Projections of Solids , Sectional Views
And
Intersections of Solids
Projection of Solids
Projections of Planes
Solid
A solid is a 3-D object having
length, breadth and thickness and bounded by surfaces which may be
either plane or curved, or combination of the
two.
Solids are classified under two main
headings
A regular polyhedron is solid bounded only by plane surfaces (faces). Its faces are formed by regular polygons of same size and all dihedral angles are equal to one another. when faces of a polyhedron are not formed by equal identical faces, they may be classified into prisms and pyramids.
Five regular polyhedral are shown in figure 1
Figure 1: Five regular polyhedra
Prism
Prisms are polyhedron
formed by two equal parallel regular polygon, end faces connected
by side faces which are either rectangles or
parallelograms.
Different types of prisms are shown
in figure 2
Figure 2. Various types of prisms generally encountered in engineering applications
Some definitions regarding prisms
Base and lateral faces. When the prism is placed vertically on one of its end faces, the end face on which the prism rests is called the base. The vertical side faces are the lateral faces, as shown in Figure 3.
Figure 3. Shows the base and lateral face of a prism.
Base edge/Shorter edge: These are the sides of the end faces, as shown in figure 4.
Figure 4. showing the base edge or shorter edge of a pentagonal prism.
Axis – it is the imaginary line connecting the end faces is called axis and is shown in figure 5.
Figure 5 showing the Axis of a triangular prism.
Longer edge/lateral edges: These are the edges connecting the respective corners of the two end faces. The longer edge of a square prism is illustrated in figure 6.
Figure 6. illustrating the longer edge of a square prism.
Right prism – A prism whose axis is perpendicular to its end face is called as a right prism .Prisms are named according to the shape of their end faces, i.e, if end faces are triangular, prism is called a triangular prism.
Oblique prism: It is the prism in which the axis is inclined to its base.
Pyramids
Pyramid is a polyhedron formed by a
plane surface as its base and a number of triangles as its side
faces, all meeting at a point, called vertex or apex.
Axis – the imaginary line connecting the apex and the center of the base.
Inclined/slant faces – inclined triangular side faces.
Inclined/slant/longer edges – the edges which connect the apex and the base corners.
Right pyramid – when the axis of the pyramid is perpendicular to its base.
Oblique pyramid – when the axis of the pyramid is inclined to its base.
Solids of revolution
when some of the plane figures are revolved about one of their sides – solids of revolution is generated some of the solids of revolution are:
Figure 7. Illustrates some examples of truncated / frustrum of solids.
Visibility
When drawing the orthographic views
of an object, it will be required to show some of the hidden
details as invisible. To distinguish the invisible portions
from the visible ones, the invisible edges of the object are shown
on the orthographic views by dashed lines. However, in practice,
these lines of dashes conveniently and colloquially, but wrongly
called as dotted lines. To identify the invisible portions of the
object, a careful imaginative thinking is
essential.
Rules of
visibility
When viewing an object, the outline
of the object is visible. Hence the outlines of all the views are
shown by full lines. All the visible edges will be shown as solid
lines as shown in figure 8. Figure shows the frustum of a
pentagonal pyramid.
Figure 8. Front view of the object. The visible edges
are shown as solid lines and the hidden edges are shown as dashed
lines.
Figure 9 shows the projections of the object. In the top view, the highest portions of the object are visible. The top face ABCDE is at the top and is completely visible in the top view. In the top view, edges ab, bc, cd, de and ea are shown as full lines. The bottom pentagonal faces A1B1C1D1E1 is smaller than the top face, hence invisible. The slant edges AA_{1}, BB_{1}, CC_{1}, DD_{1} and EE_{1} are invisible in the top view, hence they are shown as lines of dashes. The line connecting a visible point and an invisible point is shown as an invisible line of dashes unless they are out lines.
In the front view, the front faces of the object are shown as visible. The faces ABB1A1 and BCC1B1 are the front faces. Hence in the front view, the corners a, b, c and a_{1}, b_{1}, c_{1} are visible to the observer. Hence in the front view, the lines a’a’_{1}, b’b’_{1} and c’c’_{1} are shown as full lines. However the corners d, e, d_{1} and e_{1} are invisible in the front view. The lines, e’e’_{1}, d’d’_{1} are invisible, hence shown as dashed lines. The top rear edges a’e’, e’d’ and d’c’ coincide with the top front visible edges a’b’ and b’c’.
Figure 9. Projections of the frustrum of a pentagonal pyramid.
In the side view - the face lying on that side are visible. As seen in the left side view, the corners e, a, b and e_{1}, a_{1}, b_{1} lie on left side and are visible in the left view. Hence the lines, e”e_{1}”, a”a_{1}” and b”b_{1}” are shown as full lines. The edges d”d_{1}”, c”c_{1}” coincide with the visible edges e”e_{1}” and a”a_{1}” respectively.
Projections of solids placed in different positions
The solids may be placed on HP in various positions
Axis of the solid perpendicular to HP
A solid when placed on HP with its axis perpendicular to it, then it will have its base on HP. This is the simplest position in which a solid can be placed. When the solid is placed with the base on HP position, in the top view, the base will be projected in its true shape. Hence, when the base of the solid is on HP, the top view is drawn first and then the front view and the side views are projected from it. Figure 10 shows a cylinder with its axis perpendicular to HP. There is only one position in which a cylinder or a cone may be placed with its base on HP.
Figure 10. Front view and top view of a cylinder and cone
For prisms, there are 4 positions it may be placed with its base on HP.These positions are illustrated in figure 11.
Figure 11. Projections of a triangular prism resting on its base on HP with different positions.
There are 4 positions in which pyramids may be placed with its base on HP. These positions are shown in figure 12.
Figure 12. Projections of a triangular Pyramid resting on its base on HP with different positions.
