EXTRA EXERCISE FROM JOHN BIRDS BOOK

484 Engineering Mathematics

of the curve can be produced. When x = 1, y = −9, body moves from the position where x = 1 m
showing that the part of the curve between x = 0 and
x = 4 is negative. A sketch of y = x3 − 2x2 − 8x is shown to that where x = 3 m. [29.33 Nm]

in Fig. 55.10. (Another method of sketching Fig. 55.10 4. Find the area between the curve y = 4x − x2

would have been to draw up a table of values.) and the x-axis. [10.67 square units]

5. Determine the area enclosed by the curve
y = 5x2 + 2, the x-axis and the ordinates x = 0
and x = 3. Find also the area enclosed by the

curve and the y-axis between the same limits.

[51 sq. units, 90 sq. units]

6. Calculate the area enclosed between
y = x3 − 4x2 − 5x and the x-axis.

[73.83 sq. units]

7. The velocity v of a vehicle t seconds after a

Figure 55.10 certain instant is given by: v = (3t2 + 4) m/s.
Shaded area =
0 Determine how far it moves in the interval from
=
(x3 − 2x2 − 8x) dx t = 1 s to t = 5 s. [140 m]

−2

4 8. A gas expands according to the law pv =

− (x3 − 2x2 − 8x) dx constant. When the volume is 2 m3 the pres-

0 sure is 250 kPa. Find the work done as the gas

x4 − 2x3 − 8x2 0 expands from 1 m3 to a volume of 4 m3 given
v2
43 2 −2
that work done = p dv [693.1 kJ]

− x4 − 2x3 − 8x2 4 v1

43 20

= 2 − −42 2 55.4 The area between curves
6
33

1 The area enclosed between curves y = f1(x) and y = f2(x)
= 49 square units (shown shaded in Fig. 55.11) is given by:

3

Now try the following exercise shaded area = bb
=
f2(x) dx − f1(x) dx
Exercise 192 Further problems on areas
under curves aa

In Problems 1 and 2, find the area enclosed between b
the given curves, the horizontal axis and the given
ordinates. [ f2(x) − f2(x)] dx

1. y = 2x3; x = −2, x = 2 [16 square units] a

Section 9 2. xy = 4; x = 1, x = 4 [5.545 square units] y
y ϭ f1(x)
3. The force F newtons acting on a body at y ϭ f2(x)
a distance x metres from a fixed point is
given by: F = 3x + 2x2. If work done = 0 xϭa xϭb x
Figure 55.11
x2

F dx, determine the work done when the

x1

Areas under and between curves 485

Problem 11. Determine the area enclosed Problem 12. (a) Determine the coordinates of the
between the curves y = x2 + 1 and y = 7 − x points of intersection of the curves y = x2 and
y2 = 8x. (b) Sketch the curves y = x2 and y2 = 8x on
At the points of intersection, the curves are equal. the same axes. (c) Calculate the area enclosed by
Thus, equating the y-values of each curve gives: the two curves
x2 + 1 = 7 − x, from which x2 + x − 6 = 0. Factorising
gives (x − 2)(x + 3) = 0, from which, x = 2 and x = −3. (a) At the points of intersection the coordinates of the
By firstly determining the points of intersection the curves are equal. When y = x2 then y2 = x4.
range of x-values has been found. Tables of values are
produced as shown below. Hence at the points of intersection x4 = 8x, by
equating the y2 values.
x −3 −2 −1 0 1 2
Thus x4 − 8x = 0, from which x(x3 − 8) = 0, i.e.
y = x2 + 1 10 5 2 1 2 5 x = 0 or (x3 − 8) = 0.

x −3 0 2 Hence at the points of intersection x = 0 or x = 2.

y=7−x 10 7 5 When x = 0, y = 0 and when x = 2, y = 22 = 4.

A sketch of the two curves is shown in Fig. 55.12. Hence the points of intersection of the curves
y = x2 and y2 = 8 x are (0, 0) and (2, 4)

(b) A sketch of y = x2 and y2 = 8x is shown in
Fig. 55.13

y
10 y ϭ x 2 ϩ 1

5
yϭ7Ϫx

Ϫ3 Ϫ2 Ϫ1 0 1 2x

Figure 55.12

22 Figure 55.13

Shaded area = (7 − x)dx − (x2 + 1)dx 2 √ x2
{ 8x
−3 −3 (c) Shaded area = − }dx

2 0

= [(7 − x) − (x2 + 1)]dx = 2 {(√8)x1/2 − x2}dx Section 9

−3 0

2 = √ x3/2 x3 2
( 8) −
= (6 − x − x2)dx ( 3 ) 3
2 0
−3 √√
8 8 8
x2 x3 2 = − 3 − {0}
= 6x − − ( 3 )
2
2 3 −3

= 12 − 2 − 8 − −18 − 9 + 9
32

= 1 − −13 1 = 16 − 8 = 8
7 3 33
32

= 20 5 square units = 2 2 square units
6 3

486 Engineering Mathematics

Problem 13. Determine by integration the area Now try the following exercise
bounded by the three straight lines y = 4 − x, y = 3x
and 3y = x Exercise 193 Further problems on areas
between curves
Each of the straight lines is shown sketched in
Fig. 55.14. 1. Determine the coordinates of the points of
intersection and the area enclosed between the
Figure 55.14 parabolas y2 = 3x and x2 = 3y.
[(0, 0) and (3, 3), 3 sq. units]

2. Sketch the curves y = x2 + 3 and y = 7 − 3x
and determine the area enclosed by them.
[20.83 square units]

3. Determine the area enclosed by the curves
y = sin x and y = cos x and the y-axis.
[0.4142 square units]

4. Determine the area enclosed by the three
straight lines y = 3x, 2y = x and y + 2x = 5
[2.5 sq. units]

Shaded area = 1x
3x − dx
=
= 03

3x
+ (4 − x) − dx

13

3x2 x2 1 x2 x2 3
− + 4x − −
2 60 2 61

3 − 1 − (0)
26

+ 12 − 9 − 9 − 4 − 1 − 1
26 26

= 1 + 6 − 31
1
33

= 4 square units

Section 9

Chapter 56

Mean and root mean
square values

56.1 Mean or average values half a cycle’, since the mean value over a complete
cycle is zero.
(i) The mean or average value of the curve shown in
Fig. 56.1, between x = a and x = b, is given by: Problem 1. Determine, using integration, the
mean or average value, mean value of y = 5x2 between x = 1 and x = 4
area under curve
y= Mean value,
length of base
1 41 4
y y ϭ f (x) y= y dx =
4−1 1 31 5x2 dx
y
=1 5x3 4 = 5 [x3]14 = 5 − 1) = 35
3 3 1 9 (64

9

Problem 2. A sinusoidal voltage is given by
v = 100 sin ωt volts. Determine the mean value of
the voltage over half a cycle using integration

0 xϭa xϭb x Half a cycle means the limits are 0 to π radians.

Mean value,

Figure 56.1 1π
v = π − 0 0 v d(ωt)

(ii) When the area under a curve may be obtained by =1 π 100 −cos ωt π
integration then: π π 0
mean or average value, 100 sin ωt d(ωt) =

0

b = 100 [(−cos π) − (−cos 0)]
π
y dx
y= a = 100 [(+1) − (−1)] = 200
ππ
b−a

1b = 63.66 volts

i.e. y = b−a a f (x) dx [Note that for a sine wave,
2
(iii) For a periodic function, such as a sine wave, the
mean value is assumed to be ‘the mean value over mean value = × maximum value
π

488 Engineering Mathematics (b) By integration, mean value

In this case, mean value = 2 × 100 = 63.66 V] =1 3 = 1 3
π 3−0
y dx (3x2 + 2)dx
Problem 3. Calculate the mean value of 03
y = 3x2 + 2 in the range x = 0 to x = 3 by (a) the 0
mid-ordinate rule and (b) integration
(a) A graph of y = 3x2 over the required range is = 1 [x3 + 2x]30 = 1 + 6) − (0)]
3 [(27
shown in Fig. 56.2 using the following table:
x 0 0.5 1.0 1.5 2.0 2.5 3.0 3
y 2.0 2.75 5.0 8.75 14.0 20.75 29.0
= 11
y
y ϭ 3x 2 ϩ 2 The answer obtained by integration is exact;
greater accuracy may be obtained by the mid-
30 ordinate rule if a larger number of intervals are
selected.
20
Problem 4. The number of atoms, N, remaining
10
in a mass of material during radioactive decay after
2 time t seconds is given by: N = N0e−λt, where N0
0 1 2 3x and λ are constants. Determine the mean number of

atoms in the mass of material for the time period
t = 0 and t = 1

λ

Mean number of atoms

= 1 1/λ 1 1/λ
1 −0 1
N dt = N0e−λt dt

0 0

λ λ

= λN0 1/λ e−λt 1/λ
−λ 0
e−λt dt = λN0

0

= −N0[e−λ(1/λ) − e0] = −N0[e−1 − e0]

= +N0[e0 − e−1] = N0[1 − e−1] = 0.632 N0

Figure 56.2 Now try the following exercise

Section 9 Using the mid-ordinate rule, mean value Exercise 194 Further problems on mean or
average values
= area under curve
length of base √
3x
= sum of mid-ordinates 1. Determine the mean value of (a) y = from
number of mid-ordinates
x = 0 to x = 4 (b) y = sin 2θ from θ = 0 to
Selecting 6 intervals, each of width 0.5, the mid- θ = π (c) y = 4et from t = 1 to t = 4
ordinates are erected as shown by the broken lines
in Fig. 56.2. 4
2

(a) 4 (b) or 0.637 (c) 69.17
π

2.2 + 3.7 + 6.7 + 11.2

Mean value = + 17.2 + 24.7 2. Calculate the mean value of y = 2x2 + 5 in the
6
range x = 1 to x = 4 by (a) the mid-ordinate

65.7 rule, and (b) integration. [19]
= = 10.95

6

Mean and root mean square values 489

3. The speed v of a vehicle is given by: = 1 4 4 x5 4

4x4 dx =
v = (4t + 3) m/s, where t is the time in seconds. 31 3 51

Determine the average value of the speed from 4√
= (1024 − 1) = 272.8 = 16.5
t = 0 to t = 3 s. [9 m/s]
15

4. Find the mean value of the curve Problem 6. A sinusoidal voltage has a maximum
value of 100 V. Calculate its r.m.s. value
y = 6 + x − x2 which lies above the x-

axis by using an approximate method. Check

the result using integration. [4.17]

5. The vertical height h km of a missile varies A sinusoidal voltage v having a maximum value of 100 V
may be written as: v = 100 sin θ. Over the range θ = 0
with the horizontal distance d km, and is given to θ = π,
by h = 4d − d2. Determine the mean height of
the missile from d = 0 to d = 4 km. r.m.s. value

[2.67 km] = 1 π

π−0 0 v2 dθ

6. The velocity v of a piston moving with simple = 1 π
harmonic motion at any time t is given by:
v = c sin ωt, where c is a constant. Determine π0 (100 sin θ)2 dθ
the mean velocity between t = 0 and t = π
ω = 10 000 π
2c
π0 sin2 θ dθ
π
which is not a ‘standard’ integral. It is shown in Chapter

27 that cos 2A = 1 − 2 sin2 A and this formula is used

56.2 Root mean square values whenever sin2 A needs to be integrated.

Rearranging cos 2A = 1 − 2 sin2 A gives

sin2 A = 1 (1 − cos 2A)
2
The root mean square value of a quantity is ‘the sqaure
root of the mean value of the squared values of the quan- Hence 10 000 π
tity’ taken over an interval. With reference to Fig. 56.1,
the r.m.s. value of y = f (x) over the range x = a to x = b π0 sin2 θ dθ
is given by:
10 000 π 1
(1
= π 02 − cos 2θ) dθ

r.m.s. value = 1 b sin 2θ π

b−a a y2 dx 10 000 1

= θ− 20
π2

One of the principal applications of r.m.s. values is with = 10 000 1 π − sin 2π − 0 − sin 0 Section 9
alternating currents and voltages. The r.m.s. value of π2 2 2
an alternating current is defined as that current which
will give the same heating effect as the equivalent direct = 10 000 1 [π] = 10 000
current.

π2 2

Problem 5. Determine the r.m.s. value of y = 2x2 = 1√00 = 70.71 volts
between x = 1 and x = 4 2

R.m.s. value [Note that for a sine wave,
r.m.s. value = √1 × maximum value.
= 1 4 1 4 2

y2 dx = (2x2)2 dx In this case, r.m.s. value = √1 × 100 = 70.71 V]
2

4−1 1 31

490 Engineering Mathematics

Problem 7. In a frequency distribution the 2. Calculate the r.m.s. values of:
(a) y = sin 2θ from θ = 0 to θ = π
average distance from the mean, y, is related to the 4
variable, x, by the equation y = 2x2 − 1. Determine, (b) y = 1 + sin t from t = 0 to t = 2π
correct to 3 significant figures, the r.m.s. deviation (c) y = 3 cos 2x from x = 0 to x = π
from the mean for values of x from −1 to +4

R.m.s. deviation (Note that cos2 t = 1 + cos 2t), from Chap-
(1
2
ter 27).
4
= 1 (a) √1 or 0.707 (b) 1.225 (c) 2.121
y2 dx
4 − −1 −1 2

1 4 3. The distance, p, of points from the mean value

= (2x2 − 1)2dx of a frequency distribution are related to the
5 −1
variable, q, by the equation p = 1 + q. Deter-
= 1 4 q

(4x4 − 4x2 + 1)dx mine the standard deviation (i.e. the r.m.s.

