Chapter 5

Metrology and Inspection

Computer Integrated
Manufacturing
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Limits, Fits and Tolerances without causing any functionable trouble, when assembled
with its mating part and put into actual service.
An article manufactured consists of assembly of a number
of components. Thus a component manufactured should
Tolerance
be matching with some other mating component. This is
Upper limit
important for the proper functioning and prolonged life of
the product. Lower limit
In the production of a component, it is not possible to
make any part precisely to a given dimension, due to vari-
ability of elements of production processes. If attempts
are made to achieve the perfect size, the cost of produc-
tion will increase tremendously. But for practical purposes,
perfect fitting of the mating components are not necessary.
Slight dimensional variations are acceptable for the proper Systems of Writing Tolerances
functioning.
The allowable variation in the basic size required in pro- 1. Unilateral system
duction is called tolerance. 2. Bilateral system
Larger and smaller dimensions allowable are called lim-
its– the high limit and low limit. In unilateral system dimension of a part is allowed to vary
Thus difference between high and low limits is the toler- only on one side of the basic size i.e., tolerance zone is
ance. It is the margin allowed for variation in workmanship. either above or below the basic size line.
Tolerance can also be defined as the amount by which
the job is allowed to go away from accuracy and perfectness
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.859

Tolerance of metal 24.98 mm will be minimum or least metal limit

(LML) as shaft at this limit will have the least possible
amount of metal

Hole LML
LML
Basic size Tolerance MML
MML
Shaft

Unilateral tolerance

Terminology for Limits and Fits

With a 25 mm of basic size examples of unilateral toler- Shaft: It is the external dimension of a component (need
ance are not be a circular shaft)
+ 0.02 + 0.01 Hole: It is the internal dimension of a component (need not
25+ 0.01 , 25− 0.00 , be a circular hole)
etc
+ 0.00 − 0.02
25− 0.02 , 25− 0.03 Basic or Nominal Size
It is the standard size of a part with reference to which limits
The first case means upper limit is 25.02 mm and lower of dimensions are determined. Basic size is same for hole
limit is 25.01 mm and shaft. In design calculations basic size is used.
\ Tolerance = 25.02 – 25.01 Actual sizes is the measured size of a manufactured part.
= 0.01 mm Zero line: It is the horizontal line representing the basic
In the fourth case, size.
Upper limit = 24.98 Deviation is the algebraic difference between the size
Lower limit = 24.97 (maximum, minimum, actual etc) with the basic size.
Tolerance = 24.98 – 24.97 Deviations are shown with respect to the zero line or datum
= 0.01 mm line.
In bilateral system, the dimension of the part is allowed to Upper deviation is a positive quantity when maximum
vary on both sides of the basic size. limit of size is greater than the basic size. It is negative when
maximum limit of size is less than the basic size. For hole it
Tolerance
denoted by ES and for shaft it is es
25 ±0.02
Lower deviation is a positive quantity when the lower
+0.01
−0.02 limit of size is greater than the basic size and negative when
25
it is less than the basic size. It is denoted by EI and ei for
hole and shaft respectively.
Basic size Tolerance is represented by IT which stands for interna-
tional tolerance grade.

Relationship of  Tolerance and Deviations

+ 0.01 For shaft, IT = es – ei
25± 0.02 25− 0.02 etc are examples of bilateral tolerance. (upper deviation – lower deviation)
In the first case, For hole, IT = ES – EI
Upper limit = 25.02 mm
Fundamental Deviation
Lower limit = 24.98 mm
Tolerance = 25.02 – 24.98 It is the deviation (either upper or lower) nearest to the basic
= 0.04 mm line. It fixes the position of the tolerance zone in relation to
the zero line.
Maximum or Minimum Metal Limits Basic shaft is the shaft whose upper deviation is zero.
Basic size and maximum size is same in this case. It is
If the tolerance for the shaft is given as 25± 0.02 the shaft is
denoted by ‘h’.
said to have maximum metal limit (MML) of 25.02 mm.
Basic hole is the hole whose lower deviation is zero. Its
Since at this limit, the shaft has maximum possible amount
low limit is same as the basic size. It is denoted by letter ‘h’.
3.860 | Part III • Unit 7 • Manufacturing Technology

Tolerance In Interference fit, the minimum permissible diameter

zones of the shaft is larger than the maximum allowable diam-
eter of the hole. After fitting the members are permanently
Zero line Zero line attached.

Transition Fit
Transition fit is in between clearance fit and interference fit.
Depending upon the actual sizes of the parts clearance or
interference may occur.
Basic hole Basic shaft
Zero line Hole tolerance

Tolerance grade is an indication of degree of accuracy and

is represented by letters IT followed by a number. For exam-
ple, IT0, IT01, IT1 etc
Shaft
tolerance
Standard Tolerance Unit
(a) (b) (c)
It serves as a basis for determining standard tolerance (IT). Clearance fit Transition fit Interference fit
It is denoted by ‘i’ and expressed in microns.
Hole Basis System for Three Types of Fits
Fits
Figure above gives hole basis system where the fundamen-
When two parts are to be assembled, the relation resulting tal deviation of hole is zero (For shaft basis system funda-
from the difference between their sizes before assembly is mental deviation of shaft is zero)
called a fit. It is the degree of tightness or looseness between Hole basis system is generally used because it is easier to
two mating parts. vary the size of the shaft and to measure it than to vary the
hole size, provided the work is not very large.
Types of Fits
Shaft basis system is useful when a number of acces-
Depending on the basis of positive, zero or negative clear- sories such as bearings, collars etc are to be fitted on same
ance values between mating parts fits are classified as under shaft.

Fits Allowance
Allowance is the intentional difference between the lower
Clearance Transition Interference limit of hole and higher limit of the shaft.
fit fit fit It is the variation given for the purpose of providing dif-
ferent classes of fit. It is the difference between maximum
material size limits of mating parts.
(a) Slide fit (a) Push fit (a) Force fit The allowance may be positive or negative. Positive
(b) easy slide fit (b) Wringing fit (b) Tight fit allowance is the minimum clearance and negative allow-
(c) Running fit (c) Shrink fit ance is the maximum interference intended between the
mating parts.
Clearance Fit
Clearance is the difference between sizes of hole and shaft. Allowance vs Tolerance
Minimum clearance is the difference between minimum Allowance is the prescribed difference between the dimen-
size of hole and maximum size of shaft. sions of two mating parts (hole and shaft). Tolerance is the
Maximum clearance is the difference between maximum permissible variation in dimension of a part (either hole or
size of hole and minimum size of the shaft. a shaft)
In clearance fits, the largest permissible shaft diameter
is smaller than the diameter of the smallest hole. So the Standard Limit Systems
shaft can rotate or slide in the hole with different degrees The aim of a standard limit system is
of freedom.
1. To select basic functional clearances and interferences
Interference Fit for a given application or type of fit
Interference is the arithmetical difference between sizes of 2. to establish tolerances which will provide a reasonable
hole and shaft, when shaft is of bigger size. and economical balance fits, consistency and cost
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.861

British standard, international standard (ISO) and Indian Designation of Holes, Shafts and Fits
standards are some of the standard limit systems A shaft or a hole is completely described if the basic size
followed by the appropriate letter and the tolerance grade
Indian Standard System of Limits and Fits is given.
(IS-919 and 2709) For example,
Indian standards are in line with ISO recommendations. The 50 H6 means 50 mm H hole with tolerance grade IT6
standards cover holes and shafts from the smallest size to 50 f7 means 50 mm f shaft with tolerance grade IT7
3150 mm. For any size over this range there is a wide choice
of fits available and for each of the fits there is a series of
Tolerance and Fundamental Deviation
tolerance grades from very fine to wide tolerances.
Standard tolerance and fundamental deviations are used for Larger Sizes
in the limit system. The finer tolerance grades IT01 to IT5 are not provided for
18 grades of fundamental tolerances are used. These are sizes above 500 mm.
designated as IT01, IT0, IT1 to IT16 For size above 500 mm and up to 3150 mm
Fundamental deviations are indicated by 25 letters A to IS : 2101 specify various grades
ZC for holes and letters a to zC for shafts. (For holes: A, B, C, I (in m) = 0.004D + 2.1
D, E, F, G, H, JS, J, K, M, N, P, R, S, T, U, V, X, Y, Z, ZA, ZB,
ZC. For shafts corresponding small letters are used) Tolerance Analysis in Manufacturing
For A to H holes lower deviation is above zero line and and Assembly
for J to ZC it is below the zero line. Tolerance can be defined as the amount of variation permis-
For shafts a to h upper deviation is below the zero line sible from accuracy and perfectness of the dimension of a
and for j to zc it is above the zero line. component without causing any functional trouble. Human
Standard tolerances are expressed in terms of standard failure and machine limitations prevent the achievement of
tolerance unit, i. It is given by ideal dimensions of the part during fabrication.
By providing tolerance the cost of production can be
i = 0.45 3 D + 0.001D microns reduced; without sacrificing the functional requirement.
The difference between upper limit and lower limit of a
Where D is the geometric mean upper and lower values of
dimension is called the tolerance zone. It is the margin for
a diameter step in which the diameter lies. IS–919 specifies
variation in workmanship. Selection of tolerance is based
the following diameter steps.
on the following.
1–3, 3–6, 6–10, 10–14, 14–18, 18–24, 24–30, 30–40, 40–50,
etc upto 180–200 mm. 1. Functional requirement : With the permitted tolerance,
Values of tolerances for tolerance grades IT5 to IT16 are assembly should be possible and the equipment
obtained from the following table. should be able to perform the required function.
2. Standardisation: Standardisation of the parts are
required for interchangeability which is essential for
Grade IT5 IT6 IT7 IT8 IT9 mass production.
3. Manufacturing needs
Value 7i 10i 16i 25i 40i When the functional requirement is not so rigid tolerance
choice may be influenced and determined by factors like
methods of tooling, equipment available etc.

Need for Tolerance

IT 10

IT 11

IT 12

IT 13

IT 14

IT 15

IT 16

1. Different materials have different properties. Variation

in properties cause errors during machining.
1000i

2. Production machines may have inherent inaccuracies

100i

160i

250i

400i

640i
64i

and have limitation to produce parts with perfect

dimensions.
3. Machine operators have limitation to make perfect
Other tolerances– settings. There are chances of errors in setting up
For IT01 : 0.3 + 0.08 D machines, adjusting tools and work piece on the
For IT0 : 0.5 + 0.12 D machine etc.
For IT1 : 0.8 + 0.02 D
Aiming at ideal conditions of dimensions will result in
For IT2 to IT4 values are approximated between IT1 and
exhorbitant costs. If the components are made with large
IT5
3.862 | Part III • Unit 7 • Manufacturing Technology

tolerances without affecting the functional requirement, different craftsmen or operators in different batches in dif-
cost of production can be reduced. So the tolerance selected ferent machines. In this case the parts are produced within
should be just enough to the required job and not better. specified tolerance limits so that a component selected at
So it can be said that tolerance is a compromise between random will assemble correctly with any other mating com-
accuracy required for proper functioning and ability for ponent, selected at random. By this method manufacturing
economic production of this accuracy. cost is considerably reduced.

