chapter 9 first and second moments of area AWS
CHAPTER 9 FIRST AND SECOND MOMENTS OF AREA EXERCISE 55, Page 130
1. Find the position of the centroid of the area bounded by the curve y = 2x, the x-axis and the ordinates x = 0, x = 3
A sketch of the area is shown below.
3
2x 3 3 3 3 2 3 2 3 − 0 (27) xy dx ∫ x ( 2x ) dx ∫ 2x 2 dx 3 ∫ 0 0 0 0 3 = 3= 2 (3) = 2 = x = = = = 3 3 3 2 3 (9) 3 32 − 0 2x y dx 2xdx 2x dx ∫0 ∫0 ∫0 2 0 1 3 2 1 3 2 3 y dx 2x ) dx ( ∫ ∫ 1 3 2 1 4x 3 4 3 2 0 0 2 2 3 − 0= = y = = 4x dx = = (27) = 2 3 ∫ 0 9 18 18 3 0 54 27 y dx ∫ 0
Hence, the centroid lies at (2, 2)
2. Find the position of the centroid of the area bounded by the curve y = 3x + 2, the x-axis and the ordinates x = 0, x = 4
A sketch of the area is shown below.
161 © John Bird & Carl Ross Published by Taylor and Francis
4
3x 3 2x 2 + 3 xy dx ∫ x ( 3x + 2 ) dx ∫ (3x 2 + 2x)dx 3 2 0 42 ∫ 64 + 16 80 0 0 0 4 + = = = = = = = x 4 4 4 4 2 3 + 8 32 24 2 ∫0 y dx ∫0 (3x + 2)dx ∫0 (3x + 2) dx 3x + 2x 2 (4) + 2(4) 2 0 4
4
4
= 2.50 1 4 2 1 4 2 4 y dx ( 3x + 2 ) dx 1 4 2 ∫ ∫ 1 9x 3 12x 2 1 0 0 2 2 = = = = + + 4x= y (9x + 12x + 4)dx [192 + 96 + 16] 4 ∫ 32 64 0 64 3 2 64 y dx 0 ∫ 0
=
1 (304) = 4.75 64
Hence, the centroid lies at (2.50, 4.75) 3. Find the position of the centroid of the area bounded by the curve y = 5x2 ; x = 1, x = 4
A sketch of the area is shown below.
4
5x 4 xy dx ∫ x ( 5x 2 ) dx ∫ 5x 3dx 4 ∫ 1 1 1 1 x = = = 4= = 4 4 3 4 2 2 ∫1 y dx ∫1 5x dx ∫1 5x dx 5x 3 1 4
4
4
5 4 4 5 4 − 1 (255) 4 = 4= 318.75 = 3.036 5 3 3 5 105 4 − 1 (63) 3 3
2 1 4 2 1 4 4 y dx 5x 2 ) dx ( ∫ ∫ 1 4 1 25x 5 5 5 1 1 4 5 2 2 45 − 1= = y = = 25x = dx = (1023) 4 ∫ 1 105 210 210 5 1 210 210 y dx ∫ 1
= 24.36 Hence, the centroid lies at (3.036, 24.36)
162 © John Bird & Carl Ross Published by Taylor and Francis
4. Find the position of the centroid of the area bounded by the curve y = 2x3 , the x-axis and the ordinates x = 0, x = 2
A sketch of the area is shown below.
2
2x 5 2 2 2 xy dx ∫ x ( 2x 3 ) dx ∫ 2x 4 dx 5 ∫ 0 0 0 0 x = = = 2 = = 2 2 4 2 3 3 ∫0 y dx ∫0 2x dx ∫0 2x dx 2x 4 0
2 5 2 2 − 0 (32) 5 = 5= 2 (4) = 1.60 1 4 8 5 2 − 0 2
2 1 2 2 1 2 2 y dx 2x 3 ) dx ( ∫ ∫ 1 2 6 1 x7 1 7 1 0 0 2 2 2 − 0= y 4x dx (128) = 4.57 = = = = = 2 ∫ 8 16 0 4 7 0 28 28 y dx ∫ 0
Hence, the centroid lies at (1.60, 4.57)
5. Find the position of the centroid of the area bounded by the curve y = x(3x + 1), the x-axis and the ordinates x = - 1, x = 0
0
3x 4 x 3 + 2 3 2 xy dx ∫ x ( 3x + x ) dx ∫ (3x + x )dx 4 3 −1 ∫ −1 −1 −1 = = = = x = 0 0 0 3 2 0 2 2 ∫ −1 y dx ∫ −1 (3x + x)dx ∫ −1 (3x + x) dx 3x + x 2 −1 3 0
0
0
3 (−1)3 4 − − + (0) ( 1) 3 4 1 3 2 (0) − (−1) + 2 (−1)
3 1 5 0− − − 4 3 = 12 = − 5 = - 0.833 = 1 0 − ( −1 + 0.5 ) 6 2 163 © John Bird & Carl Ross Published by Taylor and Francis
2 1 0 2 1 0 2 y dx 3x + x dx ( ) ∫ −1 ∫ −1 y= 2 0 = 2 = 1 y dx ∫ −1 2
0
9x 5 6x 4 x 3 (9x + 6x + x )dx = 5 + 4 + 3 ∫ −1 −1 0
4
3
2
1 = (0) − −1.8 + 1.5 − = 0.633 3 Hence, the centroid lies at (- 0.833, 0.633)
164 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 56, Page 131 1. Determine the position of the centroid of a sheet of metal formed by the curve y = 4x - x2 which lies above the x-axis. y = 4x – x2 = x(4 – x) i.e. when y = 0, x = 0 and x = 4. The area of the sheet of metal is shown sketched below.
