E n g in e e r in g D r a w in g
w it h W o r k e d E x a m p le s 2
T h ir d e d it io n
M. A. Parker, TEng. (CEI), MIMGTech.E, and
F. Pickup, CEng., MIProd.E
'I
Stanley Thornes (Publishers)
Ltd
F. Pickup and M. A Parker 1960, 1970 and 1981
Illustrations © M. A Parker 1981
©
C o n te n ts
All rights reserved. No part of this publication may be reproduced or
transmitted in any form or by any means, electronic or mechanical.
including photocopy, recording, or any information storage and
retrieval system, without permission in writing from the publisher or
under licence from the Copyright Licensing Agency Limited. Further
details of such licences (for reprographic reproduction) may be
obtained from the Copyright Licensing Agency Limited, of 90
Tottenham Court Road, London WIP 9HE.
Originally published in 1960 by Hutchinson Education
Reprinted 1961, 1%2, 1963, 1964 and 1966
Second (metric) edition 1970
Reprinted 1972, 1973, 1974, 1975, 1977, 1978, 1979 and 1980
Third edition 1981
Reprinted 1982, 1984, 1985, 1986, 1988, 1989
Reprinted 1991 by
Stanley Thomes (Publishers) Ltd
Old Station Drive
Leckhampton
CHELTENHAM GL53 ODN
England
Reprinted
P r e fa c e
4
1
Auxiliary projection
5
2
Interpenetration of surfaces
38
3
Conic sections
77
4
Development
Parallel line - Rildialline - Triangulation - Panel
development - Approximate development of spheres
88
5
Cams
Types of cams - Followers - Cam profiles - Displacement
curves - Cylindrical cams
6
Involute gears
154
Spur gears - Racks - Helical gears - Bevel gears - Worm gearing
132
1992
British library Cataloguing in Publication Data
Pickup, Fred
Engineering drawing with worked examples.-3rd
ed. 2
1. Engineering drawings
2. Mechanical drawing
I. Title
II. Parker, Maurice Arthur
604'.2'4
T353
ISBN 0 7487 1014 0
Printed and bound in Great Britain by
Scotprint Ltd, Musselburgh
- Hypoid gears
7
Vector geometry
173
8
Traces
204
9
Machine drawing:
231
T a b le s
268
P r e fa c e
1 A u x ilia r y p r o j e c t io n
The changes introduced in the 1972 revision ofBS 308, Engineering
drawing practice, have made a new edition of this book necessary.
The general plan of the book, however, remains unchanged. The
text has been kept to a minimum sufficient to outline the general
principles of the subject, and worked examples have been freely
used to enlarge on it. Each example shows the method of obtaining
the solution, together with additional explanatory notes. For some
topics where a solution on one drawing would have been difficult to
understand, the solution has been drawn in step-by-step form. The
number of such solutions has been increased in this edition, and
additional problems have also been provided.
The drawings have been completely redrawn and conform to the
recommendations of BS 308: 1972. To mark the equal status given
to first and third angle projection in this Standard, equal use has
been made of the two systems. New material has been added to the
chapters on auxiliary projection, interpenetration of surfaces, development and involute gears to make the treatment of these topics
more complete.
Several people have made suggestions for improvements in the
book and have pointed out errors in previous editions. My thanks
are due to them for their interest. I also acknowledge with thanks
the permission given by the British Standards Institution for extracts from some of their Standards to be reprinted.
Cases arise in practice where views of an object projected on to the
principal planes of projection are either insufficient to describe the
object or are difficult to draw or dimension. Such cases include
objects with inclined faces of a complex nature and are best drawn
using auxiliary views. An auxiliary view is one which is drawn on a
plane other than a principal plane of projection. An auxiliary view
which is projected from a normal elevation or plan is called a first
auxiliary elevation or plan. Other auxiliary views may be projected
from first auxiliary views. These are called second auxiliary elevations or plans. It should be noted that an elevation can only be
projected from a plan and vice versa.
London 1981
M.A.P.
F ir s t a u x ilia r y e le v a t io n s
Figure 1(a) illustrates the method of projecting these views, using
first angle projection. The standard elevation and plan are first
drawn with an XY line or datum line between them. It may be
convenient to use the centre line of the plan or the base of the
elevation as the XY line. The first auxiliary elevation is required in
the direction of arrow Q, so points on the plan view are projected
1
parallel to the arrow to cross the new datum line X1Y at right
angles. This new datum line may be placed in any convenient
position. The heights above the XY line, a, band c, of points in the
original elevation, are then transferred to the appropriate
projectors above the new X1Y 1 line and the view is lined in. To
avoid confusion between full lines and hidden detail lines it is better
to complete first those faces which, by inspection, are seen to be
visible in the auxiliary view.
Figure l(b) shows the same views drawn in third angle projection.
The method of using the same heights in both elevations is identical
with views drawn in first angle projection.
S
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
First auxiliary plan views
The method for these views is similar to that for first auxiliary
elevations and is shown, using first angle projection, in Figure 2(a).
Projectors from points on the normal elevation are drawn parallel
to the new direction of viewing, given by arrow Q, and cross the new
datum X1Yl at right angles. Depths wand z, from the XY line to
points in the original plan, are then transferred to the appropriate
projectors from the X1Y1 line and the view is completed.
Figure 2(b), in third angle projection, shows the method to be the
same in this system of projection.
Second auxiliary elevations
These views are projected from first auxiliary plans and the method
is shown in Figure 3( a). The first auxiliary plan view is first drawn as
described above and projectors from points on this view are drawn
parallel to the direction of viewing and crossing the new datum xzyz
at right angles. Heights above X1Y 1, such as h, of points in the
original elevation R are transferred above xzy z on the appropriate
projectors. To complete the second auxiliary view in the available
space it is sometimes necessary to move X1Yl to the position KL, as
. has been done in Figure 3(a).
It is important to realize that the original plan view S is not
required in the projection of the second auxiliary elevation. This is
illustrated in Figure 3(b) which shows the original elevation and first
auxiliary plan drawn in the conventional positions relative to a
horizontal ground line. The projection of the second auxiliary elevation Q is identical with the projection of a first auxiliary elevation as
outlined above and the original plan view S is not needed.
Figures 4(a) and (b) show the above construction in third angle
projection. As before, the use of third angle projection makes no
difference to the method.
Second auxiliary plan views
These views are projected from first auxiliary elevations as shown in
Figures 5(a) and (b). From the normal elevation R and plan S the
first auxiliary elevation P is projected as described above. Projectors parallel to the direction of viewing and crossing the new
datum line xzyz at right angles are drawn from this auxiliary elevation. Depths w from datum X1Y1 to the original plan are transferred
to the projectors from the new datum xzy z . Note that to save space
X1Yl has been moved to KL. The original elevation R is ignored in
8
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
the projection of the second auxiliary plan, this being demonstrated
in Figure 5(b).
The above construction in third angle projection is shown in
Figures 6( a) and (b).
Projection of arcs of circles and other curves
This is performed by projecting a series of points on the curve in the
same way that the points at the ends of straight lines are projected.
The resulting curves in the auxiliary views can then be filled in with
french curves. If too many points are projected the work becomes
tedious and there is no increase in accuracy.
The solution of problems
The solutions of auxiliary projection problems are obtained more
easily if the following points are observed.
All XY lines should be marked in some way to distinguish them
from each other. Note, however, that XY lines are not shown on
industrial drawings.
The XY line for the first auxiliary view can be moved to a new
position to reduce to a manageable size the distances to points in the
second auxiliary view. Also, this will often enable the second
auxiliary view to be drawn in the available space. The new position
of the XY line must, of course, be parallel to its original position.
Views of symmetrical details are drawn more quickly if centre
lines are used as XY lines.
When distances to points on an object are transferred from one
XY line to another, they must be laid off in the same direction
relative to the linking view. For example, in Figure 5(a) the first
auxiliary elevation links the normal plan and the second auxiliary
plan. Dimension w on the normal plan is laid off from XIYl away
from the first auxiliary elevation. When it is transferred to x z y z it is
again laid off away from the first auxiliary elevation. Observance of
this rule will prevent views being drawn upside down or reversed.
It is unnecessary to project all the points in an auxiliary view since
lines which are parallel in the normal elevation and plan remain
parallel in the auxiliary views. Thus a few lines can be projected in
an auxiliary view and the view completed by making the remainder
parallel to them.
Some worked examples follow, together with examples of the uses
of auxiliary views. Further problems will be found on pages 33-7.
12
A u x ilia r y p r o je c tio n
A u x ilia r y
p r o j e c t io n
p r o b le m s
Scale full size unless otherwise stated. Use the same projection
angle for the solutions as has been used for the given views. Hidden
detail is to be shown in first auxiliary views only.
1
Draw the given views of the hexagonal prism and project a first
auxiliary plan on X1Y1 and a first auxiliary elevation on X2Y2•
2
Views are given of a small cast iron bracket. Draw the given
elevation and plan, and project a first auxiliary plan on XIY·
and a first auxiliary elevation on X 2Y 2•
3
The given views show a right hexagonal pyramid which is cut by
the plane SS. Draw the lower portion of the elevation below
the plane and project from it a plan. Using these views project a
first auxiliary plan to show the true shape of the cut face and a
first auxiliary elevation on X1Yl.
4
A length of
section bar is shown. It is inclined at 45° to the
horizontal plane with the face F parallel to the vertical plane.
Draw in third angle projection the given elevation, a plan
projected from it and a first auxiliary elevation on X1Yl .
5
A small bracket is shown in plan and elevation. Draw the given
views and project from them a first auxiliary elevation on X1Y1
and a second auxiliary plan on X 2 Y2.
6
The plan and elevation of a machined block are given. Draw
these views and project from them a first auxiliary plan on X1Y 1
and a first auxiliary elevation on X 2y 2•
7
A length of extruded bar is shown in plan and elevation. Draw
the given views and project from them a first auxiliary sectional
elevation on SS.
8
Draw the given views of a prism and project a complete first
auxiliary plan to show the true shape of the face A.
9
The plan and elevation of a block are given. Draw these views
and project from them a first auxiliary plan to show the true
shape of the sloping face in the elevation. From the first
auxiliary plan project a second auxiliary elevation, viewing the
first auxiliary plan in the direction of arrow P.
10
A slide block is shown in plan and elevation. Draw these views
and project a first auxiliary elevation viewing the plan in the
direction of arrow Q.
I
33
E n g in e e r in g
D r a w in g
w ith
W orked
E x a m p le s
2
11
Draw the given views of the block shown and project a first
auxiliary plan viewing the elevation in the direction of arrow R.
12
The given views show a component in plan and elevation.
Draw these views and project from them a first auxiliary elevation on Xl Yl and a second auxiliary plan on xzyz.
13
Views are given of a machined detail. Draw them and project a
first auxiliary elevation on X1Y 1• From this view project a
second auxiliary plan on xzyz.
14
Details for a small assembly are given. Assemble the parts as
shown in the end view of the support plate and draw an elevation and plan of the assembly. The face B of the support plate is
to be parallel to the horizontal plane with the edge A positioned relative to the XY line as shown. Face C of the angle
plate is to be visible in the elevation. Hidden detail is not
required.
15
The given views show the plan and elevation of the centre line
AB of a cylinder 10 mm diameter and the centre C of a 15 mm
diameter sphere. Find the clearance between the cylinder and
sphere.
16
A belt running from left to right passes round a pulley
45 mm diameter. Draw the pulley in the given views, neglecting
its thickness.
17
Using auxiliary views find the true shapes ofthe given aircraft
window panels. Also find the true angles between the panels at
the joints AB and CD.
18
For the given lines AB and CD find (a) the common perpendicular, (b) the true length of a line which joins them vertically,
and (c) the true length of a horizontal line which joins them and
passes through C.
34
2 I n t e r p e n e t r a t io n o f
su r fa c e s
Surface intersections appear on many drawings and, apart from
those occurring in the development of sheet metal work which must
be plotted accurately, the draughtsman usually approximates the
curves which they produce. An example of this practice is the use of
circular arcs to represent the chamfer curves on hexagon nuts,
instead of plotting them exactly as hyperbolas.
Approximate curves are satisfactory provided they follow the
true curves closely. Inaccurate approximations can be avoided if the
draughtsman plots a few points to determine the general shape of
the curve. To do this he must understand the method used to
construct the curves, which is a repetitive process employing simple
sections.
S im p le s e c t io n s
o f g e o m e t r ic a l
s o lid s
The solids which occur most often are shown in Figures 1 to 4,
together with various sections taken through them.
R ig h t c y lin d e r
Refer to Figure 1.
S e c t i o n A A . This is taken at right angles to the axis and produces a
circle in the adjacent view. Thus a series of sections at right angles to
the axis will all give the same circle, which is the adjacent view.
S e c t i o n B B . Provided these sections are viewed along the cylinder
axis they produce the same circle as section AA.
S e c t i o n Cc. Sections parallel to the axis produce rectangular
flats, the width of the flat depending on the position of the section
relative to the axis. For the section shown, the width of the flat must
be found from the plan view. If, however, the section is parallel to
the horizontal centre line in the plan, the flat may be projected
directly to the elevation.
S e c t i o n D D . Such sections viewed normally produce ellipses, and
38
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
the construction becomes tedious if several have to be drawn.
Compare these sections with sections BB.
Hence, when it is necessary to choose the simplest section to take
through a right cylinder, the section should be either AA or BB.
O b liq u e c y lin d e r
Sections parallel to the base
shape and size as the base.
rectangles similar to section
example those at right angles
cated and should be avoided.
of an oblique cylinder have the same
Sections parallel to the axis produce
CC in Figure 1. Other sections, for
to the axis, are generally more compli-
R ig h t c o n e
Refer to Figure 2.
S e c t i o n E E . Sections at right angles to the axis give circles in the
adjacent circular view.
S e c t i o n F F . This section gives an ellipse in the adjacent view, and
is relatively complex to draw.
S e c t i o n GG. Any section through the apex of the cone produces a
triangular flat in an adjacent view.
S e c t i o n H H . A hyperbola results in the end view, which, like the
ellipse, is time-consuming to plot.
S e c t i o n JJ. The section is parallel to a generator giving a parabola
in the view and plan. Again, this takes time to draw.
It can be seen from the above analysis that if a number of sections
are required through a right cone, for ease of drawing the sections
should be taken at right angles to the axis, or through the apex .
•••
O b liq u e c o n e
As with the oblique cylinder, sections parallel to the base
oblique cone have the same shape as the base, but the size
section will decrease as the cutting plane approaches the
Sections through the apex result in triangles. Sections in
positions are usually complicated and too time-consuming
considered.
of an
of the
apex.
other
to be
E lb o w s
Refer to Figure 3.
S e c t i o n K K . This section produces a curved flat in the adjacent
view, the radii rand R being found as shown. These radii vary
40
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
depending on the distance from the cutting plane to the centre line
of the elbow.
S e c t i o n L L . Provided these sections are viewed normally they
appear as circles, all with the same diameter.
S e c t i o n M M . These sections need considerable construction and
should be avoided.
For interpenetrations involving elbows, section KK is generally
used. However, cases do arise which can only be solved by using
cutting cylinders instead of cutting planes.
Spheres
Any section through a sphere projects as a circle when viewed
normally. Other directions of viewing, except for edge views, result
in ellipses. Thus sections NN and PP are equally satisfactory for use
in interpenetration problems. See Figure 4.
Application of simple sections to interpenetration problems
The line of intersection between two solids is the line which is
common to both their surfaces. If a cutting plane is passed through
the solids in the area of the line of intersection, it will produce a
section on both solids. The points where the two sections meet lie on
the intersection line. This is illustrated in Figure 5 on page 43. The
process is repeated until sufficient points on the intersection line
have been plotted for it to be drawn accurately. The points may be
projected as necessary to complete other views on the drawing.
Ideally the cutting planes used should produce the simplest section on both solids. This cannot often be arranged, and generally
the simplest section results on one solid and a less simple section on
the other. Sometimes a choice of cutting planes is possible, each
having an advantage over the other. To illustrate this, two methods
of solution have been given for Examples 1 and 2.
Limits of the interpenetration curve
Before starting to solve an interpenetration problem the views
should be examined to determine the limits ofthe curve. These are,
for example, the highest and lowest points and the points where the
curve is tangential to the profiles of the intersecting solids. Sometimes the limits may be found by direct projection from view to
view, as in Examples 1 to 5, otherwise the limits may be found by
taking appropriate sections. Most of the following worked examples
show how this is done. When the limits have been fixed, the curve
42
In te r p e n e tr a tio n
o f s u r fa c e s
can be completed by taking additional sections between them. It is
unnecessary to space these additional sections closely except where
the interpenetration curve changes direction abruptly.
