PP7305:1991
STANDARDS
Exercise Sheet 1 of 34
Exercise 1
Complete the following sentence by writing in the missing words.
Mating features whose suifaces are cylindrical, spherical or a pair of plane parallel suifaces are
calle<l~~~~~~~~~
Exercise 2
The part shown is toleranced with independency of size and forro, but for the two diameters the boundary
of perfect forro at the maximum material limit is not to be crossed. Add the appropriate symbol to the
drawing so that this condition will be met.
~
30.021
30 000
¡3 X 45º
BOTH ENOS
/
"!
..-- o
'
1
C'JN
rn rn
•
1
1
1
""
~
-
/
1
20.035
20.022
Exercise 3
Geometrical tolerancing should be applied fully to all drawings to avoid any parts being rejected on
inspection.
TRUE/FALSE
PP7305: 1991
-
Exercise Sheet 2 of 34
Exercise 4
Complete the table by writing the number of each feature against its description.
Description
Feature number
Face
Axis
Edge
Median plane
External cylindrical surlace
Internal cylindrical surlace
Exercise 5
In the right-hand column of the table write the letter F if the characteristic refers to the form of single
features, the letter O if it refers to the orientation of related features and the letter L if it refers to the location
of related fe atures.
Characteristic
Letter
Characteristic
Angularity
Perpendicularity
Flatness
Coaxiality
Parallelism
Circularity
Position
Symmetry
Straightness
Pro:file of a surlace
Pro:file of a line
Cylindricity
Letter
STANDARD$
PP7305:1991
STANDARDS
Exercise Sheet 3 of 34
Exercise 6
A run-out tolerance can only be applied to related features.
TRUE/FALSE
Exercise 7
Write against each symbol in the table the letter for the characteristic that it represents.
Symbol
Letter
Characteristic
o
A
Perpendicularity
L_
B
Position
o
e
Circular run-out
/:/
11
D
Straightness
E
Coaxiality
-
F
Profile of a line
J_
G
Cylindricity
o
H
Angularity
I
J
Flatness
@
K
Circularity
-$-
L
Total run-out
(\
M
Symmetry
ti
N
Parallelism
p
Profile of a surface
-
-
-
Exercise 8
A geometrical tolerance applied to a feature defines the size and shape of a tolerance zone within which
the feature is to lle.
TRUE/FALSE
PP7305: 1991
STANDARDS
Exercise Sheet 4: of 34
Exercise 9
Place a tick by those dimensions on the drawing that are theoretically exact.
NO
cO a:i
l.()l.()
R 10.5
10.0
80.2
80.0
15.5
R 15.0 .
FEATURE CONTROLLED
Exercise 10
Place a tick in the box that describes the feature controlled by each leader line.
!ll
Qi
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d)
-==
Qi
~
a)
/
A•
B~
A
l
1
1
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1
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1
1
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___
B
D
E
-
E
b)
A
1
1
•O
e
B
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-
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t
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o=
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rJJ.
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=
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Qi
Qi
~e
Qi
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Qi
'OC
-=
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rJJ.
=e,¡
.e~i
!ll
~
=
FEATURE CONTROLLED
Exercise 1 O ( continued)
Place a tick in the box that describes the feature controlled by each leader line.
]
""'
Q.I
"CI
~
Q.I
~
1
c)
l
o
1(
~~)
E
~
•B
1
'A
1
-
1
1
1
A
1
1
B
e
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1
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E
t
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r
d)
1
-
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--'---
t
1
-+\,
/
A
/
\{
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e
B
1
r
1
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e
1
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E
F
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u
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fil
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rxi
=
=
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o =
o~~
u e
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Q.I
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~=
Q.I
"1
Q.I
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"CI
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~
rxi
=
Q.I
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.5
~
"1
fil
~
PP7305: 1991
STANDARDS
Exercise Sheet 7 of 34
Exercise 11
(a) A face of a component is required to be parallel witlún 0.2 to a second face wlúch is identified as datum
feature D. Make a sketch of the tolerance frame for tlús requirement.
(b) Datum feature D is to be flat witlún O. l. Make a sketch of the tolerance frame for tlús requirement.
(c) The axis of a bore is required to be square to datum feature D witlún a cylindrical tolerance zone
0.05 diameter. Make a sketch of the tolerance frame for tlús requirement.
Exercise 12
Complete the following sentence by writing in the missing words.
A _ _ _ _ _ _ _ _ _ _ _ _ _ is indicated by a leader from the tolerance frame terminating in an
equilateral triangle.
Exercise 13
Complete where necessary the tolerance frames for the drawing shown.
i..no
NO
ºº
'°'°
(T) (T)
'&
11 0.1
Exercise 14
In the right hand column of the table write the letter for the correct description of the datum designation.
Letter
Datum
1
1
1DE 1
íl
1D1E1
1 1
1 D-E
1
Description
A A single datum established by two datum features
B A multiple datum ~ system consisting of two datums
C Multiple datums, the datums to be applied in a
particular order.
