Chapter 15
SECTIONS OF SOLIDS
An object is difficult to visualize from its orthographic views if its internal
structure is complicated. In such a case, the concept of ‘sectioning the object’
helps to interpret and visualize the object easily. The object is assumed to be
cut by an imaginary plane to reveal internal details. The imaginary plane which
cuts the object is called the cutting plane (or section plane). The new
imaginary face generated on the object is called the section.
THEORY OF SECTIONING
Whenever a section plane cuts a solid, it intersects (and or coincides with) the
edges of the solids. The point at which the section plane intersects an edge of
the solid is called the point of intersection (POI). In case of the solids having a
curved surface, viz., cylinder, cone and sphere, POIs are located between the
cutting plane and the lateral lines. A section plane will cut a minimum of three
edges of the polyhedra, creating three POIs. The maximum number of POIs
depends on the type of polyhedron, i.e., prism or pyramid, as shown in Table
15.1.
True Shape of a Section
A section will show its true shape when viewed in normal direction.
Obviously, to find the true shape of a section, it must be projected on a
plane parallel to the section plane.
For polyhedra, the true shape of the section depends on the number of
POIs. The shape of the section will be a polygon of the sides equal to the
number of POIs. The true shape of the section of a sphere is always a
circle. The sections of prisms and pyramids are straight line segmented
curves. The sections of cylinders and cones will mostly have smooth
curves.
The true shapes of the sections of the cylinder and cone, depending on the
position of the cutting plane, are explained in Table 15.2.
Locating the Section Plane When the True Shape of the Section is Known
Locating the section plane when the true shape of the section is given is a
reverse process. For polyhedra, the cutting plane must cut the solid at the points
equal to the number of corners in the section. For example, if the true shape of a
section of a cube is a hexagon, then the cutting plane must create 6 POIs in FV or
TV. To locate the cutting plane, it is a usual practice to draw the true shape of the
section in auxiliary view and then, projecting all the corners in FV or TV.
SECTIONS OF PRIMS AND CUBES
Example 15.1 A triangular prism with a base side of 50 mm and an axis length
of 70 mm is resting on its rectangular face on the HP with the axis
perpendicular to the VP. The prism is cut by a horizontal section plane
passing through the axis. Draw FV and sectional TV of the prism.
Solution Refer Fig. 15.1.
1. Draw FV and TV of the prism as shown.
2. In FV, draw the section plane, parallel to XY and passing through the axis
as shown. The direction of the arrowheads should be towards XY (since it is a
first angle method of projection).
3. Locate POIs 1’, 2’, 3’ and 4’ in FV. 1’, 2’, 3’ and 4’ represent the intersections
of the cutting plane with a’ c’, b’ c’, b1’ c1’ and a1’ c1’ respectively.
4. Project 1’, 2’, 3’ and 4’ to 1, 2, 3 and 4 on the corresponding edges in TV.
Join 1–2–3–4 and hatch the area to indicate the section. 1–2–3–4 represents
the true shape of the section (since the cutting plane is parallel to XY).
Example 15.4 A triangular prism, with a base side of 50 mm and an axis
length of 70 mm, is resting on a rectangular face on the HP, the axis being
parallel to the VP. An AIP inclined at 45° to the HP cuts the prism. The cutting
plane intersects the axis at a distance of 30 mm from one end of the prism.
Draw FV, sectional TV and sectional SV of the prism.
Solution Refer Fig. 15.4.
1. Draw SV, FV and TV of the prism as shown.
2. In FV, draw the section plane, inclined at 45° to XY and intersecting the
axis at 30 mm from an end of the prism, as shown.
3. Locate 1’, 2’ and 3’ in FV at the intersections of the cutting plane with a’ a1’,
b’ b1’ and c’ c1’ respectively.
4. Project 1’, 2’ and 3’ to 1, 2 and 3 on corresponding edges in TV. Join 1–2–3
and hatch the section.
5. Obtain section 1”–2”–3” in SV by projecting POIs from FV and TV. 1”, 2”
and 3” will obviously lie at a”, b” and c” respectively.
Example 15.6 A cube of 50 mm side length rests on an edge on the HP. The
edge is parallel to the VP and the two faces sharing the edge are equally
inclined to the HP. An AIP, inclined at 47° to the HP and passing through one
of the top corners of the cube, cuts the cube. Draw FV, sectional TV and
sectional SV. Also, draw the true shape of the section.
Solution Refer Fig. 15.6.