By symmetry,
x=2 4
1 16x 3 8x 4 x 5 1 4 2 1 4 1 4 2 2 2 3 4 y dx 4x − x ) dx 16x − 8x + x ) dx 2 3 − 4 + 5 ( ( ∫ ∫ ∫ 0 0 0 0 2 = y 2= = 2 = 4 4 4 4 2 2 2 x3 ∫0 y dx ∫0 ( 4x − x ) dx ∫0 ( 4x − x ) dx 2x − 3 0
1 ( 341.333 − 512 + 204.8 ) − (0) 17.0665 2 = = 1.6 = 64 10.666 32 − − (0) 3 Hence, the co-ordinates of the centroid are (2, 1.6)
2. Find the coordinates of the centroid of the area that lies between the curve
y = x - 2 and the x
x-axis.
y = x – 2 i.e. y = x(x – 2) = x 2 − 2x . The area is shown sketched below. x
165 © John Bird & Carl Ross Published by Taylor and Francis
2
x 4 2x 3 − xy dx ∫ x ( x 2 − 2x ) dx ∫ ( x 3 − 2x 2 ) dx 4 3 0 ∫ 0 0 0 = = = = x = 2 2 2 3 2 2 2 2 x 2x − − y dx x 2x dx x 2x dx ( ) ( ) ∫0 ∫0 ∫0 3 − 2 0 2
2
2
− (0) ] [(4 − 5.3333)= [(2.6667 − 4) − (0)]
−1.3333 −1.3333 =1
2
1 x 5 4x 4 4x 3 2 1 2 2 1 2 2 1 2 4 3 2 y dx x − 2x ) dx x − 4x + 4x ) dx 2 5 − 4 + 3 ( ( ∫ ∫ ∫ 0 0 0 0 2 = = 2 = y 2= 2 2 2 2 3 2 2 2 x 2x ∫0 y dx ∫0 ( x − 2x ) dx ∫0 ( x − 2x ) dx 3 − 2 0
1 [(6.4 − 16 + 10.6667) − (0)] 0.53335 2 = = - 0.4 = −1.3333 [(2.6667 − 4) − (0)]
Hence, the co-ordinates of the centroid are (1, - 0.4) 3. Determine the coordinates of the centroid of the area formed between the curve y = 9 - x2 and the x-axis. The area is shown sketched below where by symmetry, x = 0
3
2 1 3 2 1 3 1 3 2 4 y dx 9 − x 2 ) dx ( ∫ ∫ ∫ −3 (81 − 18x + x ) dx −3 −3 2 2 2 = y = = = 3 3 3 2 2 ∫ y dx ∫ ( 9 − x ) dx ∫ ( 9 − x ) dx −3
−3
−3
1 18x 3 x 5 81x − + 2 3 5 −3 3
x3 9x − 3 −3
1 [(243 − 162 + 48.6) − (−243 + 162 − 48.6)] 129.6 = 2 = 3.6 = 36 [(27 − 9) − (−27 + 9)]
Hence, the co-ordinates of the centroid are (0, 3.6)
166 © John Bird & Carl Ross Published by Taylor and Francis
4. Determine the centroid of the area lying between y = 4x2, the y-axis and the ordinates y = 0 and y = 4. The area is shown sketched below.
4
1 y2 1 1 y 2 ∫ 0 x dy 2 ∫ 0 4 dy 2 8 0 1 2 = = = From Section= 9.4, x = 0.375 4 4 8 4 y x dy ∫1 ∫ 0 4 dy 1 y3/2 3 3 2 2 0 4
4
4
∫ 0 xy dy y = and = 4 ∫ x dy 4
−
0
∫
4 0
1 y5/2 3/2 4 y 2 5 y 32 (y) dy ∫ dy 2 0 0 4= 2 = = 5 = 2.40 8 8 8 8 3 3 3 3
Hence the position of the centroid is at (0.375, 2.40)
5. Find the position of the centroid of the area enclosed by the curve y = 5x , the x-axis and the ordinate x = 5. The area is shown sketched below.
167 © John Bird & Carl Ross Published by Taylor and Francis
5
5
xy dx ∫0 = = x 5 ∫ y dx 0
∫ ( 5
0
x
∫
5 0
)
5x dx = 5x dx
5 x2 5 3 5 2 5 ∫ x dx 2 0 0 = = 1 5 5 3 5 ∫ x 2 dx 2 x 0 3 2 0
2 52 5 − 0 5 = 3 2 2 5 − 0 3
2 (55.9017) 22.3607 5 = = 3.0 2 (11.1802) 7.45356 3
5
(
)
2 1 5 2 1 5 y dx 5x dx ∫ ∫ 0 2 0 = y 2= = 5 5 y dx ∫ ∫ 5x dx 0
0
1 5 5x dx 0 2 ∫=
( 5)∫
5 0
1 2
x dx
1 5x 2 1 125 − 0 2 2 0 31.25 2 2 = 1.875 = = 16.66667 16.66667 5 7.45356
( )
Hence, the centroid lies at (3.0, 1.875) 6. Sketch the curve y2 = 9x between the limits x = 0 and x = 4. Determine the position of the centroid of this area. The curve y2 = 9x is shown sketched below.