Exercise 15
Place a tick in the box that describes the datum feature denoted by each datum triangle.
DATUM FEATURE
~
'Q
l'll
·a
ª
'.S
o=
io
ª
~
u
~
1
a)
1
~/
1
A
B
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B
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-
1
O.a.
-
LJ
---
%/~((////
_/
-
e
D
E
.
,...e
1
b)
F
.4 ..
F
1
1
1
1
A
/
%1
-------
J~F
'~ J
A
~
-------______
,___
-------
--------
,
~~
.n
B
-
1
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1
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--
,.
-$1
B
E
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F
1
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3
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o
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=:!
PP7305:1991
STANDARDS
Exercise Sheet 9 of 34
Exercise 16
Make dimensioned sketcbes of the geometrical reference frame for:
(a) the axes of the four 10.0/10.1 diameter boles;
(b) the axes of the group of two boles 12.0/12.1 diameter and the 20.0/20.1 diameter bole;
ee)
the axes of the six 8. 0/8.1 diameter boles.
196.0
195.5
20.1
115.0
114.5
1X~20.0
12.1
10.1
4x ~ 10.0
2x~12.0
-$-
50.0
49.5
250
249
~ 0.2 M
8.1
6 X ~ 8.0 1 EQU 1 SP
1
PP7305:1991
Exercise Sheet 10 of 34
Exercise 17
A feature of a component is to be toleranced for straightness. Over the whole length of the feature the
tolerance is to be 0.05. Over any length of 50 the tolerance is to be reduced to 0.02. Make a sketch of the
tolerance frame.
Exercise 18
A feature of a component is 700 long and is to be toleranced for parallelism with reference to a datum
surface A. The functional requirements will be satisfied if any line 100 long on the fe ature is parallel to
datum A within 0.03.
(a) Make a sketch of the tolerance frame.
(b) To what maximum value could the parallelism tolerance accumulate over the whole length of
the feature?
In exercises 19 to 35, add to the drawings tolerance frames, leaders, datums and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 19
The three scribed lines are to be straight within a tolerance of 0.02.
STANDARDS
PP7305:1991
STANDARDS
Exercise Sheet 11 of 34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, datums and notes as 'necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 20
The axis of the central hole is to be contained within a cylindrical tolerance zone of 0.03 diameter.
LnC>
OC>
c:) c:)
NN
'&
Exercise 21
The surface of the washer face on the blanking plug is to be flat within a tolerance of 0.04 and is not to
be concave.
.....---
1
"
J
r
l
l.
J
'
j
1
PP7305:1991
STANDARDS
Exercise Sheet 12 of 34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, daturns and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 22
The conical surface of the needle valve is to be circular within a tolerance of 0.01 and the right-hand
cylindrical surface is to be circular within a tolerance of 0.05.
Exercise 23
The profile of the component in any section parallel to the plane of projection of the drawing is to be
contained within a bilateral tolerance zone 0.4 wide.
3 PLACES
PP7305:1991
STANDARDS
Exercise Sheet 13 of 34
In exercises 19 to 35, add to the drawings tolerance frarnes, leaders, datums and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 24
The bore of the bearing block is 25.00/25.02 diarneter. Its axis is to be contained between two planes
0.01 apart that are parallel to the plane that contains the bottom surface of the component.
Bore
PP7305:1991
STANDARDS
Exercise Sheet 14 of34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, datums and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 25
In the sectional view, the axis of the upper hole is to lie between two horizontal planes 0.2 apart that are
parallel to the cornmon axis of the two lower holes. In the other view, the axis of the upper hole is to lie
between two vertical planes 0.1 apart that again are parallel to the cornmon axis of the two lower holes.
1
coo
~o
00
~
PP7305:1991
STANDARD$
Exercise Sheet 15 of 34
11) exercises 19 to 35, add to the drawings tolerance frames, leaders, datums and notes as necessary so that
the toleranced features wi1l comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 26
(a) Each side of the slot is to be square to the chamfered face within 0.2.
(b) The maximum material principle could, if desired, be applied to the datum feature.
TRUE/FALSE
01.I'l
~o
cri cri
NN
_____l/
Exercise 27
The common axis of the branch pipe is to lie between two straight lines 0.12 apart that are inclined ata
true angle of 60º to the common axis of the main pipe.
I'\
/
/
'
"
/
I
--
~~
l..i'
r
PP7305:1991
Exercise Sheet 16of34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, datums and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 28
The angled face is to lie between two parallel planes 0.1 apart that are inclined ata true angle of 30º to the
axis of the 8.0/7.9 diameter hole.
