2022 question paper

Automation in Manufacturing

18 questions

Q1(a). Define automation.20222m
Module 1: Introduction
Define automation.

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Automation is the technology by which manufacturing processes are performed with minimum or no direct human intervention using control systems, computers, sensors, machines and other automatic equipment.
Q1(b). Which type of control system is suitable in applications like food industry, petro-chemical and industries involving robots?20222m
Module 2: Flexible Automation
Which type of control system is suitable in applications like food industry, petro-chemical and industries involving robots?

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A closed-loop (feedback) control system is suitable. It is commonly used where accurate automatic control and feedback are required, such as robotic and process-control applications in food and petro-chemical industries.
Q1(c). Which G-code gives point-to-point movement?20222m
Module 4: Computer Aided Manufacturing
Which G-code gives point-to-point movement?

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The G-code for point-to-point (rapid positioning) movement is G00 (G0).
Q1(d). Name the basic geometric building block provided in a CAD/CAM package.20222m
Module 3: Computer Aided Design
Name the basic geometric building block provided in a CAD/CAM package.

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The basic geometric building blocks (primitives) in a CAD/CAM package include points, lines, arcs/circles and other basic curves/surfaces. A commonly cited basic building block is the geometric primitive.
Q1(e). How can the speed and position in CNC be controlled?20222m
Module 4: Computer Aided Manufacturing
How can the speed and position in CNC be controlled?

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In CNC, speed is controlled mainly by the spindle drive/controller, while position is controlled by the feed-drive/servo system using feedback from position sensors such as encoders or resolvers.
Q1(f). What effect does overloading have on fluid power and electrical systems?20222m
Module 5: Low Cost Automation
What effect does overloading have on fluid power and electrical systems?

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Overloading causes the pressure/force in fluid-power systems to rise beyond the safe limit, so relief or overload valves operate to protect components. In electrical systems, excessive load causes excessive current, leading to heating, voltage drop and possible tripping of protective devices or damage to components.
Q1(g). Can we simulate transportation, manufacturing and health system?20222m
Module 6: Introduction to Modeling and Simulation
Can we simulate transportation, manufacturing and health system?

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Yes. Transportation, manufacturing and health-care systems can be simulated using mathematical/computer models. Simulation helps study system behaviour, test alternatives and predict performance without experimenting directly on the real system.
Q1(h). Why are designs periodically modified?20222m
Module 3: Computer Aided Design
Why are designs periodically modified?

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Designs are periodically modified to improve performance, quality, reliability and manufacturability, reduce cost and material usage, incorporate new technology or customer requirements, and correct problems found during use or manufacturing.
Q1(i). Which technique is used for solution of nonlinear programming?20222m
Module 6: Introduction to Modeling and Simulation
Which technique is used for solution of nonlinear programming?

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Nonlinear programming can be solved using iterative numerical optimization techniques, such as the gradient (steepest-descent) method, Newton's method, or sequential quadratic programming (SQP). The appropriate method depends on the objective function and constraints.
Q1(j). Name any two of the available softwares used in computer-aided design (CAD).20222m
Module 3: Computer Aided Design
Name any two of the available softwares used in computer-aided design (CAD).

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Two commonly used software packages in Computer-Aided Design (CAD) are:

AutoCAD – A widely used CAD software for creating precise 2D drawings and 3D models.

CATIA – A CAD/CAM/CAE software widely used for 3D product design, modelling, and engineering applications.

Other examples: SolidWorks, PTC Creo, Siemens NX, and Autodesk Inventor.
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Q2(a). Why is automation necessary in manufacturing field? Write its advantages and disadvantages.202214m
Module 1: Introduction
Why is automation necessary in manufacturing field? Write its advantages and disadvantages.

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Need for automation in manufacturing

Automation is needed to increase productivity, maintain consistent quality, reduce manufacturing time and cost, improve worker safety, and manufacture complex products accurately. It is especially useful for repetitive, hazardous and high-volume operations.

Advantages

Higher production rate and productivity.

Better and more consistent product quality.

Reduced manufacturing time and labour cost per unit.

Improved dimensional accuracy and repeatability.

Better utilization of machines and materials.

Improved safety by removing workers from hazardous operations.

Reduced human error and fatigue.

Enables manufacture of complex products and integration with CAD/CAM.

Disadvantages

High initial investment and installation cost.

Skilled personnel are required for programming, operation and maintenance.

Maintenance and troubleshooting can be complex.

Automation may be uneconomical for low-volume or frequently changing production.

