pertemuan 2

Chapter 38

Some applications ofintegration

38.1 Introduction

There are a number of applications of integral calculusin engineering. The determination of areas, mean andr.m.s. values, volumes, centroids and second momentsof area and radius of gyration are included in thischapter.

38.2 Areas under and between curves

In Fig. 38.1,

total shaded area =∫ b

af (x)dx−

∫ c

bf (x)dx

+∫ d

cf (x)dx

E

0 F

G

y

a b c d x

y 5 f (x)

Figure 38.1

Problem 1. Determine the area between the curvey = x3 − 2x2 − 8x and the x-axis.

y = x3−2x2 − 8x = x(x2−2x − 8) = x(x + 2)(x − 4)

When y = 0, x = 0 or (x + 2) = 0 or (x − 4) = 0, i.e.when y = 0, x = 0 or −2 or 4, which means that thecurve crosses the x-axis at 0, −2, and 4. Since the curveis a continuous function, only one other co-ordinatevalue needs to be calculated before a sketch of thecurve can be produced. When x = 1, y = −9, show-ing that the part of the curve between x = 0 and x = 4is negative. A sketch of y = x3 − 2x2 − 8x is shown inFig. 38.2. (Another method of sketching Fig. 38.2 wouldhave been to draw up a table of values.)

y

x

210

10

220

22 4

y 5 x32 2x22 8x

221 0 1 3

Figure 38.2

Shaded area

=∫ 0

−2(x3 − 2x2 − 8x)dx −

∫ 4

0(x3 − 2x2 − 8x)dx

=[

x4

4− 2x3

3− 8x2

2

]0

−2−

[x4

4− 2x3

3− 8x2

2

]4

0

=(

62

3

)−

(−42

2

3

)=49

13

square units

376 Higher Engineering Mathematics

Problem 2. Determine the area enclosed betweenthe curves y = x2 + 1 and y = 7 − x .

At the points of intersection the curves are equal. Thus,equating the y values of each curve gives:

x2 + 1 = 7 − x

from which, x2 + x − 6 = 0

Factorizing gives (x − 2)(x + 3)= 0

from which x = 2 and x = −3By firstly determining the points of intersection therange of x-values has been found. Tables of values areproduced as shown below.

x −3 −2 −1 0 1 2

y = x2 + 1 10 5 2 1 2 5

x −3 0 2

y = 7 − x 10 7 5

A sketch of the two curves is shown in Fig. 38.3.

y

10

21 0 1 2 x2223

5y 5 7 2 x

y 5 x 211

Figure 38.3

Shaded area =∫ 2

−3(7 − x)dx −

∫ 2

−3(x2 + 1)dx

=∫ 2

−3[(7 − x)− (x2 + 1)]dx

=∫ 2

−3(6 − x − x2)dx

=[

6x − x2

2− x3

3

]2

−3

=(

12 − 2 − 8

3

)−

(−18 − 9

2+ 9

)

=(

71

3

)−

(−13

1

2

)

= 2056

square units

Problem 3. Determine by integration the areabounded by the three straight lines y = 4 − x ,y = 3x and 3y = x .

Each of the straight lines are shown sketched inFig. 38.4.

1 2 3 x4

y 5 3xy 5 4 2 x

y

4

2

0

3y 5 x (or y 5 x )3

Figure 38.4

Shaded area

=∫ 1

0

(3x − x

3

)dx +

∫ 3

1

[(4 − x)− x

3

]dx

=[

3x2

2− x2

6

]1

0+

[4x − x2

2− x2

6

]3

1

=[(

3

2− 1

6

)− (0)

]+

[(12 − 9

2− 9

6

)

−(

4 − 1

2− 1

6

)]

=(

11

3

)+

(6 − 3

1

3

)= 4 square units

Now try the following exercise

Exercise 147 Further problems on areasunder and between curves

1. Find the area enclosed by the curvey = 4 cos3x , the x-axis and ordinates x = 0

and x = π

6[1 1

3 square units]

Some applications of integration 377

2. Sketch the curves y = x2 + 3 and y = 7 − 3xand determine the area enclosed by them.

[20 56 square units]

3. Determine the area enclosed by the threestraight lines y =3x , 2y = x and y +2x =5.

[2 12 square units]

38.3 Mean and r.m.s. values

With reference to Fig. 38.5,

mean value, y = 1b−a

∫ b

aydx

and r.m.s. value =√√√√{

1b−a

∫ b

ay2 dx

}

0 x 5 a x 5 b

y

x

y

y 5 f(x)

Figure 38.5

Problem 4. A sinusoidal voltage v= 100 sinωtvolts. Use integration to determine over half a cycle(a) the mean value, and (b) the r.m.s. value.