By symmetry, y = 0 4
5 3x 2 5 3 4 4 4 6 5 6 5 6 2 x 4 − 0 (32) xy dx ∫ x 3 x dx ∫ 3x dx 2 ∫ 0 0 0 0 5= 5 = 5 x = = = = = 4 4 1 4 3 3 4 ∫0 y dx ∫0 3 x dx ∫0 3x 2 dx 3x 32 2 x 2 4 − 0 2(8) 3 2 0 38.4 = = 2.4 16
(
)
Hence, the centroid is at (2.4, 0) 168 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 57, Page 137
1. Determine the second moment of area and radius of gyration for the rectangle shown below about (a) axis AA (b) axis BB, and (c) axis CC
From Table 9.1, page 133 of textbook, the second moment of area about axis AA, Similarly, and
I AA = k AA =
I CC = Radius of gyration, k CC =
(8.0)(3.0)3 db3 = = 72 cm4 3 3 b 3.0 = = 1.73 cm 3 3 bd 3 (3.0)(8.0)3 = = 512 cm 4 3 3
d 8.0 = = 4.62 cm 3 3
The second moment of area about the centroid of a rectangle is
bd 3 when the axis through the 12
centroid is parallel with the breadth b. Hence
I BB =
and
k BB =
(3.0)(8.0)3 bd 3 = = 128 cm4 12 12
d 8.0 = = 2.31 cm 12 12
2. Determine the second moment of area and radius of gyration for the triangle shown below about (a) axis DD (b) axis EE, and (c) an axis through the centroid of the triangle parallel to axis DD
169 © John Bird & Carl Ross Published by Taylor and Francis
From Table 9.1, page 133 of textbook:
(12.0 )( 9.0 )
b h3 I DD = (a) Second moment of area about DD, = 12 Radius of gyration about DD, k= DD
12
= 729 cm 4
h 9.0 = 3.67 cm = 6 6
b h3 I EE = (b) Second moment of area about EE, = 4 Radius of gyration about EE, k= EE
3
(12.0 )( 9.0 )
3
4
= 2187 cm 4
h 9.0 = = 6.36 cm 2 2
b h3 = = (c) Second moment of area about axis through centroid, 36 Radius of gyration about axis through centroid, =
h = 18
(12.0 )( 9.0 )
3
= 243 cm 4
36
9.0 = 2.12 cm 18
3. For the circle shown below, find the second moment of area and radius of gyration about (a) axis FF and (b) axis HH
π r 4 π ( 4.0 ) = = = 201 cm 4 4 4 4
(a) Second moment of area coinciding with the diameter, I FF
Radius of gyration coinciding with the diameter, k FF =
r 4.0 = = 2.0 cm 2 2
170 © John Bird & Carl Ross Published by Taylor and Francis
5π r 4 5π ( 4.0 ) = (b) Second moment of area coinciding with a tangent, I HH = = 1005 cm 4 4 4 4
5 5 r= (4.0) = 4.47 cm 2 2
Radius of gyration coinciding with a tangent, k HH =
4. For the semicircle shown below, find the second moment of area and radius of gyration about axis JJ
π r 4 π (10.0 ) = = = 3927 mm 4 8 8 4
Second moment of area coinciding with the diameter, I JJ
Radius of gyration coinciding with the diameter, k JJ =
r 10.0 = = 5.0 mm 2 2
5. For each of the areas shown below determine the second moment of area and radius of gyration about axis LL, by using the parallel axis theorem.
bl3 (3.0)(5.0)3 + (3.0)(5.0)(2.5 + 2.0) 2 (a) Second moment of area, I LL =IGG + Ad 2 = + Ad 2 = 12 12 = 31.25 + 303.75 = 335 cm 4
k LL I LL = Ak LL 2 from which, radius of gyration,=
I LL = area
335 = 4.73 cm 15.0 171
© John Bird & Carl Ross Published by Taylor and Francis
(b) Second moment of area, I= IGG + Ad 2 LL
I= GG
bh 3 (18)(12)3 = = 864 cm 4 where h = 152 − 92 = 12 cm, as shown in the diagram below. 36 36
1 (18)(12) = 108 cm 2 2 2 12 2 =IGG + Ad =864 + 108 10 + =864 + 108(14) 2 = 22032 cm 4 = 22032 cm 4 3 correct to 4 significant figures.
Hence, area of triangle, A = Thus, I LL
k LL I LL = Ak LL 2 from which, radius of gyration,=
22032 = 14.3 cm 108
I LL = area
πr 4 4 π(2.0) 4 (c) Second moment of area, I LL = IGG + Ad= + ( πr 2 ) 5 + = + π(2.0) 2 (7) 2 4 2 4 2
2
= 12.57 + 615.75 = 628 cm 4
k LL I LL = Ak LL 2 from which, radius of gyration,=
I LL = area
628 = 7.07 cm 2 π(2.0)
6. Calculate the radius of gyration of a rectangular door 2.0 m high by 1.5 m wide about a vertical axis through its hinge.
From Table 9.1, page 133 of textbook, radius of gyration of rectangle, coinciding with length (i.e. height in this case) =
b 1.5 = 3 3 = 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.
172 © John Bird & Carl Ross Published by Taylor and Francis
From Table 9.1, page 133 of textbook,
5π r 4 5πd 4 5π (1.0 ) = 0.245 m 4 second moment of area coinciding with the tangent = = = 4 64 64 4
and radius of gyration of rectangle coinciding with tangent =
5 5 1.0 r= = 0.559 m 2 2 2
8. A circular cover, centre 0, has a radius of 12.0 cm. A hole of radius 4.0 cm and centre X, where 0X = 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 0X.
Second moment of area about diameter, i.e. axis CC in the diagram below
ICC =
=
πr 4 πr 4 − I DD = − IGG + Ad 2 4 4
π(12) 4 π(4.0) 4 − + π(4.0) 2 (6.0) 2 = 16286 – [201 + 1810] = 14275 = 14280 cm 4 , correct 4 4 to 4 significant figures.
