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).

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).

865.3 | |

2429.3 | |

2145.6 | |

1254.5 |

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}

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 |

In a direct current arc welding process, the power
source has an open circuit voltage of 100 V and
short circuit current of 1000 A. Assume a linear
relationship between voltage and current. The
arc voltage (V) varies with the arc length (l) as
V = 10 + 5l, where V is in volts and l is in mm. The
maximum available arc power during the process is
_________ kVA (in integer).

80 | |

120 | |

180 | |

25 |

Question 2 Explanation:

Given, OCV = 100 V and SCC = 1000 A

Voltage Arc length characteristic V_{arc}=10+5l

We have,

\begin{aligned} V_P&=OCV-\frac{OCV}{SCC}I\\ V_P&=100-\frac{100}{1000}I\\ _P&=100-\frac{I}{10} \end{aligned}

For stable arc

\begin{aligned} V_{arc}&=V_P\\ 10+5l&=100-\frac{I}{10}\\ I&=900-50l\\ \text{Arc Power} (P)&=V \times I\\ P&=(10+5l)(900-50l)\\ P&=9000+4000l-250l^2 \end{aligned}

For Maximum Arc Power

\begin{aligned} \frac{dP}{dl}&=0\\ \frac{d}{dl}(9000+400l-250l^2)&=0\\ l=4000/500&=8mm \end{aligned}

Hence, for Maximum arc power, arc length is 8 mm

Voltage at l=8mm

\begin{aligned} V&=10+5l\\ V&=10+( 5 \times 8)\\ V&=50 \;Volt \end{aligned}

Current at l=8mm

\begin{aligned} I&=900-50l\\ I&=900-( 50 \times 8)\\ I&=500 \;A \end{aligned}

Maximum Arc Power

\begin{aligned} P_{max}&= V \times I\\ P_{max}&=50 \times 500\\ &=25000VA\\ &=25 kVA \end{aligned}

Voltage Arc length characteristic V_{arc}=10+5l

We have,

\begin{aligned} V_P&=OCV-\frac{OCV}{SCC}I\\ V_P&=100-\frac{100}{1000}I\\ _P&=100-\frac{I}{10} \end{aligned}

For stable arc

\begin{aligned} V_{arc}&=V_P\\ 10+5l&=100-\frac{I}{10}\\ I&=900-50l\\ \text{Arc Power} (P)&=V \times I\\ P&=(10+5l)(900-50l)\\ P&=9000+4000l-250l^2 \end{aligned}

For Maximum Arc Power

\begin{aligned} \frac{dP}{dl}&=0\\ \frac{d}{dl}(9000+400l-250l^2)&=0\\ l=4000/500&=8mm \end{aligned}

Hence, for Maximum arc power, arc length is 8 mm

Voltage at l=8mm

\begin{aligned} V&=10+5l\\ V&=10+( 5 \times 8)\\ V&=50 \;Volt \end{aligned}

Current at l=8mm

\begin{aligned} I&=900-50l\\ I&=900-( 50 \times 8)\\ I&=500 \;A \end{aligned}

Maximum Arc Power

\begin{aligned} P_{max}&= V \times I\\ P_{max}&=50 \times 500\\ &=25000VA\\ &=25 kVA \end{aligned}

Question 3 |

The best size wire is fitted in a groove of a metric
screw such that the wire touches the flanks of the
thread on the pitch line as shown in the figure. The
pitch (p) and included angle of the thread are 4 mm
and 60^{\circ} , respectively. The diameter of the best size
wire is ___________ mm (round off to 2 decimal
places).

1.12 | |

2.31 | |

4.25 | |

3.36 |

Question 3 Explanation:

Pitch (p) = 4 mm

Thread angle 2\alpha =60^{\circ}

Semi Thread Angle \alpha =30^{\circ}

Diameter of best size wire

\begin{aligned} d_w&=\frac{p}{2}\sec \alpha \\ d_w&=\frac{4}{2} \sec 30^{\circ}\\ d_w&=2.039 mm=2.31mm \end{aligned}

Thread angle 2\alpha =60^{\circ}

Semi Thread Angle \alpha =30^{\circ}

Diameter of best size wire

\begin{aligned} d_w&=\frac{p}{2}\sec \alpha \\ d_w&=\frac{4}{2} \sec 30^{\circ}\\ d_w&=2.039 mm=2.31mm \end{aligned}

Question 4 |

In an orthogonal machining operation, the cutting
and thrust forces are equal in magnitude. The uncut
chip thickness is 0.5 mm and the shear angle is
15^{\circ} . The orthogonal rake angle of the tool is 0^{\circ} and
the width of cut is 2 mm. The workpiece material
is perfectly plastic and its yield shear strength is
500 MPa. The cutting force is _________ N (round
off to the nearest integer).