1. Draw SV, FV and TV of the cube as shown.
2. In FV, draw the section plane passing through c1’ and inclined at 47° to XY.
3. Locate 1’, 2’, 3’, 4’ and 5’ in FV at the intersections of the cutting plane with
the edges. (1’ coincides with c1’.)
4. Project 2’ and 5’ to 2 and 5 on the corresponding edges in TV. Project 1’, 3’
and 4’ to 1”, 3” and 4” on the corresponding edges in SV. (1” coincides with
c1”.)
5. Project 1”, 3” and 4” to 1, 3 and 4 in TV. (1 coincides with c1.) Join 1–2–3–4–
5 and hatch the area.
6. Project 2 and 5 to 2” and 5” in SV. (2 and 5 coincide with b” ( b1”) and d”
(d1”) respectively.) Join 1”–2”–3”–4”–5” and hatch the area.
7. To draw the true shape of the section, draw X1 Y1 parallel to the cutting
plane. Project 1’, 2’, 3’, 4’ and 5’ on X1 Y1 and draw auxiliary view 11–21–
31–41–51. (Distances of 11, 21, 31, 41 and 51 from X1 Y1 = Distances of 1, 2,
3, 4 and 5 from XY.)
Example 15.7 A triangular prism with a base side of 50 mm and a 70 mm
height stands on its base on the HP with a rectangular face perpendicular to
the VP. It is cut by different AIPs such that the true shape of the section is
(i) an isosceles triangle of 42 mm base and 38 mm height
(ii) an isosceles triangle of maximum size
(iii) a trapezium of parallel sides of 50 mm and 22 mm
Locate the cutting plane and the draw FV, TV and the true shape of the section
in each case.
Solution For the true shape of the section to be a triangle, the section plane
must cut 3 edges of the prism. Similarly, the section plane must cut 4 edges
for the section to be trapezium.
Draw TV and FV of the prism as shown in Fig. 15.7.
Case (i): True shape: Isosceles triangle of 42 mm base and 38 mm height For
the isosceles triangle of given size, the section plane must cut two adjacent
edges of the top at points 42 mm apart. The third point can be located on the
vertical edge (emerging from the intersection of the two top edges) such that
the distance of that point from the line joining the first two points will be equal
to the altitude of the triangle.
1. In TV, locate 1 and 2 such that 1–2 = 42 mm. Project 1 and 2 to 1’(2’) in FV.
2. With 1’(2’) as a centre and radius = 38 mm, cut an arc on the vertical edge at
3’. Draw the cutting plane A– A through 1’(2’)–3’.
3. Project 1’, 2’ and 3’ on the cutting plane A– A to obtain auxiliary view 1–2–3
revealing true shape of the section.
Case (ii): True shape: Isosceles triangle of maximum size
For the maximum size of the isosceles triangle, the cutting plane must pass
through a top edge and the base corner opposite to that edge.
1. Draw the cutting plane B– B through 4’(5’)–6’.
2. Project 4’, 5’ and 6’ on the cutting plane B– B to obtain the auxiliary view 4–
5–6, indicating the true shape of the section.
Case (iii): True shape: Trapezium of parallel sides 50 mm and 22 mm
For trapezium, the cutting plane must cut both the ends of the prism. The
distance between the intersecting points on each face should be equal to the
parallel side of the trapezium. In this case, one of the parallel sides is 50 mm
long. Hence, the cutting plane must pass through a base edge.
1. In TV, locate 7 and 8 on two adjacent sides such that 7–8 = 50 mm.
2. Project 7 and 8 to 7’(8’) in FV.
3. Draw the cutting plane C– C through 7’ (8’)–9’ (10’). Project 7’ (8’) and 9’ (10’)
on C– C and obtain the true shape of the section 7–8–9–10.
Example 15.9 A square prism, base side 40 mm and a 75 mm axis length rests on
its base on the HP with all the sides of the base equally inclined to the VP.
Different AIPs cut the prism in such way that the true shape of the section is
(i) a rhombus of maximum size
(ii) a rhombus of longest diagonal 65 mm
(iii) a trapezium of parallel sides 50 mm and 15 mm
(iv) a pentagon of 25 mm base side and maximum altitude
Locate the cutting plane and draw FV and the true shape of the section in each
case.
Solution The section plane must cut 4 edges of the prism for a quadrilateral true
shape and 5 edges for a pentagonal true shape of the section.
Refer Fig. 15.9. Draw TV and FV of the prism.
Case (i): True shape: Rhombus of maximum size
For the rhombus of maximum size, the cutting plane must pass through two
opposite corners of the end faces. The longest diagonal of the rhombus will be
equal to the shortest distance between these points, i.e., the solid diagonal of the
prism. The shortest diagonal will, always, be equal to the diagonal of the end face.