Solved Examples Selective Assembly

Example 1: What is meant by 20 H 7 f8 in Indian standard When high degree of accuracy is required in mass produc-
limit system? tion, interchangeable assembly also become not economi-
Solution: It means a 20 mm basic size H-hole with toler- cal. In this case selective assembly method can be used. In
ance grade IT7 is fitted with a f-shaft of tolerance grade IT8. selective assembly parts are produced with wider tolerances
so that the cost is reduced. The components produced are
0.050
classified into groups according to their sizes by automatic
Example 2: A hole is specified as 400.000 mm. The mating
gauging. Assemblies can be made only with matched sets of
shaft has a clearance fit with minimum clearance 0.02 mm.
parts such that the clearances are within limits
The tolerance on the shaft is 0.04 mm. Maximum clearance
An example for selective assembly is the assembly of
between hole and shaft is
pistons with cylinder bores.
Solution: Low limit of hole: 40.000 mm If the basic size is 50 mm for bore and 49.88 mm for
  High limit of hole: 40.050 mm piston and required clearance is 0.12 mm, these can be pro-
duced with a wide tolerance of 0.04 mm.
40.050
40.000 Hole Then dimension of bore = 50±0.02
Dimension of piston = 49.88±0.02
In interchangeable assembly the clearance can go upto
39.980 50.02 – 49.86 = 1.16 mm which is not acceptable.
Shaft But the components are grouped in match sets as follows
39.940
the required clearance of 0.12 mm can be achieved.
 
High limit of shaft  = 40.000 – 0.02
     = 39.980 mm
Cylinder bore 49.98 50.00 50.02
 
Low limit of shaft    = 39.980 – 0.04
     = 39.940 mm Piston 49.86 49.88 49.90
Maximum clearance = 40.050 – 39.940
 
     = 0.110 mm
Gauges
Types of Assemblies In mass production where a large number of similar compo-
In assembly of components the mating parts can be made to nents are produced, inspection by measuring each compo-
fit in three ways. These are nent will be time consuming and not economical. Therefore
1. By trial and error the conformance of each part is checked with the tolerance
2. Interchangeable assembly specification using tools called gauges.
3. Selective assembly Gauges are inspection tools of rigid design without a
scale, which serve to check the dimensions of manufactured
Assembly by Trial and Error parts. They check only whether the inspected parts are made
When the assemblies required are only small in number and within the specified limits. Plain gauges are used for check-
they are similar, these can be produced by trial and error ing plain (unthreaded) holes and shafts.
by same craftsman. In this method one part is made to its Classification––
nominal size as accurately as possible and the mating part is
made to suit it by trial and error. 1. According to type
(a) standard gauges
Interchangeable Assembly (b) limit gauges
2. According to purpose
In mass production the quantities required are very large and
(a) Workshop
it is not economical to produce it by trial and error by same
(b) Inspection
person. In mass production the components are produced by
(c) reference or master gauges
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.863

3. According to form of tested surface Gauge Maker’s Tolerance or Gauge

(a) plug gauge Tolerance
(b) snap and ring gauges
Gauges cannot be manufactured to the exact sizes. If closer
4. According to their design
limits are held, the gauge become expensive. Therefore
(a) Single limit and double limit
some allowance is given in the manufacturing of gauges
(b) Single ended and double ended
This is known as gauge tolerance.
(c) Fixed and adjustable gauges
Limit gauges are usually provided with a tolerance of
Standard gauge is an exact of copy of their mating part. 10% of the work tolerance. For inspection gauges it is 5%
For example a shaft and a bush. The shaft can be inspected of work tolerance.
with a gauge which is an exact copy of the bush.
Limit gauges are used to check whether the parts are within Gauge tolerance
the specified limits. HL NO HL GO
Limit plug gauges are used for checking holes. The GO plug GO
gauge has the size of low limit of the hole while NO GO plug
Work
gauge has the size of high limit of the hole. If the inspected tolerance
hole is within their tolerance limits the GO gauge will enter the
hole while the NO GO gauge cannot enter the hole. NO
GO LL
Plug gauges can be plain or screwed. GO
LL
Plain plug gauges are generally double ended type for Plug gauge Snap gauge
sizes up to 63 mm and single ended for sizes above 63 mm
size. Wear Allowance
Ring gauges are used to check shafts. The measuring surfaces of GO gauges which constantly
Plain ring gauges are made of suitable wear resisting rub against the surface of parts in inspection are conse-
steel and gauging surfaces are hardened. GO and NO GO quently subjected to wear and lose their initial size, while
type ring gauges are available. the size of snap GO gauges are increased. Therefore a wear
Snap and gap gauges are also used to check shafts. The allowance is added in a direction opposite to the wear. Wear
jaws of the gauge is made use of for checking. Double ended allowance is taken as 10% of the gauge tolerance. Wear
type plate snap gauges are used for sizes in the range of 2 to allowance is applied to a normal GO gauge diameter before
100 mm and single ended progressive type in the range of gauge tolerance is applied. As per British standards when
100 to 250 mm. In adjustable type gap gauges, the gauging the work tolerance is more than 0.09 mm, wear allowance
anvils are adjustable end wise in the horse shoe frame. is provided.
These gauges are set by means of slip gauges to any par- Inspection gauges are used by inspectors for final
ticular limit required. inspection of the manufactured parts. These have slightly
larger tolerance than work shop gauges. This is to ensure
Combined limit gauges are used for gauging of cylindrical
that work which passes the work shop gauges will be
holes. In these a single gauge is used to check both upper
accepted by inspection gauges. The tolerance on inspection
and lower limits.
gauges fall outside the work tolerance.
Contour gauges are employed for checking the dimen-
sional accuracy and shapes of irregular work. Radius gauge
is an example. Taylor’s Principle for The Design
Taper plug gauge is used for checking tapered holes and
of Limit Gauge
taper ring gauge is used for checking shafts. These check 1. GO gauges should inspect all the features of a
the diameter at the bigger end and the change in diameter component at a time and should be able to control
per unit length. the maximum metal limit, or in other words
Slip gauges or gauge blocks are used as standards of the maximum metal limit of as many related
measurement. It is the universally accepted end standard dimensions as possible should be incorporated in
of length. These are rectangular blocks of high grade steel the GO gauge.
with exceptionally close tolerances. Faces are round and 2. NOT GO gauge should check only one element at a
lapped. These are available in standard sets. The blocks are time for the minimum metal limit.
placed one over other by the phenomenon of wringing to get Example 3: What are the upper and lower limits of the
a specified length. Slip gauges are also made from Tungsten shaft represented by 60f 8. Diameter 60 mm lies in diameter
carbide which is extremely hard and wear resistant. They step of 50 –80 mm. Fundamental deviation of f shaft is –5.5
are carefully finished by high grade lapping to a high degree D0.41 mm.
of finish flatness and accuracy.
3.864 | Part III • Unit 7 • Manufacturing Technology

Example 6: For the following hole and shaft assembly find

Solution: Geometric mean diameter = 50 × 80
shaft tolerance, hole tolerance and state the type of fit.
= 63.246 mm
+0.05 +0.05
Fundamental tolerance unit
Hole: 60 +0.002 shaft: 60 +0.005
i = 0.45 3 D + 0.001D Solution: HL of hole: 60.025 mm
3
= 0.45 63.246 + 0.001 × 63.246 LL of hole: 60.00
 
Hole tolerance : 60.025 – 60.000 = 0.025 mm
= 1.859 mm
HL of shaft : 60.05
= 0.00186 mm
LL of shaft : 60.005
IT8 = 25i
Shaft tolerance = 60.05 – 60.005 = 0.045 mm
   = 25 × 0.00185
Allowance = LL of hole – HL of shaft
  = 0.04646 mm
= 60.00 – 60.05
Fundamental deviation of f shaft
= –0.05 mm (interference)
= –5.5 D0.41
HL of hole – LL of shaft = 60.025 – 60.005
= –5. 5 [63.246]0.41 mm
= 0.025 mm (clearance)
= –0.03012 mm
\ The type of fit is transition fit
Upper limit of shaft = 60 – 0.03012
= 59.97 mm Example 7: In a limit system, the following limits are
Lower limit of shaft = 59.97 – 0.0465 specified to give a clearance fit between shaft and hole
= 59.924 mm. −0.006

Example 4: Find the values of allowance and tolerances Shaft: 40 −0.020 mm

for hole and shaft assembly for the following mating parts +0.30
−0.03 +0.25 Hole: 40 −0.000 mm
Hole: 25−0.05 shaft: 60 +0.00
Determine–
Solution:
(A) basic size
(a) Hole
(B) shaft and hole tolerances
Tolerance = 25.04 – 25.00
(C) Minimum clearance
= 0.04 mm
(D) Maximum clearance
(b) Shaft
Tolerance = HL – LL Solution:
HL = 25.00 – 0.03 (a) Basic size : 40 mm
= 24.97 (b) Shaft tolerance:
LL = 25.00 – 0.05 (40 – 0.006) – (40 – 0.02)
= 24.95 = (–0.006) – (–0.020) = 0.014 mm
Tolerance = 24.97 – 24.95 Hole tolerance: (+0.030) – (0.00)
= 0.02 mm = 0.030 mm
(c) Allowance = LL of hole – HL of shaft (c) Minimum clearance = LL of hole – HL of shaft
= 25.00 – 24.97 = 50 – (50 – 0.006)
= 0.03 mm. = 0.006 mm
(d) Maximum clearance
Example 5: A 60 mm diameter shaft is made to rotate in
HL of hole – LL of shaft
the bush. Tolerance for both bush and shaft are 0.050 mm.
= (50 + 0.030) – (50 – 0.020)
Determine dimension shaft and bush to give a minimum
= 0.050 mm
clearance 0.075 mm with hole basis system
In an assembly of two parts with 50 mm nominal diameter,
Solution: the lower deviation of the hole is zero and the higher is 5
microns. For the shaft lower and higher deviations are –8
0.05 Zero line and –4 microns respectively. Determine the allowance and
0.75
type of fit of the assembly.
0.05 Hole

0.005
Zero line
0.004
0.008
HL of hole = LL + tolerance = 60 + 0.050 = 60.050 mm
50.000
HL of shaft = LL of hole – Allowance
= 60.00 – 0.075 = 59.925 Shaft
LL of shaft = 59.925 – 0.050 = 59.875
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.865

