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Engineering Drawing and Graphics / Module 10 /

Problem Set

1. Develop the lateral surface of a pentagonal prism of side of base 25 mm and height 50 mm.
2. A pentagonal prism, side of base 25 mm and axis 55 mm long, rests with its base on HP and
an edge of the base is inclined at 450 to VP. It is cut by a plane perpendicular to VP inclined at
300 to HP and passing through a point on the axis at a distance of 35 mm from the base.
Develop the lateral surface of the truncated prism.
3. A cube of 40 mm edge stands on one of its face on HP with a vertical face making 450 to VP.
A horizontal hole of 30 mm diameter is drilled centrally through the cube such that the hole
passes through the opposite vertical edge of the cube. Obtain the development of the lateral
surface of the cube with the hole.
4. Draw the development of the surface as shown.

Examples:

1. Draw the development of the lateral surface of a right square prism of edge of base 30 mm
and axis 50 mm long

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Steps:
1. Draw the top and front views of the prism and name the corners.
To draw the development of the lateral surface of the prism:

2. It consists of four equal rectangles of size 50 mm x 30 mm in contact and in sequence.

Note: In the development of the lateral surface, the starting and closing edges should be
the same (viz. AA1) to obtain the closed object.

2. Draw the development of the outside case and tray of match box of size 45 mm x 33 mm x 16
mm.

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Steps:
1. Development of the outside case:
The outside case has four sides ABFE, BCGF, etc. the development of which is shown in
fig. i.
2. Development of the tray:
The tray has five sides and the sixth side is open. Fig. ii shows the development of the
tray.

3. Draw the development of the complete surface of a G.I cylinder drum with lid. Diameter is 30
cm and the height is 1.6 times the diameter.

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Steps:
1. Draw the front view and top view.
To draw the development of the lateral surface of the prism:

2. It consists of a rectangle having length 3.14 x 30 draw a rectangle A1A1AA such that
A1A1= perimeter of the circle(3.14 x 30) and AA1=its height.
3. On the line A1 A1, mark eight equal divisions A1,B1,B1C1, etc. .
4. Erect perpendiculars at B1,C1 and D1. Darken the rectangle which give the
development of the lateral surface of the Cylinder.

5. It is advisable to draw the top and bottom circle at the extreme end as shown, to facilitate
economical way of cutting the sheet metal.

Note: In the development of the lateral surface, the starting and closing edges should be the
same (viz. AA1) to obtain the closed object.

4. A hexagonal prism, edge of base 20 mm and axis 50 mm long. Rests with its base on HP
such that one of its rectangular face is parallel to VP. It is cut by a plane perpendicular to VP,
inclined at 450 to HP and passing through the right corner of the top face of the prism. Draw
the sectional TV and develop the lateral surface of the truncated prism.

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Steps:
1. Draw the top and front views of the prism for the given position.
2. Draw the section plane in front view and mark the section points.
3. Draw the sectional top view as shown and hatch it.
4. Draw two stretch-out lines AA and A1A1 each equal to the perimeter of the base (120
mm) of the prism.
5. Divide A1A1 into six equal parts and draw six equal rectangles to represent the
development of the lateral surface of the prism.

To draw the development of the lateral surface of the truncated prism:

6. From the section point 1 draw a horizontal line and mark 1 on AA1. Similarly obtain the
other points 2,3,...6 in the development.
7. Join 1-2,2-3,3-4,....6-1 as straight lines and darken the development of the truncated
prism as shown. [Top]

5. A vertical chimney of 70 cm diameter joins a roof sloping at an angle of 350 with the
horizontal. The shortest portion over the roof is 32 cm. Determine the shape of the sheet metal
from which the chimney can be fabricated. Take a scale 1:20

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Steps:
1. Draw the top view of the chimney as a circle of diameter 35 mm scale 1:20. Divide the
circle into 8 equal parts and project the generators to obtain the front view.
2. Draw a line inclined at 350 to XY as shown. This line represents the roof line and is
considered as a section plane.
3. Mark a height of 32 mm on the left extreme generator shortest height of the chimney and
complete the front view of the chimney.
4. Draw a stretch – out line. AA=∏ x 70 mm. Divide it into 8 equal parts and show the
generators in the development.Project the section points on the corresponding
generators in the development.
6. Draw the development of the lateral surface of the part A of the cylinder shown in Fig. (i)

