GATE Past Year Questions: Machining Processes | Manufacturing Engineering - Mechanical Engineering PDF Download

Q1: A flat surface of a C60 steel having dimensions of 100 mm (length) × 200 mm (width) is produced by a HSS slab mill cutter. The 8-toothed cutter has 100 mm diameter and 200 mm width. The feed per tooth is 0.1 mm, cutting velocity is 20 m/min and depth of cut is 2 mm. The machining time required to remove the entire stock is ______ minutes (rounded off to 2 decimal places). [GATE ME 2024 ]Ans: 2.2 to 2.65
Data given:
Z = 8, D = 100 mm, w = 200 mm 
V = 20 m/min, d = 2 mm, f_t = 0.1 mm/ tooth
N = ?

Le = effective length of approach = QR + L = 14 + 100 = 114 mm
= For removal of entire length

So, feed = f_t × NZ = 0.1 × 63.662 × 8 = 50.93 mm/min
t_m/c = Le/F = 114/50.93 = 2.24 min

Q2: A cutting tool provides a tool life of 60 minutes while machining with the cutting speed of 60 m/min. When the same tool is used for machining the same material, it provides a tool life of 10 minutes for a cutting speed of  100 m/min. If the cutting speed is changed to 80 m/min for the same tool and work material combination, the tool life computed using Taylor's tool life model is _____ minutes (rounded off to 2 decimal places).  [GATE ME 2024]
Ans: 21 to 22.5
Taylor Tool Life Equation
VTⁿ = Constant

Given : V₁ = 60(m/min)⁡, T₁ = 60 min
V₂ = 100(m/min)⁡, T₂ = 10 min
So, to find out

InV₁ + n nnT₁ = lnV₂ + nlnT₂
lnV₁ − InV₂ = n[lnT₂ − InT₁]

Now, V₃ = 80 m/min, T₃ = ?

So, =100 × 10^0.285
⇒ T₃ = 21.87 min

Q3: The grinding wheel used to provide the best surface finish is  [GATE ME 2024]
(a) A36L5V
(b) A54L5V
(c) A60L5V
(d) A80L5V
Ans: (d)
Grinding Wheel Components :
Abrasive Type A60-M5-V
Abrasive size A60-M5-V
Grade/Hardness A60-M5-V
Structure A60-M5-V
Bond A60-M5-V

Fig. Specification of grinding wheel
Fine and superfine grain size give best surface finish.

Q1: A solid part (see figure) of polymer material is to be fabricated by additive manufacturing (AM) in square-shaped layers starting from the bottom of the part working upwards. The nozzle diameter of the AM machine is a/10 mm and the nozzle follows a linear serpentine path parallel to the sides of the square layers with a feed rate of a/5 mm/min.
Ignore any tool path motions other than those involved in adding material, and any other delays between layers or the serpentine scan lines.
The time taken to fabricate this part is _______ minutes. (Answer in integer)  [GATE ME 2023]
Ans: 9000 to 9000
V = (3a)² × 1.5a + (2a)² × a + (a)² × 0.5a
= 18a³

= (18a³/a³) x 100 x 5 = 9000 min

Q2: In an ideal orthogonal cutting experiment (see figure), the cutting speed V is 1 m/s, the rake angle of the tool α = 5º, and the shear angle, ϕ, is known to be 45º.
Applying the ideal orthogonal cutting model, consider two shear planes PQ and RS close to each other. As they approach the thin shear zone (shown as a thick line in the figure), plane RS gets sheared with respect to PQ (point R₁ shears to R₂, and S₁ shears to S₂).
Assuming that the perpendicular distance between PQ and RS is δ = 25 μm, what is the value of shear strain rate (in s⁻¹) that the material undergoes at the shear zone? [GATE ME 2023]
(a) 1.84 × 10⁴
(b) 5.20 × 10⁴
(c) 0.71 × 10⁴
(d) 1.30 × 10⁴
Ans: (b)
Data given α = 5º
ϕ = 45º
ΔH = 25 μm
V =  1 m/sec