STANDARDS
PP7305:1991
Exercise Sheet 17 of 34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, daturns and notes as necessary so that
the toleranced features will cornply with the stated requirernents. The geornetrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 29
The eight holes are 5.00/5.02 diameter and are equally spaced on a 60 pitch circle diameter. Their axes are
to be contained in cylindrical tolerance zones of 0.05 diameter. The cylindrical tolerance zones are to be
square to the face of the flange adjacent to the spigot. Apply the rnaxirnurn material principle to this
geornetrical tolerance.
STANDARD$
PP7305:1991
STANDARDS
Exercise Sheet 18 of 34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, daturns and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 30
The functional requirements for the component are as follows.
(a) The axis of the 16.00/16.02 diameter bore is to be contained in a cylindrical tolerance zone of
0.01 diameter that is coaxial with the outside diameter.
(b) The axis of the 25.00/25.03 diameter counterbore is to be contained in a cylindrical tolerance zone of
0.02 diameter that is coaxial with the 16.00/16.02 diameter bore.
( c) The maximum material principle is to be applied wherever possible to the geometrical tolerances of
features and to daturn features.
rno
~~
NO
OC>
'S.
'°'°
'S.
LrH.n
NN
NO
OC>
o...:t...:t
o
~~
'S.
Exercise 31
(a) The axes ofthe three grooves are to be contained within one cylindrical tolerance zone of 0.03 diameter.
(b) Can the maximum material principle be applied to this geometrical tolerance?
.__
-
YES/NO
['\
l/)
UJ
>
o
o
--
~
'---'--
~
rn
I/
-
-
-
PP7305:1991
STANDARDS
Exercise Sheet 19 of 34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, datums and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 32
The median plane of the slot is to lle between two parallel planes 0.03 apart that are symmetrically disposed
about the common median plane of the 15.00/14.97 wide tongues.
OC--
~~
lf"l...:t-
lf"IO
~~
OOCO
Exercise 33
The median planes of each of the two widths of the tee slot and the median plane of the dovetail are all
to lie between two parallel planes 0.05 apart that are equally disposed about the median plane of the
45.00/44.98 block width. Apply the maximum material principle to the geometrical tolerances and to the
datum feature.
20.1
20.0
10.1
1·
10.0
---~~~~-;-~~
-1
-~--~~~~---.
25.00
24.98
45.00
44.98
PP7305:1991
STANDARD$
Exercise Sheet 20 of 34
In exercises 19 to 35, add to the drawings tolerance frames, leaders, datums and notes as necessary so that
the toleranced features will comply with the stated requirements. The geometrical tolerances are to apply
regardless of feature size (RFS) except where stated otherwise.
Exercise 34
During one complete revolution of the component about the mean axis of the 30.00/29.98 and 38.00/37.98
diameters, the circular run-out of the 42.00/41.98 diameter is not to exceed 0.08 and the radial run-out of
the right-hand face of the 50.00/49.98 diameter collar is not to exceed 0.02.
Ln
-:t-:t
'$.
'$.
---
oco
º"'
crir.-:
(Y)(Y)
'$.
•
-
Exercise 35
While the component is revolving about the axis of the 35.018/35.002 diameter:
(a) the total radial run-out of the surface of the 54.970/54.930 diameter is not to exceed 0.05;
(b) the total axial nm-out of the right-hand face of the 54.970/54.930 diameter is not to exceed 0.08.
CON
~o
ºº
u-\L.ri
(Y) (Y)
Q
-
..__ - - - - - - - - f - + - -
PP7305:1991
STANDARDS
Exercise Sheet 21 of 34
Exercise 36
Place ticks in the appropriate places in the YES and NO colunms.
Toleranced characteristic
Maximum material principie may be applied
YES
Flatness of a face
Pet"pendicularity of a bore axis to a datum plane
Angularity of a surface to a datum plane
Parallelism of a bore axis to a datum axis
Roundness of a conical surface
Position of a spigot axis
Cylindricity of a bore
Axial run-out of a face
Straightness of the generator of a cylinder
Symmetry of the median plane of a slot
Profile of a surface
Squareness of a face to a datum plane
Coaxiality of a bore axis to a datum axis
Parallelism of a surface to a datum surface
Profile of a line
Angularity of a bore axis to a datum plane
NO
PP7305:1991
STANDARDS
Exercise Sheet 22 of 34
Exercise 37
The drawings show two mating components of an assembly.
(a) For component 1 write down:
(1) the MMC size of the tenon;
(2) the virtual size of the tenon.
(b) For component 2 write down:
(1) the MMC size of the mating slot for the tenon;
(2) the virtual size of the mating slot for the tenon.
ee)
What is the maximum permitted v~lue of the symmetry tolerance:
e1) on the tenon;
(2) on the mating slot for the tenon?
49.97
49.95
-r~
...---~~~~--~~~~~~
Component 1
24.95
24 93
- --·----------.ij
50.02
50.00
= O. O5 @ A@
1
~~~~~~~~~~--------1B
25.02
25.00
Component 2
1
1
PP7305:1991
STANDARDS
Exercise Sheet 23 of 34
Exercise 38
For the component shown:
(a) What are the MMC sizes of the two holes?