1. Draw the cutting plane A– A through 1’–3’.
2. Locate 2’(4’) at the intersections of the cutting plane with the
intermediate edges.
3. Locate 1, 2, 3 and 4 in TV at the corners.
4. Project 1’, 2’, 3’ and 4’ on the cutting plane A– A to obtain true shape of
the section 1–2–3–4.
Case (ii): True shape: Rhombus of longest diagonal 65 mm The cutting
plane, in this case, will cut two opposite vertical edges such that the
distance between the intersecting points will be 65 mm.
1. With any suitable point on an extreme vertical edge, say 5’, as a centre
and radius = 65 mm, cut an arc at 7’ on another extreme vertical edge.
2. Draw the cutting plane B– B through 5’–7’. Locate 6’ (8’) at the
intersections of the cutting plane with the intermediate vertical edges.
3. In TV, 5, 6, 7 and 8 will be seen at the corners.
4. Obtain the true shape of the section 5–6–7–8 in a similar way explained
in Case (i) above.
Case (iii): True shape: Trapezium of parallel sides 50 mm and 15 mm
The cutting plane must cut both the ends of the prism on one side of the
axis. The lines joining the intersecting points on the corresponding
faces will represent parallel sides of the trapezium.
1. In TV, locate 9, 10, 11 and 12 on two adjacent sides such that 9–10 = 15
mm and 11–12 = 50 mm.
2. Project 9, 10, 11 and 12 to 9’, 10’, 11’ and 12’. 9’ and 10’ lie on the
bottom end while 11’ and 12’ lie on the top end.
3. Draw the cutting plane C– C through 9’ (10’)–12’(11’) and obtain the true
shape of the section, 9– 10–11–12, as explained in case (i) above.
Case (iv): True shape: Pentagon of 25 mm base side and maximum altitude
The cutting plane must pass through two edges of the base, two vertical
edges and a corner of the top. The line joining the two intersecting points
on the base will decide the base of the pentagon. The shortest distance
between this line and the top corner will represent the altitude of the
pentagon.
1. In TV, locate 13 and 14 on two adjacent sides such that 13–14 = 25 mm.
2. Project 13 and 14 to 14’(13’) on the bottom end of the prism.
3. Draw the cutting plane D–D through 14’(13’)–1’. Locate 15’(16’) at the
intersections of the cutting plane and the vertical edges.
4. Mark 15 and 16 in TV. Obtain the true shape of the section 13–14–15–1–16
as explained earlier.
Example 15.10 A cube of 70 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 Refer Fig. 15.10.
1. Draw TV and FV of the cube as shown.
As the true shape of the section is a hexagon, the cutting plane must cut the
prism at 6 points. Obviously, the cutting plane will cut two edges of the top,
two edges of the base and two vertical edges. The POIs 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).
2. 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.
3. In FV, 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.
4. 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.
5. 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.
SECTIONS OF PYRAMIDS AND TETRAHEDRON
Example 15.13 A triangular pyramid with a base side of 50 mm and a slant
height of 70 mm, rests on the base on the HP with a side of base
perpendicular to the VP. It is cut by an AIP inclined at 30° to the HP and
bisecting the axis and a profile section plane intersecting the AIP at the
edge of the pyramid parallel to the VP. Draw FV, sectional TV and
sectional SV.
Solution Refer Fig. 15.13.
1. Draw TV, FV and SV of the pyramid.
2. In FV, draw the cutting plane inclined at 30° to XY and passing through the
midpoint of the axis. Locate POIs 1’, 2’ and 3’ as shown. Through 1’, draw
a vertical cutting plane cutting the base at 4’ and 5’.
3. Project 1’, 2’, 3’, 4’ and 5’ to 1, 2, 3, 4 and 5 in FV and 1”, 2”, 3”, 4” and 5”
in SV on the corresponding edges. (4” and 5” are projected from TV.).
Join 1–2–3, 1”–2”–3” and 1”–4”–5” and hatch the areas.
Example 15.15 A tetrahedron of side 50 mm rests on a face on the HP. One of
the edges other than those on the HP is parallel to the VP. Different section
planes cut the tetrahedron in such a way that the true shape of the section is
(i) an isosceles triangle of base 18 mm and maximum height
(ii) an isosceles triangle of maximum base and 37 mm, height
(iii) an equilateral triangle of 18 mm side.
Draw FV and TV and locate the cutting planes. Also, draw true shape of the
section in each case.
Solution For the true shape of the section to be a triangle, the section plane
must cut 3 edges of the tetrahedron.
