Engineering Mathematics 4 Episode 13 pps

[1, 7 and 3, 23, 1, 10.20] 57.7 Theorem of Pappus A theorem of Pappus states: ‘If a plane area is rotated about an axis in its own plane but not intersecting it, the volume of the solid

5 0



5x2

2 

x33

5 0

D

625

3 

6254125

2 

1253D

625121256D

5

0125

D2.5

Figure 57.6

Hence the centroid of the area lies at (2.5, 2.5)

(Note from Fig 57.6 that the curve is symmetrical

about x D 2.5 and thus x could have been

deter-mined ‘on sight’)

Problem 5 Locate the centroid of the areaenclosed by the curve y D 2x2, the y-axisand ordinates y D 1 and y D 4, correct to 3decimal places

From Section 57.4,

x D

12

 4

1

x2d y

 4 1

x d y

D

12

y

2d y

D

12



y24

4 1



2y3/2

3p2

4 1D

15814

3p2

D

 4 1

y3/2p

2 d y14

3p2

D

1p2





y

5/252

3p2

D

2

5p2 3114

The value of y is given by the height of the typical

strip shown in Fig 55.7, i.e y Dp8x  x2 Hence,

x44

p

2552

4

223

Care needs to be taken when finding y in such

examples as this From Fig 57.7, y Dp8x  x2and

8x  x2) The perpendicular distance from

centroid C of the strip to Ox is 1

2

p8x  x2 C x2.Taking moments about Ox gives:

(total area) y DxD2

xD0 (area of strip) lar distance of centroid of strip to Ox)

(perpendicu-Hence (area) yD

 p8x  x2

 1

2

p8x  x2 C x2

x22

d x

D

 2 0

8x

2 

x42

2 0D

8  315

223

D1.8

Thus the position of the centroid of the shaded area in Fig 55.7 is at (0.9, 1.8)

Now try the following exercise

Exercise 191 Further problems on

cen-troids of simple shapes

1 Determine the position of the centroid of

a sheet of metal formed by the curve

y D4x  x2 which lies above the x-axis

[(2, 1.6)]

2 Find the coordinates of the centroid of thearea that lies between the curve y

x Dx 2and the x-axis [(1, 0.4)]

3 Determine the coordinates of the centroid

of the area formed between the curve

y D9  x2 and the x-axis [(0, 3.6)]

4 Determine the centroid of the area lyingbetween y D 4x2, the y-axis and theordinates y D 0 and y D 4

[(0.375, 2.40]

5 Find the position of the centroid of thearea enclosed by the curve y Dp5x, thex-axis and the ordinate x D 5

[(3.0, 1.875)]

6 Sketch the curve y2 D 9x between thelimits x D 0 and x D 4 Determine theposition of the centroid of this area

[(2.4, 0)]

7 Calculate the points of intersection of the

curves x2 D 4y and y

2

4 D x, and mine the position of the centroid of the

deter-area enclosed by them

[(0, 0) and (4, 4), (1.8, 1.8)]

8 Determine the position of the centroid of

the sector of a circle of radius 3 cm whose

angle subtended at the centre is 40°

On the centre line, 1.96 cm

from the centre



9 Sketch the curves y D 2x2C5 and

y  8 D x x C 2 on the same axes

and determine their points of intersection

Calculate the coordinates of the centroid

of the area enclosed by the two curves

[(1, 7) and (3, 23), (1, 10.20)]

57.7 Theorem of Pappus

A theorem of Pappus states:

‘If a plane area is rotated about an axis in its own

plane but not intersecting it, the volume of the solid

formed is given by the product of the area and the

distance moved by the centroid of the area’

With reference to Fig 57.8, when the curve

y D f xis rotated one revolution about the x-axis

between the limits x D a and x D b, the volume V

generated is given by:

volume V D A 2y, from which,

Problem 7 Determine the position of the

centroid of a semicircle of radius r by using

the theorem of Pappus Check the answer byusing integration (given that the equation of

a circle, centre 0, radius r is x2Cy2Dr2)

A semicircle is shown in Fig 57.9 with its diameterlying on the x-axis and its centre at the origin Area

of semicircle D r

2

2 When the area is rotated aboutthe x-axis one revolution a sphere is generated ofvolume 4

3r

3

y

y C

0

x 2 +y 2 = r 2

Figure 57.9

Let centroid C be at a distance y from the origin

as shown in Fig 57.9 From the theorem of Pappus,volume generated D area ð distance moved through

 r

r

r2x2 d x

r22

D

12



r2x x33

r

r

r22

D

12



r3r33



r3Cr33



r22

D 4r3

Hence the centroid of a semicircle lies on the axis

of symmetry, distance 4r

3p (or 0.424 r) from its

diameter.