(a) Half a cycle means the limits are 0 to π radians.

Mean value, y = 1

π − 0

∫ π

0vd(ωt)

= 1

π

∫ π

0100 sinωt d(ωt)

= 100

π[−cosωt ]π0

= 100

π[(−cosπ)− (−cos 0)]

= 100

π[(+1)− (−1)] = 200

π

= 63.66 volts

[Note that for a sine wave,

mean value= 2π

×maximum value

In this case, mean value= 2

π× 100 = 63.66 V]

(b) r.m.s. value

=√{

1

π − 0

∫ π

0v2 d(ωt)

}

=√{

1

π

∫ π

0(100 sinωt)2 d(ωt)

}

=√{

10000

π

∫ π

0sin2 ωt d(ωt)

},

which is not a ‘standard’ integral.

It is shown in Chapter 17 thatcos2A=1 − 2 sin2 A and this formula is usedwhenever sin2 A needs to be integrated.

Rearranging cos2A = 1 − 2 sin2 A gives

sin2 A = 1

2(1 − cos2A)

Hence

√{10000

π

∫ π

0sin2 ωt d(ωt)

}

=√{

10000

π

∫ π

0

1

2(1 − cos2ωt)d(ωt)

}

=√{

10000

π

1

2

[ωt − sin 2ωt

2

]π

0

}

=

√√√√√√√⎧⎪⎪⎨⎪⎪⎩

10000

π12

[(π − sin 2π

2

)−

(0 − sin 0

2

)]⎫⎪⎪⎬⎪⎪⎭

=√{

10000

π

1

2[π]

}

=√{

10000

2

}= 100√

2= 70.71 volts

[Note that for a sine wave,

r.m.s. value= 1√2

× maximum value.


In this case,

r.m.s. value = 1√2

× 100 = 70.71 V]


Exercise 148 Further problems on meanand r.m.s. values

1. The vertical height h km of a missile varieswith the horizontal distance d km, and is givenby h =4d −d2. Determine the mean height ofthe missile from d = 0 to d = 4 km.

[2 23 km].

2. The distances of points y from the mean valueof a frequency distribution are related to the

variate x by the equation y = x + 1

x. Deter-

mine the standard deviation (i.e. the r.m.s.value), correct to 4 significant figures forvalues of x from 1 to 2. [2.198]

3. A current i = 25 sin100π t mA flows in anelectrical circuit. Determine, using integralcalculus, its mean and r.m.s. values each cor-rect to 2 decimal places over the range t = 0to t = 10 ms. [15.92 mA, 17.68 mA]

4. A wave is defined by the equation:

v = E1 sinωt + E3 sin 3ωt

where E1, E3 and ω are constants.Determine the r.m.s. value of v over the

interval 0≤ t ≤ π

ω. ⎡

⎣√

E21 + E2

3

2

⎤⎦

38.4 Volumes of solids of revolution

With reference to Fig. 38.6, the volume of revolution,V , obtained by rotating area A through one revolutionabout the x-axis is given by:

V =∫ b

aπy2 dx

If a curve x = f ( y) is rotated 360◦ about the y-axisbetween the limits y = c and y = d then the volume

0 x 5 a x 5 b

y 5 f (x)y

x

A

Figure 38.6

generated, V , is given by:

V =∫ d

cπx2 dy

Problem 5. The curve y = x2 +4 is rotated onerevolution about the x-axis between the limits x = 1and x = 4. Determine the volume of solid ofrevolution produced.

Revolving the shaded area shown in Fig. 38.7, 360◦about the x-axis produces a solid of revolution given by:

Volume =∫ 4

1πy2 dx =

∫ 4

1π(x2 + 4)2 dx

=∫ 4

1π(x4 + 8x2 + 16)dx

= π

[x5

5+ 8x3

3+ 16x

]4

1

= π[(204.8 + 170.67 + 64)

− (0.2 + 2.67 + 16)]

= 420.6π cubic units

0 1 2 x

45

30

20BA

D C10

y

43 5

y 5 x21 4

Figure 38.7


Problem 6. Determine the area enclosed by thetwo curves y = x2 and y2 = 8x . If this area isrotated 360◦ about the x-axis determine the volumeof the solid of revolution produced.

At the points of intersection the co-ordinates of thecurves are equal. Since y = x2 then y2 = x4. Henceequating the y2 values at the points of intersection:

x4 = 8x

from which, x4 − 8x = 0

and x(x3 − 8)= 0

Hence, at the points of intersection, x = 0 and x = 2.When x =0, y =0 and when x =2, y =4. The points

of intersection of the curves y = x2 and y2 =8x aretherefore at (0,0) and (2,4). A sketch is shown inFig. 38.8. If y2 =8x then y =√

8x .