Draw TV and FV of the tetrahedron as shown in Fig. 15.15.
Case (i): True shape: Isosceles triangle of 18 mm base and maximum height
Solution For isosceles triangle of 18 mm base, the section plane must cut
two adjacent edges of base at the points, 18 mm apart. For maximum altitude,
the section plane must pass through the apex.
1. In TV, locate 1 and 2 on two adjacent sides such that 1–2 = 18 mm. Project 1
and 2 to 1’(2’) in FV.
2. In FV, mark 3’ at the apex. Draw the cutting plane A– A through 1’(2’)–3’.
3. Project 1’, 2’ and 3’ on the cutting plane A– A to obtain the true shape 1–2–3.
Case (ii): True shape: Isosceles triangle of maximum base and height 37 mm
For a maximum base of the isosceles triangle, the cutting plane must pass
through a base edge. For height of the triangle to be 37 mm, the cutting plane
must pass through a point on the opposite edge such that the distance of that
point from the base edge is equal to 37 mm.
1. In FV, locate 4’(5’) at the base corner. With 4’(5’) as a centre and radius = 37
mm, cut an arc on the opposite edge at 6’.
2. Draw the cutting plane B– B through 4’(5’)–6’.
3. Project 4’, 5’ and 6’ on the cutting plane B– B to obtain the true shape 4–5–6.
Case (iii) True shape: Equilateral triangle of side 18 mm
For an equilateral triangle, the cutting plane must cut parallel to any triangular
face.
1. In TV, locate 7 and 8 such that 7–8 = 18 mm. Project 7 and 8 to 7’ (8’) in FV.
2. Through 7’ (8’), draw a cutting plane C– C parallel to base face intersecting
other edge at 9’.
3. Project 7’ (8’) and 9’ on the cutting plane C– C to obtain the true shape of
the section 7–8–9.
SECTIONS OF CYLINDERS
Example 15.19 A cylinder with a 60 mm diameter and a 100 mm height
stands on its base on the HP. It is cut by two section planes,
(i) an AIP inclined at 60° to the HP and intersecting an extreme generator at a
point 36 mm from the base and,
(ii) an AVP inclined at 75° to the VP and 21 mm away from the axis of the
cylinder.
Draw the sectional TV and sectional FV.
Solution
Refer Fig. 15.19.
1. Draw TV and FV of the cylinder. Obtain 12 division points in TV and
corresponding lateral lines in FV.
2. In FV, draw the cutting plane inclined at 60° to XY and passing through 5’,
36 mm above the base. Mark POIs 1’, 2’, 3’, etc., between the cutting plane
and the lateral lines.
3. Project 1’(9’) to 1 and 9 in TV. Other points will lie on the periphery of the
circle. Join 1– 9 and hatch the section.
4. In TV, draw the cutting plane inclined at 75° to XY and 21 mm away from
the centre of the circle cutting the circle at a( d) and b( c).
5. Project a(d) and b(c) to a’, d’, b’ and c’ in FV. Join a’– b’– c’– d’ and hatch
the area.
SECTIONS OF CONES
The sections of a cone give special curves called conic sections.
Example 15.22 A cone with a base diameter of 75 mm and an axis length of
100 mm is lying in space with its axis parallel to both the RPs. An AVP inclined
at 30° to the VP and passing through a point on the axis, 32 mm from the base
cuts the cone. Draw TV, sectional FV and sectional SV. The part of the cone
containing the apex is retained.
Solution
Refer Fig.15.22.
1. Draw SV, FV and TV of the cone as shown. Divide SV into 12 equal parts and
obtain the lateral lines in FV and TV.
2. In TV, locate the cutting plane inclined at 30° to XY and passing through
axis, 32 mm away from the base. Mark 1, 2, 3, etc., at the intersections of the
cutting plane with the lateral lines.
3. Project 1, 2, 3, etc., in FV and in SV on the corresponding lateral lines. 1 and
9 are projected to 1” and 9” in SV on the base and then to 1’ and 9’ in FV.
Similarly, 2 and 8 are projected to 2’ and 8’ in FV and then to 2” and 8” in SV.
4. Join 1’, 2’, 3’, etc., and 1”, 2”, 3”, etc., and hatch the areas.
Example 15.25 A cone with a 70 mm diameter of base and 90 mm length of
axis, rests on its base on the HP. Two section planes—a profile section
plane and a horizontal section plane—cut the cone. Both the section
planes pass through the midpoint of an extreme generator. Draw FV,
section planes, sectional TV and sectional SV. Also, draw an auxiliary TV
on a plane parallel to the extreme generator of the cone at which the two
cutting planes intersect.