5 −1 value), correct to 3 significant figures, for

values from q = 1 to q = 3. [2.58]

= 1 4x5 − 4x3 + x 4 4. A current, i = 30 sin 100πt amperes is applied
55 3 across an electric circuit. Determine its mean
−1 and r.m.s. values, each correct to 4 significant
figures, over the range t = 0 to t = 10 ms.
1 4 (4)5 − 4 (4)3 + 4 [19.10 A, 21.21 A]
=5 5 3

− 4 (−1)5 − 4 (−1)3 + (−1) 5. A sinusoidal voltage has a peak value of 340 V.
53

Calculate its mean and r.m.s. values, correct to

= 1 [(737.87) − (−0.467)] 3 significant figures. [216 V, 240 V]
5
6. Determine the form factor, correct to 3 sig-
1
= [738.34] nificant figures, of a sinusoidal voltage of

5 maximum value 100 volts, given that form
√
= 147.67 = 12.152 = 12.2, factor = r.m.s. value [1.11]
average value

correct to 3 significant figures. 7. A wave is defined by the equation:

Now try the following exercise v = E1 sin ωt + E3 sin 3ωt

Section 9 where, E1, E3 and ω are constants.

Exercise 195 Further problems on root Determine the r.m.s. value of v over the interval
mean square values 0≤t≤ π
ω ⎡⎤
1. Determine the r.m.s. values of:
(a) y = 3x from x = 0 to x = 4 ⎣ E12 + E32 ⎦
(b) y = t2 from t = 1 to t = 3 2
(c) y = 25 sin θ from θ = 0 to θ = 2π

(a) 6.928 (b) 4.919 (c) √25 or 17.68
2

Chapter 57

Volumes of solids of
revolution

57.1 Introduction (i) Let the area shown in Fig. 57.1(a) be divided into
a number of strips each of width δx. One such strip
If the area under the curve y = f (x), (shown in is shown shaded.
Fig. 57.1(a)), between x = a and x = b is rotated 360◦
about the x-axis, then a volume known as a solid of (ii) When the area is rotated 360◦ about the x-axis,
revolution is produced as shown in Fig. 57.1(b). each strip produces a solid of revolution approx-
imating to a circular disc of radius y and thick-
y ness δx. Volume of disc = (circular cross-sectional
area) (thickness) = (πy2)(δx)
y ϭ f (x)
(iii) Total volume, V , between ordinates x = a and
x = b is given by:

x=b b
Volume V = limit πy2δx = πy2 dx
y

δx→0 x=a a

0 xϭa xϭb x If a curve x = f (y) is rotated about the y-axis 360◦
y dx between the limits y = c and y = d, as shown in
Fig. 57.2, then the volume generated is given by:
(a)
y=d d
y ϭ f (x)
dx

Volume V = limit πx2δy = πx2 dy
δy→0 y=c
c

y bx y
0a yϭd

x x ϭ f (y)
dy
(b) yϭc
0 x
Figure 57.1
Figure 57.2
The volume of such a solid may be determined precisely
using integration.

492 Engineering Mathematics

57.2 Worked problems on volumes of = 500π = 2 units
solids of revolution 166 π cubic
33

y

Problem 1. Determine the volume of the solid of 10 y ϭ 2x

revolution formed when the curve y = 2 is rotated 5 10
360◦ about the x-axis between the limits x = 0

to x = 3

When y = 2 is rotated 360◦ about the x-axis between 0 1 2 3 45x
x = 0 and x = 3 (see Fig. 57.3): Ϫ5
Ϫ10
volume generated
Figure 57.4
33
[Check: The volume generated is a cone of radius 10
= πy2 dx = π(2)2 dx and height 5. Volume of cone

00 = 1 πr2h = 1 π(10)25 = 500π
3 33 3

= 4π dx = 4π[x]30 = 12π cubic units

0

[Check: The volume generated is a cylinder of radius 2 2
and height 3. = 166 π cubic units.]
Volume of cylinder = πr2h = π(2)2(3) = 12π cubic
units.] 3

y yϭ2 Problem 3. The curve y = x2 + 4 is rotated one
2 revolution about the x-axis between the limits
x = 1 and x = 4. Determine the volume of the
1 solid of revolution produced

0 12 3 x y
Ϫ1 30

Ϫ2

Figure 57.3 20 A y ϭ x2 ϩ 4
B

Section 9 Problem 2. Find the volume of the solid of 10 C
revolution when the cure y = 2x is rotated
one revolution about the x-axis between the limits D
x = 0 and x = 5 5
4

0 12345x

When y = 2x is revolved one revolution about the Figure 57.5
x-axis between x = 0 and x = 5 (see Fig. 57.4) then:

volume generated Revolving the shaded area shown in Fig. 57.5 about the
x-axis 360◦ produces a solid of revolution given by:

55 Volume = 44
=
= πy2dx = π(2x)2dx πy2 dx = π(x2 + 4)2dx

00 11

5 x3 5 4

= 4πx2dx = 4π π(x4 + 8x2 + 16) dx
0 30
1

Volumes of solids of revolution 493

x5 8x3 4

= π + + 16x enclosed by the given curves, the y-axis and the
53 given ordinates through one revolution about the
1 y-axis.

= π[(204.8 + 170.67 + 64) − (0.2 + 2.67 + 16)]

= 420.6π cubic units 6. y = x2; y = 1, y = 3 [4π]

Problem 4. If the curve in Problem 3 is revolved 7. y = 3x2 − 1; y = 2, y = 4 [2.67π]
about the y-axis between the same limits, determine
the volume of the solid of revolution produced 8. y = 2 y = 1, y = 3 [2.67π]
;
x

The volume produced when the curve y = x2 + 4 is 9. The curve y = 2x2 + 3 is rotated about (a) the
rotated about the y-axis between y = 5 (when x = 1) x-axis between the limits x = 0 and x = 3,
and y = 20 (when x = 4), i.e. rotating area ABCD of
Fig. 57.5 about the y-axis is given by: and (b) the y-axis, between the same limits.

20 Determine the volume generated in each case.

volume = πx2 dy [(a) 329.4π (b) 81π]

5 57.3 Further worked problems on
volumes of solids of revolution
Since y = x2 + 4, then x2 = y − 4
Problem 5. The area enclosed by the curve
20 y2 20
Hence volume = π(y − 4)dy = π − 4y x

5 25 y = 3e 3, the x-axis and ordinates x = −1 and x = 3
is rotated 360◦ about the x-axis. Determine the
= π[(120) − (−7.5)] volume generated

= 127.5π cubic units y
8x
Now try the following exercise
y ϭ 3e3
Exercise 196 Further problems on volumes
of solids of revolution 4

(Answers are in cubic units and in terms of π). Ϫ1 0 1 2 3x

In Problems 1 to 5, determine the volume of the Section 9
solid of revolution formed by revolving the areas
enclosed by the given curve, the x-axis and the
given ordinates through one revolution about the
x-axis.

1. y = 5x; x = 1, x = 4 [525π] Figure 57.6

2. y = x2; x = −2, x = 3 [55π] x

3. y = 2x2 + 3; x = 0, x = 2 [75.6π] A sketch of y = 3e 3 is shown in Fig. 57.6. When the
shaded area is rotated 360◦ about the x-axis then:
4. y2 = x; x = 1, x = 5 [48π]
4 [1.5π] 3

5. xy = 3; x = 2, x = 3 volume generated = πy2 dx

In Problems 6 to 8, determine the volume of the −1
solid of revolution formed by revolving the areas 3 x2

= π 3e 3 dx

−1
3 2x

= 9π e 3 dx

−1

494 Engineering Mathematics

⎡ 2x ⎤3 y
x 2ϩy 2 ϭ 42
= 9π⎣⎢ e3 ⎦⎥
2 Ϫ4 Ϫ2 0 1 2 3 4 x

3 −1

27π e2 − e− 2
= 3
2

= 92.82π cubic units

Problem 6. Determine the volume generated Figure 57.8

when the area above the x-axis bounded by the The volume of a frustum of a sphere may be determined
curve x2 + y2 = 9 and the ordinates x = 3 and by integration by rotating the curve x2 + y2 = 42 (i.e.
x = −3 is rotated one revolution about the x-axis
a circle, centre 0, radius 4) one revolution about the
Figure 57.7 shows the part of the curve x2 + y2 = 9 lying x-axis, between the limits x = 1 and x = 3 (i.e. rotating
above the x-axis, Since, in general, x2 + y2 = r2 repre-
sents a circle, centre 0 and radius r, then x2 + y2 = 9 the shaded area of Fig. 57.8).
represents a circle, centre 0 and radius 3. When the
semi-circular area of Fig. 57.7 is rotated one revolution 3
about the x-axis then:
Volume of frustum = πy2 dx
3
1
volume generated = πy2dx
3
−3
= π(42 − x2) dx
3
1
= π(9 − x2) dx
x3 3
−3 = π 16x −

x3 3 31
= π 9x −
= π (39) − 2
3 −3 15
3
= π[(18) − (−18)]
= 23 1 π cubic units
= 36π cubic units 3

y Problem 8. The area enclosed between the two
x 2ϩy 2ϭ9 parabolas y = x2 and y2 = 8x of Problem 12,
Chapter 55, page 485, is rotated 360◦ about the

x-axis. Determine the volume of the solid produced

Section 9 Ϫ3 0 3x The area enclosed by the two curves is shown in

Figure 57.7 Fig. 55.13, page 485. The volume produced by revolv-

ing the shaded area about the x-axis is given by:
[(volume produced by revolving y2 = 8x) − (volume
produced by revolving y = x2)]

(Check: The volume generated is a sphere of 22
radius 3. Volume of sphere = 4 πr3 = 4 π(3)3 =
i.e. volume = π(8x) dx − π(x4) dx
33
36π cubic units.) 00

2 8x2 x5 2
−
=π (8x − x4) dx = π
0 2 50

Problem 7. Calculate the volume of a frustum of = π 16 − 32 − (0)
a sphere of radius 4 cm that lies between two 5
parallel planes at 1 cm and 3 cm from the centre and
on the same side of it = 9.6π cubic units

Volumes of solids of revolution 495

Now try the following exercise

Exercise 197 Further problems on volumes 6. The area enclosed between the two curves
of solids of revolution x2 = 3y and y2 = 3x is rotated about the x-axis.
Determine the volume of the solid formed.
(Answers to volumes are in cubic units and in
terms of π.) [8.1π]

7. The portion of the curve y = x2 + 1 lying
x

In Problems 1 and 2, determine the volume of the between x = 1 and x = 3 is revolved 360◦ about
solid of revolution formed by revolving the areas
enclosed by the given curve, the x-axis and the the x-axis. Determine the volume of the solid
given ordinates through one revolution about the
x-axis. formed. [57.07π]

8. Calculate the volume of the frustum of a sphere

of radius 5 cm that lies between two parallel

1. y = 4ex; x = 0; x = 2 [428.8π] planes at 3 cm and 2 cm from the centre and
2. y = sec x; x = 0, x = π [π]
on opposite sides of it. [113.33π]
4
9. Sketch the curves y = x2 + 2 and y − 12 = 3x

In Problems 3 and 4, determine the volume of the from x = −3 to x = 6. Determine (a) the co-
solid of revolution formed by revolving the areas
enclosed by the given curves, the y-axis and the ordinates of the points of intersection of the
given ordinates through one revolution about the
y-axis. two curves, and (b) the area enclosed by the

two curves. (c) If the enclosed area is rotated
360◦ about the x-axis, calculate the volume of

3. x2 + y2 = 16; y = 0, y = 4 [42.67π] the solid produced ⎡ (a) (−2, 6) and (5, 27) ⎤
4. x√y = 2; y = 2, y = 3 [1.622π]
⎣ (b) 57.17 square units ⎦

(c) 1326π cubic units

5. Determine the volume of a plug formed by the

frustum of a sphere of radius 6 cm which lies

between two parallel planes at 2 cm and 4 cm

from the centre and on the same side of it.