HL of hole = 50 + 0.005 IT7 = 16 × i = 20.918 microns

= 50.005 = 0.021 mm
LL of hole = 50.000 IT8 = 25 × i = 32.69 microns
HL of shaft = 50.000 – 0.004 = 49.996 mm = 0.033 mm
LL of shaft: 50.000 – 0.008 = 49.992 mm Upper deviation of shaft
Allowance = LL of hole – HL of shaft = –5.5 × (23.24)0.41
= 50.000 – 49.996 = –20 microns
= 0.004 mm = –0.020 mm
(Clearance fit) Fundamental or lower deviation of hole = 0
+0.021
Example 8: Between two mating parts of 90 mm basic size
\ Limits for 20 H7 = 20 +0.000
the actual interference fit is to be from 0.05 mm to 0.12 mm.
Limits for shaft f8
For shaft and hole tolerance is same. Determine the shaft
HL for shaft = 20 – 0.020
size and hole size in hole basis unilateral system.
LL for shaft = 20 – 0.02 – 0.033 = 20 – 0.053
Shaft HL of hole = 20.021 NO GO

0.12 mm 0.05 mm

Zero line
Hole
90 mm Gauge tolerance
Wear allowance
GO
Solution:
LL of hole = 90.00 mm LL of hole = 20.00
HL of shaft = 90.000 + 0.12
= 90.12 mm −0.020

Tolerance of shaft = Tolerance of hole \ Shaft limits are 20 −0.053

=
(0.12 − 0.05) Tolerance for plug gauge for gauging the hole = 10% of
work tolerance = 0.021 × 0.1 = 0.0021 mm
2
Wear allowance = 10% of gauge tolerance
= 0.035 mm = 0.1 × 0.0021
\ HL of hole = 90 + 0.035 = 90.035 mm = 0.0002 mm
LL of shaft = 90.035 + 0.05 = 90.085 mm Upper limit of GO plug gauge = 20 + 0.0002 + 0.0021
+0.12 = 20.0023 mm
\ Size of shaft = 90 +0.085
Low limit of GO gauge = 20 + 0.0002 = 20.0002
+0.035 +0.0023
Size of hole = = 90 +0.000 Limits of GO gauge = 20 +0.0002
Example 9: Design the general GO and NO GO gauge for
hole of the assembly 20 H7 f 8 fit. The following data can Linear and Angular Measurements
be used
For linear measurements various standards followed are
(a) i = 0.001 D + 0.45 3 D microns 1. Line standard
(b) Upper deviation of f shaft 2. End standard
= −5.5 D
0.41
3. Wavelength standard
(c) Diameter step for 20 mm When length is measured between two engraved lines it is
= 18 –30 called line standard.
(d) IT7 = 16i When length is expressed as the distance between two
(e) IT8 = 25i flat parallel faces, it is called end standard. In wavelength
(f) Wear allowance = 10% of gauge tolerance standard, wavelength of mono chromatic light is used as
Solution: unit of length.
Imperial standard yard (England) and international
D = 18 × 30 = 23.24 mm standard meter (France) are line standards.
i = 0.001 × 23.24 + 0.45 3 23.24 Metre in wavelength standard is defined as 1650763.73
= 1.3074 microns wavelengths of orange radiation in vacuum of krypton 86
isotope.
3.866 | Part III • Unit 7 • Manufacturing Technology

Any system of measurement must be related to a known Straight Edge

standard for maintaining uniformity of measurement Straight edges are used for checking straightness and flat-
throughout the world. ness of work pieces with the help of spirit levels. Types of
Standards can be classified as international standard, straight edges are
national standard, national reference standard working
standard, Laboratory reference standard, Laboratory work- 1. Tool maker’s straight edge
ing standard and shop floor standard. 2. Wide edged straight
3. Angle straight edge
4. Box straight edge
Linear Measurements
Linear measurement means measurement of lengths diam- Spirit Level
eters, heights, thicknesses etc. It can be external or internal.
Spirit levels are used for measuring small angles or inclina-
For linear measurements the instruments used are, non-
tions, to determine position of a surface relative to horizon-
precision instruments, precision instruments, comparators,
tal position and for establishing horizontal datum.
measuring machines etc.
In spirit level a sealed glass tube is mounted on a base. A
Non-precision instruments are steel rule, calipers, divid-
scale is engraved on top of the glass tube. The tube contains
ers, telescopic gauges, depth gauges etc
ether or alcohol with a bubble. For measuring flatness or
Precision instruments are micrometers, vernier calipers,
angle the position of the bubble is made use of
height gauges, slip gauges etc.
Linear measuring instruments can also be classified as
Frame Level
1. Direct measuring instruments
Frame level is used for checking vertical surfaces. In frame
2. Indirect measuring instruments
levels base and side edge is exactly at 90°. A glass tube with
Direct measuring instruments can be graduated or non ether and bubble is fixed on the base. Side edge of the level
graduated. is kept in contact with the vertical edge. Position of the bub-
Rules, vernier calipers, micrometers, dial indications etc ble in the glass tube is noted for checking the verticality.
are examples of graduated instruments.
Callipers, trammels, straight edges, wire gauges, screw Vernier Instruments
pitch gauges, radius gauges, thickness gauges, slip gauges
Vernier instruments consists of a main scale and vernier
etc are examples of non-graduated instruments.
scale. The vernier scale can be moved on the main scale.
Callipers The difference between value of vernier scale division and
a main scale division forms the basis for precision meas-
Callipers are used when direct linear measurement is not urements. Vernier callipers and vernier height gauges are
possible. Callipers consist of two legs, hinged at top. The examples of vernier instruments. Least count of a vernier
leg ends are used for measurement. For example measure- instrument is the difference in value of a vernier scale divi-
ment of diameter of a rod. Value of the measurement is sion and a main scale division. For example if the value of
obtained with the help of a rule. The hinge joint can be firm one main scale division is 0.5 mm and 24 main scale divi-
joint or spring loaded. sions coincide with 25 vernier scale divisions,
Outside Calliper, inside calliper, Transfer calliper and
hermophrodite calliper are examples of callipers. ⎡ 24 ⎤
Least count = ⎢1 − ⎥ × 0.5
Hermophrodite calliper is also called odd leg calliper. This ⎣ 25 ⎦
is a scribing tool having one leg bent and the other leg equipped 1
with a scriber. This is used for scribing distances from the edge = × 0.5
25
of a work piece. The curved leg has a notch at its end.
1
= mm
Surface Plate 50
Surface plate is a flat plate on which measurement of parts = 0.02 mm.
are carried out. They are used in workshops and metrologi- Main scale reading + coinciding division on the vernier
cal laboratories. Surface plate form a datum surface for test- scale × least count gives the measurement value.
ing flatness of surfaces.
Micrometers (Screw Gauges)
V-block Micrometers are used when more accuracy in measure-
V-blocks are used in workshops as a support for round ment is required. Least count of micrometers is in the order
shaped work pieces. The work pieces are placed on this of 0.01 mm where as that of a vernier caliper is 0.02 mm.
V-shaped groove on top of the block. The angle of the Micrometer works on the principle of a screw and nut. For
groove is generally 90°. V-block is used for checking round- one revolution of the screw in the nut the axial distance
ness of rods, marking centres accurately etc. moved by it is equal to the pitch of the screw.
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.867

Micrometer consists of a U-shaped frame, spindle, bar- Sine bars are graded as A grade and B grade. A grade sine-
rel and thimble. By rotating the thimble on the barrel the bars are made with an accuracy of 0.01 mm/m of length
spindle advances. The object to be measured is placed in and B grade sine bars with an accuracy of 0.02 mm/m of
between spindle end and an anvil at the end of the frame. length
Main scale is marked on the barrel. The thimble has equal
divisions around its periphery, usually 50 nos. Therefore for Clinometers
rotation through one division of the thimble, axial distance Clinometer is a spirit level mounted as a rotary member car-
1 ried in a housing. One face of the housing forms the base
moved is × pitch of the screw.
50 of the instrument. There is a circular scale on the housing.
If pitch of the screw corresponds to one division on the Clinometers can be used to measure the included angle
1 between two adjacent slanted sides of a work piece.
main scale and if it is 0.5 mm, the least count is 0.5 × Different types are vernier clinometers, micrometer cli-
= 0.01 mm. 50
nometers and dial clinometers.
Measurement value using a micrometer is obtained simi-
lar to that of a vernier calliper i.e Main scale reading + read- Comparators
ing on the thimble × least count.
A comparator is a precision instrument employed to com-
A comparator is a precision instrument used for compar-
pare the dimension of a given component with a standard,
ing the dimension of a given component with a standard
such as slip gauges. It does not measure the actual dimen-
such as slip gauges. It does not measure the exact dimen-
sion but indicates how much it differs from the basic speci-
sion, but indicates the difference from the basic dimension.
men. The indicated difference is usually small and suitable
Dial gauge is the simplest form of mechanical comparator.
magnification device is provided.
Apart from mechanical comparators, mechanical optical
comparators, electrical and electronic comparators, pneu- Optimeters or Optical Comparators
matic comparators etc are available.
In these comparators the fundamental optical law is made
Measuring machines are generally used for measurement
use of If a ray of light falls on a mirror and is reflected and
of length over the outer faces of a length–bar or any other
the mirror is titled by an angle a, the reflected light moves
long member. These are generally used for measurement of
through an angle 2a. In optimeters the mirror is titled by
considerably greater dimension.
a measuring plunger movement and the movement of the
Angular Measurements reflected light is recorded as an image on a screen. The
shadow of the object is projected on to a curved graduated
Various instruments used for angular measurements are, scale to indicate in comparison measurement. Optical com-
bevel protractors, angle gauges, sinebars, clinometers, auto parators which make use of the enlarged image principle
collimators etc. are commonly known as optical projectors or optical pro-
Angle Gauges jection comparators. Optical projector is used for checking
the shape or profile of relatively small engineering com-
Angle gauges are taper pieces of hardened and stabilized
ponents with an accurate standard or drawing. It enables a
steel. The measuring pieces are lapped and polished to a
magnified image of part of a component to be projected on
high degree of accuracy and flatness. These pieces can be
to a screen where it is compared with an enlarged profile
wrung together just like slip gauges. The pieces are 75 mm
drawing. The degree of magnification may range from five
long and 16 mm wide. Angle gauges are generally available
to hundred.
in 2 sets of 12 and 13 numbers and a square block. In the 13
piece set the following angles are available – Tool Maker’s Microscope
1°, 3°, 9°, 27°, 41° This is used for measurement of small and delicate parts.
1′, 3′, 9′, 27′ It is used for complex measurement of profile of exter-
3″, 6″, 18″ , and 30″ nal threads as well as tools, templates and gauges, centre
In a 12 piece set all the above except 3″ is available. to center distance of holes in a plane etc. and for accurate
Selecting and placing the gauge pieces one over the other angular measurements.
different angles can be set for the measurement
Interferometry
Sine Bars
The advantage and peculiar property of monochromatic
Sine bar is a precision instrument used along with slip light is that its wave length has precise value until the pri-
gauges for the measurement of angles. Their general uses are mary colours which have ill defined wavelengths and the
1. To measure angles very accurately monochromatic light such as from mercury 198 or krypton
2. To locate a work to a given angle within very close 86 are exactly reproducible.
limits
3.868 | Part III • Unit 7 • Manufacturing Technology