Projections of a solid with the axis perpendicular to VP
When a solid is placed with
its axis perpendicular to VP, the base of the solid will always be
perpendicular to HP and parallel to VP. Hence in the front
view, base will be projected in true shape. Therefore, when the
axis of the solid is perpendicular to VP, the front view is drawn
first and then the top and side views are drawn from it. When
a cylinder rests on HP with its axis perpendicular to VP, one
of its generators will be on HP.
Figure 13 shows the Front view and
Top view of a cylinder and cone resting on HP with their axes
perpendicular to VP. In this case one of the points on the
circumference of the base will be on XY.
Figure 13 shows the Front view and Top view of a cylinder and cone
Prism may be placed with their axis perpendicular to VP in three different positions. The different positions are shown in figure 14.
Figure 14. Projections of a pentagonal prism resting on HP and axis perpendicular to VPwith different positions.
As shown in Figure 15, pyramid may be placed with their axis perpendicular to VP in three different positions.
Figure 15. Projections of a pentagonal pyramid resting on HP and axis perpendicular to VP with different positions.
Axis of the solid inclined to HP and parallel to VP
When a solid is placed on HP with
its axis inclined to HP, the elemental portion of the solid that
lies on HP depends upon the type of the solid.
When a prism is placed on HP with
its axis inclined to it, then it will lie either on one of its base
edges or on one of its corners on
HP.
When a pyramid is placed on HP with
its axis inclined to HP, then we will have one of its base edges on
HP or one of its base corners on HP or one of its slant edges
on HP or one of its triangular faces on HP or an apex on
HP.
Methods of Projection of Solids
Methods of drawing the
projections of solids
These are two methods for drawing
the projections of solids:
Change of position
method
In this method,
the solids are placed first in the simple
position and then tilted successively in two or three stages to
obtain the final position. The following are some of the
examples.
a. Method of obtaining the top and the front views of the pyramid when it lies on HP on one of its base edges with its axis or the base inclined to HP.
If the solid is required to be
placed with an edge of the base on HP, then initially the solid has
to be placed with its base on HP such that an edge of the base is
perpendicular to VP, i.e., to XY line in top view preferably to lie
on sthe right side.
When a pentagonal prism has to be
placed with an edge of base on HP such that the base or axis is
inclined to HP, then initially, the prism is placed with its base
on HP with an edge of the base perpendicular to VP and the lying on
the right side. In this position, the first set of top and front
views are drawn with the base edges (c_{1})(d_{1})
perpendicular to XY line in the top view. In the front view,
this edge c_{1}’(d_{1}’) appears as a
point.
Since the prism has to lie with an
edge of the base on HP, the front view of the prism is tilted on
the edge c_{1}’(d_{1}’) such that the axis is
inclined at q to HP.
Redraw the first front view in the
tilted position. Whenever the inclination of axis q with HP is
given, first the base is drawn at (90- θ) in the front view,
otherwise improper selection of the position of the axis may result
in the base edge c_{1}’(d_{1}’) lying above
or below the XY line. The second top view is projected by drawing
the vertical projectors from the corners of the second front
view and the horizontal projectors from the first top view.
Figure 1 shows the sequence in obtaining the projection of
the solid for the above case.
Figure 1. Illustrating the sequence for obtaining the projections of a pentagonal prism placed with an edge of base on HP such that the base or axis is inclined to HP
Following the above procedure, the top and front views of the pyramid when it rests on HP on one of its base corners such that the two base edges containing the corner on which it rests make equal inclinations with HP is shown in figure 3.
Figure 3 showing the projection of a pyramid resting on HP on one of its base corners with two base edges containing the corner on which it rests make equal inclinations with HP
c. Projections of a pyramid lying on one of its triangular faces on HP
If a pyramid has to be placed on one of its triangular faces on HP, then initially let the pyramid be placed with its base on HP such that the base edge containing that face is perpendicular to VP. (i.e. perpendicular to XY line). Figure 4 illustrates the sequence in obtaining the projections of the pyramid. In the first front view, the inclined line, o’c’(d’) represents a triangular face. Redraw the front view such that the line representing the triangular face o’c’(d’) lies on HP. Project the top view in this position.
Figure 4 Illustrates the sequence in obtaining the projections a pyramid lying on one of its triangular faces on HP
d. Projections of a pyramid lying on one of its slant edge HP
The sequence of obtaining the projections of a pyramid lying on one of its slant edge on HP is shown in figure 5. In step-1, The FV and TV of the pyramid in the simple projection is drawn such that in the, top view the slant edge (line cd) on which it will rest on ground is drawn parallel to HP ( parallel to XY line) in the. In the front view this edge will be line c’ o’. In step-2, the object is then rotated such that the pyramid lies with its edge o’c’ on HP. i.e. in the front view, o’c’ lies on XY line. Project the Top view from this Front view.
Figure 5. Projections of a pyramid lying on one of its slant edges on HP
Examples In Projection of Solids
Problem
1.
A cube of 30 mm sides is held on one
of its corners on HP such that the bottom square face containing
that corner is inclined at 30^{0} to HP. Two of its
adjacent base edges containing the corner on which it rests are
equally inclined to VP. Draw the top and front views of the
cube.
Solution:
The procedure of obtaining the
projections is shown in figure 6. InStep-1, the projections of the
cube is drawn in the simple position. The cube is assumed to lie
with one of its faces completely on HP such that two vertical faces
make equal inclinations with VP. Draw a square abcd to represent
the top view of the cube such that two of its sides make equal
inclinations with the XY line, i.e., with VP. Let (a_{1}),
(b_{1}), (c_{1}) and (d_{1}) be the four
corners of the bottom face of the cube which coincide in the top
view with the corners a, b, c and d of the top face. Project the
front view of the cube. The bottom face
a_{1}’b_{1}’c_{1}’(d_{1}’) in the
front view coincide with the XY line. Now the cube is tilted
on the bottom right corner c_{1}’ (step-2) such that
the bottom face
a_{1}’b_{1}’c_{1}’(d_{1}’) is
inclined at 30^{0} to HP. Reproduce the front
view with face
a_{1}’b_{1}’c_{1}’(d_{1}’) inclined
at 30^{0} to the XY line.