Problem 8 Calculate the area bounded by

the curve y D 2x2, the x-axis and ordinates

x D0 and x D 3 (b) If this area is revolved

(i) about the x-axis and (ii) about the y-axis,

find the volumes of the solids produced

(c) Locate the position of the centroid using

(i) integration, and (ii) the theorem of Pappus



2x33

3 0

D18 square units

(b) (i) When the shaded area of Fig 57.10 is

revolved 360° about the x-axis, the volume

3 0

D4

2435

D194.4p cubic units

(ii) When the shaded area of Fig 57.10 is

revolved 360° about the y-axis, the volume

generated D (volume generated by x D 3 

18 0

D81pcubic units

(c) If the co-ordinates of the centroid of the shaded

area in Fig 57.10 are (x, y) then:



2x44

3 0

 3

0 2x22d x

18

D

12

 3

04x4d x

12



4x55

3

0

18 D5.4(ii) using the theorem of Pappus:

Volume generated when shaded area isrevolved about Oy D area 2x)

i.e 81 D 18 2x,from which, x D 81

36 D2.25Volume generated when shaded area isrevolved about Ox D area 2y

i.e 194.4 D 18 2y,from which, y D 194.4

36 D5.4

Hence the centroid of the shaded area in Fig 55.10 is at (2.25, 5.4)

Problem 9 A cylindrical pillar of diameter

400 mm has a groove cut round itscircumference The section of the groove is asemicircle of diameter 50 mm Determine thevolume of material removed, in cubiccentimetres, correct to 4 significant figures

A part of the pillar showing the groove is shown inFig 57.11

The distance of the centroid of the semicircle fromits base is 4r

3 see Problem 7 D

4 253 D

1003 mm.The distance of the centroid from the centre of thepillar D

200 1003

mm

From the theorem of Pappus,

volume D area ð distance moved by centroid

Hence the volume of material removed is

1168 cm3 correct to 4 significant figures.

Problem 10 A metal disc has a radius of

5.0 cm and is of thickness 2.0 cm A

semicircular groove of diameter 2.0 cm is

machined centrally around the rim to form a

pulley Determine, using Pappus’ theorem,

the volume and mass of metal removed and

the volume and mass of the pulley if the

density of the metal is 8000 kg m3

A side view of the rim of the disc is shown in

Fig 57.12

When area PQRS is rotated about axis XX the

volume generated is that of the pulley The centroid

of the semicircular area removed is at a distance of

4r

3 from its diameter (see Problem 7), i.e.

4 1.03 ,i.e 0.424 cm from PQ Thus the distance of the

centroid from XX is (5.0  0.424), i.e 4.576 cm

The distance moved through in one revolution by

Figure 57.12

By the theorem of Pappus, volume generated

Darea ð distance moved by centroidD

Volume of pulley D volume of cylindrical disc

volume of metal removed

D 5.02 2.0  45.16 D 111.9 cm3Mass of pulley D density ð volume

D8000 kg m3ð 111.9

106 m3

D0.8952 kg or 895.2 g Now try the following exercise

Exercise 192 Further problems on the

theorem of Pappus

1 A right angled isosceles triangle having

a hypotenuse of 8 cm is revolved onerevolution about one of its equal sides asaxis Determine the volume of the solidgenerated using Pappus’ theorem

[189.6 cm3]

2 A rectangle measuring 10.0 cm by 6.0 cmrotates one revolution about one of itslongest sides as axis Determine the

volume of the resulting cylinder by using

the theorem of Pappus [1131 cm2]

3 Using (a) the theorem of Pappus, and

(b) integration, determine the position of

the centroid of a metal template in the

form of a quadrant of a circle of radius

4 cm (The equation of a circle, centre 0,

radius r is x2Cy2 Dr2)

On the centre line, distance 2.40 cm

from the centre, i.e at coordinates

(1.70, 1.70)