Shaded area

=∫ 2

0

(√8x − x2

)dx =

∫ 2

0

(√8)

x12 − x2

)dx

=⎡⎣(√

8) x

32

32

− x3

3

⎤⎦2

0

={√

8√

832

− 8

3

}−{0}

= 16

3− 8

3= 8

3= 2

23

square units

y

x0 1

2

4

2

y 5 x2

y25 8x(or y 5ŒŒ(8x)

Figure 38.8

The volume produced by revolving the shaded areaabout the x-axis is given by:

{(volume produced by revolving y2 = 8x)

− (volume produced by revolving y = x2)}

i.e. volume =∫ 2

0π(8x)dx −

∫ 2

0π(x4)dx

= π

∫ 2

0(8x − x4)dx = π

[8x2

2− x5

5

]2

0

= π

[(16 − 32

5

)− (0)

]

= 9.6π cubic units


Exercise 149 Further problems on volumes

1. The curve x y=3 is revolved one revolutionabout the x-axis between the limits x =2 andx =3. Determine the volume of the solidproduced. [1.5π cubic units]

2. The area betweeny

x2 =1 and y + x2 =8 is

rotated 360◦ about the x-axis. Find the vol-ume produced. [170 2

3π cubic units]

3. The curve y = 2x2 + 3 is rotated about (a) thex-axis between the limits x = 0 and x = 3,and (b) the y-axis, between the same limits.Determine the volume generated in each case.

[(a) 329.4π (b) 81π]

4. The profile of a rotor blade is bounded by thelines x =0.2, y =2x, y=e−x , x =1 and thex-axis. The blade thickness t varies linearlywith x and is given by: t =(1.1− x)K, whereK is a constant.

(a) Sketch the rotor blade, labelling the limits.

(b) Determine, using an iterative method, thevalue of x , correct to 3 decimal places,where 2x =e−x

(c) Calculate the cross-sectional area of theblade, correct to 3 decimal places.

(d) Calculate the volume of the blade in termsof K, correct to 3 decimal places.

[(b) 0.352 (c) 0.419 square units(d) 0.222 K]


38.5 Centroids

A lamina is a thin flat sheet having uniform thickness.The centre of gravity of a lamina is the point whereit balances perfectly, i.e. the lamina’s centre of mass.When dealing with an area (i.e. a lamina of negligiblethickness and mass) the term centre of area or centroidis used for the point where the centre of gravity of alamina of that shape would lie.

If x and y denote the co-ordinates of the centroid Cof area A of Fig. 38.9, then:

x=

∫ b

axydx∫ b

aydx

and y=12

∫ b

ay2 dx∫ b

aydx

0

Area A

x 5 a x 5 b

y

x

y 5 f(x)

yx

C

Figure 38.9

Problem 7. Find the position of the centroid ofthe area bounded by the curve y = 3x2, the x-axisand the ordinates x = 0 and x = 2.

If (x , y) are co-ordinates of the centroid of the givenarea then:

x =

∫ 2

0x y dx∫ 2

0y dx

=

∫ 2

0x(3x2)dx∫ 2

03x2 dx

=

∫ 2

03x3 dx∫ 2

03x2 dx

=

[3x4

4

]2

0

[x3]20

= 12

8= 1.5

y =1

2

∫ 2

0y2 dx∫ 2

0y dx

=1

2

∫ 2

0(3x2)2 dx

8

=1

2

∫ 2

09x4 dx

8=

9

2

[x5

5

]2

0

8

=9

2

(32

5

)8

= 18

5= 3.6

Hence the centroid lies at (1.5, 3.6)

Problem 8. Determine the co-ordinates ofthe centroid of the area lying between the curvey =5x − x2 and the x-axis.

y =5x − x2 = x(5− x). When y=0, x =0 or x =5.Hence the curve cuts the x-axis at 0 and 5 as shownin Fig. 38.10. Let the co-ordinates of the centroid be(x , y) then, by integration,

x =

∫ 5

0x y dx∫ 5

0y dx

=

∫ 5

0x(5x − x2)dx∫ 5

0(5x − x2)dx

=

∫ 5

0(5x2 − x3)dx∫ 5

0(5x − x2)dx

=[

5x3

3 − x4

4

]5

0[5x2

2 − x3

3

]5

0

8

C

6

4

2

1 2 3 4 5 x

y

y 5 5x2 x2

y

x

0

Figure 38.10


=625

3− 625

4125

2− 125

3

=625

12125

6

=(

625

12

)(6

125

)= 5

2= 2.5

y =1

2

∫ 5

0y2 dx∫ 5

0y dx

=1

2

∫ 5

0(5x − x2)2 dx∫ 5

0(5x − x2)dx

=1

2

∫ 5

0(25x2 − 10x3 + x4)dx

125

6

=

1

2

[25x3

3− 10x4

4+ x5

5

]5

0125

6

=1

2

(25(125)

3− 6250

4+ 625

)125

6

= 2.5

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

(Note from Fig. 38.10 that the curve is symmetricalabout x = 2.5 and thus x could have been determined‘on sight’.)