Draw the vertical projectors through
all the corners in the reproduced front view and horizontal
projectors through the corners of the first top view. These
projectors intersect each other to give the corresponding corners
in the top view
Figure 1. The projections of the cube of problem 1.
Problem-2.
A cube of 30 mm side rests with one
of its edges on HP such that one of the square faces
containing that edge is inclined at 30^{0} to HP and
the edge on which it rests being inclined to 60^{0} to
VP. Draw its projections.
Solution.
The procedure of obtaining the
projections is shown in figure 7. First the TV and FV of the
cube is drawn with the cube in the simple position. The edge bc is
drawn perpendicular to the XY line. In step2, the cube is tilted
such that the base of the cube is inclined at
30^{0} to HP. The front view is reproduced with b1’
c1’a1’d1’ inclined at 30^{0} to XY. The top view of
the cube in step-2 is obtained by drawing projectors mentioned in
problem 1. In step-3, the top view in step-2 is rotated such
that line c1 b1 is inclined at 60^{0} to XY
line. The front view in step-2 is obtained by drawing
projectors from the top view in step-3 and Front view in
Step-2.
Figure 2. The projections of the cube of problem 2.
Problem
3
An equilateral triangular prism 20
mm side of base and 50 mm long rests with one of its shorter edges
on HP such that the rectangular face containing the edge on which
the prism rests is inclined at 30^{0} to HP. The
edge on which prism rests is inclined at 60^{0} to
VP. Draw its projections.
Solution: The procedure of
obtaining the projections is shown in figure 8. The prism rests
with one of its shorter edges, i.e., triangular or base edge on HP
such that the rectangular face containing that edge is
inclined at 30^{0} to HP.
Draw the simple views of the prism
when it rests with one of its triangular faces, i.e., base
completely lying on HP and also with one of its shorter edges
perpendicular to VP, i.e., to XY line. The shorter edge
(b_{1})(c_{1}) is perpendicular to the XY
line. The rectangular face containing the edge
b_{1}’(c_{1}’) is
b_{1}’b’(c’)(c_{1}’).
Now tilt the prism on the edge
b_{1}’(c_{1}’) such that the rectangular face
b_{1}’b’(c’)(c_{1}’) is inclined at
30^{0} to the XY line. In this tilted position,
project the top view.
It is seen that the edge
b_{1}c_{1} in the top view shown is
perpendicular to VP, i.e, to XY line. But the edge
b_{1}c_{1} has to be inclined at
60^{0} to VP, i.e, to XY line. Therefore,
reproduce the top view with the edge
b_{1}c_{1} inclined at 60^{0} to
the XY line as shown in the top view.
Project the reproduced top view to
get the front view.
Figure 3. The projections of the triangular prism of problem-3
Problem-4
A hexagonal pyramid has an altitude
of 60 mm and side base 30mm. The pyramid rests on one of its
side of the base on HP such that the triangular face containing
that side is perpendicular to HP. Draw the front and top
views.
Solution: The solution the problem
is shown in figure 9. In step-1, the pyramid is drawn in the simple
position with base edge cd perpendicular to XY line. In
Step-2, the Front view is tilted about cd such that line o’c’d’ is
made perpendicular to XY line. The top view is obtained by drawing
projectors from the top view of step 1 and front view in
step-2.
Figure 4. The projections of the hexagonal pyramid of problem-4.
Problem-5
Problem Draw the top and
front views of a rectangular pyramid of sides of base 40x 50 mm and
height 70 mm when it lies on one of its larger triangular faces on
HP. The longer edge of the base of the triangular face lying
on HP is inclined at 60^{0} to VP in the top view with
the apex of the pyramid being nearer to VP.
Solution
:
The solution the problem is shown in
figure 5. The projectors are obtained in 3-steps as illustrated in
the figure. In the first step, the solid is projected in the
simple position with base BC perpendicular to VP. In the second
step, the solid is tilted about the edge BC such that the face BCO
is made to lie on the ground. The front view is rotated and the top
view is projected from the front view and the top view in the first
step. In step-3, the top view is rotated such that edge BC is
inclined at 60° to XY line . The Front view is projected using this
top view and Front view of Step-2.
Figure 5. The projections of the rectangular pyramid of problem-5.
Problem-6
A cone of base 80 mm diameter and
height 100 mm lies with one of its generators on HP and the axis
appears to be inclined to VP at an angle of 40^{0} in
the top view. Draw its top and front views.
Solution:
Figure 6 illustrates the procedure
for obtaining the projections of the cone. Three steps are
involved. In step-1, the Top view and Front View of the cone
is drawn in the simple position. The base circle is divided in to
12 equal parts. These points ate joined with the apex to obtain the
respective generators. Instep 2, the cone is tilted such that
the cone lies on one of its generator in the HP. i.e. the generator
g’o’ is made to coincide with the XY line. The top view of
the object in this condition is drawn by drawing projectors.
In step-3, the cone is titled such that in the top view the axis is
inclined at 400 to the XY line. The front view of the object is
obtained by projection technique.
Figure 6. The projections of the cone of problem-6.
Problem-7
Draw the top and the front views of
a right circular cylinder of base 45 mm diameter and 60 mm long
when it lies on HP such that its axis is inclined
at 35^{0} to HP and the axis appears to
be perpendicular to VP in the top view.
Solution:
The solution to the problem is
illustrated in figure-7 . Three steps are involved as shown in the
figure. In Step-1, the cylinder is drawn in the simple
position (resting on the base on HP). The circle in the top
view is dived in to 12 equal parts and then projected in to the
front view. In step-2, The Front view is rotated about g1 such that
the axis is inclined at 35° to HP (or XYline). The top view is
projected from this front view with thehelp of Top view in step-1.
. In step 3, the top view is rotated such that axis is
perpendicular to XY line. The front view is then projected from the
top view.
