which may include the use of transmission and scanning- electron microscopy.

3. Light scattering from a laser that illuminates a sample, consisting of particles suspended in a liquid medium; the particles cause the light to be scattered, and a detector then digitizes the signals and computes the particle-size distribution.

4. Optical methods, such as particles blocking a beam of light, in which the particle is sensed by a photocell.

5. Suspending particles in a liquid and detecting particle size and distribution by electrical sensors.

Question 5 |

A 200 mm wide plate having a thickness of 20 mm is fed through a rolling mill with two rolls. The radius of each roll is 300 mm. The plate thickness is to be reduced to 18 mm in one pass using a roll speed of 50 rpm. The strength coefficient (K) of the work material flow curve is 300 MPa and the strain hardening exponent, n is 0.2. The coefficient of friction between the rolls and the plate is 0.1. If the friction is sufficient to permit the rolling operation then the roll force will be _____kN (round off to the nearest integer).

780 | |

652 | |

365 | |

985 |

Question 5 Explanation:

Given: Width of plate, w = 200 mm

Thickness of plate, t_0 = 20 mm

Radius of roller, R = 300 mm

Reduced thickness of plate in one pass,

t_f = 18 mm

Rollers speed, N = 50 rpm

Strength coefficient, (K) = 300 MPa

Strain hardening exponent, n = 0.2

Coefficient of friction, \mu = 0.1

Find: Roll force (F)

\Delta h=h_0-h_f=20-18=2mm

Contact length, L=\sqrt{R\Delta h}=\sqrt{300 \times 2 }=24.49mm

True strain, \varepsilon _T=\ln \left ( \frac{t_0}{t_f} \right )=\ln\left ( \frac{20}{18} \right )=0.1054

Average Flow stress, \bar{\sigma _f}=\frac{K(\varepsilon )^n}{1+n}=\frac{300 \times 0.1054^{0.2}}{1.2}=159.4MPa

Rolling force, F=\sigma _f(Lw)=159.4 \times 24.49 \times 200=780.776kN

Thickness of plate, t_0 = 20 mm

Radius of roller, R = 300 mm

Reduced thickness of plate in one pass,

t_f = 18 mm

Rollers speed, N = 50 rpm

Strength coefficient, (K) = 300 MPa

Strain hardening exponent, n = 0.2

Coefficient of friction, \mu = 0.1

Find: Roll force (F)

\Delta h=h_0-h_f=20-18=2mm

Contact length, L=\sqrt{R\Delta h}=\sqrt{300 \times 2 }=24.49mm

True strain, \varepsilon _T=\ln \left ( \frac{t_0}{t_f} \right )=\ln\left ( \frac{20}{18} \right )=0.1054

Average Flow stress, \bar{\sigma _f}=\frac{K(\varepsilon )^n}{1+n}=\frac{300 \times 0.1054^{0.2}}{1.2}=159.4MPa

Rolling force, F=\sigma _f(Lw)=159.4 \times 24.49 \times 200=780.776kN

Question 6 |

There are two identical shaping machines S_1 and S_2. In machine S_2, the width of the
workpiece is increased by 10% and the feed is decreased by 10%, with respect to that
of S_1. If all other conditions remain the same then the ratio of total time per pass in
S_1 and S_2 will be __________ (roundoff to one decimal place).

1.2 | |

0.6 | |

0.8 | |

0.2 |

Question 6 Explanation:

\begin{aligned} t_{1} &=\frac{B_{1}}{f_{1} N_{1}} \\ t_{2} &=\frac{B_{2}}{f_{2} N_{2}}=\frac{1.1 B_{1}}{0.9 f_{1} \times N_{2}} \\ \frac{t_{1}}{t_{2}} &=\frac{B_{1} / f_{1} N}{1.1 B_{1}}=\frac{0.9}{1.9 f_{1} N}=0.8 \end{aligned}

Question 7 |

A slot of 25 mm x 25 mm is to be milled in a workpiece of 300 mm length using a
side and face milling cutter of diameter 100 mm, width 25 mm and having 20 teeth.

For a depth of cut 5 mm, feed per tooth 0.1 mm, cutting speed 35 m/min and approach and over travel distance of 5 mm each, the time required for milling the slot is_______ minutes (round off to one decimal place).

For a depth of cut 5 mm, feed per tooth 0.1 mm, cutting speed 35 m/min and approach and over travel distance of 5 mm each, the time required for milling the slot is_______ minutes (round off to one decimal place).