(b) When both holes are at their MMC sizes what is the diameter of the coaxiality tolerance zone for the
16.06/16.00 diameter hole?
(e) What is the diameter of the coaxiality tolerance zone when the hole diameters are:
(1) 20.00 and 16.01
(2) 20.02 and 16.03
(3) 20.04 and 16.06?
1
1
1
1
1
1
1
1
1
$--
1
1
-$-
1
1
1
1
1
1
1
1
1
1
1
1
1
1
l
1
1
1
PP7305:1991
STANDARDS
Exercise Sheet 24 of 34
Exercise 39
(a) What is the MMC size of datum bore C?
(b) What is the MMC size of the four holes?
( c) A cornponent is inspected and the individual diarneters of the four holes are found to be:
(1) 12.00
(2) 12.01
(3) 12.02
(4) 12.03
Complete the following table.
Hole diameter
Positional tolerance zone diameter
12.00
12.01
12.02
12.03
( d) Place a tick in the box against a correct staternent and a cross in the box against an incorrect staternent.
D
D
D
D
Datum axis
e is to be perpendicular to datum face D.
The axes of the four holes are to be perpendicular to datum face D.
Datum face D is to be flat.
The axes of the tolerance zones for the four holes are to be perpendicular to datum face D.
e
e) If datum bore e is 40.02 diarneter, by what rnaxirnurn arnount rnay its axis be displaced frorn the centre
of the geornetrical reference frarne of the four holes?
-.:to
00
ºº
-.:t-.:t
'&
PP7305:1991
STANDARDS
Exercise Sheet 25 of 34
Exercise 40
(a) Write down the sizes of the four diameters at which they are all to be perfectly coaxial and have their
axes perfectly square to datum face E.
(b) Write down the virtual size of the bore.
(e) The larger outside diameter is measured and found to be 44.99. What is the diarneter of the cylindrical
tolerance zone within which its axis is required to lie?
º'°
º"'
LI1 _j
,.,,,.,,
rno
00
CX)CX)
(Y'l (Y'l
'S.
'S.
~~
~~
~o
NN
º'°
º"'
LI1
_j
'S.
'S.
...:t...:t
E
PP7305:1991
STANDARDS
Exercise Sheet 26 of 34
Exercise 41
The four holes are 12.000/12.062 diameter equally spaced on the 70 diameter pitch circle. The posicional
tolerance zone for their axes is to be 0.08 diameter, posicioned relacive to datums F and G. The median plane
of the pair of flats is to have a posicional tolerance of 0.04 relacive to the centre of the geometrical reference
frame of the four holes and datum F. Complete the drawing showing these requirements, applying the
maximum material principle wherever possible.
®
0
l.
!
G
lrr--
80.0
79.9
(
1
I
39.975
~ 39.950
1
1
~
PP7305: 1991
STANDARDS
Exercise Sheet 27 of34
Exercise 42
(a) The six 12 diameter holes are to carry an H7 size tolerance. Relative to datums R, S and T their axes
are to be contained within 0.25 diameter tolerance zones. In addition, each hole axis is to lie within a
0.08 diameter tolerance zone, the axis of which is within the 0.25 diameter tolerance zone and perpendicular
to the primary datum. Complete the tolerance frame to indicate these requirements.
(b) What is the shape, size and position of the virtual condition for each hole?
Cb
$
1
1
~
~
~[ill
1
1
1
1
$
~
$
[ill
HTI
PP7305:1991
STANDARDS
Exercise Sheet 28 of 34
Exercise 43
The positional tolerance zone for the axes of the four M16-6H tappings is to be 0.08 diameter relative to
datums D and E. The tolerance zone is to be projected a distance of 50 from the primary datum. Show these
requirements on the drawing, without using the maximum material principle.
¡-n--r--rg-¡t----J
1"7"'""TT"""r;]
_I
1
-+----!
1
..
-0.025
1
~so -o.oso_~
.
PP7305:1991
STANDARDS
Exercise Sheet 29of34
Exercise 44
~a) Components A and B are to be fastened together using the stud shown at figure (b). A projected
tolerance zone is required for the axis of the tapping in component A. What length should be specified for
the projected tolerance zone?
(b) If the design is changed and the set bolt at figure ee) is substituted for the stud, what length should now
be specified for the projected tolerance zone?
(o)
.,.
so :t 0.1
Component
7S:t0.1
A_/
Component B
(b)
'
65 :t 0.2
letal end
11S:t0.2
-
-
160:t0.2
, (e)
11S:t0.2
--,
~
¡---1
j
[
50 ±0.2
_¡
-
'
PP7305:1991
Exercise Sheet 30 of 34
Exercise 45
Three datum target indications are shown. Write down the meaning of each of them.