Shear strain rate = = 5.20 x 10⁴s<sup>-1

Q3: A cuboidal part has to be accurately positioned first, arresting six degrees of freedom and then clamped in a fixture, to be used for machining. Locating pins in the form of cylinders with hemi-spherical tips are to be placed on the fixture for positioning. Four different configurations of locating pins are proposed as shown. Which one of the options given is correct?  [GATE ME 2023]</sup>(a) Configuration P1 arrests 6 degrees of freedom, while Configurations P2 and P4 are over-constrained and Configuration P3 is under-constrained.
(b) Configuration P2 arrests 6 degrees of freedom, while Configurations P1 and P3 are over-constrained and Configuration P4 is under-constrained.
(c) Configuration P3 arrests 6 degrees of freedom, while Configurations P2 and P4 are over-constrained and Configuration P1 is under-constrained.
(d) Configuration P4 arrests 6 degrees of freedom, while Configurations P1 and P3 are over-constrained and Configuration P2 is under-constrained.
Ans: (a) 
3-2-1 principle of location
The 3-2-1 principle of location (six point location principle) is used to constrain the movement of workpiece along the three axes XX, YY and ZZ.
This is achieved by providing six locating points, 3-pins in base plate, 2-pins in vertical plane and 1-pin in a plane which is perpendicular to first two planes.

Q1: 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 15º. The orthogonal rake angle of the tool is 0º 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).  [GATE ME 2022 SET-2]
Ans: 2700 to 2750
Given, F_C = F_T
Uncut chip thickness (t₁) = 0.5mm 
Shear angle (ϕ) = 15º
Orthogonal rake angle α = 0º
Width of cut w = 2mm 
Shear strength (τ_S) = 500MPa
Cutting force f_c = ____ N
F_S = F_C cosϕ − F_T sinϕ
= F_C cosϕ − F_C sinϕ
= F_C (cosϕ − sinϕ)

= 2732 N.

Q2: 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 = Kt_cw, where specific cutting force K = 10N/mm², w is the width of cut, and t_c is the uncut chip thickness.
The depth of cut (dd) is D/2, and hence the assumption of d/D << 1 is invalid. The maximum cutting force required is __________ kN (round off to one decimal place).  [GATE ME 2022 SET-2]
Ans: 2.4 to 2.6
Given data, No. of teeth (n) = 4 
Diameter of cutter (D) = 200mm
Rotational speed (N) = 100rpm 
Linear feed to work piece = 1000 mm/min
Width of work piece (w) = 100mm
Cutting force/tooth = F = Kt_cw
Specific cutting force = K = 10N/mm²
Depth of cut (d) = D/2 
Feed (f) = 1000/100 = 10mm/rev
Uncut chip thickness = t_c 
Maximum uncut chip thickness (t_c)_max

Maximum force
(F)_max = K(t_c)_max·ω
= 10 x 2.5 x 100
= 2500 N = 2.5 kN

Q3: Which of these processes involve(s) melting in metallic workpieces?  [GATE ME 2022 SET-2]
(a) Electrochemical machining
(b) Electric discharge machining
(c) Laser beam machining
(d) Electron beam machining
Ans: (b, c, d)
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.

Q4: A 1 mm thick cylindrical tube, 100 mm in diameter, is orthogonally turned such that the entire wall thickness of the tube is cut in a single pass. The axial feed of the tool is 1 m/minute and the specific cutting energy (u) of the tube material is 6 J/mm³. Neglect contribution of feed force towards power. The power required to carry out this operation is _________ kW (round off to one decimal place). [GATE ME 2022 SET-1]
Ans: 30 to 32
Given data, cylindrical tube thickness (t) = 1 mm
Diameter (D) = 100 mm
Orthogonal cutting such that entire wall thickness of tube is cut in single pass.
Therefore, Tube thickness = depth of cut = d = 1mm
Axial feed of the tool = 1m/min = f.N
Specific cutting Energy (U) = 6J/mm³
Power _________ kW
Specific cutting energy(U) = 6J/mm³ = Power/MRR
Power = 6 x MRR
= 6 x f.d.v
= 6 x f.d.π.D.N
= 6 × d × π × D × f.N
= 6 x 1 × π ×100 x 1 x (1000/60)
= 31415.9 J/sec or Watt
= 31.4 kW