ICC = Ak CC 2 from which, radius of gyration,
= k CC
ICC = area
14275 = 2 2 π(12.0) − π(4.0)
14275 = 5.96 cm 128π
9. For the sections shown below, find the second moment of area and the radius of gyration about axis XX
173 © John Bird & Carl Ross Published by Taylor and Francis
(a) For rectangle A in the diagram below,
bl3 (18.0)(3.0)3 = = 40.5 m 4 second moment of area about C= A 12 12 Hence, I XXA = 40.5 + Ad 2 = 40.5 + (3.0)(18.0)(12.0 + 1.5) 2 = 40.5 + 9841.5 = 9882 mm 4
For rectangle B, second moment of area about C= B
bl3 (4.0)(12.0)3 = = 576 m 4 12 12
Hence, I XXB = 576 + (4.0)(12.0)(6.0) 2 = 576 + 1728 = 2304 mm 4 Thus, total second moment of area about XX, I XXT = 9882 + 2304 = 12186 = 12190 mm 4 , correct to 4 significant figures. Radius of gyration about XX,= k XX
I XXT = area
12186 = 10.9 mm 54 + 48
(b) Rectangle A (see diagram below): Second moment of area about C= A
bl3 (6.0)(2.0)3 = = 4 cm 4 12 12
I XXA = 4 + (2.0)(6.0)(6.0) 2 = 4 + 432 = 436 cm 4
174 © John Bird & Carl Ross Published by Taylor and Francis
Rectangle B:
bl3 (2.5)(3.0)3 = = 5.625cm 4 12 12
Second moment of area about C= B
= I XXB 5.625 + (2.5)(3.0)(3.5) 2 = 97.5 cm 4 Rectangle C:
bl3 (6.0)(2.0)3 = = 4 cm 4 12 12
Second moment of area about C= C
I XXC = 4 + (6.0)(2.0)(1.0) 2 = 16 cm 4 Hence, total second moment of area about axis XX, I XXI = 436 + 97.5 + 16 = 549.5 cm 4 Radius of gyration about XX,= k XX
I XXI = area
549.5 = 4.18 cm 12 + 7.5 + 12
10. Determine the second moments of areas about the given axes for the shapes shown below (In diagram (b), the circular area is removed.)
(a) For rectangle A in the diagram below,
db3 (9.0)(4.0)3 = = = 192 cm 4 second moment of area about AA, 3 3
175 © John Bird & Carl Ross Published by Taylor and Francis
= For rectangle P, second moment of area about its centroid
db3 (3.0)(12.0)3 = = 432 cm 4 12 12
432 + (3.0)(12.0)(10.0) 2 = 432 + 3600 = 4032 cm 4 Hence, I AAQ = Thus, total second moment of area about AA, I= 4032 + 192 = 4224 cm 4 AA T
db3 (24.0)(10.0)3 = = = 8000 cm 4 (b) Second moment of area about BB, 3 3 πr 4 π ( 3.5 ) = Second moment of area about diameter of circle, I D = = 117.86 cm 4 4 4 4
Second moment of area of circle about BB = I D + AH 2 = 117.86 + ( π(3.5) 2 ) ( 5.5 )
2
= 117.86 + 1164.16 = 1282 cm 4 Thus, total second moment of area about BB, I= 8000 − 1282 (negative since circle is removed) BBT
= 6718 cm 4
= Second moment of area about CC,
b d 3 (10.0)(24.0)3 = = 46080 cm 4 3 3
Second moment of area of circle about CC = I D + AH 2 = 117.86 + ( π(3.5) 2 ) (15 )
2
= 117.86 + 8659.01 = 8777 cm 4 Thus, total second moment of area about CC, = ICCT 46080 − 8777 (negative since circle is removed)
= 37300 cm 4 176 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 58, Page 141
1. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
π ( d 2 4 − d14 ) π ( 304 − 264 ) = 17329 mm 4 Second moment= of area, I = 64 64
2. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
BD3 bd 3 40 × 303 36 × 263 − = − Second moment of area, I = 12 12 12 12 = 90000 – 52728 = 37272 mm 4
3. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
177 © John Bird & Carl Ross Published by Taylor and Francis
I Second moment of area,=
BD3 bd 3 20 × 303 9 × 263 − 2× = − 2× 12 12 12 12 = 45000 – 26364 = 18636 mm 4
4. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
BD3 bd 3 30 × 203 14 ×163 I − 2× = − 2× Second moment of area,= 12 12 12 12 = 20000 – 9557.3 = 10443 mm 4
5. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
Second moment of area, I =
BD3 bd 3 25 × 403 23 × 363 − = − 12 12 12 12 = 133333.3 – 89424 = 43909 mm 4
6. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
178 © John Bird & Carl Ross Published by Taylor and Francis
Section
a
y
ay
ay 2
1 2
40 56 96
29 14
1160 784 1944
33640 10976 44616
∑
ay ∑ = ∑a
= y = I XX
i 13.3 3659 3672
1944 = 20.25 mm 96
∑ ay + ∑ i 2
= 44616 + 3672 = 48288 mm 4
I NA = I XX − y 2 ∑ a = 48288 – 20.25 2 × 96 = I = 8922 mm 4
7. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
179 © John Bird & Carl Ross Published by Taylor and Francis
Section
a
y
ay
ay 2
i
1 2
60 36 96
19 9
1140 324 1464
21660 2916 24576
20 972 992
∑
ay ∑ = ∑a
= y = I XX
1464 = 15.25 mm 96
∑ ay + ∑ i 2
= 24576 + 992 = 25568 mm 4
I NA = I XX − y 2 ∑ a = 25568 – 15.25 2 × 96 = I = 3242 mm 4
8. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
Section 1 2 3
∑ = y = I XX
ay ∑ = ∑a
a
y
ay
ay 2
40 52 80 172
29 15 1
1160 780 80 2020
33640 11700 80 45420
i 13.3 2929.3 26.7 2969
2020 = 11.74 mm 172
∑ ay + ∑ i 2
= 45420 + 2969 = 48389 mm 4
I NA = I XX − y 2 ∑ a = 48389 – 11.74 2 × 172 = I NA = 24683 mm 4 180 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 59, Page 142 Answers found from within the text of the chapter, pages 127 to 141.
EXERCISE 60, Page 142 1. (c) 2. (b) 3. (d) 4. (a) 5. (b) 6. (c) 7. (a) 8. (c) 9. (d) 10. (a)
181 © John Bird & Carl Ross Published by Taylor and Francis
1. Find the position of the centroid of the area bounded by the curve y = 2x, the x-axis and the ordinates x = 0, x = 3
A sketch of the area is shown below.