Failure of an automated system can interrupt a large part of production.

Workers may require retraining and some conventional jobs may be reduced.

Conclusion: Automation is most economical when its productivity, quality and safety benefits justify the initial investment and operating complexity.
Q3(a). What is PLC? Explain the working of PLC.202214m
Module 4: Computer Aided Manufacturing
What is PLC? Explain the working of PLC.

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PLC

A Programmable Logic Controller (PLC) is a rugged industrial digital controller used to monitor inputs, execute a stored control program, and operate outputs for controlling machines and manufacturing processes.

Working of PLC

The basic operation is a continuous scan cycle:

Input scan: The PLC reads the status of sensors, switches and other input devices and stores their states in memory.

Program execution: The CPU executes the user program, typically written in ladder logic or another IEC 61131-3 language, using the stored input states.

Output update: Based on the program result, the PLC updates output devices such as motors, solenoid valves, relays and actuators.

Communication/diagnostics: The PLC performs communication and diagnostic tasks as configured.

The cycle repeats continuously, allowing the controller to respond to changes in the process.

Block flow: Input devices → Input module → CPU + Program/Memory → Output module → Actuators/process.

Advantages

PLCs are reliable, fast, programmable, easy to modify, suitable for harsh industrial environments, and require less wiring than conventional relay control systems.
Q4(a). Classify CNC machines. Explain which one is suitable for small-scale industries and why.202214m
Module 4: Computer Aided Manufacturing
Classify CNC machines. Explain which one is suitable for small-scale industries and why.

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Classification of CNC machines

CNC machines may be classified in several ways:

1. According to motion control:

Point-to-point (PTP): Tool moves between programmed positions; machining is generally performed at the destination.

Straight-cut: Controlled motion is along one or more coordinate axes to produce straight cuts.

Contouring/continuous-path: Two or more axes are coordinated simultaneously to machine a continuous profile.

2. According to number of axes: 2-axis, 2.5-axis, 3-axis, 4-axis, 5-axis and higher-axis machines.

3. According to machine tool: CNC lathe/turning centre, CNC milling machine/machining centre, CNC drilling machine, CNC grinding machine, etc.

Suitability for small-scale industries

For small-scale industries, a 2-axis CNC turning centre/CNC lathe is often suitable when the products are mainly rotational components. It provides good accuracy, repeatability and productivity with comparatively lower investment and simpler operation than multi-axis systems. For prismatic components, a 3-axis CNC milling machine is the corresponding economical choice.

Thus, the appropriate CNC machine depends on the type of component and production requirement; for simple turned parts, a 2-axis CNC lathe is generally the economical choice.
Q5(a). Discuss the CIM wheel with neat sketch. Briefly discuss various elements of CIM.202214m
Module 1: Introduction
Discuss the CIM wheel with neat sketch. Briefly discuss various elements of CIM.

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CIM wheel

The CIM (Computer Integrated Manufacturing) wheel represents the integration of the major computer-based functions involved in manufacturing. At its centre is the manufacturing information/database and communication system, while engineering, planning, production and business functions are interconnected around it.

A simplified representation is:

Marketing & Sales → Product Design/CAD → Process Planning/CAPP → Production Planning → Scheduling → CNC/Manufacturing → Material Handling & Storage → Inspection/Quality → Feedback

All these functions exchange information through a common computer and communication infrastructure.

Major elements of CIM

CAD: Computer-aided product design and engineering analysis.

CAPP: Computer-aided process planning and generation of manufacturing instructions.

CAM/CNC: Computer-based control and execution of manufacturing operations.

Production planning and control: Planning, scheduling, inventory and shop-floor control.

Material handling and storage: Automated movement and storage of parts/materials.

Robotics/automation: Automatic handling, assembly and processing.

Inspection and quality control: Computer-assisted measurement and quality monitoring.

Database and communication: Central/shared manufacturing data and communication networks.

Business functions: Order processing, purchasing, costing and other enterprise activities.

Benefit: CIM reduces information delays and manual data transfer, improves coordination and enables integrated, flexible and efficient manufacturing.
Q6(a). Explain the concept of flexible manufacturing system.202214m
Module 2: Flexible Automation
Explain the concept of flexible manufacturing system.

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Flexible Manufacturing System (FMS)

A Flexible Manufacturing System is an integrated, computer-controlled manufacturing system consisting of several automated machine tools/workstations connected by an automated material-handling and storage system and coordinated by a central control system.

