10.2 Method of Composite Areas
10.2 Method of Composite Areas Example 1, page 1 of 2
1. Determine the moment of inertia of the crosshatched region
about the x axis.
y
50 mm
x
20 mm
y
1 Consider the crosshatched region to
be composed of the difference of
two circular regions.
y
y
=
x
x
x
10.2 Method of Composite Areas Example 1, page 2 of 2
2
A table of properties of planar regions gives the
information shown below.
3
For our particular problem,
1
(50 mm)4
4
= 4.9087 106 mm4
1
Small circle Ix =
(20 mm)4
4
= 0.1257 106 mm4
Large circle Ix =
Ix = 1 r4
4
Iy = 1 r4
4
Moment of Inertia
y
For the composite region, subtracting gives
r
C
x
Ix = Large circle Ix
= 4.9087
= 4.78
Circle
Small circle Ix
106 mm4
106 mm4
0.1257
106 mm4
Ans.
10.2 Method of Composite Areas Example 2, page 1 of 2
2. The figure shows the cross section of a
beam made by gluing four planks together.
Determine the moment of inertia of the
cross section about the x axis.
y
150 mm 150 mm
60 mm
200 mm
x
200 mm
60 mm
60 mm
1
60 mm
Consider the crosshatched region to be consist of a
small rectangle subtracted from a large rectangle.
y
60 mm + 200 mm + 200 mm + 60 mm = 520 mm y
200 mm + 200 mm = 400 mm
y
=
x
x
520 mm
300 mm
420 mm
60 mm + 150 mm + 150 mm + 60 mm = 420 mm
Large rectangle
x
400 mm
150 mm + 150 mm = 300 mm
Small rectangle
10.2 Method of Composite Areas Example 2, page 2 of 2
y
2 A table of properties of planar regions
gives the information shown below.
3
I x = bh
12
3
For our particular problem,
3
Large rectangle Ix = bh
12
Moment of Inertia
(420 mm)(520 mm)3
=
12
y
3
Iy = hb
12
b
2
b
2
= 4.9213
x
520 mm
109 mm4
h
2
420 mm
x
y
C
h
2
4
For our particular problem,
3
Small rectangle Ix = bh
12
Rectangle
x
400 mm
3
=
(300 mm)(400 mm)
12
= 1.6000
109 mm4
300 mm
5
For the composite region, subtracting gives
Ix = Large rectangle Ix
= 4.9213
= 3.32
109 mm4
109 mm4
Small rectangle Ix
1.6000
109 mm4
Ans.
10.2 Method of Composite Areas Example 3, page 1 of 3
3. Determine the moment of inertia of the
beam cross section about the x centroidal
axis.
y
80 mm 80 mm
20 mm
120 mm
x
120 mm
20 mm
20 mm
10.2 Method of Composite Areas Example 3, page 2 of 3
1
Consider the cross section to be composed of a large rectangle minus two small rectangles.
y
y
y
120 mm + 120 mm = 240 mm
80 mm 80 mm
20 mm
120 mm
y
=
x
x
280 mm
x
240 mm
x
240 mm
120 mm
20 mm
20 mm
80 mm
180 mm
80 mm
80 mm + 20 mm + 80 mm = 180 mm
20 mm + 120 mm + 120 mm + 20 mm = 280 mm
y
2
y
=
x
2
x
Two regions of the same size and same position
relative to the x axis can be combined into 2
times a single region.
10.2 Method of Composite Areas Example 3, page 3 of 3
3
3
Large rectangle Ix = bh
12
y
(180 mm)(280 mm)3
=
12
= 3.2928
108 mm4
x
280 mm
180 mm
y
3
4 Small rectangle Ix = bh
12
240 mm
(80 mm)(240 mm)3
=
12
= 0.9216
x
108 mm4
5
80 mm
For the composite region, subtracting gives
Ix = Large rectangle Ix
= 3.2928
= 1.450
108 mm4
108 mm4
2
Small rectangle Ix
2(0.9216
108 mm4)
Ans.