5247 | |

1245 | |

2732 | |

3214 |

Question 4 Explanation:

Given, F_C=F_T

Uncut chip thickness (t_1)=0.5mm

Shear angle (\phi )=15^{\circ}

Orthogonal rake angle \alpha =0^{\circ}

Width of cut w=2mm

Shear strength (\tau _S) =500MPa

Cutting force f_c= ____N

\begin{aligned} F_S&=F_C \cos \phi -F_T \sin \phi \\ &=F_C \cos \phi -F_C \sin \phi \\ \\ &= F_C(\cos \phi - \sin \phi)\\ F_C&=\frac{F_S}{\cos \phi - \sin \phi} \\ &= \frac{\tau _S\cdot w\cdot t_1}{\sin \phi (\cos \phi - \sin \phi)}\\ &=\frac{500 \times 2 \times 0.5}{\sin 15^{\circ}(\cos 15^{\circ}-\sin 15^{\circ})}\\ &=2732 N \end{aligned}

Uncut chip thickness (t_1)=0.5mm

Shear angle (\phi )=15^{\circ}

Orthogonal rake angle \alpha =0^{\circ}

Width of cut w=2mm

Shear strength (\tau _S) =500MPa

Cutting force f_c= ____N

\begin{aligned} F_S&=F_C \cos \phi -F_T \sin \phi \\ &=F_C \cos \phi -F_C \sin \phi \\ \\ &= F_C(\cos \phi - \sin \phi)\\ F_C&=\frac{F_S}{\cos \phi - \sin \phi} \\ &= \frac{\tau _S\cdot w\cdot t_1}{\sin \phi (\cos \phi - \sin \phi)}\\ &=\frac{500 \times 2 \times 0.5}{\sin 15^{\circ}(\cos 15^{\circ}-\sin 15^{\circ})}\\ &=2732 N \end{aligned}

Question 5 |

A straight-teeth horizontal slab milling cutter is
shown in the figure. It has 4 teeth and diameter (D)
of 200 mm. The rotational speed of the cutter is
100 rpm and the linear feed given to the workpiece
is 1000 mm/minute. The width of the workpiece
(w) is 100 mm, and the entire width is milled in
a single pass of the cutter. The cutting force/tooth
is given by F = K t_cw, where specific cutting force
K = 10 N/mm^2
, \; w is the width of cut, and t_c
is the
uncut chip thickness.

The depth of cut (d) is D/2, and hence the assumption of \frac{d}{D} \lt \lt 1 is invalid. The maximum cutting force required is __________ kN (round off to one decimal place).

The depth of cut (d) is D/2, and hence the assumption of \frac{d}{D} \lt \lt 1 is invalid. The maximum cutting force required is __________ kN (round off to one decimal place).

1.8 | |

2.5 | |

3.2 | |

2.9 |

Question 5 Explanation:

Given data, No. of teeth (n)=4

Diameter of cutter (D) = 200 mm

Rotational speed (N) = 100 rpm

Linear feed to work piece = 1000 mm/min

Width of work piece (w) = 100 mm

Cutting force/tooth = F = K t_c w

Specific cutting force = K = 10 N/mm^2

Depth of cut (d)= D/2

Feed(f) = 1000/100 = 10 mm/rev

Uncut chip thickness =t_c

Maximum uncut chip thickness (t_c)_{max}

\begin{aligned} &=\frac{2f}{n}\sqrt{d/D(1-d/D)}\\ &=\frac{2 \times 10}{4}\sqrt{\frac{D/2}{D}\left ( 1- \frac{D/2}{D}\right )}\\ &=5\sqrt{1/2(1-1/2)}=2.5mm\\ &\text{Maximum force }\\ (F)_{max}&=K(t_c)_{max}\cdot w\\ &=10 \times 2.5 \times 100\\ &=2500N=2.5kN \end{aligned}

