Engineering Mechanics

Engineering Mechanics: STATICS Anthony Bedford and Wallace Fowler SI Edition Teaching Slides Chapter 8: Moments of Inertia

Chapter Outline   

Introduction Definitions Parallel-Axis Theorems

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8.1 Introduction

Tarbela Dam

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8.1 Introduction

Hoover Dam (C) 2005 Pearson Education South Asia Pte Ltd

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8.1 Introduction

When forces are distributed continuously over an area on which they act, it is often necessary to calculate the moment of these forces about some axis either in or perpendicular to the plane of the area.  Frequently the intensity of the force (pressure or stress) is proportional to the distance of the line of action of the force from the moment axis. 

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8.1 Introduction



The elemental force acting on an element of area is proportional to distance times differential area, and the elemental moment is proportional to distance squared times differential area.

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2

∫ (dis tan ce) d (area)

8.1 Introduction 

Total moment involves an integral of form 2

∫ (dis tan ce) d (area) This integral is called the moment of inertia or the second moment of the area.  The integration is a function of the geometry of the area and occurs frequently in the applications of mechanics. 

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



The second moment of area, also known as the area moment of inertia or second moment of inertia is a property of a shape that can be used to predict the resistance of beams to bending and deflection. The deflection of a beam under load depends not only on the load, but also on the geometry of the beam's cross-section.

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©2001 Brooks/Cole, a division of Thomson Learning, Learning Inc. isThomson a trademark used herein under license. ™

8.1 Introduction

8.1 Introduction 



This is why beams with higher area moments of inertia, such as I-beams, are so often seen in building construction as opposed to other beams with the same area. It is analogous to the polar moment of inertia, which characterizes an object's ability to resist torsion.

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9 ©2001 Brooks/Cole, a division of Thomson Learning, Inc. ThomsonLearning" is a trademark used herein under license.

8.1 Introduction The surface area ABCD is subjected to a distributed pressure p whose intensity is proportional to the distance y from the axis AB. The moment about AB due to the pressure on the element of area dA is py dA=ky2dA 2 Total Moment = 

M = k ∫ y dA

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8.1 Introduction The word inertia appears in the terminology by reason of the similarity between the mathematical form of the integrals for second moments of areas and those for the resultant moments of the so-called inertia forces in the case of rotating bodies.  The moment of inertia of an area is a purely mathematical property of the area and in itself has no physical significance. 

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8.2 Definition 

Consider an area A in the x-y plane:

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8.2 Definition



4 moments of inertia of A are defined: 1. Moment of inertia about the x axis:

I x = ∫ A y 2 dA

(8.1)

where y is the y coordinate of the differential element of area dA

This moment of inertia is sometimes expressed in terms of the radius of gyration about the x axis, kx, which is defined by: (8.2) I x = k x2 A

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8.2 Definition

2. Moment of inertia about the y axis: I y = ∫ A x 2 dA

(8.3)

where x is the x coordinate of the element dA

The radius of gyration about the y axis, ky, is defined by: 2 I y = ky A (8.4) 3. Product of inertia: (C) 2005 Pearson Education South Asia Pte Ltd

I xy = ∫ A xy dA

(8.5)

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8.2 Definition 4. Polar moment of inertia: J O = ∫ A r 2 dA

(8.6)

where r is the radial distance from the origin of the coordinate system to dA The radius of gyration about the origin, kO, is defined by: J O = kO2 A (8.7) The polar moment of inertia is equal to the sum of moments of inertia about the x & y axes:J O = ∫ A r 2dA = ∫ A y 2 + x 2 dA = I x + I y

(

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8.2 Definition Substituting the expressions for the moments of inertia in terms of the radii of gyration into this equation, we obtain: kO2 = k x2 + k y2  The

dimensions of the moments of inertia of an area are (length)4 & the radii of gyration have dimensions of length  Notice that the definitions of the moments of inertia Ix, Iy & JO & the radii of gyration imply that they have positive values for any area  They cannot be negative or zero (C) 2005 Pearson Education South Asia Pte Ltd

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8.2 Definition 

If an area A is symmetric about the x axis, for each element dA with coordinates (x, y), there is a corresponding element dA with coordinates (x, −y):

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8.2 Definition  The

contributions of these 2 elements to the product of inertia Ixy of the area cancel:

xy dA + (−xy) dA = 0  This

means that the product of inertia of the area is zero  The same kind of argument can be used for an area that is symmetric about the y axis  If an area is symmetric about either the x or y axis, its product of inertia is zero (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.1 Moments of Inertia of a Triangular Area Determine Ix, Iy & Ixy for the triangular area in Fig. 8.1.