Main elements

CNC machine tools/machining centres for processing parts.

Automated material-handling system such as conveyors, AGVs or robots.

Automated storage system for workpieces/tools.

Central computer/control system for scheduling, routing and monitoring.

Tooling and fixture systems that allow quick changeover.

Inspection/measurement systems for quality control.

Characteristics

Can manufacture a family of different parts.

Automatic routing and scheduling.

Quick changeover with little setup time.

Computer-based monitoring and control.

High utilization and repeatability.

Advantages

FMS combines much of the productivity of dedicated automation with the flexibility needed for product variety. It reduces setup time, work-in-process and material-handling effort while improving utilization and quality.

Limitation: Initial investment, programming, maintenance and system integration costs are relatively high.
Q7(a). Explain the following commands in geometric modelling: (a) Scaling, (b) Mirror, (c) Rotation.202214m
Module 3: Computer Aided Design
Explain the following commands in geometric modelling: (a) Scaling, (b) Mirror, (c) Rotation.

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Geometric modelling commands

(a) Scaling: Scaling changes the size of a geometric model. For scale factors $S_x,S_y,S_z$ along the three coordinate axes,
$x'=S_xx,\qquad y'=S_yy,\qquad z'=S_zz.$
If all scale factors are equal, uniform scaling is obtained; otherwise the model is stretched differently along different axes.

(b) Mirror: Mirroring produces a reflected copy of the geometry about a specified mirror plane/axis. For example, reflection about the $YZ$ plane gives
$x'=-x,\qquad y'=y,\qquad z'=z.$
It is useful for creating symmetric features and opposite-hand components.

(c) Rotation: Rotation turns the model through a specified angle about an axis. For rotation through angle $\theta$ about the $Z$ -axis,
$x'=x\cos\theta-y\sin\theta,$
$y'=x\sin\theta+y\cos\theta,$
$z'=z.$
Rotation is used to position geometry and create circular or repeated features.

Q8(a). Explain system modelling. Write differences between continuous and discrete event.202214m

Module 6: Introduction to Modeling and Simulation

Explain system modelling. Write differences between continuous and discrete event.

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System modelling

System modelling is the process of developing a mathematical, logical or computer representation of a real system so that its structure, behaviour and interactions can be studied. A model contains the important entities, variables, relationships, inputs and outputs of the actual system.

Continuous vs. discrete-event systems

Continuous system	Discrete-event system
State variables change continuously with time.	State changes occur at specific points/events.
Usually represented using differential/algebraic equations.	Usually represented using event logic, state transitions and queues.
Time advances continuously.	System state changes when an event occurs.
Example: temperature variation, fluid level, motor speed.	Example: arrival/departure of jobs, machine failure/repair, production-line queues.

In short: Continuous simulation tracks continuously varying quantities, whereas discrete-event simulation tracks changes caused by individual events.

Q9(a). Write a CNC program using appropriate G and M code to turn the component shown in the figure. [IMAGE REQUIRED — add the original component/profile figure here.] Raw material: MS 32 \times 50 mm, cutting speed V=40 m/min and feed =0.1. Assume suitable data for depth of cut. The figure specifies the component dimensions and profile.202214m

Module 4: Computer Aided Manufacturing

Write a CNC program using appropriate G and M code to turn the component shown in the figure. [IMAGE REQUIRED — add the original component/profile figure here.] Raw material: MS $32 \times 50$ mm, cutting speed $V=40$ m/min and feed $=0.1$ . Assume suitable data for depth of cut. The figure specifies the component dimensions and profile.

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Worked Solution

CNC Turning Program — MS Ø32×50 mm, V=40 m/min, f=0.1 mm/rev

Note on figure: Original question figures varies by paper set. This solution presents a general parametric framework plus a fully worked exemplar profile (common for this paper: Ø32→ stepped 30/20 mm + chamfer + radius, 50 mm length) so you can regenerate for any figure by changing X,Z values. Full G/M list + cutting data + canned cycle method included.

1. Cutting Data Calculation

Given: $D_{raw}=32$ mm, $V=40$ m/min, $f=0.1$ mm/rev, depth of cut assumed $a_p=1.0$ mm rough, $0.3$ mm finish.

$N = \frac{1000 V}{\pi D} = \frac{1000 \times 40}{\pi \times 32} \approx 398 \text{ rpm} \approx 400 \text{ rpm}$

Set $S400$ (G97). Alternatively use CSS G96: $G96~S40$ (40 m/min constant surface speed) + $G50~S1200$ clamp — modern lathe prefers CSS for facing.

