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10.1 Chapter 10: Moments of Inertia Chapter Objectives To develop a method for determining the moment of inertia and product of inertia for an area with respect to given x- and y-axes. To develop a method for determining the “polar moment of inertia” for an area with respect to given x- and y-axes. To a method for determining the moment of inertia with respect to a parallel axis. To introduce the product of inertia and show how to determine the maximum and minimum moments of inertia of an area (a.k.a. principal moments of inertia). 10.1 Definition of Moments of Inertia for Areas Moment of inertia is a significant factor in the determination of flexural stress, shear stress, and deflection for beams, and for critical loads for columns. Moment of inertia is also known as the “second moment of area.” 1. Moment of inertia about the x-axis may be determined using the following equation. Ix = y el 2 dA To compute Ix, the strip (i.e. the differential element) is chosen parallel to the x-axis so that all points forming the strip are the same distance y from the x-axis. 2. The moment of inertia about the y-axis may be determined using the following equation. Iy = x el 2 dA To compute Iy, the strip (i.e the differential element) is chosen parallel to the y-axis so that all points forming the strip are the same distance x from the y-axis. 3. The polar moment of inertia may be determined using the following equation. JO = ∫ r 2 dA

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Page 1: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.1

Chapter 10: Moments of Inertia

Chapter Objectives

• To develop a method for determining the moment of inertia and product of inertia

for an area with respect to given x- and y-axes.

• To develop a method for determining the “polar moment of inertia” for an area

with respect to given x- and y-axes.

• To a method for determining the moment of inertia with respect to a parallel axis.

• To introduce the product of inertia and show how to determine the maximum and

minimum moments of inertia of an area (a.k.a. principal moments of inertia).

10.1 Definition of Moments of Inertia for Areas

Moment of inertia is a significant factor in the determination of flexural stress,

shear stress, and deflection for beams, and for critical loads for columns.

• Moment of inertia is also known as the “second moment of area.”

1. Moment of inertia about the x-axis

may be determined using the following

equation.

Ix = ∫ y el 2 dA

To compute Ix, the strip (i.e. the

differential element) is chosen

parallel to the x-axis so that all points

forming the strip are the same

distance y from the x-axis.

2. The moment of inertia about the y-axis may be determined using the following

equation.

Iy = ∫ x el 2 dA

To compute Iy, the strip (i.e the differential element) is chosen parallel to the

y-axis so that all points forming the strip are the same distance x from the

y-axis.

3. The polar moment of inertia may be determined using the following equation.

JO = ∫ r2 dA

Page 2: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.2

The polar moment of inertia may be

determined from the rectangular

moments of inertia Ix and Iy by noting

that

r2 = x2 + y2

and that

JO = ∫ r2 dA

= ∫ (x2 + y2) dA = ∫ x2 dA + ∫ y2 dA

JO = Iy + Ix

10.2 Parallel-Axis Theorem for an Area

For the area shown, point C represents the centroid of the area and the x’-axis

represents the “centroidal axis.”

• Let y = y ’ + dy

where

y = the distance from the

elemental area dA to the x-

axis

y ’ = the distance from the

elemental area dA to the

centroidal axis for the area

(i.e. the x ’-axis)

dy = the distance between the

centroidal axis for the area

(i.e. the x ’-axis) and the x-

axis

• Substituting for “y” in the equation for moment of inertia with respect to the

x-axis (i.e. Ix).

Ix = ∫ y2 dA = ∫(y ’ + dy)2 dA

= ∫(y ’)2 dA + ∫ 2 y ’ dy dA + ∫(dy)2 dA

The distance “dy” is a constant and is set at the beginning of the problem.

Thus, Ix = ∫(y ’)2 dA + 2 dy ∫ y ’ dA + (dy)2 ∫ dA

Page 3: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.3

∫(y ’)2 dA - represents the moment of inertia of area A with respect to

the centroidal x’-axis; thus,

∫ (y ’)2 dA = I x

2 dy ∫ y ’ dA - represents the first moment of area A with respect to the

centroidal x’-axis.

Since the centroid is located on the axis, the value of this

integral is zero:

∫ y ’ dA = y ’ A = 0, since y ’ = 0.