Diameter of cutter (D) = 200 mm

Rotational speed (N) = 100 rpm

Linear feed to work piece = 1000 mm/min

Width of work piece (w) = 100 mm

Cutting force/tooth = F = K t_c w

Specific cutting force = K = 10 N/mm^2

Depth of cut (d)= D/2

Feed(f) = 1000/100 = 10 mm/rev

Uncut chip thickness =t_c

Maximum uncut chip thickness (t_c)_{max}

\begin{aligned} &=\frac{2f}{n}\sqrt{d/D(1-d/D)}\\ &=\frac{2 \times 10}{4}\sqrt{\frac{D/2}{D}\left ( 1- \frac{D/2}{D}\right )}\\ &=5\sqrt{1/2(1-1/2)}=2.5mm\\ &\text{Maximum force }\\ (F)_{max}&=K(t_c)_{max}\cdot w\\ &=10 \times 2.5 \times 100\\ &=2500N=2.5kN \end{aligned}

Question 6 |

Which of these processes involve(s) melting in
metallic workpieces?

**MSQ**Electrochemical machining | |

Electric discharge machining | |

Laser beam machining | |

Electron beam machining |

Question 6 Explanation:

Non traditional thermal energy process includes electrical discharge machining (EDM), electron beam machining (EBM), and laser beam machining (LBM). EDM removes metal by a series of discrete electrical discharge (sparks) that cause localized temperatures high enough to melt or vaporize the metal in the immediate vicinity of the discharge. EBM adopts a high velocity stream of electrons focused on the workpiece surface to remove material by melting and vaporization. LBM employs the light energy from a laser to remove material by vaporization and ablation.

Question 7 |

Consider sand casting of a cube of edge length a. A
cylindrical riser is placed at the top of the casting.
Assume solidification time, t_s\propto V/A , where V isthe
volume and A is the total surface area dissipating
heat. If the top of the riser is insulated, which of the
following radius/radii of riser is/are acceptable?

**MSQ**\frac{a}{3} | |

\frac{a}{2} | |

\frac{a}{4} | |

\frac{a}{6} |

Question 7 Explanation:

Riser should take more time for solidification than
casting \left ( \frac{V}{A} \right )_r\geq\left ( \frac{V}{A} \right )_c

For top riser bottom area is not cooling surface and it is given that top cross section is also insulated. So only lateral area is cooling area.

\frac{\frac{\pi}{4}D^2H}{\pi DH}\geq \frac{a^3}{6a^2}\Rightarrow \frac{D}{4}\geq \frac{a}{6}\Rightarrow D\geq \frac{2a}{3}

Therefore, Radius R\geq \frac{a}{3} .

only answer is available \frac{a}{2}.

For top riser bottom area is not cooling surface and it is given that top cross section is also insulated. So only lateral area is cooling area.

\frac{\frac{\pi}{4}D^2H}{\pi DH}\geq \frac{a^3}{6a^2}\Rightarrow \frac{D}{4}\geq \frac{a}{6}\Rightarrow D\geq \frac{2a}{3}

Therefore, Radius R\geq \frac{a}{3} .

only answer is available \frac{a}{2}.

Question 8 |

Which one of the following CANNOT impart linear
motion in a CNC machine?

Linear motor | |

Ball screw | |

Lead screw | |

Chain and sprocket |

Question 8 Explanation:

Chain and Sprocket mechanism is not used in CNC
Machines.

Linear Motors Recently linear motors are being increasingly considered for use in high performance CNC machine tools. The linear motor consists of a series of magnets attached to the machine base and a set of electrical coils potted around a steel laminate core attached to the moving slide.

The fact that there are no mechanical parts in contact means that there is no wear periodic maintenance required. Linear motors are not limited in travel like ball screws. Larger bare required to achieve high velocity, with a longer travel to prevent undue vibration.

This larger ball screw results in a higher inertia. This means a larger motor with more torque is required (introduction inertia) and the responsiveness and bandwidth of the system is reduced, resulting in poor servo performance.