(The equation of a circle, centre 0, radius r is

x2 + y2 = r2). [53.33π]

Section 9

Chapter 58

Centroids of simple shapes

58.1 Centroids the strip is approximately rectangular and is given
y
A lamina is a thin flat sheet having uniform thickness.
The centre of gravity of a lamina is the point where by yδx. The centroid, C, has coordinates x, .
it balances perfectly, i.e. the lamina’s centre of mass. 2
When dealing with an area (i.e. a lamina of negligible
thickness and mass) the term centre of area or centroid y y ϭ f (x)
is used for the point where the centre of gravity of a x R
lamina of that shape would lie. S

y C (x, y )
2

58.2 The first moment of area P Q x
0 xϭa xϭb
The first moment of area is defined as the product of
the area and the perpendicular distance of its centroid Figure 58.2 dx
from a given axis in the plane of the area. In Fig. 58.1,
the first moment of area A about axis XX is given by (ii) First moment of area of shaded strip about axis
(Ay) cubic units.

Oy = (yδx)(x) = xyδx.

Area A Total first moment of area PQRS about axis
C
Oy = limit x=b xyδx = b xy dx
x=a a
δx→0
(iii) First moment of area of shaded strip about axis

y Ox = (yδx) y = 1 y2x.
22

Total first moment of area PQRS about axis

X X Ox = limit x=b 1 y2δx = 1 b
x=a 2 2
δx→0 y2 dx

Figure 58.1 b a
a
(iv) Area of PQRS, A = y dx (from Chapter 55)

(v) Let x¯ and y¯ be the distances of the centroid

58.3 Centroid of area between a of area A about Oy and Ox respectively then:
curve and the x-axis
(x¯)(A) = total first moment of area A about axis
(i) Figure 58.2 shows an area PQRS bounded by the
curve y = f (x), the x-axis and ordinates x = a and Oy = b xy dx
x = b. Let this area be divided into a large number a
of strips, each of width δx. A typical strip is shown
shaded drawn at point (x, y) on f (x). The area of b

xy dx

from which, x¯ = a
b

y dx

a

Centroids of simple shapes 497

and (y¯)(A) = total moment of area A about axis Let a rectangle be formed by the line y = b, the x-axis
and ordinates x = 0 and x = l as shown in Fig. 58.4. Let
Ox = 1 b y2 dx the coordinates of the centroid C of this area be (x¯, y¯).
2 a

1 b l l

2 y2 dx xy dx (x)(b) dx

from which, y¯ = a By integration, x¯ = 0 = 0
b l l

y dx y dx b dx

a 0

0

58.4 Centroid of area between x2 l bl 2
a curve and the y-axis b
20 =1
= [bx]0l = 2 2
bl
If x¯ and y¯ are the distances of the centroid of area EFGH
in Fig. 58.3 from Oy and Ox respectively, then, by 1 l 1 l
similar reasoning as above:
y¯ = 2 y2 dx b2 dx
=2 0
and 0 bl
l
y=d d
x =1 y dx
(x¯)(total area) = limit xδy x2dy
δy→0 y=c 2 2c 0

1 d 1 [b2x]l0 b2l b
2 2
x¯ = 2 x2dy = = =

from which, c bl bl 2
d

x dy y
yϭb
c
b
y=d d xC

and (y¯)(total area) = limit (xδy)y = xy dy y
δy→0 y=c
c 0 lx

d

xy dy

from which, y¯ = c
d

x dy Figure 58.4

c lb
i.e. the centroid lies at ,
y which is at the
22
yϭd E F intersection of the diagonals.
dy x ϭ f (y)
yϭc H x
Problem 2. Find the position of the centroid of Section 9
y C(x2, y) the area bounded by the curve y = 3x2, the x-axis
G
and the ordinates x = 0 and x = 2

0x If, (x¯, y¯) are the co-ordinates of the centroid of the given

area then:

Figure 58.3 2 2

xy dx x(3x2)dx

x¯ = 0 = 0
2 2
58.5 Worked problems on centroids y dx 3x2dx
of simple shapes
00

2 3x4 2

Problem 1. Show, by integration, that the 3x3dx 40 12
centroid of a rectangle lies at the intersection of [x3]02 8
the diagonals = 0 = = = 1.5
2
3x2dx

0

498 Engineering Mathematics

1 2 1 2 that the centroid of a triangle lies at one-third of the
perpendicular height above any side as base.
2 y2dx (3x2)2dx
=2 0
y¯ = 0 8
2

ydx

0 Now try the following exercise

1 2 = 9 x5 29 32 Exercise 198 Further problems on
=2 2 5 0= 2 5 centroids of simple shapes
9x4dx 8
In Problems 1 to 5, find the position of the centroids
0 of the areas bounded by the given curves, the x-axis
and the given ordinates.
88

18
= = 3.6

5

Hence the centroid lies at (1.5, 3.6)

Problem 3. Determine by integration the position 1. y = 2x; x = 0, x = 3 [(2, 2)]
of the centroid of the area enclosed by the line
y = 4x, the x-axis and ordinates x = 0 and x = 3 2. y = 3x + 2; x = 0, x = 4

[(2.50, 4.75)]

y 3. y = 5x2; x = 1, x = 4
12
y ϭ 4x [(3.036, 24.36)]
x
0A D 4. y = 2x3; x = 0, x = 2

C [(1.60, 4.57)]
yB
3 x 5. y = x(3x + 1); x = −1, x = 0
[(−0.833, 0.633)]

Figure 58.5

Let the coordinates of the area be (x¯, y¯) as shown in 58.6 Further worked problems on
Fig. 58.5. centroids of simple shapes

33

xy dx (x)(4x)dx

Then x¯ = 0 =03 Problem 4. Determien the co-ordinates of the
3
centroid of the area lying between the curve
y dx 4x dx y = 5x − x2 and the x-axis

00

3 4x3 3

4x2dx 30 36 y = 5x − x2 = x(5 − x). When y = 0, x = 0 or x = 5,
[2x2]03 18 Hence the curve cuts the x-axis at 0 and 5 as shown in
= 0 = = = 2 Fig. 58.6. Let the co-ordinates of the centroid be (x¯, y¯)
3 then, by integration,

Section 9 4x dx

0

1 3 1 3

y2dx (4x)2dx
2 =2 0 5 5
y¯ = 0 18
3 xy dx x(5x − x2)dx

y dx x¯ = 0 = 0
5 5
0 y dx (5x − x2)dx

1 3 1 16x3 3 00
=2 2 3
16x2dx = 0 = 72 = 4 5 5x3 − x4 5
3 40
0 (5x2 − x3)dx 5x2 − x3 5

18 18 18

Hence the centroid lies at (2, 4). = 0 =
In Fig. 58.5, ABD is a right-angled triangle. The cen- 5
(5x − x2)dx
troid lies 4 units from AB and 1 unit from BD showing 0 2 30

Centroids of simple shapes 499

625 − 625 625 From Section 58.4,

= 3 4 = 12 1 4 1 4y
125 125 125 dy
− x2dy
23 6 2 21 2
x¯ = 1 = y
= 625 6 = 5 = 2.5 4 4

x dy dy
12 125 2 1 12

1 5 1 5 1 y2 4 15

2 y2dx 2 (5x − x2)2dx 24
2y√3/2
y¯ = 0 = 0 = 32 1 = 8 = 0.568
5 5 4 1√4
y dx (5x − x2) dx 32

00 1

1 5 4 4y

=2 0 (25x2 − 10x3 + x4) dx xy dy (y)d y

125 and y¯ = 1 =1 2
4 14
√
6 x dy 32

1 25x3 10x4 x5 5 1 ⎡ ⎤4

=2 3 −+ 50 √1 y5 /2
4 2 ⎢⎣ 5 ⎦⎥
125 4 y√3/2 dy

6 = 1 2 = 21
14 1√4
1 25(125) − 6250 + 625 √
32 32
=2 34 = 2.5
125 √2 (31)

6 =5 2 = 2.657
1√4
32

Hence the position of the centroid is at (0.568, 2.657)

Problem 6. Locate the position of the centroid
enclosed by the curves y = x2 and y2 = 8x

Figure 58.7 shows the two curves intersection at (0, 0)
and (2, 4). These are the same curves as used in
Problem 12, Chapter 55 where the shaded area was

Figure 58.6 y y 2 ϭ 8x Section 9
y ϭx 2 (or y ϭ 8x)
Hence the centroid of the area lies at (2.5, 2.5)
(Note from Fig. 58.6 that the curve is symmetrical about 4y x
x = 2.5 and thus x¯ could have been determined ‘on 32
sight’).
2y C
Problem 5. Locate the centroid of the area
enclosed by the curve y = 2x2, the y-axis and 1
ordinates y = 1 and y = 4, correct to 3 decimal x2
places
0 12

Figure 58.7

500 Engineering Mathematics

calculated as 2 2 square units. Let the co-ordinates of Thus the position of the centroid of the enclosed area
3 in Fig. 58.7 is at (0.9, 1.8)
centroid C be x¯ and y¯.

2

xy dx

By integration, x¯ = 0 Now try the following exercise
2

y dx

0 Exercise 199 Further problems on centroids
The value of y is given byy=th√e 8hxei−ghxt2o.fHthenectey,pical strip of simple shapes
shown in Fig. 58.7, i.e.

2√ 2√ 1. Determine the position of the centroid of a
x( 8x − x2)dx ( 8 x3/2 − x3)
sheet of metal formed by the curve y = 4x − x2
x¯ = 0 =0
2 2 which lies above the x-axis. [(2, 1.6)]

22
33
⎡ ⎤2 ⎛ √ 2. Find the coordinates of the centroid of the
area that lies between curve y = x − 2 and the
⎢⎣√8 x5/2 x4 √ 25 ⎞
5 4 = ⎜⎜⎜⎝⎜⎜⎜⎜ 8
− ⎥⎦ 5 − 4 ⎟⎟⎟⎠⎟⎟⎟⎟ x-axis. x [(1, −0.4)]

= 20 2 3. Determine the coordinates of the centroid of
2 2
2 the area formed between the curve y = 9 − x2
2 3
3 and the x-axis. [(0, 3.6)]

2 2 4. Determine the centroid of the area lying

= 5 = 0.9 between y = 4x2, the y-axis and the ordinates
2
2 y = 0 and y = 4. [(0.375, 2.40]

3 5. Find the position of the c√entroid of the area
enclosed by the curve y = 5x, the x-axis and
Care needs to be taken when findin√g y¯ in such
examples as this. From Fig. 58.7, y = 8x − x2 and
1√ the ordinate x = 5. [(3.0, 1.875)]
y ( 8x x2).
= − The perpendicular distance from
22
centroid 1 √ − x2 + x2. 6. Sketch the curve y2 = 9x between the limits
( 8x
C of the strip to Ox is 2 ) x = 0 and x = 4. Determine the position of the

Taking moments about Ox gives: centroid of this area. [(2.4, 0)]

(total area) (y¯) = x=2 (area of strip) (perpendicular 7. Calculate the points of intersection of the
x=0

distance of centroid of strip to Ox) curves x2 = 4y and y2 = x, and determine the
4
Hence (area) (y¯)
position of the centroid of the area enclosed by
√ 1 √
= 8x − x2 ( 8x − x2 ) + x2 dx them. [(0, 0) and (4, 4), (1.80, 1.80)]

Section 9 2

2 2√ √ x2 8. Determine the position of the centroid of the
i.e. 2 (y¯) = 8x − x2 8x + dx sector of a circle of radius 3 cm whose angle
subtended at the centre is 40◦.
30 22
On the centre line, 1.96 cm
= 2 8x − x4 dx = 8x2 − x5 2 from the centre

02 2 4 10 0 9. Sketch the curves y = 2x2 + 5 and
y − 8 = x(x + 2) on the same axes and
= 8 − 3 1 − (0) = 4 4 determine their points of intersection. Calcu-
55 late the coordinates of the centroid of the area
enclosed by the two curves.
4 4 [(−1, 7) and (3, 23), (1, 10.20)]

Hence y¯ = 5 = 1.8
2
2
3

Centroids of simple shapes 501

58.7 Theorem of Pappus volume generated = area × distance moved through by
centroid i.e.
A theorem of Pappus states:
‘If a plane area is rotated about an axis in its own plane 4 πr3 = πr2 (2πy¯)
but not intersecting it, the volume of the solid formed is 32
given by the product of the area and the distance moved
by the centroid of the area’. Hence y¯ = 4 πr3 = 4r
3 3π
With reference to Fig. 58.8, when the curve y = f (x) π2r2
is rotated one revolution about the x-axis between the
limits x = a and x = b, the volume V generated is By integration,
given by:
volume V = (A)(2πy¯), from which, 1 r

y¯ = V y2dx
2πA y¯ = 2 −r
area

y ϭ f (x) 1 r 1 r2x − x3 r
y =2 2 3 −r
(r2 − x2)dx
= πr2
−r
Area A
C πr2

22

y 1 r3 − r3 − −r3 + r3

xϭa xϭb x =2 3 3 = 4r
πr2 3π
Figure 58.8
2

Problem 7. Determine the position of the Hence the centroid of a semicircle lies on the axis
centroid of a semicircle of radius r by using the 4r
theorem of Pappus. Check the answer by using
integration (given that the equation of a circle, of symmetry, distance (or 0.424 r) from its
centre 0, radius r is x2 + y2 = r2) 3π

diameter.

