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)}

Note: Average Flow Stress = Material Strength Coefficient \times \frac{(True \; Strain)^n}{(1+n)}

1167.26 | |

2512.36 | |

3621.47 | |

1124.36 |

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 125cm^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)

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)

5 | |

0 | |

4.2 | |

6.8 |

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}

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.

2.26 | |

4.25 | |

1.265 | |

0.124 |

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}

\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).

0.98 | |

1.25 | |

1.68 | |

2.24 |

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.

11 | |

19 | |

16 | |

12 |

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}

Question 6 |

In a sheet metal of 2 mm thickness a hole of 10 mm diameter needs to be punched. The yield strength in tension of the sheet material is 100 MPa and its ultimate shear strength is 80 MPa. The force required to punch the hole (in kN) is __________

4.025kN | |

2.087kN | |

1.098kN | |

5.024kN |

Question 6 Explanation:

Punching force,

\begin{aligned} F_{\max } &=\pi d t\tau_{\text {pers. }} \\ &=\pi \times 10 \times 2 \times 80=5.024 \mathrm{kN} \end{aligned}

\begin{aligned} F_{\max } &=\pi d t\tau_{\text {pers. }} \\ &=\pi \times 10 \times 2 \times 80=5.024 \mathrm{kN} \end{aligned}

Question 7 |

Internal gears are manufactured by

Internal gears are manufactured by | |

shaping with pinion cutter | |

shaping with rack cutter | |

milling |

Question 7 Explanation:

Internal gears are manufactured by gear Broaching and shaping with pinion cutter only, whereas shaping with pinion cutter is used for both external and internal gears.

Question 8 |

The value of true strain produced in compressing a cylinder to half its original length is

0.69 | |

-0.69 | |

0.5 | |

-0.5 |

Question 8 Explanation:

\epsilon_{T}=\ln \left(\frac{L}{L_{0}}\right)=\ln \left(\frac{(L / 2)}{L}\right)=\ln 1 / 2=-0.69

Question 9 |

A 300 mm thick slab is being cold rolled using roll of 600 mm diameter. If the coefficient of friction is 0.08, the maximum possible reduction (in mm) is __________

5.36mm | |

3.25mm | |

1.92mm | |

2.35mm |

Question 9 Explanation:

\Delta H_{\max }=\mu^{2} R=(0.08)^{2} \times 300=1.92 \mathrm{mm}

Question 10 |

In a rolling process, the maximum possible draft, defined as the difference between the initial and the final thickness of the metal sheet, mainly depends on which pair of the following parameters?

P: Strain

Q: Strength of the work material

R: Roll diameter

S: Roll velocity

T: Coefficient of friction between roll and work

P: Strain

Q: Strength of the work material

R: Roll diameter

S: Roll velocity

T: Coefficient of friction between roll and work

Q, S | |

R, T | |

S, T | |

P, R |

Question 10 Explanation:

\left(t_{i}-t_{f}\right)_{\max }=\mu^{2} R

where \mu is coefficient of friction

R is roll radius.

where \mu is coefficient of friction

R is roll radius.

There are 10 questions to complete.

Questions 15 is wrong

It’s a common data question.