# Forming Process

 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).

 A 865.3 B 2429.3 C 2145.6 D 1254.5
GATE ME 2022 SET-2   Manufacturing Engineering
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}
 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, $\mu =0.1$. The flow stress ($\sigma$) of the material in MPa is $\sigma=207+414\varepsilon$ , 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.
 A 0.12 B 0.35 C 0.55 D 0.85
GATE ME 2022 SET-1   Manufacturing Engineering
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}

 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).
 A 3254.2 B 2453.6 C 4521.3 D 3883.5
GATE ME 2021 SET-2   Manufacturing Engineering
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}
 Question 4
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 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.
 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).
 A 780 B 652 C 365 D 985
GATE ME 2021 SET-1   Manufacturing Engineering
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, $F=\sigma _f(Lw)=159.4 \times 24.49 \times 200=780.776kN$

There are 5 questions to complete.

### 1 thought on “Forming Process”

1. There is a correction in GATE ME 2022 set-1 question
Mistake : μ=0.01
correction : μ=0.1

Mistake 2 (in the same question)
Mistake : 204+414ε (flow stress equation)
Correction : 207+414ε