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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.

4 Ix = 1 r 4 4 Iy = 1 r 4

Moment of Inertia y

3

For our particular problem, 1 Large circle Ix = 4 (50 mm)4 = 4.9087 106mm4 1 Small circle Ix = 4 (20 mm)4 = 0.1257 106mm4 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

520 mm

420 mm 60 mm + 150 mm + 150 mm + 60 mm = 420 mm Large rectangle

x

x

400 mm

300 mm 150 mm + 150 mm = 300 mm Small rectangle

10.2 Method of Composite Areas Example 2, page 2 of 2 2

y

A table of properties of planar regions gives the information shown below. 3 Ix = bh 12

3

For our particular problem, 3 Large rectangle Ix = bh 12

Moment of Inertia y

3

Iy = hb 12

b 2

= b 2

(420 mm)(520 mm)3 12

= 4.9213

x

520 mm

109mm4

h 2

420 mm x

y

C

h 2

4 Rectangle

For our particular problem, 3 Small rectangle Ix = bh 12

x

400 mm

(300 mm)(400 mm)3 = 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 Two regions of the same size and same position relative to the x axis can be combined into 2 times a single region.

y

=

x

2

x

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

108mm4 x

280 mm

180 mm

y

3 4 Small rectangle Ix = bh 12

240 mm 3

=

(80 mm)(240 mm) 12

= 0.9216

x

108 mm4 5 80 mm

For the composite region, subtracting gives I x = 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 y

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.

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

y

+2

+4

x

x 2

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: I x = 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) = 4.6273 8

= 0.202

107 mm4

= 1.3643

y

16.2 mm

107 mm4

(3)

16.2 mm

Angle

C

106 mm4 + (123.8 mm)2(877 mm2)

xc

140 mm

A = 877 mm2 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 I y for the rectangle

3

Since the centroid C' does not lie on the x axis, we have to use the parallel axis theorem to calculate Ix.

y

Rectangle Ix = I xc' + d2A 3 in.

=

+ (3 in.)2[(3 in. + 3 in.)(6 in.)]

xc'

C'

= 432 in4

3 in.

6 in.

(6 in.)(3 in. + 3 in.)3 12

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

A table of properties of planar regions gives the information below.

bh3 Ixc' = 36

For our particular triangle y yc'

= 3 Ixx = bh 36 3 IBB = hb 12

Moment of Inertia

= 6 in4

6 in. xc'

C'

bh Area = 2 2h 3 x

x C

h 3

1 in.

x

B

B

b Triangle

(1 in.)(6 in.)3 36

Iyc' = =

b'(h')3 36 (6 in.)(1 in.)3 36

= 0.1667 in4

7 Parallel axis theorem applied to triangle y y c' Triangle Ix = Ixc' + dx2 A = 6 in4 + (2 in.)2(3 in2 )

1 Area, A = (1 in.)(6 in.) 2 = 3 in2

1 dy = 3 in. + in. = 3.3333 in. 3

= 18 in4

6 in. x c' 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 4 Iy = 1 r 4

C'

Moment of Inertia

r = 0.8 in. x

y 5

For our particular problem, 1 Iyc' = 4

r C

(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 I y = 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. r Ix = 8

4

y

B

r4 8 4r yc = 3

C'

Iy = I yc' + d2A

yc

IBB = r x

4r 3

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 4 in.

Semicircle

We would like to apply the parallel-axis theorem:

x

y

C

8 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 ( y c') axis : I BB = Iyc' + d2A or, r4 4r 2 r2 8 = Iyc' + ( 3 ) ( 2 ) Solving gives I yc' = (

8

8 4 9 )r

(4)

10.2 Method of Composite Areas Example 6, page 5 of 5 y c'

y 4r 3

r = 2 in. x

C'

4 in. 10 Eqs. 3 and 4 can now be applied to the semicircular region Semicircle I y = Iyc' + d2A =(

8 )r4+ (4 in. + 4r )2( r2) 3 2 9

8

Substituting r = 2 in. and evaluating the resulting expression gives Semicircle I y = 149.4808 in4

(5)

11 For the composite region, using Eqs. 1, 2, and 5 gives Iy = Rectangle Iy 4

= 170.6667 in = 405 in4

2

Circle Iy + 2

Semicircle I y

4

2(32.4913 in ) + 2(149.4808 in4) Ans.

y

y +2

2 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

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 4 Iy = 1 r 4

C'

Moment of Inertia

r = 0.8 in. x

y 5

For our particular problem, 1 Iyc' = 4

r C

(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 = I yc' + 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. r Ix = 8

4

y

B

r4 8 4r yc = 3

C' B 4 in.

yc

IBB = r x

Semicircle

We would like to apply the parallel-axis theorem: Iy = I yc' + d2A

x

y

C

8 r = 2 in.

Moment of Inertia

Iy =

yc'

4r 3

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'.

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 ( y c') axis : I BB = Iyc' + d2A or, r4 4r 2 r2 + ( )( = I ) yc' 8 2 3

Solving gives I yc' = (

10.2 Method of Composite Areas Example 6, page 5 of 5 y c'

y 4r 3

r = 2 in. C'

4 in.

x

8

8 4 9 )r

(4)

10 Eqs. 3 and 4 can now be applied to the semicircular region Semicircle Iy = I yc' + 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 I y

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 =

y cA A

(1)

where y c 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 3

y

Upper rectangle

C' (centroid)

75 mm

A = (1200 mm)(75 mm) =9

75 mm = 37.5 mm 2

104 mm2

y c' = 425 mm + 37.5 mm

yc'

425 mm

= 462.5 mm x 2 4

y

Lower rectangle

(250 mm + 50 mm + 300 mm) = 1200 mm

y c'

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 yc'A ( mm3 )

A ( mm2 ) y c' ( mm ) 9.000 2(2.125 A = 13.250

104 104 ) 104

462.5 212.5

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 Yc = A 50.656 106 =

13.250

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 Ixc' = 12 (1200 mm)(75 mm)3 = 12

x c' 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

107mm4 + (80.19 mm)2(9

= 6.2093

108mm4

104 mm2) (2)

10.2 Method of Composite Areas Example 7, page 5 of 5 8

y

I x of lower rectangle

yc'

3 Ixc' = bh 12 (50 mm)(425 mm)3 = 12

= 3.1986

8

C (centroid of entire section) xc d

4

10 mm

d = 382.31 mm

xc'

425 mm

425 mm = 212.5 mm 2

212.5 mm

= 169.81 mm Y c = 382.31 mm

Parallel axis theorem: Lower rectangle Ixc = Ixc' + d2 A

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

50 mm

Lower rectangle Ixc

108 mm4 + 2(9.3261

107 mm4

108 mm4) Ans.

104 mm2)

x C' (centroid of rectangle)...