Machines built with linear motors and all-digital drive systems can produce parts with higher accuracy and tighter tolerances at higher feeds and speeds. Also, they reduce significantly the non-machining time with high acceleration and deceleration rates.

Linear Motors Recently linear motors are being increasingly considered for use in high performance CNC machine tools. The linear motor consists of a series of magnets attached to the machine base and a set of electrical coils potted around a steel laminate core attached to the moving slide.

The fact that there are no mechanical parts in contact means that there is no wear periodic maintenance required. Linear motors are not limited in travel like ball screws. Larger bare required to achieve high velocity, with a longer travel to prevent undue vibration.

This larger ball screw results in a higher inertia. This means a larger motor with more torque is required (introduction inertia) and the responsiveness and bandwidth of the system is reduced, resulting in poor servo performance.

Machines built with linear motors and all-digital drive systems can produce parts with higher accuracy and tighter tolerances at higher feeds and speeds. Also, they reduce significantly the non-machining time with high acceleration and deceleration rates.

Question 9 |

Match the additive manufacturing technique in
Column I with its corresponding input material in
Column II.

Additive manufacturing technique (Column I)

P. Fused deposition modelling

Q. Laminated object manufacturing

R. Selective laser sintering

Input material (Column II)

1. Photo sensitive liquid resin

2. Heat fusible powder

3. Filament of polymer

4. Sheet of thermoplastic or green compacted metal sheet

Additive manufacturing technique (Column I)

P. Fused deposition modelling

Q. Laminated object manufacturing

R. Selective laser sintering

Input material (Column II)

1. Photo sensitive liquid resin

2. Heat fusible powder

3. Filament of polymer

4. Sheet of thermoplastic or green compacted metal sheet

P-3, Q-4, R-2 | |

P-1, Q-2, R-4 | |

P-2, Q-3, R-1 | |

P-4, Q-1, R-4 |

Question 9 Explanation:

Fused-deposition modeling consists of a computercontrolled extruder, through which a polymer
filament is deposited to produce a part slice by slice.

Laminated-object manufacturing uses a laser beam or vinyl cutter to first cut the slices on paper or plastic sheets (laminations); then it applies an adhesive layer, if necessary and finally stacks the sheets to produce the part.

Selective laser sintering uses a high-powered laser beam to sinter powders or coatings on the powders in a desired pattern. Selective laser sintering has been applied to polymers, sand, ceramics, and metals.

Laminated-object manufacturing uses a laser beam or vinyl cutter to first cut the slices on paper or plastic sheets (laminations); then it applies an adhesive layer, if necessary and finally stacks the sheets to produce the part.

Selective laser sintering uses a high-powered laser beam to sinter powders or coatings on the powders in a desired pattern. Selective laser sintering has been applied to polymers, sand, ceramics, and metals.

Question 10 |

A shaft of diameter 25^{^{-0.04}}_{-0.07} mm is assembled in a
hole of diameter 25^{^{+0.02}}_{0.00} mm.

Match the allowance and limit parameter in Column I with its corresponding quantitative value in Column II for this shaft-hole assembly.

Allowance and limit parameter (Column I)

P. Allowance

Q. Maximum clearance

R. Maximum material limit for hole

Quantitative value (Column II)

1. 0.09 mm

2. 24.96 mm

3. 0.04 mm

4. 25.0 mm

Match the allowance and limit parameter in Column I with its corresponding quantitative value in Column II for this shaft-hole assembly.

Allowance and limit parameter (Column I)

P. Allowance

Q. Maximum clearance

R. Maximum material limit for hole

Quantitative value (Column II)

1. 0.09 mm

2. 24.96 mm

3. 0.04 mm

4. 25.0 mm

P-3, Q-1, R-4 | |

P-1, Q-3, R-2 | |

P-1, Q-3, R-4 | |

P-3, Q-1, R-2 |

Question 10 Explanation:

(1) Allowance = Lower limit of hole - upper limit of shaft

Allowance = 25.00 - 24.96 = 0.04 mm

(2) Maximum clearance C_{max} = Upper limit of hole - lower limit of shaft

C_{max} = 25.02 - 24.93 = 0.09 mm

(3) Maximum material limit for hole = minimum size of hole = 25.00

There are 10 questions to complete.

question number -12 is mostly to engineering mechanics than metrology, manufacturing