(a)
GD
\F9
(b)
~
~
(e)
Q
(2
Exercise 46
Pairs of pipe flanges, as shown, are assembled by bolting together with M8 bolts and nuts. The limits of size
for the diameter of the unthreaded shank of an M8 bolt are 8.00 and 7.78.
Complete the tolerance frame on the drawing by inserting the maximum value of the positional tolerance
for the hole axes.
STANDARDS
PP7305:1991
STANDARDS
Exercise Sheet 31 of34
Exercise 47
The cover is assembled to the housing using M8 hexagon nuts. The limits of size for the plain portion of an
M8 stud are 8.00 and 7.85.
Complete the drawings for the cover and housing by inserting in the tolerance frames suitable maximum
yalues for the positional tolerances for the hole and stud axes.
70 PCO
COVER
3xMBx1.25-6g 1 EQUI SPI
~~
70 PCO
00
+
lf'I
o
HOUSING
w
PP7305:1991
STANDARDS
Exercise Sheet 32 of 34
Exercise 48
The block is secured to the slideway by two M8 screws which pass through the slots in the flange of
the block.
Complete the drawings for the slideway and block by inserting in the tolerance frames suitable maximum
yalues for the symmetry tolerances for the median planes of the slideway slot and the tongue of the block.
t-1
A
SLIDEWAY
r---------¡
1
1
1
1
1
1
1
1
B
BLOCK
PP7305:1991
STANDARDS
Exercise Sheet 33 of 34
Exercise 49
The canister is pushed into the bore in the body so that its flange seats in the counterbore below the thread.
The canister is retained in the body by a screwed plug.
Complete the drawings for the canister and body by inserting in the tolerance frames suitable maximum
values for the coaxiality tolerances for the canister flange and its mating counterbore.
MA@
A
CANISTER
BOOY
PP7305:1991
STANDARDS
Exercise Sheet 34 of 34
Exercise 50
The lever is secured to the spindle by an M12 hexagon nut and a plain washer.
Complete the drawings for the lever and spindle by inserting in the tolerance :frames suitable maximum
values for the positional tolerances for the median planes of the pairs of flats.
@X M
y
</>
19.980
19. 959
pp 7305
Solutions to Exercises in
Geometrical Tolerancing
1991 edition
........
___....,._
...
....._.
---~­
STANDARDS
PP7305: 1991
First published 1983 as PD 7305
First revision 1991
BSI Catalogue number PP 7305
ISBN 0-580-20393-X
UDC 744:003.63:621.753.1:514
STANDARDS
STANDARDS
PP7305:1991
Foreword
PP 7305 consists of this booklet and 34 exercise sheets.
The exercises are designed for students taking courses in
engineering drawing and design that require a knowledge
of geornetrical tolerancing, to give practice in the
principies of the subject. The principies are those set out
in BS 308 Engineering drawing practice:Part 3:1990
Geometrical tolerancing and PP 7304: 1991 Introduction
to geometrical tolerancing. This is a cornpanion
publication to PP 7305.
In this edition the nurnber of exercises has been increased
to take account of the changes in the subject introduced
in BS 308:Part 3: 1990. It is intended that students will
work the exercises directly on the exercise sheets, and
with this in rnind the sheets have been designed for
easy reproduction. Perrnission is given for educational
establishrnents to reproduce thern for the use of lecturers
and their students within the one establishment.
The solutions in this booklet rnay not be photocopied
or otherwise reproduced without the prior perrnission
in writing of BSI. The booklet is not included in any
national licence scherne nor in BSI's special scherne
for educational establishrnents which relates to
British Standards.
©
British Standards Institution 1991
STANDARDS
PP7305:1991
Introduction
Linear dimensions and tolerances on the drawings are in
rnillimetres. They are examples only and are not quoted
as recornmended practices.
The drawings are in first angle projection. They are not
to be considered as design examples and are complete
only in so far as is necessary to enable the question to
which they refer to be answered.
STANDARDS
PP7305:1991
Solution 5 continued
Solution 1
'features of size'
Characteristic
Letter
Perpendicularity
o
Coaxiality
L
Circularity
F
Syrnrnetcy
L
Pro:file of a surface
F
Cylindricity
F
Solution 2
30 .021
(1130000
®
r
3 x45°
BOTH ENOS
/
"¡
1
1
~o
NN
,..,..,,..,..,
Solution 6
TRUE. The toleranced surface needs to be related to a
datum axis.