Using random sections, without first finding the limits of the
interpenetration curve, is time-wasting and often results in an inaccurate solution.
H id d e n
d e ta il
Unless they are made from sheet metal, interpenetrating solids
should be considered as one-piece castings and not as one solid
pushed into the other. This will reduce the hidden detail in the
solution, but all hidden parts of the interpenetration curve must be
shown.
C u ttin g s p h e r e m e th o d
Most interpenetration problems can be solved by using cutting
planes which produce simple sections. However, there are problems involving two solids of revolution whose axes are not at right
angles, for which cutting planes do not give simple sections on both
solids. See problem 32 on page 76. Problems of this type may be
solved by using cutting spheres instead of cutting planes.
The method, illustrated in Figure 7 on page 68, depends on the
fact that a sphere with its centre on the axis of a solid of revolution
cuts circles on the solid. In a view in which the axis appears true
length these circles will appear in edge view. Figures 7(a) and (b)
show such a sphere cutting a right cylinder and a right cone. In
Figure 7(c) the axes of the solids intersect and both appear true
length. Therefore, the circles cut on the solids by a sphere centred at
the intersection of the axes, both appear in edge view. The points
where the circles intersect are common to the cylinder and cone and
so lie on the interpenetration curve.
If the axis of one solid is not true length, the circles cut on that
solid will appear as ellipses. If the axes do not intersect, no spheres
can be drawn which cut edge view circles on both solids. In this case
a solution can be obtained using cutting cylinders, provided a view is
available showing both axes true length.
The cutting sphere method has the advantage that the solution is
completed using only one view. If other views are required they are
often those which show one set of circles cut by the sphere in true
shape. Points on the interpenetration curve will lie on these circles,
and can be found by direct projection from the first view.
53
In te r p e n e tr a tio n
o f s u r fa c e s
Examples 1,2 and 8, for which solutions are given using cutting
planes, can also be solved by the cutting sphere method.
I n t e r p e n e t r a t io n
p r o b le m s
Scale full size unless otherwise stated. Use the same projection
angle for the solutions as has been used for the given views. Hidden
detail must be shown.
1
Figure 1 shows a cylinder penetrated by a hexagonal prism.
Draw the given views, completing the elevation, and add an
end view looking in the direction of arrow R.
2
Figure 2 shows a cylinder penetrated by a square prism. Draw
the given views and complete the elevation with the interpenetration curve. Add an end view looking in the direction of
arrow S.
3
The plan and elevation of a right hexagonal prism penetrated
by an oblique cylinder are given in Figure 3. Draw the given
views, constructing the interpenetration curve in the elevation,
and project an end view in the direction of arrow T.
4
Figure 4 shows the plan and elevation of a cylinder penetrated
by a triangular prism. Draw the given views and complete the
elevation with the interpenetration curve. Add an end view
positioned on the left of the elevation.
5
Two interpenetrating oblique cylinders are shown in plan and
elevation in Figure 5. Draw these views and complete them
with the interpenetration curves.
6
The elevation of a milled cylinder is shown in Figure 6. Copy
this view and project an end view in the direction of arrow U.
7
Figure 7 shows an elevation and plan of the boss for a built-up
handwheel. The boss is to be drilled for four 50 mm diameter
equally spaced spokes, the centre line for one spoke being
given. Complete the given views with the interpenetration
curves produced by the hole for this spoke, but do not show the
spoke.
8
Complete the elevation given in Figure 8 of the cylinder pierced
by a conical hole, and project an end view looking in the
direction of arrow V.
69
In te r p e n e tr a tio n
o f s u r fa c e s
9
Figure 9 shows the end view of a cone pierced by a square hole.
Draw this view, project a front elevation looking in the direction of arrow W, and a plan.
10
A cone and cylinder interpenetration is shown in Figure 10.
Obtain the curves of interpenetration in the two given views.
11
An open-topped conical hopper penetrated by an inclined
circular pipe is shown in Figure 11. Draw the given views to a
scale of 1:50 and complete them with the interpenetration
curves. Use the cutting sphere method.
12
The plan and elevation of a cone and a vertical triangular prism
are given in Figure 12. Complete the elevation and project an
end view looking in the direction of arrow R.
13
Figure 13 shows the plan and elevation of a cone penetrated by
a vertical hexagonal prism. Complete the elevation and draw
an end view looking in the direction of arrow S.
14
The elevation and plan of two intersecting cones are given in
Figure 14. Draw and complete these views with the interpenetration curves, and project an end view looking in the
direction of arrow T.
15
Two views of the interpenetration of a cone and sphere are
shown in Figure 15. Complete these views with the interpenetration curves.
16
Figure 16 shows a sphere penetrated by an oblique cone, the
apex of the cone apparently being on the surface of the sphere
in the elevation. Complete the given views and add an end view
looking in the direction of arrow U.
17
Two views of an oblique cone penetrated by an inclined square
prism are shown in Figure 17. Plot the interpenetration curves
on both views.
18
The elevation and plan of a milled triangular pyramid are
shown in Figure 18. Complete the plan and add an end view
looking in the direction of arrow V.
19
An elevation and end view of a barrel-shaped solid of revolution penetrated by a cylinder are given in Figure 19. Construct
the curves of interpenetration for both views.
71
In te r p e n e tr a tio n
o f s u r fa c e s
20
An auxiliary elevation and plan of a hemisphere and interpenetrating cylinder are shown in Figure 20. Draw these views,
completing the plan, and project an elevation looking in the
direction of arrow R, and an end view placed on the right of the
elevation.
21
Views are given in Figure 21 of a hemisphere pierced by a
cylinder. Draw the given views, complete them with the interpenetration curves, and add an end view looking in the direction of arrow S.
22
Figure 22 shows the plan and incomplete elevation of a hemisphere pierced by a hexagonal hole. Draw these views and
complete the elevation.
23
A plan and auxiliary elevation of a hemisphere pierced by a
triangular hole are given in Figure 23. Draw these views,
project an elevation looking in the direction of arrow T, and an
end view placed on the right of the elevation.
24
Figure 24 shows the plan and elevation of a solid of revolution
penetrated by a cylinder. Draw these views and complete them
with the interpenetration curves.
25
A sheet metal hopper feeding into a reducing neck is shown in
Figure 25. Complete the given view with the interpenetration
curve without using any other view.
26
Two views of an elbow piece with a branch cylinder are shown
in Figure 26. Complete these views with the curves of interpenetration.
27
Construct the curve of interpenetration in the front elevation
of the elbow and branch cylinder shown in Figure 27.
28
Figure 28 shows incompletely a plan and elevation of an elbow
with an oblique conical branch. Construct the interpenetration
curves for both views, showing the complete branch in the
plan.
29
The sectional elevation of a control rod end is shown in Figure
29. Draw this view and project from it a plan.
30
A plan and incomplete elevation of an eccentrically turned
square bar are shown in Figure 30. Draw the given views and
complete the elevation with the interpenetration curves.
73
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
31
Figure 31 shows two views of a needle valve. The elevation
shows the valve after being turned, and the plan shows it after
being ground. Draw the given views and construct the interpenetration curves in the elevation produced by the grinding
operation. Scale 5:1.
32
A solid of revolution pierced by a cylinder is shown in Figure
32. Use the cutting sphere method to plot the interpenetration
curves in the given view and project a plan.
33
Half views of a connecting rod end are shown in Figure 33.
Draw these views and complete them with the interpenetration
curves produced by milling the pair of vertical flats, and by
spotfacing the 12 mm diameter holes. Add an end view positioned on the left of the elevation.
3 C o n ic s e c t io n s
Conic sections are produced when a plane intersects a right circular
cone. They may be considered as loci, which was done in Chapter 4
of Volume 1, or, as here, they may be considered as curves of
interpenetration.
The surface of a right circular cone can be produced by the
rotation of two straight lines which cross at some acute angle equal
to half the apex angle of the cone. One line, which is shown vertical
in Figure 1, is the axis; the other is the generator. Successive
positions of the generator as it sweeps out the conic surface are
called elements of the cone. The conic surface theoretically extends
to infinity in both directions from the apex but practically is limited
by planes at right angles to the axis. Each half of the double cone is
called a nappe.
Depending on the position of the cutting plane several sections
are possible. These are illustrated in Figures 1 to 5.
(a) C i r c l e (Figure 2). Produced when the plane is at right angles to
the axis and cuts all the elements on one side of the apex.
(b) E l l i p s e (Figure 3). Produced by a plane which cuts all the
elements on one side of the apex, but is not at right angles to
the axis.
(c) P a r a b o l a (Figure 4). Produced by a plane parallel to one
element of the cone.
(d) H y p e r b o l a (Figure 5). Produced by a plane cutting both
nappes but not passing through the apex.
If the plane cuts both nappes and passes through the apex two
isosceles triangles are produced.
T h e e llip s e
Figure 6 shows a right circular cone cut by a plane XX. This plane
produces an ellipse since it is not at right angles to the axis but cuts
all the elements on one side of the apex. The ellipse appears in true
76
77
C o n ic
s e c tio n s
size in an auxiliary view, the projectors for which are at right angles
to XX.
To draw the ellipse cutting planes such as AA are used. These are
at right angles to the axis and so produce circles as shown in the
plan. Where planes XX and AA intersect, the widths across them
are equal. This width a is found from the plan view and transferred
to the auxiliary view to plot two points on the ellipse. Additional
cutting planes parallel to AA give more points on the ellipse. Note
that the curve in the plan is also an ellipse. Thus any view of plane
XX, except an edge view, is an ellipse.
By using focal spheres the focal points pi and Pz and the directrices may be established. The focal spheres are inscribed in the
cone so that the conical surface and the cutting plane XX are
tangential to them. The centres of the focal spheres projected onto
the axis in the auxiliary view position the foci pi and PZ. The axis of
the ellipse is the projection in the auxiliary view of the axis of the
cone. The lines of intersection between XX and horizontal planes
through the tangent circles of the focal spheres and cone give the
directrices when projected into the auxiliary view.
To draw the tangent to the ellipse at any point P on the curve,
bisect the exterior angle between the focal radii pip and PZP. To
draw the normal, bisect the supplementary angle of the exterior
angle. The normal is perpendicular to the tangent at the point of
contact. This is true for all the conics.
The ratio of the distances focus to vertex and vertex to directrix is
called the eccentricity of the curve. For the ellipse the eccentricity is
less than unity. It is constant for any point on the curve.
T h e p a r a b o la
The cutting plane YY in Figure 7 is parallel to an element of the
cone so the conic section is a parabola. It is shown in true shape in an
auxiliary view projected from YY at right angles. The curve produced by YY in the plan is also a parabola, as is any view of YY,
except an edge view.
As with the ellipse, cutting planes at right angles to the axis are
used to draw the true shape view. The focus is the projection on the
axis of the centre of the focal sphere, which has the conical surface
and YY tangential to it. The directrix is the line of intersection
between YY and a horizontal plane through the tangent circle of the
focal sphere and cone, projected into the auxiliary view. The axis is
the projection of the cone axis in the auxiliary view.
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C o n ic s e c tio n s
Constructions to find the tangent and normal to the parabola at
any point P on the curve are also shown in Figure 7. The lines FP are
focal radii. The other lines through P parallel to the axis are diameters.
For the parabola the eccentricity is unity and is the ratio FV and
YD. It is constant for any point on the curve.
The hyperbola
The plane ZZ in Figure 8 cuts both nappes of the double cone and so
produces a hyperbola. The plane is shown parallel to the cone axis
but need not be, since any plane cutting both nappes without
passing through the apex will produce a hyperbola.
Cutting planes such as AA, at right angles to the cone axis, are
used as before to plot points on the true shape view of the hyperbola. The foci F I and P are the projections on the axis of the centres
of the focal spheres. These spheres have the conical surface and ZZ
tangential to them. The directrices are the intersection lines
between ZZ and planes through the tangent circles of the focal
spheres and cone. The asymptotes are the lines which would be
tangential to the curves at infinity. They are the projections of the
outer elements of the cone. To establish the asymptotes draw a
circle with centre C through the vertices VI and V 2• This circle cuts
the directrices at four points which lie on the asymptotes as shown.
As with the ellipse, bisecting the angles between focal radii at a
point on the curve enables the tangent and normal at that point to be
found.
The eccentricity of the hyperbola is greater than unity and is the
ratio PV l to VID 1 or PV 2 to V 2D 2• It is constant for any point on
the hyperbola.
V 1V 2, the distance between the vertices, is called the transverse
axis and is equal to the difference between the focal radii of a point.
Conic section problems
Scale full size unless otherwise stated. Use first or third angle
projection as required by the question. Show hidden detail where
specified.
1
A right cone with a base diameter of 100 mm and an apex angle
of 60° is cut by a plane inclined at 4 5 ° to the horizontal. The
plane cuts the cone axis 50 mm up from the base. Draw the
views of the cone as shown in Figure 6 on page 80, adding an
end view on the right of the front elevation of the part of the
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
cone below the plane. Establish the foci and directrices on the
true shape view of the ellipse. At any point P, other than the
vertex, draw the tangent and nonnal. Use third angle projection.
2
A right cone with the same dimensions as that in Question 1 is
cut by a plane parallel to an element. The plane cuts the base
32 mm to the right of the axis. Using first angle projection draw
the views of the cone as shown in Figure 7 on page 81. Add an
end view on the left of the front elevation of the part of the cone
below the plane. On the true shape view of the parabola
position the focus and directrix. At any point P, other than the
vertex, draw the tangent and normal.
3
A double right cone has base diameters of 100 mm and an apex
angle of 60°. Its axis is horizontal and the cone is cut by a plane
parallel to the axis and 20 mm above it. Draw the views of the
cone as shown in Figure 8 on page 82, using first angle projection. In the true shape view of the hyperbola establish the foci,
directrices and asymptotes. Draw the tangent and normal at
any point P, other than a vertex.
4
A flood light which projects a right circular conical beam is
mounted on a pylon 7 metres above ground level. The axis of
the beam makes an angle of 45° with the ground and the apex
angle of the beam is 60°. Using a scale of 1:10, determine the
shape of the illuminated ground area. Use third angle projection.
5
Using first angle projection, draw the elevation of the right
cone shown in the figure, and project a plan of the portion
below the plane AA. Add an end view of the lower portion in
the direction of arrow R.
6
The figure shows the elevation of part of a cone after being cut
by two planes. Draw this view and project from it a plan and an
end view in the direction of arrow S. Use third angle projection.
7
Copy the elevation of the part cone shown and draw a plan and
an auxiliary plan in the direction of arrow T to show the true
view of the sloping face which is parallel to an element. Use
first angle projection.
8
Draw the elevation of the cone given in the figure and project a
plan, using third angle projection.
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9
Copy the given elevation of the part cone shown and, using
third angle projection, add a plan.
10
The figure shows a view of a right conical hopper fed by an
inclined chute. The hopper and chute have open tops. Draw
the true cross-section of the chute using third angle projection.
11
Draw the given views of the conical spacer shown and complete
the elevation, using first angle projection.
12
Views of a conical stop are shown. Using first angle projection
draw these views, making the elevation a full view instead of a
half view. Complete the elevation and project a plan.
13
The figure shows views of a vee support. Draw these views in
first angle projection, making the plan a full view. Complete
the plan and add an end view. Show hidden detail.
14
The elevation of a casting is given in the figure. Draw this view
and project an end view looking on the 16 mm diameter boss.
Use third angle projection.
15
Views of the end of a turnbuckle are shown. Draw the views
and complete the elevation, using third angle projection.
16
The figure shows one view of a jig detail. Draw this view and
project from it, using third angle projection, a view in the
direction of arrow R.
17
The standard lamp shown in the figure illuminates part of the
wall and part of the sloping ceiling. Using first angle projection
draw views of the illuminated areas in the directions of arrows
A and B.
87
D e v e lo p m e n t
4 D e v e lo p m e n t
Development is a term used in sheet metal work and means the
unfolding or unrolling of a surface into a plane. The resulting shape
is called a pattern. Only ruled, single-curved surfaces and those
composed of planes can be developed exactly.
A ruled surface is generated by the motion of a straight line and
may be a plane, a single-curved surface or a warped surface. To
produce a single-curved surface the generating line always touches a
curve and moves so that any two successive positions either
intersect or are parallel. A conical surface is produced if successive
positions of the generating line intersect, the intersection point
being the vertex. If successive positions are parallel a cylindrical
surface results.
Warped surfaces comprise those ruled surfaces for which
successive positions of the generating line neither intersect nor are
parallel. A common example is the flank of a vee thread which is an
oblique helicoid. The flanks of a square thread are right helicoids.
In both cases the directing curves are helices. For a right helicoid the
generating line is perpendicular to the axis of the helix, whilst for an
oblique helicoid it is at a constant angle to the axis.