Q5: Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 5º. The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 45º and 25º, respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).  [GATE ME 2022 SET-1]
Ans: 2570 to 2576
Cutting speed (V) = 4m/sec
Orthogonal rake angle α = 5º
Uncut chip thickness (t₁) = 0.2 mm
Width of cut (w) = 3mm
Friction angle β = 45º
Shear angle (ϕ) = 25º
Shear strength (τ₀) = 1000MPa
Cutting force (F_c) =____N
F_S = Rcos(ϕ + β − α)


= 2573.4 N

Q6: Electrochemical machining operations are performed with tungsten as the tool, and copper and aluminum as two different workpiece materials. Properties of copper and aluminum are given in the table below.
Ignore overpotentials, and assume that current efficiency is 100% for both the workpiece materials. Under identical conditions, if the material removal rate (MRR) of copper is 100 mg/s, the MRR of aluminum will be ________________ mg/s (roundoff to two decimal places). [GATE ME 2022 SET-1]
Ans: 27 to 30
Copper At. Wt = 63, Valency = 2

Aluminium At. Wt = 27, Valency = 3
e =27/3 = 9
Material Removal Rate in g/s is given by
MRR = e.I/F
e = Chemical equivalent
I = Current
F = Faraday constant
At Constant current & Faraday's constant
MRR ∝ e

MRR_Al = 28.57 mg/s

Q7: Which of the following additive manufacturing technique(s) can use a wire as a feedstock material?  [GATE ME 2022 SET-1]
(a) Stereolithography
(b) Fused deposition modeling
(c) Selective laser sintering
(d) Directed energy deposition processes
Ans: (b, d)
Fused-deposition modeling consists of a computercontrolled extruder, through which a polymer filament is deposited to produce a part slice by slice.
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.
Stereolithography involves a computer-controlled laser-focusing system, that cures a liquid thermosetting polymer containing a photosensitive curing agent.
Directed energy deposition (DED) processes enable the creation of parts by melting material as it is being deposited. Although this basic approach can work for polymers, ceramics, and metal matrix composites, it is predominantly used for metal powders. Thus, this technology is often referred to as "metal deposition" technology.

Q8: A CNC worktable is driven in a linear direction by a lead screw connected directly to a stepper motor. The pitch of the lead screw is 5 mm. The stepper motor completes one full revolution upon receiving 600 pulses. If the worktable speed is 5 m/minute and there is no missed pulse, then the pulse rate being received by the stepper motor is  [GATE ME 2022 SET-1]
(a) 20 KHz
(b) 10 kHz
(c) 3 kHz
(d) 15 kHz
Ans: (b)
No. of steps required for one full revolution of stepper motor shaft or lead screws n_S = 600
Pitch (p) = 5mm
Linear table speed V_table = 5m/min = 5000mm/min
RPM of lead Screw (N_S) = V_table/p = 1000 rpm
We have equation of frequency of pulse generator
f_p = N_s x n_S
f_p = 1000 × 600 = 600,000 pulses/min
f_p = 600000/60 pulses/sec
f_p = 10000 pulses/sec or Hz
f_p = 10 kHz

Q1: In a pure orthogonal turning by a zero rake angle single point carbide cutting tool, the shear force has been computed to be 400 N. The cutting velocity, V_c = 100m/min, depth of cut, t = 2.0 mm, feed, s₀ = 0.1mm/revolution and chip velocity, V_f = 20m/min, the shear strength, τ_s of the material will be _______MPa (round off to two decimal places). [GATE ME 2021 SET-2]
Ans: 388 to 400
α = 0º
F_s = 400 N
Cutting velocity (V) = 100 m/min (V_c)
d = 2.0 mm(t)
f = 0.1 mm/rev (S_o)
Chip velocity (V_c) = 20 m/min (V_f)
(V_c = chip velocity; V = cutting velocity)
= 20/100 = 0.2