Exercise 150 Further problems on centroids

In Problems 1 and 2, find the position of the cen-troids of the areas bounded by the given curves, thex-axis and the given ordinates.

1. y = 3x + 2 x = 0, x = 4 [(2.5, 4.75)]

2. y = 5x2 x = 1, x = 4 [(3.036, 24.36)]

3. Determine the position of the centroid of asheet of metal formed by the curvey = 4x − x2 which lies above the x-axis.

[(2, 1.6)]

4. Find the co-ordinates of the centroid of the areawhich lies between the curve y/x = x − 2 andthe x-axis. [(1, −0.4)]

5. Sketch the curve y2 = 9x between the limitsx = 0 and x = 4. Determine the position of thecentroid of this area.

[(2.4, 0)]

38.6 Theorem of Pappus

A theorem of Pappus states:

‘If a plane area is rotated about an axis in its own planebut not intersecting it, the volume of the solid formed isgiven by the product of the area and the distance movedby the centroid of the area’.With reference to Fig. 38.11, when the curve y = f (x)

is rotated one revolution about the x-axis betweenthe limits x = a and x = b, the volume V generatedis given by:

volume V = (A)(2π y ), from which, y= V2πA

y

C

Area A

xx 5 bx 5 a

y 5 f(x)

y

Figure 38.11

Problem 9. (a) Calculate the area bounded by thecurve y = 2x2, the x-axis and ordinates x = 0 andx = 3. (b) If this area is revolved (i) about thex-axis and (ii) about the y-axis, find the volumes ofthe solids produced. (c) Locate the position of thecentroid using (i) integration, and (ii) the theoremof Pappus.

(a) The required area is shown shaded in Fig. 38.12.

Area =∫ 3

0y dx =

∫ 3

02x2 dx

=[

2x3

3

]3

0= 18 square units


y

0

6

12

18

1 2 3 x

y 5 2x2

x

y

Figure 38.12

(b) (i) When the shaded area of Fig. 38.12 isrevolved 360◦ about the x-axis, the volumegenerated

=∫ 3

0πy2 dx =

∫ 3

0π(2x2)2 dx

=∫ 3

04πx4 dx = 4π

[x5

5

]3

0

= 4π

(243

5

)=194.4πcubic units

(ii) When the shaded area of Fig. 38.12 isrevolved 360◦ about the y-axis, the volumegenerated

= (volume generated by x = 3)

− (volume generated by y = 2x2)

=∫ 18

0π(3)2 dy −

∫ 18

0π

( y

2

)dy

= π

∫ 18

0

(9 − y

2

)dy = π

[9y − y2

4

]18

0

=81π cubic units

(c) If the co-ordinates of the centroid of the shadedarea in Fig. 38.12 are (x, y) then:(i) by integration,

x =

∫ 3

0x y dx∫ 3

0y dx

=

∫ 3

0x(2x2)dx

18

=

∫ 3

02x3 dx

18=

[2x4

4

]3

0

18

= 81

36=2.25

y =1

2

∫ 3

0y2 dx∫ 3

0y dx

=1

2

∫ 3

0(2x2)2 dx

18

=1

2

∫ 3

04x4 dx

18=

1

2

[4x5

5

]3

0

18= 5.4

(ii) using the theorem of Pappus:

Volume generated when shaded area isrevolved about OY=(area)(2πx).

i.e. 81π = (18)(2πx ),

from which, x = 81π

36π= 2.25

Volume generated when shaded area isrevolved about OX=(area)(2π y).

i.e. 194.4π = (18)(2π y),

from which, y = 194.4π

36π= 5.4

Hence the centroid of the shaded area inFig. 38.12 is at (2.25, 5.4).

Problem 10. A metal disc has a radius of 5.0 cmand is of thickness 2.0 cm. A semicircular groove ofdiameter 2.0 cm is machined centrally around therim to form a pulley. Determine, using Pappus’theorem, the volume and mass of metal removedand the volume and mass of the pulley if the densityof the metal is 8000 kg m−3.