Solution Refer Fig. 15.25.
As the profile cutting plane is parallel to the axis, the true shape of the section
will be a rectangular hyperbola. For the horizontal section plane, the
section will be a circle.
1. Draw TV, FV and SV of the cone. Obtain 12 division points in TV and the
lateral lines in FV and TV.
2. Draw the profile section plane A– A and a horizontal section plane B– B
through midpoint 1’ of o’ a’, cutting the base at 3’(4’) and the other extreme
generator at 6’ respectively.
3. Mark 2’(5’) at the intersections of A– A with lateral lines. Project 1’, 2’(5’)
and 3’(4’) to 1”, 2”, 5”, 3”, and 4” in SV. 3’(4’) are projected to 3 and 4 in
TV and then to 3” and 4” in SV. Join 3”–2”–1”–5”–4” by smooth curve and
hatch the area to represent section hyperbola.
4. Mark r’(s’) at the intersections of B– B with intermediate lateral lines.
Project r’(s’) to r”and s” in SV. Join r”– s”.
5. Project 6’ to 6 in TV. With o as a centre and radius = 0–6, draw a circle.
Hatch the circle to represent the section. Also, join 3–4.
6. Draw X2 Y2 parallel to o’a’. Project the FV on X2 Y2 to obtain the
auxiliary TV showing the sections.
SECTIONS OF SPHERES
The section of a sphere when cut by any section plane is always a circle.
However, in FV, TV or SV, the section may be seen as an ellipse depending
on the type of the cutting plane.
Example 15.26 A sphere of diameter 75 mm rests on the HP. It is cut by an
AIP inclined at 50° to the HP and 18 mm away from the centre of the
sphere. Draw FV, sectional TV and true shape of the section.
Solution
Refer Fig. 15.26.
1. Draw FV and TV of the sphere.
2. In FV, draw a circle A of radius 18 mm concentric to the sphere-circle.
Draw cutting plane tangent to this circle and inclined at 50° to XY.
3. In FV, draw a few more circles, say circle B, circle C, etc., of suitable
radius concentric with the sphere-circle. Mark 1’, 2’, 3’, etc., at the
intersections of the cutting plane and the circles.
4. In TV, the circles will be seen as lines parallel to XY. Hence, project the
circles to draw lines A– A, B– B, etc., in TV. Project 1’, 2’, 3’, etc., to 1, 2, 3,
etc., on the corresponding lines in TV.
5. Join 1, 2, 3, etc., by a smooth curve and hatch the area.
6. To draw the true shape of the section, project o’ perpendicular to the
cutting plane and locate o on it at a suitable distance. With o as a centre
and radius = ½ (1’–7’), draw a circle. Hatch the circle.
SECTIONS OF SOLIDS IN COMBINATION AND COMPOSITE SOLIDS
Example 15.28 A frustum of a pentagonal pyramid having a base edge of 50 mm,
top edge of 30 mm and a height of 50 mm is placed on its base on the HP with
a base edge AB perpendicular to the VP. A semicylinder of 25 mm radius is
placed symmetrically on its flat face on the top of the frustum. The axis of the
semicylinder is parallel to the VP. The end faces of the semicylinder lie on the
planes drawn through AB and the corner of the base opposite to AB. The
combination is cut by an AVP inclined at 54° to the VP and passing through A.
Draw the sectional elevation and the true shape of the section.
Solution
Refer Fig. 15.28.
1. Draw TVs and FVs of the frustum of the pyramid and the semicylinder. ab is
drawn perpendicular to XY.
2. Draw the cutting plane passing through a and inclined at 54° to XY. Locate c, d
and e at the intersections of the cutting plane with the edges of the frustum.
3. Project c, d and e to c’, d’ and e’ on the corresponding edges of the frustum in
FV. Join a’c’d’e’ and hatch the area.
4. Draw an end view of the semicylinder. Obtain 6 division points on it. Draw
the corresponding lateral lines in FV and TV.
5. In TV, locate 1, 2, 3, etc., at the intersections of the cutting plane with the
lateral lines. (1 and 6 coincide with a and d respectively.)
6. Project 1, 2, 3, etc., to 1’, 2’, 3’, etc., on the corresponding lateral lines in
FV. Join these points and hatch the area to indicate the section of the
semicylinder.
7. Draw X1 Y1 parallel to the cutting plane. Project a, c, d, e, 1, 2, 3, etc., on it
and draw the true shape of the section by the auxiliary plane projection
method.