Computer Integrated Manufacturing

 Question 1
Which one of the following CANNOT impart linear motion in a CNC machine?
 A Linear motor B Ball screw C Lead screw D Chain and sprocket
GATE ME 2022 SET-2   Manufacturing Engineering
Question 1 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.
 Question 2
Match the additive manufacturing technique in Column I with its corresponding input material in Column II.

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
 A P-3, Q-4, R-2 B P-1, Q-2, R-4 C P-2, Q-3, R-1 D P-4, Q-1, R-4
GATE ME 2022 SET-2   Manufacturing Engineering
Question 2 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.
 Question 3
In a CNC machine tool, the function of an interpolator is to generate
 A signal for the lubrication pump during machining B error signal for tool radius compensation during machining C NC code from the part drawing during post processing D reference signal prescribing the shape of the part to be machined
GATE ME 2021 SET-2   Manufacturing Engineering
Question 3 Explanation:
In contouring systems the machining path is usually constructed from a combination of linear and circular segments. It is only necessary to specify the coordinates of the initial and final points of each segment, and the feed rate. The operation of producing the required shape based on this information is termed interpolation and the corresponding unit is the "interpolator". The interpolator coordinates the motion along the machine axes, which are separately driven, by providing reference positions instant by instant for the position-and velocity control loops, to generate the required machining path. Typical interpolators are capable of generating linear and circular paths.
 Question 4
The XY table of a NC machine tool is to move from P(1,1) to Q(51,1); all coordinates are in mm. The pitch of the NC drive leadscrew is 1 mm. If the backlash between the leadscrew and the nut is $1.8^{\circ}$, then the total backlash of the table on moving from P to Q is _______mm (round off to two decimal places).
 A 0.15 B 0.25 C 0.55 D 0.85
GATE ME 2021 SET-1   Manufacturing Engineering
Question 4 Explanation:

P(1, 1) to P(51, 1) is 50 mm distance. As pitch of the leadscrew is 1 mm. It has to move 50 rotation and in each rotation, backlash is 1.8 degree. Therefore total backlash
= 1.8 x 50 degree.
In one rotation i.e. $360^{\circ}$ rotation - 1 mm movement take place
In $1.8 \times 50^{\circ}=\frac{1.8 \times 50}{360} \mathrm{~mm}=0.25 \mathrm{~mm}$
 Question 5
In modern CNC machine tools, the backlash has been eliminated by
 A preloaded ballscrews B rack and pinion C ratchet and pinion D slider crank mechanism
GATE ME 2021 SET-1   Manufacturing Engineering
Question 5 Explanation:

 Question 6
A point 'P' on a CNC controlled XY-stage is moved to another point 'Q' using the coordinate system shown in the figure below and rapid positioning command (G00).

A pair of stepping motors with maximum speed of 800 rpm, controlling both the X and Y motion of the stage, are directly coupled to a pair of lead screw, each with a uniform pitch of 0.5 mm. The time needed to position the point 'P' to the point 'Q' is _______ minutes. (round off to 2 decimal places).
 A 0.75 B 1.5 C 2 D 1.2
GATE ME 2020 SET-2   Manufacturing Engineering
Question 6 Explanation:
\begin{aligned} N&=800 \mathrm{rpm}, P=0.5 \mathrm{mm} / \mathrm{rev} \\ V &=N \times P\\ &=\mathrm{rev} / \mathrm{min} \times \mathrm{mm} / \mathrm{rev}\\&=400 \mathrm{mm} / \mathrm{min} \\ \Delta t_{x} &=\frac{600}{400}=1.5 \mathrm{min} \\ \Delta t_{y} &=\frac{300}{400}=0.75 \mathrm{min} \end{aligned}
There are two stepper motor so both will work till 0.75 min then y axis motor will stop then only x axis motor will run for 0.75 more, so total time will be 1.5 min.
 Question 7
Interpolator in a CNC machine
 A controls spindle speed B coordinates axes movements C operates tool changer D commands canned cycle
GATE ME 2018 SET-1   Manufacturing Engineering
Question 7 Explanation:
As interpolator provides two functions:
1. It computes individual axis velocities to drive the tool along the programmed path at given feed rate.
2. It generates intermediate coordinate positions along the programmed path.
 Question 8
Circular arc on a part profile is being machined on a vertical CNC milling machine. CNC part program using metric units with absolute dimensions is listed below:
------------------------------
N60 G01 X 30 Y 55 Z 5 F
50 N70 G02 X 50 Y 35 R 20
N80 G01 Z 5
--------------------------------
The coordinates of the centre of the circular arc are :
 A (30, 55) B (50, 55) C (50, 35) D (30, 35)
GATE ME 2017 SET-1   Manufacturing Engineering
Question 8 Explanation:

 Question 9
A point P(1,3,-5) is translated by $2\hat{i}+3\hat{j}-4\hat{k}$ and then rotated counter clockwise by $90^{\circ}$ about the z-axis. The new position of the point is
 A (-6, 3, -9) B (-6, -3, -9) C (6, 3, -9) D (6, 3, 9)
GATE ME 2016 SET-3   Manufacturing Engineering
Question 9 Explanation:
Initial point is (1, 3, -5)
After tranlated by 2i+3j-4k, it becomes
=(1+2, 3+3, -5-4)= (3, 6, -9)

Rotation by 90 CCW about z axis (NOTE: : when rotated about z-axis the z component will remain same.)

Hence the point (3, 6, -9) changes to (-6, 3, -9).
 Question 10
Match the following part programming codes with their respective functions

 A P- II, Q- I, R- IV, S- III B P- IV, Q- II, R- III, S- I C P- IV, Q- III, R- II, S- I D P- III, Q- IV, R- II, S- I
GATE ME 2016 SET-3   Manufacturing Engineering
Question 10 Explanation:
G01 is used for linear interpolation, G03 is used for circular interpolation counter clockwise, M03 for spindle rotation clockwise and M05 for spindle top.
There are 10 questions to complete.