10.2 Method of Composite Areas Example 4, page 1 of 3
4. A composite beam is constructed
from three plates and four standard
rolled-steel angles. Determine the
moment of inertia of the cross
section about the x centroidal axis.
y
110 mm 110 mm
10 mm
C
140 mm
C
xc
A = 877 mm2
Ixc = 0.202
16.2 mm
x
140 mm
10 mm
10 mm
1
Consider the cross section to be composed of four angles and three rectangles.
y
y
y
=
x
+2
+4
x
x
2
y
The top and bottom
plates are identical.
x
3 The four angles
are identical.
106 mm4
10.2 Method of Composite Areas Example 4, page 2 of 3
4
y
Middle rectangle
3
Middle rectangle Ix = bh
12
140 mm
3
=
(10 mm)(140 mm + 140 mm)
12
= 1.8293
107 mm4
x
C'
(1)
140 mm
10 mm
5
Upper rectangle
Use parallel axis theorem:
Ix = Ixc' + d2A
(2)
Area
Distance between xc' and x
Moment of inertia
about axis xc' through
centroid of rectangle
6
Here,
y
10 mm
= 5 mm
2
xc'
110 mm 110 mm
C'
140 mm
Ixc' =
(110 mm + 110 mm)(10 mm)3
bh3
=
12
12
= 18333 mm4
d = 140 mm + 5 mm = 145 mm
10 mm
A = (110 mm + 110 mm)(10 mm)
x
= 2200 mm2
10.2 Method of Composite Areas Example 4, page 3 of 3
7 The parallel axis theorem, Eq. 2, now gives
9
Upper rectangle Ix = Ixc' + d2A
The parallel axis theorem gives
Angle Ix = Ixc + d2A
= 18333 mm4 + (145 mm)2(2200 mm2)
107 mm4
= 4.6273
8
= 0.202
= 1.3643
y
107 mm4
(3)
16.2 mm
Angle
C
106 mm4 + (123.8 mm)2(877 mm2)
xc
140 mm
A = 877 mm2
16.2 mm
Ixc = 0.202
C'
d = 140 mm
x
106 mm4
10 For the composite region, adding gives
Ix = Middle rectangle Ix + 2
= 1.8293
= 165.4
Upper rectangle Ix + 4
107 mm4 + 2(4.6273
106 mm4
Angle Ix
107 mm4) + 4(1.3643
Ans.
107 mm4)
16.2 mm = 123.8 mm
10.2 Method of Composite Areas Example 5, page 1 of 4
5. Determine the moment of
inertia of the trapezoidal region
about the x and y axes.
y
3 in.
3 in.
6 in.
x
4 in.
1
4 in.
Consider the trapezoid to be the sum of a rectangle and two triangles.
y
y
3 in.
3 in.
3 in.
y
3 in.
=
6 in.
6 in.
x
4 in.
4 in.
+2
6 in.
x
x
3 in.
1 in.
10.2 Method of Composite Areas Example 5, page 2 of 4
2
Ix and Iy for the rectangle
Since the centroid C' does not lie on the x axis, we have to use
the parallel axis theorem to calculate Ix.
3
y
Rectangle Ix = Ixc' + d2A
(6 in.)(3 in. + 3 in.)3
=
12
3 in.
+ (3 in.)2[(3 in. + 3 in.)(6 in.)]
xc'
C'
= 432 in4
3 in.
6 in.
x
4
(1)
Since the centroid C' lies on the y axis, we do not have to
use the parallel axis theorem for Iy.
Rectangle Iy = Iyc'
(3 in. + 3 in.)(6 in.)3
=
12
= 108 in4
(2)
10.2 Method of Composite Areas Example 5, page 3 of 4
5
Ix and Iy for the triangle
6
bh3
Ixc' =
36
For our particular triangle
y
yc'
A table of properties of planar regions gives the
information below.
=
3
Ixx = bh
36
3
IBB = hb
12
Moment of Inertia
= 6 in4
6 in.