3
2x 3 3 3 3 2 3 2 3 − 0 (27) xy dx ∫ x ( 2x ) dx ∫ 2x 2 dx 3 ∫ 0 0 0 0 3 = 3= 2 (3) = 2 = x = = = = 3 3 3 2 3 (9) 3 32 − 0 2x y dx 2xdx 2x dx ∫0 ∫0 ∫0 2 0 1 3 2 1 3 2 3 y dx 2x ) dx ( ∫ ∫ 1 3 2 1 4x 3 4 3 2 0 0 2 2 3 − 0= = y = = 4x dx = = (27) = 2 3 ∫ 0 9 18 18 3 0 54 27 y dx ∫ 0
Hence, the centroid lies at (2, 2)
2. Find the position of the centroid of the area bounded by the curve y = 3x + 2, the x-axis and the ordinates x = 0, x = 4
A sketch of the area is shown below.
161 © John Bird & Carl Ross Published by Taylor and Francis
4
3x 3 2x 2 + 3 xy dx ∫ x ( 3x + 2 ) dx ∫ (3x 2 + 2x)dx 3 2 0 42 ∫ 64 + 16 80 0 0 0 4 + = = = = = = = x 4 4 4 4 2 3 + 8 32 24 2 ∫0 y dx ∫0 (3x + 2)dx ∫0 (3x + 2) dx 3x + 2x 2 (4) + 2(4) 2 0 4
4
4
= 2.50 1 4 2 1 4 2 4 y dx ( 3x + 2 ) dx 1 4 2 ∫ ∫ 1 9x 3 12x 2 1 0 0 2 2 = = = = + + 4x= y (9x + 12x + 4)dx [192 + 96 + 16] 4 ∫ 32 64 0 64 3 2 64 y dx 0 ∫ 0
=
1 (304) = 4.75 64
Hence, the centroid lies at (2.50, 4.75) 3. Find the position of the centroid of the area bounded by the curve y = 5x2 ; x = 1, x = 4
A sketch of the area is shown below.
4
5x 4 xy dx ∫ x ( 5x 2 ) dx ∫ 5x 3dx 4 ∫ 1 1 1 1 x = = = 4= = 4 4 3 4 2 2 ∫1 y dx ∫1 5x dx ∫1 5x dx 5x 3 1 4
4
4
5 4 4 5 4 − 1 (255) 4 = 4= 318.75 = 3.036 5 3 3 5 105 4 − 1 (63) 3 3
2 1 4 2 1 4 4 y dx 5x 2 ) dx ( ∫ ∫ 1 4 1 25x 5 5 5 1 1 4 5 2 2 45 − 1= = y = = 25x = dx = (1023) 4 ∫ 1 105 210 210 5 1 210 210 y dx ∫ 1
= 24.36 Hence, the centroid lies at (3.036, 24.36)
162 © John Bird & Carl Ross Published by Taylor and Francis
4. Find the position of the centroid of the area bounded by the curve y = 2x3 , the x-axis and the ordinates x = 0, x = 2
A sketch of the area is shown below.
2
2x 5 2 2 2 xy dx ∫ x ( 2x 3 ) dx ∫ 2x 4 dx 5 ∫ 0 0 0 0 x = = = 2 = = 2 2 4 2 3 3 ∫0 y dx ∫0 2x dx ∫0 2x dx 2x 4 0
2 5 2 2 − 0 (32) 5 = 5= 2 (4) = 1.60 1 4 8 5 2 − 0 2
2 1 2 2 1 2 2 y dx 2x 3 ) dx ( ∫ ∫ 1 2 6 1 x7 1 7 1 0 0 2 2 2 − 0= y 4x dx (128) = 4.57 = = = = = 2 ∫ 8 16 0 4 7 0 28 28 y dx ∫ 0
Hence, the centroid lies at (1.60, 4.57)
5. Find the position of the centroid of the area bounded by the curve y = x(3x + 1), the x-axis and the ordinates x = - 1, x = 0
0
3x 4 x 3 + 2 3 2 xy dx ∫ x ( 3x + x ) dx ∫ (3x + x )dx 4 3 −1 ∫ −1 −1 −1 = = = = x = 0 0 0 3 2 0 2 2 ∫ −1 y dx ∫ −1 (3x + x)dx ∫ −1 (3x + x) dx 3x + x 2 −1 3 0
0
0
3 (−1)3 4 − − + (0) ( 1) 3 4 1 3 2 (0) − (−1) + 2 (−1)
3 1 5 0− − − 4 3 = 12 = − 5 = - 0.833 = 1 0 − ( −1 + 0.5 ) 6 2 163 © John Bird & Carl Ross Published by Taylor and Francis
2 1 0 2 1 0 2 y dx 3x + x dx ( ) ∫ −1 ∫ −1 y= 2 0 = 2 = 1 y dx ∫ −1 2
0
9x 5 6x 4 x 3 (9x + 6x + x )dx = 5 + 4 + 3 ∫ −1 −1 0
4
3
2
1 = (0) − −1.8 + 1.5 − = 0.633 3 Hence, the centroid lies at (- 0.833, 0.633)
164 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 56, Page 131 1. Determine the position of the centroid of a sheet of metal formed by the curve y = 4x - x2 which lies above the x-axis. y = 4x – x2 = x(4 – x) i.e. when y = 0, x = 0 and x = 4. The area of the sheet of metal is shown sketched below.