4 (a) Determine the area bounded by the

curve y D 5x2, the x-axis and the

ordinates x D 0 and x D 3

(b) If this area is revolved 360° about

(i) the x-axis, and (ii) the y-axis, find

the volumes of the solids of revolution

produced in each case

(c) Determine the co-ordinates of the troid of the area using (i) integralcalculus, and (ii) the theorem ofPappus

cen-

(a) 45 square units (b) (i) 1215cubic units (ii) 202.5 cubic units(c) (2.25, 13.5)

5 A metal disc has a radius of 7.0 cm and

is of thickness 2.5 cm A semicirculargroove of diameter 2.0 cm is machinedcentrally around the rim to form a pulley.Determine the volume of metal removedusing Pappus’ theorem and express this

as a percentage of the original volume

of the disc Find also the mass of metalremoved if the density of the metal is

7800 kg m3

[64.90 cm3, 16.86%, 506.2 g]

Second moments of area

58.1 Second moments of area and

radius of gyration

The first moment of area about a fixed axis of a

lamina of area A, perpendicular distance y from the

centroid of the lamina is defined as Ay cubic units

The second moment of area of the same lamina

as above is given by Ay2, i.e the perpendicular

distance from the centroid of the area to the fixed

axis is squared Second moments of areas are usually

denoted by I and have units of mm4, cm4, and so on

Radius of gyration

Several areas, a1, a2, a3, at distances y1, y2,

y3, from a fixed axis, may be replaced by a single

area A, where A D a1Ca2Ca3C Ð Ð Ðat distance k

from the axis, such that Ak2 D

The second moment of area is a quantity much

used in the theory of bending of beams, in the

torsion of shafts, and in calculations involving water

planes and centres of pressure

58.2 Second moment of area of regular

sections

The procedure to determine the second moment of

area of regular sections about a given axis is (i) to

find the second moment of area of a typical element

and (ii) to sum all such second moments of area by

integrating between appropriate limits

For example, the second moment of area of the

rectangle shown in Fig 58.1 about axis PP is found

by initially considering an elemental strip of width

υx, parallel to and distance x from axis PP Area of

shaded strip D bυx Second moment of area of the

shaded strip about PP D x2

The second moment of area of the whole rectangle

about PP is obtained by summing all such strips

Figure 58.1

between x D 0 and x D l, i.e
xDl

xD0x2bυx It is afundamental theorem of integration that

limitυx!0

l 0

58.3 Parallel axis theorem

In Fig 58.2, axis GG passes through the centroid C

of area A Axes DD and GG are in the same plane,are parallel to each other and distance d apart Theparallel axis theorem states:

I DD=I GGYAd2

Using the parallel axis theorem the second moment

of area of a rectangle about an axis through the

58.4 Perpendicular axis theorem

In Fig 58.4, axes OX, OY and OZ are mutually

perpendicular If OX and OY lie in the plane of

area A then the perpendicular axis theorem states:

I OZ =I OX YI OY

58.5 Summary of derived results

A summary of derive standard results for the second

moment of area and radius of gyration of regular

sections are listed in Table 58.1

l p 3 length l

(2) Coinciding with l lb

3 3

b p 3 breadth b

(3) Through centroid, parallel to b

bl 3 l2

l p 12 (4) Through centroid,

parallel to l

lb312

b p 12

Triangle (1) Coinciding with b bh

3 12

h p 6 Perpendicular

(2) Through centroid, parallel to base

bh336

h p 18 height h

(3) Through vertex, parallel to base

bh 3 4

h p 2 base b

Circle (1) Through centre,

perpendicular

to plane (i.e.

polar axis)

