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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 withrespect 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 andminimum moments of inertia of an area (a.k.a. principal moments of inertia).


10.1 Definition of Moments of Inertia for AreasMoment of inertia is a significant factor in the determination of flexural stress, shearstress, and deflection for beams, and for critical loads for columns.• Some common formulas using moment of inertia (I) include the following.

Flexural (bending) stress: σb (sigma) = fb = Mc/I

Shear stress: τ (tau) = fv = VQ/It

Deflection at mid-span for a simplysupported beam with a uniform load: ∆ = 5 w L4/384 EI

Critical axial load for a column: Pcr = π2 EI/L2


• 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= ∫ yel 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 = ∫ xel 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


The polar moment of inertia may be determined from the rectangular moments ofinertia Ix and Iy by noting that

r2= x2 + y2

and thatJO = ∫ r2 dA

= ∫ (x2 + y2) dA = ∫ x2 dA + ∫ y2 dA

JO = Iy + Ix


10.2 Parallel-Axis Theorem for an AreaFor the area shown, point C represents the centroid of thearea and the x’-axis represents the “centroidal axis.”• Let y = y’ + dy

wherey = the distance from the elemental area dA to the x-axis

y’ = the distance from the elemental area dA to thecentroidal 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


∫(y’)2 dA - represents the moment of inertia of area A with respect to the centroidal x’-axis; thus,

∫(y’)2 dA = Īx

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 x’-axis, the value of this integral is zero:

∫ y’ dA = y ’ A = 0, since y ’ = 0.

(dy)2 ∫ dA = (dy)2 A


Thus, Ix = Īx + (dy)2 A

The moment of inertia (Ix) of an area about a parallel axis is equal to the moment ofinertia of that area about its centroidal axis (Īx) 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 = Īy + (dx)2A


10.3 Radius of Gyration of an AreaConsider an area A that has a moment of inertia Ix with respect tothe 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 inertiaas the original area, the thin strip would have to be placed adistance 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.

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 = ky

2 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 = ko

2 A, and

ko = (JO/A)½

where

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

Also observe that kx2 + ky

2 = ko2.


Examples – Moments of Inertia for an Area by Integration

Given: Rectangular area shown.

Find: The moment of inertia with respectto the x-axis (i.e. Ix).


Applicable equation:Ix = ∫ yel

2 dA yel = y dA = b dy

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

Ix = ∫ yel2 dA = ∫

0h y2 b dy = b ∫

0h y2 dy

= b (y3/3) |0

h

Ix = bh3/3

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


Given: Rectangular area shown.

Find: The moment of inertia of a rectangular area about an axis through the centroid.


Applicable equation:Ix = ∫ yel

2 dA yel = y dA = b dy


Ix = ∫ yel2 dA = y2 b dy = b y2 dy

= b (y3/3) | = (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.

2

2

/h

/h


Moment of Inertia by Alternative AnalysisThe previous definition for moment of inertia requires the use ofparallel “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 = ∫ xel 2 dA

where xel = 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 xel or yel may be either positive or negative.- In either case, (xel

2) or (yel2) are used in the expression, and the result will always

be positive.


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 = ∫ yel

2 dA dA = x dy

Iy = ∫ xel2 dA cannot be used with xel = x/2

the horizontal element. yel = y

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


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


Ix = ∫ yel2 dA = ∫

0h y2 (b/h) (h – y) dy = (b/h) ∫

0h y2 (h – y) dy

= (b/h) ∫0

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

h

= (b/h)(h4/3 – h4/4) = (b/h)(h4/12)

Ix= bh3/12


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) ∫

0h (b/h)3 (h – y)3 dy

= (1/3) (b/h)3 ∫0

h (h – y)3 dy

= (1/3) (b/h)3 (- 1) [(h - y)4/4] |0

h

= 0 – [(1/3)(b/h)3 (- 1) (h4/4)]

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

Iy = b3 h/12


Given: The area shown.




2 dA dA = (a – x) dy

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

dIy = (1/3) x3 dy will be used to determine Iy. yel = 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 = ∫ yel2 dA = ∫

0b y2 (a - √y) dy = ∫

0b (a y2 – y5/2) dy

Ix = a y3/3 – (2/7) y7/2 |0

b = ab3/3 – (2/7) b7/2

Format the solution in the form f(a, b3)

Ix = a b3/3 – (2/7) b1/2 b3

Ix = 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 =∫ xel2 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.