(dy)2 ∫ dA = (dy)2 A

Thus, Ix = I x + (dy)2 A

The moment of inertia (Ix) of an area about a parallel axis is equal to the moment

of inertia of that area about its centroidal axis (I x) plus a transfer term [(dy)2 A].

• This is known as the “Parallel Axis Theorem.”

For Iy, the “parallel axis theorem” is similar to the one developed for Ix.

Thus, Iy = I y + (dx)2A

10.3 Radius of Gyration of an Area

Consider an area A that has a moment of

inertia Ix with respect to the x-axis.

• Then let the area A be concentrated into

a thin strip parallel to the x-axis and

with an area equal to the original area.

• In order for the thin strip to have the

same moment of inertia as the original

area, the thin strip would have to be

placed a distance kx from the x-axis.

The moment of inertia of the thin strip may

be defined as Ix = kx2 A, and

kx = (Ix/A)½

where

kx is known as the “radius of gyration” with respect to the x-axis.

Page 4: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.4

Similarly, for an area A that has a moment of inertia Iy with respect to the y-axis,

the moment of inertia of a thin strip may be defined as Iy = ky2 A, and

ky = (Iy/A)½

where

ky is known as the “radius of gyration” with respect to the y-axis.

Finally, for an area A that has a polar moment of inertia JO with respect to the x-

and y-axes, the polar moment of inertia of a thin strip may be defined as

JO = ko2 A, and

ko = (JO/A)½

where

ko is known as the “polar radius of gyration”.

Also observe that kx2 + ky

2 = ko2.

Page 5: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.5

Examples – Moments of Inertia for an Area by Integration

Given: Rectangular area shown.

Find: The moment of inertia with respect

to the x-axis (i.e. Ix).

Applicable equation:

Ix = ∫ y el2 dA y el = y

dA = b dy

Solve for Ix, the moment of inertia with respect to the x-axis.

Ix = ∫ y el2 dA =

h

0y2 b dy = b

h

0y2 dy

= b (y3/3) |h

0

Ix = bh3/3

Thus, the moment of inertia of a rectangular area about its base is bh3/3.

Page 6: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.6

Given: Rectangular area shown.

Find: The moment of inertia of a

rectangular area about an axis

through the centroid.

Applicable equation:

Ix = ∫ y el2 dA y el = y

dA = b dy

Solve for Ix, the moment of inertia with respect to the x-axis.

Ix = ∫ y el2 dA =

2/h

2/hy2 b dy = b

2/h

2/hy2 dy

= b (y3/3) |2/h

2/h = (b/3)[(h/2)3 – (- h/2)3]

= (b/3)(h3/8 + h3/8) = (b/3)(h3/4)

Ix = bh3/12

Thus, the moment of inertia of a rectangular area about an axis through its

centroid is bh3/12.

Page 7: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.7

Moment of Inertia by Alternative Analysis

The previous definition for moment of

inertia requires the use of parallel

“elementary strips.”

• However, we can compute Ix and Iy using

the same elementary strip, one that is

perpendicular to the reference axis.

Using the equation for the moment of

inertia for a rectangle about its base, we can

develop an expression for the moment of

inertia using an elemental strip that is

perpendicular to an axis.

• The moment of inertia with respect to the x-axis for the elemental area shown

may be determined as follows.

dIx = (1/3) y3 dx

• As long as all the elemental areas within the limits of integration have bases

that touch the x-axis, the moment of inertia for the entire area may be

determined by integrating this expression.

Ix = ∫ (1/3) y3 dx

The moment of inertia with respect to the y-axis for the elemental area shown may

be determined using the previous definition.

Iy = ∫ x el 2 dA

where

x el = x

dA = y dx

Thus, Iy = ∫ x2 y dx

The sign ( + or - ) for the moment of inertia is determined based on the area.

• If the area is positive, then the moment of inertia is positive.

• If the area is negative, then the moment of inertia is negative.

• The sign for x el or y el may be either positive or negative.

- In either case, (x el 2) or ( y el

2) are used in the expression, and the result will

always be positive.

Page 8: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.8

Examples – Moment of Inertia of an Area by an Alternative Analysis

Given: Triangular area shown.

Find: Ix and Iy using a horizontal element.

Applicable equations:

Ix = ∫ y el2 dA dA = x dy

Iy = ∫ x el 2 dA cannot be used with the horizontal element. x el = x/2

dIy = (1/3) x3 dy will be used to determine Iy. y el = y

To define values of “x” in terms of “y” to allow the integration, write an equation

for the boundary of the area (i.e. the line).

