# 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.