Figure 7. The projections of the cylinder as per problem-7.
Sections of Solids
Section Views
Introduction
In engineering industries, when the internal structure of an object
is complicated, it is very difficult to visualize the object from
its orthographic views since there will be several hidden
lines. In such case, the internal details are shown by
sectional views. Sectional views are an important
aspect of design and documentation since it is used to improve
clarity and reveal interior features of parts.
Sectional drawings are multi-view technical drawings that
contain special views of a part or parts, that reveal interior
features. A primary reason for creating a section view is the
elimination of hidden lines, so that a drawing can be more easily
understood or visualized. Traditional section views are based
on the use of an imaginary cutting plane that cuts through the
object to reveal interior features. This imaginary cutting
plane is controlled by the designer and are generally represented
by any of the following:
(a) Full section view, where the section plane go completely
through the object. Example shown in figure 1.
(b) Half section view, where the section plane go half-way
through the object. Example shown in figure 2.
(c) Offset section, where the sectional plane bent through
the features that are not aligned. Example shown in figure 3.
(d) Broken-out section where the section go through part of the
object . Example shown in figure 4.
Figure 1. Illustrates a full Section view
Figure 2. Illustrating a half section view
Figure 3. Illustrating an offset section
Figure 4. Illustrating a Broken-out Section
An important reason for using section views is to reduce the number of hidden lines in a drawing. A section view reveals hidden features without the use of hidden lines. Adding hidden lines to a section view complicates the drawing defeating the purpose of clarifying with a section. As illustrated in figure 5, there are times, when a minimum number of hidden lines are needed to represent features other than the primary one shown by the section. In the figure, the through and through cavity may not be represented properly since it could be mistaken with a circular hole.
Figure 5 Illustration of the need of section view.
Figure 6 illustrates a full section view of an object. As
shown in the correct drawing, the section lined areas are bounded
by visible lines, never by hidden lines, because the bounding lines
are visible in the section view Visible surfaces and edges that
represent a change of planes behind the cutting plane are drawn as
lines in a section view.
Figure 6 shows a section view for which the cutting plane
passes through the center of a counter bored hole. A line
represents the change of planes between the drilled and counter
bored holes and is an example of a visible feature that is behind
the cutting plane.
Figure 6 Shows a correct and incorrect representation along with the multiview drawing
Visualization of Section Views
Figure 7 is multi-view drawing of a part that may be difficult to
visualize in its 3-D form, because of the many hidden lines.
Figure 7 A multiview drawing of an object
A section view is created by passing an imaginary cutting plane vertically through the center of the part. Figure 8 is a 3D representation of the part after it is sectioned. The section view more clearly shows the interior features of the part. In the left corner of the figure, the cutting plane arrows, in the front view, point to the left, to represent the direction of sight for producing a right side view in full section. The direction of the arrow can also be thought of as pointing toward the half of the object being kept. The right half of the object is "removed" to reveal the interior features of the part. The line of sight for the section view is perpendicular to the cut surfaces, which means they are drawn true size and shape in the section view. Also, no hidden lines are drawn and all visible surfaces and edges behind the cutting plane are drawn as object lines. The corners of the section view are numbered as shown in the right hand figure so that they can be compared with the orthographic section view.
Figure 9 showing a full section and the physically sectioned plane of the object
The representation of the section view of the object shown in figure 9 is shown as (b) in figure 10. The section view in figure (a) shows only those surfaces touched by the cutting plane. Since conventional practice requires that features behind the cutting plane be represented, the change of planes between the two holes in the counter bored hole are shown in figure (b). If the section is viewed along the line of sight identified by the arrows in figure (c), arcs A, B, and C will be visible and should be represented as lines. In figure (b), the lines are 2-7,4-5,15-14. The counter bore and through holes are represented as rectangular features 2-7-6-3, and 4-5-14-15. All the surfaces touched by the cutting plane are marked with section lines. Because all the surfaces are the same part, the section lines are identical and are drawn in thesame direction. The center line is added to the counter bored hole to complete the section view.
Figure 10. showing the object, orthographic view, sectioned surface and sectional view.
CUTTING PLANE LINES: The cutting plane line show the exact line along which the cutting plane passes through the object. This represent the edge view of the cutting plane and are drawn in the view(s) adjacent to the section view. This is represented in figure 11. In the figure the cutting plane line is drawn in the top view, which is adjacent to the sectioned front view.
Figure 11 showing the representation of a cutting plane line
Cutting plane lines are thick (0.7 mm) dashed lines, that extend past the edge of the object 6 mm and have line segments at each end drawn at 90 degrees and terminated with arrows. The arrows represent the direction of the line of sight for the section view and they point away from the sectioned view. Two types of lines are acceptable for cutting plane lines in multi-view drawings. The normal representation of cutting plane lines are shown in figure 12. Line B-B is composed of alternating long and two short dashes, which is one of the two standard methods. The length of the long dashes varies according to the size of the drawing, and is approximately 20 to 40 mm. For a very large section view drawing, the long dashes are made very long to save drawing time. The short dashes are approximately 3 mm long. The open space between the lines is approximately 1.5 mm. Capital letters are placed at each end of the cutting plane line, for clarity or when more than one cutting plane is used on a drawing. The second method used for cutting plane lines is shown by line C-C, which is composed of equal-length dashed lines. Each dash is approximately 6 mm long, with a 1.5 mm space between.
Figure 12 Normal representation of cutting plane lines.
Placement of Cutting Plane Lines
Cutting plane lines are only added to a drawing for clarity. If the
position of the cutting plane is obvious, the line need not be
drawn. Also, if the cutting plane line is in the same position as a
center line, the cutting plane line has precedence.
In figure 10, the cutting plane appears as an edge in the top view
and is normal in the front view; therefore, it is
a frontal cutting
plane. The front half of the object is
"removed" and the front view is drawn in section.
If the cutting plane appears as an edge in the front view and is
normal in the top view, it is a horizontal cutting plane. The
top half of the object is "removed" and the top view is drawn in
section. Figure 13 shows a horizontal cutting plane.