• 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,

isIy1 = 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) ∫0

b y3/2 dy = (1/3) (2/5) y5/2) |0

b

= (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



2 dA dA = (x1 – x2) dy

Iy = ∫ xel 2 dA cannot be used with xel = ½ (x1 + x2) the horizontal element. yel = y

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

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 = ∫ yel2 dA = ∫

04 y2 (y – y2/4) dy = ∫

04 (y3 – y4/4) dy

= (y4/4 – y5/20) |0

4 = 64.0 – 51.2

Ix = 12.8 in4


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

Iy =∫ xel2 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.


• The moment of inertia for the area is

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

= 4 (4)3/12 – (1/3) ∫ (y2/4)3 dy = 21.33 - (1/3) ∫ (y6/64) dy

= 21.33 – (1/3) (1/64) (y7/7) |0

4 = 21.33 – 12.19

Iy = 9.14 in4


10.4 Moments of Inertia for Composite AreasThe 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

Īx = bh3/12


• The moment of inertia with respect to the x-axis can be determined using the Parallel Axis Theorem.

Ix = Īx + y2A

= 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

Īx = Ix - y2A = b h3/12 - (h/3)2 (b h/2)= b h3/12 – b h3/18

Īx = b h3/36

Īy = Iy - x2A = h b3/12 - (b/3)2 (b h/2) = h b3/12 – h b3/18

Īy = h b3/36


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 = Īx1 + ( y2A)1

Ix2 = Īx2 + ( y2A)2

Then summing the moments of inertia for each part of the composite area Ix = Ix1 + Ix2 + • • •

Ix = Īx1 + ( y2A)1 + Īx2 + ( y2A)2 + • • •

or Ix = ∑ Īxi + ∑ yi2Ai


• Similarly, for the moment of inertia with respect to the y-axis, the moment of inertia may be found as follows.

Iy1 = Īy1 + ( x2A)1

Iy2 = Īy2 + ( x2A)2

Then summing the moments of inertia for each part of the composite area Iy = Iy1 + Iy2 + • • •

Iy = Īy1 + ( x2A)1 + Īy2 + ( x2A)2 + • • •

or Iy = ∑ Īyi + ∑ xi2Ai

A tabular solution is illustrated for the following example.


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 i2 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

x3 = 4R/3π = 4(6)/3π = 2.55 = y3 y4 = - 4R/3π = - 4(2)/3π = - 0.85Īx1 = bh3/12 = 3(9)3/12 = 182.25 Īy1 = bh3/12 = 9(3)3/12 = 20.25Īx2 = bh3/36 = 6(9)3/36 = 121.50 Īy2 = bh3/36 = 9(6)3/36 = 54.00Īx3 = 0.0549R4 = 0.0549(6)4 = 71.15 Īy3 = Īx3 = 71.15Īx4 = 0.109R4 = 0.109(2)4 = 1.74 Īy4 = (π/8)R4 = (π/8)(2)4 = 6.28

Use Īx4 = - 1.74 (negative area) Use Īy4 = - 6.28 (negative area)



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

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

determined as follow.Ix = ∑ Īxi + ∑ yi

2Ai = 373.16 + 240.04 = 613.20 in4

Iy = ∑ Īyi + ∑ xi2Ai = 139.12 + 863.06 = 1002.18 in4



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

• The location of the centroid is determined as follows.

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

y = ∑ yiAi/∑Ai = 117.93/75.99 = 1.55”


Ix = 613.20 in4

Iy = 1002.18 in4

x = - 1.11”

y = 1.55”

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

Īx = Ix – y2A = 613.20 – (1.55)2 75.99 = 430.63 in4

Īy = Iy – x2A = 1002.18 – (- 1.11)2 75.99 = 908.55 in4


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 yi 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


10.5 Product of Inertia for an AreaThe 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 TheoremFor 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 + Īxy

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 = Īxy


Example – Product of Inertia by Integration


Find: Ixy and Īxy


Applicable equations:Ixy = ∫ xel yel dA Using a vertical element:

Ixy = Īxy + x y A, dA = (y1 – y2) dx

so Īxy = Ixy – x y A xel = x

yel = ½ (y1 + y2)

Define dA and yel in terms of “x”.dA = (y1 – y2) dx = (3x – x3/3) dxyel = 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 = ∫xel yel dA = ∫

03 x [½ (3x + x3/3)] (3x – x3/3) dx

= ½ ∫0

3 x (9x2 – x6/9) dx

= ½ ∫0

3 (9x3 – x7/9) dx

= ½ (9x4/4 – x8/72) |0

3

= ½ (729/4 – 6561/72)

= ½ (182.25 – 91.125)

Ixy = 45.56 in4


Next, solve for Īxy using the Parallel Axis Theorem.