For interference to occur, two conditions are necessary

Optical flat
1. Light rays are obtained by division from a single
source
2. Rays before being combined at the eye must travel h
paths whose lengths differ by an odd number of half
wavelengths. C G
L
For interference of light waves, the light should be from
coherent sources. i.e., the two sources should continuously x mm
have light rays of same wavelength and equal intensity
and maintain same phase differences. Further two sources The component C and the standard gauge block G are
should be very narrow and close to each other. placed on a reference surface at a known distance x. The
optical surface is placed over them as shown. The optical
Optical Flat flat will be inclined at a small angle producing an air wedge.
Optical flat is a circular piece of optical glass or fused quartz If N number of dark fringes are observed over the width L
having its two plain faces flat and parallel and the surfaces mm of the block G,
are finished to an optical degree of flatness. The difference in height
Optical flats are used to test flatness of lapped surfaces
such gauge blocks, gauges, micrometer anvils etc. When an λ xN
optical piece placed on a work piece surface, it will not be per- h= ⋅
2 L
fectly parallel to the surface, but will be at slight inclination
to the surface, forming an air wedge between the surfaces. Where l = wavelength
When the optical flat is illuminated by monochromatic If x = 50 mm
source of light interference fringes will be obtained. These L = 25 mm
are produced by the interference of light rays reflected from N=5
bottom face of optical flat and the top piece of the work λ
piece. If they are in phase, it will result brightness. If the rays = 0.000333909 mm
2
are out of phase, it will cause dark fringes. Each adjacent
fringe represent a change in elevation by half wavelength. h = 0.00333909 mm
There are four possible cases, when the contact between Interferometers are optical instruments used for measuring
optical flat and work surface is at one point only flatness and determining the length of slip gauges. These are
1. If the surfaces are perfectly wrung together no air gap based on interference principle and employs wavelengths of
exist and no fringe pattern will be observed. light as measuring units. Disadvantages of a optical flat are
2. If angle is increased, fringes are brought closer overcome in this by providing refined arrangement
together
3. If angle is decreased, fringes spacing increases. Screw Thread Measurement
4. When angle is made too large, then fringes will be Measurement of effective diameter.
closely spaced, as to be undistinguishable and no Effective diameter is the most important of all thread ele-
observable pattern will be visible. ments. It plays a big role in perfect fitting of the mating
threads.
• Eye
Effective diameter or pitch diameter is the diameter of an
imaginary co-axial cylinder which intersects the flanks of
the threads, such that the width of the threads (metal) and
Optical
flat widths of the spaces between the threads are equal, each
being-half the pitch.
Air wedge Effective diameter can be measured by the following
Surface to be methods.
tested 1. Thread micrometer method
2. One wire, two wire or three wire methods
Method of Checking Height of
a Component with the Help of Three Wire Method
Optical Flat In this method, three wires of equal and precise diameter
are placed in the groves of the threads, two on one side and
By using an optical flat the height of a given component can
one on the other side and the outer distance M is measured.
be checked against a known standard.
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.869

Surface Finish Measurement

The irregularity on the surface which is in the form of
hills and valleys varying in height and spacing are usually
termed as surface roughness, surface finish, surface texture
M or surface quality. If the respective or random deviations
from the nominal surface which forms pattern on the sur-
face. Surface texture, includes roughness, waviness, lays
and flaws.
Surface irregularities of small wavelength are called
primary texture or roughness. These are caused by direct
Effective diameter = E action of the cutting elements.
E=M–Q The surface irregularities of considerable wavelength of a
Where M = Micrometer reading across the top of wires. periodic character are called secondary texture or waviness.
Q = a constant depending up on wire diameter and flank Lay is the predominant direction or pattern of the surface
angle. texture.
P
= W (1 + cosecq) − cot q Flaws are surface irregularities or imperfections which
2
occur at infrequent and at random intervals. Cracks,
Where W = wire diameter
scratches, inclusions and similar defects are included in
P = pitch
this.
Best Wire Size
Evaluation of Surface Finish
Best size of wire is the size of wire of such diameter that
Surface finish is evaluated by the following methods.
it makes contact with the flanks of the thread on the pitch
line. Depending up on the pitch the best wire size varies. 1. Peak to valley height method
Measured with any other wire size will include error in the 2. The average roughness
measurement. 3. Form factor or bearing curves

B
Peak to Valley Height Method
(Rt Measurement)
r O
Pitch line It is the maximum peak to valley height within the assess-
Q A P ment length. The draw back of the method is that it may give
p same Rt value for two largely different texture.
4

Effective x Average Roughness

diameter
1. CLA method
p
2. RMS method
2
3. Ten point height method
Line OP is perpendicular to the flank position of the thread.
Let x = half the included angle. CLA (Centre Line Average) Method
In triangle OAP In this average values of ordinates from the mean line,
AP regardless of the arithmetic signs of the ordinates (average
Sin POA =
OP deviations from the nominal surface) are measured
AP CLA value (Ra)
i.e., sin(90 – x) =
OP h1 + h2 + h3 +  hn
=
AP AP n
\ OP = = = APsecx
sin (90 − x ) cos x A1 + A2 + A3 +  An
=
But OP = r, the radius of wire L
\ Best wire diameter
=
∑ A
p L
db = 2APsecx = 2 sec x
4 1 L
= ∫ h dx over 2 to 20 consecutive sampling
p L 0
i.e., db = 2 sec x lengths
3.870 | Part III • Unit 7 • Manufacturing Technology

RMS (Root Mean Square) Method In a CIM system a group of NC machines are connected
In this method also, the roughness is measured as the aver- together by an automated material handling system and
age deviations from the nominal surface. operating under computer control.
A computer integrated manufacturing system incorpo-
h12 + h22 +  hn2 rate many of the individual CAD/CAM technologies and
RMS value = concepts such as
n
Computer numerical control or CNC
L
1 2 Direct Numerical control or DNC
L ∫0
= h dx Computer aided process planning CAPP
Computer integrated production management industrial
Ten Point Height Method (Rz) roots.
Ten point height of irregularities is defined as the average
difference between the five highest peaks and five deepest Numerical Control
valleys within the sampling length measured from a line, Numerical control (NC) refers to the operations of machine
parallel to the mean line and not crossing the profile. tools from numerical data. Data for operations may be
stored on paper tape, magnetic tape, magnetic discs etc. As
2 1 numerical information is used, it is called numerical con-
trol. Machine tools and other machines are operated by a
3
series of loaded instructions. If the machine tool works with
4 a built in computer controlling, the system is known com-
5
puter numerical control (CNC)
The basic components of a NC system are
10 9
8 6 7 1. A program. i.e., a set of instructions
R2 R1 2. A machine control unit (MCU)
R6
3. The machine tool
L The MCU is further divided into two elements: The data
processing unit (DPU) and control loops unit (CLU). Data
1 process unit processes the coded data read from storage
Rz = [(R1 + R2 + R3 + R4 + R5) – (R6 + R7 + R8 + R9 + R10)] ×
5 devices and passes information such as position of each
axis, required direction of motion, speed, feed etc and aux-
This method is relatively simple, but does not account for iliary function control signals to CLU.
the frequency of the irregularities and profile shape. It is A typical program may contain an instruction like x +
used when cost is to be controlled and for checking rough 100, y + 50, s + 90
machining. This instruction is interpreted as to move a distance
100 mm in the x positive direction, 50 mm in the y +
Computer Integrated ­direction and rotate the spindle at 90 rpm clockwise.
The information pieces are decoded by the DPU and sent
Manufacturing System (CIMS) to CLU. The CLU operates the drive mechanisms as per
CIMS is a production system consisting of a group of the instructions, then receive feed back signals regarding
Numerical controlled (NC) machines connected together by the actual positions velocity etc. When one instruction is
an automated material handling system and operating under executed another is read.
computer control. Many of the individual CAD/CAM tech-
nologies are incorporated in CIM. Other concepts incor- Advantages of NC Systems
porated in CIM are Computer Numerical control (CNC), 1. High machine utilization
Direct Numerical Control (DNC), Computer Aided Process 2. Need for special tooling is mostly eliminated
planning (CAPP), Computer integrated production man- 3. High quality products can be manufactured
agement and Industrial robots. 4. Consistency in quality
CIMS depend upon the production requirements. Each 5. Quality is not dependent on operator skill
system vary according to the requirements. 6. Cost of production is less
Computer integrated manufacturing (CIM) is a recent 7. Minimum scrap
technology. It comprises of a combination of software and 8. In process inventory is less
hardware for product design, production planning, produc- 9. Higher productivity
tion control, production equipment and production process. 10. Reduced set up time
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.871