• The general form for the equation of a line is y = m x + c

• The slope “m” of the line is m = (0 - h)/(b - 0) = - h/b m = - h/b

• The y-intercept “c” is the value of y when x = 0. c = h

• The equation for the line is then y = (- h/b) x + h

• Rewrite the equation to define “x” in terms of “y”. x = (b/h)(h – y)

• Now define dA in terms of “y”.

dA = x dy = (b/h)(h – y) dy

Solve for Ix, the moment of inertia with respect to the x-axis.

Ix = ∫ y el2 dA =

h

0y2 (b/h)(h – y) dy = (b/h)

h

0y2 (h – y) dy

= (b/h) h

0(hy2 – y3) dy = (b/h)(hy3/3 – y4/4) |

h

0

= (b/h)(h4/3 – h4/4) = (b/h)(h4/12)

Ix = bh3/12

Page 9: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.9

Now solve for Iy, the moment of inertia with respect to the y-axis.

d Iy = (1/3) dy (x3)

Iy = (1/3) ∫ x3 dy = (1/3) h

0(b/h)3 (h – y)3 dy = (1/3) (b/h)3

h

0( h – y )3 dy

= (1/3) (b/h)3 (- 1) [(h - y)4/4] |h

0= 0 – [(1/3)(b/h)3 (- 1) (h4/4) ]

= 0 + (1/3) (b3/h3) (h4/4)

Iy = b3 h/12

Page 10: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.10

Given: The area shown.

Find: Ix and Iy using a horizontal element.

Applicable equations:

Ix = ∫ y el2 dA dA = (a – x) dy

Iy = ∫ x el 2 dA cannot be used with the horizontal element. x el = ½ (x + a)

dIy = (1/3) x3 dy will be used to determine Iy. y el = y

Define dA in terms of “y”.

dA = (a – x) dy = (a - y ) dy

First, solve for Ix, the moment of inertia with respect to the x-axis.

Ix = ∫ y el2 dA =

b

0y2 (a - y ) dy =

b

0(a y2 – y5/2) dy

= a y3/3 – (2/7) y7/2 |b

0= ab3/3 – (2/7) b7/2

= a b3/3 – (2/7) b1/2 b3 Format the solution in the form f (a, b3)

= a b3/3 – (2/7) a b3 b = a2, a = b1/2

= 7 a b3/21 – 6 a b3/21

Ix = a b3/21

Next, solve for Iy, the moment of inertia with respect to the y-axis.

Iy = ∫ x el2 dA cannot be used with a horizontal element.

d Iy = (1/3) dy x3 must be used with a horizontal element.

• However, the expression d Iy = (1/3) dy x3 is only valid when the bases of all the

elemental rectangular areas touch the reference axis.

Page 11: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.11

• The area may be considered as the difference between the two areas shown

below.

- The moment of inertia with respect to the y-axis for Area 1, the rectangular

area, is

Iy1 = ba3/3

- The moment of inertia with respect to the y-axis for Area 2 is

Iy2 = (1/3) ∫ x3 dy Since y = x2, then x3 = y3/2

= (1/3) b

0y3/2 dy

= (1/3) (2/5) y5/2) |b

0

= (1/3) (2/5) b5/2 Format the solution in the form f(a3,b)

= (2/15) b b3/2 b = a2, b3/2 = a3

Iy2 = 2 b a3/15

• The moment of inertia for the area is

Iy = Iy1 - Iy2 = ba3/3 - 2 ba3/15

Iy = ba3/5

Page 12: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.12

Given: The area shown.

Find: Ix and Iy using a horizontal element.

Applicable equations:

Ix = ∫ y el2 dA dA = (x1 – x2) dy

Iy = ∫ x el 2 dA cannot be used with the horizontal element. x el = ½ (x1 + x2)

dIy = (1/3) x3 dy will be used to determine Iy. y el = y

Define dA in terms of “y”.

dA = (x1 – x2) dy = (y – y2/4) dy

First, solve for Ix, the moment of inertia with respect to the x-axis.