Figure 13. A horizontal cutting plane.
If the cutting plane appears as an edge in the top and front
views and is normal in the profile view, it is a profile cutting
plane. The left (or right) half of the object
is "removed" and the left (or right) side view is drawn in section.
A profile cutting plane is shown by BB in figure 14.
Multiple sections can be done on a single object, as shown in the
figure 14. In this example, two cutting planes are used: one
a horizontal and the other a profile cutting plane. Both cutting
planes appear on edge in the front view, and are represented by
cutting plane lines A-A and B-B, respectively. Each cutting plane
will create a section view, and each section view is drawn as if
the other cutting plane did not exist.
Figure 14. shows a profile cutting plane.
Section Line Practices
Section lines or cross-hatch lines are added to a section view to
indicate the surfaces that are cut by the imaginary cutting
plane.
Different section line symbols can be used to represent various
types of materials.
However, there are so many different materials used in engineering
design that the general symbol (i.e., the one used for cast iron)
may be used for most purposes on engineering drawings.
The actual type of material required is then noted in the title
block or parts list or as a note on the drawing. The angle at
which lines are drawn is usually 45° degrees to the horizontal, but
this can be changed for adjacent parts shown in the same
section. Also the spacing between section lines is uniform on
a section view.
Material Symbols
The type of section line used to represent a surface varies
according to the type of material. Symbols generally used for
various materials are shown in figure 15. However, the general
purpose section line symbol used in most section view drawings is
that of cast iron. The specific type of steel to
be used will be indicated in the title block or parts list.
Occasionally, with assembly section views, material symbols are
used to identify different parts of the assembly.
Figure 15. General symbols used to represent various materials in section view.
The general purpose cast iron section line is drawn at a 45° and spaced 1.5 mm to 3 mm or more, depending on the size of the drawing. As a general rule, use 3mm spacing. Section lines are drawn as thin (.35 mm) black lines. The section lines should be evenly spaced and of equal thickness, and should be thinner than visible lines . Also, do not run section lines beyond the visible outlines or stop them too short. The correct and incorrect method of drawing section lines are shown in figure 16.
Figure 16. Representation of section line.
Section lines should not run parallel or perpendicular to the visible outline. If the visible outline to be sectioned is drawn at 45°, the section lines are drawn at a different angle, such as 30° as shown in figure 17.
Figure 17 showing the direction of section lines with respect to the visible outline of the object.
Avoid placing dimensions or notes within the section lined areas.
If the dimension or note must be placed within the sectioned area,
omit the section lines in the area of the note as shown in figure
17.
Figure 18 Placement of dimensions or notes within the section lines.
An outline section view is created by drawing partial section outlines adjacent to all object lines in the section view as shown in figure 19. For large parts, outline sectioning may be used to save time.
Figure 19. shows an outline section view.
Very thin parts such as washers and gaskets are not easily represented with section lines, so conventional practice calls for representing the thin part in solid black. Figure 20 shows the section of a gasket drawn as solid black.
Figure 20. Representation of section view of a very thin part (e.g. gasket)
Full Section View
A full section view is made by passing the imaginary cutting plane
completely through the object. As shown in figure 21, all the
hidden features intersected by the cutting plane are represented by
visible lines in the section view. Surfaces touched by the cutting
plane have section lines drawn at a 45-degree angle to the
horizontal. Hidden lines are omitted in all section views
unless they must be used to provide a clear understanding of the
object. The top view of the section drawing shows the cutting plane
line, with arrows pointing in the direction of line of sight to
view the sectioned half of the object. In a multi-view drawing, a
full-sectioned view is placed in the same position that an
un-sectioned view would normally occupy, I.e., a front section view
would replace the traditional front view.
Figure 21 shows a full section view of an object.
Half Section view
Half sections are created by passing an imaginary cutting plane
only halfway through an object. Hidden lines are omitted on both
halves of the section view. Hidden lines may be added to the
un-sectioned half, for dimensioning or for clarity. External
features of the part are drawn on the un-sectioned half of the
view. A center line, not an object line, is used to separate the
sectioned half from the un-sectioned half of the view. The cutting
plane line shown in the top view. The cutting plane line in the top
view is bent at 90° and one arrow is drawn to represent the line of
sight needed to create the front view in section. Half section
views are used most often on parts that are symmetrical, such as
cylinders. Also, half sections are commonly used in assembly
drawings when external features are also to be shown. Figure 22
shows a half section view of an object.
Figure 22 shows the cutting plane passing halfway through an object and one quarterof the object being removed
Broken-Out Sections
A broken-out section is used when only a portion of the object
needs to be sectioned. A broken-out section view of an object is
shown in figure 23. A break line separates the sectioned
portion from the un-sectioned portion of the view. A break
line is drawn freehand to represent the jagged edge of the break.
No cutting plane line is drawn. Hidden lines may be omitted
from the un-sectioned part of the view, unless necessary for
clarity. A broken-out section is also used instead of a half or
full section view to save time.
Figure 22 shows the broken out section of an object. Majority of the technical details are also shown.
Revolved Sections And Sections of Simple Solids
Revolved Sections
Revolved sections show the shape of an object's cross-section
superimposed on a longitudinal view as shown infigure 1.
Figure 1. Showing the revolved sections of (a) two hollow points joined together, (b) a Hand wheel (c) a long I- Joist.
While drawing the revolved sections of hand wheels, any part with an odd number of spokes or ribs will give an unsymmetrical and misleading section if the principle of true projections are strictly adhered to. Therefore the normal practice is that the the spoke is rotated to the path of the vertical cutting plane and then projected on the side view as shown in figure 2 .Neither of the spokes should be sectioned (hatched).
Figure 2. illustrates the practice of drawing revolved section view of a hand wheel.