Disadvantages of NC System Advantages of CNC System over

1. Initial investment is very high for the specialized Conventional NC System
equipment 1. As the computer can be easily and readily
2. Redundancy of labour reprogrammed the system is very flexible. The
3. Down time is highly expensive machine can manufacture a part followed by other
4. Special skill is required for programming and parts of different designs.
operating the equipment 2. Editing and debugging programs, reprogramming
and plotting and printing part shapes etc are simpler.
CNC Retrofitting 3. Manufacturing programme for a component can be
Retrofitting means adding accessories to a given object to easily called by computer.
improve its performance. High initial investment of the This saves time and eliminates errors due to tape reading
4. Greater accuracy
CNC machines make it not affordable to small scale indus-
tries which are back bone of our economy. It is possible 5. Ease of operation
to modernize the existing conventional machines to CNC 6. Trouble shooting is easier as the microprocessors
have self diagnostic features.
machines by adding accessories and making slight design
modification. This is known as CNC retrofitting. Even
though the retrofitted machines are not as good as CNC Adaptive Control Systems (ACS)
machines, their performance is far better than the original ACS is a logical extension of CNC system in the sense that
machines. So when there is a budgetary constraint retrofit- in CNC system operating conditions are specified by the
ting can be resorted to. user in the form of a program. But in ACS calculation and
setting up of operating conditions like speed, feed, depth
Direct Numerical Control (DNC) of cut etc are done during machining by the control system
DNC is a manufacturing system in which a number of itself.
For example in drilling the torque on the drill is measured
machines are inter connected using a computer through
direct connection in real time. In the case of NC or CNC and feed and speed or both are adjusted within programmed
systems one computer is used to control one machine tool. limits . Adaptive control is still in its infancy, since the effect
of various process variables on the finished part is still rela-
But in the case of DNC system one computer can be used to
control more than 100 machines. One computer is designed tively unknown.
to provide instructions to each machine tool on demand.
The components of a DNC system are Basic Concepts of CAD and CAM
1. Central computer CAD/CAM means computer aided design and computer
2. Bulk memory which stores the CNC programs aided manufacturing. It is the technology concerned with
3. Tele communication lines the use of digital computers in design and production
4. Machine tolls Therefore CAD is the use of computer systems to assist
In DNC, the computer calls the program instruction from in the creation, modification, analysis or optimization of a
bulk storage and sends to machines as the need arises. It design.
also receives feed back from the machines. CAM is the use of computer systems to plan, manage
DNC system has the main draw back that if the central and control the operations of a manufacturing unit through
computer goes down, all the machines become in operative. direct or indirect computer interface with the plants pro-
This draw back is overcome by using Distributed Numerical duction resources.
control system in place of direct numerical control. In this A digital computer is the essential ingredient of a CAD/
type even if there is a central computer, the individual NC CAM.
machines are not directly controlled by this central com- A digital computer can be used for image processing,
puter. Each NC machine has its own dedicated on – board real time process control and for solving complex problem
microcomputer just like a CNC system. in a few seconds.
With the development of dedicated min- computers the A digital computer consists of the following three
benefits of DNC system can be realized in CNC system also. components
More over with the availability of small computers 1. Central processing unit (CPU)
with large memory, microprocessors and program edit- 2. Memory
ing capabilities CNC machines are widely used at pre- 3. Input/output section
sent. Also the availability of low cost programmable logic A modern CAD system can perform graphics and non-
controlled has helped in the successful implementation of graphics is function. It is based on interactive computer
CNC system. graphics (ICG)
3.872 | Part III • Unit 7 • Manufacturing Technology

Computer Aided Process Planning (CAPP) 1. Process rationalization

Process planning means the determination of the sequence 2. Increased productivity of process planners
of individual manufacturing operations required to produce 3. Reduced turn around time
a given part or product. In CAPP production planning is 4. Improved legibility
done with the use of computers. The operation sequence is 5. Incorporation of other application programs.
documented on a route sheet. The advantages of CAPP are

Exercises
Practice Problems 1
+0.003 +0.000
1. A 60 mm diameter shaft is made to rotate in a bush. (A) 25.000 −0.000 mm (B) 25.000 +0.003 mm
Tolerances for both bush and shaft are 0.050 mm. The +0.003 +0.000
dimensions of the bush to give a minimum clearance of (C) 25.030 −0.000 mm (D) 25.030 −0.003 mm
0.075 in the shaft basis system is
−0.075 +0.125 +0.05
(A) 60 −0.125 (B) 60 +0.075 7. A hole of size 30 −0.03 mm is to be checked by workshop
+0.125 −0.125
GO and NO GO plug gauges. Assuming wear allow-
(C) 59+0.025 (D) 60 −0.075 ance and gauge allowance as 10% of work tolerance,
size of the NO GO gauge will be
2. The following is the hole and shaft dimensions of an
+0.014 +0.014
assembly
(A) 30 −0.022 mm (B) 30 +.0.014 mm
+0.04
Hole: 30 +0.00 mm −0.014 −0.014

+0.06
(C) 30 +.0.022 mm (D) 30 −.0.022 mm
Shaft: 30 +0.04 mm 8. Determine the size of the general type NO GO plug
The type of fit is gauge for checking hole of a 30 H7/f8.
(A) Transition (B) Clearance
Given: i = 0.453 D + 0.001D microns (D in mm)
(C) Interference (D) Running
3. In a hole and shaft assembly of 30 mm nominal size Upper deviation of shaft = –5.5 D0.41
with following dimensions. Maximum and minimum Diameter step for 30 mm = 18 – 30 mm
(Least) metal limits of the shaft are −0.0231 +0.0210
+0.02 (A) 30 −0.0231 (B) 30 +0.0189
Hole: 30 −0.00 mm +0.0231 +0.0210
−0.040 (C) 30 +0.0210 (D) 30 −0.0189
Shaft: 30 −0.070 mm 9. Cold drawn shafts upto accuracy ±0.01 mm are avail-
(A) 29.960, 29.930 mm (B) 30.02, 30 mm able. An interference fit is to be designed for a 50 mm
(C) 30.07, 30.04 mm (D) 29.950, 29.930 mm basic size hole. Maximum and minimum interferences
4. A hole and mating shaft have nominal size of 50 mm. are 0.02 mm and 0.01 mm respectively. Tolerance for
Maximum clearance is 0.15 mm and minimum clear- hole will be
ance is 0.05 mm. Hole tolerance is 1.5 times the shaft (A) 0.04 mm (B) 0.06 mm
tolerance. Limits for hole in a shaft basis system is (C) 0.05 mm (D) 0.03 mm
(A) 49.02, 49.08 mm (B) 51.04, 51.10 mm 10. For a 90 H8 e9 hole shaft assembly the GO gauge for
(C) 49.05, 49.11 mm (D) 50.05, 50.11 mm shaft will be
5. The interference between two mating parts of basic size (given:
100 mm is to be from 0.05 mm to 0.12 mm. Tolerance IT8: 25i
of shaft and hole are same. The hole size in a shaft basis T9: 40i
system is
Fundamental deviation
(A) 100.035, 100.000 mm
(B) 100.020, 100.085 mm For ‘e’ type shaft
(C) 100.00, 99.965 mm = –11D0.41
(D) 99.915, 99.88 mm (Assume hole basis system)
6. A 25 mm H8 hole is to be checked using a plug gauge. 0.0693 +0.0693
The hole high limit is 25.030 mm. Taking gauge mak- (A) 90 −0.677 (B) 90 −0.0777
er’s tolerance as 10% of work tolerance dimensions of +0.0693 −0.0693
the GO plug gauge will be (C) 90 +0.677 (D) 90 −0.777
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.873

11. For the above problem the size of NO gauge for shaft +0.040
15. In an interchangeable assembly shafts of size 30 −0.010
will be +0.030
+0.1524 −0.1524 mm mates with holes of size 30 +0.020 The maximum
(A) 90 −0.1608 (B) 90 −0.1608
interference in microns in the assembly is
+0.0693 +0.1524
(C) 90 −0.677 (D) 90 −0.1608 (A) 16 microns
(B) 18 microns
Direction for questions 12 and 13: (C) 22 microns
0.000 (D) 20 microns
40 −0.009 16. Two slip gauges of 10 mm width measuring 1.000
Q Datum
mm and 1.015 mm are kept side by side in contact
with each other lengthwise. An optical flat is kept
40 f7 resting on them and inspected using monochromatic
+1.00 light of wavelength 0.0058928 mm. The total number
75 0.000 of straight fringes that can be observed on both slip
gauges will be
(A) 4 (B) 6
(C) 3 (D) 8
17. In surface roughness measurement for a sampling
0.010 0.005 length of 0.8 mm, the graph is drawn to a vertical mag-
0.000 0.000 nification of 15000 and horizontal magnification of
φ32 φ27
100 and areas above and below datum line are 160, 90,
Refer the figure given above 170, 150 mm2 respectively. CLA value for this surface
Given IT7 = 16i diameter steps for 40 mm = 30 and 50 mm. is
Fundamental deviation (A) 0.6 mm (B) 0.9 mm
for f shaft = –5.5 D0.41 micron (C) 0.8 mm (D) 0.5 mm
12. Size 40 f 7 will be 18.
+0.0245 +0.0140
(A) 40 +0.0494 (B) 40 +0.0480 Milled
−0.0245 −0.0140 2.0
(C) 40 −0.0494 (D) 40 −0.0480
13. Dimension of length Q will be
+0.009 +0.009 5.0
(A) 34 −0.100 (B) 35−0.100
+0.100 0.100 0.5
(C) 34 −0.009 (D) 35+0.009 mm
14. In a drawing the machined surface was represented as
shown above. The machining allowance of the surface is
(A) 0.5 mm (B) 5.0 mm
(C) 2.0 mm (D) Not shown
19. Match the following:

P. Limits 1. Algebraic difference between the

P Q R actual size and the corresponding
basic size
S Q. Fits 2. Permissible variation in size

The part shown in figure is machined to the sizes given below: R. Tolerance 3. Prescribed difference between
S = 35.00 ± 0.08 mm the dimensions of mating parts to
perform specific function
P = 12.00 ± 0.02 mm
Q = 9.99 ± 0.03 mm S. Allowance 4. The ranges of permissible variation
in dimensions of a part.
Dimension R will have specifications
+0.04 +0.04 5. Degree of lightness and looseness
(A) 13−0.02 (B) 12 −0.02 between the mating parts
+0.04 +0.02
(C) 13−0.01 (D) 12 −0.04
3.874 | Part III • Unit 7 • Manufacturing Technology

(A) P–4, Q–5, R–2, S–3 (B) P–2, Q–3, R–4, S–1 4 slip gauges totalling a height of 54.464 mm was
(C) P–3, Q–4, R–2, S–5 (D) P–1, Q–2, R–3, S–4 inserted below the top roller. The angle set up was
20. A sine bar was used to set up an angle with the help of (A) 30° (B) 25°
slip gauges. Distance between roller centres is 100 mm. (C) 33° (D) 15°

Practice Problems 2 (A) Transition fit (B) Interference fit

(C) Clearance fit (D) Shrink fit
1. Allowance and maximum clearance of a 30 H7/h8 fit
will be 8.
(A) 0.0054, 0.001 mm (B) 0,0.0054 mm 118.3 +0.08
−0.09
(C) 0.0054, 0.005 mm (D) 0,0.054 mm
2. It is possible to drill a 25 mm nominal size hole to an
+0.02 T
accuracy of 25− 0.02 mm using a standard drill. A shaft
is to be machined to obtain a clearance fit in the above
hole such that allowance should be 0.01 mm and maxi-
mum clearance should not be more than 0.08 mm. The
tolerance on the shaft will be
(A) 0.04 mm (B) 0.05 mm 32.7 ± 0.02
(C) 0.03 mm (D) 0.06 mm
3. A hole and shaft assembly has the following dimensions 10.0+0.02 25.2+0.01
50H8/C8. Multiplier for grade IT8 is 25. Fundamental −0.01 −0.02
deviation for shaft C for D > 40 mm is (–95 + 0.8D) 30.4±0.01
and diameter step for 50 mm is 50 to 80 mm. Maximum
In the above sketch the dimension T will be
size of the shaft will be
+0.02 +0.03
(A) 49.8544 mm (B) 51.9072 mm (A) 19 −0.01 mm (B) 20 −0.02 mm
(C) 50.9623 mm (D) 48.8233 mm +0.02
+0.02