By symmetry,
x=2 4
1 16x 3 8x 4 x 5 1 4 2 1 4 1 4 2 2 2 3 4 y dx 4x − x ) dx 16x − 8x + x ) dx 2 3 − 4 + 5 ( ( ∫ ∫ ∫ 0 0 0 0 2 = y 2= = 2 = 4 4 4 4 2 2 2 x3 ∫0 y dx ∫0 ( 4x − x ) dx ∫0 ( 4x − x ) dx 2x − 3 0
1 ( 341.333 − 512 + 204.8 ) − (0) 17.0665 2 = = 1.6 = 64 10.666 32 − − (0) 3 Hence, the co-ordinates of the centroid are (2, 1.6)
2. Find the coordinates of the centroid of the area that lies between the curve
y = x - 2 and the x
x-axis.
y = x – 2 i.e. y = x(x – 2) = x 2 − 2x . The area is shown sketched below. x
165 © John Bird & Carl Ross Published by Taylor and Francis
2
x 4 2x 3 − xy dx ∫ x ( x 2 − 2x ) dx ∫ ( x 3 − 2x 2 ) dx 4 3 0 ∫ 0 0 0 = = = = x = 2 2 2 3 2 2 2 2 x 2x − − y dx x 2x dx x 2x dx ( ) ( ) ∫0 ∫0 ∫0 3 − 2 0 2
2
2
− (0) ] [(4 − 5.3333)= [(2.6667 − 4) − (0)]
−1.3333 −1.3333 =1
2
1 x 5 4x 4 4x 3 2 1 2 2 1 2 2 1 2 4 3 2 y dx x − 2x ) dx x − 4x + 4x ) dx 2 5 − 4 + 3 ( ( ∫ ∫ ∫ 0 0 0 0 2 = = 2 = y 2= 2 2 2 2 3 2 2 2 x 2x ∫0 y dx ∫0 ( x − 2x ) dx ∫0 ( x − 2x ) dx 3 − 2 0
1 [(6.4 − 16 + 10.6667) − (0)] 0.53335 2 = = - 0.4 = −1.3333 [(2.6667 − 4) − (0)]
Hence, the co-ordinates of the centroid are (1, - 0.4) 3. Determine the coordinates of the centroid of the area formed between the curve y = 9 - x2 and the x-axis. The area is shown sketched below where by symmetry, x = 0
3
2 1 3 2 1 3 1 3 2 4 y dx 9 − x 2 ) dx ( ∫ ∫ ∫ −3 (81 − 18x + x ) dx −3 −3 2 2 2 = y = = = 3 3 3 2 2 ∫ y dx ∫ ( 9 − x ) dx ∫ ( 9 − x ) dx −3
−3
−3
1 18x 3 x 5 81x − + 2 3 5 −3 3
x3 9x − 3 −3
1 [(243 − 162 + 48.6) − (−243 + 162 − 48.6)] 129.6 = 2 = 3.6 = 36 [(27 − 9) − (−27 + 9)]
Hence, the co-ordinates of the centroid are (0, 3.6)
166 © John Bird & Carl Ross Published by Taylor and Francis
4. Determine the centroid of the area lying between y = 4x2, the y-axis and the ordinates y = 0 and y = 4. The area is shown sketched below.
4
1 y2 1 1 y 2 ∫ 0 x dy 2 ∫ 0 4 dy 2 8 0 1 2 = = = From Section= 9.4, x = 0.375 4 4 8 4 y x dy ∫1 ∫ 0 4 dy 1 y3/2 3 3 2 2 0 4
4
4
∫ 0 xy dy y = and = 4 ∫ x dy 4
−
0
∫
4 0
1 y5/2 3/2 4 y 2 5 y 32 (y) dy ∫ dy 2 0 0 4= 2 = = 5 = 2.40 8 8 8 8 3 3 3 3
Hence the position of the centroid is at (0.375, 2.40)
5. Find the position of the centroid of the area enclosed by the curve y = 5x , the x-axis and the ordinate x = 5. The area is shown sketched below.
167 © John Bird & Carl Ross Published by Taylor and Francis
5
5
xy dx ∫0 = = x 5 ∫ y dx 0
∫ ( 5
0
x
∫
5 0
)
5x dx = 5x dx
5 x2 5 3 5 2 5 ∫ x dx 2 0 0 = = 1 5 5 3 5 ∫ x 2 dx 2 x 0 3 2 0
2 52 5 − 0 5 = 3 2 2 5 − 0 3
2 (55.9017) 22.3607 5 = = 3.0 2 (11.1802) 7.45356 3
5
(
)
2 1 5 2 1 5 y dx 5x dx ∫ ∫ 0 2 0 = y 2= = 5 5 y dx ∫ ∫ 5x dx 0
0
1 5 5x dx 0 2 ∫=
( 5)∫
5 0
1 2
x dx
1 5x 2 1 125 − 0 2 2 0 31.25 2 2 = 1.875 = = 16.66667 16.66667 5 7.45356
( )
Hence, the centroid lies at (3.0, 1.875) 6. Sketch the curve y2 = 9x between the limits x = 0 and x = 4. Determine the position of the centroid of this area. The curve y2 = 9x is shown sketched below.