A = ∫dA = ∫0

3 (3x – x3/3) dx

= (3x2/2 – x4/12) | 0

3 = (27/2 – 81/12)

A = 6.75 in2

x A = ∫0

3 x (3x – x3/3) dx

= ∫0

3 (3x2 – x4/3) dx = (3x3/3 – x5/15) | 0

3

= (81/3 – 243/15) = 27.0 – 16.2

x A = 10.80 in3


y A = ∫0

3 [½ (3x + x3/3)] (3x – x3/3) dx

= ½ ∫0

3 (9x2 – x6/9) dx

= ½ (9x3/3 – x7/63) | 0

3

= 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”

Īxy = Ixy – x y A = 45.56 – (1.60)(3.43) 6.75

Īxy = 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

xel = ½ (x2 + x1)

yel = yDefine dA and xel in terms of “y”.

dA = (x2 – x1) dy = ( - y/3) dy

xel = ½ (x2 + x1) = ½ ( + y/3)


Ixy = ∫ xel yel dA

= ∫0

9 ½ ( + y/3) y ( - y/3) dy

= ½ ∫0

9 y [(3y)2/3 – y2/9] dy

= ½ ∫0

9 (2.08 y5/3 – y3/9) dy

= ½ [2.08 (3/8) y8/3 – y4/36] | 0

9

= ½ (273.36 – 182.25)

Ixy = 45.56 in4


10.6 Moments of Inertia for an Area about Inclined AxesFor 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. v = y cos θ – x sin θ

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. u = x cos θ + y sin θ

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

• Observe the following relationship.

Ix + Iy = Iu + Iv


Principal Moments of InertiaTwo 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 Īmax and Īmin 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.


Examples – Principal Moments of Inertia

Given: Area shown.

Find: Imax and Imin


Part Ai xi yi Ai xi2 Ai yi

2 Ai xi yi Īxi Īyi Īxyi

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 = ∑ Īxi + ∑ Ai yi2 = 1.187 + 9.18 = 10.37 in4

Iy = ∑ Īyi + ∑ Ai xi2 = 2.28 + 4.68 = 6.96 in4

Ixy = ∑ Īxyi + ∑ Ai xi yi = 0 + (- 6.56 ) = - 6.56 in4


Ix = 10.37 in4 Iy = 6.96 in4 Ixy = - 6.56 in4


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

θ = + 37.7°


Ix = 10.37 in4 Iy = 6.96 in4 Ixy = - 6.56 in4

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


Given: Area shown.(Extend previous example.)

Find: a) Imax and Imin

b) Īmax and Īmin

From previous work:a) Ix = 613.20 in4

Iy = 1002.18 in4

b) Īx = 430.63 in4

Īy = 908.55 in4


Īxy2 = b2 h2/72 = (6)2 (9)2/72 = + 40.50

Īxy3 = - 0.0165 R4 = - 0.0165 (6)4 = - 21.38

Part Ai xi yi Īxyi

1 - 60.75 02 0 40.503 183.83 - 21.384 - 16.01 0

107.07 19.12


a) Find Imax and Imin Ix = 613.20 in4 Iy = 1002.18 in4

Ixy = ∑ Īxyi + ∑ xi yi Ai = 19.12 + 107.07 = 126.19 in4


= 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°


b) Find Īmax and Īmin Īx = 430.63 in4 Īy = 908.55 in4

Īxy = Ixy – x y A = 126.19 – (- 1.11) (1.55) 75.99Īxy = 256.93 in4

Īmax,min = ½ (Īx + Īy) ± {[(Īx – Īy )/2]2 + Īxy2 }½

= ½ (430.63 + 908.55) ± {[(430.63 – 908.55)/2]2 + (256.93)2 }½ = 669.59 ± 350.88

Īmax = 1020.47 in4

Īmin = 318.71 in4


Īx = 430.63 in4 Īy = 908.55 in4 Īxy = 256.93 in4

Īmax = 1020.47 in4 Īx = 430.63 in4

Īmin = 318.71 in4 Īy = 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°

Chapter 10 Moments of Inertia 10 PP Moments of... · &kdswhu ² 0rphqwv ri ,qhuwld 0rphqw ri...

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Transcript of Chapter 10 Moments of Inertia 10 PP Moments of... · &kdswhu ² 0rphqwv ri ,qhuwld 0rphqw ri...