A double-curved surface is produced by the motion of a curved
generating line, as with the sphere, which may be formed by
rotating a semicircle about its diameter. Other examples are the
paraboloid, which is generated by revolving a parabola about its
axis, and the torus, which is formed by moving a circle in a circular
path about an axis outside the circle but in the same plane as the
circle. There are no straight lines on a double-curved surface.
Warped and double-curved surfaces are theoretically not
developable. However, they can be developed approximately by
dividing them into small sections which are developable surfaces.
Alternatively a plane may be pressed, stamped or otherwise
deformed into the required shape.
88
When the pattern for a sheet metal detail is laid out, extra
material must be allowed for bends, since sharp comers cannot be
produced because of the material thickness. Allowance must also
be made for seams or laps and for finishing raw edges by hems. If the
bases of the part are closed their true shapes will be needed. For
simplicity these aspects have been ignored in the following worked
examples.
Generally the shortest seam or joint line is used because this gives
the greatest economy in time and material in making the joint.
Sometimes, however, using the shortest seam may increase the area
of sheet metal needed, or may make it impossible to fit the pattern
on to a standard sized sheet.
There are three methods of pattern development: p a r a l l e l l i n e ,
r a d i a l l i n e and t r i a n g u l a t i o n .
These methods are shown pictorially in
Figure 1 on page 90. Before considering them in detail the student
must remember that all lines on a pattern should be true lengths or
the object made from the pattern will not be the correct size or shape.
Parallel line d e v e lo p m e n t
This method is used for objects which have a constant cross-section
throughout their length, for example, prisms and cylinders. Parallel
lines, parallel to the axis of the detail, are drawn on the surface in a
view which shows them as true lengths. These true lengths are
projected or transferred with dividers to the pattern so that they
remain true lengths. When the object is a prism, its comers will
serve as surface lines, as in Examples 1 to 4. The true distances
between the comers are also needed and these are available in a
view which shows the right cross-section of the prism. This is the
cross-section in a plane at right angles to the axis and comers. On
the right cross-section the prism comers appear as points. Thus the
pattern length, called the stretch-out or girth, is the perimeter of the
right cross-section.
When the object is cylindrical, as in Examples 5 to 9, surface lines
are obtained by dividing the circumference of the right cross-section
into a number of equal parts, often twelve for convenience. The
stretch-out, which is the circumference of the right cross-section, may
be found approximately by setting off the chordal distance a twelve
times along the stretch-out line. For a more accurate pattern calculate
the circumference of the right cross-section and divide it into twelve
parts geometrically. At each division draw a perpendicular representing the surface line and project or transfer its true length.
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For intersecting sheet metal components, such as those shown in
Examples 8 and 9, accurate interpenetration curves must be drawn
before the patterns can be laid out.
P a r a lle l lin e d e v e lo p m e n t a p p lie d to o b liq u e c y lin d e r s
Parallel line development can be used to develop oblique cylinders,
but before considering this the following definitions should be
noted, in conjunction with Figures 2 and 3 on page 99.
A right circular cylinder is one in which all sections at right angles
to the axis are circular. Thus, when dimensioned, its diameter is
shown at right angles to the axis.
An oblique circular cylinder is one in which the axis is inclined at
some angle to the plane of its circular base. Therefore its right
cross-section is elliptical. When dimensioned, its diameter is shown
at some angle to the axis.
It will be seen therefore, that oblique cylinders have a constant
cross-section for their full length and so can be developed by the
parallel line method. However, a slight modification is necessary
which will be apparent from the following explanation.
Referring to the solution to Example 10, divide the circular base
into twelve equal parts and draw surface lines in the elevation as for
a right cylinder. These surface lines will be true lengths and can be"
projected into the pattern. The stretch-out of the pattern will not
now be equal to the circumference of the circular base. Instead, the
circumference will equal the length of the curved edge in the pattern.
Project the ends of the surface lines into the pattern with projectors
at right angles to them, and position the joint line in the pattern.
From the ends of the joint line strike arcs a , equal to one-twelfth of
the circumference of the circular base, to cut the projectors from the
surface line nearest the joint line. This will fix two points, one on
each edge of the pattern. From these points again strike arcs a to cut
the projectors from the next surface line, and so on. This procedure
ensures that the length of the curved pattern edges equals the
circumference ofthe circular base ofthe cylinder. It should be noted
that the distances between surface lines in the pattern are not equal.
An alternative method is to use the elliptical right cross-section of
the oblique cylinder, and this is shown in part in the solution to
Example 11. The right cross-section will then develop as a straight
line perpendicular to the surface lines and equal in length to the
circumference of the ellipse. The surface lines can be positioned
along this stretch-out line by chordal distances taken from the right
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
cross-section view, but note that if the surface lines are equally
spaced round the circular base they will not be equally spaced round
the ellipse.
Radial line d e v e lo p m e n t
This method is used for objects which taper to an apex, such as
cones and pyramids. For cones radial lines are drawn on the surface
from the apex to the base. For pyramids the sloping edges are used
as surface lines.
R ig h t c o n e s
The development of any right cone is a sector of a circle since the
radial surface lines are all the same true length. The angle at the
centre of the sector depends on the base radius of the cone and the
slant height.
Referring to the solution to Example 14, let the radius of the base
of the cone be r, the slant height be R and the angle at the centre of
the pattern 9. Then the ratio of 9 to 360 equals the ratio of the
length of the sector arc to the circumference of a circle of radius R.
0
0
For a complete cone, shown in Example 14, or a frustum, shown in
Example 15, the pattern can be laid out using 9 and the slant heights
Rand Rl only. That is, surface lines need not be drawn in the
orthographic views or in the pattern. The solution to Example 15
shows that the pattern for a frustum is a sector with a smaller sector
removed from it. However, for truncated cones, such as those in
Examples 16 and 17, surface lines are needed. In such cases, instead
of calculating 9, it is more convenient to draw in the pattern the arc
representing the base of the cone, and to step off round it twelve
chords a each equal to the chord of one-twelfth of the base circumference of the cone. This method fixes the size of the pattern and
positions the surface lines in one operation and avoids having to
divide an awkward angle into twelve parts. The size ofthe pattern is
not quite as accurate as that obtained by using 9 since the chord a is
shorter than the arc of one-twelfth of the base circumference, but
this is offset by the convenience of the method.
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2
The student should realize that the only surface lines in the
orthographic views of the cone which are true lengths are the
outside elements in the elevation. So, when using the other radial
lines to obtain the truncated edges in the pattern, their ends are
projected to the outside of the cone to find the true lengths. This is
illustrated in the solutions to Examples 16 and 17.
It is sometimes necessary to use radial lines additional to the
twelve equally spaced ones, as in the solution to Example 18. These
are drawn as required in the elevation and the positions x, y and z of
their ends are found in the plan by projection. Points x, y and z are
plotted in the pattern using arcs 2-x, 3-y and 4-z transferred from
the plan.
When developing any cone or pyramid always layout the pattern
for the complete object before plotting the edges produced by
truncations or other cuts.
O b liq u e
cones
An oblique cone has its axis at some angle to the plane of its base
and so radial surface lines are not all the same true length. Therefore the development is not a sector of a circle. The surface lines
divide the cone into approximate triangles and the development
consists of finding the true lengths of the sides of the triangles and
then laying out the triangles side by side in order.
Referring to the solution to Example 20, the plan of the circular
base is divided into twelve equal parts which are used to position
radial surface lines in the elevation and plan. As with a right cone
only the outside elements in the elevation, A--{) and A-6, appear i~
true length. The true lengths of h~e remainder are found by rotating
their plans about the apex untIl they meet the vertical plane and
joining these points to the apex in the elevation. Note that to avoid
confusion between surface lines in the elevation and true lengths,
the true length diagram is constructed outside the elevation.
Start the pattern by positioning A-6, the surface line in the
centre. From 6 strike arcs a , equal to the chord of one-twelfth of the
base circumference, on both sides of A-6. Intersect these arcs with
arcs struck from A, equal to the true length of A-5. This completes
the two triangles A56. R~
p eat this proce.dure for the remaining
triangles until the pattern IScompleted. Smce the pattern is symmetrical about A-6, it is quicker to work outwards from the centre
rather than to begin at the seam.
It will be realized that this method ensures that the length round
106
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
the curved edge of the pattern equals the circumference of the
circular base of the cone. Compare this with the development of
oblique cylinders in Examples 8 to 13.
When an oblique cone is truncated, as in Example 21, the method
for obtaining the true lengths of parts of the surface lines is similar to
that used for a right cone. The end of each part surface line in the
elevation is projected parallel to the base of the cone to the a p p r o p r i a t e true length line in the true length diagram.
Additional surface lines may be employed where required exactly
as for a right cone, such as lines A-x in the solutions to Examples 22
and 23.
Although the worked examples above are of oblique circular
cones, the same method may be used if the base is elliptical.
P y r a m id s
The lateral surfaces of a pyramid are triangular and the pattern
consists of these triangles drawn in order, side by side, and in true
shape. Thus the general method which is used for cones can be
applied to pyramids.
Example 24 is a right pyramid and, therefore, all the sloping
edges are the same true length. To find this true length, rotate the
plan of a sloping edge h as 02b 2 until it is parallel to the vertical
plane. Project to the elevation and join to the vertex in the same
way as for finding the true length of a surface line on an oblique
cone. The true lengths of the part sloping edges are found by
projecting their ends in the elevation to the true length line. The
true length of the seam ON is available directly in the elevation and
the true lengths of the base edges appear in the plan.
Start the pattern by laying off the true length of ON in a convenient position. Complete the half side ONA by drawing an arc
n 2 a 2centred at N, to cut an arc equal to the true length ofthe sloping
edges, centred at O. This arc may be extended round the pattern as
all the sloping edges have this true length. Fix the points D, C and B
on this arc by stepping round it a 2 d 2 ,d 2 c 2and c 2 b 2 .Join D, C and B
to 0 and add the other half side ONB to give the pattern for the
complete pyramid. Mark off on the bending lines the appropriate
true lengths to fix M, P, S, Rand Q on the cut-out.
This pattern can also be drawn by beginning at the centre and
working outwards in both directions.
A right triangular pyramid is shown in Example 25. In this case
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the true length of the sloping edge OC differs from the true length of
the other edges OA and OB. It may be found by rotating its plan
02C 2as before, or by projecting an end view, when 03C 3is the true
length. The end view will also enable S2to be positioned. Rotation
of 02b 2 or 02a 2 and projection to the elevation constructs their true
length. The pattern is drawn in a similar way to the previous
example but note that the pattern for the full pyramid is laid out
before attempting to cut off the base.
Example 26 shows an oblique rectangular pyramid. Here the true
lengths of the sloping edges are all different and each must be found
by rotating the plan length and projecting to the elevation. Example
27 is similar, but since 02a 2 and 02b 2 are parallel to the vertical plane
their true lengths appear directly in the elevation.
D e v e lo p m e n t
b y t r ia n g u la t io n
this method of development the surface of the object is divided
into a number of triangles. The true sizes of the triangles are found
and they are drawn in order, side by side, to produce the pattern. It
will be apparent that to find the true sizes of the triangles it is first
necessary to find the true lengths of their sides. Figures 4 and 5 on
page 116 show how this is done.
Figure 4 shows a rectangle to round transition piece having the
top and bottom edges in planes parallel to the horizontal plane. This
is the simplest case for a triangulation problem. Two surface lines
AC and DE are shown in plan and elevation. Their true lengths are
found by placing their plan lengths at right angles to their vertical
heights. Then the hypotenuses of the right angled triangles so
formed are the true lengths of the surface lines.
This method is essentially the same as that used to find the true
lengths of surface lines on pyramids and oblique cones. However, if
plan lengths are rotated and projected to the elevation in triangulation problems, the many construction lines used make the drawing
very diffiicult to follow and may lead to mistakes being made. Also
the true lengths often appear on the elevation and are difficult to
distinguish from the original surface lines. For these reasons a
separate true length diagram is preferred.
Since in Figure 4 the top and bottom edges of the transition are
parallel to the horizontal plane, distances on these edges may be
taken directly from the plan, as in this view they are true lengths.
Using these lengths and the true lengths of the surface lines from the
In
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E n g in e e r in g
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2
true length diagram, the triangles into which the object is divided
may be laid out in order and the pattern obtained.
When, as in Figure 5, the object does not lie between planes
parallel to the horizontal plane, the procedure is similar to that
outlined above. Now, however, the distances ST, TU, UV, etc., on
the top of the object are not true lengths in either view. Their true
lengths must be found in the same way that the true lengths of the
surface lines are found. Note too, that the vertical heights of the
surface lines now vary. It is convenient to construct the true lengths
of the divisions of the circular edge and the true lengths of the
surface lines on opposite sides of the vertical reference line in the
true length diagram. Reference to Figure 5 will make these points
clear.
When dividing the surface of an object into triangles prior to
developing it, care should be taken so that surface lines lie as near as
possible on the surface of the object and are straight. For example,
in Figure 5 a line from A to T could only lie on the surface if it were
curved. If it were straight it would only touch the surface at its ends.
P a n e l d e v e lo p m e n t
Example 32 shows a sheet metal detail composed of eight similar
panels. The development of one panel may be obtained as follows.
Draw an elevation and plan, positioning the detail so that the
horizontal edges of one panel are parallel to the vertical plane.
Obtain the profile of this panel in the elevation by taking a series of
horizontal sections such as XX. Project the width waf the panel at
XX back to the elevation from the plan. Width w is a true length.
The stretch-out of the panel is the length of the profile of the
elevation.
Start the pattern by drawing a centre line and mark off along it
lengths such as a and b from the profile of the elevation. These
lengths are the distances between successive horizontal sections.
Through the points so obtained on the pattern centre line draw
horizontals and mark off on them the appropriate panel widths.
Complete the panel by lining in the profile.
Example 33 is similar to the previous example but now horizontal
sections such as YY produce hexagons in the plan. The pattern is
drawn in the same way as for Example 32 except that an end view is
projected by transferring widths such as t from the sections in the
plan.
120
D e v e lo p m e n t
S p h e r e d e v e lo p m e n t
A sphere is a double-curved surface and is therefore non-developable. However, an approximate development may be obtained by
dividing the spherical surface into either gores or zones. A gore is
that part of the spherical surface which is contained between two
planes which pass through the axis. A zone lies between two planes
which are at right angles to the axis.
When gores are used cylindrical surfaces are substituted for the
spherical surfaces of the gores, hence the term polycylindric
method. This is shown in Figure 6 on page 124. The polyconic
method, shown in Figure 7(page 125), is so named because the
zones are replaced by conic frusta.
The surface of any solid of revolution may be developed by either
of these methods. Examination of the solutions to Examples 32 and
33 will show that the panels are essentially gores.
D e v e lo p m e n t
p r o b le m s
Develop the details full size except where a different scale is shown
on the drawing. Make no allowance for metal thickness, seams or
hems. Give full patterns unless otherwise stated.
1-4
These are prisms, open at both ends.
5
Develop the vertical and horizontal pipes of the smoke
cowl and one of the inclined pipes.
6
Develop both 32 mm diameter pipes and show the hole
shape required in the pattern for the square duct. The
complete pattern for the square duct is not required.
Complete the elevation with the interpenetration
curves.
7
Layout the patterns for both branch pipes and construct
the hole shapes required in the pattern for the main
pipe. Do not give a full pattern for the main pipe. Show
in the plan the interpenetration curve for the 375 mm
diameter branch.
8
Using 30° segments construct a lobster-back bend to
connect the two given pipes with (a) right cylindrical
segments (b) oblique cylindrical segments. Develop one
segment in each case. Refer to Figures 2 and 3 on page
99.
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
9, 10
These are combinations of right cylinders and prisms,
open at both ends.
11-14
These are oblique cylinders.
In problem 12 develop the gusset piece A and the
horizontal branch pipe but show only the shape required
in the pattern for the vertical branch.
15-19
These are right cones, open at both ends.
In problem 17 the 30 mm diameter hole passes
through one side only of the cone.
In problem 18 obtain the interpenetration curve between the hopper and the discharge pipe and develop
both components.
In problem 19 find the interpenetration curve between the hopper and the rectangular trunking and lay
out the pattern for the hopper.
20-23
These are oblique cones, open at both ends.
In problem 22 layout the developed hole shape in the
80 mm diameter pipe as well as constructing the pattern
for the cone.
In problem 23 develop the cone and the 76 mm diameter cylinder.
24-28
These are right and oblique pyramids, open at both
ends.
29-38
These are transition pieces to be developed by triangulation.
In problem 34 the area of the circular end is to be the
same as that of the rectangular end. Calculate the diameter D and layout the pattern.