A semicircle is shown in Fig. 58.9 with its diame- Problem 8. Calculate the area bounded by the Section 9
ter lying on the x-axis and its centre at the origin. curve y = 2x2, the x-axis and ordinates x = 0 and
Area of semicircle = πr2 . When the area is rotated x = 3. (b) If this area is revolved (i) about the x-axis
and (ii) about the y-axis, find the volumes of the
2
about the x-axis one revolution a sphere is generated solids produced. (c) Locate the position of the
of volume 4 πr3.
centroid using (i) integration, and (ii) the theorem
3
of Pappus
y
y y ϭ 2x 2
x 2ϩy 2 ϭ r 2 18

C 12
y 6x
y
Ϫr 0 rx 0 1 2 3x

Figure 58.9

Let centroid C be at a distance y¯ from the origin Figure 58.10
as shown in Fig. 58.9. From the theorem of Pappus,

502 Engineering Mathematics

(a) The required area is shown shaded in Fig. 58.10. (ii) using the theorem of Pappus:
Volume generated when shaded area is revolved
Area = 3 3 2x3 3 about Oy = (area)(2πx¯)

y dx = 2x2 dx = 30

00 i.e. 81π = (18)(2πx¯),

= 18 square units

(b) (i) When the shaded area of Fig. 58.10 is revolved from which, x¯ = 81π = 2.25
360◦ about the x-axis, the volume generated 36π

33 Volume generated when shaded area is revolved
about Ox = (area)(2πy¯)
= πy2dx = π(2x2)2dx
i.e. 194.4π = (18)(2πy¯),
00

= 3 x5 3 243 194.4π
y¯ = = 5.4
4πx4dx = 4π = 4π
0 50 5 from which, 36π

= 194.4π cubic units Hence the centroid of the shaded area in
Fig. 58.10 is at (2.25, 5.4)
(ii) When the shaded area of Fig. 58.10 is revolved
360◦ about the y-axis, the volume generated = Problem 9. A cylindrical pillar of diameter
400 mm has a groove cut round its circumference.
(volume generated by x = 3) − (volume generated The section of the groove is a semicircle of diameter
by y = 2x2) 50 mm. Determine the volume of material removed,
in cubic centimetres, correct to 4 significant figures
= 18 18 y
π dy A part of the pillar showing the groove is shown in
π(3)2dy − Fig. 58.11.
0 02 The distance of the centroid of the semicircle from
its base is 4r (see Problem 7) = 4(25) = 100 mm.
= π 18 9 − y dy = π 9y − y2 18
02 40 3π 3π 3π
The distance of the centroid from the centre of the
= 81π cubic units pillar = 200 − 100 mm.

(c) If the co-ordinates of the centroid of the shaded 3π
area in Fig. 58.10 are (x¯, y¯) then:

(i) by integration,

3 3

xy dx x(2x2) dx

x¯ = 0 =0
3 18

y dx

Section 9 0

= 3 2x4 3 50 mm 400 mm
4 0 = 81 = 2.25
2x3dx
0=
18 18 36

1 3 1 3

y2dx (2x2)2dx
2 =2 0
y¯ = 0 18
3 200 mm

y dx

0

1 3 = 1 4x5 3
2 5
2 4x4dx 0 = 5.4

= 0

18 18 Figure 58.11

Centroids of simple shapes 503

The distance moved by the centroid in one revolution Area of semicircle

= 2π 200 − 100 = 400π − 200 mm. = πr2 = π(1.0)2 = π cm2
3π 3 2 22

From the theorem of Pappus, By the theorem of Pappus, volume generated
volume = area × distance moved by centroid = area × distance moved by centroid
= π (2π)(4.576)
= 1 π252 400π − 200 = 1168250 mm3 2
23
i.e. volume of metal removed = 45.16 cm3
Hence the volume of material removed is 1168 cm3 Mass of metal removed = density × volume
correct to 4 significant figures.
= 8000 kg m−3 × 45.16 m3
Problem 10. A metal disc has a radius of 5.0 cm 106
and is of thickness 2.0 cm. A semicircular groove of
diameter 2.0 cm is machined centrally around the = 0.3613 kg or 361.3 g
rim to form a pulley. Determine, using Pappus’
theorem, the volume and mass of metal removed Volume of pulley = volume of cylindrical disc
and the volume and mass of the pulley if the density − volume of metal removed
of the metal is 8000 kg m−3
= π(5.0)2(2.0) − 45.16 = 111.9 cm3

Mass of pulley = density × volume

= 8000 kg m−3 × 111.9 m3
106

= 0.8952 kg or 895.2 g

A side view of the rim of the disc is shown in Fig. 58.12.

Now try the following exercise

2.0 cm Exercise 200 Further problems on the
PQ theorem of Pappus

5.0 cm 1. A right angled isosceles triangle having a

hypotenuse of 8 cm is revolved one revolution

SR about one of its equal sides as axis. Deter-
X
X mine the volume of the solid generated using

Pappus’ theorem. [189.6 cm3]

2. A rectangle measuring 10.0 cm by 6.0 cm

rotates one revolution about one of its longest Section 9

sides as axis. Determine the volume of the

Figure 58.12 resulting cylinder by using the theorem of

Pappus. [1131 cm2]

When area PQRS is rotated about axis XX the vol- 3. Using (a) the theorem of Pappus, and (b) inte-

ume generated is that of the pulley. The centroid of gration, determine the position of the centroid

the semicircular area removed is at a distance of 4r of a metal template in the form of a quadrant

4(1.0) 3π of a circle of radius 4 cm. (The equation of a
circle, centre 0, radius r is x2 + y2 = r2).
from its diameter (see Problem 7), i.e. 3π , i.e.
⎡⎤
0.424 cm from PQ. Thus the distance of the centroid On the centre line, distance 2.40 cm

from XX is (5.0 − 0.424), i.e. 4.576 cm. The distance ⎣ from the centre, i.e. at coordinates ⎦

moved through in one revolution by the centroid is (1.70, 1.70)

2π(4.576) cm.

504 Engineering Mathematics

4. (a) Determine the area bounded by the curve 5. A metal disc has a radius of 7.0 cm and is
(b) y = 5x2, the x-axis and the ordinates of thickness 2.5 cm. A semicircular groove of
(c) x = 0 and x = 3. diameter 2.0 cm is machined centrally around
If this area is revolved 360◦ about (i) the the rim to form a pulley. Determine the vol-
ume of metal removed using Pappus’ theorem
x-axis, and (ii) the y-axis, find the vol- and express this as a percentage of the origi-
nal volume of the disc. Find also the mass of
umes of the solids of revolution pro- metal removed if the density of the metal is
7800 kg m−3.
duced in each case. [64.90 cm3, 16.86%, 506.2 g]

Determine the co-ordinates of the cen-

troid of the area using (i) integral calcu-

lus, and (ii) the theorem of Pappus
⎡⎤
(a) 45 square units (b) (i) 1215π
⎣ cubic units (ii) 202.5π cubic units ⎦

(c) (2.25, 13.5)

Section 9

Chapter 59

Second moments of area

59.1 Second moments of area and strip = bδx. Second moment of area of the shaded strip
radius of gyration about PP = (x2)(bδx).

The first moment of area about a fixed axis of a lamina Figure 59.1
of area A, perpendicular distance y from the centroid
of the lamina is defined as Ay cubic units. The second The second moment of area of the whole rectangle
moment of area of the same lamina as above is given by
Ay2, i.e. the perpendicular distance from the centroid of about PP is obtained by summing all such strips between
the area to the fixed axis is squared. Second moments of x=l
areas are usually denoted by I and have limits of mm4, x = 0 and x = l, i.e. x=0 x2bδx. It is a fundamental
cm4, and so on.
theorem of integration that
Radius of gyration
x=l l
Several areas, a1, a2, a3, . . . at distances y1, y2, y3, . . . limit x2bδx = x2b dx
from a fixed axis, may be replaced by a single area δx→0 x=0
A, where A = a1 + a2 + a3 + · · · at distance k from the 0
axis, such that Ak2 = ay2. k is called the radius
of gyration of area A about the given axis. Since Thus the second moment of area of the rectangle about
Ak2 = ay2 = I then the radius of gyration, k = I
PP = b l x3 l = bl3
A
The second moment of area is a quantity much used in x2dx = b
the theory of bending of beams, in the torsion of shafts, 0 30 3
and in calculations involving water planes and centres Since the total area of the rectangle, A = lb, then
of pressure.
Ipp = (lb) l2 = Al2
59.2 Second moment of area of 3 3
regular sections
Ipp = Akp2p thus kp2p = l2
The procedure to determine the second moment of area 3
of regular sections about a given axis is (i) to find the
second moment of area of a typical element and (ii) i.e. the radius of gyration about axes PP,
to sum all such second moments of area by integrating
between appropriate limits. kpp = l2 l
=√
For example, the second moment of area of the rect- 33
angle shown in Fig. 59.1 about axis PP is found by
initially considering an elemental strip of width δx, par-
allel to and distance x from axis PP. Area of shaded

506 Engineering Mathematics

59.3 Parallel axis theorem Figure 59.4

In Fig. 59.2, axis GG passes through the centroid C of 59.5 Summary of derived results
area A. Axes DD and GG are in the same plane, are
parallel to each other and distance d apart. The parallel A summary of derive standard results for the sec-
axis theorem states: ond moment of area and radius of gyration of regular
sections are listed in Table 59.1.
IDD = IGG + Ad2

Using the parallel axis theorem the second moment of
area of a rectangle about an axis through the centroid
may be determined. In the rectangle shown in Fig. 59.3,

bl3
Ipp = 3 (from above). From the parallel axis theorem

l2
Ipp = IGG + (bl) 2

Table 59.1 Summary of standard results of the sec-
ond moments of areas of regular sections

Shape Position of axis Second Radius of
moment gyration, k
of area, I

Figure 59.2 Rectangle (1) Coinciding with b bl3 √l
i.e. 3 3
length l (2) Coinciding with l lb3 b
from which, breadth b 3
(3) Through centroid, bl3 √
bl3 bl3 parallel to b l2 3
3 = IGG + 4 lb3 l
(4) Through centroid, 12
IGG = bl3 − bl3 = bl3 parallel to l √
3 4 12 12

P IG I Triangle (1) Coinciding with b bh3 √b
22 Perpendicular 12 12
height h (2) Through centroid, bh3
base b parallel to base 36 √h
bh3 6
Section 9 C b (3) Through vertex, 4 h
x parallel to base
√
dx 18
PG
√h
2

Circle (1) Through centre, πr4 √r
radius r perpendicular 2 2
to plane (i.e.
Figure 59.3 polar axis) πr4 r
4 2
59.4 Perpendicular axis theorem (2) Coinciding 5πr4 √
with diameter 4
In Fig. 59.4, axes OX, OY and OZ are mutually perpen- 5
dicular. If OX and OY lie in the plane of area A then the (3) About a tangent r
perpendicular axis theorem states:
2
IOZ = IOX + IOY
Semicircle Coinciding πr4 r
radius r with diameter 8 2

Second moments of area 507

59.6 Worked problems on second lb3
moments of area of regular IGG = 12 where l = 40.0 mm and b = 15.0 mm
sections
Hence IGG = (40.0)(15.0)3 = 11 250 mm4
12

Problem 1. Determine the second moment of From the parallel axis theorem, IPP = IGG + Ad2,
area and the radius of gyration about axes AA, BB where A = 40.0 × 15.0 = 600 mm2 and d = 25.0 +
and CC for the rectangle shown in Fig. 59.5.
7.5 = 32.5 mm, the perpendicular distance between

GG and PP.

l ϭ12.0 cm A Hence, IPP = 11 250 + (600)(32.5)2
= 645 000 mm4
CC
b ϭ 4.0 cm IPP = AkP2P

BB IPP
A area

Figure 59.5 from which, kPP =

From Table 59.1, the second moment of area about axis = 645 000 = 32.79 mm
bl3 (4.0)(12.0)3 600
3 3
AA, IAA = = = 2304 cm4 Problem 3. Determine the second moment of
area and radius of gyration about axis QQ of the
Radius of gyration, triangle BCD shown in Fig. 59.7

kAA = l = 1√2.0 = 6.93 cm
√ 33

lb3 (12.0)(4.0)3 256 cm4 B
3 3
Similarly, IBB = = =

and kBB = √b = √4.0 = 2.31 cm 12.0 cm G G
33

The second moment of area about the centroid of a CD
bl3 8.0 cm 6.0 cm

rectangle is when the axis through the centroid is QQ
12
Figure 59.7
parallel with the breadth, b. In this case, the axis CC is
parallel with the length l.