Fig. 8.1 (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.1 Moments of Inertia of a Triangular Area Strategy Eq. (8.3) for the moment of inertia about the y axis is very similar to the equation for the x coordinate of the centroid of an area & it can be evaluated for this triangular area in exactly in the same way: by using a differential element of area dA in the form of a vertical strip of width dx. Then show that Ix & Ixy can be evaluated by using the same element of area.

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Example 8.1 Moments of Inertia of a Triangular Area Solution Let dA be the vertical strip. The equation describing the triangular area’s upper boundary is f(x) = (h/b)x, so dA = f(x) dx = (h/b)x dx.

To integrate over the entire area, we must integrate with respect to x from x = 0 to x = b. (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.1 Moments of Inertia of a Triangular Area Solution Moment of Inertia About the y Axis:

I y = ∫ x 2dA = ∫ x 2 f ( x ) dA A

A

b 2 h  x  x  dx 0 b 

=∫

4 b

h x 1 3 =   = hb b  4 0 4

Moment of Inertia About the x Axis: 1st, determine the moment of inertia of the strip dA about the x axis while holding x & dx fixed. (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.1 Moments of Inertia of a Triangular Area Solution In terms of the element area dAs = dx dy:

( I x ) strip = ∫

y dAs = ∫ 2

strip

3  f ( x)

y =   3 0

f ( x)

0

( y 2dx ) dy

1 dx = [ f ( x ) ] 3 dx 3

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Example 8.1 Moments of Inertia of a Triangular Area Solution Integrating this expression with respect to x from x = 0 to x = b, we obtain the value of Ix for the entire 3 area: b1 b 1 h  3 I x = ∫ [ f ( x ) ] dx = ∫  x  dx 0 3 0 3 b  4 b

h x 1 3 = 3   = bh 3b  4  0 12 3

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Example 8.1 Moments of Inertia of a Triangular Area Solution Product of Inertia: 1st evaluate the product of inertia of the strip dA, holding x & dx fixed: f ( x)

( I xy )strip = ∫strip xy dAs = ∫0 2  f ( x)

y =   2 0

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( xy dx ) dy

1 2 [ ] x dx = f ( x ) x dx 2

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Example 8.1 Moments of Inertia of a Triangular Area Solution Integrate this expression with respect to x from x = 0 to x = b to obtain the value of Ixy for the entire area: 2 b1 b 1h  2 I xy = ∫ [ f ( x ) ] x dx = ∫  x  x dx 0 2 0 2 b  4 b

h x 1 2 2 = 2  = b h 2b  4  0 8 2

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Example 8.1 Moments of Inertia of a Triangular Area Critical Thinking 



This example is chosen so that you can confirm that we obtain the results tabulated for a triangular area in Appendix B Notice that the same procedure can be used to obtain the moments of inertia of other areas whose boundaries are described by the functions of the form y = f(x)

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Example 8.2 Moments of Inertia of a Circular Area Determine the moments of inertia & radii of gyration of the circular area in Fig. 8.2.

Fig. 8.2

Strategy 1st , determine the polar moment of inertia JO by integrating in terms of polar coordinates. We know from the symmetry of the area that Ix = Iy & since Ix + Iy = JO, the moments of inertia of Ix & Iy are each equal to ½ JO. We also know from the symmetry of the area that Ixy = 0. (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.2 Moments of Inertia of a Circular Area Solution By letting r change by an amount dr, we obtain an annular element of area dA = 2π r dr. The polar moment of inertia is: R

J O = ∫ r dA = ∫ 2πr 3 dxr 2

A

0

4 R

r 1 4 = 2π   = πR  4 0 2

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Example 8.2 Moments of Inertia of a Circular Area Solution And the radius of gyration about O is: kO =

JO = A

(1 2) πR 4 πR 2

1 = R 2

The moments of inertia about the x & y axes are: 1 1 4 I x = I y = J O = πR 2 4 and the radii of gyration about the x & y axes are: I kx = k y = x = A (C) 2005 Pearson Education South Asia Pte Ltd

