CHAPTER 76 SECOND MOMENTS OF AREA
EXERCISE 293 Page 800
1. Determine the second moment of area and radius of gyration for the rectangle shown about
(a) axis AA (b) axis BB, and (c) axis CC.
From Table 76.1, page 797:
l b3
(a) Second moment of area about AA, I=
=
AA
3
Radius of gyration about AA, k=
AA
3
l
=
12
( 3.0 )(8.0 )
= 72 cm 4
3
12
= 128 cm 4
8.0
= 2.31 cm
12
b l3
(c) Second moment of area about axis CC, I=
=
CC
3
Radius of gyration about CC, k=
CC
3
b
3.0
=
= 1.73 cm
3
3
b l3
(b) Second moment of area about BB, I=
=
BB
12
k BB
Radius of gyration about BB, =
(8.0 )( 3.0 )
( 3.0 )(8.0 )
3
3
= 512 cm 4
l
8.0
=
= 4.62 cm
3
3
2. Determine the second moment of area and radius of gyration for the triangle shown about
(a) axis DD (b) axis EE and (c) an axis through the centroid of the triangle parallel to axis DD.
1182
© 2014, John Bird
From Table 76.1, page 797:
(12.0 )( 9.0 )
b h3
(a) Second moment of area about DD, I=
=
DD
12
Radius of gyration about DD, k=
DD
= 729 cm 4
12
h
9.0
=
= 3.67 cm
6
6
b h3
(b) Second moment of area about EE, =
I 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 )
36
3
= 243 cm 4
9.0
= 2.12 cm
18
3. For the circle shown, find the second moment of area and radius of gyration about (a) axis FF,
and (b) axis HH.
Second moment of area about diameter FF =
Radius of gyration about diameter FF, k=
π r4
4
=
π ( 4.0 )
4
4
= 201 cm 4
r 4.0
= 2.0 cm
=
2
2
5π r 4 5π ( 4.0 )
Second moment of area about a tangent =
= 1005 cm 4
=
4
4
4
Radius of gyration about tangent,
=
k
5
=
r
2
5
(4.0) = 4.47 cm
2
1183
© 2014, John Bird
4. For the semicircle shown, find the second moment of area and radius of gyration about axis JJ.
From Table 76.1, page 797:
Second moment of area about axis JJ, =
I JJ
Radius of gyration about JJ, k JJ=
π r4
=
8
π (10.0 )
8
4
= 3927 mm 4
r 10.0
= 5.0 mm
=
2
2
5. For each of the areas shown, determine the second moment of area and radius of gyration about
axis LL, by using the parallel axis theorem.
(3.0)(5.0)3
bl 3
(a) Second moment of area, I LL =I GG + Ad 2 = + Ad 2 =
+ (3.0)(5.0)(2.5 + 2.0) 2
12
12
= 31.25 + 303.75 = 335 cm 4
I LL = Ak LL 2 from which, radius of gyration,
=
k LL
I LL
=
area
 335 

 = 4.73 cm
 15.0 
(b) Second moment of area, I=
I GG + Ad 2
LL
I=
GG
bh3 (18)(12)3
=
= 864 cm 4 where h = 152 − 92 = 12 cm, as shown in the diagram below.
36
36
1184
© 2014, John Bird
Hence, area of triangle, A =
1
(18)(12) = 108 cm 2
2
2
Thus, I LL
12 

=I GG + Ad =864 + 108 10 +  =864 + 108(14) 2 = 22032 cm 4 = 22 030 cm 4
3

correct to 4 significant figures
2
I LL = Ak LL 2 from which, radius of gyration,
=
k LL
(c) Second moment of area, I LL
 22 032 

 = 14.3 cm
 108 
4  π (2.0) 4

= I GG + Ad =
+ (π r )  5 +  =
+ π (2.0) 2 (7) 2
4
2
4


2
π r4
I LL
=
area
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 
(2.0)
π


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.
k
From Table 76.1, radius of gyration about length of rectangular door,=
b 1.5
= = 0.866 m
3
3
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.
From Table 76.1, page 797:
5π r 4 5π ( 0.5 )
Second moment of area about a tangent =
= 0.245 m 4
=
4
8
4
Radius of gyration about tangent,
=
k
5
=
r
2
5
(0.5) = 0.559 m
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
1185
© 2014, John Bird
I CC =
=
π r4
4
− I DD =
π (12) 4
4
π r4
4
− [ I GG + Ad 2 ]
 π (4.0) 4

−
+ π (4.0) 2 (6.0) 2  = 16 286 – [201 + 1810] = 14 275 = 14 280 cm 4 ,
 4

correct to 4 significant figures
I CC = AkCC 2 from which, radius of gyration,
=
kCC
I CC
=
area


