Basic Engineering Mathematics, Fifth Edition

252 Basic Engineering Mathematics

12. Find the volume (in cm^3 ) of the die-casting
    shown in Figure 27.19. The dimensions are
    in millimetres.

60

30 rad

25

50

100

Figure 27.19

13. The cross-section of part of a circular ven-
    tilation shaft is shown in Figure 27.20, ends
    ABandCDbeing open. Calculate
    (a) the volume of the air, correct to
    the nearest litre, contained in the
    part of the system shown, neglect-
    ing the sheet metal thickness (given
    1litre=1000cm^3 ).
    (b) the cross-sectional area of the sheet
    metal used to make the system, in
    square metres.
    (c) thecost ofthesheet metal ifthematerial
    costs £11.50 per square metre, assum-
    ing that 25% extra metal is required due
    to wastage.

500mm

A

B

2m

1.5m

1.5m

800mm

C D

Figure 27.20

27.5 Volumes and surface areas of

frusta of pyramids and cones

Thefrustumof a pyramidor cone is the portionremain-

ing when a part containing the vertex is cut off by a

plane parallel to the base.

Thevolume of a frustum of apyramid or coneis given

by the volume of the whole pyramid or cone minus the

volume of the small pyramid or cone cut off.

Thesurface area of the sides of a frustum of a pyra-

mid or coneis given by the surface area of the whole

pyramid or cone minus the surface area of the small

pyramid or cone cut off. This gives the lateral surface

area of the frustum. If the total surface area of the frus-

tum is required then the surface area of the two parallel

ends are added to the lateral surface area.

There is an alternative method for finding the volume

and surface area of afrustum of a cone. With reference

to Figure 27.21,

h

R

I

r

Figure 27.21

Volume=

1

3

πh(R^2 +Rr+r^2 )

Curved surface area=πl(R+r)

Total surface area=πl(R+r)+πr^2 +πR^2

Problem 24. Determine the volume of a frustum

of a cone if the diameter of the ends are 6.0cmand

4 .0cm and its perpendicular height is 3.6cm

(i) Method 1

A section through the vertex of a complete cone is

shown in Figure 27.22.

Using similar triangles,

AP

DP

=

DR

BR

Hence, AP

2. 0

=

3. 6

1. 0

from which AP=

( 2. 0 )( 3. 6 )

1. 0

= 7 .2cm