Forming Process

Question 1
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).
A
3254.2
B
2453.6
C
4521.3
D
3883.5
GATE ME 2021 SET-2   Manufacturing Engineering
Question 1 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}
Question 2
The size distribution of the powder particles used in Powder Metallurgy process can be determined by
A
Laser scattering
B
Laser reflection
C
Laser absorption
D
Laser penetration
GATE ME 2021 SET-2   Manufacturing Engineering
Question 2 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.
Question 3
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).
A
780
B
652
C
365
D
985
GATE ME 2021 SET-1   Manufacturing Engineering
Question 3 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
Question 4
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).
A
1.2
B
0.6
C
0.8
D
0.2
GATE ME 2020 SET-2   Manufacturing Engineering
Question 4 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 5
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).
A
1.61
B
8.1
C
40.5
D
16.5
GATE ME 2020 SET-1   Manufacturing Engineering
Question 5 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}
Question 6
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
A
0.45 and 38.84
B
0.39 and 38.84
C
0.39 and 44.72
D
0.45 and 44.72
GATE ME 2020 SET-1   Manufacturing Engineering
Question 6 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 7
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
A
Roll forming
B
Strip rolling
C
Knurling
D
Chamfering
GATE ME 2019 SET-2   Manufacturing Engineering
Question 7 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 8
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).
A
220
B
158.85
C
316.25
D
452.15
GATE ME 2018 SET-2   Manufacturing Engineering
Question 8 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}
Question 9
The percentage scrap in a sheet metal blanking operation of a continuous strip of sheet metal as shown in the figure is _______ (correct to two decimal places).
A
12.28
B
53.25
C
68.75
D
78.55
GATE ME 2018 SET-1   Manufacturing Engineering
Question 9 Explanation: 
This rectangle ABCD will be repeated again and again.
\begin{aligned} A_{t}&=\text { Total Area }=\frac{7}{5} D \times \frac{6}{5} D=\frac{42}{25} D^{2} \\ A_{u}&=\text { Area of blanking Disc }=\frac{\pi D^{2}}{4}\\ \% \text { of scrap } &=\frac{A_{t}-A_{v}}{A_{t}} \times 100 \% \\ &=\left[1-\frac{\left(\frac{\pi}{4}\right)}{\left(\frac{42}{25}\right)}\right] \times 100 \%=53.25 \% \end{aligned}
Question 10
The maximum reduction in cross-sectional area per pass ( R ) of a cold wire drawing process is R=1-e^{-(n+1)} where n represents the strain hardening coefficient. For the case of a perfectly plastic material, R is
A
0.865
B
0.826
C
0.777
D
0.632
GATE ME 2018 SET-1   Manufacturing Engineering
Question 10 Explanation: 
\sigma_{d}=\sigma_{0} \ln \frac{A_{0}}{A_{f}}
For maximum reduction
\begin{aligned} \sigma_{d} &=\sigma_{0} \\ \sigma_{0} &=\sigma_{0} \ln \frac{A_{0}}{A_{f, \min }} \\ \Rightarrow \qquad 1&=\ln \left(\frac{A_{0}}{A_{f, \min }}\right) \\ \Rightarrow \qquad \frac{A_{0}}{A_{f, \min }} &=e=2.71828 \end{aligned}
Maximum reduction in Area
R=\frac{A_{0}-A_{f, \min }}{A_{0}}=\left(1-\frac{1}{e}\right)=\left(1-\frac{1}{2.71828}\right)=0.632
There are 10 questions to complete.

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