Ribs, lugs and holes often occur in odd numbers and should be aligned to show the true relationship of elements. Figure 3 shows the normal practice of drawing revolved sections of ribs, holes, and lugs. True projection of the ribs (A) show the pair on the right shortened suggesting that they does not extend to the outer edge. Section (B) shows symmetry. Rib part is not sectioned. At (C), the lugs and holes are aligned showing the holes at their true radial distance from the axis. Ribs are never shown sectioned (hatched).
Figure 3. Practice of drawing revolved sections of ribs, holes, and lugs
The practice of drawing revolved sections of pulleys having web and spokes are shown in figure 4. . Pulley A has a solid web connecting the hub and rim. Pulley B has four spokes. Even though the cutting plane passes through two of the spokes, the sectional view of B must be made with out hatching the spokes in order to avoid the appearance of a solid web.
Figure 4. representation of pulleys having we and spokes.
Few examples of representing revolved sections are shown in figure 5.
Figure 5 some examples of revolved sections of objects.
Section of simple solids
The subsequent paragraphs explains the sections of simple solids
when the section plane is oriented differently with respect to the
projection planes.
To draw the projection of the solid when sectioned by a plane whichis parallel to the VP, first the projections of the solid is drawn without section plane. (i.e. top view and front view according to the given conditions). Then introduce the section plane in the top view. As the section plane is parallel to the VP, is seen as a line in top view. Project the points of intersection of the section plane (line HT in the Top view in figure 6) to the respective points in the front view. E.g. point 4 , whichis the intersection of line HT with edge ef in the top view is projected to point 4’ . This is obtained by the intersection of the vertical projector from point 4 and the edge represented by e’f’ in the front view. Siimilarly all other points like 3’, 2’ and 1’ is obtained. Since the portiaon away from the VP (i.e. line XY) is removed, the shaded area represents the sectional front view of the object. Though the section are is to be drawn by hatched line, it is shown as colored green for convenience. The same would be followed.
Figure 6. Sectional view of a square prism sectioned by a plane parallel to VP.
Problem -1: A triangular prism, side of base 30 mm and axis 50 mm long is lying on the HP on one of its rectangular faces with its axis inclined at 30° to the VP. It is cut by a horizontal section plane at a distance of 12 mm above the ground. Draw its front view, side view and sectional top view
Solution: The Top view and front view of the triangular prism with respect to the projection planes is drawn first. In the Front view, the section plane (line V-T) which is parallel to HP will be a line parallel to and 12 mm above XY line. It is evident from figure 7. that the line VT intersects with the edges a’-b’ at 1’, a’-c’ at 2’, d’-e’ at 4’ and d’-f’ at 3’. Draw verticalprojectors from this points to meet the edges a-b at 1, a-c at 2, e-d at 4 and d-f at 3 respectively. When the top portion of the prism is removed, the top view will beobtained as a rectangle, i.e the hatched region bounded by 1-2-3-4.
Figure 7. Sectional view of the object in problem-1.
Problem-2: A pentagonal pyramid, side of
base 30 mm and axis 65 mm long, has its base horizontal and an edge
of the base parallel to the VP. A horizontal section plane cuts it
at a distance of 25 mm above the base. Draw its front view and
sectional top view.
Solution : Figure 8 shows the solution of the
problem. The sectional view is obtained in the same manner as
in probel-1. Here the intersection the slant edges of the pyramid
with the section plane is projected on to the respective
edges in the top view.
Figure 8. Sectional view for the problem-2.
Problem-3 A pentagonal pyramid has its
base on the HP. Base of the pyramid is 30 mm in side, axis 50 mm
long. The edge of the base nearer to VP is parallel to it. A
vertical section plane, inclined at 45° to the VP, cuts the pyramid
at a distance of 6 mm from the axis. Draw the top view,
sectional front view and the auxiliary front view on an AVP
parallel to the section plane.
Solution:
The solution is shown in figure 9. The Front view and top view of
the pyramid is first drawn. The vertical section which is inclined
at 45° to VP is shown by the line HT inclined at 45° to XY line.
Since the section plane passes through a point 6 mm from the axis
of the pyramid, a circle of radius 6 mm is drawn in the top view.
The line HT is drawn such that it is tangent to this circle.
The sectional front view of the object is obtained by drawing
vertical projectors and its intersection with the
respective slant edges in the front view. Since point 2 (in
the top view) is in the line ob, which is vertical, the projection
of point 2’ is obtained in the following manner. With centre O and
radius o-2, draw an arc to intersect the line o-c at
2_{1}. Project point 2_{1} in the top
view to point 2’ in the Front view. Draw a horizontal projector
from 2’ to intersect the line o’-b’ at 2’. The section front
view is shown by the hatched area in the front view.
The auxiliary front view is drawn to obtain the true shape of the
section. This is obtained by the auxiliary projection method shown
in figure 9 .
Figure 9. Solution to problem – 3.
Problem-4
A cube of 80 mm long edges has its vertical faces equally inclined
to the VP. It is cut by an AIP in such a way that the true shape of
the cut part is a regular hexagon. Determine the inclination of the
cutting plane with the HP. Draw FV, sectional TV and true shape of
the section.
Solution:
The solution is shown in figure 10. Draw TV and FV of the cube.
As the true shape of the section is a hexagon, the cutting
plane must cut the prism at 6 points. The cutting plane will cut
two edges of the top, two edges of the base and two vertical edges.
The Points of intersection at two vertical edges will be farthest
from each other. These points will represent the two opposite
corners of the hexagon and the distance between them will be equal
to b( b1)– d( d1).
Draw a line 3–6 = b( b1)– d( d1). Draw a circle with 3–6 as a diameter. Inscribe a hexagon 1– 2–3–4–5–6 in it as shown. Measure the distance between 1–2 and 4–5, i.e., PQ. In the front view, locate 3’ at the midpoint of b’( d’)– b1’( d1’). With 3’ as a centre and radius = ½( PQ), cut arcs on a’ b’ and b1’ c1’ to locate 1’ and 4’ respectively. Join 1’–4’ for the required cutting plane. Measure θ.