Area =
xc'
C'
bh
2
2h
3
x
x
C
h
3
1 in.
x
B
B
b
Triangle
(1 in.)(6 in.)3
36
b'(h')3
Iyc' =
36
=
(6 in.)(1 in.)3
36
= 0.1667 in4
7 Parallel axis theorem applied to triangle
y
yc'
Triangle Ix = Ixc' + dx2A
= 6 in4 + (2 in.)2(3 in2)
Area, A =
6 in.
1
(1 in.)(6 in.)
2
= 3 in2
1
dy = 3 in. +
in. = 3.3333 in.
3
= 18 in4
xc'
1
in.
3
3 in.
C'
dx =
1 in.
6 in.
3 = 2 in.
x
(3)
Triangle Iy = Iyc' + d2A
= 0.1667 in4
+ (3.3333 in.)2(3 in2)
= 33.4994 in4
(4)
10.2 Method of Composite Areas Example 5, page 4 of 4
8 For the composite region, using Eqs. 1, 2, 3, and 4 gives
Ix = Rectangle Ix + 2
Triangle Ix
= 432 in4 + 2(18 in4)
= 468 in4
`
Iy = Rectangle Iy + 2
Ans.
Triangle Iy
= 108 in4 + 2(33.4994 in4)
= 175 in4
Ans.
10.2 Method of Composite Areas Example 6, page 1 of 5
6. Determine the moment of inertia of the
crosshatched region about the y axis.
2 in.
y
4 in.
4 in.
0.8 in.
2 in.
0.8 in.
x
1
Consider the crosshatched region to be composed of a rectangle
minus two circular regions plus two semicircular regions.
y
y
=
x
x
y
y
+2
2
x
x
10.2 Method of Composite Areas Example 6, page 2 of 5
2
Iy for rectangle
y
4 in.
4 in.
2 in.
x
C'
2 in.
3
Since the y axis passes through the centroid of the
rectangle, the parallel axis theorem is not needed.
Rectangle Iy = Iyc'
(2 in. + 2 in.)(4 in. + 4 in.)3
=
12
= 170.6667 in4
(1)
10.2 Method of Composite Areas Example 6, page 3 of 5
4
For the circular region, a table of properties of planar
regions gives the information shown below.
yc'
y
d = 4 in.
Ix = 1 r4
4
Iy = 1 r4
4
C'
Moment of Inertia
r = 0.8 in.
x
y
5
For our particular problem,
Iyc' =
r
C
1
4
(0.8 in.)4
= 0.3217 in4
x
Area, A = r2
= (0.8 in.)2
Circle
= 2.0106 in2
6
Applying the parallel axis theorem gives
Circle Iy = Iyc' + d2A
= (0.3217 in4) + (4 in.)2(2.0106 in2)
= 32.4913 in4
(2)
10.2 Method of Composite Areas Example 6, page 4 of 5
7
For the semicircle, a table of properties of
planar regions gives the information shown
below.
Ix =
r4
8
y
B
r4
8
4r
yc =
3
8
4 in.
4r
3
C
IBB =
r
x
Semicircle
r4
8
9
(3)
to compute Iy for the semicircle.
Unfortunately, the table gives us the
moment of inertia with respect to the
base, BB, of the semicircle, not with
respect to the axis through the
centroid yc'.
B
y
Iy = Iyc' + d2A
x
C'
yc
We would like to apply the
parallel-axis theorem:
r = 2 in.
Moment of Inertia
Iy =
yc'
But we can still make use of the result IBB from the table by
applying the parallel axis theorem between the BB axis and
the yc ( yc' ) axis :
IBB = Iyc' + d2A
or,
r4
4r 2 r2
=
I
+
(
)(
)
yc'
8
3
2
Solving gives
Iyc' = (
8
8 4
)r
9
(4)
10.2 Method of Composite Areas Example 6, page 5 of 5
yc'
y
4r
3
r = 2 in.
x
C'
4 in.