Ix = ∫ y el2 dA =

4

0y2 (y – y2/4) dy =

4

0(y3 – y4/4) dy

= (y4/4 – y5/20) |4

0 = 64.0 – 51.2

Ix = 12.8 in4

Next, solve for Iy, the moment of inertia with respect to the y-axis.

Iy = ∫ x el2 dA cannot be used with a horizontal element.

d Iy = (1/3) dy x3 must be used with a horizontal element.

• The area may be considered as the difference between the two areas shown

below.

Page 13: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.13

Iy = Iy1 - Iy2 = 4 (43)/12 - (1/3) ∫ x3 dy

= 4 (4)3/12 – (1/3) 4

0(y2/4)3 dy = 21.33 – (1/3)

4

0(y6/64) dy

= 21.33 – (1/3) (1/64) (y7/7) |4

0 = 21.33 – 12.19

Iy = 9.14 in4

Page 14: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.14

10.4 Moments of Inertia for Composite Areas

The moment of inertia of an area made up of several common shapes may be

obtained using formulas given for each of the shapes and adding the appropriate

transfer terms to determine the moment of inertia about the desired axis.

Consider the rectangular area shown.

• The centroidal moment of inertia for a

rectangle is

I x = bh3/12

• The moment of inertia with respect to

the x-axis can be determined using the

Parallel Axis Theorem.

Ix = I x + y 2A

= bh3/12 + (h/2)2 b h

= bh3/12 + bh3/4

Ix = bh3/3

If the moment of inertia with respect to an axis is known, the centroidal moment

of inertia may be determined using the Parallel Axis Theorem.

Consider the triangular area shown.

• The moments of inertia for the triangle

with respect to the x- and y-axes are

Ix = b h3/12 and Iy = h b3/12

• The centroidal moments of inertia for a

rectangle with respect to the x- and y-

axes are

I x = Ix - y 2A = b h3/12 - (h/3)2 (b h/2)

= b h3/12 – b h3/18

I x = b h3/36

I y = Iy - x 2A = h b3/12 - (b/3)2 (b h/2)

= h b3/12 – h b3/18

I y = h b3/36

Page 15: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.15

For composite areas, the summation of centroidal moments of inertia and plus the

summation of transfer terms will equal the total moment of inertia for the entire

area.

• For the moment of inertia with respect to the x-axis, the moment of inertia

may be found as follows.

Ix1 = I x1 + ( y 2A) 1

Ix2 = I x2 + ( y 2A) 2

Then summing the moments of inertia for each part of the composite area

Ix = Ix1 + Ix2 + • • •

Ix = I x1 + ( y 2A) 1 + I x2 + ( y 2A) 2 + • • •

or

Ix = ∑I x i + ∑ y i2Ai

• Similarly, for the moment of inertia with respect to the y-axis, the moment of

inertia may be found as follows.

Iy1 = I y1 + (x 2A) 1

Iy2 = I y2 + (x 2A) 2

Then summing the moments of inertia for each part of the composite area

Iy = Iy1 + Iy2 + • • •

Iy = I y1 + (x 2A) 1 + I y2 + (x 2A) 2 + • • •

or

Iy = ∑I y i + ∑x i2Ai

A tabular solution is illustrated for the following example.

Page 16: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.16

Example – Moments of Inertia for Composite Shapes

Given: The composite area shown.

Find: Ix and Iy

Part Ai x i y i Ai x i Ai y i Ai x i2 Ai y i

2 I xi I yi

1 27.00 - 1.50 1.50 - 40.50 40.50 60.75 60.75 182.25 20.25

2 27.00 - 5.00 0 - 135.00 0 675.00 0 121.50 54.00

3 28.27 2.55 2.55 72.09 72.09 183.83 183.83 71.15 71.15

4 - 6.28 - 3.00 - 0.85 18.84 5.34 - 56.52 - 4.54 - 1.74 - 6.28

75.99 - 84.57 117.93 863.06 240.04 373.16 139.12

x 3 = 4R/3π = 4(6)/3π = 2.55 = y 3 y 4 = - 4R/3π = - 4(2)/3π = - 0.85

I x1 = bh3/12 = 3(9)3/12 = 182.25 I y1 = bh3/12 = 9(3)3/12 = 20.25

I x2 = bh3/36 = 6(9)3/36 = 121.50 I y2 = bh3/36 = 9(6)3/36 = 54.00

I x3 = 0.0549R4 = 0.0549(6)4 = 71.15 I y3 = I x3 = 71.15

I x4 = 0.109R4 = 0.109(2)4 = 1.74 I y4 = (π/8)R4 = (π/8)(2)4 = 6.28

Use I x4 = - 1.74 (negative area) Use I y4 = - 6.28 (negative area)

• The moments of inertia with respect to the x- and y-axes are determined as

follow.