(1 4) πR 4 1 πR

2

= R 2 30

Example 8.2 Moments of Inertia of a Circular Area Solution The product of inertia is zero:

I xy = 0 Critical Thinking   

The symmetry of this example saved us from having to integrate to determine Ix, Iy & Ixy Be alert for symmetry that can shorten your work In particular, remember that Ixy = 0 if the area is symmetric about either the x or the y axis

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8.2 Parallel-Axis Theorems 



The values of the moments of inertia of an area depend on the position of the coordinate system relative to the area In some situations the moments of inertia of an area are known in terms of a particular coordinate system but we need their values in terms of a different coordinate system

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8.2 Parallel-Axis Theorems 

When the coordinate systems are parallel, the desired moments of inertia can be obtained using the parallel-axis theorems:  Possible to determine the moments of inertia of a composite area when the moments of inertia of its parts are known

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8.2 Parallel-Axis Theorems 

Suppose that we know the moments of inertia of an area A in terms of a coordinate system x’y’ with its origin at the centroid of the area & we wish to determine the moments of inertia in terms of a parallel coordinate system xy:

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8.2 Parallel-Axis Theorems  Denote

the coordinates of the centroid of A in the xy coordinate system by (dx, dy) & 2 d = d x2 + d is y the distance from the origin of the xy coordinate system to the centroid

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8.2 Parallel-Axis Theorems  In

terms of the x’y’ coordinate system, the coordinates of the centroid of A are: d = d x2 + d y2

∫ A x′dA x′ = , ∫ A dA

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∫ A y′dA y′ = ∫ A dA

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8.2 Parallel-Axis Theorems  But

the origin of x’y’ coordinate system is located at the centroid of A, so x ′ = 0 & y ′ = 0.  Therefore,

∫ A x′dA x′ = , ∫ A dA



∫ A y′dA y′ = ∫ A dA

(8.8)

Moment of Inertia About the x Axis:  In

terms of the xy coordinate system, the moment of inertia of A about the x axis is: 2

I x = ∫ A y dA

(8.9)

where y is the y coordinate of the element dA relative to the xy coordinate system (C) 2005 Pearson Education South Asia Pte Ltd

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8.2 Parallel-Axis Theorems  From

the figure, y = y’ + dy, where y’ is the coordinate of dA relative to the x’y’ coordinate system  Substituting this expression into Eq. (8.9), we obtain:

I x = ∫ ( y′ + d y ) 2 dA = ∫ ( y′) 2 dA + 2d y ∫ y′ dA + d y2 ∫ dA A

A

A

A

 The

1st integral on the right is the moment of inertia of A about the x’ axis  From Eq. (8.8) the 2nd integral on the right equals zero (C) 2005 Pearson Education South Asia Pte Ltd

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8.2 Parallel-Axis Theorems  Therefore,

we obtain: I x = I x′ + d y2 A

(8.10)

 This

is a parallel-axis theorem:  It relates the moment of inertia of A about the x’ axis through the centroid to the moment of inertia about the parallel axis x

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8.2 Parallel-Axis Theorems 

Moment of Inertia About the y Axis:  In

terms of the xy coordinate system, the moment of inertia of A about the y axis is: I y = ∫ x 2 dA = ∫ ( x′ + d x ) 2 dA A

=∫

A

A

2 2 ′ ′ ( x ) dA + 2d x ∫ x dA + d x ∫ dA A

A

 From

Eq. (8.8), the 2nd integral on the right equals zero

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8.2 Parallel-Axis Theorems  Therefore,

the parallel-axis theorem that relates the moment of inertia of A about the y’ axis through the centroid to the moment of inertia about the parallel axis y is: I y = I y ′ + d x2 A

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(8.11)

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8.2 Parallel-Axis Theorems 

Product of Inertia:  In

terms of the xy coordinate system, the product of inertia is: I xy = ∫ xy dA = ∫ ( x′ + d x ) ( y′ + d y ) dA A

A

= ∫ x′y′ dA + d y ∫ x′ dA + d x ∫ y′ dA + d x d y ∫ dA A

A

A

A

 The

2nd & 3rd integrals equal zero from Eq. (8.8)  The parallel-axis theorem for the product of inertia is: (8.12) I xy = I x′y ′ + d x d y A (C) 2005 Pearson Education South Asia Pte Ltd