10 Eqs. 3 and 4 can now be applied to the semicircular region
Semicircle Iy = Iyc' + d2A
=(
8 )r4 + (4 in. + 4r )2( r2)
3
9
2
8
Substituting r = 2 in. and evaluating the resulting expression gives
Semicircle Iy = 149.4808 in4
(5)
11 For the composite region, using Eqs. 1, 2, and 5 gives
Iy = Rectangle Iy
= 170.6667 in4
= 405 in4
2
Circle Iy + 2
Semicircle Iy
2(32.4913 in4) + 2(149.4808 in4)
Ans.
10.2 Method of Composite Areas Example 7, page 1 of 5
7. A precast concrete floor beam has the cross section shown.
Locate the centroid of the section and determine the moment of
inertia about a horizontal axis through the centroid.
y
250 mm
300 mm
300 mm
250 mm
75 mm
425 mm
50 mm
50 mm
x
1
Definition of centroid
Xc = 0, by symmetry
Yc =
ycA
A
(1)
where yc is the centroidal coordinate of the
region with area A.
10.2 Method of Composite Areas Example 7, page 2 of 5
2
Consider the section to be composed of a horizontal
rectangle and and two identical vertical rectangles.
y
y
=
x
x
y
+2
x
10.2 Method of Composite Areas Example 7, page 3 of 5
y
Upper rectangle
3
C' (centroid)
75 mm
A = (1200 mm)(75 mm)
=9
75 mm
= 37.5 mm
2
104 mm2
yc' = 425 mm + 37.5 mm
yc'
425 mm
= 462.5 mm
x
2
4
yc'
y
Lower rectangle
(250 mm + 50 mm + 300 mm) = 1200 mm
A = (425 mm)(50 mm)
C' (centroid)
4
2
= 2.125 10 mm
425 mm
yc' =
2
425 mm
yc'
= 212.5 mm
x
50 mm
5
Set up table
Region
upper rectangle
lower rectangles
2 lower rectangles
A ( mm2 ) yc' ( mm )
9.000
2(2.125
104
104)
A = 13.250
104
462.5
212.5
yc'A ( mm3 )
41.625 106
9.031 106
yc'A = 50.656 106
10.2 Method of Composite Areas Example 7, page 4 of 5
6
Eq. 1 gives the distance to the centroid of the entire cross section.
ycA
A
50.656
=
13.250
Yc =
106
104
= 382.31 mm
75 mm = 37.5 mm
2
Ans.
y
7
Ix of upper rectangle
C' (centroid of rectangle)
75 mm
3
bh
12
(1200 mm)(75 mm)3
=
12
Ixc' =
xc'
d
107 mm4
= 4.2188
xc
C (centroid of entire section)
425 mm
Yc = 382.31 mm
d = 75 mm + 425 mm
(37.5 mm + 382.31 mm)
x
= 80.19 mm
Parallel axis theorem:
Upper rectangle Ixc = Ixc' + d2A
Area A was calculated previously (See the table).
= 4.2188
107 mm4 + (80.19 mm)2(9
= 6.2093
108 mm4
104 mm2)
(2)
10.2 Method of Composite Areas Example 7, page 5 of 5
y
Ix of lower rectangle
8
yc'
bh3
12
(50 mm)(425 mm)3
=
12
C (centroid of entire section)
Ixc' =
= 3.1986
8
xc
d
4
10 mm
d = 382.31 mm
425 mm
425 mm
= 212.5 mm
2
212.5 mm
x
50 mm C' (centroid of rectangle)
= 169.81 mm
Yc = 382.31 mm
Parallel axis theorem:
Lower rectangle Ixc = Ixc' + d2A
9
Area A was calculated previously.
= 3.1986
108 mm4 + (169.81 mm)2(2.125
= 9.3261
108 mm4
(3)
For the composite region, using Eqs. 1and 2 gives
Ixc = Upper rectangle Ixc + 2
= 6.2093
= 249
Lower rectangle Ixc
108 mm4 + 2(9.3261
107 mm4
xc'
108 mm4)
Ans.