39,40
These are examples of details built up from panels.
Problem 39 is an exhaust flare made from six similar
panels. Develop one panel.
Problem 40 is a bowl made from eight panels and all
horizontal sections are octagons. Develop panels A, B
and C.
128
5
C am s
A cam is a device having a profile or groove machined on it, which
gives to a 'follower' an irregular or special motion. The type of
follower and the motion required of it decide the shape of the profile
or groove.
Types of earn
Cams fall into two main classes: radial, disc, edge or plate cams, and
cylindrical cams. The follower working with a radial cam reciprocates or oscillates in a plane at right angles to the carn axis. With a
cylindrical cam the follower moves in a plane parallel to the carn
axIS.
The left-hand drawing in Figure l(a) shows the most elementary
type of cam. A block with a sloping top face has a follower, which is
a knife edged rod, resting on it. When the block is reciprocated in a
horizontal plane, the follower is caused to reciprocate in its guide in
a vertical plane. In the centre drawing (Figure 1(a) ) the cam rotates
about a centre, usually at constant speed, and the follower reciprocates up and down. In the right-hand drawing the follower is pivoted
at one end and oscillates about the pivot as the cam rotates. These
are all examples of radial cams. Occasionally the carn may oscillate
instead of rotating.
A cylindrical cam is illustrated in Figure l(b). Here a groove is cut
in the surface of the cylinder, which reciprocates the follower as the
cylinder rotates. Sometimes a radial follower, oscillating about a
pivot, is used. Again, the cam may oscillate instead of rotating.
Instead of cutting a groove in a cylinder, the end may be
machined to a special form when the cam is called an end carn. In
this case thrust is applied to the follower in only one direction,
whereas the groove applies it in both directions.
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
Types of follower
These are illustrated in Figure 2. The simplest type is one having a
knife edge or point which works on the profile of the cam. It is not
often used as it wears rapidly but it has the advantage that any form
can be given to the cam profile.
With the roller follower the rate of wear is reduced, but the
profile of the carn must not have any concave portions with a radius
smaller than the roller radius.
The flat follower is sometimes used but here the carn profile must
have no concave portions. All three types of follower may have
their axes offset from the cam axis.
Followers to work with radial cams are generally spring loaded to
keep them always in contact with the cam profile.
F o llo w e r m o tio n s
When studying the motions which are given to a follower by the
profile of a radial carn, it is convenient to use a cam graph or
displacement diagram in which the rise and fall of the follower is
plotted against the angular displacement of the carn. The 360°
rotation of the carn is laid off in 15° steps along the horizontal axis,
to a convenient scale, and the rise and fall of the follower is shown
on the vertical axis, full size.
There are three motions which may be given to the follower:
u n i f o r m v e l o c i t y , s i m p l e h a r m o n i c m o t i o n and u n i f o r m a c c e l e r a t i o n
a n d r e ta r d a tio n .
For the follower to have uniform velocity, it must move through
equal increments of rise or fall as the cam turns through equal
angles. Thus the graph is a straight line, as shown in Figure 3(a). It is
usual in practice to smooth out the abrupt changes of direction at
each end of the motion by radii. If this is not done large accelerations are given to the follower which tends to jump on the carn
profile.
Simple harmonic motion when plotted on the cam graph is a sine
curve and may be drawn as in Figure 3(b). A semicircle is drawn on
the rise or fall of the follower and divided into a number of equal
parts. The cam displacement is divided into the same number of
equal parts and at these points verticals are erected. Horizontal
lines drawn through the points on the semi-circle to cut these
verticals give points on the graph.
Figure 3(c) shows the construction for a curve of uniform
acceleration and retardation. In this case the curve is parabolic in
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
form. The angular displacement of the cam and the rise or fall of the
follower are each divided into the same even number of equal
divisions. During the first half of the motion the follower is accelerated uniformly; during the second half it is retarded. The construction, which is the circumscribing rectangle method of drawing a
parabola, should be clear from the figure.
In the following worked examples it has been assumed that the
follower is the part which is to be given the special motion by the
cam. This is not always the case. A third part of the mechanism may
require the special motion, the follower being only an intermediary
between this part and the cam. If this is so, the motion required by
the third part should be plotted on the line along which it moves and
transferred to the cam profile through the follower. Problems 10, 16
and 17 on pages 150 and 153 are illustrations of this point.
To construct
a r a d ia l carn p r o f ile
The method for drawing the profile of a radial carn to operate a
point or knife edged follower is shown on page 137. The same
general method is used for all radial cams but is varied by the type of
follower and its position relative to the cam axis.
First position the cam centre and the nearest approach of the
follower to it. From this point layoff the maximum lift of the
follower. Draw circles through these two points about the cam
centre and divide them into 15° divisions. The circle through the
lowest position of the follower is called the base circle. Construct
the cam graph, positioning the angular displacement axis on a level
with the lowest position of the follower end. Now imagine the cam
to be held stationary and the follower to be moved round it. To
obtain the same motion as with the carn rotating, the follower must
be moved in the opposite direction to the cam rotation. When the
follower has moved through 15° its height will be given by the point
on the cam graph at the 15° station. Project this point to the line of
action of the follower, which in this case is the vertical centre line of
the cam, and swing it round the cam centre with compasses to cut
the 15° line on the cam. Succeeding points are obtained in the same
way. Where the follower remains at the same height, or 'dwells', the
cam radius is constant. When points on all the 15° radial lines have
been plotted, a smooth curve through them will be the cam profile.
Complete the carn with an arrow showing the direction of rotation.
In the worked examples 15° angular displacement divisions have
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
been used, and these usually give a sufficiently accurate profile.
However, cams for automatic machine tools are divided into 100
parts.
The construction of a radial cam to work with a flat follower is
shown on page 139. As before, points on the cam profile are
positioned on the radial displacement lines by projection from the
cam graph. These points lie on the face of the flat follower. Lines
representing the follower face are drawn through them at right
angles to the angular displacement lines. The cam profile must be
tangential to these lines.
It should be noticed that the point of contact between the follower face and the cam profile is in most cases a considerable
distance from the angular displacement line. This point must be
borne in mind when deciding the size of the follower face.
The example on page 140 is a plate cam to operate a roller
follower. Here points projected from the cam graph to the radial
displacement lines on the cam are positions of the centre of the
roller follower. The roller must be drawn at each ofthese positions,
the cam profile being tangential to the rollers. The line through the
roller centres is the pitch curve of the cam.
Note, that in examples involving roller followers the angular
displacement axis of the cam graph is positioned on a level with the
start position of the roller centre. This ensures that lift points on the
cam graph can be projected directly to the cam.
The previous examples of cam design have all had in-line followers, that is, the line of action of the follower has been vertically
above the cam centre. Sometimes, due to lack of space for the
mechanism, this cannot be arranged and the line of action must be
offset to one side. The cam with a roller follower on page 141 is an
example of this.
First fix the highest and lowest positions of the roller relative to
the cam centre. Draw the offset circle, that is, the circle having the
line of action of the follower tangential to it, and divide it into 15°
divisions. At each division draw a tangent, and number them opposite to the cam rotation. These tangents represent successive positions of the line of action as it is moved round the cam. Construct the
cam graph, placing the angular displacement axis on the same level
as the initial position of the roller centre. Project points on the cam
graph to the line of action and then swing them round the cam centre
to the appropriate tangent. As in the previous example these points
on the tangents are positions of the roller centre. The roller is drawn
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E n g in e e r in g
D r a w in g
w ith
W orked
E x a m p le s
2
at each position and the cam profile drawn tangential to the rollers.
I n some applications, for example on automatic machine tools, a
radial arm or oscillating follower is more suitable than a reciprocating follower. A radial arm follower is pivoted at one end. The
other end, which usually carries a roller, works on the cam profile.
The method for constructing the cam profile in this case is shown on
page 143.
Set out the cam centre, pivot centre and extreme positions of the
roller centre. Draw circles about the cam centre through the pivot
centre and the roller centres. Divide the circle through the pivot
into 1 5 ° divisions, starting at the pivot, and number them opposite
to the cam rotation. If the cam is held stationary and the follower
moved round it, the pivot centre will move to each ofthese points in
turn. So, with these points as centres, draw arcs with a radius equal
to the length of the follower arm. The follower is to move outward
through 6 0 ° with uniform angular velocity during the first half
revolution of the cam, that is, the first twelve 1 5 ° divisions. Therefore, divide the 6 0 ° movement of the follower into twelve equal
parts. Project these points to the appropriate follower arc, thus
obtaining the first twelve positions of the roller centre. Positions of
the roller centre during the return motion of the follower are
obtained in the same way. Draw the roller at each position of the
centre and draw the cam profile tangential to the rollers.
T o d r a w t h e c a r n g r a p h f o r a g iv e n c a r n p r o f ile
This is the reverse of the previous examples. The example on page
144 illustrates the method for an offset roller follower. The method
for in-line followers of other types can easily be determined by the
student.
Start by drawing the cam profile and plot round it the pitch curve.
Since this is the locus of the roller centre it will be parallel to the
profile and the roller radius away from it. Position the line of action
of the follower and draw the offset circle with the line of action
tangential to it. Divide the offset circle into 1 5 ° divisions and at each
draw a tangent, numbering them opposite to the cam rotation.
Where the tangents cut the pitch curve are positions of the roller
centre. Swing these points round the cam centre to the line of
actionof the follower and project them into the cam graph to the
appropriate displacement line.
The cam graph may have a time base as shown, instead of an
angular displacement base. To determine the time base calculate
142
C am s
the time taken by the cam to make one revolution and divide it by
twenty-four. This gives the time for the cam to turn through 15°,
enabling the graph to be related to the roller centre positions on the
cam.
Cylindrical cams
The drawing of a cylindrical cam, illustrated on page 146, requires a
development of the cam surface. From this development a template
is made which is used in the manufacture of the cam.
Start by drawing an elevation and end view of the cam blank,
dividing both into a number of equal parts. Twelve have been used
in the illustration for clarity but the motion of the follower may
require 15° divisions or others. From the elevation project the
development of the cam surface and divide it into the same number
of equal parts as the elevation and end view. Layoff on the development the extreme positions of the centre of the roller follower and
plot its motion between them in the same way that motions are
plotted on a plate cam graph. Draw the roller at each point where
the curve crosses an angular displacement line, and draw the groove
profile tangential to the rollers. The elevation can then be completed by projection from the development.
A roller follower is commonly used to work with a cylindrical cam
but the roller must be conical or it will slip on the groove profile.
This happens because points on the outside of the cam move faster
than points at the bottom of the groove. For simplicity a cylindrical
roller has been assumed for the cam in the worked example and in
the problems which follow.
Cam problems
All solutions are to be drawn full size.
1
A plate cam rotating clockwise is to give an in-line point
follower the following motion.
0°-120° lift 32 mm with uniform velocity
120°-180° dwell
180°-360° fall 32 mm with simple harmonic motion. Draw the
cam profile if the minimum cam radius is 38 mm and the
camshaft diameter is 24 mm.
2
Draw the profile of a radial cam to operate an in-line knife edge
follower, the follower motion being:
0°_ 60° dwell
145
C am s
60°-180° lift 38 mm with uniform acceleration and retardation
180°-240° dwell
240°-360° fall 38 mm with simple harmonic motion
The cam rotates anti-clockwise, the least thickness of metal
round the cam centre is 32 mm and the camshaft diameter is
22mm.
3
An in-line flat follower is to be given the motion described
below by a plate cam rotating anti-clockwise. The nearest
approach of the follower to the cam centre is to be 38 mm and
the camshaft diameter 24 mm. Set out the profile of the cam.
0°-150° lift 44 mm with uniform acceleration and retardation
150°-210° dwell
210°-270° fall 20 mm with uniform velocity
270°-360° fall 24 mm with simple harmonic motion
4
Layout the profile of a radial cam to give an in-line roller
follower, 16 mm diameter, the following motion:
0°_ 90° dwell
90°-180° lift 38 mm with simple harmonic motion
180°-270° dwell
270°-360° fall 38 mm with uniform velocity
Cam rotation clockwise, minimum cam radius 56 mm, camshaft diameter 24 mm.
5
A plate cam rotating anti-clockwise at 10 rev/min is to give an
in-line roller follower the following motion in one revolution.
Lift 38 mm with simple harmonic motion in 1.5 seconds
Dwell for 1 second
Fall 20 mm with uniform velocity in 1.25 seconds
Fall 18 mm with simple harmonic motion in 1.5 seconds
Dwell for the remainder of the revolution
Draw the cam profile making the minimum distance from the
cam centre to the roller centre 50 mm, the roller diameter
20 mm and the camshaft diameter 32 mm. Use a scale of 20 mm
to 1 second on the horizontal axis of the cam graph.
6
An offset roller follower, 20 mm diameter, is to be given the
motion set out below by a plate cam rotating clockwise. Construct the profile of the cam.
Lift 38 mm with uniform velocity in 120° of cam rotation
Dwell for 60° of cam rotation
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
Fall 38 mm with simple harmonic motion in 90° of cam
rotation
Dwell for the remainder of the revolution
The line of action of the follower is offset 32 mm to the left of
the vertical centre line of the cam. The minimum distance from
the cam centre to the roller centre is 56 mm. The camshaft
diameter is 24 mm.
7
Draw the profile of a plate cam rotating clockwise which gives
an 18 mm diameter offset roller follower the following motion.
0°_120° lift 24 mm with uniform acceleration and retardation
120°-180° dwell
180°-240° lift 12 mm with uniform velocity
240°-360° fall 36 mm with simple harmonic motion
The minimum distance from cam centre to roller centre is
50 mm, the camshaft diameter is 20 mm and the line of action of
the follower is offset 24 mm to the right of the vertical centre
line of the earn.
8
A radial carn rotating anti-clockwise is to operate an offset flat
follower giving it the following motion.
0°-180° lift 36 mm with uniform acceleration and retardation
180°-270° dwell
270°-360° fall 36 mm with simple harmonic motion
Construct the cam profile if the follower is offset 22 mm to the
right of the vertical centre line of the carn. The nearest
approach of the follower to the horizontal centre line of the
cam is 40 mm and the camshaft diameter is 32 mm.
9
Figure 1 on page 149 shows a radial arm roller follower which is
to be operated by a plate carn rotating clockwise. The motion
given to the follower is to be:
Dwell for first 45° of cam rotation
Move outward through 30° with uniform angular velocity
during the next 90° of carn rotation
Dwell for the next 45° of carn rotation
Move inward through 15° with uniform angular velocity
during next 60° of cam rotation
Dwell for the next 60° of cam rotation
Move inward through 15° with uniform angular velocity
during the remainder of the revolution of the cam
Construct the cam profile.
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
10
The guide block in the mechanism shown in Figure 2 on page
149 is reciprocated along the line AB by a plate cam rotating
clockwise. The carn operates a radial arm roller follower which
is linked to the guide block as shown. The block is to move 54
mm to the left with simple harmonic motion while the cam
rotates through 90°. It is then to remain stationary for 180° of
cam rotation, returning to its starting position with simple
harmonic motion during the final 90° rotation of the cam. If the
starting position is as shown, construct the cam profile.
11
Figure 3 on page 149 shows the profile of a plate carn to operate
an in-line roller follower. If the cam rotates clockwise determine the displacement curve for the follower on an angular
base, using a scale of 6 mm to 15°. Assume the starting position
of the carn to be shown.
12
A plate cam to operate an offset 24 mm diameter roller follower is shown in Figure 4 on page 149. If the line of action of
the follower is 24 mm to the right of the vertical centre line of
the cam, draw the cam displacement curve on a time base when
the cam makes 62.5 rev/min. 1 5 ° angular divisions of cam
rotation must be used with a scale of 6 mm to one-twenty-fifth
of a second on the time axis of the displacement curve. The cam
rotates anti-clockwise.
13
The displacement curve for a radial cam to rotate clockwise at
40 rev/min is shown incompletely in Figure 5 on page 151.
Complete the graph to the given scale and construct the carn
profile if the roller follower is 20 mm diameter with its line of
action offset 32 mm to the left of the vertical centre line of the
carn. The nearest approach of the roller centre to the carn
centre is to be 40 mm and the camshaft diameter is to be 25 mm.
14
Figure 6 on page 151 shows a blank for a cylindrical carn.
Complete the given view adding a development of the cam
surface, if the follower moves 76 mm to the right with simple
harmonic motion during one-third of a revolution of the carn,
dwells for the next one-sixth of a revolution, and returns to the
start position with simple harmonic motion during the remainder of the revolution. The cam rotates clockwise when
viewed from the left of the given view.