1
1
"-
1
1
/
20.035
(1120.000
®
Solution 3
FALSE
Solution 7
Symbol
Solution 4
Description
Feature number
Face
1, 9
Axis
2, 7
Edge
4
Median plane
3
Externa! cylindrical surface
5
Interna! cylindrical surface
6, 8
Solution 5
Letter
o
J
L
H
o
K
jj
G
11
N
-
D
J_
A
o
p
!
e
@
E
~
B
Characteristic
Letter
(\
F
Angularity
o
ti
L
Flatness
F
---
M
Parallelisrn
o
Position
L
Straightness
F
Pro:file of a surface
F
Solution 8
TRUE
1
STANDARDS
PP7305:1991
Solution 9
Solution 10 continued
FEATURE CONTROLLED
Cll
Q,l
;§
·~
= !S
~
o
o
""'
"CI
Q,l
=
Q,l
(e)
NO
Cll
·~
~
~
Q,l
¡..;:¡
u
A
v
=rl:i
Q,l
u
e
D
Q,)
"CI
rl:i
s
Q,l
s
Q,l
Q,l
Q,l
Q,l
'b«I
"CI
~
R 15.5
15.0
80 .2
80.0
F
(d)
A
e
D
FEATURE CONTROLLED
Cll
Q,l
;§
""'Q,l
"CI
=
Q,l
¡..;:¡
(a)
·~
Cll
·~
= !S
~
o
o
~
u
rl:i
~
Q,l
=-
Q,l
E
Q,l
....=
=
o ;=-
=
'a
~ =
~:9
~:e
o
u s :s
Q,l
Q,l
Q,l
v
v
v
Q,l
'btl
"CI
~
Solution 11
v
(a)
v
D
v
E
v
v
(b)
v
v
e
D
E
F
¡..;:¡
v
e
B
v
Cll
·rr.i
v
A
A
=~
v
v
v
~ .$~~=
~
B
(b)
F
Q,l
Q,l
¡..;:¡
v
B
Solution 10
Cll
rl:i
v
v
E
R 10.5
10.0
=~
~ .$~~=
~
v
B
cx:icx:i
LJ"lLJ"l
Q,l
.3 .3==
=o =
~ :9 ~.3
o
=-
1010.1 I
v
v
(e)
l_Ll!ZSo.oslol
Solution 12
'datum feature'
2
'I
\,
STANDARDS
PP7305:1991
Solution 13
Solution 15
DATUM FEATURE
-Q
=
:s
Cl.l
ª
~
=
= !E
C.I
i
Q
u
C.I
Q,l
~
Q,l
=
'a
=
i~ ~~ t
u e :e -=
Q
00
Q
Q,l
C.I
Q,l
Q,l
~
.¡
E
.¡
A
.¡
B
Solution 14
.¡
e
Letter
Datum
E
B
1 1
1DE1
1 1
1D1E1
e
1 1
1D-E1
A
.¡
D
1
.¡
.¡
F
Solution 16
{al
{b)
1
o
rn
o
rn
30
1
30
f--+---1
L_~ºLJ
{e)
~
3
1
p;o~T-!
70
C.I
Cl.l
00
.¡
D
Q,l
~ .5...;:¡~~=
.¡
e
(b)
=
.¡
B
11 0.1 L
Q,l
C.I
.¡
A
(a)
=
='a
=
·~
Q
~
Q,l
Cl.l
·~
lfl
rn
r
1
j
1
lfl
rn
1
1
+
STANDARD$
PP7305:1991
Solution 17
Solution 20
0.05
0.02/ 50
Solution 18
o
o o
c::ic::i
IJ)
NN
(a)
'&
l//I0.03/1001
Al
(b) Since the feature is 700 long, the maximum value to
which the parallelism tolerance could accumulate over
its whole length will be 7 x 0.03 = 0.21.
-
f/}Q.03
Care should be taken in checking the solutions to
exercises 19 to 35 to ensure that:
e
a) the leader from the tolerance frame always carries
an arrowhead that touches the toleranced feature in
such a direction that the orie.:;i.tation of the tolerance zone
is correct;
Solution 21
(b) the arrowhead is positioned correctly in relation to
the dimension lines depending on whether the tolerance
refers to a surface, face, edge or line; or to the axis or
median plane of the dimensioned feature only; or to the
common axis or common median plane;
ec) datum triangles are positioned correctly using the
same criteria as in (b);
ed)
where the tolerance zone is circular or cylindrical,
0 appears on the tolerance frame before the tolerance
valu e.
Solution 19
Solution 22
Note that the leader from the left-hand tolerance frame
is to be normal to the axis and not normal to the
conical surface.
4
STANDARDS
PP7305:1991
Solution 25
Solution 23
+
1
Solution 24
11 0.01 A
Solution 26
NO
OC>
LrÍU"Í
NN
"Q
(a)
..l 0.02 A
The dimension line for the bore diameter is to be
positioned as shown so that the direction of the
arrowhead on the leader frorn the tolerance frame
o~entates the tolerance zone parallel to the daturn
surface.
Note that the arrowheads on the leaders frorn the
tolerance frame are not to be in line with the
dimension line.
(b) FALSE. Since the daturn feature is aplane surface,
the rnaximurn material principle cannot be applied to it.
5
STANDARD$
PP7305:1991
Solution 27
Solution 29
L 0.12 A
A positional tolerance could be used instead of an
angularity tolerance.
;
Solution 28
1
Solution 30
B
NO
00
00
...t...t
"Q
A
A
A positional tolerance could be used instead of an
angularity tolerance.