Draw X1 Y1 parallel to 1’–4’. Redraw hexagon 1–2–3–4–5–6 as 11–21–31–41–51–61 such that pq is parallel to X1 Y1. Project all the corners of the hexagon in FV. 2’, 6’ and 5’ will coincide with 1’, 3’ and 4’ respectively. Project 1’, 2’, 3’, etc., to 1, 2, 3, etc., on the corresponding edges in TV to obtain the section. 3 and 6 will coincide with d( d1) and b( b1) respectively
Figure 10. Solution to problem - 4
Intersections of Solids
Intersection of solids
Whenever two or more solids combine, a definite curve is seen at their intersection. This curve is called the curve of intersection (COI). Lines of intersection are a common feature in engineering applications or products. Figure 1 shows few examples of intersection lines frequently observed in chemical plants, domestic appliances, pipe joints, etc. Curves of intersections are important from the point of view of production of components for engineering applications.
Figure 1. photographs of engineering components showing cures of intersection.
Cases of Intersection
The cases of intersection depend on the type of intersecting solids
and the manner in which they intersect. Two intersecting solids may
be of the same type (e.g., prism and prism) or of different types
(e.g., prism and pyramid). The possible combinations are
shown in table 1.:
Table 1. Different cases of intersection
The two solids may intersect in different ways. The axes of the
solids may be parallel, inclined or perpendicular to each other.
The axes may be intersecting, offset or coinciding. Therefore, the
following sub-cases exist:
(i) Axes perpendicular and intersecting
(ii) Axes perpendicular and offset
(iii) Axes inclined and intersecting
(iv) Axes inclined and offset
(v) Axes parallel and coinciding
(vi) Axes parallel and offset
The type of intersection created depends on the types of
geometric forms, which can be two- or three- dimensional.
Intersections must be represented on multiview drawings
correctly and clearly. For example, when a conical and a
cylindrical shape intersect, the type of intersection that occurs
depends on their sizes and on the angle of intersection relative to
their axes as well as relative position of their axes.
The line of intersection is determined using auxiliary views and
cutting planes.
Methods – (1) Line and (2) Cutting-plane methods
Line method:A number of lines are drawn on the lateral surface of one of the solids and in the region of the line of intersection. Points of intersection of these lines with the surface of the other solid are then located. These points will lie on the required line of intersection. They are more easily located from the view in which the lateral surface of the second solid appears edgewise (i.e. as a line). The curve drawn through these points will be the line of intersection.
Cutting-plane method: The two solids are assumed to be cut by a series of cutting planes. The cutting planes may be vertical (i.e. perpendicular to the H.P.), edgewise (i.e. perpendicular to the V.P.) or oblique. The cutting planes are so selected as to cut the surface of one of the solids in straight lines and that of the other in straight lines or circles.
Intersection of two prisms
Prisms have
plane surfaces as their faces. The line of intersection between two
plane surfaces is obtained by locating the positions of points at
which the edges of one surface intersect the other surface and then
joining the points by a straight line. These points are called
vertices. The line of intersection between two prisms is therefore
a closed figure composed of a number of such lines meeting at the
vertices. The method of obtaining intersection lines are
shown by means of problem.
Problem 1. A vertical square prism, base
50 mm side, is completely penetrated by a horizontal square prism,
base 35 mm side, so that their axes intersect. The axis of the
horizontal prism is parallel to the prism, while the faces of the
two prisms are equally inclined to the prism. Draw the projections
of the solids, showing lines of intersection. (Assume suitable
lengths for the prisms.)
Solution
The intersection obtained is shown in figure
2. Draw the projections of the prisms in the
required position. The faces of the vertical prism are seen as
lines in the top view. The points of intersection in this
view can be located by the following method.
Lines 1-1 and 3-3 intersect the edge of the vertical prism at
points p_{1} and p_{3} (coinciding
with a).Lines 2-2 and 4-4 intersect the faces
at p_{2} and p_{4} respectively.
The exact positions of these points along the length of the
prism may now be determined by projecting them on corresponding
lines in the front view. For
example, p_{2 }is projected
to p_{2}' on the line 2'2'. Note
that p4' coincides
with p_{2}'.Similarly other points are
obtained. Draw lines p_{1}’p_{2}'
and p_{2}‘p_{3}'. Lines
p_{1}‘p_{4}' and
p_{3}‘p_{4}' coincide with the
front lines. These lines show the line of intersection. Lines
q_{1}'q_{2}' and
q_{2}‘q_{3}' on the other side
are obtained in the same manner. Note that the lines for the hidden
portion of the edges are shown as dashed lines. The green lines in
the figure represents construction lines. The portions
p_{1}’p_{3}' and
q_{1}’q_{3}' of vertical edges
a'a' and c'c' do not exist and hence, must be removed or kept
fainter.
Figure 2 The lines of intersection obtained for problem 1.
Intersection of Cylinder and Cylinder
The line of intersection between cylinders will be curved since the
lateral surfaces of cylinders are curved. Points on the
intersection line can be located by either line method or cutting
plane method. For plotting an accurate curve, the points at which
the curve changes direction must also be located. These
points lie on the outermost or extreme lines of each cylinder
pierce the surface of the other cylinder. For obtaining the
intersection of cylinder and cylinder, cutting plane method is more
useful. In this technique, a series of horizontal cutting
planes are assumed to be passing through the lines of the
horizontal cylinder and these planes cuts both cylinders as shown
in figure 3. After each sectioning, the top view of the section of
the horizontal cylinder will be a rectangle with its width
depending on the position of the cutting plane, where as the top
view of the vertical cylinder will be a circle with diameter equal
the diameter of the vertical cylinder. The points at which
the sides of the rectangles intersect the circle will lie on the
curve of intersection. The procedure is illustrated through
problem 2.
Figure 3 illustrating the cutting plane method for obtaining the curve of intersection.