15
The cylindrical carn shown in Figure 7 on page 151 is to move
the follower 64 mm to the left with simple harmonic motion
150
Cams
during the first half revolution of the cam, and return it to the
start position with uniform acceleration and retardation during
the second half revolution. Draw the cam with a development
of its surface, if the rotation is clockwise when viewed from the
left of the given view.
16
Figure 8 on page 152 shows two blocks, A and B, which are to
be reciprocated by an oscillating plate cam and a system of
levers. If the initial position of the system is as shown draw the
cam profile from the following information.
Block A is to move 48 mm to the left with simple harmonic
motion, and block B is to move 64 mm to the left with uniform
velocity while the cam rotates anti-clockwise through 120°.The
blocks are to return to their starting positions with the same
motions while the cam rotates clockwise through 120°.
The non-working portions of the cam profile may be shown
as circular arcs about the cam centre, blending into the working
portions with 6 mm fillet radii, as shown.
17
A plate carn rotating anti-clockwise is to move the tappet in
Figure 9 (page 152) 36 mm to the left with uniform acceleration
and retardation in half a revolution. In the second half revolution the tappet is to return to its start position, again with
uniform acceleration and retardation. If the start position is as
shown, layout the cam profile.
153
6 I n v o lu t e g e a r s
gears
Imagine two cylinders in contact at a common point on their circumferences. If one cylinder is turned on its axis the other will be
turned by the friction between the cylinders at the point of contact.
The driven cylinder will rotate in the opposite direction to the
driving cylinder and the ratio of the angular velocities will equal the
inverse ratio of the diameters. It is obvious that very little power can
be transmitted by this system before the cylinders slip on each
other. To avoid this slipping, slots are cut in the cylinders with
projections added between them. These slots and projections form
teeth and the cylinders become spur gears. To keep the ratio of the
angular velocities constant the teeth must have profiles of either
cycloidal or involute form. The involute form is much commoner
than the cycloidal, mainly because it is easier to manufacture. Only
involute gears will be considered here.
Spur
I n v o lu t e
c o n s t r u c t io n
Figure 1 shows the method for constructing an involute to a circle.
The involute is generated by rolling a straight line round the circle,
when points on the line will trace out involutes. Draw the circle,
which is called the base circle, and divide its circumference into a
number of equal parts. Through the points so obtained on the
circumference draw tangents. Mark off on these tangents the
lengths of corresponding arcs on the base circle. The ends of the
tangents will then be points on the involute.
Figure 2 shows the effect on the involute of varying the base circle
diameter. If it is increased the involute becomes straighter, until
when the base circle diameter is infinitely large the involute is a
straight line. Thus an involute rack, which may be thought of as a
gear with an infinitely large base circle diameter, has straight sided
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
teeth. This property of an involute rack is made use of in the cutting
of involute spur gear teeth on hobbing machines, the Sunderland
gear planer and the Maag grinder.
I n v o lu t e
sp u r g e a r te r m s
The names of the various parts and dimensions of spur gears and
their teeth are illustrated in Figure 3.
The p i t c h c i r c l e is the circle representing the original cylinder
which transmitted motion by friction, and its diameter the p i t c h
c ir c le d ia m e te r .
The c e n t r e d i s t a n c e of a pair of meshing spur gears is the sum of
their pitch circle radii. One of the advantages of the involute system
is that small variations in the centre distance do not affect the
correct working of the gears.
The p i t c h p o i n t is the point of contact between the pitch circles of
two gears in mesh.
The l i n e o f a c t i o n . Contact between the teeth of meshing gears
takes place along a line tangential to the two base circles. This line
passes through the pitch point and is called the l i n e o f a c t i o n .
The p r e s s u r e a n g l e . The angle between the line of action and the
common tangent to the pitch circles at the pitch point is the p r e s s u r e
a n g l e or a n g l e o f o b l i q u i t y .
The
circle.
addendum
is the radial height of a tooth above the pitch
The d e d e n d u m is the radial depth below the pitch circle.
The c l e a r a n c e is the difference between the addendum and the
dedendum.
The w h o l e
dedendum.
d e p th
of a tooth is the sum of the addendum and
The w o r k i n g d e p t h of a tooth is the maximum depth that the tooth
extends into the tooth space of a mating gear. It is the sum of the
addenda of the gears.
The a d d e n d u m c i r c l e is that which contains the tops of the teeth
and its diameter is the o u t s i d e or b l a n k d i a m e t e r .
The d e d e n d u m or r o o t c i r c l e is that which contains the bottoms of
the tooth spaces and its diameter is the r o o t d i a m e t e r .
The t o o t h f a c e is the surface of a tooth above the pitch circle,
parallel to the axis ofthe gear.
The t o o t h f l a n k is the tooth surface below the pitch circle, parallel
to the axis of the gear. If any part of the flank extends inside the base
circle it cannot have involute form. It may have any other form
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
which does not interfere with mating teeth, and is usually a straight
radial line .
C i r c u l a r t o o t h t h i c k n e s s is measured on the tooth around the pitch
circle, that is, it is the length of an arc. C h o r d a l t o o t h t h i c k n e s s is the
chord of this arc.
The
teeth.
m o d u le
is the pitch circle diameter divided by the number of
D i a m e t r a l p i t c h is the number of teeth per inch of pitch circle
diameter. This is a ratio.
C i r c u l a r p i t c h is the distance from a point on one tooth to the
corresponding point on the next tooth, measured round the pitch
circle.
If contact between the teeth of meshing gears does not take place
on the line of action i n t e r f e r e n c e may occur. This is often the case
when a pinion with a small number of teeth is in mesh with a wheel
with a large number of teeth, the faces of the wheel teeth fouling the
flanks of the pinion teeth. If interference is allowed the pinion teeth
will be undercut at the roots.
The term p i n i o n is applied to the smaller of two mating gears.
Proportions and relationships of standard involute spur gear teeth
mo d u Ie m
,
pitch circle diameter, PCD
__
number of teeth, T
=~-------
hence, PCD = m x T
circular pitch, p = 'IT X m
In v o lu te g e a r s
Approximate construction for the spur gear teeth
It is usual to represent the involute curves on the teeth by circular
arcs, since to draw them accurately would take too long. The
method shown on page 160 is Unwin's construction.
Draw the pitch, addendum and dedendum circles and the base
circle tangential to the line of action. On a convenient radial line fix
a point A on the addendum circle and a point E on the base circle.
Divide AE into three equal parts and through B, the division
nearest A, draw a tangent to the base circle at D. Divide BD into
four equal parts and through F, the division nearest D, draw a
circle. The centres for the profile arcs lie on this circle and F is the
centre for the profile arc passing through B. From C, the intersection of this profile arc and the pitch circle, mark off tooth
thicknesses round the pitch circle. Draw profile arcs, radius FB,
through these points, with their centres on the circle through F. The
fillet radius at the roots of the teeth is approximately one-seventh of
the widest tooth space.
On a working drawing of a spur gear it is usually unnecessary to
show more than two or three teeth. The working drawing consists
generally of two views, a circular view on which the addendum
circle is shown as a thick, continuous line and the pitch circle as a
chain dot line, and a sectional view taken through the centre line of
the gear. On the sectional view the hatching does not cross the
teeth. The data relating to the teeth are generally shown in tabular
form on the drawing.
Figure 10 on page 169 shows the conventional representations of
the commoner type of gears.
circular tooth thickness = ~
addendum = module
clearance = 0.25 x module
dedendum = addendum + clearance
Pressure angle, 'P = 14.5 or 20 The British Standard recommendation is 20 This value reduces the possibility of interference
and gives the tooth a wider root.
Base circle diameter, BCD = PCD cos 'P . (Refer to Figure 3, page
157.)
0
0
•
0
•
Involute gears will work correctly together if they have the same
pressure angle and module.
158
Involute racks
As explained previously, involute racks have straight sided teeth.
The sides of the teeth are normal to the line of action, therefore,
they are inclined to the vertical at the pressure angle.
Figure 4 on page 161 shows the basic rack tooth profile for unit
normal metric module. To obtain the dimensions for the profile if
the module is, say, 4, the given dimensions must be multiplied by 4.
In practice the basic rack is usually modified as shown in Figure 5
but the modifications are so small that for drawing purposes they
can be ignored.
Figure 6 illustrates the point that for a gear and rack in mesh the
pitch line of the rack is tangential to the pitch circle of the gear.
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
H e lic a l
gears
These are a development of spur gears. Instead of the teeth being
parallel to the axis they lie on helices as shown in Figure 7 (page
163). This means that contact between teeth in mesh takes place
progressively along a diagonal line across the faces and flanks of the
teeth. Before contact ceases between one pair of teeth the next pair
begin to engage and so engagement is continuous. Because of this
the load on the teeth is distributed over a larger area than is the case
with spur gears, and the shock loads which are present when spur
gears transmit large torques are reduced.
Helical gears run smoothly and quietly under heavy loads and,
compared with spur gears, backlash is smaller. Since helical teeth
are longer than spur gear teeth, for the same thickness of gear, the
strength of the teeth is greater.
Figure 7 illustrates the terms l e a d , h e l i x a n g l e and l e a d a n g l e as
applied to helical gears and shows how meshing single helical gears
produce an end thrust which must be absorbed with a thrust
bearing. The end thrust is analogous to the axial movement of a
screw as it is turned in a nut. The direction of the end thrust depends
on the hand of the helix on the gear and its direction of rotation.
End thrust may be eliminated by using double helical gears as
shown in Figure 7. These may be cut on the same wheel or on
separate wheels. The end thrusts from each wheel act in opposite
directions and so cancel each other out.
Helical gears can be used to connect parallel shafts and shafts
with any angle between them up to 90°. When the shaft angle is 90°
single helical gears have the same helix angle and the same hand.
B evel
gears
This type of gearing is used to transmit power between shafts in the
same plane whose axes would intersect if produced. The angle
between the shafts is usually a right angle, but it may have any value
up to 180°.
When the transmission of motion by spur gears was considered,
the starting point was two cylinders in contact at a point on their
circumferences. With bevel gears the starting point is two cones in
contact along a pair of generators, the apices of the cones being
coincident, as shown in Figure 8. These cones are the pitch cones of
the gears. If they roll on each other without slipping the velocity
ratio is the inverse ratio of the diameters of their bases. These
diameters are the pitch circle diameters.
162
In v o lu te g e a r s
The teeth on a bevel w heel lie on the curved surface of a second
cone having the sam e base as the pitch cone and w ith its generators
at right angles to the generators of the pitch cone. This cone is the
back cone of the gear.
If the curved surface of the back cone is view ed norm ally the teeth
have the sam e profiles as the teeth on a spur gear. The addendum
and dedendum have the sam e proportions as spur gear teeth but are
m easured above and below the pitch circle, parallel to the back cone
generator. Pressure angles for bevel gears are usually 14.5° or 20° as
for spur gears.
B e v e l g e a r te r m s
In addition to those term s used for spur gears the follow ing are also
used and are illustrated in Figure 8.
The p i t c h a n g l e is the angle betw een the axis of the gear and the
pitch cone generator. W hen the pitch angle is 45° the gear is a m i t r e
g e a r . W hen the pitch angle is 90° the gear is a c r o w n g e a r .
The c u t t i n g a n g l e or r o o t a n g l e is the angle betw een the gear axis
and the root cone generator.
The f a c e a n g l e is the angle betw een a line at right angles to the axis
and the top surfaces of the teeth.
The e d g e a n g l e is the angle betw een a line at right angles to the
axis and the generator of the back cone.
The a d d e n d u m a n g l e is the angle betw een the pitch cone generator and the top surfaces of the teeth.
The d e d e n d u m a n g l e is the angle betw een the pitch cone generator and the bottom s of tooth spaces.
The w i d t h of the tooth face is the w idth m easured parallel to the
pitch cone generator.
The a n g u l a r a d d e n d u m is the addendum m easured on a line at
right angles to the gear axis.
The o u t s i d e d i a m e t e r is the diam eter of the gear at the tops of the
teeth.
The v e r t e x d i s t a n c e is the height of a cone w ith the outside
diam eter as base.
A p p r o x im a t e
c o n s t r u c t io n
fo r b e v e l g e a r te e th
This is show n on page 166, and is due to Tredgold. First draw the
gear centre line w ith the pitch circle diam eter crossing it at right
. angles. O n the pitch diam eter layout the pitch cone, using the pitch
.'angle. D raw the back cone w ith the generators at right angles to the
.
"
165
In v o lu te
gears
pitch cone generators. O n the back cone set out the addendum and
dedendum and joint these points to the vertex. M ark off the face
w idth of the teeth on the pitch cone and com plete the sectional view
of the tooth. D raw lines across the view at right angles to the axis
through the tip and root points and points on the pitch cone at each
end of the teeth. N ow develop the back cone by sw inging arcs about
a from the addendum , dedendum and pitch points on the back
cone. These arcs fix the addendum , dedendum and pitch circles of
the developed teeth. U sing the sam e construction as for a spur gear,
layout on these circles the profiles of one or tw o teeth. In the end
view set out the centre lines of the teeth and draw six circles
representing the addendum , dedendum and pitch circles at each
end of the teeth. Fix three points on the profiles of teeth at the large
end by transferring the w idths x, y and z from the developed view of
the teeth. The view m ay now be com pleted by filling in the teeth
profiles w ith a french curve or w ith circular arcs. A ll straight lines on
the teeth are radial and this enables three points on the profile of the
teeth at the sm all end to be fixed. The elevation can be com pleted by
projection from the end view . A ll straight lines on the teeth in this
view pass through the vertex if produced.
It w ill be seen from the illustration that w hen the back cone is
developed a spur gear is form ed w ith a pitch circle diam eter equal to
tw ice the length of the back cone generator. This spur gear is called
the virtual spur gear and it is used in the m anufacturing calculations
for the bevel w heel.
W orm
g e a r in g
W orm gearing, consisting of a w orm and w orm w heel, connects
shafts at right angles w hich lie in different planes. A n exam ple and
the com m oner term s used in w orm gearing are show n in Figure 9 on
page 168
The w orm is essentially a screw , w hich m ay be single or m ultithreaded, and it engages w ith teeth cut on the w orm w heel. O n an
axial section through the w orm the teeth have the form and proportions of a standard involute rack, w hilst on a sim ilar section through
the w orm w heel they are sim ilar to those on an involute spur gear.
N orm ally the w orm drives the w orm w heel and in one revolution
w ill turn the w heel pitch circle through a circum ferential distance
equal to the lead of the w orm . H igh velocity ratios can be achieved.
For exam ple, a single-start w orm driving a 40 tooth w heel w ill give a
167
E n g in e e
r in g D r a w in g
w ith W o r k e d
E x a m p le s
2
In v o lu te
velocity ratio of 40 and the sam e w heel driven by a double-start
w orm w ill give a velocity ratio of20.
For high efficiences the pitch diam eter of the w orm should be
sm all and the lead angle large. A s the lead angle of a single-start
w orm is sm all it is relatively inefficient but, on the other hand, there
is little tendency for the w heel to drive the w orm , that is, for the
drive to be reversible. H ow ever, w hen the lead angle exceeds about
9° the drive is reversible. A t lead angles of about 45° m axim um
efficiency occurs.
diam eter 112 m m , four equi-spaced lightening holes 22 m m
diam eter, through the w eb on a 78 m m diam eter pitch circle.
Fillet radii 3 m m .
D raw full size in first angle projection a front elevation
show ing three teeth draw n approxim ately, and a sectional end
view . In the end view the boss is positioned sym m etrically
about the centre line. Show a table giving the gear data.
5
D raw full size in third angle projection a dim ensioned front
elevation and sectional end view of a 30 tooth spur gear,
m odule 6, 20° pressure angle. Show three teeth on the front
elevation draw n approxim ately and a table of gear data. The
face w idth of the gear is 25 m m , rim diam eter 158 m m , boss
diam eter 34 m m , ream ed 24 m m diam eter, boss length 50 m m .
The boss is supported by four straight spokes of elliptical
cross-section, the m inor diam eter being constant at 10 m m and
the m ajor diam eter tapering from 24 m m at the boss to 22 m m
at the rim . Fillet radii 5 m m . The boss is sym m etrical about the
centre line in the end view . Show a revolved section of one
spoke on the elevation.
6
D raw full size one view of a pair of involute spur gears in m esh
show ing five teeth on each gear. The gears have 32 teeth and 24
teeth of m odule 6 and a 20° pressure angle.
The 32 tooth gear has a rim diam eter of 170 m m , boss
diam eter of 48 m m and a shaft diam eter of28 m m . There are six
lightening holes 35 m m diam eter on a pitch circle of 105 m m
diam eter.
O n the 24 tooth gear the rim diam eter is 120 m m , the boss
diam eter is 35 m m , the shaft diam eter is 22 m m and there are
three 32 m m diam eter lightening holes on a pitch circle
diam eterof75 m m .