4. For the shaft assembly 90 H8 e9 the minimum size of (C) 19 −0.03 mm (D) 20 −0.03 mm
the shaft is 9. In an interchangeable assembly shaft of size
(Given: value of tolerance for IT8 = 25i 20 −−00..040 +0.020
0100 mm mate with holes of size 20 −0.000 . The
IT9 = 40i minimum clearance in the assembly will be
Value of fundamental deviation for ‘e’ type shaft = (A) 10 microns (B) 15 microns
–11D0.41 (C) 8 microns (D) 12 microns
(A) 84.7476 mm (B) 86.7475 mm 10. GO and NO GO plug gauges are to be designed for a
(C) 92.3456 mm (D) 89.8476 mm hole of 20 ++00..050
010 mm . Gauge tolerance can be taken as
5. For the above problem GO plug gauge size will be 10% of the hole tolerance. Following ISO system of
+0.0005 +0.0057 gauge design, sizes of GO and NO GO gauge will be
(A) 90 − 0.0057 (B) 90 +0.0005 respectively
+0.0057 +0.0005
(A) 20.134 mm, 20.164 mm
(C) 89 + 0.0005 (D) 89 − 0.0025
(B) 20.014 mm, 20.054 mm
6. For the above problem NO GO plug gauge size will be (C) 21.123 mm, 21.136 mm
+0.0576 +0.0082 (D) 21.227 mm, 21.732 mm
(A) 90 − 0.0524 (B) 90 + 0.0076
+0.0576 +0.0576 11. An unknown specimen and a set of slip gauges are
(C) 90 +0.0524 (D) 90 − 0.0524 placed over flat surface at a distance x. Using cadmium
light source of wave lengths 0.509 mm, and an opti-
H −7
7. For a hole shaft assembly 60 , the type of fit will be cal flat about 4.75 fringes were observed over distance
m−6 between the slip gauges and the specimen. Difference
(Diameter step for 60 mm = 50 to 80 mm. in height between them will be
Fundamental deviation for an shaft (A) 1.3 mm (B) 1.1 mm
= + (IT7 – IT6) (C) 1.2 mm (D) 1.4 mm
IT7 = 16i 12. A threaded nut having 2 mm pitch with a pitch diam-
eter of 14.7 mm is to be checked for its pitch diameter
IT6 = 10i
using wires. Angle of threads is 60°. The diameter of
For i use the standard formula) wire used should be
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.875

(A) 1.155 mm (B) 2.255 mm Direction for questions 19 and 20: In the measurement
(C) 1.055 mm (D) 2.055 mm of surface roughness, heights of 20 successive peaks and
13. Match the following: troughs were measured from a datum and these were
35, 25, 40, 20, 35
Features to be inspected Instrument 18, 42, 25, 35, 22
P. Pitches and angle 1. Auto collimator 36, 18, 42, 22, 32
errors of screw thread 21, 37, 18, 35, 20 microns
Q. latness error of surface 2. Optical interferometer These measurements were obtained over a length of 18 mm
R. Alignment error of 3. Deviding head and dial 19. Approximate CLA or Ra value will be
machine slide way gauge (A) 27 micron (B) 29 micron
(C) 31 micron (D) 26 micron
S. Profile of a cam 4. Spint level
20. Approximate RMS value will be
5. Sine bar
(A) 29.32 micron (B) 32.73 micron
6. Toolmaker’s microscope (C) 31.18 micron (D) 28.87 micron
21. In the measurement of surface roughness the height of
(A) P–2, Q–3, R–4, S–1 (B) P–3, Q–4, R–5, S–2 10 successive peaks and valleys over datum line over a
(C) P–4, Q–5, R–2, S–3 (D) P–5, Q–2, R–1, S–6 specified sampling length were found to be
−0.008
14. A shaft has a dimension, f 40 −0.025 . The respective val-
Peaks 45 42 40 35 35 mm
ues of fundamental deviation and tolerance are.
(A) –0.008, –0.025 Valleys 30 25 25 24 18 mm
(B) –0.008, –0.017
(C) –0.017, 0.008 The Rz value of the surface will be
(D) –0.017, +0.025 (A) 15 mm (B) 20 mm
+0.018
15. A hole is of dimension f 90 +0.000 mm and the corre- (C) 12 mm (D) 18 mm
+0.012 22. Figure given below shows the dimension obtained on a
sponding shaft is of dimension 9+0.001 mm . When they component by a certain instrument
are assembled they will form a The instrument is
(A) Interference fit (B) Clearance fit (A) Precise but not accurate
(C) Transition fit (D) Running fit (B) Accurate but not precise
16. A 30 h7 shaft has the dimensional limits (C) Neither precise nor accurate
(A) 30.000, 29.979 (D) Sensitive
(B) 30.000, 30.021
(C) 30.000, 30.007
(D) 30.000, 29.993 Average
× × × × ×
17. Essential condition for an interference fit is that the × × × value
lower limit of the shaft should be
(A) Lesser than upper limit of the hole
(B) Greater than the lower limit of the hole True value
(C) Lesser than the lower limit of the hole
(D) Greater than the upper limit of the hole
18. In measuring the surface roughness of an object, a
graph was drawn to a vertical magnification of 10000 23. The reflector combined with auto collimater can be
and a horizontal magnification of 100 and the areas used for checking
above and below the datum lines were (A) parallelism (B) Circularity
(C) Surface finish (D) Alignment
Above 150 80 170 40 mm2 24. According to Taylor’s principle, NO GO gauge checks
Below 80 60 150 120 mm2 (A) Only important dimensions at a time
(B) All the dimensions at a time
The sampling length was 0.8 mm. (C) Only one feature at a time
Ra value (CLA) of the surface is (D) Only related dimensions at a time
(A) 0.08 mm (B) 1.53 mm 25. Expressing a dimension as 42.5/42.3 mm is the case of
(C) 0.85 mm (D) 1.06 mm (A) Unilateral tolerance (B) Bilateral tolerance
(C) Limiting dimensions (D) None of the above
3.876 | Part III • Unit 7 • Manufacturing Technology

26. Most accurate instrument is 3 blocks B1, B2 and B3 are to be inserted in a channel
(A) Vernier caliper of width S maintaining a minimum gap of width T =
(B) Screw gauge 0.125 mm, as shown in figure. For P = 18.75 ± 0.08, Q
(C) Slip gauge = 15.00 ± 0.12, R = 18.125 + 0.1 and S = 52.35 + x the
(D) Optical projector tolerance x is
27. Figure shows 3 wire method of inspecting screw threads. (A) –0.05 mm
The screw thread is ISO metric M 16 with pitch 2 mm (B) +0.05 mm
and effective diameter = 14.701 mm. Diameter of roller (C) –0.35 mm
used for measurement is 1.155 mm and corresponds to (D) +0.35 mm
best wire diameter (i.e., touches at points of effective 29. For measuring taper of a ring gauge, two balls of diam-
diameter) eter 30 mm and 15 mm were used. During inspection
the ball of 30 mm diameter was protruding by 2.5 mm
above top surface of the ring. This surface was located
at a height of 50 mm from the top of the 15 mm diam-
eter ball. The taper of the angle is
S (A) 25° (B) 20°
(C) 15° (D) 18°
30. A fits is specified as 25 H8/e8. The tolerance value
Over write measurement S will be for a basic diameter 25 mm in IT8 is 33 microns and
(a) 17.6 mm (B) 18.8 mm fundamental deviation for the shaft is –40 microns. The
(C) 18.2 mm (D) 19.4 mm maximum clearance of the fit in microns is
(A) 66 microns
28.
S (B) 73 microns
(C) 33 microns
R Q P
T (D) 106 microns
B3 B2 B1

Previous Years’ Questions

1. In an interchangeable assembly, shafts of size (A) 20.010 mm and 20.050 mm
+0.040
(B) 20.014 mm and 20.046 mm
25.000 −0.0100 mm mate with holes of size
+0.020 (C) 20.006 mm and 20.054 mm
25.000 −0.000 mm . the maximum possible clearance in (D) 20.014 mm and 20.054 mm
the assembly will be [2004] 4. In a 2-D CAD package, clockwise circular arc of
(A) 10 microns (B) 20 microns radius 5, specified from P1(15,10) to P2(10, 15) will
(C) 30 microns (D) 60 microns have its center at [2004]
2. During the execution of a CNC part program block (A) (10, 10) (B) (15, 10)
NO20 GO2 X45.0 Y25.0 R5.0 the type of tool motion (C) (15, 15) (D) (10, 15)
will be [2004] 5. Match the following: [2004]
(A) circular interpolation – clockwise
Feature to be Inspected Instrument
(B) circular interpolation – counter clockwise
(C) linear interpolation P. Pitch and Angle errors of 1. Auto Collimator
screw thread
(D) rapid feed
Q. Flatness error of a 2. Optical Interferometer
3. GO and No-Go plug gages are to be designed for Surface plate
+0.050

a hole 20.000 +0.010 mm. Gauge tolerances can be R. Alignment error of a 3. Dividing Head and
machine slideway Dial gauge
taken as 10% of the hole tolerance. Following ISO S. Profile of cam 4. Spirit Level
5. Sine bar
system of gauge design, sizes of GO and NO-GO
6. Tool makes’s
gauge will be respectively [2004] microscope
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.877

(A) P-6 Q-2 R-4 S- 6 (B) P-6 Q-4 R-1 S-3 (A) AW, LC and M
(C) P-5 Q-2 R-1 S-6 (D) P-1 Q-4 R-4 S-2 (B) AW, D, LC and M
6. In order to have interference fit, it is essential that the (C) D, LC, P and SW
lower limit of the shaft should be [2005] (D) D, LC and SW
(A) greater than the upper limit of the hole 9. NC contouring is an example of  [2006]
(B) lesser than the upper limit of the hole (A) continuous path positioning
(C) greater than the lower limit of the hole (B) point-to-point positioning
(D) lesser than the lower limit of the hole (C) absolute positioning
7. The tool of an NC machining has tomove along a cir- (D) incremental positioning
cular arc from (5, 5) to(10, 10) while performing an 10. A ring gauge is used to measure [2006]
operation. The center of the arc is at (10, 5). Which (A) outside diameter but not roundness
one of the following NC tool path commands per- (B) roundness but not outside diameter
forms the above mentioned operation? [2005] (C) both outside diameter and roundness
(A) N010 G02 X10 Y10 X5 Y5 R5 (D) only external threads
(B) N010 G03 X10 Y10 X5 Y5 R5 11. Which type of motor is NOT used in axis or spindle
(C) N010 G01 X5 Y5 X10 Y10 R5 drives of CNC machine tools?[2007]
(D) N010 G02 X5 Y5 X10 Y10 R5 (A) induction motor (B) dc servo motor
8. Which among the NC operations given below are (C) stepper motor (D) linear servo motor
0.050
continuous path operations?[2005]
12. A hole is specified as 400.00 mm. The mating shaft
Arc welding (AW) Milling (M) has a clearance fit with minimum clearance of 0.01
Drilling (D) Punching in sheet mm. The tolerance on the shaft is 0.04 mm. The maxi-
metal (P) mum clearance in mm between the hole and the shaft
Laser cutting of Spot welding (SW) is  [2007]
(A) 0.04 (B) 0.05
Sheet Metal(LC)
(C) 0.10 (D) 0.11