6
PP7305:1991
Solution 31
Solution 33
(a)
20 .1
20 .0
10.1
10.0
25 .00
24.98
45 .00
____ _ _ _4_4_.9_8_ __ _ _ _ __,A
(b) YES. The rnaximum material principle rnay be applied
to the geornetrical tolerance because the fe ature to which
the tolerance applies incorporates an axis.
Solution 34
Solution 32
O<D
--++---
º"'
---tt- a:i,....:
rnrn
Q
-::- 0.03 A-B
B
or-
º"'
0i
UlO
00
cricri
Solution 35
B
U 0.05 A
CDN
~o
ºº
uiui
rnrn
Q
A
U 0.08 A
7
STANDARDS
PP7305: 1991
Solution 36
Toleranced characteristic
Maximum material principie may be applied
YES
NO
v
Flatness of a face
Squareness of a bore axis to a datum plane
v
v
Angularity of a surface to a datum plane
Parallelism of a bore axis to datum axis
v
v
Roundness of a conical surface
Position of a spigot axis
v
v
v
v
Cylindricity of a bore
Axial run-out of a face
Straightness of the generator of a cylinder
Symmetry of the median plane of a slot
v
v
v
Profile of a surface
Squareness of a face to a datum plane
'
Coaxiality of a bore axis to a datum axis
v
v
v
Parallelism of a surface to a datum surface
Profile of a line
Angularity of a bore axis to a datum plane
v
Solution 37
(e) ( 1) Be cause the maximum material principle is
applied to the geometrical tolerance on the tenon
and to the datum to which this geometrical tolerance
is referred, the maximum permitted value of the
symmetry tolerance on the tenon is the stated
symmetry tolerance plus the size tolerance on the
tenon plus the size tolerance on datum feature A.
(a) (1) 24.95
(2) Virtual size of tenon
MMC size + stated geometrical tolerance
24.95 + 0.05
25.00
Therefore, the maximum permitted value of the
symmetry tolerance on the tenon is 0.05 + 0.02 + 0.02
= 0.09.
(b) (1) 25.00
(2) Virtual size of slot
MMC size - stated geometrical tolerance
25.00 - 0.03
24.97
(2) Similarly, the maximum permitted value of the
symmetry tolerance on the slot is 0.03 + 0.02 + 0.02
= 0.07.
8
STANDARD$
PP7305:1991
Solution 38
(a) 20.00 and 16.00
Solution 40
(a) Since the geometrical tolerances are speci:fied as zero
at MMC, the sizes of the four diameters at which they are
all to be perfectly coaxial and have their axes perfectly
square to datum face E are the MMC sizes:
(1) 35.00
(2) 18.00
(3) 32.00
(4) 45.00
(b) 0.02
ec) The diameter of the coaxiality tolerance zone is the
sum of the stated coaxiality tolerance and the departures
of the hole diameters from their maximum material limits
of size.
Therefore, coaxiality tolerance zone diameters are:
(1) 0.02 +
o+
0.01
=
(b) The virtual size of the bore
0.03
(2) 0.02
+ 0.02 + 0.03 = 0.07
(3) 0.02
+ 0.04 + 0.06
=
MMC size - stated geometrical tolerance
18.00 - o
18.00
0.12.
ec) The diameter of the cylindrical tolerance zone for the
larger outside diameter is the departure of the feature
from its maximum material limit of size.
Solution 39
(a) 40.00
Cylindrical tolerance zone diameter
MMC size - actual finished size
45.00 - 44.99
0.01
(b) 12.00
ec) The diameter of the positional tolerance zone is the
sum of the stated positional tolerance and the departure
of the hole from its maximum material limit of size.
Solution 41
Hole diameter
Positional tolerance
zone diameter
12.00
0.01
12.01
0.01
+ 0.01 = 0.02
12.02
0.01
+ 0.02 = 0.03
12.03
0.01
+ 0.03 = 0.04
( d) [8) Datum axis C is to be perpendicular to datum
face D.
[8] The axes of the four holes are to be perpendicular
to datum face D.
[8] Datum face D is to be flat.
[2] The axes of the positional tolerance zones are to
be perpendicular to datum face D.
ee) The axis of datum bore e is required to lie in a
cylindrical tolerance zone whose diameter is the
departure of the bore from its maximum material limit
of size and whose axis coincides with the centre of the
geometrical reference frame of the four holes. When
the datum bore is 40.02 diameter the tolerance zone
for its axis is 0.02 diameter. Therefore the maximum
displacement of the datum bore axis from the centre of
the geometrical reference frame for the four holes is 0.01,
that is, the radius of the tolerance zone.
39 .975
G ~-91~39~.9::....:5=0-.J
9
STANDARDS
PP7305:1991
Solution 44
(a) The mínimum specified length of the projected
tolerance zone for the tapping in component A when the
stud is used is the maximum projection of the stud from
the joint face of component A.