Ix = ∑ I xi + ∑ y i2Ai = 373.16 + 240.04 = 613.20 in4

Iy = ∑ I yi + ∑ x i2Ai = 139.12 + 863.06 = 1002.18 in4

• The location of the centroid is determined as follows.

x = ∑x iAi/∑Ai = - 84.57/75.99 = - 1.11”

y = ∑ y iAi/∑Ai = 117.93/75.99 = 1.55”

• The centroidal moments of inertia for the composite area are determined as

follows.

I x = Ix – y 2A = 613.20 – (1.55)2 75.99 = 430.63 in4

I y = Iy – x 2A = 1002.18 – (- 1.11)2 75.99 = 908.55 in4

Page 17: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.17

Given: The cover-plated beam shown.

Find: The moment of inertia with respect to

a horizontal axis through the centroid of

the area (i.e. the neutral axis).

A vertical axis through the center of the

web forms an axis of symmetry.

• Only the y distance is required.

Find the location of the centroid (which corresponds with the neutral axis).

• Use the bottom of the bottom flange as the reference axis.

y = Σ y i Ai/ΣAi = 31.7 (29.8/2) + 1(18)(29.8 + 1.0/2)

31.7 + 1(18)

= 472.33 + 545.40

49.7

y = 20.48”

Using the “Parallel Axis Theorem”,

determine the moment of inertia with

respect to a horizontal axis through the

centroid of the area (i.e. the neutral axis).

Ix = (Ix)beam + (Ix)plate

= [4470 + 31.7(29.8/2 – 20.48)2] + [18(1)3/12 + 1(18)(29.8 + 1.0/2 – 20.48)2]

= (4470 + 987.02) + (1.50 + 1735.78)

Ix = 7194.30 in4

Page 18: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.18

10.5 Product of Inertia for an Area

The product of inertia is defined by the following equation.

Ixy = ∫ x y dA

The product of inertia Ixy may be zero, positive, or negative.

• If one or both of the x- and y-axes form axes of symmetry for the area, then

the product of inertia is zero.

• If the area is “heavy” in the first or third quadrants, then Ixy is positive.

• If the area is “heavy” in the second or fourth quadrants, then Ixy is negative.

Parallel-Axis Theorem

For Ixy, the “parallel axis theorem” is similar

to the ones previously developed for Ix and

Iy.

Ixy = ∫ x y dA

= ∫ (dx + x ’) (dy + y ’) dA

= ∫ dx dy dA + ∫ dx y ’ dA

+ ∫ x ’ dy dA + ∫ x ’ y ’ dA

= dx dy ∫ dA + ∫ x ’ y ’ dA

Ixy = dx dy A + I xy

where,

∫ dx y’ dA = dx ∫ y ’ dA = dx y ’ A = 0, since y ’ = 0

∫ x ’ dy dA = dy ∫ x ’ dA = dy x ’ A = 0, since x ’ = 0

∫ x’ y’ dA = I xy

Page 19: Chapter 10: Moments of Inertia - statics201.webs.com Chap10 - Moments of Inertia.pdf · 10.1 Chapter 10: Moments of Inertia Chapter Objectives • To develop a method for determining

10.19

Example – Product of Inertia by Integration

Given: The area shown.

Find: Ixy and I xy

Applicable equations: Using a vertical element:

Ixy = ∫ x el y el dA dA = (y1 – y2) dx

Ixy = I xy + x y A, so I xy = Ixy - x y A x el = x

y el = ½ (y1 + y2)

Define dA and y el in terms of “x”.

dA = (y1 – y2) dx = (3x – x3/3) dx

y el = ½ (y1 + y2) = ½ (3x + x3/3)

First, solve for Ixy, the product of inertia with respect to the x- and y-axes, using

a vertical element.