A side view of the rim of the disc is shown in Fig. 38.13.

5.0 cm

2.0 cm

XX

S R

QP

Figure 38.13


When area PQRS is rotated about axis XX the vol-ume generated is that of the pulley. The centroid of the

semicircular area removed is at a distance of4r

3πfrom its

diameter (see ‘Engineering Mathematics 6th edition’,

Chapter 58), i.e.4(1.0)

3π, 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 thecentroid is 2π(4.576)cm.

Area of semicircle= πr2

2= π(1.0)2

2= π

2cm2

By the theorem of Pappus,volume generated = area × distance moved by

centroid =(π

2

)(2π)(4.576).

i.e. volume of metal removed=45.16 cm3

Massofmetal removed = density × volume

= 8000 kg m−3×45.16

106m3

= 0.3613 kg or 361.3 g

volume of pulley=volume of cylindrical disc

−volume of metal removed

= π(5.0)2(2.0)− 45.16

= 111.9 cm3

Mass of pulley= density×volume

= 8000 kg m−3 × 111.9

106m3

= 0.8952 kg or 895.2 g


Exercise 151 Further problems on thetheorem of Pappus

1. A right angled isosceles triangle having ahypotenuse of 8 cm is revolved one revolutionabout one of its equal sides as axis. Deter-mine the volume of the solid generated usingPappus’ theorem. [189.6 cm3]

2. Using (a) the theorem of Pappus, and (b) inte-gration, determine the position of the centroidof a metal template in the form of a quadrant

of a circle of radius 4 cm. (The equation of acircle, centre 0, radius r is x2 + y2 = r2).⎡

⎢⎢⎢⎣On the centre line, distance

2.40 cm from the centre,

i.e. at co-ordinates

(1.70,1.70)

⎤⎥⎥⎥⎦

3. (a) Determine the area bounded by the curvey =5x2, the x-axis and the ordinatesx =0 and x =3.

(b) If this area is revolved 360◦ about (i) thex-axis, and (ii) the y-axis, find the vol-umes of the solids of revolution producedin each case.

(c) 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)

⎤⎥⎥⎥⎦

4. A metal disc has a radius of 7.0 cm and isof thickness 2.5 cm. A semicircular groove ofdiameter 2.0 cm is machined centrally aroundthe rim to form a pulley. Determine the vol-ume of metal removed using Pappus’ theoremand express this as a percentage of the origi-nal volume of the disc. Find also the mass ofmetal removed if the density of the metal is7800 kg m−3.

[64.90 cm3, 16.86%, 506.2 g]

For more on areas, mean and r.m.s. values, volumes andcentroids, see ‘Engineering Mathematics 6th edition’,Chapters 55 to 58.

38.7 Second moments of area ofregular sections

The first moment of area about a fixed axis of a laminaof area A, perpendicular distance y from the centroidof the lamina is defined as Ay cubic units. The secondmoment of area of the same lamina as above is givenby Ay2, i.e. the perpendicular distance from the centroidof the area to the fixed axis is squared.


Second moments of areas are usually denoted by I andhave 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 areaA, where A = a1 + a2 + a3 + · · · at distance k from theaxis, such that Ak2 = ∑

ay2.k is called the radius of gyration of area A about thegiven axis. Since Ak2 = ∑

ay2 = I then the radius ofgyration,

k =√

I

A

The second moment of area is a quantity much used inthe theory of bending of beams, in the torsion of shafts,and in calculations involving water planes and centresof pressure.

The procedure to determine the second moment ofarea of regular sections about a given axis is (i) to find thesecond moment of area of a typical element and (ii) tosum all such second moments of area by integratingbetween appropriate limits.

For example, the second moment of area of the rect-angle shown in Fig. 38.14 about axis PP is found byinitially considering an elemental strip of width δx , par-allel to and distance x from axis PP. Area of shadedstrip = bδx .

b

l

x

P

P

�x

Figure 38.14

Second moment of area of the shaded strip aboutPP = (x2)(bδx).The second moment of area of the whole rectangle aboutPP is obtained by summing all such strips between x =0 and x = l, i.e.

∑x=lx=0 x2 bδx .