104 mm2)
10.2 Method of Composite Areas Example 8, page 1 of 6
8. A beam is built up from two standard rolled-steel channels and a
cover plate. Locate the centroid of the section and determine the
moments of inertia with respect to horizontal and vertical axes through
the centroid.
y
100 mm
100 mm
yc
Centroid of channel
40 mm
127 mm
254 mm
A = 3780 mm2
C
xc
127 mm
x
15.3 mm
1
Definition of centroid
Xc = 0, by symmetry
Yc =
ycA
A
(1)
where yc is the centroidal coordinate of the region with area A.
Ixc = 32.6
106 mm4
Iyc = 1.14
106 mm4
10.2 Method of Composite Areas Example 8, page 2 of 6
2
Consider the cross section to be composed of a rectangle and two channels.
y
y
y
=
x
+2
x
x
y
Centroid of channel
127 mm
C'
3
127 mm
yc' = 127 mm
x
Note that yc' is known.
10.2 Method of Composite Areas Example 8, page 3 of 6
4
y
Rectangle
C'
Area, A = (40 mm)(100 mm + 100 mm)
= 8000 mm2
100 mm
40 mm
100 mm
40 mm
= 20 mm
2
yc'
5 Locate centroid of rectangle:
yc' = 254 mm + 20 mm
127 mm + 127 mm = 254 mm
= 274 mm
6 Set up table
A ( mm2 ) yc' ( mm )
Region
Channel
Rectangle
7
x
2 channels (area was given)
2(3780)
8000
A = 15560
127
274
yc'A ( mm3 )
0.9601 106
2.1920 106
yc'A = 3.1521 106
Eq. 1 gives the distance to the centroid of the entire cross section:
ycA
A
3.1521 106
=
15560
Yc =
= 202.58 mm
(Eq. 1 repeated)
Ans.
10.2 Method of Composite Areas Example 8, page 4 of 6
8
Ixc and Iyc of channels (Ixc', Iyc', and A are given.)
y yc'
Centroid of entire beam section
Use the parallel axis theorem.
Channel Ixc = Ixc' + (dx)2A
= 32.6
= 54.193
106 mm4 + (202.58 mm
C
(2)
Yc = 202.58 mm
106 mm4 + (15.3 mm)2(3780 mm2)
106 mm4
xc'
x
dy = 15.3 mm
= 2.025
C'
127 mm
Channel Iyc = Iyc' + (dy)2A
= 1.140
xc
dx
127 mm)2(3780 mm2)
106 mm4
Centroid of
channel
(3)
10.2 Method of Composite Areas Example 8, page 5 of 6
9
y
Ixc and Iyc of upper rectangle
Ixc' =
bh3
12
100 mm
40 mm
3
(100 mm + 100 mm)(40 mm)
=
12
= 1.067
100 mm
40 mm
= 20 mm
2
C'
xc'
d
C
106 mm4
d = (40 mm + 254 mm)
(20 mm + 202.58 mm)
254 mm
xc
Yc = 202.58 mm
= 71.42 mm
Use the parallel axis theorem.
Area A was calculated previously.
x
2
Rectangle Ixc = Ixc' + d A
= 1.067
= 41.874
Rectangle Iyc =
=
Centroid of entire beam section
106 mm4 + (71.42 mm)2(8000 mm2)
106 mm4
(4)
bh3
12
(40 mm)(100 mm + 100 mm)3
12
= 26.667
106 mm4
(5)
10.2 Method of Composite Areas Example 8, page 6 of 6
10 For the composite region,
Ixc = Rectangle Ixc + 2
by Eq. 4
= 41.874
= 150.3
106 mm4)
106 mm4
by Eq. 5
= 30.7
by Eq. 2
106 mm4 + 2(54.193
Iyc = Rectangle Iyc + 2
= 26.667
Channel Ixc
Ans.
Channel Iyc
by Eq. 3
106 mm4 + 2(2.025
106 mm4
106 mm4)
Ans.