14 275
=


2
2
 π (12.0) − π (4.0) 
1186
14 275
= 5.96 cm
128π
© 2014, John Bird
EXERCISE 294 Page 802
1. For the sections shown, find the second moment of area and the radius of gyration about axis
XX.
(a)
(b)
(a) For rectangle A in the diagram below,
second moment of area about C
=
A
bl 3 (18.0)(3.0)3
=
= 40.5 m 4
12
12
Hence, I XX A = 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
bl 3 (4.0)(12.0)3
=
= 576 m 4
12
12
Hence, I XX B =
576 + (4.0)(12.0)(6.0) 2 =
576 + 1728 = 2304 mm 4
Thus, total second moment of area about XX, I XX T = 9882 + 2304 = 12 186 = 12 190 mm 4 ,
correct to 4 significant figures
Radius of gyration about XX,=
k XX
I XX T
=
area
12186
= 10.9 mm
54 + 48
(b) Rectangle A (see diagram below):
Second moment of area about C
=
A
bl 3 (6.0)(2.0)3
=
= 4 cm 4
12
12
I XX A =
4 + (2.0)(6.0)(6.0) 2 =
4 + 432 = 436 cm 4
1187
© 2014, John Bird
Rectangle B:
Second moment of area about C
=
B
bl 3 (2.5)(3.0)3
=
= 5.625cm 4
12
12
=
I XX B 5.625 + (2.5)(3.0)(3.5) 2 = 97.5 cm 4
Rectangle C:
Second moment of area about C
=
C
bl 3 (6.0)(2.0)3
=
= 4 cm 4
12
12
I XX C = 4 + (6.0)(2.0)(1.0) 2 = 16 cm 4
Hence, total second moment of area about axis XX, I XX I = 436 + 97.5 + 16 = 549.5 cm 4
Radius of gyration about XX,=
k XX
I XX I
=
area
549.5
= 4.18 cm
12 + 7.5 + 12
2. Determine the second moments of areas about the given axes for the shapes shown below.
(In Figure (b), the circular area is removed.)
(a) Shape A is divided into two rectangles, X and Y as shown below
1188
© 2014, John Bird
For rectangle X in the diagram ,
second moment of area about C=
X
bl 3 (3.0)(16.0 − 4.0)3
=
= 432 cm 4
12
12
Hence, I AAX =432 + Ad 2 =432 + (12.0)(3.0)(4.0 + 8.0) 2 =432 + 3600 = 4032 cm 4
For rectangle Y, second moment of area about C
=
Y
bl 3 (9.0)(4.0)3
=
= 48cm 4
12
12
Hence, I AAY =
48 + (4.0)(9.0)(2.0) 2 =
48 + 144 = 192 cm 4
Thus, total second moment of area about AA, =
I AAT 4032 + 192 = 4424 cm 4 ,
(b) For diagram B,
second moment of area of rectangle about BB
=
and second moment of area of circle about BB =
l b3 (9.0 + 15.0)(10.0)3
=
= 8000 cm 4
3
3
π r4
4
π ( 3.5 )
+ Ad
=
2
4
4
+ (π × 3.52 )(10.0 − 4.5 )
2
= 117.86 + 1164.16 = 1282 cm 4
Since the circular area is removed, second moment of area about BB,
I BBT = 8000 – 1282 = 6718 cm 4
Second moment of area of rectangle about BB
=
and second moment of area of circle about BB =
b l 3 (10.0)(24.0)3
=
= 46 080 cm 4
3
3
π r4
4
π ( 3.5 )
+ Ad
=
2
4
4
+ (π × 3.52 )(15.0 )
2
= 117.86 + 8659.01 = 8777 cm 4
Since the circular area is removed, second moment of area about BB,
I CCT = 46 080 – 8777 = 37 300 cm 4
1189
© 2014, John Bird
3. Find the second moment of area and radius of gyration about the axis XX for the beam section
shown below.
Second moment of area about axis XX,
I XX = I XX A + I XX B + I XX C
 ( 6.0 )(1.0 )3
  ( 2.0 )( 8.0 )3
  (10.0 )( 2.0 )3

2
2
= 
+ (6.0)(10.5)  + 
+ (16)(6.0)  + 
+ (20)(1.0) 2 
12
12
12

 
 

= 662 + 661.33 + 26.66 = 1350 cm 4
Radius of gyration about XX,=
k XX
I XX
=
area
1350
= 5.67 cm
6 + 16 + 20
1190
© 2014, John Bird