ϕ = 11.31º

t = f sin 90º
b = d sin 90º
bt = fd

= 392.23 MPa

Q2: A surface grinding operation has been performed on a Cast Iron plate having dimensions 300 mm (length) x 10 mm (width) x 50 mm (height). The grinding was performed using an alumina wheel having a wheel diameter of 150 mm and wheel width of 12 mm. The grinding velocity used is 40 m/s, table speed is 5 m/min, depth of cut per pass is 50 μm and the number of grinding passes is 20. The average tangential and average normal force for each pass is found to be 40 N and 60 N respectively. The value of the specific grinding energy under the aforesaid grinding conditions is __________J/mm³ (round off to one decimal place). [GATE ME 2021 SET-2]
Ans: 38 to 39
Power = F_c·V
= 40 N × 40 m/s = 1600 W
MRR = 10 x 0.050 x (5000/60) mm³/s
= 41.667 mm³/s


Q3: The machining process that involves ablation is  [GATE ME 2021 SET-2]
(a) Abrasive Jet Machining
(b) Chemical Machining
(c) Electrochemical Machining
(d) Laser Beam Machining
Ans: (d) 
Laser beam machining (LBM) is a nonconventional machining process, which broadly refers to the process of material removal, accomplished through the interactions between the laser and target materials. The processes can include laser drilling, cutting, grooving, writing, scribing, ablation, welding, cladding, milling, and so on. LBM is a thermal process, and unlike conventional mechanical processes, LBM removes material without mechanical engagement. In general, the workpiece is heated to melting or boiling point and removed by melt ejection, vaporization, or ablation.

Q4: In a grinding operation of a metal, specific energy consumption is 15 J/mm³. If a grinding wheel with a diameter of 200 mm is rotating at 3000 rpm to obtain a material removal rate of 6000 mm³/min, then the tangential force on the wheel is _______N (round off to two decimal places). [GATE ME 2021 SET-1]
Ans: 45 to 50
E_sp = 15 J/mm³
MRR = 6000 mm³/min
N = 3000 rpm
D = 200 mm

Q5: An orthogonal cutting operation is performed using a single point cutting tool with a rake angle of 12º on a lathe. During turning, the cutting force and the friction force are 1000 N and 600 N, respectively. If the chip thickness and the uncut chip thickness during turning are 1.5 mm and 0.75 mm, respectively, then the shear force is _______N (round off to two decimal places).  [GATE ME 2021 SET-1]
Ans: 625 to 750
F_c = 1000 N, F = 600 N, t = 0.75 mm, t_c = 1.5 mm, α = 12º
This is orthogonal turning operation

⇒ ϕ = 28.63º
Method - 1: Using force relations
F = F_c sin α + F_t cos α
600 = 1000 sin 12º + f_t cos12º
or F_t = 400.85 N
∴ F_s = F_c cosϕ − F_t sinϕ
= 1000 cos 28.63º − 400.85 sin28.63º
= 685.66 N
Method - 2 : Using Merchant circle
F_c = Rcos(β − α)
F = R sinβ
or

0.6 cos β cos12º + 0.6 sin β sin12º = sinβ
0.586888 cos β = (1 − 0.6 sin 12º) sinβ = 0.875253 sin β

⇒ β = 33.84º
∴ F = R sin β
600 = R sin 33.84º
⇒ R = 1077.44 N
∴ F_s = R cos (ϕ + β - α)
= 1077.44 cos (28.63º + 33.84º - 12º)
= 685.77 N

Q6: The correct sequence of machining operations to be performed to finish a large diameter through hole is  [GATE ME 2021 SET-1]
(a) drilling, boring, reaming
(b) boring, drilling, reaming
(c) drilling, reaming, boring
(d) boring, reaming, drilling
Ans: (a) 
Drilling: to produce a hole, which then may be followed by boring it to improve its dimensional accuracy and surface finish.
Boring: to enlarge a hole or cylindrical cavity made by a previous process or to produce circular internal grooves.
Reaming: is an operation used to (a) make an existing hole dimensionally more accurate than can br achived by drilling alone and (b) improve its surface finish. The most accurate holes in workpieces generally are produced by the following sequence of operation.
Centering -> Drilling -> Boring -> Reaming.