Direction for questions 13 and 14:

f Table
Pulse Stepper Gear box
generator motor
U Nut

Lead screw

In the feed drive of a Point–to–Point open CNC drive, 14. A customer insists on a modification to change the
a stepper motor rotating at 200 steps/rev drives a table BLU of the CNC drive to 10 microns without chang-
through a gear box and lead screw-nut mechanism ing the table speed. The modification can be accom-
(pitch = 4 mm, number of starts = 1). The gear ratio = plished by [2008]
⎛ Output rotational speed ⎞ 1 1 f
=⎜ ⎟ is given by U = . The (A) Changing U to and reducing f to
⎝ Input rotational speed ⎠ 4 2 2
stepper motor (driven by voltage pulses from a pulse gen- 1
(B) Changing U to and increasing f to 2f
erator) executes 1 step/pulse of the pulse generator. The 8
frequency of the pulse train from the pulse generator is f = 1
10,000 pulses per minute. (C) Changing U to and keeping f unchanged
2
13. The Basic Length Unit (BLU), i.e., the table move- (D) Keeping U unchanged and increasing f to 2f
ment corresponding to 1 pulse of the pulse generator, 15. Which of the following is the correct data structure
is [2008] for solid models? [2009]
(A) 0.5 microns (A) solid part → faces → edges → vertices
(B) 5 microns (B) solid part → edges → faces → vertices
(C) 50 microns (C) vertices → edges → faces → solid parts
(D) 500 microns (D) vertices → faces → edges → solid parts
3.878 | Part III • Unit 7 • Manufacturing Technology

16. Match the following: [2009]

NC Code Definition
Z = 40
P M05 1. Absolute coordinate system
Q G01 2. Dwell
Z = 20
R G04 3. Spindle stop
S G90 4. Linear interpolation
Z=0

(A) P-2, Q-3, R-4, S-1

(B) P-3, Q-4, R-1, S-2
(C) P-3, Q-4, R-2, S-1 (A) 13.334 (B) 15.334
(D) P-4, Q-3, R-2, S-1 (C) 15.442 (D) 15.542
+0.015
17. What are the upper and lower limits of the shaft rep- 21. A hole is of dimension ϕ 9+0 mm. . The correspond-
resented by 60 f8? +0.010
Use the following data: ing shaft is of dimension ϕ 9+0.001 mm.
The resulting
Diameter 60 lies in the diameter step of 50–80 mm assembly has [2011]
Fundamental tolerance unit, i, in mm = 0.45D1/3 + (A) loose running fit (B) close running fit
0.001D, where D is the representative size in mm; (C) transition fit (D) interference fit
Tolerance value for IT8 = 25i. Fundamental deviation 22. In an interchangeable assembly, shafts of
for ‘f ’ shaft = –5.5D0.41 [2009] +0.040
(A) Lower limit = 59.924 mm, Upper Limit = 59.970 size 25.000 −0.020 mm mate with holes of size
+0.030
mm
25.000 −0.020 mm. The maximum interference (in
(B) Lower limit = 59.954 mm, Upper Limit = 60.000
microns) in the assembly is [2012]
mm
(A) 40 (B) 30 (C) 20 (D) 10
(C) Lower limit = 59.970 mm, Upper Limit = 60.016
mm 23. A metric thread of pitch 2 mm and thread angle 60° is
(D) Lower limit = 60.000 mm, Upper Limit = 60.046 inspected for its pitch diameter using 3-wire method.
mm The diameter of the best size wire in mm is[2013]
−0.009 (A) 0.866 (B) 1.000
18. A shaft has a dimension, ϕ 35−0.025 . The respective (C) 1.154 (D) 2.000
values of fundamental deviation and tolerance are +0.020

[2010] 24. Cylindrical pins of 25+0.010 mm diameter are elec-

(A) −0.025, ±0.008 troplated in a shop. Thickness of the plating is 30 ±2.0
micron. Neglecting gauge tolerances, the size of the
(B) −0.025, 0.016
GO gauge in mm to inspect the plated components is
(C) −0.009, ±0.008
 [2013]
(D) −0.009, 0.016 (A) 25.042 (B) 25.052
19. In a CNC program block, N002 G02 G91 X40 Z40…, (C) 25.074 (D) 25.084
G02 AND G91 refer to[2010] 25. In a CAD package, mirror image of a 2D point P(5,10)
(A) circular interpolation in counterclockwise direc- is to be obtained about a line which passes through
tion and incremental dimension the origin and make an angle of 45° counterclockwise
(B) circular interpolation in counterclockwise direc- with the x-axis. The coordinates of the transformed
tion and absolute dimension point will be [2013]
(C) circular interpolation in clockwise direction and (A) (7.5, 5) (B) (10, 5)
incremental dimension (C) (7.5, –5) (D) (10, –5)
(D) circular interpolation in clockwise direction and
absolute dimension 26. For machining a rectangular island represented by
coordinates P(0, 0), Q(100, 0), R(100, 50) and S(0, 50)
20. A taper hole is inspected using a CMM, with a probe on a casting using CNC milling machine, an end mill
of 2 mm diameter. At a height, Z = 10 mm from the with a diameter of 16 mm is used. The trajectory of the
bottom, 5 points are touched and a diameter of cir- cutter centre to machine the island PQRS is[2014]
cle (not compensated for probe size) is obtained as (A) (–8, –8), (108, –8), (108, 58), (–8, 58), (–8, –8)
20 mm. Similarly, a 40 mm diameter is obtained at a (B) (8, 8), (94, 8), (94, 44), (8, 44), (8, 8)
height Z = 40 mm. The smaller diameter (in mm) of (C) (–8, 8), (94, 0), (94, 44), (8, 44), (–8, 8)
hole at Z = 0 is [2010] (D) (0, 0), (100, 0), (100, 50), (50, 0), (0, 0)
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.879

27. Which one of the following instruments is widely H1

used to check and calibrate geometric features of H2 d1 Diameter
machine tools during their assembly? [2014] C
(A) Ultrasonic probe
H
(B) Coordinate Measuring Machine (CMM) B
(C) Laser interferometer A
R
(D) Vernier callipers
Recessed Ring
28. For the given assembly : 25 H7/g8, match Group A
with Group B [2014]
d2 Diameter D
Group A Group B
(P) H (I) Shaft Type The distance H2 = 35.55 mm and H1 = 20.55 mm. The
diameter (D, in mm) of the ring gauge is ____
(Q) IT8 (II) Hole Type
33. A GO–No GO plug gauge is to be designed for meas-
(R) IT7 (III) Hole Tolerance Grade
uring a hole of nominal diameter 25 mm with a hole
(S) g (IV) Shaft Tolerance Grade tolerance of ±0.015 mm. Considering 10% of work
tolerance to be the gauge tolerance and no wear con-
(A) P-I, Q-III, R-IV, S-II dition, the dimension (in mm) of the GO plug gauge
(B) P-I, Q-IV, R-III, S-II as per the unilateral tolerance system is [2014]
(C) P-II, Q-III, R-IV, S-I +0.003 +0.000
(D) P-II, Q-IV, R-III, S-I (A) 24.985−0.003 (B) 25.015−0.006
+0.03 +0.003
29. The flatness of a machine bed can be measured using (C) 24.985−0.03 (D) 24.985−0.000
 [2014]
(A) Vernier callipers 34. Holes of diameter 25.0 ++00..040
020 mm are assembled inter-
(B) Auto collimator changeably with the pins of diameter 25.0 +−00..005
008 mm.
(C) Height gauge The minimum clearance in the assembly will be:
(D) Tool maker’s microscope  [2015]
30. A robot arm PQ with end coordinates P(0, 0) and (A) 0.048 mm (B) 0.015 mm
Q(2, 5) rotates counter clockwise about P in the XY (C) 0.005 mm (D) 0.008 mm
plane by 90°. The new coordinate pair of the end point 35. The function of interpolator in a CNC machine con-
Q is  [2014] troller is to [2015]
(A) (–2, 5) (B) (–5, 2) (A) control spindle speed
(C) (–5, –2) (D) (2, –5) (B) coordinate feed rates of axes
31. For the CNC part programming, match Group A with (C) control tool rapid approach speed
Group B : [2014] (D)  perform Miscellaneous (M) functions (tool
change, coolant control etc.)
Group A Group B 36. In the assembly shown below, the part dimensions
circular interpolation, counter
(P) (I) G02 are: [2015]
clockwise
(Q) dwell (II) G03 L1 = 22.0±0.01 mm,
circular interpolation, L2 = L3 = 10.0±0.005 mm.
(R) (III) G04
clockwise
(S) point to point countering (IV) G00 Assuming the normal distribution of part dimensions, the
dimension L4 in mm for assembly condition would be:
(A) P-II, Q-III, R-I, S-IV
(B) P-I, Q-III, R-II, S-IV
(C) P-I, Q-IV, R-II, S-III
(D) P-II, Q-I, R-III, S-IV
32. The diameter of a recessed ring was measured by
using two spherical balls of diameter d2 = 60 mm and L4
L2 L3
d1= 40 mm as shown in the figure. [2014] L1

(A) 2.0±0.008 (B) 2.0±0.012

(C) 2.0±0.016 (D) 2.0±0.020
3.880 | Part III • Unit 7 • Manufacturing Technology

37. A triangular facet in a CAD model has vertices: 42. Match the following: [2016]
P1(0,0,0); P2(1,1,0) and P3(1,1,1). The area of the
facet is: [2015]   P. Feeler gauge    I. Radius of an object
(A) 0.500 (B) 0.707 Q. Fillet gauge II. Diameter within limits by
(C) 1.414 (D) 1.732 comparison
  R. Snap gauge III. Clearance or gap
38. Which one of the following statements is TRUE?
between components
 [2015]
S. Cylindrical plug gauge   IV. Inside diameter of
(A) The ‘GO’ gage controls the upper limit of a hole.
straight hole
(B) The ‘NO GO’ gage controls the lower limit of a
shaft. (A) P-III, Q-I, R-II, S-IV
(C) The ‘GO’ gage controls the lower limit of a hole. (B) P-III, Q-II, R-I, S-IV
(D) The ‘NO GO’ gage controls the lower limit of a (C) P-IV, Q-II, R-I, S-III
hole. (D) P-IV, Q-I, R-II, S-III
39. A project consists of 7 activities. The network along 43. The figure below represents a triangle PQR with
with the time durations (in days) for various activities initial coordinates of the vertices as P(1, 3), Q(4,
is shown in the figure. 5) and R(5, 3, 5). The triangle is rotated in the
X-Y plane about the vertex P by angle q in clock-
12 11 10 wise direction. If sin q = 0.6 and cos q = 0.8,
1 3 5 6
the new coordinates of the vertex Q are: [2016]
14 12
9 Y
Q(4, 5)
7
2 4