r 4 2

r p 2 radius r

(2) Coinciding with diameter

r44

r 2 (3) About a tangent 5r

4 4

p 5

2 r

Semicircle Coinciding

with diameter

r48

r 2 radius r

58.6 Worked problems on second moments of area of regular sections

Problem 1 Determine the second moment

of area and the radius of gyration about axes

AA, BB and CC for the rectangle shown inFig 58.5

A B

3 D2.31 cmThe second moment of area about the centroid of a

rectangle is bl

3

12 when the axis through the centroid

is parallel with the breadth b In this case, the axis

CCis parallel with the length l

12 D1.15 cm

Problem 2 Find the second moment of

area and the radius of gyration about axis PP

for the rectangle shown in Fig 58.6

40.0 mm

15.0 mm G

D645 000 mm 4

IPPDAkPP2from which, k PP D



IPPareaD



645 000

600 D32.79 mm

Problem 3 Determine the second moment

of area and radius of gyration about axis QQ

of the triangle BCD shown in Fig 58.7

B

G G

i.e bh3

36 D

36 D384 cm

4, A is the area ofthe triangle D 12bh D 12 2 and d

is the distance between axes GG and

QQ D6.0 C13 Hence the second moment of area about axisQQ,



5184

48 D10.4 cm

Problem 4 Determine the second moment

of area and radius of gyration of the circle

shown in Fig 58.8 about axis YY

Problem 5 Determine the second moment

of area and radius of gyration for the

semicircle shown in Fig 58.9 about axis XX

15.0 mm 10.0 mm

Figure 58.9

The centroid of a semicircle lies at 4r

3 from itsdiameter

Using the parallel axis theorem: IBB DIGGCAd2,where IBBD r4

IXXDIGGCA 2

58 179 D 59 276 mm4 or 59 280 mm4, correct to 4significant figures

Radius of gyration, k XX D



IXXareaD



59 276157.1

D19.42 mm

Problem 6 Determine the polar secondmoment of area of the propeller shaftcross-section shown in Fig 58.10

is given by the polar second moment of area ofthe 7.0 cm diameter circle minus the polar secondmoment of area of the 6.0 cm diameter circle.Hence the polar second moment of area of the

cross-section shown D 

2



7.02

4

2



6.02

4

D235.7  127.2 D 108.5 cm 4

Problem 7 Determine the second moment

of area and radius of gyration of a

rectangular lamina of length 40 mm and

width 15 mm about an axis through one

corner, perpendicular to the plane of the

lamina

The lamina is shown in Fig 58.11

Figure 58.11

From the perpendicular axis theorem:

IZZDIXXCIYY

D24.7 mm or 2.47 cm

Now try the following exercise

Exercise 193 Further problems on second

moments of area of regular sections

1 Determine the second moment of area and

radius of gyration for the rectangle shown

in Fig 58.12 about (a) axis AA (b) axis

in Fig 58.13 about (a) axis DD (b) axis

EE, and (c) an axis through the centroid

of the triangle parallel to axis DD

5 For each of the areas shown in Fig 58.16

determine the second moment of area and

radius of gyration about axis LL, by using

the parallel axis theorem

6 Calculate the radius of gyration of a

rect-angular door 2.0 m high by 1.5 m wide

about a vertical axis through its hinge

[0.866 m]

7 A circular door of a boiler is hinged

so that it turns about a tangent If its

diameter is 1.0 m, determine its second

moment of area and radius of gyration

about the hinge

[0.245 m4, 0.559 m]

8 A circular cover, centre 0, has a radius

of 12.0 cm A hole of radius 4.0 cm and

centre X, where OX D 6.0 cm, is cut in

the cover Determine the second moment

of area and the radius of gyration of

the remainder about a diameter through

0 perpendicular to OX

[14 280 cm4, 5.96 cm]

58.7 Worked problems on second

moments of areas of composite

areas

Problem 8 Determine correct to 3

significant figures, the second moment of

area about axis XX for the composite area

2.0 cm

CT

4.0 cm

1.0 cm 2.0 cm

Figure 58.17

For the semicircle, IXX D r4

8 D

 48

D100.5 cm4For the rectangle, IXX D bl3

3 D

3

D1024 cm4For the triangle, about axis TT through centroid CT,ITT Dbh3

D 100.5 C 1024 C 3060 D 4184.5 D 4180 cm 4,correct to 3 significant figures

Problem 9 Determine the second moment

of area and the radius of gyration about axis

XXfor the I-section shown in Fig 58.18

Figure 58.18

The I-section is divided into three rectangles, D, E

and F and their centroids denoted by CD, CE and

Now try the following exercise

Exercise 194 Further problems on second

moment of areas of ite areas

compos-1 For the sections shown in Fig 58.19, find

the second moment of area and the radius

of gyration about axis XX



Figure 58.20

3 Find the second moment of area andradius of gyration about the axis XX forthe beam section shown in Fig 58.21

[1351 cm4, 5.67 cm]