Solution 42
(a)
12 .018
6 X 12.000
-$-
9l O. 25 T S R
9l 0.08 T
Therefore, the mínimum specified length should be 115.2.
(b) The mínimum specified length when the set bolt is
used is the maximum thickness of component B.
~
cD
ED
1
1
1
i
1
1
1
1
-$---$--
@]
~
[l9J
Therefore, the mínimum specified length should be 75.1.
Solution 45
(a) The datum target is circular with a diameter of 6. It
is the third target for establishing datum A.
$
[§]
R
hTI
(b) The datum target is square with a side of 5. It is the
first target for establishing datum B.
(b) The virtual condition for each hole is a geometrically
perfect cylinder whose axis is at the theoretically exact
position of the hole axis.
(c) The datum target is square with a side of 20. lt is the
second target for establishing datum C.
The size of the virtual condition is the maximum material
size of the hole miuus the stated positional tolerance for
its axis.
Solution 46
This is an example of the 'floating fastener' case. The
maximum value of the positional tolerance for the hole
axes is the difference between the maximum material
limits of size of the holes and the unthreaded shanks of
the holes. Therefore, the maximum positional tolerance
for the hole axes is 8.40 - 8.00 = 0.4.
12.000 - 0.25 diameter
11.975 diameter
That is, virtual size
Solution 43
~
[
@
n:
1
1
Solution 47
This is an example of the 'fixed fastener' case. The sum
of the positional tolerances for the hole axes and the
stud axes should not exceed the difference between the
maximum material limits of size of the holes and the plain
portions of the studs. Therefore, the sum of the positional
tolerances for the hole and stud axes is 8.40 - 8.00 = 0.4.
1
iii
~!I
-0.025
9' 50 -0.050
1
1
~
_l
m
This sum can be shared in· any proportion between the
positional tolerances for the hole and stud axes, say as
0.2 for each.
Note that the studs and holes are groups of three features.
Therefore the clearance at MMC between the datums V
and W cannot be used as additional positional tolerance
for the hole axes and stud axes.
Since the whole of the positional tolerance which is
theoretically available has been allocated between
the hole ancr stud axes, it follows that in the most
unfavourable assembly conditions the hole and stud will
touch, giving a 'size-to-size' assembly. In the light of
measurement uncertainty this could be considered
10
STANDARD$
PP7305:1991
Solution 49
This is an example of the 'fixed fastener' case. The sum
of the coaxiality tolerances for the canister flange and its
mating counterbore should not exceed the sum of the
clearances at MMC between datums A and B and between
the canister flange and the counterbore.
undesirable. A size-to-size assembly can be avoided by
reducing the tolerance zone diameter for either the hole
or the stud axes, which will give clearance between the
studs and holes under all assembly conditions.
Note. This also applies to the symmetry, coaxiality
and positional tolerances in exercises 48, 49 and 50
respectively.
Clearance
Solution 48
This is an example of the 'fixed fastener' case. The sum
of the symmetry tolerances for the median planes of the
slideway slot and the tongue of the block should not
exceed the clearances at MMC between the slot and the
tongue and between the slideway datum and the block
datum. Since the slot and tongue are single features the
clearance at MMC between the datums A and B can be
used as part of the symmetry tolerance for the median
planes of slot and tongue. This also applies to the
coaxiality tolerance in exercise 49 and the positional
tolerance in exercise 50.
MMC of slideway datum
MMC of block datum
75.00
74.95
Clearance
0.05
MMC of slideway slot
MMC of block tongue
MMC of counterbore
MMC of canister flange
Clearance
Total clearance
0.1
Total clearance
0.05
0.15
0.08
51.00
50.95
0.05
0.08 + 0.05
0.13
This total clearance can be shared in any proportion
between the coaxiality tolerances, sayas 0.07 for the
canister flange and 0.06 for the mating counterbore.
Note. See also solution 4 7.
Solution 50
This is an example of the 'fixed fastener' case. The sum
of the positional tolerances for the median planes of the
pairs of flats should not exceed the sum of the clearance
at MMC of the spindle and lever bore and the clearance
at MMC of the pairs of flats.
40.1
40.0
Clearance
41.00
40.92
MMC of datum B
MMC of datum A
MMC of lever bore
MMC of spindle
+ 0.1
Clearance
This total clearance can be shared in any proportion
between the symmetry tolerances, sayas 0.08 for the
median plane of the slideway slot and 0.07 for the median
plane of the tongue of the block.
Note. See also solution 47.
20.00
19.98
0.02
MMC across flats on lever
MMC across flats on spindle
15.000
14.984
Clearance
0.016
Total clearance
0.02 + 0.016
0.036
This total clearance can be shared in any proportion
between the positional tolerances, sayas 0.02 for the
median plane of the flats in the lever and 0.016 for the
median plane of the flats on the spindle.
Note. See also solution 47.
11