Q7: In a machining operation, if a cutting tool traces the workpiece such that the directrix is perpendicular to the plane of the generatrix as shown in figure, the surface generated is  [GATE ME 2021 SET-1]
(a) plane
(b) cylindrical
(c) spherical
(d) a surface of revolutio
Ans: (b)

Q1: Bars of 250 mm length and 25 mm diameter are to be turned on a lathe with a feed of 0.2 mm/rev. Each regrinding of the tool costs Rs. 20. The time required for each tool change is 1 min. Tool life equation is given as VT^0.2 = 24 (where cutting speed V is in m/min and tool life T is in min). The optimum tool cost per piece for maximum production rate is Rs. ________ (round off to 2 decimal places).  [GATE ME 2020 SET-2]
Ans: 26 to 28
Optimum tool life


= 24/4^0.2 = 18.19 m/min
V = πDN
⇒ 18.19 = π x 0.025 x N
⇒ N = 231.6 rpm
Maching time per piece

Number of tool needed per piece work = 5.397/4 piece
∴ The optimum tool cost per piece = (5.397/4) x 20 = 26.985

Q2: A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0°−9°−7°−8°−15°−30°−0.05 inch (Coordinate system, ASA) resulting in a cutting force F_c1. If the tool geometry is changed to 0°−9°−7°−8°−15°−0°−0.05  inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to F_c2. Specific cutting energy (in J/mm³) is U_c = U₀(t₁) − 0.4, where U₀  is the specific energy coefficient, and t₁ is the uncut thickness in mm. The value of percentage change in cutting force Fc₂, i.e. is _______ (round off to one decimal place). [GATE ME 2020 SET-2]
Ans: -5.8 to -5.5
C_s1 = 30°
∴ λ₁ = 90 - 30 = 60°
C_s2 = 0°
∴ λ₂ =90 - 0 = 90°
We know that specific energy consumption

Q3: The process, that uses a tapered horn to amplify and focus the mechanical energy for machining of glass, is  [GATE ME 2020 SET-2]
(a) electrochemical machining
(b) electrical discharge machining
(c) ultrasonic machining
(d) abrasive jet machining
Ans: (c)
In Ultrasonic machining, the function of horn (also called concentrator, it is a tapered metal bar) is to amplify and focus vibration of the transducer to an adequate intensity for driving the tool to fulfll the cutting operation.

Q4: In a turning process using orthogonal tool geometry, a chip length of 100 mm is obtained for an uncut chip length of 250 mm.
The cutting conditions are cutting speed = 30 m/min. rake angle = 20º.
The shear plane angle is _________ degrees (round off to one decimal place).  [GATE ME 2020 SET-1]
Ans: 22 to 25

ϕ = tan^-1(0.4354) = 23.5º.

Q5: The following data applies to basic shaft system:
tolerance for hole = 0.002 mm,
tolerance for shaft = 0.001 mm,
allowance = 0.003 mm,
basic size = 50 mm.
The maximum hole size is ______ mm (round off to 3 decimal places).  [GATE ME 2020 SET-1]
Ans: 50.005 to 50.005
Using diagram method
UL of hole = BS + 0.003 + 0.002 mm = 50.005 mm

Q6: The base of a brass bracket needs rough grinding. For this purpose, the most suitable grinding wheel grade specification is  [GATE ME 2020 SET-1]
(a) C30Q12V
(b) A50G8V
(c) C90J4B
(d) A30D12V
Ans: (a)
Base of brass bracket need rough grinding, for this purpose the most suitable grinding wheel grade specifcation is C30Q12V.
Because 30 is a grain number.