Show the gear data in tabular form .
7
A 300 m m pitch circle diam eter involute spur gear is to m esh
w ith a pinion w ith a 200 m m pitch circle diam eter. If the gears
have a pressure angle of 20° and a m odule of 8, draw full size
four teeth on each gear in m esh at the pitch point. G ive a table
of gear data.
8
The centre distance of an involute spur gear and pinion is 175
m m and the speed reduction is 2.5:1. A ssum ing a m odule of 5
and a pressure angle of 20° calculate the follow ing:
M ulti-start w orm s have large lead angles and so give high
efficiencies. W aisted w orm s, w hich envelop the w orm w heel, also
increase the efficiency com pared to parallel w orm s.
W orm gearing has m any applications and com m on exam ples are
indexing m echanism s, slow -m otion drives and torque converters.
H ypoid gears
These gears are sim ilar to bevel gears but the basic surface on w hich
the teeth are cut are hyperboloids instead of cones. A hyperboloid is
the solid of revolution generated by rotating a hyperbola about its
directrix. The teeth of hypoid gears are helical and the axes of the
shafts do not intersect w hen produced. H ypoid gears are com m only
used in the rear axles of cars.
Involute gear problem s
1
Layout tw ice full size five teeth of a spur gear w ith 35 teeth of
involute form , m odule 5 and 20° pressure angle. U se the
approxim ate construction show n on page 160. G ive a table
show ing the num ber of teeth, m odule, pressure angle, pitch
circle diam eter, circular pitch, addendum and dedendum .
2
D raw tw ice full size five teeth of a 15 tooth involute spur gear,
m odule 6 and 20° pressure angle. Show in tabular form the
im portant dim ensions of the gear.
3
A 16 tooth pinion w ith a m odule of 10 has a 20° pressure angle.
Set out tw ice full size five teeth on the pinion. A dd a table
show ing the im portant gear dim ensions.
4
M ake a dim ensioned w orking draw ing of an 18 tooth pinion,
m odule 4, 20° pressure angle using the follow ing inform ation.
Face w idth 38 m m , central w eb 12 m m thick, boss length
50 m m , boss diam eter 45 m m , boss bored 32 m m , inside rim
170
gears
171
E n g in e e r in g
(a)
(b)
(c)
(d)
(e)
(f)
9
the
the
the
the
the
the
D r a w in g
w ith W o r k e d
E x a m p le s
2
pitch circle diam eters of gear and pinion
base circle diam eters of gear and pinion
num ber of teeth on the gear and pinion
circular pitch of the teeth
addendum
dedendum
A pinion w ith 20 teeth is to m esh w ith a rack w hose teeth have a
pressure angle of 20° and an addendum of 6 m m . The travel of
the rack is to be 125 m m .
D raw tw ice full size all the teeth on the rack and five teeth on
the pinion. Show the rack and pinion in m esh; tabulate the data.
10
A m achine w ork table is driven by an involute rack w hose teeth
have a pitch of 31.42 m m and a pressure angle of 20°. The rack
is operated by a pinion and interm ediate w heel, the gear train
giving a speed reduction of 3: 1. The pitch circle diam eter of the
pinion is 120 m m .
D raw full size the interm ediate gear and rack in m esh show ing five teeth on each. Tabulate data for tw o gears and the rack.
11
12
7 V e c to r g e o m e tr y
A pair of bevel gears is to connect 38 m m diam eter shafts
having an included angle of 90°. The velocity ratio of the drive
shaft to the driven shaft is to be 3:2, the m odule 8, and the face
w idth 55 m m . The pinion is to have 16 teeth.
D raw full size a half sectional elevation and half end view of
each gear, sim ilar to the view s show n on page 166. A ll other
dim ensions necessary to com plete these view s are to be supplied by the student.
A reduction gear in the form of a pair of bevel gears connects
30 m m diam eter shafts having an included angle of 90°. The
pinion, w hich is the driver, revolves at 125 r.p.m . and has a
90 m m pitch diam eter, a face w idth of 20 m m and a m odule of 5.
The bevel w heel is to revolve at 75 r.p.m . D raw the follow ing
view s of the assem bly full size.
(a) A front elevation, sim ilar to that on page 164, the gears
being show n in half section w ith the pinion axis vertical
(b) A n outside plan view of the pinion only
(c) A n outside end view on the right of the front elevation
show ing the w heel only
D im ensions for the hubs, etc., are to be supplied by the
student.
172
M any problem s in engineering involve quantities such as (a) d i s - change of position, (b) v e l o c i t y - rate of change of
displacem ent w ith respect to tim e, (c) f o r c e - the action of one body
p la c e m e n t
on another.
These quantities have m agnitude, direction and line of action and
m ay be represented on a draw ing by a line, draw n to scale in the
stated direction w ith an arrow indicating the sense of the line. This
line is called a v e c t o r .
Typical vector lines are illustrated in Figure 1. a b represents a
vector 3.5 units long directed upw ards to the right at 45° to the
horizontal and c d represents a vector 5 units long directed horizontally from left to right.
A ddition of vector quantities
W hen adding vector quantities w e have to take into account not
only their num erical value but also their direction. To add tw o or
m ore vectors proceed as follow s. D raw the first vector in the direction of its arrow , continue the second one on the end of the first, the
third on the end of the second and so on. The sum of the vectors is
the vector joining the beginning of the first to the end of the last in
the series. The final diagram w ill be a closed polygon.
Figure 2 (a) show s four vector quantities A , B, C and D . To
determ ine the sum of these quantities proceed by draw ing a b equal
and parallel to vector A ; on the end b draw b c equal and parallel to
vector B; on the end c draw c d equal and parallel to vector C and
from d draw d e equal and parallel to vector D . The sum of A , B, C
and D is the vector a e , and its direction is indicated by the arrow .
See Figure 2(b).
Subtraction of vector quantities
A gain considering the vector quantities in Figure 2 (a), if w e w ish to
subtract the vector C from the sum of the vectors A , Band D then
173
V e c to r g e o m e tr y
w e m ay add -C to A + B + D . A m inus vector is a positive one w ith
the direction of its arrow reversed.
Solving the problem in the sam e order as before w e now have
A + B + (-C) + D (see Figure 3).
V ector exam ples
1 A dd together the follow ing vectors; 6 units vertically upw ards,
to 4 units horizontal left to right, to 4.5 units dow nw ards, 10° to
the right of the vertical, to 3 units vertically dow nw ards (Figure
4).
2
A vector a b 15 units long, left to right, 45° above the horizontal,
represents the sum of tw o vectors, a c and c b . The angle betw een
a c and c b is 30°, and a c is upw ards 30° to the right of the vertical.
Find the values of the vectors a c and c b (Figure 5).
3
A dd together the follow ing vectors; 4 units vertically upw ards,
to 5 units horizontal right to left, to 4 units dow nw ards 15° to the
left of the vertical, and from the sum of these vectors subtract a
vector 3 units upw ards 30° to the right of the vertical (Figure 6).
V ectors applied to the solution of problem s in statics
Statics is that part of m echanics w hich deals w ith system s of forces in
equilibrium . The term static im plies a state of rest for the bodies on
w hich the forces act, as for exam ple, in the case of roof trusses,
structural m em bers of supporting fram es, colum ns, bridge sections
etc.
Force m ay be represented vectorially since the length of the
vector m ay represent the am ount of the force, the inclination of the
vector m ay represent Its line of action and the arrow head w ill show
its sense or direction.
Concurrent co-planar forces
Concurrent forces are those w hose lines of action intersect at a
com m on point. Co-planar forces lie in the sam e plane.
Figure 7 (a) illustrates a system of five concurrent co-planar
forces. The forces along O P, O Q and O R are 5, 7 and 6 units of
force respectively. W e are required to find the forces along O S and
O T so that equilibrium m ay exist.
The condition for equilibrium is that the vectors w hen draw n
should form a closed polygon.
175
V e c to r g e o m e tr y
S o lu tio n
1
Reproduce the space diagram as illustrated in Figure 7 (b), and
place the letters A , B, C, D and E in the spaces betw een the
forces in accordance w ith Bow 's notation.
2
Com m encing at a , draw a b to scale to represent the 5 units of
force O P in m agnitude and direction.
3
From b draw b e to represent the 7 units of force O Q in m agnitude and direction.
4
From c draw c d to represent the 6 units of force O R in m agnitude and direction.
5
Through d draw a line parallel to the force as. The length of this
line w ill be unknow n since the m agnitude of the force as is
unknow n.
6
Through a draw a line parallel to the force aT so that this line
intersects the line draw n through d (stage 5) in point e .
7
The vectors d e and e a represent in m agnitude and direction the
forces as and aT.
N ote that the arrow s m ust follow through in direction in the force
polygon as show n in Figure 8.
T o d e te r m in e th e e q u ilib r a n t a n d r e s u lta n t o f a s y s te m o f c o n c u r r e n t
c o -p la n a r fo r c e s
Three electric w ires attached to the top of a pole are illustrated in
Figure 9(a). The w ires have the follow ing pulls and directions:
(a) 500 units of force due north
(b) 480 units of force due east
(c) 540 units afforce south-east
D eterm ine the equilibrant of this system of forces.
S o lu tio n
1
Produce a space diagram as illustrated in Figure 9 (b) and place
the letters A , B, C and D betw een the forces.
2
Com m encing at a , draw a b to scale to represent the 500 unit
force in m agnitude and direction.
3
From b draw b e to represent the 480 unit force in m agnitude and
direction.
177
V e c to r g e o m e tr y
4
From c draw
direction.
5
The line joining d to a w ill represent the equilibrant in m agnitude and direction, see Figure 10.
cd
to represent the 540 unit force in m agnitude and
N ote that the polygon closes and the equilibrant acts from d to a ,
the arrow follow ing in the sam e direction as those on the other
vector quantities. The m agnitude of the resultant of this system of
forces w ould have the sam e value as the equilibrant but its sense
w ould be opposite to the equilibrant, that is, it w ould act from a to
d.
T o determ ine the resultant of a system of non-concurrent co-planar
forces
N on-concurrent forces are those w hose lines of action do not intersect at a com m on point.
Figure 11 illustrates a system of three forces A , Band C w hich are
all acting in one plane but w hich do not pass through the sam e point.
Let the values of A , Band C be 8, 6 and 4 units offorce respectively.
The m agnitude and direction of the resultant can be determ ined by
the usual m ethod of draw ing the force polygon, Figure 12 (a). This,
how ever, w ill not give the com plete answ er to the problem since w e
do not know the position in w hich the resultant acts. Its line of
action m ust still be determ ined.
In order to determ ine the line of action of the resultant it is
necessary to draw another diagram w hich is know n as the l i n k or
fu n ic u la r p o ly g o n .
T h e c o n s tr u c tio n o f th e lin k p o ly g o n
1
D raw the force polygon, p q r s , in the usual w ay, and obtain the
m agnitude and direction of the resultant as represented by the
vector p s .
2
Choose any point 0 either inside or outside the force polygon,
and join 0 to each corner of the force polygon. Figure 12 (b).
3
Referring to Figure 13, from any point 1 in the line of action of
force A , draw a line in space P parallel to the line o p .
4
From the point 1, draw the line 1 to 2 in space Q parallel to o q
and cutting the line of action of force B in point 2. It m ay be
necessary to produce the line of action of f
necessary intersection.
179
V e c to r g e o m e tr y
5
From point 2, draw the line 2 to 3 in space R parallel to line or
and cutting the line of action of force C in point 3. A gain the line
of action of force C m ay have to be produced in order to obtain
the intersection point.
6
From point 3, draw the line 3 to 4 in the space S parallel to the
line o s w hich w ill intersect the line parallel to o p , draw n through
point 1, in 4.
7
The resultant of the forces w ill pass through point 4. Through 4
draw a line parallel to p s to represent the resultant force R.
The diagram 1234 is the link or funicular polygon.
N o t e : The technique carried out to produce the link polygon
involves replacing each of the original forces A , Band C by tw o
forces.
Force A represented by p q is replaced by p o and o q .
Force B represented by q r is replaced by q o and o r .
Force C represented by r s is replaced by r o and o s .
The resultant force represented by p s is replaced by p o and o s .
Therefore the intersection of p o and o s w ill give the position w here
the resultant acts.
Figure 14 (a) indicates forces of 8,6,4 and 5 units w hich act in the
directions CA , D B, BC and A B respectively, on a square of 50 m m
side. The problem is to determ ine the m agnitude, direction and
position relative to A , of the resultant of these forces.
S o lu tio n
1
Produce a new space diagram , Figure 14 (b), show ing the
forces w ith their lines of action term inating at the corners of the
square. This w ill enable the lettering of the diagram for Bow 's
notation to be done m ore easily.
2
D raw the force polygon p q r s t , Figure 15 (a). The vector p t
draw n from the starting point to the finishing point closes the
polygon and represents the resultant of the given forces in
m agnitude and direction.
3
To determ ine the position of the resultant it is now necessary to
draw the link polygon. Choose any point 0 inside or outside the
force polygon and draw lines o p , o q , o r , o s and o t . Figure 15
(b).
181
V e c to r g e o m e tr y
4
From any point 1 on the line of action of the 8 unit force draw a
line in space P parallel to the line o p .
5
From 1 draw the line 1 to 2 in space Q parallel to the line
cutting the line of action of the 6 unit force in point 2.
6
From 2 draw the line 2 to 3 in space R parallel to the line or and
cutting the line of action of the 5 unit force in point 3.
7
From 3 draw the line 3 to 4 in space S parallel to the line o s and
cutting the line of action of the 4 unit force in point 4.
8
From point 4, draw the line 4 to 5 in the space T parallel to the
line a t , w hich w ill intersect the line parallel to o p , draw n
through point 1, in 5.
9
The resultant of the forces w ill pass through point 5. Through 5
draw a line parallel to p t to represent the resultant force. Figure
16 show s the com plete link polygon.
10
M easure the distance from the intersection of the line of action
of the resultant and the square to point A .
To determ ine
oq
and
the resultant and reactions of a parallel system of
forces acting on a beam
Figure 17 (a) indicates a loaded beam supported at its extrem ities.
The problem is to determ ine the m agnitude and position of the
single force w hich could replace the three given forces, and to find
the reactions R and Rl at the extrem e ends of the beam in order to
m aintain equilibrium .
S o lu tio n
1
Reproduce the given space diagram w ith Bow 's notation as
illustrated in Figure 17 (b).
2
D raw the force diagram a b c d for the three parallel forces. This
w ill be a straight line, the m agnitude of the resultant being
represented by the vector a d (Figure 18).
3
Choose any point 0 , w hich m ust be outside the force diagram ,
and join a o , b o , and d o (Figure 19).
183
V e c to r g e o m e tr y
4
See Figure 20. From any point 1 on the line of action of the
reaction R, draw a line in space A , parallel to o a , and cutting
the line of action of the 2 unit force in point 2.
5
From 2 draw the line 2 to 3 in space B, parallel to
the line of action of the 6 unit force in point 3.
ob
and cutting
6
From 3 draw the line 3 to 4 in space C, parallel to
the line of action of the 4 unit force in point 4.
oc
and cutting
7
From 4 draw the line 4 to 5 in space D parallel to o d and cutting
the line of action of the reaction Rl in point 5. N ote the
relationship betw een space A and o a , space Band o b , space C
and o c , and space D and o d .
8
Produce the line 1 to 2 and the line 4 to 5 to intersect in 6. The
resultant of the given system of forces w ill pass through point 6.
9
Join the points 1 to 5, and through 0 in Figure 19, draw a line
parallel to the line 1 to 5, cutting the vector a d in e .
oe
Then a e represents the m agnitude of the reaction Rand
represents the m agnitude ofthe reaction RIo
ed
10
F o r c e s in f r a m e w o r k s
The sim ple triangular fram e form s the basis of the m ore com plex
fram ew orks used in structural engineering. Roof trusses, bridges,
crane supports are but a few of the exam ples of the use of fram ew orks. Figure 21 gives a few illustrations of fram ew orks and it w ill
be seen that each one is m ade up of a num ber of sim ple triangular
fram es. It is essential to know the forces in each of the m em bers of
the fram e in order to produce a satisfactorily designed fram ew ork
capable of supporting the loads it has to carry. A ny diagram
show ing the layout of a fram ew ork is in fact a space diagram of a
series of forces. These forces w ould consist of (a) external forces
such as loads and reactions, (b) internal forces w ithin the m em bers
ofthe actual fram ew ork.
Since the com plete fram ew ork is in a state of equilibrium it
follow s that each of the junctions of the fram e m ust also be in a state
of equilibrium . Thus a separate vector diagram could be draw n for
each of the junctions of a fram ew ork, as show n in Figure 22.