Problem 2
A vertical cylinder of 80 mm diameter is completely penetrated by
another cylinder of 60 mm diameter, their axes bisecting each other
at right angles. Draw their projections showing curves of
penetration, assuming the axis of the penetrating cylinder to be
parallel to the VP.
Solution:
The solution is shown in figure 3. The Front view, top view and
side view of the two cylinders are drawn with out the intersection
lines. Divide the circumference of the circle in the side view in
to 12 equal parts. Draw horizontal projectors from these
points on to the front view. Project the same points from the
side view on to the top view and obtain lines 1-1, 2-2,
3-3 , etc. Let us now consider a horizontal sectio plane
passing through points 2-2 and 12-12 (shown in figure
3).
In the front view, it will be seen as a line coinciding with
line 2’ 2'. In the top view, the section of the horizontal cylinder
will be a rectangle of width (i.e. the line 2-12). The section of
the vertical cylinder will be a circle.
Points p_{2} and p_{12} at
which the sides (2-2 and 12-12) of the rectangle cuts the circle
will lie on the curve of intersection. This point is first obtained
in the top view by the intersecting point of line 12-12
and 2-2 with the circle. Vertical projector lines are drawn
from these points to the front view so as to intersect with
the horizontal projectors drawn through points 2 and 12 in the side
view to obtain P2’ and P12’ in the front view.
Other cutting planes are also assumed passing through 3-11, 4-10,
etc and the procedure repeated to obtain other points p3’, p4’,
p5’, etc. Similar procedure is adopted to obtain points q1’, q2’,
etc on the right hand side. Since the axis to the two cylinders
intersect, points p2’ and p12’ will coincide and hence cannot
be shown in the figure.
Figure 4. solution to problem 2.
Intersection of Cone and Cylinder
Figure 5 illustrate the cutting plane method for obtaining the curve of intersection. In this the top view of the cone after sectioning will be a circle and that of the cylinder will be a rectangle with width W. The diameter of the circle and the width of the rectangle in the top view will depend on the height of the section from the cone base. At each height, the intersection points P2, P3, P4 … are obtained and finally these points are joined by a smooth curve in the front view.
Figure 5 illustrating the cutting plane method for obtaining the curve of intersection.
Problem 3
Example - A vertical cone, diameter of base 75 mm and axis 100 mm
long, is completely penetrated by a cylinder of 45 mm diameter. The
axis of the cylinder is parallel to HP and VP and intersects the
axis of the cone at a point 22 mm above the base. Draw the
projections of the solids showing curves of
intersection.
Solution
Draw lines dividing the surface of the cylinder into twelve equal
parts.
Assume a horizontal cutting plane passing through say, point 2. The
section of the cylinder will be a rectangle of width w (i.e. the
line 2-12), while that of the cone will be a circle of
diameter ee. These two sections intersect at points
p_{2} and p_{12}. These sections are clearly
indicated in the top view by the rectangle 2-2-12-12 and the circle
of diameter ee. In the front view, the cutting plane is seen as a
line coinciding with 2' 2’. Points p_{2} and
p_{12} when projected on the line 2' 2’ (with which
the line 12'-12' coincides) will give a point p2' (with which p12'
will coincide). Then p_{2}' and p_{12}' are the
points on the curve of intersection. To obtain the points
systematically, draw circles with centre 0 and diameters dd, ee,
ff,etc. cutting lines through 1, 2 and 12, 3 and 11 etc. at points
p_{1}, p_{2} and p_{12},
p_{3} and p_{ll} etc . Project
these points to the corresponding lines in the front view.
Figure 6. solution to problem 3.
Examples In Lines In Lines of Intersections of Solids
Intersection of Prism and Pyramid
A square pyramid with a base side of 55 mm and an axis length of 80
mm stands on its base on the HP with the sides of base equally
inclined to the VP. A triangular prism with a base side of 34 mm
and length of axis 100 mm, penetrates the pyramid completely. The
axis of the prism is perpendicular to the VP and intersects the
axis of pyramid at 25 mm from the HP. One of the lateral faces of
the prism is perpendicular to the HP. Draw the three views of the
solids showing LOI.
Intersection of Prism and Cylinder
A vertical cylinder with a 60 mm diameter is penetrated by a
horizontal square prism with a 40 mm base side, the axis of which
is parallel to the VP and 10 mm away from the axis of the cylinder.
A face of the prism makes an angle of 30° with the HP. Draw their
projections showing curves of intersection.
Intersection of Cylinder and Cone
A cone with a base diameter of 64 mm and an axis length of 70 mm is
kept on its base on the HP. A cylinder of diameter 30 mm and length
90 mm penetrates the cone horizontally. The axis of the cylinder is
20 mm above the base of the cone and 5 mm away fromthe axis of the
latter. Draw the three views of the solids showing curve of
intersection.
A vertical pentagonal prism 30 mm edge of base and height 100 mm has one of its rectangular faces parallel to VP and nearer to it. It is penetrated by a rectangular prism of side 40mm x 20 mm and 100 mm high, with its front largest lower front rectangular face inclined at 60° to HP. The axis of the rectangular prism is inclined at 30° to HP and parallel to VP, 5 mm infront of the axis of the pentagonal prism and appears to bisect it in the front view. Draw the interpenetration line.
Problem: A square prism of 40 mm edge of base and 90 mm high rests vertically with its base on HP such that the front right vertical rectangular face is inclined at 60° to VP. This prism is penetrated by another horizontal square prism whose rectangular faces make equal inclination with both HP and VP. The axis of the horizontal prism is passing at the mid height at a distance of 10 mm infront of the vertical prism. The horizontal square prism is of the same dimensions as that of the vertical square prism. Draw the lines of intersection.
A vertical cylinder of 40 mm diameter and 80 mm high is intersected by another cylinder of 35 mm diameter and 80 mm long. The axis of the penetrating cylinder is inclined at 30° to HP, parallel to VP, 6 mm infront of the vertical cylinder and appears to bisect it in front view. Draw the intersection curve.