Figure 58.21

Assignment 15

This assignment covers the material in

Chapters 54 to 58 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 distance x metres from a fixed point is

given by: F D 2x C 3x2 If work done D

x2

x 1 F d x, determine the work done when

the body moves from the position when

x D1 m to that when x D 4 m (4)

2 Sketch and determine the area enclosed

by the curve y D 3 sin#

2, the #-axis andordinates # D 0 and # D 2

3 Calculate the area between the curve

y D x3x26x and the x-axis (10)

4 A voltagevD25 sin 50t volts is applied

across an electrical circuit Determine,

using integration, its mean and r.m.s

val-ues over the range t D 0 to

t D 20 ms, each correct to 4 significant

5 Sketch on the same axes the curves

x2D2y and y2D16x and determine the

co-ordinates of the points of intersection

Determine (a) the area enclosed by the

curves, and (b) the volume of the solid

produced if the area is rotated one

revo-lution about the x-axis (13)

6 Calculate the position of the centroid of

the sheet of metal formed by the x-axis

and the part of the curve y D 5x  x2

which lies above the x-axis (9)

7 A cylindrical pillar of diameter 500 mm

has a groove cut around its circumference

3, use the theorem

of Pappus to determine the volume ofmaterial removed, in cm3, correct to 3

8 For each of the areas shown in Fig A15.2determine the second moment of area andradius of gyration about axis XX (15)

of gyration about the hinge (5)

Part 10 Further Number and

Algebra

59

Boolean algebra and logic circuits

59.1 Boolean algebra and switching

circuits

A two-state device is one whose basic elements can

only have one of two conditions Thus, two-way

switches, which can either be on or off, and the

binary numbering system, having the digits 0 and

1 only, are two-state devices In Boolean algebra,

if A represents one state, then A, called ‘not-A’,

represents the second state

The or-function

In Boolean algebra, the or-function for two elements

Aand B is written as A C 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 59.1(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 59.1(b), all the possible switch combinations

are shown in columns 1 and 2, in which a 0

repre-sents a switch being off and a 1 reprerepre-sents the switch

being on, these columns being called the inputs

Col-umn 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

ele-ments A and B is written as A Ð 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 59.2(a)the lamp will be on only when both A and B are

on The truth table for a two-input and-function is

shown in Fig 59.2(b)

Figure 59.1

Figure 59.2

The not-function

In Boolean algebra, the not-function for element A

is 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 truth table for the not-function

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 ACBCC.

Similarly, a three-input and-function is written as

A Ð B Ð C The equivalent electrical circuits and

truth tables for three-input or and and-functions are

shown in Figs 59.3(a) and (b) respectively

Figure 59.3

To achieve a given output, it is often necessary

to use combinations of switches connected both in

Figure 59.4

series and in parallel If the output from a switchingcircuit is given by the Boolean expression Z DAÐBCAÐB, the truth table is as shown in Fig 59.4(a)

In this table, columns 1 and 2 give all the possiblecombinations of A and B Column 3 corresponds

to A Ð B and column 4 to A Ð B, i.e a 1 output isobtained when A D 0 and when B D 0 Column 5 is

the or-function applied to columns 3 and 4 giving

an output of Z D A Ð B C A Ð B The correspondingswitching circuit is shown in Fig 59.4(b) in which

Aand B are connected in series to give A Ð B, A and

Bare connected in series to give A Ð B, and A Ð B and

A Ð Bare connected in parallel to give A Ð B C A Ð B.The circuit symbols used are such that A means theswitch is on when A is 1, A means the switch is onwhen A is 0, and so on

Problem 1 Derive the Boolean expressionand construct a truth table for the switchingcircuit shown in Fig 59.5

Figure 59.5

The switches between 1 and 2 in Fig 59.5 are inseries and have a Boolean expression of B Ð A Theparallel circuit 1 to 2 and 3 to 4 have a Boolean

expression of B Ð A C B The parallel circuit can

be treated as a single switching unit, giving the

equivalent of switches 5 to 6, 6 to 7 and 7 to 8

in series Thus the output is given by:

Z =A · B · AYB / · B

The truth table is as shown in Table 59.2 Columns 1

and 2 give all the possible combinations of switches

A and B Column 3 is the and-function applied to

columns 1 and 2, giving BÐA Column 4 is B, i.e., the

opposite to column 2 Column 5 is the or-function

applied to columns 3 and 4 Column 6 is A, i.e the

opposite to column 1 The output is column 7 and is

obtained by applying the and-function to columns

Problem 2 Derive the Boolean expression

and construct a truth table for the switching

circuit shown in Fig 59.6

Figure 59.6

The parallel circuit 1 to 2 and 3 to 4 gives A C B

and this is equivalent to a single switching unit

between 7 and 2 The parallel circuit 5 to 6 and

7 to 2 gives C C A C B and this is

equiva-lent to a single switching unit between 8 and 2

The series circuit 9 to 8 and 8 to 2 gives the

output

Z =B · [C Y AYB /]

The truth table is shown in Table 59.3 Columns

1, 2 and 3 give all the possible combinations of

A, B and C Column 4 is B and is the

oppo-site to column 2 Column 5 is the or-function

applied to columns 1 and 4, giving A C B

Col-umn 6 is the or-function applied to colCol-umns 3

and 5 giving C C A C B The output is given

in column 7 and is obtained by applying the

and-function to columns 2 and 6, giving Z D B Ð [C C

The three terms joined by or-functions, C, indicate

three parallel branches,having: branch 1 A and Cin series

branch 2 A and Bin seriesand branch 3 A and B and Cin series

Figure 59.7

Hence the required switching circuit is as shown in

Fig 59.7 The corresponding truth table is shown in

Column 4 is C, i.e the opposite to column 3

Column 5 is A Ð C, obtained by applying the

and-function to columns 1 and 4

Column 6 is A, the opposite to column 1

Column 7 is A Ð B, obtained by applying the

and-function to columns 2 and 6

Column 8 is A Ð B Ð 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

Problem 4 Derive the Boolean expression

and construct the switching circuit for the

truth table given in Table 59.5

Examination of the truth table shown in Table 59.5

shows that there is a 1 output in the Z-column in

rows 1, 3, 4 and 6 Thus, the Boolean expression

and switching 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 0 and B is 0 and C is 0 and this

corresponds to the Boolean expression A Ð B Ð C In

row 3, A is 0 and B is 1 and C is 0, i.e the Boolean

expression in A Ð B Ð C Similarly in rows 4 and 6,the Boolean expressions are A Ð B Ð C and A Ð B Ð Crespectively Hence the Boolean expression is:

Z=A · B · C YA · B · C

YA · B · CYA · B · C

The corresponding switching circuit is shown in

Fig 59.8 The four terms are joined by or-functions,

C, and are represented by four parallel circuits

Each term has three elements joined by an

and-function, and is represented by three elements nected in series

con-Figure 59.8

Now try the following exercise

Exercise 195 Further problems on

Boo-lean algebra and switching circuits

In Problems 1 to 4, determine the Booleanexpressions and construct truth tables for theswitching circuits given

1 The circuit shown in Fig 59.9

C Ð A Ð B C A Ð B;

see Table 59.6, col 4



Figure 59.9

8 Table 59.7, column 4[A Ð B Ð C C A Ð B Ð C; see Fig 59.16]

59.2 Simplifying Boolean expressions

A Boolean expression may be used to describe

a complex switching circuit or logic system If

the Boolean expression can be simplified, then the

number of switches or logic elements can be reduced

resulting in a saving in cost Three principal ways

of simplifying Boolean expressions are:

(a) by using the laws and rules of Boolean algebra(see Section 59.3),

(b) by applying de Morgan’s laws (see Section 59.4),and

(c) by using Karnaugh maps (see Section 59.5)

59.3 Laws and rules of Boolean algebra

A summary of the principal laws and rules ofBoolean algebra are given in Table 59.8 The way inwhich these laws and rules may be used to simplifyBoolean expressions is shown in Problems 5 to 10

4r

3 from its diameter (see Problem 7), i.e.

4 1.03 ,i.e 0 .42 4 cm from PQ Thus the distance of the

centroid from XX is (5.0  0 .42 4), i.e 4. 576... about Ox D area 2y

i.e 1 94. 4 D 18 2y,from which, y D 1 94. 4

36 D5.4< /sup>

Hence the centroid... through centroid CT,ITT Dbh3

D 100.5 C 10 24 C 3060 D 41 84. 5 D 41 80 cm 4< /small>,correct to significant figures

Problem Determine the second