H ow ever, norm al procedure is to draw a vector diagram for the
com plete fram ew ork and the m ethod of doing this w ill appear from
the follow ing exam ples.
185
V e c to r
D ir e c tio n
g e o m e tr y
o f r e a c tio n s
It is com m on practice to support one end of a fram ew ork by a hinge
and the other end upon rollers. Thus allow ances can be m ade for
expansion and contraction of the fram ew ork due to variations in
tem perature.
O ne assum ption m ade is that the reaction at the roller support
w ill alw ays be perpendicular to the line joining the centres of the
rollers. The reaction at the hinge can be in any direction, but it is
governed by the conditions required to m aintain the w hole fram ew ork in a state of equilibrium .
E x t e r n a L L o a d s v e r t i c a L , r o l l e r r e a c t i o n v e r t i c a L - the line of action
of the hinge reaction w ill be vertical (Figure 23 (a)).
R e s u L t a n t o f e x t e r n a L L o a d s i n c l i n e d , r o l l e r r e a c t i o n v e r t i c a L - the
line of action of the hinge reaction m ust pass through the point of
intersection of the resultant and the line of action of the reaction of
the roller support. (Figure 23 (b)).
T ie s a n d s tr u ts
The t i e exerts an inw ard pull at each end of the m em ber to prevent
the m em ber being pulled apart by tensile loads w hilst the s t r u t
exerts an outw ard force at each end of the m em ber to prevent the
m em ber being crushed by com pressive loads. The difference
betw een them is show n in Figure 24. It is assum ed that the force in
any m em ber of a fram ew ork acts in the direction of that m em ber of
the fram ew ork, as show n in Figure 25.
To determ ine the m agnitudes of the forces in the m em bers of a
fram ew ork
Figure 26 show s a pin jointed structure w hich is sim ply supported at
Rand RI' For the loaded structure determ ine (a) the m agnitude of
the reactions at Rand Rl and (b) the m agnitudes of the forces in
each m em ber of the fram ew ork. State for each m em ber w hether it is
a tie or a strut.
S o L u tio n
1
Choose a suitable scale for the space diagram and draw the
fram ew ork w ith the external forces, reactions and Bow 's notation as show n in Figure 27.
2
Choose a suitable scale for the force diagram , and draw the line
a b c d to represent the external forces as in Figure 28.
187
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
4
Through b draw b e parallel to BE and through a draw a e parallel
to A E. The intersection gives point e .
5
Through e draw e d parallel to ED and through a draw a d parallel
to A D . The intersection gives point d.
N ote that the line c d should be parallel to D C.
6
M easure the lengths of the lines in the force diagram and hence
obtain the m agnitudes of the forces in the m em bers of the fram e.
7
The directions of the forces acting in the m em bers are obtained
as in the previous exam ple (See Figure 35). Tabulate the forces
in the m em bers as show n.
T o d e t e r m in e
t h e r e a c t io n s
c a r r y in g a g iv e n lo a d
A pin jointed structure is show n in Figure 36. D eterm ine the m agnitudes and directions of the reactions w hen the structure is carrying
the loading show n. D eterm ine also the forces in the m em bers,
distinguishing betw een tension and com pression.
S o lu tio n
1
Choose a suitable scale for the space diagram and draw the
fram ew ork w ith the external forces, reactions and Bow 's notation as show n in Figure 38. N ote that the 2 and 3 units of force
have been replaced by a single inclined force acting at P. Figure
37 show s the construction for determ ining the value of this
single force.
2
Choose a suitable scale for the force diagram and draw the lines
and b e to represent the external forces. The line a e w ill be
the resultant of the external forces (Figure 39).
ab
adjacent t o a b and b e and join a o , b o and c o
3
Choose any point
(Figure 39).
4
To construct the link polygon (Figure 38). Choose any point 1
on the line of action of the inclined force and draw a line
through this point in space A , parallel to a o .
5
From point 1 draw a line in space B, parallel to o b , to cut the
line of action of the 9 unit force in point 2.
6
Through point 2 draw a line in space C parallel to o e , to cut the
line through 1 in point 3.
7
The resultant of all the external forces w ill have its line of
action passing through this point 3.
0
192
E n g in e e r in g
8
D r a w in g
w ith W o r k e d
Since the line of action of each reaction
the resultant
of the external
point to m aintain
a e , to m eet
2
and the line of action of
forces m ust pass through
equilibrium ,
the vertical
E x a m p le s
the sam e
draw a line through 3, parallel to
line of the right hand reaction
in point
X.
9
Join X to T (Figure 38) to give the line of action of the reaction
R.
10
Referring
action
to Figure
39, draw a line a d , parallel
of the reaction
the reaction
R. Then a d represents
R and d e represents
to the line of
the m agnitude
the m agnitude
of
of the reaction
RI'
11
Repeat
as in the
previous
exam ple
to com plete
the
polygon (Figure 40). Figure 40 (a) show s the positions
and ties.
To determ ine
force
of struts
the centre of gravity of a given system of m asses
Figure 41 show s four m asses, 4, 2, 8 and 1 unit of m ass, having fixed
relative positions.
this system .
It is required
to determ ine
the centre of gravity of
S o lu tio n
1
The w eights of the m asses could be considered
the lines PQ ,
Q R,
RS and ST. D raw
to be acting along
the force diagram
(Figure 42) so that the vectors p q , q r , r s and s t represent
2 and 4 units of m ass respectively.
2
Choose
po,
3
any point a aw ay from the vector line p to t and connect
and a t .
qo, ro, so
The link polygon
(Figure 43) can now be constructed
w ay, and the resultant
through point N .
4
The
pqrst
the 8, 1,
m ass
system
of the four m asses
can now
be turned
in the usual
can be seen to act
through
any angle,
for
convenience one of 90°. The w eights of the m asses now act along
the lines A B, BC, CD and D E.
5
Construct
another
second resultant
polygon (Figure
force
diagram
as show n
can be seen to act through
45).
194
in Figure
44. The
point M of the link
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
6
The intersection of the tw o lines of action of the resultants w ill
give point G , the position of the centre of gravity of the system of
m asses.
To determ ine the centre of area of a given section
Figure 46 show s a plane figure and it is required to find the centre of
area of this figure.
S o lu tio n
1
D ivide the figure into three rectangles, A , Band C, as show n in
Figure 47.
2
D raw the force diagram p q r s such that the vector lengths p q , q r ,
and r s have m agnitudes proportional to the areas of the
rectangles C, B and A respectively (Figure 48).
3
The link polygon (Figure 49) w ill give the position of the
resultant, point N .
4
Com plete the construction by draw ing a second force diagram
Figure 50, and a second link polygon Figure 51, and thus obtain
a second resultant acting through point M .
5
The centre of area of the section is at G , the point of intersection
of the tw o resultants.
The vectors are taken in tw o directions; horizontally, and inclined
at approxim ately 45°. 45° w as chosen to avoid the sm all diagram that
w ould result if the vectors w ere taken at 90° to each other.
To determ ine non co-planar concurrent forces in the m em bers of a
fram ew ork
Figure 52 show s a pictorial view of a tripod arrangem ent. The
m em bers A B and A C are equal in length. It is required to determ ine the forces in each m em ber of the fram ew ork due to a 4 unit
force acting vertically at A .
S o lu tio n
1
Consider a vertical plane containing A D and the vertical 4 unit
force acting at A (Figure 53 (a».
2
D raw the space diagram (Figure 53 (b» show ing the vertical
loading, the back leg A D and another m em ber A E. The latter
196
V e c to r g e o m e tr y
m em ber is introduced to provide equilibrium and replaces the
tw o shorter legs A B and A C. The m em ber A E lies in the sam e
plane as A B and A C.
3
Figure 53 (b) illustrates three co-planar forces in equilibrium .
Choose a suitable scale and draw the force diagram for these
three forces as show n in Figure 54; pq represents the vertical
load, qr represents the force in the new m em ber A E and rp
represents the force in the m em ber A D .
4
N ow consider the inclined plane containing the m em bers A B,
A C and A E (Figure 55). A view taken norm al to this plane in the
direction of the arrow S w ill give the true inclination of A B and
A C. This view is show n in Figure 56.
5
Figure 56 again illustrates three co-planar forces in equilibrium .
The force diagram wxy representing these forces vectorially is
show n in Figure 57; x y represents the force in the m em ber A E,
y w and w x represent the forces in the m em bers A C and A B.
V e c t o r p r o b le m s
1
D eterm ine the resultant force for the given co-planar, concurrent force system show n.
2
D eterm ine the m agnitude and the direction of the equilibrant
for the given co-planar, concurrent force system show n.
3
The figure show s the directions of four concurrent forces referred to axes O X and O Y . D eterm ine the equilibrant and give
its sense and inclination to O X .
4
Three forces are located by m eans of the square show n. D eterm ine the equilibrant of this system of forces, and the perpendicular distance from the line of action of the equilibrant to the
centre of the square.
5
The figure show s a pin jointed square fram e w ith loads acting at
each com er. D eterm ine the m agnitude, direction and position
relative to A of the resultant of these forces.
6
D eterm ine the reactions at the supports for the beam show n.
7
D eterm ine the reactions at the supports for the loaded beam
show n. N ote that the beam carries tw o loads and is acted on by
an upw ard force of 4 kN .
199
V e c to r g e o m e tr y
8
The beam show n is hinged to a fixed support at the left-hand
end and is supported at a point 0.3 m from the right-hand end,
so as to be in the horizontal position. D eterm ine the reactions
at the supports for the given system of loading.
9
The figure show s a pin jointed structure loaded at point 0 and
hinged at point P. The structure is m aintained in equilibrium
w ith PO vertical by the pull in the vertical chain RT. O R is
horizontal. D eterm ine (a) the pull in the chain; (b) the m agnitude and direction of the reaction at the hinge P; (c) the forces
in the m em bers PO and SR. State w hether each m em ber is in
tension or com pression.
10
The figure show s a pin jointed structure w hich is sim ply supported at A and B. For the system of loading illustrated determ ine (a) the m agnitudes of the reactions at A and B; (b) the
m agnitude of the force in each m em ber of the structure stating
w hether the m em ber is a tie or a strut.
11
D eterm ine the forces acting in the various m em bers of the
fram ew ork show n.
12
Construct the com plete force polygon for the sim ple roof truss
show n. State the force acting in each m em ber and w hether the
m em ber is a tie or a strut.
13
A pin jointed structure is loaded and supported as show n in the
figure. D eterm ine the tension T in the bracing cable and the
m agnitude and direction of the reaction at the hinge. A ll
m em bers are of equal length. D raw the force diagram for the
structure and indicate the type of load in each m em ber. State
the m agnitudes of the loads in m em bers RJ and JK only.
14
U sing the link polygon, find the parallel equilibrants for the
system of forces show n. The equilibrants are to pass through
the points P and O . Find also the resultant of the given system
offorces.
15
A roof truss w ith a 10 m etre span and a rise of 4 m etres is loaded
as show n in the figure. The end X is hinged and the end Y rests
on a support. D eterm ine the m agnitude and the direction of
the resultant reactions at X and Y . Find also the m agnitudes of
the forces in m em bers G R, RJ, K L and LG . State w hether the
m em ber is in tension or com pression.
201
V e c to r g e o m e tr y
16
The figure show s a loaded roof truss hinged at the left-hand
support and supported on rollers at the right-hand support.
D raw the com plete force diagram for the truss and tabulate the
forces acting in the m em bers, indicating w hich are in tension
and w hich are in com pression.
17
D eterm ine the centre of gravity of the given system of m asses
show n. State its position relative to the lines A B and CD .
18,19
D eterm ine the centres of area of the figures show n.
20
Tw o legs of a tripod are each 400 cm long, the third leg is 396 cm
long. If the feet of the legs lie at the corners of an equilateral
triangle of side 300 cm , lying in a horizontal plane, find the
force in each leg w hen a m ass of 2000 kilogram m es is suspended from the apex of the tripod.
21
In a tripod the length of each of the legs O A , O B and O C is 6
m etres. The lines joining their feet are A B = 4 m etres, BC =
3.5 m etres, and A C = 3 m etres. If a m ass of 5000 kilogram m es
is suspended from the apex 0 , determ ine the force in each leg
of the tripod.
203
8 T races
Traces of lines
Figure
1 on page 205 show s the principal
first quadrant.
Figure
2 show s
planes of projection
the sam e
planes
in the
w ith a line A B
positioned
betw een them . The line is projected on to each plane
giving the elevation albion
the vertical plane and the plan a 2 b 2 on
the horizontal plane.
In Figure 3 the line is produced to pierce the vertical and horizontal planes. The point w here the line, produced if necessary,
pierces
w here
the vertical
plane is the v e r t i c a l t r a c e of the line. The point
the line, produced
is called the h o r i z o n t a l
H T are com m only
if necessary,
Figure
and illustrates
and
XY
is therefore
line, or the elevation
horizontal
the vertical
The intersection
trace
so the intersection
XY represents
of the elevation
Project
of the
This is the vertical trace of A B. The
to the elevation,
trace in the elevation.
to
of the
of the line in plan.
to cut the elevation
trace is found in a sim ilar w ay. Produce
m eet Xy. Relative
plane,
plane.
into the elevation
produced.
in ortho-
the plan to m eet X Y . Relative
the vertical
this intersection
of A B draw n
how the traces of a line are found.
trace, produce
the plan, X Y represents
Project
and sim ilarly that the H T of A B
4 show s the plan and elevation
projection
To find the vertical
plan
plane
V T and
used. It can be seen from Figure 3 that the V T of
A B lies on the elevation produced
lies on the plan produced.
graphic
pierces the horizontal
of the line. The abbreviations
tr a c e
the elevation
to
the horizontal
w ith X Y is the horizontal
this intersection
into the plan to cut
the plan of the line, produced if necessary. This is the horizontal
trace of A B. A com parison of Figures 3 and 4 w ill m ake the m ethod
clear.
In general,
the vertical
and the horizontal
the line relative
trace of a line appears
trace below it. H ow ever,
to the principal
planes
204
above the X Y line
it is possible to position
so that both traces appear
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
above, or both traces appear below , the X Y line. This is illustrated
in Figures 5 to 8.
Figure 5 show s pictorially a line A B w hose horizontal trace lies in
the second quadrant. W hen the principal planes are rabatted and
orthographic view s of this line are draw n as in Figure 6, the horizontal trace w ill lie above X Y w ith the vertical trace.
In Figure 7 the vertical trace of A B lies in the fourth quadrant.
Therefore, after rabatting the principal planes, the orthographic
view s show n in Figure 8 w ill have the vertical trace below X Y w ith
the horizontal trace.
Four w orked exam ples on traces of lines follow on pages 209 to
211. These exam ples involve the points set out above. In addition,
the m ethod of finding the true angles of inclination of the line to the
principal planes by using auxiliary view s is show n in each case. This
is revision of the principles given in Chapter 13 of Book 1.
Traces of planes
The lines in w hich a plane m eets the principal planes of projection
are the traces of the plane. The line w here the plane m eets the
horizontal plane is the horizontal trace. The line w here the plane
m eets the vertical plane is the vertical trace. A s w ith traces of a line,
the abbreviations H T and V T are usually used.
Planes are of tw o m ain types, p e r p e n d i c u l a r or o b l i q u e . Perpendicular planes are perpendicular to one or both of the principal
planes. They m ay be horizontal, vertical or inclined, as show n in
Figure 9 on page 212.
Figure 9 (a) show s a plane P w hich is perpendicular to both
principal planes. Its traces V T and H T m eet on X Y . This is true for
all traces of planes provided they are not parallel to X Y . The lines
A B and CD are draw n on the plane P and therefore their traces
m ust lie in the traces of the plane. This is alw ays true.
In Figure 9 (b) the plane P is parallel to the horizontal plane.
Therefore it cannot have any horizontal trace and neither can the
line A B draw n on the plane. The vertical trace of P is parallel to X Y .
The plane P in Figure 9( c) has no vertical trace because it is
parallel to the vertical plane. Sim ilarly the line A B contained by the
plane has no vertical trace.
In Figure 9( d) plane P is perpendicular to the vertical plane and
inclined to the horizontal plane. N ote that its traces intersect on
X Y , and the traces of lines A B and CD lie in the traces of the plane.
It w ill be seen from Figure 9 that w ith all perpendicular planes at
206
T races
in the auxiliary elevation and transfer the height h 2 from the front
elevation to the auxiliary elevation. Project p3 on to the inclined
plane, m aking the projector perpendicular to the plane. N ow
project the projection of p3 back to the plan view . The projector
from pi to the oblique plane in this view m ust be at rightangles to
H T, so the position of the projection of the point on the oblique
plane can be found. Project it to the elevation and, to find its
position in this view , transfer h I from the auxiliary view .