It is a fundamental theorem of integration that

limitδx→0

x=l∑x=0

x2bδx =∫ l

0x2b dx

Thus the second moment of area of the rectangleabout PP

= b∫ l

0x2 dx = b

[x3

3

]l

0= bl3

3

Since the total area of the rectangle, A = lb, then

Ipp = (lb)

(l2

3

)= Al2

3

Ipp = Ak2pp thus k2

pp = l2

3

i.e. the radius of gyration about axes PP,

kpp =√

l2

3= l√

3

Parallel axis theorem

In Fig. 38.15, axis GG passes through the centroid Cof area A. Axes DD and GG are in the same plane, areparallel to each other and distance d apart. The parallelaxis theorem states:

IDD = IGG + Ad2

Using the parallel axis theorem the second moment ofarea of a rectangle about an axis through the centroid

d

G

C

Area A

G

D

D

Figure 38.15


bC

x

P G

GP

�x

l2

l2

Figure 38.16

may be determined. In the rectangle shown in Fig. 38.16,

Ipp = bl3

3(from above).

From the parallel axis theorem

Ipp = IGG + (bl)

(1

2

)2

i.e.bl3

3= IGG + bl3

4

from which, IGG = bl3

3− bl3

4= bl3

12

Perpendicular axis theorem

In Fig. 38.17, axes OX , OY and OZ are mutually per-pendicular. If OX and OY lie in the plane of area A thenthe perpendicular axis theorem states:

IOZ = IOX + IOY

Z

Y

X

O

Area A

Figure 38.17

A summary of derived standard results for the secondmoment of area and radius of gyration of regularsections are listed in Table 38.1.

Problem 11. Determine the second moment ofarea and the radius of gyration about axes AA, BBand CC for the rectangle shown in Fig. 38.18.

A

AB

b 5 4.0 cm

l 5 12.0 cm

B

CC

Figure 38.18

From Table 38.1, the second moment of area aboutaxis AA,

IAA = bl3

3= (4.0)(12.0)3

3= 2304 cm4

Radiusofgyration,kAA = l√3

= 12.0√3

= 6.93 cm

Similarly, IBB = lb3

3= (12.0)(4.0)3

3= 256 cm4

and kBB = b√3

= 4.0√3

= 2.31 cm

The second moment of area about the centroid of a

rectangle isbl3

12when the axis through the centroid is

parallel with the breadth b. In this case, the axis CC isparallel with the length l.

Hence ICC = lb3

12= (12.0)(4.0)3

12= 64 cm4

and kCC = b√12

= 4.0√12

= 1.15 cm


Table 38.1 Summary of standard results of the second moments of areas of regular sections

Shape Position of axis Second moment Radius of

of area, I gyration, k

Rectangle (1) Coinciding with bbl3

3

l√3

length l, breadth b

(2) Coinciding with llb3

3

b√3

(3) Through centroid, parallel to bbl3

12

l√12

(4) Through centroid, parallel to llb3

12

b√12

Triangle (1) Coinciding with bbh3

12

h√6

Perpendicular height h,

base b (2) Through centroid, parallel to basebh3

36

h√18

(3) Through vertex, parallel to basebh3

4

h√2

Circle (1) Through centre, perpendicular toπr4

2

r√2

radius r plane (i.e. polar axis)

(2) Coinciding with diameterπr4

4

r

2

(3) About a tangent5πr4

4

√5

2r

Semicircle Coinciding with diameterπr4

8

r

2radius r

Problem 12. Find the second moment of area andthe radius of gyration about axis PP for therectangle shown in Fig. 38.19.

PP

40.0 mm

15.0 mm

25.0 mm

GG

Figure 38.19

IGG = lb3

12where 1 = 40.0 mm and b = 15.0 mm

Hence IGG = (40.0)(15.0)3

12= 11250 mm4

From the parallel axis theorem, IPP = IGG + Ad2 ,where A = 40.0 × 15.0 = 600 mm2 andd =25.0+7.5=32.5 mm, the perpendiculardistance between GG and PP. Hence,

IPP = 11 250+ (600)(32.5)2

= 645000 mm4


IPP = Ak2PP , from which,

kPP =√

IPP

area=

√(645000

600

)= 32.79 mm

Problem 13. Determine the second moment ofarea and radius of gyration about axis QQ of thetriangle BCD shown in Fig. 38.20.

Q Q

DC

6.0 cm

12.0 cm

8.0 cm

G

B

G

Figure 38.20

Using the parallel axis theorem: IQQ = IGG + Ad2,where IGG is the second moment of area about thecentroid of the triangle,

i.e.bh3

36= (8.0)(12.0)3

36= 384 cm4,

A is the area of the triangle,

= 12 bh = 1

2 (8.0)(12.0) = 48 cm2

and d is the distance between axes GG and QQ,

= 6.0 + 13 (12.0) = 10 cm.

Hence the second moment of area about axis QQ,

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

Radius of gyration,

kQQ =√

IQ Q

area=

√(5184

48

)= 10.4 cm

Problem 14. Determine the second moment ofarea and radius of gyration of the circle shown inFig. 38.21 about axis YY .