Hence ICC = lb3 = (12.0)(4.0)3 = 64 cm4
12 12
Using the parallel axis theorem: IQQ = IGG + Ad2,
and kCC = b = √4.0 = 1.15 cm where IGG is the second moment of area about the
√ centroid of the triangle,
12 12
Section 9
Problem 2. Find the second moment of area and i.e. bh3 = (8.0)(12.0)3 = 384 cm4, A is the area of
the radius of gyration about axis PP for the 36 36
rectangle shown in Fig. 59.6 1 1 cm2
the triangle = 2 bh = 2 (8.0)(12.0) = 48 and d is the
40.0 mm
GG distance between axes GG and QQ = 6.0 + 1 (12.0) =
3
15.0 mm 10 cm.

25.0 mm Hence the second moment of area about axis QQ,

IQQ = 384 + (48)(10)2 = 5184 cm4

Radius of gyration,

P P IQQ = 5184 = 10.4 cm
Figure 59.6 area 48
kQQ =

508 Engineering Mathematics Using the parallel axis theorem: IBB = IGG + Ad2,

Problem 4. Determine the second moment of where IBB = πr4
area and radius of gyration of the circle shown in and (from Table 59.1)
Fig. 59.8 about axis YY 8

r ϭ 2.0 cm = π(10.0)4 = 3927 mm4,
GG 8

A = πr2 = π(10.0)2 = 157.1 mm2
22

d = 4r = 4(10.0) = 4.244 mm
3π 3π

3.0 cm Hence 3927 = IGG + (157.1)(4.244)2

i.e. 3927 = IGG + 2830,

Y Y from which, IGG = 3927 − 2830 = 1097 mm4
Figure 59.8

In Fig. 59.8, IGG = πr4 = π (2.0)4 = 4π cm4. Using Using the parallel axis theorem again:
4 4 IXX = IGG + A(15.0 + 4.244)2
the parallel axis theorem, IYY = IGG + Ad2, where i.e. IXX = 1097 + (157.1)(19.244)2
= 1097 + 58 179 = 59 276 mm4 or 59 280 mm4, cor-
d = 3.0 + 2.0 = 5.0 cm. rect to 4 significant figures.

Hence IYY = 4π + [π(2.0)2](5.0)2
= 4π + 100π = 104π = 327 cm4
Radius of gyration, kXX = IXX = 59 276
area 157.1
Radius of gyration,
= 19.42 mm

kYY = IYY = 104π = √ = 5.10 cm Problem 6. Determine the polar second moment
area π(2.0)2 26 of area of the propeller shaft cross-section shown in
Fig. 59.10

Problem 5. Determine the second moment of
area and radius of gyration for the semicircle shown
in Fig. 59.9 about axis XX

Section 9
6.0 cm
7.0 cm
G 10.0 mm G
B B

15.0 mm Figure 59.10

X X πr4
Figure 59.9 The polar second moment of area of a circle = .

4r 2
The centroid of a semicircle lies at from its diameter. The polar second moment of area of the shaded area is
given by the polar second moment of area of the 7.0 cm
3π diameter circle minus the polar second moment of area
of the 6.0 cm diameter circle. Hence the polar second

Second moments of area 509

moment of area of the

cross-section shown = π 7.0 4 − π 6.0 4 Fig. 59.12 about (a) axis AA (b) axis BB, and

(c) axis CC. ⎡ (a) 72 cm4, 1.73 cm ⎤
⎣ (b) 128 cm4, 2.31 cm ⎦
22 22

= 235.7 − 127.2 = 108.5 cm4 (c) 512 cm4, 4.62 cm

Problem 7. Determine the second moment of B C
area and radius of gyration of a rectangular lamina 8.0 cm
of length 40 mm and width 15 mm about an axis
through one corner, perpendicular to the plane of AA
the lamina 3.0 cm

BC

The lamina is shown in Fig. 59.11. Figure 59.12

Z Y I ϭ 40 mm ϭ 15 mm 2. Determine the second moment of area and
X
b radius of gyration for the triangle shown in

X Y Z Fig. 59.13 about (a) axis DD (b) axis EE, and
Figure 59.11
(c) an axis through the centroid of the triangle

parallel to axis DD⎡. (a) 729 cm4, 3.67 cm ⎤
⎣ (b) 2187 cm4, 6.36 cm ⎦
(c) 243 cm4, 2.12 cm

From the perpendicular axis theorem:

EE

IZZ = IXX + IYY

IXX = lb3 = (40)(15)3 = 45 000 mm4 9.0 cm
3 3

and IYY = bl3 = (15)(40)3 = 320 000 mm4 D D
Hence 3 3 12.0 cm

IZZ = 45 000 + 320 000 Figure 59.13

= 365 000 mm4 or 36.5 cm4 3. For the circle shown in Fig. 59.14, find the

Radius of gyration, second moment of area and radius of gyration

kZZ = IZZ = 365 000 about (a) axis FF, and (b) axis HH.
area (40)(15) (a) 201 cm4, 2.0 cm
(b) 1005 cm4, 4.47 cm
rϭ
= 24.7 mm or 2.47 cm H 4.0 cm
H
Section 9
Now try the following exercise
FF

Exercise 201 Further problems on second Figure 59.14
moments of area of regular
sections 4. For the semicircle shown in Fig. 59.15, find the

1. Determine the second moment of area and second moment of area and radius of gyration
radius of gyration for the rectangle shown in
about axis JJ. [3927 mm4, 5.0 mm]

510 Engineering Mathematics

rϭ 4.0 cm
10.0 mm

JJ X 1.0 cm X
1.0 cm
2.0 cm 8.0 cm

Figure 59.15 T 2.0 cm

5. For each of the areas shown in Fig. 59.16 deter- CT

mine the second moment of area and radius of T
6.0 cm
gyration about axis LL, by using the parallel

axis theorem. ⎡ 335 cm4, ⎤
(a)
4.73 cm

⎣ (b) 22 030 cm4, 14.3 cm ⎦

(c) 628 cm4, 7.07 cm Figure 59.17

(a) (b) (c) For the semicircle, IXX = πr4 = π(4.0)4
3.0 cm Dia=4.0 cm 8 8

5.0 cm 15 cm 15 cm 5 cm = 100.5 cm4
L
18 cm 10 cm For the rectangle, IXX = bl3 = (6.0)(8.0)3
33
2.0 cm = 1024 cm4

L

Figure 59.16 For the triangle, about axis TT through centroid CT ,

6. Calculate the radius of gyration of a rectan- ITT = bh3 = (10)(6.0)3 = 60 cm4
gular door 2.0 m high by 1.5 m wide about a 36 36
vertical axis through its hinge. [0.866 m]
By the parallel axis theorem, the second moment of area
7. A circular door of a boiler is hinged so that
it turns about a tangent. If its diameter is of the triangle about axis XX
1.0 m, determine its second moment of area
and radius of gyration about the hinge. = 60 + 1 (10)(6.0) 8.0 + 1 (6.0) 2 = 3060 cm4.
[0.245 m4, 0.599 m] 2 3

8. A circular cover, centre 0, has a radius of Total second moment of area about XX.
12.0 cm. A hole of radius 4.0 cm and centre X, = 100.5 + 1024 + 3060 = 4184.5 = 4180 cm4, correct
where OX = 6.0 cm, is cut in the cover. Deter-
mine the second moment of area and the radius to 3 significant figures
of gyration of the remainder about a diameter
through 0 perpendicular to OX. Problem 9. Determine the second moment of
[14 280 cm4, 5.96 cm] area and the radius of gyration about axis XX for
the I-section shown in Fig. 59.18

Section 9 S

8.0 cm

CD 3.0 cm

59.7 Worked problems on second 3.0 cm 7.0 cm
moments of area of composite C
areas CE
4.0 cm
Problem 8. Determine correct to 3 significant C CF X
figures, the second moment of area about XX for y 15.0 cm
the composite area shown in Fig. 59.17
X

S

Figure 59.18

Second moments of area 511

The I-section is divided into three rectangles, D, E (a) (b)
and F and their centroids denoted by CD, CE andCF 18.0 mm 6.0 cm
respectively.
3.0 mm 2.5 cm
For rectangle D:
The second moment of area about CD (an axis through 12.0 mm 2.0 cm
CD parallel to XX) 3.0 cm

= bl3 = (8.0)(3.0)3 = 18 cm4 X 4.0 mm X 2.0 cm
12 12 X
X

Using the parallel axis theorem: IXX = 18 + Ad2 where Figure 59.19
A = (8.0)(3.0) = 24 cm2 and d = 12.5 cm
2. Determine the second moment of area about
Hence IXX = 18 + 24(12.5)2 = 3768 cm4 the given axes for the shapes shown in
Fig. 59.20. (In Fig. 59.20(b), the circular area
For rectangle E: is removed.)
The second moment of area about CE (an axis through IAA = 4224 cm4, IBB = 6718 cm4,
CE parallel to XX) ICC = 37 300 cm4

= bl3 = (3.0)(7.0)3 = 85.75 cm4 3.0 cm
12 12

Using the parallel axis theorem: B
IXX = 85.75 + (7.0)(3.0)(7.5)2 = 1267 cm4 4.5 cm

For rectangle F: 16.0 cm 9.0 cm

bl3 (15.0)(4.0)3 320 cm4 4.0 cm Dia ϭ 7.0 cm
3 3
IXX = = = A A 15.0 cm
9.0 cm

Total second moment of area for the I-section about CC
B 10.0 cm
axis XX,
IXX = 3768 + 1267 + 320 = 5355 cm4 (a) (b)

Total area of I-section Figure 59.20
= (8.0)(3.0) + (3.0)(7.0) + (15.0)(4.0) = 105 cm2.
3. Find the second moment of area and radius of
Radius of gyration, gyration about the axis XX for the beam section
shown in Fig. 59.21. [1350 cm4, 5.67 cm]
kXX = IXX = 5355 = 7.14 cm
area 105

Now try the following exercise 6.0 cm 1.0 cm Section 9
2.0 cm 8.0 cm

Exercise 202 Further problems on section X 2.0 cm
moment of areas of composite 10.0 cm X
areas
Figure 59.21
1. For the sections shown in Fig. 59.19, find
the second moment of area and the radius of
gyration about axis XX.
(a) 12 190 mm4, 10.9 mm
(b) 549.5 cm4, 4.18 cm

Revision Test 16

This Revision test covers the material contained in Chapters 55 to 59. The marks for each question are shown in
brackets at the end of each question.

1. The force F newtons acting on a body at a dis-

tance x metres from a fixed point is given by:
x2
F = 2x + 3x2. If work done = x1 F dx, deter- 500 mm
40 mm
mine the work done when the body moves from
250 mm
the position when x = 1 m to that when x = 4 m.

(4)

2. Sketch and determine the area enclosed by the

curve y=3 sin θ the θ-axis and ordinates θ =0
,
2
and θ = 2π .
3 (4)

3. Calculate the area between the curve

y = x3 − x2 − 6x and the x-axis. (10)

Figure R16.1

4. A voltage v = 25 sin 50πt volts is applied across of Pappus to determine the volume of material
an electrical circuit. Determine, using integration, removed, in cm3, correct to 3 significant figures.
its mean and r.m.s. values over the range t = 0 to
t = 20 ms, each correct to 4 significant figures. (8)
(12)
8. For each of the areas shown in Fig. R16.2 deter-

5. Sketch on the same axes the curves x2 = 2y and mine the second moment of area and radius of

y2 = 16x and determine the co-ordinates of the gyration about axis XX. (15)

points of intersection. Determine (a) the area

enclosed by the curves, and (b) the volume of the (a) (b) (c)
48 mm
solid produced if the area is rotated one revolution
70 mm
about the x-axis. (13) 15.0 mm 15.0 mm
25 mm
X DIA = 5.0 cm

6. Calculate the position of the centroid of the sheet 4.0 cm 18.0 mm
5.0 mm
of metal formed by the x-axis and the part of the
curve y = 5x − x2 which lies above the x-axis. X

(9)

7. A cylindrical pillar of diameter 500 mm has a Figure R16.2

Section 9 groove cut around its circumference as shown in 9. A circular door is hinged so that it turns about a

Fig. R16.1. The section of the groove is a semicir- tangent. If its diameter is 1.0 m find its second

cle of diameter 40 mm. Given that the centroid of moment of area and radius of gyration about the
4r
hinge. (5)
a semicircle from its base is , use the theorem
3π

Section 10

Further Number and
Algebra

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Chapter 60

Boolean algebra and
logic circuits

60.1 Boolean algebra and switching when both A and B are on. The truth table for a two-input
circuits and-function is shown in Fig. 60.2(b).

A two-state device is one whose basic elements can Figure 60.1
only have one of two conditions. Thus, two-way
switches, which can either be on or off, and the binary Figure 60.2
numbering system, having the digits 0 and 1 only, are
two-state devices. In Boolean algebra, if A represents The not-function
one state, then A, called ‘not-A’, represents the second In Boolean algebra, the not-function for element A is
state. written as A, and is defined as ‘the opposite to A’. Thus if
A means switch A is on, A means that switch A is off. The
The or-function truth table for the not-function is shown in Table 60.1.