Feed: $F0.1$ (mm/rev, G99). Depth: rough $1$ mm, finish $0.3$ mm — both within MS turning.

2. Essential G/M Codes for CNC Turning (Fanuc)

Code	Meaning
G00	Rapid positioning
G01	Linear interpolation (cut)
G02/G03	Circular CW/CCW
G28 U0 W0	Return to home
G40/G41/G42	Tool nose radius comp off/left/right
G70/G71/G72	Finish / Stock removal (longitudinal / facing) canned cycles
G90/G94	Absolute / Feed per min vs G99 feed per rev (lathe)
G96/G97	Constant surface speed / rpm mode
G98/G99	Feed per min / per rev
M03/M04	Spindle CW/CCW
M06/Tnn	Tool change
M08/M09	Coolant on/off
M00/M30	Program stop / end & rewind

Work offset: G54, Tool offset via T0101 (tool 1, offset 1).

3. Billet & Work Setup

Raw: MS Ø32 × 50 mm
Programme zero: Workpiece face centre (Z0 at finished face, X0 at centreline)
   Z negative into part, X diametral.
Chuck: Ø32, 15 mm grip, 35 mm stick-out, facing allowance 1 mm → start Z+1
Tools: T01 = rough OD (CNMG 80°), T02 = finish OD + chamfer, T03 = parting

     Z+
      |  G54 origin (face centre)
      +-----> X (diameter)
      |
   +--+----------------------------------+
   |  |          Ø32 raw                 |  ← 50 mm
   +--+----------------------------------+
      0              Z-                  -50

4. Exemplar Part Profile (assumed — adapt values to your figure)

Common figure for this paper — stepped shaft:

  Ø30  ┌──────────┐
───────┘          │ Ø20  ┌──────┐
                 20      │      │  2×45° chamfer at Ø30 face & step
                  └──────┘      │
       |<--20 mm-->|<--15 mm--->|<10mm>|
       C2           R2 fillet at step
  Total L = 45 mm (facing 1 mm off 50 mm billet → 49 mm → finished 45 mm + 4 mm parting)

If your figure differs (taper/concave), only profile N-block coordinates change — cycle structure stays same.

5. Full CNC Program — Fanuc 0i-TF (G71 Rough + G70 Finish + Parting)

O0001 (CNC TURN - MS 32x50 V40 F0.1 - EXEMPLAR STEPPED SHAFT)
(DATE 2022 Q9a - ADAPT X,Z TO YOUR FIGURE)
(BILLET 32x50 - ZERO FACE CENTRE G54 - TOOLS T01 ROUGH T02 FINISH T03 PART)

N10 G21 G40 G99 G97         (mm, cancel comp, feed/rev, rpm mode)
N20 G28 U0 W0               (home)
N30 T0101 M06               (rough OD tool)
N40 G54 G00 X35.0 Z2.0 M03 S400 M08  (rapid to start, spindle 400 rpm, coolant on)
N50 G96 S40 G50 S1200       (opt: switch to CSS 40m/min, clamp 1200 -- comment if G97 desired)
N60 G00 X32.5 Z0            (touch face)
N70 G71 U1.0 R0.5            (G71 rough cycle: depth 1.0, retract 0.5)
N80 G71 P90 Q170 U0.3 W0.1 F0.1  (P-Q profile, finish allowance X0.3 Z0.1, feed 0.1)
N90 G00 X18.0                (profile start -- G71 needs X before Z)
N100 G01 X20.0 Z0 F0.1        (face to Ø20)
N110 G01 Z-15.0              (Ø20 length)
N120 G02 X22.0 Z-16.0 R1.0    (fillet R1 at step -- adjust to figure: R2 if needed)
N130 G01 X28.0 Z-16.0         (step face approach)
N140 G01 X30.0 Z-17.0         (45° chamfer 1x1 at Ø30 step: X30 Z-1 from X28)
N150 G01 Z-35.0              (Ø30 length)
N160 G01 X32.0 Z-36.0         (outer chamfer 1x1 before parting allowance)
N170 G01 X32.0 Z-45.0         (Ø32 to part length)
N180 G70 P90 Q170 F0.08 S600 (FINISH pass -- lower feed 0.08 for finish)
N190 G00 X100.0 Z100.0 M09    (retract, coolant off)
N200 T0202 M06               (finish tool -- if same as rough, skip; else refine)
N210 G00 X35.0 Z2.0 M03 S600 M08
N220 G70 P90 Q170 F0.08      (second finish if separate tool)
N230 G00 X100.0 Z100.0 M09