The minimum time (in days) for completion of the R(5, 3, 5)

project is _______. [2015] P(1, 3)
40. A drill is positioned at point P and it has to proceed to
point Q. The coordinates of point Q in the incremental
O X
system of defining position of a point in CNC part
program will be [2015]
(A) (4.6, 2.8) (B) (3.2, 406)
Y
(C) (7.9, 5.5) (D) (5.5, 7.9)
44. For the situation shown in the figure below the expres-
Q sion for H in terms of r, R and D is : [2016]
(A) H = D + r 2 + R 2
P
12 (B) H = (R + r) + (D + r)
5
D
X
3 4

(A) (3, 12) (B) (5, 7) r

(C) (7, 12) (D) (4, 7)
41. In a CNC milling operation, the tool has to machine
H
the circular arc from point (20, 20) to (10, 10) at a R
sequence number 5 of the CNC part program. If the
center of the arc is at (20, 10) and the machine has
incremental mode of defining position coordinates,
the correct tool path command is: [2015]
(A) N 05 G90 G01 X-10 Y-10 R10
(B) N 05 G91 G03 X-10 Y-10 R10
(C) N 05 G90 G03 X20 Y20 R10 (C) H = (R + r) + D 2 − R 2
(D) N 05 G91 G02 X20 Y20 R10
(D) H = (R + r) + 2 D ( R + r ) − D 2
 Chapter 5 • Metrology and Inspection Computer Integrated Manufacturing | 3.881

45. Match the following part programming codes with Monochromatic light
their respective functions: [2016]
Part Programming
Functions
Codes
  P. G01    I.   Spindle stop
Q. G03 II. Spindle rotation,
clockwise
  R. M03 III. C
 ircular interpolation, θ
anticlockwise
S. M05   IV. Linear interpolation
If the wavelength of light used to get a fringe spacing
(A) P–II, Q–I, R–IV, S–III of 1 mm is 450 nm, the wavelength of light (in mm) to
(B) P–IV, Q–II, R–III, S–I get a fringe spacing of 1.5 mm is ________. [2016]
(C) P–IV, Q–III, R–II, S–I
(D) P–III, Q–IV, R–II, S–I 47. A point P (1, 3, –5) is translated by 2iˆ + 3 ˆj − 4 kˆ
and then rotated counter clockwise by 90o about the
46. Two optically flat plates of glass are kept at a small z-axis. The new position of the point is: : [2016]
angle θ as shown in the figure. Monochromatic light is
incident vertically. (A) (–6, 3, –9) (B) (–6, –3, –9)
(C) (6, 3, –9) (D) (6, 3, 9)

Answer Keys
Exercises
Practice Problems 1
1. B 2. C 3. A 4. D 5. D 6. B 7. D 8. C 9. A 10. D
11. B 12. C 13. D 14. A 15. D 16. B 17. C 18. A 19. A 20. C

Practice Problems 2
1. B 2. C 3. A 4. D 5. B 6. C 7. A 8. D 9. A 10. B
11. C 12. A 13. D 14. B 15. C 16. A 17. D 18. D 19. B 20. A
21. A 22. A 23. D 24. C 25. C 26. D 27. C 28. A 29. B 30. D

Previous Years’ Questions

1. D 2. A 3. D 4. C 5. C 6. A 7. A 8. B 9. A 10. A
11. C 12. C 13. B 14. D 15. C 16. C 17. A 18. D 19. C 20. A
21. C 22. C 23. C 24. D 25. B 26. A 27. C 28. D 29. B 30. B
31. A 32. 91 to 94 33. D 34. B 35. B 36. D 37. B 38. C
39. 39 to 40 40. D 41. B 42. A 43. A 44. D 45. C 46. 675 47. A
3.882 | Part III • Unit 7 • Manufacturing Technology

Test

Manufacturing Technology Time: 60 Minutes

Direction for questions 1 to 25: Select the correct a­ lternative 8. A rolling mill has rollers of 400 mm diameter. The
from the given choices coefficient of friction is 0.15. In order to reduce the
1. The condition for interference fit is that the lower limit thickness from 150 mm to 10 mm for a strip, the num-
of the shaft ber of passes required are
(A) Should be greater than the upper limit of the hole (A) 4 (B) 5
(B) Should be greater than the lower limit of the hole (C) 6 (D) 7
(C) Should be less than the upper limit of the hole 9. Consider tungsten, aluminium, copper and titanium.
(D) Should be less than the lower limit of the hole If they are arranged in the decreasing order of magni-
2. In an assembly of shaft and hole, shaft size is specified tude of forgeability
as (A) Copper, tungsten, aluminium, titanium
(B) Aluminium, titanium, tungsten, copper
+0.010 -0.000
20 -0.040 mm and hole size as 20 +0.020 mm. (C) Aluminium, titanium, copper, tungsten
(D) Aluminium, copper, titanium, tungsten
The maximum clearance possible in the assembly
10. The pattern allowance for a cylindrical casting of diam-
(A) 20 microns (B) 30 microns
eter 100 mm and length 150 mm is specified as follows:
(C) 40 microns (D) 10 microns
shrinkage allowance is 2 in 50 and machining allow-
3. Cold working of steel means: ance is 2 mm/side. The pattern size is
(A) Mechanical working of steel below the lower criti- (A) d = 110.16 mm, d = 170.16 mm
cal temperature (B) d = 715.32 mm, d = 178.56 mm
(B) Mechanical working of steel below the recrystal- (C) d = 108.16 mm, d = 160.16 mm
lisation temperature (D) d = 112.5 mm, d = 163.52 mm
(C) Mechanical working below 2/3 of the melting
11. Which of the following gating ratio indicates
­temperature
a pressurised system
(D) Mechanical working of steel below the upper criti-
(A) 4:8:3 (B) 1:3:3
cal temperature
(C) 1:2:4 (D) 3:3:4
4. In punching operation the clearance is given:
12. If ‘a’ is the rate rake angle of the tool and ‘f’ the shear
(A) On the die
angle, then shear strain ‘e’ is given by
(B) On the punch
(A) e = cot (f - a) + tan f
(C) In the die or punch
(B) e = cos (f - a) + tan f
(D) In the die and the punch
(C) e = cos (f - a) + cot f
5. From a sheet metal of thickness 1 mm a cup of
(D) e = tan (f - a) + cot f
diameter 30 mm and height 150 mm is to be drawn.
If limiting draw ratio is 1.8, the number of draws 13. When chip thickness ratio is 1 and tool rake angle is
required are 12°, shear angle is equal to
(A) 2 (B) 3 (C) 4 (D) 5 (A) 33° (B) 51°
(C) 57° (D) 62°
6. A lead plate is mechanically worked at room tempera-
ture. It is 14. 18–4–1 High speed tool steel has the following
(A) A cold working process composition
(B) A hot working process (A) 18% molybdenum, 4% tungsten, 1% vanadium
(C) Neither hot working nor cold working (B) 18% tungsten, 4% chromium, 1% vanadium
(D) It is not defined (C) 18% molybdenum, 4% chromium, 1% cobalt
(D) 18% tungsten, 4% molybdenum, 1% cobalt
7. A metal having recrystallisation temperature TA is cold
worked. The recrystallisation temperature of this cold 15. Ceramic tools generally have
worked item is TA′. Then (A) Positive rake angle
(A) TA = TA′ (B) TA > TA′ (B) Negative rake angle
(C) Zero rake angle
(C) TA < TA′ (D) Cannot say from this data
(D) Zero or positive rake angle
 Test | 3.883

16. Aspiration effect in gating system is 21. Two grades of tools (A and B) are used to machine a
(A) Due to pressure difference, air flows from the gat- steel piece. The cutting speed per minute of tool life is
ing system to outside 80 for the first tool (A) and cutting speed per minute of
(B) It is the intake of air from outside atmosphere to tool life for second tool (B) is 100. If Tailors index for
the gate due to pressure difference first tool (A) is 0.2 and that for second tool (B) is 0.25,
(C) It is the oxidation of molten metal during pouring the tool which is giving maximum tool life is
(D) It is the purposeful admission of air into the mould (A) Tool A (B) Tool B
17. Non-consumable electrodes are used in (C) Both A & B (D) Data is insufficient
(A) TIG welding 22. During a machining process with 12° rake tool, the
(B) MIG welding chip thickness ratio is found to be 0.4. The shear angle
(C) Submerged arc welding is
(D) Resistance projection welding (A) 23.1° (B) 31.2°
18. Two cutting tools are designated as (C) 34.5° (D) 36.2°
A: 5, 6, 8 – 3 - 7 - 25 -0.2 mm 23. In an operation the ratio between thrust force and cut-
B: 5 - 6 - 8 - 3 - 5 - 30 - 0.2 mm ting force is found to be 2.8. Then the operation is
For the same speed and feed, which of the cutting tool (A) Drilling (B) Turning
gives better surface finish (C) Grinding (D) Milling
(A) B has better surface finish
Direction for questions 24 and 25:
(B) A produces better surface finish
(C) Both A & B give the same surface finish A cylindrical rod of 120 mm diameter is forged from 60
mm height to 40 mm height at 900°C. The flow stress of the
(D) Data is insufficient
material is 75 MPa
19. Out of the following cutting tools, the one which is
hardest next to diamond 24. The work of deformation will be
(A) Cemented carbide (B) HSS (A) 51 kNm (B) 61 kNm
(C) 71 kNm (D) 81 kNm
(C) Cubic boron nitride (D) Ceramics
20. In orthogonal cutting operation the cutting force and 25. If a drop hammer weighing 16 kN is used for forging to
thrust force are 950 N and 450 N respectively. The rake be done in one blow, the height of fall for the hammer
is
angle is zero. The coefficient of friction is
(A) 2.5 m (B) 3.2 m
(A) 0.21 (B) 0.32
(C) 4.8 m (D) 5.6 m
(C) 0.4 (D) 0.47

Answer Keys
1. A 2. B 3. B 4. A 5. B 6. B 7. B 8. C 9. D 10. C
11. A 12. D 13. B 14. B 15. B 16. B 17. A 18. A 19. C 20. D
21. A 22. A 23. C 24. A 25. B