Y

3.0 cm

Y

GG

r 5 2.0 cm

Figure 38.21

In Fig. 38.21, IGG = πr4

4= π

4(2.0)4 = 4π cm4.

Using the parallel axis theorem, IYY = IGG + Ad2,where d =3.0+2.0=5.0 cm.

Hence IYY = 4π + [π(2.0)2](5.0)2

= 4π + 100π = 104π = 327 cm4

Radius of gyration,

kYY =√

IY Y

area=

√(104π

π(2.0)2

)= √

26 = 5.10 cm

Problem 15. Determine the second moment ofarea and radius of gyration for the semicircle shownin Fig. 38.22 about axis XX .

XX

GG

BB

10.0 mm

15.0 mm

Figure 38.22


The centroid of a semicircle lies at4r

3πfrom its

diameter.

Using the parallel axis theorem:

IBB = IGG + Ad2,

where IBB = πr4

8(from Table 38.1)

= π(10.0)4

8= 3927 mm4,

A = πr2

2= π(10.0)2

2= 157.1 mm2

and d = 4r

3π= 4(10.0)

3π= 4.244 mm

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

i.e. 3927= IGG + 2830,

from which, IGG = 3927 − 2830 = 1097 mm4

Using the parallel axis theorem again:

IXX = IGG + A(15.0 + 4.244)2

i.e. IXX = 1097 + (157.1)(19.244)2

= 1097 + 58 179

= 59276 mm4 or 59280 mm4,correct to 4 significant figures.

Radius of gyration, kXX =√

IXX

area=

√(59 276

157.1

)

= 19.42 mm

Problem 16. Determine the polar second momentof area of the propeller shaft cross-section shown inFig. 38.23.

6.0

cm

7.0

cm

Figure 38.23

The polar second moment of area of a circle= πr4

2The polar second moment of area of the shaded area is

given by the polar second moment of area of the 7.0 cmdiameter circle minus the polar second moment of areaof the 6.0 cm diameter circle.

Hence the polar second moment of area of the cross-section shown

= π

2

(7.0

2

)4

− π

2

(6.0

2

)4

= 235.7 − 127.2 = 108.5 cm4

Problem 17. Determine the second moment ofarea and radius of gyration of a rectangular laminaof length 40 mm and width 15 mm about an axisthrough one corner, perpendicular to the plane ofthe lamina.

The lamina is shown in Fig. 38.24.

l 5 40 mm

b 5 15 mmX

X

ZY

YZ

Figure 38.24

From the perpendicular axis theorem:

IZZ = IXX + IYY

IXX = lb3

3= (40)(15)3

3= 45000 mm4

and IYY = bl3

3= (15)(40)3

3= 320000 mm4

Hence IZZ = 45 000 + 320 000

= 365000 mm4 or 36.5 cm4

Radius of gyration,

kZZ =√

IZ Z

area=

√(365 000

(40)(15)

)= 24.7 mm or 2.47 cm


Problem 18. Determine correct to 3 significantfigures, the second moment of area about axis XXfor the composite area shown in Fig. 38.25.

4.0cm

1.0 cm1.0 cm

8.0 cm

CT

XX

2.0 cm

6.0 cm

2.0 cm

TT

Figure 38.25

For the semicircle,

IXX = πr4

8= π(4.0)4

8= 100.5 cm4

For the rectangle,

IXX = bl3

3= (6.0)(8.0)3

3= 1024 cm4

For the triangle, about axis TT through centroid CT ,

ITT = bh3

36= (10)(6.0)3

36= 60 cm4

By the parallel axis theorem, the second moment of areaof the triangle about axis XX

=60 + [12 (10)(6.0)

] [8.0 + 1

3 (6.0)]2 = 3060 cm4.

Total second moment of area about XX

= 100.5 + 1024 + 3060

= 4184.5

= 4180 cm4, correct to 3 significant figures.

Problem 19. Determine the second moment ofarea and the radius of gyration about axis XX for theI -section shown in Fig. 38.26.

CF

CE

CD

3.0 cm7.0 cm

4.0 cm

3.0 cm

y

C C

X X

S

S

15.0 cm

8.0 cm

Figure 38.26

The I -section is divided into three rectangles, D, Eand F and their centroids denoted by CD , CE and CF

respectively.

For rectangle D:The second moment of area about CD (an axis throughCD parallel to XX )

= bl3

12= (8.0)(3.0)3

12= 18 cm4


IXX = 18 + Ad2

where A = (8.0)(3.0) = 24 cm2 and d = 12.5 cm

Hence IXX = 18 + 24(12.5)2 = 3768 cm4.