The principles set out above for converting an oblique plane to an
inclined plane and projecting a point on to an oblique plane are
illustrated in the follow ing w orked exam ples.
Exam ple 5
The traces V TH of an oblique plane are given w ith the plan and
elevation of a line A B. It is required to project this line on to the
oblique plane and to draw the plan and elevation of the projection.
This is m erely an extension of the projection of a point on to an
oblique plane. First convert the oblique plane to an inclined plane
by draw ing an auxiliary elevation and project the line into this view .
Project the ends of the line on to the inclined plane and project
these points back to the plan. Points c 2 and d 2 on the plan of the line
on the oblique plane can be found by projecting from a 2 and b 2 in the
given plan to cross H T at right angles. The sam e principle m ay be
applied to find c l and d l on the elevation of the line on the oblique
plane.
Exam ple 6
H ere the triangle A BC lies in the oblique plane V TH . The plan of
the triangle is given. The elevation and true shape of the triangle are
required. Start by converting the oblique plane to an inclined plane,
and project the corners a 2 , b 2 and c 2 of the given plan of the triangle
on to the inclined plane. The true shape of the triangle is found by
projecting a second auxiliary plan view ing the inclined plane
norm ally. The elevation of the triangle is draw n by transferring the
heights p, q and rfrom the auxiliary elevation to the front elevation.
Exam ple 7
The plan and elevation of a triangle A BC are given w ith the traces
V TH of an oblique plane. The plan and elevation of the triangle
after it has been projected on to the oblique plane are required.
Convert the oblique plane to an inclined plane by projecting an
217
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
auxiliary elevation. Project the triangle into this view and project
the corners on to the inclined plane. Project these points back to the
plan and fix a2 , b 2 and c 2 in the projected plan by projectors at right
angles to H T from the corners of the given plan. Project a2 , b 2 and c 2
to the elevation and find them by transferring heights p, q and r from
the auxiliary elevation to the appropriate projector.
Tw o exam ples are now given of solids cut by oblique planes.
These can be solved using the principles already given.
Exam ple 8
The plan and elevation of a square prism are given w ith the traces
V TH of an oblique plane. This plane is to cut the prism , passing
through point P. D eterm ine the plan and elevation of the cut prism
and the true shape of the section produced by the oblique plane.
First convert the oblique plane to an inclined plane and project the
prism into this auxiliary view . M ove V ITI parallel to itself, to pass
through p3 on the auxiliary elevation of the prism . The plan of the
cut prism can now be com pleted by projecting from the cut auxiliary
elevation, and then the front elevation can be com pleted by projection from the plan. The true shape of the section is found by
projecting a second auxiliary plan norm al to V ITI and transferring
w idths such as w from the norm al plan.
Exam ple 9
This exam ple, illustrated on pages 225 and 226, is sim ilar to Exam ple 8, but now the position of the oblique plane V TH relative to
the solid is given instead of the plane being required to pass through
a specified point on the solid. Find the plan and elevation of the cut
solid and project a view to show the true shape of the cut surface.
Proceed as before to convert the oblique plane to an inclined plane
by draw ing an auxiliary elevation. Project the solid into this view .
Project back to the plan the points w here the inclined plane V ITI
cuts the edges of the solid. Com plete the elevation by projecting
from the plan, using heights from the auxiliary elevation to locate
the points on the projectors. A second auxiliary plan, projected
from the auxiliary elevation, w ill give the true shape of the cut
surface. W idths across this view are the sam e as corresponding
w idths w across the norm al plan.
222
T races
P r o b le m s
o n tr a c e s
U nless otherw ise stated use first angle projection and draw full size.
Construction lines should not be erased. H idden detail should be
show n.
1-4
D raw the given view s of the lines and determ ine their
traces, true angles of inclination to the horizontal and
vertical planes, and true lengths. Q uestion 4 is to be
draw n in third angle projection.
5
The elevation a l b l of a line A B is given. The end A is
positioned in the plan at a 2 • If the true length of the line
is 118 m m , com plete the plan, find the traces of the line,
and its true inclinations to the principal planes. U se third
angle projection.
6
The figure show s the plan of a line A B w ith the position
b l of B in the elevation. The line m akes an angle of 3 0 °
w ith the horizontal plane, w ith A nearer the plane than
B. Com plete the elevation and find the true length of the
line. A lso find its traces and true inclinations to the
principal planes. U se third angle projection.
7
The plan view a 2 b 2 of a line A B lying in an oblique plane
V TH is given. Project the elevation of the line and find
its true length. A lso find the true inclination of the
oblique plane to the horizontal.
8
A triangle is show n in plan and elevation in the figure,
together w ith its position relative to an oblique plane
V TH . Produce the plan and elevation of the triangle
w hen it is projected on to the oblique plane, and determ ine the true shape of the projection. A lso find the true
inclination of the oblique plane to the vertical plane.
9
The plan a 2 b 2 c 2 of a triangle and the traces V TH of a
plane containing it are given. D raw the plan and elevation of the triangle and find its true shape.
10
The plan a 2 b 2 of a line A B, w hich lies in an oblique
plane, is show n, together w ith the horizontal trace H Tof
the plane. The true inclination of A B to the horizontal
plane is 5 5 ° . A second line CD lies in the sam e oblique
plane and intersects A B at P. D eterm ine the true lengths
of both lines and the true angles betw een them .
227
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
11-12
Both questions show the plan and elevation of a triangle
A BC. The triangles lie in oblique planes. D eterm ine the
traces of these planes and their true inclinations to the
vertical and horizontal planes.
13
The triangle A BC show n in plan and elevation lies in an
oblique plane. The sam e plane passes through the given
hexagonal prism . D eterm ine the traces of the oblique
plane and project an elevation of the cut prism .
14
The elevation and incom plete plan are given of a
hexagonal prism w hich is inclined to both principal
planes. The prism is 100 m m long and m akes an angle of
30° w ith the horizontal plane. A n oblique plane V TH ,
positioned as show n, cuts the prism . D raw an elevation
of the cut prism and project from it a plan. State the true
inclination of the oblique plane to the horizontal plane.
15
16
17
Tw o view s of part of a cylindrical strut are given. The
strut is to be w elded to a plate w hich lies in the oblique
plane V TH . Find the true inclination of the plate to the
horizontal plane. D raw the given view s of the strut
show ing the strut end before w elding to the plate. A lso
project a view to show the true shape of the strut end.
The triangle A BC lies in an oblique plane. Find the
traces of this plane and its true inclination to the horizontal plane. The sam e oblique plane cuts the hexagonal
pyram id show n, passing through point P. D raw the plan
and elevation of the cut pyram id show ing the section
produced.
The traces H T and V T of an oblique plane are at 30° and
45° respectively to X Y . D eterm ine the true inclination
of this plane to the horizontal plane. A triangle A BC has
sides A B 50 m m , BC 70 m m , and CA 80 m m and lies in
this oblique plane w ith BC parallel to and 20 m m from
the vertical plane. B is the corner of the triangle nearest
X Y in the elevation. D raw the plan and elevation of the
triangle. D eterm ine the horizontal and vertical traces of
a line from A to a point 20 m m from B along BC in the
elevation.
230
9 M a c h in e d r a w in g
In the problem s w hich follow the scale is to be full size, unless
otherw ise stated. U se the sam e projection angle as the given view s
are draw n in. H idden detail and dim ensions need not be show n.
Bearing
assem bly
The details for a 76 m m diam eter bearing are show n on page 232.
D raw half full size the follow ing view s of the com plete assem bly.
(a) A sectional front elevation on BB
(b) A sectional plan view on A A
(c) A n outside end view looking on the bearing cap
A ircraft
bracket
Tw o view s of this detail are given on page 233. D raw the given front
elevation and project from it sectional end view s on A A and BB,
and an outside plan view .
V alve assem bly
V iew s of the details for a sm all valve are given on page 234. D raw
tw ice full size the follow ing view s of the com plete assem bly.
(a) A half sectional front elevation corresponding to the given
sectional elevation of the body. The right -hand half of the view
is to be in section
(b) A n outside end view in the direction of arrow A
G land packing is to be show n in the 12 m m diam eter counterbore
in the spindle guide.
The handw heel to fit the 6 m m square end of the spindle has been
om itted.
O il pum p cover
D raw tw ice full size the follow ing view s of the oil pum p cover show n
on page 235.
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
(a)
(b)
(c)
The given front elevation
A sectional end view on Q Q
A sectional auxiliary plan view on PP
Lathe tailstock
The details for a sm all lathe tailstock are show n on page 237. D raw
the follow ing view s of the assem bled tailstock, supplying the m issing details to fit hole B and lock the sleeve.
(a) A sectional front elevation corresponding to the given front
elevation of the body
(b) A half-sectional end view corresponding to the given end view
of the body, w ith the left-hand half in section and the section
plane passing through hole B.
Carburettor
body
The plan and elevation of a carburettor body are illustrated on page
238. D raw the follow ing view s of the com ponent.
(a) The given plan view
(b) A sectional front elevation on A A
(c) A sectional end view on BB
Clam p
O n page 239 are show n the details for a clam p. D raw the follow ing
view s of the assem bled com ponents.
(a) A sectional front elevation corresponding to the given elevation of the fixed jaw , item 1, the section plane passing through
the centre line
(b) A n outside plan view
(c) A n outside end view positioned on the right of the front
elevation
Show the sliding jaw , item 3, open 25 m m and the tom m y bar,
item 6, vertical.
Elbow casting
The incom plete plan and elevation of a steam valve casting are
given on page 240. D raw these view s, com pleting them w ith the
interpenetration
curves, and add a sectional end view on A A .
H idden detail m ust be show n in the front elevation but need not be
show n in the other view s.
The 6 m m diam eter holes in the branch flange need only be
indicated by their centre lines in the plan view .
236
M a c h in e
E n g in e m o u n t in g
d r a w in g
f it t in g
A detail from an aeroplane engine m ounting is illustrated on page
242.
D raw the given front elevation and project from it a plan view and
an end view looking in the direction of arrow B.
L o c a t io n
f ix t u r e
D raw the follow ing view s of the fixture show n on page 243.
(a) The given plan view
(b) A sectional front elevation on BB
(c) A half-sectional end view on A A , the left-hand half being in
section
cam
V iew s of a coupling w ith an integral cam are show n on page 244.
D raw the follow ing view s of the com ponent.
(a) A n outside front elevation in place of the given sectional front
elevation
(b) Tw o outside end view s, one looking on the coupling face, the
other on the cam
The m otion given to the follow er by the cam is to be:
C o u p lin g
a n d d is c
0° - 90° lift 38 m m w ith sim ple harm onic m otion
90" - 180° dw ell
180° - 360° fall 38 m m w ith uniform acceleration and retardation
The coupling rotates clockw ise w hen the given sectional elevation is view ed from the left. The coupling flange has tw o part conical
bosses running into the 86 m m diam eter boss. The line of action of
the follow er is on the vertical centre line of the com ponent in the
end view s.
P um p barrel
D raw half full size the follow ing view s of the pum p barrel show n on
page 245.
(a) A n outside front elevation in place of the given sectional front
elevation
(b) A sectional plan view on BB in place of the given part plan
VIew
(c)
A sectional end view on A A
241
E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
W all bracket
V iew s of a special w all bracket are show n on page 247. D raw the
given plan view and com plete the front and side elevations.
W hen the plan is view ed in the direction of arrow C, the 50 m m
radius in the side elevation appears in true shape.
Lines X X and Y Y should be positioned as show n to fit the
required view s on an A 2 size sheet.
H idden detail is required in the plan view only.
D istribution casing
Tw o elevations of this com ponent are given on page 248. D raw the
follow ing view s of the casing.
(a) A front elevation obtained by view ing the given front elevation in the direction of arrow A
(b) A sectional end view on BB, in projection w ith the front
elevation
(c) A sectional plan view through CC projected from the front
elevation
Pulley belt adjuster
D etails for a pulley belt adjuster are show n on page 249. D raw the
follow ing view s w ith all the parts assem bled.
(a) A sectional front elevation w ith the edges X X and Y Y of the
base and adjusting bracket in line. The section plane is to pass
through the centre of the adjusting screw
(b) A n outside end view obtained by view ing the front elevation
from the left
V alve casing
D raw the follow ing view s of the valve casing show n on page 250.
(a) The given plan view
(b) A sectional front elevation on BB
(c) A sectional end view on A A
(d) A n outside end view obtained by view ing the front elevation in
the direction of arrow C
Bevel pinion
A part sectional front elevation and an incom plete end view of a
bevel pinion are given on page 251.
D raw tw ice full size the given front elevation com pletely in section and com plete the given end view show ing all the teeth.
The serrations are to be show n conventionally in the end view .
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
P in io n g e a r a s s e m b ly
The com ponents for a pinion gear assem bly are detailed on pages
253, 254 and 255. D raw the follow ing view s of the assem bly.
(a) A sectional front elevation corresponding to the given halfsectional front elevation of the housing, item 2
(b) A n outside end view looking on the teeth of the pinion, item 1.
The teeth are to be show n conventionally
(c) The outside end view of the opposite side of the front elevation
The lock nut, item 6, and the lock ring, item 7, retain the bearings, item s 4 and 5, on the pinion. The lock ring, item 8, retains the
pinion sub-assem bly in the housing. W hen the housing bush, item 3,
is assem bled in the housing, the faces A are to be in line.
C o n tr o l b r a c k e t
V iew s are given on pages 256-7 of a control bracket for an aeroplane. D raw the follow ing view s of the detail.
(a) The given plan view w ith hidden detail
(b) A sectional front elevation on QQ
(c) A sectional end view on P P
D r ill j ig b o d y
V iew s of this com ponent are show n on page 258. D raw the given
front elevation and a sectional end view on CC.
C o n n e c t in g
r o d e n d a s s e m b ly
The details for this assem bly are show n on page 259 and consist of
the follow ing.
A steel connecting rod, item 1, of w hich only the end is show n;
tw o brass bearing halves, item 2; a m ild steel front plate, item 3; a
m ild steel locking plate, item 4; tw o steel bolts, item 5; tw o steel
hexagon nuts, and a hexagon head cap screw to secure the locking
plate to the front plate.
D raw the follow ing view s of the com plete assem bly.
(a) A half-sectional front view , corresponding to view A of the
connecting rod. The top half of the view is to be in section. The
visible half of the intersection curve on the connecting rod is to
be correctly projected
(b) A sectional plan view , the section plane passing through the
horizontal centre line of the front view
(c) A n outside end view looking on the hexagon nuts
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E n g in e e r in g D r a w in g w ith W o r k e d E x a m p le s 2
Bracket and casting assem bly
Page 261 show s the parts for this assem bly. W hen assem bled on the
casting face the bracket centre line CC coincides w ith cutting plane
A A . W ith the bracket in this position draw the follow ing view s of
the assem bly.
(a)
A plan view , corresponding to the given part plan view of the
casting face, show ing the bracket com pletely
(b) A front view , obtained by view ing the plan view in the direction of arrow B, show ing the bracket com pletely
(c) A n auxiliary sectional view on A A .
The fixing stud dim ensions are to be settled by the student.
Bracket and base assem bly
The parts for this assem bly are show n on page 262. A ssem ble the
parts and draw the follow ing view s of the assem bly.
(a) A plan view corresponding to the given plan view of the base
(b) A front view corresponding to the given front view of the base.
Show a broken-out section around hole X
(c) A n auxiliary sectional view on A A
Footstep bearing assem bly
This assem bly consists of the follow ing details w hich are show n on
pages 263 and 264.
A cast iron base plate, item 1; a cast iron bearing support, item 2;
a bronze bearing bush, item 3; a bronze bearing pad, item 4; four
square head bolts, item 5; a steel dow el, item 6; four hexagon nuts
and four plain w ashers.
A ssem ble the parts and draw the follow ing view s of the com plete
assem bly.
(a) A plan view
(b) A half-sectional front view on A A
(c) A half-sectional end view on BB
Carn assem bly
The carn assem bly show n on pages 265 and 266 consists of the
follow ing details.
A cast iron body, item 1; a cast iron end cover, item 2; a case
hardened m ild steel cam shaft, item 3; a case hardened m ild steel
follow er, item 4; a bronze follow er bush, item 5; tw o bronze cam shaft bushes, item 6; and three hexagon head screw s.
260
M a c h in e
d r a w in g
A ssem ble the parts and draw the follow ing view s of the com plete
assem bly.
(a) A plan view corresponding to the given plan view of the body
(b) A sectional front view on A A . Show a broken-out section on
the cam shaft around the keyw ay
(c) A n outside end view looking on the end of the cam shaft
W hen assem bled the cham fered end of the follow er bears on the
earn.