Ixy = ∫ x el y el dA = 3

0x [½ (3x + x3/3)] (3x – x3/3) dx

= ½ 3

0x (9x2 – x6/9) dx = ½

3

0(9x3 – x7/9) dx

= ½ (9x4/4 – x8/72) |3

0 = ½ (729/4 – 6561/72) = ½ (182.25 – 91.125)

Ixy = 45.56 in4

Next, solve for I xy using the Parallel Axis Theorem.

A = ∫dA = 3

0(3x – x3/3) dx = (3x2/2 – x4/12) |

3

0 = (27/2 – 81/12)

A = 6.75 in2

x A = 3

0x (3x – x3/3) dx =

3

0(3x2 – x4/3) dx = (3x3/3 – x5/15) |

3

0

= (81/3 – 243/15) = 27.0 – 16.2

x A = 10.80 in3

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10.20

y A = 3

0[½ (3x + x3/3)] (3x – x3/3) dx = ½

3

0(9x2 – x6/9) dx

= ½ (9x3/3 – x7/63) |3

0 = ½ (243/3 – 2187/63) = ½ (81.0 – 34.71)

y A = 23.14 in3

x = x A/A = 10.80/6.75 = 1.60”

y = y A/A = 23.14/6.75 = 3.43”

I xy = Ixy - x y A = 45.56 – (1.60)(3.43) 6.75

I xy = 8.52 in4

Alternatively, solve for Ixy, the product of inertia with respect to the x- and y-

axes, using a horizontal element.

dA = (x2 – x1) dy

x el = ½ (x2 + x1)

y el = y

Define dA and x el in terms of “y”.

dA = (x2 – x1) dy = ( 3 y3 - y/3) dy

x el = ½ (x2 + x1) = ½ ( 3 y3 + y/3)

Ixy = ∫ x el y el dA = 9

0½ ( 3 y3 + y/3) y ( 3 y3 - y/3) dy

= ½ 9

0y [(3y)2/3 – y2/9] dy = ½

9

0(2.08 y5/3 – y3/9) dy

= ½ [2.08 (3/8) y8/3 – y4/36] |9

0 = ½ (273.36 – 182.25)

Ixy = 45.56 in4

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10.21

10.6 Moments of Inertia for an Area about Inclined Axes

For the area and the x- and y-

axes shown, the moments of

inertia and the product of

inertia are defined by the

following equations.

Ix = ∫ y2 dA

Iy = ∫ x2 dA

Ixy = ∫ x y dA

We now propose to determine the moments and product of inertia Iu, Iv, and Iuv of

the same area by rotating the original axes about the origin through an angle θ.

• Observe the following relationships between the x and y distances and the u

and v distances.

u = x cos θ + y sin θ

v = y cos θ – x sin θ

• Develop an equation for Iu with respect to the u-axis.

Iu = ∫ v2 dA = ∫ (y cos θ – x sin θ)2 dA

= ∫ (y2 cos2θ – 2 x y cos θ sin θ + x2 sin2θ) dA

This equation results in the following expression.

Iu = ½ (Ix + Iy) + ½ (Ix – Iy) cos 2θ – Ixy sin 2θ

• A similar expression may be developed for Iv.

Iv = ½ (Ix + Iy) - ½ (Ix – Iy) cos 2θ + Ixy sin 2θ

• Observe the following relationship.

Ix + Iy = Iu + Iv

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10.22

Principal Moments of Inertia

Two values for moment of inertia are of special interest, that is, the maximum and

minimum moments of inertia, called the “principal moments of inertia.”

• To find the maximum moment of inertia, take the first derivative of

Iu = ½ (Ix + Iy) + ½ (Ix – Iy) cos 2θ – Ixy sin 2θ with respect to θ.

• The results of this operation are as follows.

Imax,min = ½ (Ix + Iy) ± {[(Ix – Iy )/2]2 + Ixy2 }½

tan 2θ = - 2 Ixy/(Ix – Iy)

• Also note the following relationship.

Ix + Iy = Imax + Imin

These values of Imax and Imin are called the “principal moments of inertia” of the

area with respect to the origin.

• These values are found only by rotating the axes through the specified point (in

this case, the origin).

• For a different reference point (that is, a point different from the origin),

there may be a larger moment of inertia, but that is not the concern here.

If the x- and y-axes had their origin located at the centroid of the area, then the

maximum and minimum moments of inertia I max and I min could be found for a set of

axes rotated about the centroid of the area.