By symmetry, y = 0 4
5 3x 2 5 3 4 4 4 6 5 6 5 6 2 x 4 − 0 (32) xy dx ∫ x 3 x dx ∫ 3x dx 2 ∫ 0 0 0 0 5= 5 = 5 x = = = = = 4 4 1 4 3 3 4 ∫0 y dx ∫0 3 x dx ∫0 3x 2 dx 3x 32 2 x 2 4 − 0 2(8) 3 2 0 38.4 = = 2.4 16
(
)
Hence, the centroid is at (2.4, 0) 168 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 57, Page 137
1. Determine the second moment of area and radius of gyration for the rectangle shown below about (a) axis AA (b) axis BB, and (c) axis CC
From Table 9.1, page 133 of textbook, the second moment of area about axis AA, Similarly, and
I AA = k AA =
I CC = Radius of gyration, k CC =
(8.0)(3.0)3 db3 = = 72 cm4 3 3 b 3.0 = = 1.73 cm 3 3 bd 3 (3.0)(8.0)3 = = 512 cm 4 3 3
d 8.0 = = 4.62 cm 3 3
The second moment of area about the centroid of a rectangle is
bd 3 when the axis through the 12
centroid is parallel with the breadth b. Hence
I BB =
and
k BB =
(3.0)(8.0)3 bd 3 = = 128 cm4 12 12
d 8.0 = = 2.31 cm 12 12
2. Determine the second moment of area and radius of gyration for the triangle shown below about (a) axis DD (b) axis EE, and (c) an axis through the centroid of the triangle parallel to axis DD
169 © John Bird & Carl Ross Published by Taylor and Francis
From Table 9.1, page 133 of textbook:
(12.0 )( 9.0 )
b h3 I DD = (a) Second moment of area about DD, = 12 Radius of gyration about DD, k= DD
12
= 729 cm 4
h 9.0 = 3.67 cm = 6 6
b h3 I EE = (b) Second moment of area about EE, = 4 Radius of gyration about EE, k= EE
3
(12.0 )( 9.0 )
3
4
= 2187 cm 4
h 9.0 = = 6.36 cm 2 2
b h3 = = (c) Second moment of area about axis through centroid, 36 Radius of gyration about axis through centroid, =
h = 18
(12.0 )( 9.0 )
3
= 243 cm 4
36
9.0 = 2.12 cm 18
3. For the circle shown below, find the second moment of area and radius of gyration about (a) axis FF and (b) axis HH
π r 4 π ( 4.0 ) = = = 201 cm 4 4 4 4
(a) Second moment of area coinciding with the diameter, I FF
Radius of gyration coinciding with the diameter, k FF =
r 4.0 = = 2.0 cm 2 2
170 © John Bird & Carl Ross Published by Taylor and Francis
5π r 4 5π ( 4.0 ) = (b) Second moment of area coinciding with a tangent, I HH = = 1005 cm 4 4 4 4
5 5 r= (4.0) = 4.47 cm 2 2
Radius of gyration coinciding with a tangent, k HH =
4. For the semicircle shown below, find the second moment of area and radius of gyration about axis JJ
π r 4 π (10.0 ) = = = 3927 mm 4 8 8 4
Second moment of area coinciding with the diameter, I JJ
Radius of gyration coinciding with the diameter, k JJ =
r 10.0 = = 5.0 mm 2 2
5. For each of the areas shown below determine the second moment of area and radius of gyration about axis LL, by using the parallel axis theorem.
bl3 (3.0)(5.0)3 + (3.0)(5.0)(2.5 + 2.0) 2 (a) Second moment of area, I LL =IGG + Ad 2 = + Ad 2 = 12 12 = 31.25 + 303.75 = 335 cm 4
k LL I LL = Ak LL 2 from which, radius of gyration,=
I LL = area
335 = 4.73 cm 15.0 171
© John Bird & Carl Ross Published by Taylor and Francis
(b) Second moment of area, I= IGG + Ad 2 LL
I= GG
bh 3 (18)(12)3 = = 864 cm 4 where h = 152 − 92 = 12 cm, as shown in the diagram below. 36 36
1 (18)(12) = 108 cm 2 2 2 12 2 =IGG + Ad =864 + 108 10 + =864 + 108(14) 2 = 22032 cm 4 = 22032 cm 4 3 correct to 4 significant figures.
Hence, area of triangle, A = Thus, I LL
k LL I LL = Ak LL 2 from which, radius of gyration,=
22032 = 14.3 cm 108
I LL = area
πr 4 4 π(2.0) 4 (c) Second moment of area, I LL = IGG + Ad= + ( πr 2 ) 5 + = + π(2.0) 2 (7) 2 4 2 4 2
2
= 12.57 + 615.75 = 628 cm 4
k LL I LL = Ak LL 2 from which, radius of gyration,=
I LL = area
628 = 7.07 cm 2 π(2.0)
6. Calculate the radius of gyration of a rectangular door 2.0 m high by 1.5 m wide about a vertical axis through its hinge.
From Table 9.1, page 133 of textbook, radius of gyration of rectangle, coinciding with length (i.e. height in this case) =
b 1.5 = 3 3 = 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.
172 © John Bird & Carl Ross Published by Taylor and Francis
From Table 9.1, page 133 of textbook,
5π r 4 5πd 4 5π (1.0 ) = 0.245 m 4 second moment of area coinciding with the tangent = = = 4 64 64 4
and radius of gyration of rectangle coinciding with tangent =
5 5 1.0 r= = 0.559 m 2 2 2
8. A circular cover, centre 0, has a radius of 12.0 cm. A hole of radius 4.0 cm and centre X, where 0X = 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 0X.
Second moment of area about diameter, i.e. axis CC in the diagram below
ICC =
=
πr 4 πr 4 − I DD = − IGG + Ad 2 4 4
π(12) 4 π(4.0) 4 − + π(4.0) 2 (6.0) 2 = 16286 – [201 + 1810] = 14275 = 14280 cm 4 , correct 4 4 to 4 significant figures.