In Boolean algebra, the or-function for two elements A
and B is written as A + B, and is defined as ‘A, or B, or
both A and B’. The equivalent electrical circuit for a two-
input or-function is given by two switches connected in
parallel. With reference to Fig. 60.1(a), the lamp will
be on when A is on, when B is on, or when both A
and B are on. In the table shown in Fig. 60.1(b), all the
possible switch combinations are shown in columns 1
and 2, in which a 0 represents a switch being off and a
1 represents the switch being on, these columns being
called the inputs. Column 3 is called the output and a
0 represents the lamp being off and a 1 represents the
lamp being on. Such a table is called a truth table.

The and-function

In Boolean algebra, the and-function for two elements
A and B is written as A · B and is defined as ‘both A and
B’. The equivalent electrical circuit for a two-input and-
function is given by two switches connected in series.
With reference to Fig. 60.2(a) the lamp will be on only

516 Engineering Mathematics

Table 60.1 Output
Z=A
Input
A

01

10

In the above, the Boolean expressions, equivalent
switching circuits and truth tables for the three func-
tions used in Boolean algebra are given for a two-input
system. A system may have more than two inputs and
the Boolean expression for a three-input or-function
having elements A, B and C is A + B + C. Similarly,
a three-input and-function is written as A · B · C. The
equivalent electrical circuits and truth tables for three-
input or and and-functions are shown in Figs. 60.3(a)
and (b) respectively.

Figure 60.4

or-function applied to columns 3 and 4 giving an output
of Z = A·B + A·B. The corresponding switching circuit
is shown in Fig. 60.4(b) in which A and B are connected
in series to give A · B, A and B are connected in series
to give A · B, and A · B and A · B are connected in parallel
to give A · B + A · B. The circuit symbols used are such
that A means the switch is on when A is 1, A means the
switch is on when A is 0, and so on.

Problem 1. Derive the Boolean expression and
construct a truth table for the switching circuit
shown in Fig. 60.5.

Section 10 Figure 60.3 Figure 60.5

To achieve a given output, it is often necessary to use The switches between 1 and 2 in Fig. 60.5 are in series
combinations of switches connected both in series and in and have a Boolean expression of B · A. The parallel
parallel. If the output from a switching circuit is given by circuit 1 to 2 and 3 to 4 have a Boolean expression of
the Boolean expression Z = A · B + A · B, the truth table (B · A + B). The parallel circuit can be treated as a single
is as shown in Fig. 60.4(a). In this table, columns 1 and 2 switching unit, giving the equivalent of switches 5 to 6,
give all the possible combinations of A and B. Column 3 6 to 7 and 7 to 8 in series. Thus the output is given by:
corresponds to A · B and column 4 to A · B, i.e. a 1 output
is obtained when A = 0 and when B = 0. Column 5 is the Z = A · (B · A + B) · B

Boolean algebra and logic circuits 517

The truth table is as shown in Table 60.2. Columns 1 and Table 60.3
2 give all the possible combinations of switches A and
B. Column 3 is the and-function applied to columns 1 1234 5 6 7
and 2, giving B · A. Column 4 is B, i.e., the opposite
to column 2. Column 5 is the or-function applied to A B C B A + B C + (A + B) Z = B · [C + (A + B)]
columns 3 and 4. Column 6 is A, i.e. the opposite to
column 1. The output is column 7 and is obtained by 0001 1 1 0
applying the and-function to columns 4, 5 and 6.
0011 1 1 0

0100 0 0 0

Table 60.2 0110 0 1 1

12 3 4 5 6 7 1001 1 1 0

A B B · A B B · A + B A Z = A · (B · A + B) · B 1011 1 1 0

00 0 1 1 1 1 1100 1 1 1

01 0 0 0 1 0 1110 1 1 1

10 0 1 1 0 0

11 1 0 1 0 0 Problem 3. Construct a switching circuit to meet
the requirements of the Boolean expression:
Problem 2. Derive the Boolean expression and
construct a truth table for the switching circuit Z =A · C +A · B+A · B · C
shown in Fig. 60.6. Construct the truth table for this circuit.

The three terms joined by or-functions, (+), indicate
three parallel branches.

having: branch 1 A and C in series

branch 2 A and B in series

and branch 3 A and B and C in series

Figure 60.6

The parallel circuit 1 to 2 and 3 to 4 gives (A + B) and this Figure 60.7 Section 10
is equivalent to a single switching unit between 7 and 2.
The parallel circuit 5 to 6 and 7 to 2 gives C + (A + B) Hence the required switching circuit is as shown in
and this is equivalent to a single switching unit between Fig. 60.7. The corresponding truth table is shown in
8 and 2. The series circuit 9 to 8 and 8 to 2 gives the Table 60.4.
output
Column 4 is C, i.e. the opposite to column 3
Z = B · [C + (A + B)]
Column 5 is A·C, obtained by applying the and-function
The truth table is shown in Table 60.3. Columns 1, 2 to columns 1 and 4
and 3 give all the possible combinations of A, B and C.
Column 4 is B and is the opposite to column 2. Column Column 6 is A, the opposite to column 1
5 is the or-function applied to columns 1 and 4, giving
(A + B). Column 6 is the or-function applied to columns Column 7 is A·B, obtained by applying the and-function
3 and 5 giving C + (A + B). The output is given in col- to columns 2 and 6
umn 7 and is obtained by applying the and-function to
columns 2 and 6, giving Z = B · [C + (A + B)].

518 Engineering Mathematics

Table 60.4 0 and B is 0 and C is 0 and this corresponds to the
Boolean expression A · B · C. In row 3, A is 0 and B is
1234 5 6 7 8 9 1 and C is 0, i.e. the Boolean expression in A · B · C.
Similarly in rows 4 and 6, the Boolean expressions are
A B C C A·C A A·B A·B·C Z=A·C+A·B+A·B·C A · B · C and A · B · C respectively. Hence the Boolean
expression is:
0001 0 1 0 0 0
Z=A·B·C+A·B·C+ A·B·C+A·B·C
0010 0 1 0 0 0
The corresponding switching circuit is shown in
0101 0 1 1 1 1 Fig. 60.8. The four terms are joined by or-functions,
(+), and are represented by four parallel circuits. Each
0110 0 1 1 0 1 term has three elements joined by an and-function,
and is represented by three elements connected in
1001 1 0 0 0 1 series.

1010 0 0 0 0 0

1101 1 0 0 0 1

1110 0 0 0 0 0

Column 8 is A · B · C, obtained by applying the and-
function to columns 4 and 7

Column 9 is the output, obtained by applying the or-
function to columns 5, 7 and 8

Section 10 Problem 4. Derive the Boolean expression and Figure 60.8
construct the switching circuit for the truth table
given in Table 60.5. Now try the following exercise

Table 60.5 Exercise 203 Further problems on Boolean
algebra and switching circuits
ABCZ
100 01 In Problems 1 to 4, determine the Boolean expres-
200 10 sions and construct truth tables for the switching
301 01 circuits given.
401 11
510 00 1. The circuit shown in Fig. 60.9
610 11 C · (A · B + A · B);
711 00 see Table 60.6, col. 4
811 10

Examination of the truth table shown in Table 60.5 Figure 60.9
shows that there is a 1 output in the Z-column in rows
1, 3, 4 and 6. Thus, the Boolean expression and switch-
ing circuit should be such that a 1 output is obtained
for row 1 or row 3 or row 4 or row 6. In row 1, A is

Boolean algebra and logic circuits 519

Table 60.6 4 5 Figure 60.11
C · (A · B + A · B) C · (A · B + A) 4. The circuit shown in Fig. 60.12
123
ABC 0 0 C · [B · C · A + A · (B + C)],
000 0 1 see Table 60.6, col. 7
001 0 0
010 1 1 Figure 60.12
011 0 0 In Problems 5 to 7, construct switching circuits to
100 0 1 meet the requirements of the Boolean expressions
101 0 0 given.
110 1 0 5. A · C + A · B · C + A · B [See Fig. 60.13]
111

67 Figure 60.13 [See Fig. 60.14]
A · B · (B · C + B · C + A · B) C · [B · C · A + A · (B + C)] 6. A · B · C · (A + B + C)

00
00
00
01
00
00
10
01

2. The circuit shown in Fig. 60.10

C · (A · B + A);
see Table 60.6, col. 5

Figure 60.14 Section 10
7. A · (A · B · C + B · (A + C)) [See Fig. 60.15]

Figure 60.10 Figure 60.15

3. The circuit shown in Fig. 60.11

A · B · (B · C + B · C + A · B);
see Table 60.6, col. 6

520 Engineering Mathematics

In Problems 8 to 10, derive the Boolean expres- Figure 60.18
sions and construct the switching circuits for the
truth table stated. 60.2 Simplifying Boolean
8. Table 60.7, column 4 expressions

[A · B · C + A · B · C; see Fig. 60.16] A Boolean expression may be used to describe a com-
Table 60.7 plex switching circuit or logic system. If the Boolean
expression can be simplified, then the number of
1 2 3 456 switches or logic elements can be reduced resulting in
ABC a saving in cost. Three principal ways of simplifying
0 0 0 011 Boolean expressions are:
0 0 1 100
0 1 0 001 (a) by using the laws and rules of Boolean algebra (see
0 1 1 010 Section 60.3),
1 0 0 011
1 0 1 001 (b) by applying de Morgan’s laws (see Section 60.4),
1 1 0 100 and
1 1 1 000
(c) by using Karnaugh maps (see Section 60.5).
Figure 60.16
9. Table 60.7, column 5 60.3 Laws and rules of Boolean
algebra
A · B · C + A · B · C + A · B · C;
see Fig. 60.17 A summary of the principal laws and rules of Boolean
algebra are given in Table 60.8. The way in which
Figure 60.17 these laws and rules may be used to simplify Boolean
expressions is shown in Problems 5 to 10.
10. Table 60.7, column 6
A·B·C+A·B·C+A·B·C Problem 5. Simplify the Boolean expression:
+ A · B · C; see Fig. 60.18 P · Q + P · Q+P · Q

Section 10 With reference to Table 60.8: Reference
P · Q+P · Q+P · Q
5
= P · (Q + Q) + P · Q 10
=P · 1+P · Q 12
=P+P · Q

Boolean algebra and logic circuits 521

Table 60.8 Rule or law =P · Q+P · Q+P · Q 13
Ref. Name +P · Q · Q · P 14
7
1 Commutative A + B = B + A =P · Q+P · Q+P · Q+0 5
laws =P · Q+P · Q+P · Q 10
= P · (Q + Q) + P · Q 12
2 A·B=B·A =P · 1+P · Q
=P+P · Q
3 Associative laws (A + B) + C = A + (B + C)

4 (A · B) · C = A · (B · C)

5 Distributive laws A · (B + C) = A · B + A · C

6 A + (B · C) Problem 7. Simplify
F ·G·H +F ·G·H +F ·G·H
= (A + B) · (A + C)

7 Sum rules A+0=A

8 A+1=1 With reference to Table 60.8 Reference
F ·G·H+F ·G·H+F ·G·H
9 A+A=A 5
= F · G · (H + H) + F · G · H 10
10 A + A = 1 =F ·G·1+F ·G·H 12
=F ·G+F ·G·H
11 Product rules A · 0 = 0 = G · (F + F · H) 5

12 A · 1 = A

13 A · A = A

14 A · A = 0 Problem 8. Simplify
F ·G·H+F ·G·H+F ·G·H+F ·G·H
15 Absorption rules A + A · B = A

16 A · (A + B) = A

17 A + A · B = A + B With reference to Table 60.8: Reference

F ·G·H+F ·G·H+F ·G·H+F ·G·H

Problem 6. Simplify = G · H · (F + F) + G · H · (F + F) 5
(P + P · Q) · (Q + Q · P)
=G·H ·1+G·H ·1 10

With reference to Table 60.8: Reference =G · H +G · H 12
(P + P · Q) · (Q + Q · P)
5 = H · (G + G) 5 Section 10
= P · (Q + Q · P) 5
+ P · Q · (Q + Q · P) =H ·1=H 10 and 12

=P · Q+P · Q · P+P · Q · Q Problem 9. Simplify
+P · Q · Q · P A · C + A · (B + C) + A · B · (C + B)

using the rules of Boolean algebra.