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Steps:
1. Draw the development of the cylinder.
2. Consider the 300 line s one section plane perpendicular to V.P. and inclined at 300 to
H.P. Project the section points on the corresponding generators in the development and
complete the bottom curve.
3. Then consider the portion of a circle of radius 15 mm as another (curved) section plane
and mark the section points as shown.
4. Project all the section points on the corresponding generators in the development except
1’ and (9’). To obtain 1 and 9 in the development, project 1’ and (9’) to the top view. Mark
1 in the development such that C1=c1. Similarly obtain 9 and complete the development
as shown. [Top]

7. Draw the development of the lateral surface of a square pyramid, side of base 25 mm and
height 50 mm, resting with its base on HP and an edge of the base parallel to VP.

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Steps:
1. Draw the top and front views of the pyramid for the given position.
2. To draw its development, true length of the slant edge is required.

RULE:

If the top view of a slant edge of a pyramid is parallel to XY, then the front view of that edge
will give its true length and vice-versa.

Here, in both the views, the projections of none of the slant edges are parallel to XY. Hence
its true length can not be measured directly either front the top or front view. Therefore to
obtain the true length of a slant edge (say OA) make oa parallel to XY, i.e., with o as centre
and oa as radius draw an arc to cut the horizontal drawn from o at a1. Now o’a1’ will be the
true length of the slant edge OA.

3. With O as centre and o’a1’ as radius draw an arc. On this arc, mark 4 equal divisions,
i.e., chord AB=BC=CD=DA=25 mm.
4. Complete the triangles OAB, OBC, OCD and ODA by thick lines which will give the
development of the lateral surface of the pyramid.

Fig. iii shows the uneconomical method of development of the pyramid.

8. A cone of base circle 40 mm and height 75 rests on HP with its base. Draw its development

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Steps:
1. Draw the top and front view of the cone.
2. Divide the circle into 12 equal parts and project them in front view as 1',2'....etc.
3. Join these points 1',2'....etc. with the vertex O.(fig. 8.1)
4. Measure the slant height L from the F.V. (fig. 8.1). Take any point O1 as center and radius =
L. Draw an angle θ = 180 0 x D/L degrees, where D= diameter of the base circle i.e. 40 mm.
Mark the arc
5. Divide the arc into 12 equal parts to get the required development of the lateral surface of
the cone.(fig. 8.2)

9. Draw development of a funnel as shown.

Steps:

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The funnel is made of 2 parts, the upper part is conical and lower part is cylindrical (fig. 9.1).
The lines through the lines A and B can be extended to meet at a point O(fig. 9.2). Substract
the development of the part XYO from the complete development (A'B'O'-X'Y'O') in fig. 9.3.
angle θ = D/L x1800 = 60/72 x 1800 = 1500
OA = L = 72 mm by measurement. Take O' as center, L=72 mm radius O'A' and θ =1500 then
mark the arc A'B' fig. 9.3. Measure l = OX =OY and mark small are X'Y' to deduct the portion
OXY (O'X'Y' fig. 9.3.)
To get the development of the lower cylinder portion, draw plan and elivation. Divide the circle
into 8 parts and project them on elivation marking 1',2'....etc. (fig. 9.4)
draw holizontal projectors from X'-Y' and 1',2' .. etc. mark 8 parts with divider as 1,2,3,4,.. etc.
(fig. 9.4) now fig. 9.2 and fig 9.4 are representing the required development of the funnel.

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Notes

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Development of Surfaces
Assume object hollow and made-up of thin sheet. When its complete surface is unfolded and
laid in a plane, this is called Development of laterlal suefaces of that object or solid. The
figure thus obtained is called development of surface of soli.
Laterlal surface is the surface excluding solid’s top & base.

Note:
1. It is a shape showing AREA, means it’s a 2-D plain drawing.
2. Hence all dimensions of it must be TRUE dimensions.
3. As it is representing shape of an un-folded sheet, no edges can remain hidden and hence
DOTTED LINES are never shown on development.

Engineering application:
There are so many products or objects which are difficult to manufacture by Conventional
manufacturing processes, because of their shapes and sizes.
Those are fabricated in sheet metal industry by using Development technique. There is a vast
range of such objects.

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[Top]

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RGUKT 2010

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