ICC = Ak CC 2 from which, radius of gyration,
= k CC
ICC = area
14275 = 2 2 π(12.0) − π(4.0)
14275 = 5.96 cm 128π
9. For the sections shown below, find the second moment of area and the radius of gyration about axis XX
173 © John Bird & Carl Ross Published by Taylor and Francis
(a) For rectangle A in the diagram below,
bl3 (18.0)(3.0)3 = = 40.5 m 4 second moment of area about C= A 12 12 Hence, I XXA = 40.5 + Ad 2 = 40.5 + (3.0)(18.0)(12.0 + 1.5) 2 = 40.5 + 9841.5 = 9882 mm 4
For rectangle B, second moment of area about C= B
bl3 (4.0)(12.0)3 = = 576 m 4 12 12
Hence, I XXB = 576 + (4.0)(12.0)(6.0) 2 = 576 + 1728 = 2304 mm 4 Thus, total second moment of area about XX, I XXT = 9882 + 2304 = 12186 = 12190 mm 4 , correct to 4 significant figures. Radius of gyration about XX,= k XX
I XXT = area
12186 = 10.9 mm 54 + 48
(b) Rectangle A (see diagram below): Second moment of area about C= A
bl3 (6.0)(2.0)3 = = 4 cm 4 12 12
I XXA = 4 + (2.0)(6.0)(6.0) 2 = 4 + 432 = 436 cm 4
174 © John Bird & Carl Ross Published by Taylor and Francis
Rectangle B:
bl3 (2.5)(3.0)3 = = 5.625cm 4 12 12
Second moment of area about C= B
= I XXB 5.625 + (2.5)(3.0)(3.5) 2 = 97.5 cm 4 Rectangle C:
bl3 (6.0)(2.0)3 = = 4 cm 4 12 12
Second moment of area about C= C
I XXC = 4 + (6.0)(2.0)(1.0) 2 = 16 cm 4 Hence, total second moment of area about axis XX, I XXI = 436 + 97.5 + 16 = 549.5 cm 4 Radius of gyration about XX,= k XX
I XXI = area
549.5 = 4.18 cm 12 + 7.5 + 12
10. Determine the second moments of areas about the given axes for the shapes shown below (In diagram (b), the circular area is removed.)
(a) For rectangle A in the diagram below,
db3 (9.0)(4.0)3 = = = 192 cm 4 second moment of area about AA, 3 3
175 © John Bird & Carl Ross Published by Taylor and Francis
= For rectangle P, second moment of area about its centroid
db3 (3.0)(12.0)3 = = 432 cm 4 12 12
432 + (3.0)(12.0)(10.0) 2 = 432 + 3600 = 4032 cm 4 Hence, I AAQ = Thus, total second moment of area about AA, I= 4032 + 192 = 4224 cm 4 AA T
db3 (24.0)(10.0)3 = = = 8000 cm 4 (b) Second moment of area about BB, 3 3 πr 4 π ( 3.5 ) = Second moment of area about diameter of circle, I D = = 117.86 cm 4 4 4 4
Second moment of area of circle about BB = I D + AH 2 = 117.86 + ( π(3.5) 2 ) ( 5.5 )
2
= 117.86 + 1164.16 = 1282 cm 4 Thus, total second moment of area about BB, I= 8000 − 1282 (negative since circle is removed) BBT
= 6718 cm 4
= Second moment of area about CC,
b d 3 (10.0)(24.0)3 = = 46080 cm 4 3 3
Second moment of area of circle about CC = I D + AH 2 = 117.86 + ( π(3.5) 2 ) (15 )
2
= 117.86 + 8659.01 = 8777 cm 4 Thus, total second moment of area about CC, = ICCT 46080 − 8777 (negative since circle is removed)
= 37300 cm 4 176 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 58, Page 141
1. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
π ( d 2 4 − d14 ) π ( 304 − 264 ) = 17329 mm 4 Second moment= of area, I = 64 64
2. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
BD3 bd 3 40 × 303 36 × 263 − = − Second moment of area, I = 12 12 12 12 = 90000 – 52728 = 37272 mm 4
3. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
177 © John Bird & Carl Ross Published by Taylor and Francis
I Second moment of area,=
BD3 bd 3 20 × 303 9 × 263 − 2× = − 2× 12 12 12 12 = 45000 – 26364 = 18636 mm 4
4. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
BD3 bd 3 30 × 203 14 ×163 I − 2× = − 2× Second moment of area,= 12 12 12 12 = 20000 – 9557.3 = 10443 mm 4
5. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
Second moment of area, I =
BD3 bd 3 25 × 403 23 × 363 − = − 12 12 12 12 = 133333.3 – 89424 = 43909 mm 4
6. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
178 © John Bird & Carl Ross Published by Taylor and Francis
Section
a
y
ay
ay 2
1 2
40 56 96
29 14
1160 784 1944
33640 10976 44616
∑
ay ∑ = ∑a
= y = I XX
i 13.3 3659 3672
1944 = 20.25 mm 96
∑ ay + ∑ i 2
= 44616 + 3672 = 48288 mm 4
I NA = I XX − y 2 ∑ a = 48288 – 20.25 2 × 96 = I = 8922 mm 4
7. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
179 © John Bird & Carl Ross Published by Taylor and Francis
Section
a
y
ay
ay 2
i
1 2
60 36 96
19 9
1140 324 1464
21660 2916 24576
20 972 992
∑
ay ∑ = ∑a
= y = I XX
1464 = 15.25 mm 96
∑ ay + ∑ i 2
= 24576 + 992 = 25568 mm 4
I NA = I XX − y 2 ∑ a = 25568 – 15.25 2 × 96 = I = 3242 mm 4
8. Determine the second moments of area about a horizontal axis, passing through the centroid, for the ‘built-up’ section shown below. Dimensions are in mm and the thickness is 2 mm.
Section 1 2 3
∑ = y = I XX
ay ∑ = ∑a
a
y
ay
ay 2
40 52 80 172
29 15 1
1160 780 80 2020
33640 11700 80 45420
i 13.3 2929.3 26.7 2969
2020 = 11.74 mm 172
∑ ay + ∑ i 2
= 45420 + 2969 = 48389 mm 4
I NA = I XX − y 2 ∑ a = 48389 – 11.74 2 × 172 = I NA = 24683 mm 4 180 © John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 59, Page 142 Answers found from within the text of the chapter, pages 127 to 141.
EXERCISE 60, Page 142 1. (c) 2. (b) 3. (d) 4. (a) 5. (b) 6. (c) 7. (a) 8. (c) 9. (d) 10. (a)
181 © John Bird & Carl Ross Published by Taylor and Francis