For rectangle E:The second moment of area about CE (an axis throughCE parallel to XX )

= bl3

12= (3.0)(7.0)3

12= 85.75 cm4


IXX = 85.75 + (7.0)(3.0)(7.5)2 = 1267 cm4.


For rectangle F:

IXX = bl3

3= (15.0)(4.0)3

3= 320 cm4

Total second moment of area for the I-section aboutaxis XX,

IXX = 3768 + 1267 + 320 = 5355 cm4

Total area of I -section

= (8.0)(3.0)+ (3.0)(7.0)+ (15.0)(4.0)

= 105 cm2.

Radius of gyration,

kXX =√

IXX

area=

√(5355

105

)= 7.14 cm


Exercise 152 Further problems on secondmoment of areas of regular sections

1. Determine the second moment of area andradius of gyration for the rectangle shown inFig. 38.27 about (a) axis AA (b) axis BB and(c) axis CC. ⎡

⎣(a) 72 cm4,1.73 cm(b) 128 cm4,2.31 cm(c) 512 cm4,4.62 cm

⎤⎦

8.0 cm

B

B

C

AA

C

3.0 cm

Figure 38.27

2. Determine the second moment of area andradius of gyration for the triangle shown inFig. 38.28 about (a) axis DD (b) axis EE and(c) an axis through the centroid of the triangleparallel to axis DD.⎡

⎣(a) 729 cm4,3.67 cm(b) 2187 cm4,6.36 cm(c) 243 cm4,2.12 cm

⎤⎦

12.0 cm

9.0 cm

DD

E E

Figure 38.28

3. For the circle shown in Fig. 38.29, find thesecond moment of area and radius of gyrationabout (a) axis FF and (b) axis HH .

[(a) 201 cm4,2.0 cm

(b) 1005 cm4,4.47 cm

]

r54.0

cm

F

H

H

F

Figure 38.29

4. For the semicircle shown in Fig. 38.30, find thesecond moment of area and radius of gyrationabout axis JJ .

[3927 mm4, 5.0 mm]

JJr5

10.0

mm

Figure 38.30

5. For each of the areas shown in Fig. 38.31 deter-mine the second moment of area and radius ofgyration about axis LL , by using the parallelaxis theorem. ⎡

⎢⎣(a) 335 cm4,4.73 cm

(b) 22030 cm4,14.3 cm

(c) 628 cm4,7.07 cm

⎤⎥⎦


2.0 cm

5.0 cm

3.0 cm

(a) (b) (c)

15 cm

18 cm 10 cm

15 cm

5.0 cm

LL

Dia 5 4.0 cm

Figure 38.31

6. Calculate the radius of gyration of a rectan-gular door 2.0 m high by 1.5 m wide about avertical axis through its hinge.

[0.866 m]

7. A circular door of a boiler is hinged so thatit turns about a tangent. If its diameter is1.0 m, determine its second moment of areaand radius of gyration about the hinge.

[0.245 m4, 0.559 m]

8. A circular cover, centre 0, has a radius of12.0 cm. A hole of radius 4.0 cm and centre X ,where OX = 6.0 cm, is cut in the cover. Deter-mine the second moment of area and the radiusof gyration of the remainder about a diameterthrough 0 perpendicular to OX .

[14280 cm4, 5.96 cm]

9. For the sections shown in Fig. 38.32, findthe second moment of area and the radius ofgyration about axis XX .[

(a) 12190 mm4,10.9 mm

(b) 549.5 cm4,4.18 cm

]

18.0 mm

3.0 cm2.5 cm

2.0 cm

2.0 cm

6.0 cm

12.0 mm

3.0 mm

X X

X X4.0 mm

(a) (b)

Figure 38.32

10. Determine the second moments of areas aboutthe given axes for the shapes shown inFig. 38.33. (In Fig. 38.33(b), the circular areais removed.) ⎡

⎣IAA = 4224 cm4,

IBB = 6718 cm4,

ICC = 37300 cm4

⎤⎦

3.0 cm

16.0 cm

9.0 cm

10.0 cm

(a)

(b)

4.0 cm15.0 cm

9.0 cm

4.5 cm

A A

B

C

B

C

Dia 5 7.0 cm

Figure 38.33

11. Find the second moment of area and radiusof gyration about the axis XX for the beamsection shown in Fig. 38.34. [

1350 cm4,

5.67 cm

]

2.0 cm8.0 cm

2.0 cm

X X

1.0 cm

10.0 cm

6.0 cm

Figure 38.34

pertemuan 2

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