# Sheet Metal Operation

 Question 1
The thickness of a steel plate with material strength coefficient of 210 MPa, has to be reduced from 20 mm to 15 mm in a single pass in a two-high rolling mill with a roll radius of 450 mm and rolling velocity of 28 m/min. If the plate has a width of 200 mm and its strain hardening exponent, n is 0.25, the rolling force required for the operation is____________kN (round off to 2 decimal places).

Note: Average Flow Stress = Material Strength Coefficient $\times \frac{(True \; Strain)^n}{(1+n)}$
 A 1167.26 B 2512.36 C 3621.47 D 1124.36
GATE ME 2020 SET-1   Manufacturing Engineering
Question 1 Explanation:
\begin{aligned} \text { True strain, }\left(\epsilon_{T}\right) &=\ln \left(\frac{h_{f}}{h_{o}}\right) \\ &=\ln \left(\frac{15}{20}\right)=-0.28768 \\ \bar{\sigma}_{o} &=k\left(\frac{\epsilon_{T}^{n}}{1+n}\right) \\ &=210 \times \frac{(0.28768)^{0.25}}{1+0.25}=123.04 \mathrm{MPa} \\ F &=\bar{\sigma}_{o} \times \sqrt{R \Delta h} \times b \\ &=123.04 \times \sqrt{450 \times(20-15)} \times 200 \\ &=1167259.9 \mathrm{N}=1167.26 \mathrm{kN} \end{aligned}
 Question 2
A steel part with surface area of 125$cm^2$ is to be chrome coaled through an electroplating process using chromium acid sulphate as an electrolyte. An increasing current is applied to the part according to the following current time relation:

I = 12 + 0.2t

where, I = current (A) and t = time (minutes). The part is submerged in the plating solution for a duration of 20 minutes for plating purpose. Assuming the cathode efficiency of chromium to be 15% and the plating constant of chromium acid sulphate to be $2.50 \times 10^{-2} mm^3/A\cdot s$, the resulting coating thickness on the part surface is _________ $\mu m$ (round off to one decimal place)
 A 5 B 0 C 4.2 D 6.8
GATE ME 2020 SET-1   Manufacturing Engineering
Question 2 Explanation:
$I=12+0.2 t$
After time, 't', Next infinitly small time 'dt' let heat deposited 'dQ '.
$\therefore d Q=2.50 \times 10^{-2}\left(\mathrm{mm}^{3} / \mathrm{A} . \mathrm{s}\right) \times 12+0.2 \mathrm{t} \times d t$
As we have to convert this 's' to 'min'
$\therefore d Q=2.50 \times 10^{-2}\left(\mathrm{mm}^{3} / \mathrm{A} \times \min \right) \times 12+0.2 \mathrm{t} \times d t$
Considering cathode efficiency of 15%
\begin{aligned} d Q &=2.50 \times 10^{-2} \times 60 \times(12+0.2 t) d t \times 0.15 \mathrm{mm}^{3} \\ \therefore \ln 20 \mathrm{min}, \quad Q&=\int d Q\\&=\int_{0}^{20} 2.50 \times 10^{-2} \times 60 \times(12+0.2 t) d t \times 0.15 \\ &=0.225\left[12+0.1 t^{2}\right]_{0}^{20}\\ &=0.225\left[12 \times 20+0.1 \times 20^{2}\right] \mathrm{mm}^{3} \\ &=63 \mathrm{mm}^{3} \end{aligned}
As area os 125 $\mathrm{cm}^{2}$
$\begin{array}{l} \text { Plating thickness, } t=\frac{63}{125 \times(100)}=0.00504 \mathrm{mm}=5.04 \mu \mathrm{m} \\ (\text { As } 1 \mathrm{cm}^{2}=100 \mathrm{mm}^{2} )\end{array}$

 Question 3
A sheet metal with a stock hardness of 250 HRC has to be sheared using a punch and a die having a clearance of 1 mm between them. If the stock hardness of the sheet metal increases to 400 HRC, the clearance between the punch and the die should be_________ mm.
 A 2.26 B 4.25 C 1.265 D 0.124
GATE ME 2020 SET-1   Manufacturing Engineering
Question 3 Explanation:
Tensile strength $\propto$ Hardness number
$\therefore$ Shear strength =0.5 tensile strength (Tresca theory)
=0.577 tensile strength (Von mises theory)
Shear strength $\propto$ Hardness number
\begin{aligned} &\text{and clearance, } \\ C &=0.0032 t \sqrt{\tau} \\ C & \propto \sqrt{\text { Hardness number }} \\ \frac{C_{2}}{C_{1}} &=\sqrt{\frac{(\text { Hardness number })_{2}}{(\text { Hardness number })_{1}}} \\ \text{or}\quad C_{2} &=1 \times \sqrt{\frac{400}{250}}=1.265 \mathrm{mm} \end{aligned}
 Question 4
The thickness of a sheet is reduced by rolling (without any change in width) using 600mm diameter rolls. Neglect elastic deflection of the rolls and assume that the coefficient of friction at the roll-workpiece interface is 0.05. The sheetenters the rotating rolls unaided. If the initial sheet thickness is 2mm, the minimum possible final thickness that can be produced by this process in a single pass is ______mm(round off to two decimal places).
 A 0.98 B 1.25 C 1.68 D 2.24
GATE ME 2019 SET-2   Manufacturing Engineering
Question 4 Explanation:
$\begin{array}{l} (\Delta \mathrm{h})_{\max }=\mu^{2} \mathrm{R} \\ \mu=0.05, \mathrm{R}=300 \mathrm{mm} \\ (\Delta \mathrm{h})_{\max }=0.05^{2} \times 300=0.75 \mathrm{mm} \\ \text { Minimum possible final thickness }=\mathrm{h}_{0}-(\Delta \mathrm{h})_{\max } \\ =2-0.75=1.25 \mathrm{mm} \end{array}$
 Question 5
In a single point turning operation with cemented carbide tool and steel work piece, it is found that the Taylor's exponent is 0.25. If the cutting speed is reduced by 50% then the tool life changes by ______ times.
 A 11 B 19 C 16 D 12
GATE ME 2016 SET-3   Manufacturing Engineering
Question 5 Explanation:
\begin{aligned} V_{2} &=0.5 V_{1} \\ V_{1} T_{1}^{n} &=V_{2} T_{2}^{n} \\ \left(\frac{T_{2}}{T_{1}}\right)^{n} &=\left(\frac{V_{1}}{V_{2}}\right) \\ \frac{T_{2}}{T_{1}} &=(2)^{4}=16 \end{aligned}

There are 5 questions to complete.

### 2 thoughts on “Sheet Metal Operation”

1. Questions 15 is wrong