522 Engineering Mathematics

With reference to Table 60.8 Reference Now try the following exercise
A · C + A · (B + C) + A · B · (C + B)
5 Exercise 204 Further problems on the laws
=A·C+A·B+A·C+A·B·C the rules of Boolean algebra
+A·B·B 14
11 Use the laws and rules of Boolean algebra given in
=A·C+A·B+A·C+A·B·C Table 60.8 to simplify the following expressions:
+A·0 5
17 1. P · Q + P · Q [P]
=A·C+A·B+A·C+A·B·C
= A · (C + B · C) + A · B + A · C 5 2. P · Q + P · Q + P · Q [P + P · Q]
= A · (C + B) + A · B + A · C 5
=A·C+A·B+A·B+A·C 10 3. F · G + F · G + G · (F + F) [G]
= A · C + B · (A + A) + A · C 12
=A·C+B·1+A·C 4. F · G + F · (G + G) + F · G [F]
=A·C+B+A·C
5. (P + P · Q) · (Q + Q · P) [P · Q]

6. F · G · H + F · G · H + F · G · H
[H · (F + F · G]

7. F · G · H + F · G · H + F · G · H
[F · G · H + G · H]

8. P · Q · R + P · Q · R + P · Q · R
[Q · R + P · Q · R]

9. F · G · H + F · G · H + F · G · H + F · G · H
[G]

Problem 10. Simplify the expression 10. F · G · H + F · G · H + F · G · H + F · G · H
P · Q · R + P · Q · (P + R) + Q · R · (Q + P) [F · H + G · H]
using the rules of Boolean algebra.
11. R · (P · Q + P · Q) + R · (P · Q + P · Q)
[P · R + P · R]

With reference to Table 60.8: Reference 12. R · (P · Q + P · Q + P · Q) + P · (Q · R + Q · R)
[P + Q · R]

P · Q · R + P · Q · (P + R) + Q · R · (Q + P)

=P · Q · R+P · Q · P+P · Q · R 60.4 De Morgan’s laws

+Q·R·Q+Q·R·P 5 De Morgan’s laws may be used to simplify not-
functions having two or more elements. The laws state
=P · Q · R+0 · Q+P · Q · R+0 · R that:

+P · Q · R 14 A + B = A · B and A · B = A + B

Section 10 =P · Q · R+P · Q · R+P · Q · R 7 and 11 and may be verified by using a truth table (see Problem
11). The application of de Morgan’s laws in simplifying
=P · Q · R+P · Q · R 9 Boolean expressions is shown in Problems 12 and 13.

= P · R · (Q + Q) 5 Problem 11. Verify that A + B = A · B

=P · R · 1 10 A Boolean expression may be verified by using a truth
table. In Table 60.9, columns 1 and 2 give all the possible
=P · R 12

Boolean algebra and logic circuits 523

arrangements of the inputs A and B. Column 3 is the or- Applying de Morgan’s law to the second term gives:
function applied to columns 1 and 2 and column 4 is the
not-function applied to column 3. Columns 5 and 6 are A + B · C = A + (B + C) = A + (B + C)
the not-function applied to columns 1 and 2 respectively
and column 7 is the and-function applied to columns 5 Thus (A · B + C) · (A + B · C)
and 6.
= (A · C + B · C) · (A + B + C)
Table 60.9 =A·A·C +A·B·C +A·C ·C

12 3 4 56 7 +A·B·C +B·B·C +B·C ·C

A B A+B A+B A B A·B But from Table 60.8, A · A = A and C · C = B · B = 0
Hence the Boolean expression becomes:
00 0 1 11 1
A·C+A·B·C+A·B·C
01 1 0 10 0 = A · C(1 + B + B)
= A · C(1 + B)
10 1 0 01 0 =A·C

11 1 0 00 0 Thus: (A · B + C) · (A + B · C) = A · C

Since columns 4 and 7 have the same pattern of 0’s and Now try the following exercise
1’s this verifies that A + B = A · B.
Exercise 205 Further problems on
Problem 12. Simplify the Boolean expression simplifying Boolean
(A · B) + (A + B) by using de Morgan’s laws and the expressions using de
rules of Boolean algebra. Morgan’s laws

Applying de Morgan’s law to the first term gives: Use de Morgan’s laws and the rules of Boolean
algebra given in Table 60.8 to simplify the follow-
A · B = A + B = A + B since A = A ing expressions.
Applying de Morgan’s law to the second term gives:
1. (A · B) · (A · B) [A · B]
A+B=A·B = A·B
Thus, (A · B) + (A + B) = (A + B) + A · B 2. (A + B · C) + (A · B + C) [A + B + C]
Removing the bracket and reordering gives:
A+A·B+B 3. (A · B + B · C) · A · B [A · B + A · B · C]
But, by rule 15, Table 60.8, A + A · B = A. It follows
that: A + A · B = A 4. (A · B + B · C) + (A · B) [1]
Thus: (A · B) + (A + B) = A + B
5. (P · Q + P · R) · (P · Q · R) [P · (Q + R)]

Problem 13. Simplify the Boolean expression 60.5 Karnaugh maps Section 10
(A · B + C) · (A + B · C) by using de Morgan’s laws
and the rules of Boolean algebra. (i) Two-variable Karnaugh maps

Applying de Morgan’s laws to the first term gives: A truth table for a two-variable expression is shown in
Table 60.10(a), the ‘1’ in the third row output show-
A · B + C = A · B · C = (A + B) · C ing that Z = A · B. Each of the four possible Boolean
= (A + B) · C = A · C + B · C expressions associated with a two-variable function can
be depicted as shown in Table 60.10(b) in which one

524 Engineering Mathematics

Table 60.10 Table 60.11 Output Boolean
Inputs Z expression
cell is allocated to each row of the truth table. A matrix ABC 0
similar to that shown in Table 60.10(b) can be used to 000 1 A·B·C
depict Z = A · B, by putting a 1 in the cell corresponding 001 0 A·B·C
to A · B and 0’s in the remaining cells. This method of 010 1 A·B·C
depicting a Boolean expression is called a two-variable 011 0 A·B·C
Karnaugh map, and is shown in Table 60.10(c). 100 0 A·B·C
101 1 A·B·C
To simplify a two-variable Boolean expression, the 110 0 A·B·C
Boolean expression is depicted on a Karnaugh map, as 111 A·B·C
outlined above. Any cells on the map having either a (a)
common vertical side or a common horizontal side are
Section 10 grouped together to form a couple. (This is a coupling To simplify a three-variable Boolean expression, the
together of cells, not just combining two together). The Boolean expression is depicted on a Karnaugh map
simplified Boolean expression for a couple is given by as outlined above. Any cells on the map having com-
those variables common to all cells in the couple. See mon edges either vertically or horizontally are grouped
Problem 14. together to form couples of four cells or two cells. Dur-
ing coupling the horizontal lines at the top and bottom
(ii) Three-variable Karnaugh maps of the cells are taken as a common edge, as are the verti-
A truth table for a three-variable expression is shown cal lines on the left and right of the cells. The simplified
in Table 60.11(a), the 1’s in the output column showing Boolean expression for a couple is given by those vari-
that: ables common to all cells in the couple. See Problems
15 to 17.
Z =A·B·C+A·B·C+A·B·C
Each of the eight possible Boolean expressions asso- (iii) Four-variable Karnaugh maps
ciated with a three-variable function can be depicted
as shown in Table 60.11(b) in which one cell is allo- A truth table for a four-variable expression is shown
cated to each row of the truth table. A matrix similar in Table 60.12(a), the 1’s in the output column
to that shown in Table 60.11(b) can be used to depict:
Z = A · B · C + A · B · C + A · B · C, by putting 1’s in the
cells corresponding to the Boolean terms on the right of
the Boolean equation and 0’s in the remaining cells. This
method of depicting a three-variable Boolean expres-
sion is called a three-variable Karnaugh map, and is
shown in Table 60.11(c).

Boolean algebra and logic circuits 525

showing that:

Z =A·B·C ·D+A·B·C ·D

+A·B·C ·D+A·B·C ·D

Each of the sixteen possible Boolean expressions
associated with a four-variable function can be depicted
as shown in Table 60.12(b), in which one cell is allo-
cated to each row of the truth table. A matrix similar to
that shown in Table 60.12(b) can be used to depict

Z =A·B·C ·D+A·B·C ·D

+A·B·C ·D+A·B·C ·D

by putting 1’s in the cells corresponding to the Boolean
terms on the right of the Boolean equation and 0’s in
the remaining cells. This method of depicting a four-
variable expression is called a four-variable Karnaugh
map, and is shown in Table 60.12(c).

Table 60.12 To simplify a four-variable Boolean expression, the
Boolean expression is depicted on a Karnaugh map
Inputs Output Boolean as outlined above. Any cells on the map having com-
mon edges either vertically or horizontally are grouped
A B C D Z expression together to form couples of eight cells, four cells or two
cells. During coupling, the horizontal lines at the top
0 0 0 0 0 A·B·C·D and bottom of the cells may be considered to be com-
mon edges, as are the vertical lines on the left and the
0 0 0 1 0 A·B·C·D right of the cells. The simplified Boolean expression for
a couple is given by those variables common to all cells
0 0 1 0 1 A·B·C·D in the couple. See Problems 18 and 19.

0 0 1 1 0 A·B·C·D Summary of procedure when simplifying a Boolean
expression using a Karnaugh map
0 1 0 0 0 A·B·C·D
(a) Draw a four, eight or sixteen-cell matrix, depend-
0 1 0 1 0 A·B·C·D ing on whether there are two, three or four
variables.
0 1 1 0 1 A·B·C·D
(b) Mark in the Boolean expression by putting 1’s in
0 1 1 1 0 A·B·C·D the appropriate cells.

1 0 0 0 0 A·B·C·D (c) Form couples of 8, 4 or 2 cells having common
edges, forming the largest groups of cells possi-
1 0 0 1 0 A·B·C·D ble. (Note that a cell containing a 1 may be used
more than once when forming a couple. Also note
1 0 1 0 1 A·B·C·D that each cell containing a 1 must be used at least
once).
1 0 1 1 0 A·B·C·D Section 10
(d) The Boolean expression for the couple is given by
1 1 0 0 0 A·B·C·D the variables which are common to all cells in the
couple.
1 1 0 1 0 A·B·C·D
Problem 14. Use Karnaugh map techniques to
1 1 1 0 1 A·B·C·D simplify the expression P · Q + P · Q

1 1 1 1 0 A·B·C·D

(a)

526 Engineering Mathematics

Using the above procedure: (d) The variables common to the couple in the top row
are Y = 1 and Z = 0, that is, Y · Z and the variables
(a) The two-variable matrix is drawn and is shown in common to the couple in the bottom row are Y = 0,
Table 60.13. Z = 1, that is, Y · Z. Hence:

Table 60.13 X ·Y ·Z+X ·Y ·Z+X ·Y ·Z
+X ·Y ·Z = Y ·Z+Y ·Z

Problem 16. Using a Karnaugh map technique to
simplify the expression (A · B) · (A + B).

(b) The term P · Q is marked with a 1 in the top Using the procedure, a two-variable matrix is drawn and
left-hand cell, corresponding to P = 0 and Q = 0; is shown in Table 60.15.
P · Q is marked with a 1 in the bottom left-hand
cell corresponding to P = 0 and Q = 1. Table 60.15

(c) The two cells containing 1’s have a common hor- A · B corresponds to the bottom left-hand cell and
izontal edge and thus a vertical couple, can be (A · B) must therefore be all cells except this one, marked
formed. with a 1 in Table 60.15. (A + B) corresponds to all the
cells except the top right-hand cell marked with a 2
(d) The variable common to both cells in the couple is in Table 60.15. Hence (A + B) must correspond to the
P = 0, i.e. P thus cell marked with a 2. The expression (A · B) · (A + B)
corresponds to the cell having both 1 and 2 in
P·Q+P·Q=P it, i.e.

Problem 15. Simplify the expression (A · B) · (A + B) = A · B
X · Y · Z + X · Y · Z + X · Y · Z + X · Y · Z by using
Karnaugh map techniques. Problem 17. Simplify (P + Q · R) + (P · Q + R)
using a Karnaugh map technique.
Using the above procedure:

(a) A three-variable matrix is drawn and is shown in
Table 60.14.

Table 60.14

Section 10 (b) The 1’s on the matrix correspond to the expression The term (P + Q · R) corresponds to the cells marked
given, i.e. for X · Y · Z, X = 0, Y = 1 and Z = 0 and 1 on the matrix in Table 60.16(a), hence (P + Q · R)
hence corresponds to the cell in the two row and corresponds to the cells marked 2. Similarly, (P · Q + R)
second column, and so on. corresponds to the cells marked 3 in Table 60.16(a),

(c) Two couples can be formed as shown. The couple hence (P · Q + R) corresponds to the cells marked 4.
in the bottom row may be formed since the vertical The expression (P + Q · R) + (P · Q + R) corresponds to
lines on the left and right of the cells are taken as cells marked with either a 2 or with a 4 and is shown in
a common edge. Table 60.16(b) by X’s. These cells may be coupled as
shown. The variables common to the group of four cells
is P = 0, i.e. P, and those common to the group of two
cells are Q = 0, R = 1, i.e. Q · R.

Thus: (P + Q · R) + (P · Q + R) = P + Q · R.

Boolean algebra and logic circuits 527

Table 60.16 Table 60.18

Problem 18. Use Karnaugh map techniques to Now try the following exercise
simplify the expression:

A·B·C·D+A·B·C·D+A·B·C·D
+ A · B · C · D + A · B · C · D.