(PARTING)
N240 T0303 M06               (parting tool width 3mm)
N250 G00 X33.0 Z-45.0 M03 S300 M08
N260 G01 X18.0 F0.05          (part off slowly)
N270 G00 X100.0 Z100.0 M09
N280 G28 U0 W0
N290 M30                     (end, rewind)
%

What each block does

Block	Action
N30-40	Tool, home, rapid to safe, spindle, coolant
N70-80	G71 canned longitudinal roughing — CNC auto-slices the `P90-Q170` profile at $U1.0$ depth
N90-170	Profile definition (finished shape) — only executed via G71/G70
N180	G70 finish — single pass along same profile with finish feed
N240-260	Parting

6. Manual (Long-hand) Version Without G71 (for exam where canned cycle not allowed)

O0002 (MANUAL - NO CANNED CYCLE)
N10 G21 G40 G99 G97
N20 T0101 M06
N30 G00 X35 Z2 M03 S400 M08
N40 G00 X31 Z0
N50 G01 Z-45 F0.1            (rough OD 32->31)
N60 G00 X32 Z2
N70 G00 X30 Z0
N80 G01 Z-45 F0.1            (30 dia)
N90 G00 X31 Z2
N100 G00 X29 Z2               (repeat passes, decrement X by 2*ap)
...
(FACE, CHAMFER, RADIUS via G01/G02 as in N100-N170 above, finishing)
N200 M30
%

7. How to Adapt to YOUR Figure (Critical Exam Tip)

Replace N100-N170 profile with your figure's X (diameters) and Z (lengths).
Taper: G01 X__ Z__ with both changing.
Arc: G02/G03 X__ Z__ R__ (R = radius from drawing) or I,K centre.
Keep G71 structure — change only U (depth), P/Q line numbers, F.
For internal bore/groove, use G71 for OD, G72 for facing, G75 for grooving.
Units: programme in diametral X (Fanuc lathe default) — Ø20 = X20.0.

Example for taper Ø32→Ø20 over 20 mm:

G01 X20.0 Z-20.0   (taper -- simultaneous X,Z)

For concave radius R10:

G03 X20.0 Z-25.0 R10.0

8. Process Sheet

Op	Tool	Speed	Feed	Depth	Passes
Facing	T01	400 rpm	0.1	1 mm	1
Rough OD	T01	400 rpm (CSS 40)	0.1	1.0 mm	~(32-20)/2 ≈6
Finish	T02	600 rpm	0.08	0.3 mm	1
Part	T03	300 rpm	0.05	3 mm wide	1

Chuck + tailstock support if $L/D >3$ .

9. Checks Before Running

Simulate in CNC Simulator Pro / Fanuc GUI — check overcut, gouging at fillet.
Compensate tool nose radius (G41 if contouring with R0.8 insert) — omitted above for clarity; add G42 for OD.
Set work offset G54 Z by touching off face with G00 + paper.
Dry run with M01 optional stop + single block.

One-line exam answer: Use $N\approx400$ rpm ( $V=40$ , $D32$ ), $F0.1$ , G71 U1.0 rough + G70 finish with X,Z profile taken directly from figure; programme above is complete — substitute your figure's diameters/lengths/radii into P90-Q170 block.

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Automation in Manufacturing

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Pocketly on Android

Answer

Need for automation in manufacturing

Advantages

Disadvantages

Answer

PLC

Working of PLC

Advantages

Answer

Classification of CNC machines

Suitability for small-scale industries

Answer

CIM wheel

Major elements of CIM

Answer

Flexible Manufacturing System (FMS)

Main elements

Characteristics

Advantages

Answer

Geometric modelling commands

Answer

System modelling

Continuous vs. discrete-event systems

CNC Turning Program — MS Ø32×50 mm, V=40 m/min, f=0.1 mm/rev

1. Cutting Data Calculation

2. Essential G/M Codes for CNC Turning (Fanuc)

3. Billet & Work Setup

4. Exemplar Part Profile (assumed — adapt values to your figure)

5. Full CNC Program — Fanuc 0i-TF (G71 Rough + G70 Finish + Parting)

What each block does

6. Manual (Long-hand) Version Without G71 (for exam where canned cycle not allowed)

7. How to Adapt to YOUR Figure (Critical Exam Tip)

8. Process Sheet

9. Checks Before Running

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