1.61 | |

8.1 | |

40.5 | |

16.5 |

Question 7 Explanation:

\begin{aligned} V &=\pi D N \\ 35 &=\pi \times 0.100 \times N \\ N &=\pi \times 111.408 \mathrm{rpm} \\ &=\frac{L+\frac{D}{2}+A+O}{f Z N}=\frac{300+\frac{100}{2}+5+5}{0.1 \times 20 \times 111.408} \\ &=1.6157 \mathrm{min} \text { per pass } \end{aligned}

For 25 \mathrm{mm} cuts min 5 \mathrm{mm} depth of cut 5 pass needed

Total machining time =8.078 \mathrm{min} \simeq 8.1 \mathrm{min}

For 25 \mathrm{mm} cuts min 5 \mathrm{mm} depth of cut 5 pass needed

Total machining time =8.078 \mathrm{min} \simeq 8.1 \mathrm{min}

Question 8 |

A strip of thickness 40 mm is to be rolled to a thickness of 20 mm using a two-high
mill having rolls of diameter 200 mm. Coefficient of friction and arc length in mm,
respectively are

0.45 and 38.84 | |

0.39 and 38.84 | |

0.39 and 44.72 | |

0.45 and 44.72 |

Question 8 Explanation:

h_{0}=40 \mathrm{mm}, h_{f}=20 \mathrm{mm}, \Delta h=40-20=20 \mathrm{mm}, D=200 \mathrm{mm}, R=100 \mathrm{mm}\begin{aligned} \text { Projected length, } \quad L&=\sqrt{R \Delta h}\\ &=\sqrt{100 \times 20} \\ &=\sqrt{2000}=44.7213 \mathrm{mm} \\ \Delta h&=\mu^{2} R \\ \therefore \quad 20 &=\mu^{2} \cdot 100 \\ \therefore \quad 2.0 &=\mu^{2} \\ \mu &=0.4472 \end{aligned}

Question 9 |

The cold forming process in which a hardened tool is pressed against a workpiece (when there is relative motion between the tool and the workpiece) to produce a roughened surface with a regular pattern is

Roll forming | |

Strip rolling | |

Knurling | |

Chamfering |

Question 9 Explanation:

Knurling is the process of producing a straight angled cross lines by rolling using lathe machine.
It is done by using one or more hard rollers that contain reverse of the pattern to be imposed.

Question 10 |

A steel wire is drawn from an initial diameter(d_{i}) of 10 mm to a final diameter (d_{f}) of 7.5
mm. The half cone angle (\alpha) of the die is 5^{\circ} and the coefficient of friction (\mu) between the die and the wire is 0.1. The average of the initial and final yield stress [(\sigma_{Y}) _{avg}] is 350
MPa. The equation for drawing stress \sigma_{f}, (in MPa) is given as :

\sigma _{f}=(\sigma _{Y})_{avg}\left \{ 1+\frac{1}{\mu cot \alpha } \right \}\left [ 1-{\left (\frac{d_{f}}{d_{i}}\right )}^{2\mu cot\alpha } \right ]

The drawing stress (in MPa) required to carry out this operation is _________ (correct to two decimal places).

\sigma _{f}=(\sigma _{Y})_{avg}\left \{ 1+\frac{1}{\mu cot \alpha } \right \}\left [ 1-{\left (\frac{d_{f}}{d_{i}}\right )}^{2\mu cot\alpha } \right ]

The drawing stress (in MPa) required to carry out this operation is _________ (correct to two decimal places).

220 | |

158.85 | |

316.25 | |

452.15 |

Question 10 Explanation:

\begin{aligned} d_{i} &=10 \mathrm{mm} \\ d_{f} &=7.5 \mathrm{mm} \\ \alpha &=5^{\circ} \\ \mu &=0.1 \\ (\sigma y)_{\mathrm{avg}} &=350 \mathrm{MPa} \\ \sigma_{f} &=\left(\sigma_{y}\right)_{\mathrm{avg}} \times\left\{1+\frac{1}{\mu \cot \alpha}\right\}\left[1-\left(\frac{d_{f}}{d_{i}}\right)^{2 \mu \cot \alpha}\right] \\ &=350 \times\left[1+\frac{1}{0.1 \mathrm{cot} 5}\right]\left[1-\left(\frac{7.5}{10}\right)^{2 \times 0.1 \times \cot 5}\right] \\ &=316.2472 \mathrm{MPa} \approx 316.25 \mathrm{MPa} \end{aligned}

There are 10 questions to complete.