Using the procedure, a four-variable matrix is drawn and Exercise 206 Further problems on
is shown in Table 60.17. The 1’s marked on the matrix simplifying Boolean expres-
correspond to the expression given. Two couples can be sions using Karnaugh maps
formed as shown. The four-cell couple has B = 1, C = 1,
i.e. B · C as the common variables to all four cells and In Problems 1 to 11 use Karnaugh map techniques
the two-cell couple has A · B · D as the common vari- to simplify the expressions given.
ables to both cells. Hence, the expression simplifies to:
1. X · Y + X · Y [Y ]
B · C + A · B · D i.e. B · (C + A · D)
Table 60.17 2. X · Y + X · Y + X · Y [X + Y ]

Problem 19. Simplify the expression 3. (P · Q) · (P · Q) [P · Q]
A·B·C·D+A·B·C·D+A·B·C·D+
A · B · C · D + A · B · C · D by using Karnaugh map 4. A · C + A · (B + C) + A · B · (C + B)
techniques. [B + A · C + A · C]

The Karnaugh map for the expression is shown in Table 5. P · Q · R + P · Q · R + P · Q · R [R · (P + Q]
60.18. Since the top and bottom horizontal lines are
common edges and the vertical lines on the left and 6. P · Q · R + P · Q · R + P · Q · R + P · Q · R
right of the cells are common, then the four corner cells [P · (Q + R) + P · Q · R]
form a couple, B · D, (the cells can be considered as if
they are stretched to completely cover a sphere, as far 7. A · B · C · D + A · B · C · D + A · B · C · D
as common edges are concerned). The cell A · B · C · D [A · C · (B + D)]
cannot be coupled with any other. Hence the expression
simplifies to 8. A · B · C · D + A · B · C · D + A · B · C · D
[B · C · (A + D]
B·D+A·B·C·D
9. A · B · C · D + A · B · C · D + A · B · C · D
+A·B·C·D+A·B·C·D
[D · (A + B · C)]

10. A · B · C · D + A · B · C · D + A · B · C · D Section 10
+A·B·C·D+A·B·C·D
[A · D + A · B · C · D]

11. A · B · C · D + A · B · C · D + A · B · C · D +
A·B·C·D+A·B·C·D+A·B·C·D+
A·B·C·D
[A · C + A · C · D + B · D · (A + C)]

528 Engineering Mathematics

Section 10 60.6 Logic circuits Figure 60.19
Figure 60.20
In practice, logic gates are used to perform the and,
or and not-functions introduced in Section 60.1. Logic
gates can be made from switches, magnetic devices or
fluidic devices, but most logic gates in use are electronic
devices. Various logic gates are available. For example,
the Boolean expression (A · B · C) can be produced using
a three-input, and-gate and (C + D) by using a two-
input or-gate. The principal gates in common use are
introduced below. The term ‘gate’ is used in the same
sense as a normal gate, the open state being indicated by
a binary ‘1’ and the closed state by a binary ‘0’. A gate
will only open when the requirements of the gate are
met and, for example, there will only be a ‘1’ output on
a two-input and-gate when both the inputs to the gate
are at a ‘1’ state.

The and-gate

The different symbols used for a three-input, and-gate
are shown in Fig. 60.19(a) and the truth table is shown
in Fig. 60.19(b). This shows that there will only be a ‘1’
output when A is 1, or B is 1, or C is 1, written as:

Z =A·B·C

The or-gate

The different symbols used for a three-input or-gate are
shown in Fig. 60.20(a) and the truth table is shown in
Fig. 60.20(b). This shows that there will be a ‘1’ output
when A is 1, or B is 1, or C is 1, or any combination of
A, B or C is 1, written as:

Z =A+B+C

The invert-gate or not-gate

The different symbols used for an invert-gate are
shown in Fig. 60.21(a) and the truth table is shown in
Fig. 60.21(b). This shows that a ‘0’ input gives a ‘1’
output and vice versa, i.e. it is an ‘opposite to’ function.
The invert of A is written A and is called ‘not-A’.

The nand-gate

The different symbols used for a nand-gate are shown
in Fig. 60.22(a) and the truth table is shown in
Fig. 60.22(b). This gate is equivalent to an and-gate
and an invert-gate in series (not-and = nand) and the
output is written as:

Z =A·B·C

Boolean algebra and logic circuits 529

Figure 60.21

Figure 60.23

logic gates generally have two, three or four inputs and
are confined to one function only. Thus, for example, a
two-input, or-gate or a four-input and-gate can be used
when designing a logic circuit. The way in which logic
gates are used to generate a given output is shown in
Problems 20 to 23.

Problem 20. Devise a logic system to meet the
requirements of: Z = A · B + C

Figure 60.22 Figure 60.24 Section 10
With reference to Fig. 60.24 an invert-gate, shown as
The nor-gate (1), gives B. The and-gate, shown as (2), has inputs of
A and B, giving A · B. The or-gate, shown as (3), has
The different symbols used for a nor-gate are shown inputs of A · B and C, giving:
in Fig. 60.23(a) and the truth table is shown in
Fig. 60.23(b). This gate is equivalent to an or-gate and Z=A·B+C
an invert-gate in series, (not-or = nor), and the output
is written as: Problem 21. Devise a logic system to meet the
requirements of (P + Q) · (R + S).
Z =A+B+C

Combinational logic networks

In most logic circuits, more than one gate is needed
to give the required output. Except for the invert-gate,

530 Engineering Mathematics

Figure 60.25 of the circuit being reduced. Any of the techniques can
be used, and in this case, the rules of Boolean algebra
The logic system is shown in Fig. 60.25. The given (see Table 60.8) are used.
expression shows that two invert-functions are needed
to give Q and R and these are shown as gates (1) and Z =A·B·C +A·B·C +A·B·C
(2). Two or-gates, shown as (3) and (4), give (P + Q) = A · [B · C + B · C + B · C]
and (R + S) respectively. Finally, an and-gate, shown = A · [B · C + B(C + C)] = A · [B · C + B]
as (5), gives the required output. = A · [B + B · C] = A · [B + C]

Z = (P + Q) · (R + S) The logic circuit to give this simplified expression is
shown in Fig. 60.26.
Problem 22. Devise a logic circuit to meet the
requirements of the output given in Table 60.19, Figure 60.26
using as few gates as possible. Problem 23. Simplify the expression:

Table 60.19 Z =P ·Q·R·S +P ·Q·R·S +P ·Q·R·S
+P · Q · R · S + P · Q · R · S
Inputs Output
and devise a logic circuit to give this output.
AB C Z
The given expression is simplified using the Karnaugh
00 0 0 map techniques introduced in Section 60.5. Two couples
are formed as shown in Fig. 60.27(a) and the simplified
00 1 0 expression becomes:

01 0 0 Z =Q·R·S +P ·R
i.e. Z = R · (P + Q · S)
01 1 0 The logic circuit to produce this expression is shown in
Fig. 60.27(b).

10 0 0

10 1 1

11 0 1

11 1 1

Section 10 The ‘1’ outputs in rows 6, 7 and 8 of Table 60.19 show
that the Boolean expression is:
Figure 60.27
Z =A·B·C+A·B·C+A·B·C

The logic circuit for this expression can be built using
three, 3-input and-gates and one, 3-input or-gate,
together with two invert-gates. However, the number of
gates required can be reduced by using the techniques
introduced in Sections 60.3 to 60.5, resulting in the cost

Now try the following exercise Boolean algebra and logic circuits 531

Exercise 207 Further problems on logic Table 60.20
circuits 123 4 5 6
A B C Z1 Z2 Z3
In Problems 1 to 4, devise logic systems to meet 000 0 0 0
the requirements of the Boolean expressions given. 001 1 0 0
010 0 0 1
1. Z = A + B · C [See Fig. 60.28(a)] 011 1 1 1
100 0 1 0
2. Z = A · B + B · C [See Fig. 60.28(b)] 101 1 1 1
110 1 0 1
3. Z = A · B · C + A · B · C [See Fig. 60.28(c)] 111 1 1 1

4. Z = (A + B) · (C + D) [See Fig. 60.28(d)]

Figure 60.28 Figure 60.29 Section 10

In Problems 5 to 7, simplify the expression given 7. Column 6 of Table 60.20
in the truth table and devise a logic circuit to meet [Z3 = A · C + B, see Fig. 60.29(c)]
the requirements stated.
In Problems 8 to 12, simplify the Boolean expres-
5. Column 4 of Table 60.20 sions given and devise logic circuits to give the
[Z1 = A · B + C, see Fig. 60.29(a)] requirements of the simplified expressions.

6. Column 5 of Table 60.20 8. P · Q + P · Q + P · Q
[Z2 = A · B + B · C, see Fig. 60.29(b)] [P + Q, see Fig. 60.30(a)]

9. P · Q · R + P · Q · R + P · Q · R
[R · (P + Q), see Fig. 60.30(b)]

10. P · Q · R + P · Q · R + P · Q · R
[Q · (P + R), see Fig. 60.30(c)]

532 Engineering Mathematics a universal gate. The way in which a universal nand-
gate is used to produce the invert, and, or and nor-
Figure 60.30 functions is shown in Problem 24. The way in which
11. A · B · C · D + A · B · C · D + A · B · C · D a universal nor-gate is used to produce the invert, or,
and and nand-function is shown in Problem 25.
+A·B·C·D+A·B·C·D
[D · (A · C + B), see Fig. 60.31(a)] Problem 24. Show low invert, and, or and nor-
functions can be produced using nand-gates only.

A single input to a nand-gate gives the invert-function,
as shown in Fig. 60.32(a). When two nand-gates
are connected, as shown in Fig. 60.32(b), the out-
put from the first gate is A · B · C and this is inverted
by the second gate, giving Z = A · B · C = A · B · C i.e.
the and-function is produced. When A, B and C
are the inputs to a nand-gate, the output is A · B · C.

By de Morgan’s law, A · B · C = A + B + C =
A + B + C, i.e. a nand-gate is used to produce or-
function. The logic circuit shown in Fig. 60.32(c). If the
output from the logic circuit in Fig. 60.32(c) is inverted
by adding an additional nand-gate, the output becomes
the invert of an or-function, i.e. the nor-function, as
shown in Fig. 60.32(d).

Section 10 Figure 60.31

12. (P · Q · R) · (P + Q · R)
[P · (Q + R), see Fig. 60.31(b)]

60.7 Universal logic gates Figure 60.32

The function of any of the five logic gates in common use
can be obtained by using either nand-gates or nor-gates
and when used in this manner, the gate selected in called

Boolean algebra and logic circuits 533

Problem 25. Show how invert, or, and and nand- expression
functions can be produced by using nor-gates only.
Z =A+B+C+D
A single input to a nor-gate gives the invert-function,
as shown in Fig. 60.33(a). When two nor-gates are con- When designing logic circuits, it is often easier to start
nected, as shown in Fig. 60.33(b), the output from the at the output of the circuit. The given expression shows
first gate is A + B + C and this is inverted by the sec- there are four variables joined by or-functions. From
ond gate, giving Z = A + B + C = A + B + C, i.e. the the principles introduced in Problem 24, if a four-input
or-function is produced. Inputs of A, B, and C to a nand-gate is used to give the expression given, the input
nor-gate give an output of A + B + C.
are A, B, C and D that is A, B, C and D. However, the
By de Morgan’s law, A + B + C = A · B · C = A · B · C, problem states that three-inputs are not to be exceeded
i.e. the nor-gate can be used to produce the and- so two of the variables are joined, i.e. the inputs to the
function. The logic circuit is shown in Fig. 60.33(c). three-input nand-gate, shown as gate (1) Fig. 60.34, is
When the output of the logic circuit, shown in A, B, C, and D. From Problem 24, the and-function is
Fig. 60.33(c), is inverted by adding an additional generated by using two nand-gates connected in series,
nor-gate, the output then becomes the invert of an or- as shown by gates (2) and (3) in Fig. 60.34. The logic
function, i.e. the nor-function as shown in Fig. 60.33(d). circuit required to produce the given expression is as
shown in Fig. 60.34.

Figure 60.33 Figure 60.34 Section 10

Problem 26. Design a logic circuit, using Problem 27. Use nor-gates only to design a logic
nand-gates having not more than three inputs, circuit to meet the requirements of the expressions:
to meet the requirements of the Boolean
Z = D · (A + B + C)

It is usual in logic circuit design to start the design at the
output. From Problem 25, the and-function between D
and the terms in the bracket can be produced by using
inputs of D and A + B + C to a nor-gate, i.e. by de
Morgan’s law, inputs of D and A · B · C. Again, with
reference to Problem 25, inputs of A · B and C to a nor-
gate give an output of A + B + C, which by de Morgan’s
law is A · B · C. The logic circuit to produce the required
expression is as shown in Fig. 60.35

Figure 60.35