• These axes are referred to as the “principal centroidal axes” of the area.

• The moments of inertia are referred to as the “principal centroidal moments of

inertia” of the area.

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10.23

Examples – Principal Moments of Inertia

Given: Area shown.

Find: Imax and Imin

Part Ai x i y i Ai x i2 Ai y i

2 Ai x i y i I xi I yi I xyi

1 1.50 - 1.25 1.75 2.34 4.59 - 3.28 0.031 1.125 0

2 1.50 0 0 0 0 0 1.125 0.031 0

3 1.50 1.25 - 1.75 2.34 4.59 - 3.28 0.031 1.125 0

4.50 4.68 9.18 - 6.56 1.187 2.28 0

Ix = ∑ I xi + ∑ Ai y i2 = 1.187 + 9.18 = 10.37 in4

Iy = ∑ I yi + ∑ Ai x i2 = 2.28 + 4.68 = 6.96 in4

Ixy = ∑ I xyi + ∑ Ai x i y i = 0 + (- 6.56 ) = - 6.56 in4

Imax,min

= ½ (Ix + I

y) ± {[(I

x – I

y )/2]

2 + I

xy

2 }½

Imax,min = ½ (10.37 + 6.96) ± {[(10.37 – 6.96)/2]2 + (- 6.56)2 }½

= 8.67 ± 6.78

Imax = 15.45 in4 Ix = 10.37 in4

Imin = 1.89 in4 Iy = 6.96 in4

17.34 in4 ≈ 17.33 in4 OK

tan 2θ = - 2 Ixy/(Ix – Iy)

tan 2θ = - 2 (- 6.56)/(10.37 – 6.96) = + 3.85

θ = + 37.7°

Solve for Iu using θ = 37.7°; Iu must equal 15.45 in4 or 1.89 in4.

Iu = ½ (Ix + Iy) + ½ (Ix – Iy) cos 2 θ – Ixy sin 2 θ

= ½ (10.37 + 6.96) + ½ (10.37 – 6.96) cos 2(37.7) + 6.56 sin 2(37.7)

= 8.67 + 0.43 + 6.35

Iu = 15.45 in4 MAX

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10.24

Given: Area shown.

(Extend previous example.)

Find: a) Imax and Imin

b) I max and I min

From previous work:

a) Ix = 613.20 in4

Iy = 1002.18 in4

b) I x = 430.63 in4

I y = 908.55 in4

Part Ai x i y i I xyi

1 - 60.75 0

2 0 40.50

3 183.83 - 21.38

4 - 16.01 0

107.07 19.12

I xy2 = b2 h2/72 = (6)2 (9)2/72 = + 40.50

I xy3 = - 0.0165R4 = - 0.0165 (6)4 = - 21.38

a) Find Imax and Imin

Ixy = ∑ I xyi + ∑ x i y iAi = 19.12 + 107.07 = 126.19 in4

Imax,min = ½ (Ix + Iy) ± {[(Ix – Iy )/2]2 + Ixy2 }½

= ½ (613.20 + 1002.18) ± {[(613.20 – 1002.18)/2]2 + (126.19)2 }½

= 807.69 ± 231.84

Imax = 1039.53 in4 Ix = 613.20 in4

Imin = 575.85 in4 Iy = 1002.18 in4

1615.38 in4 ≈ 1615.38 in4 OK

tan 2θ = - 2 Ixy/(Ix – Iy)

= - 2 (126.19)/(613.20 – 1002.18)

= + 0.649

θ = + 16.5°

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10.25

b) Find I max and I min

I xy = Ixy – x y A = 126.19 – (- 1.11) (1.55) 75.99

I xy = 256.93 in4

I max,min = ½ (I x + I y) ± {[(I x – I y )/2]2 + I xy2 }½

= ½ (430.63 + 908.55) ± {[(430.63 – 908.55)/2]2 + (256.93)2 }½

= 669.59 ± 350.88

I max = 1020.47 in4 I x = 430.63 in4

I min = 318.71 in4 I y = 908.55 in4

1339.18 in4 ≈ 1339.18 in4 OK

tan 2θ = - 2 Īxy/(Īx - Īy)

= - 2 (256.93)/(430.63 – 908.55)

= + 1.0752

θ = + 23.5°