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8.2 Parallel-Axis Theorems 

Polar Moment of Inertia:  The

polar moment of inertia JO = Ix + Iy

 Summing

Eqs. (8.10) & (8.11), the parallelaxis theorem for the polar moment of inertia is: 2 2 2

(

)

J O = J O′ + d x + d y A = J O′ + d A

(8.13)

where d is the distance form the origin of the x’y’ coordinate system to the origin of the xy coordinate system (C) 2005 Pearson Education South Asia Pte Ltd

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8.2 Parallel-Axis Theorems 

To determine the moments of inertia of a composite area:  Suppose that we want to determine the moment of inertia about the y axis of the area:

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8.2 Parallel-Axis Theorems  We

can divide it into a triangle, a semicircle & a circular cutout, denoted as parts 1, 2 & 3  By using the parallel-axis theorem for Iy, we can determine the moment of inertia of each part about the y axis  E.g. the moment of inertia of part 2 (the semicircle) about the y axis is:

( I y ) 2 = ( I y′ ) 2 + ( d x ) 22 A2

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8.2 Parallel-Axis Theorems  We

must determine the values of (Iy’)2 & (dx)2

 Once

this procedure is carried out for each part, the moment of inertia of the composite area is:

I y = ( I y )1 + ( I y ) 2 − ( I y ) 3

 Notice

that the moment of inertia of the circular cutout is subtracted

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8.2 Parallel-Axis Theorems 

Determining a moment of inertia of a composite area in terms of a given coordinate system involves 3 steps:

1.Choose the parts — try to divide the composite area into parts whose moments of inertia you know or can easily determine. 2.Determine the moments of inertia of the parts — determine the moment of inertia of each part in terms of a parallel coordinate system with its origin at the centroid of the part & then use the parallelaxis theorem to determine the moment of inertia in terms of the given coordinate system.

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8.2 Parallel-Axis Theorems 3.Sum the results — sum the moments of inertia of the parts (or subtract in the case of a cutout) to obtain the moment of inertia of the composite area.

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Example 8.3 Demonstration of the Parallel-Axis Theorems The moments of inertia of the rectangular area in Fig. 8.3 in terms of the x’y’ coordinate system are

(

)

1 bh3 , I = 1 hb3 , I 1 bh3 + hb3 . ′ I x′ = 12 = 0 & J = y′ x ′y ′ O 12 12

Determine its moment of inertia in terms of the xy coordinate system.

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Fig. 8.3

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Example 8.3 Demonstration of the Parallel-Axis Theorems Strategy The x’y’ coordinate system has its origin at the centroid of the area & is parallel to the xy coordinate system. Use the parallel-axis theorems to determine the moments of inertia of A in terms of the xy coordinate system.

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Example 8.3 Demonstration of the Parallel-Axis Theorems Solution The coordinates of the centroid in terms of the xy coordinate system are dx = b/2, dy = h/2. The moment of inertia about the x axis is: Ix =

I x′ + d y2 A

1 bh3 = 12

+(

)

1 h 2 bh 2

= 13 bh3

The moment of inertia about the y axis is: 2 2 3 1 1 I y = I y ′ + d x A = 12 hb + ( 2 b ) bh = 13 hb3 The product of inertia is: I xy = I x′y ′ + d x d y A = 0 + ( 12 b )( 12 h )bh = 14 b 2 h 2 (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.3 Demonstration of the Parallel-Axis Theorems Solution The polar moment of inertia is: 1 bh3 + hb3 + J O = J O′ + d 2 A = 12

(

) [( ) + ( ) 1b 2 2

1h 2 2

] bh

Critical Thinking 

Notice that we could also have determined JO using the relation J O = I x + I y = 13 bh3 + 13 hb3

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Example 8.3 Demonstration of the Parallel-Axis Theorems Critical Thinking 



This example is designed so that you can confirm that the parallel-axis theorems yield the results in Appendix B for a rectangular area But the same procedure can be used to obtain the moments of inertia of the area in terms of any coordinate system that is parallel to the x’y’ system

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Example 8.4 Moments of Inertia of a Composite Area Determine Ix, kx & Ixy for the composite area in Fig. 8.4.

Fig. 8.4

Strategy This area can be divided into 2 rectangles. Use the parallel-axis theorems to determine Ix & Ixy for each rectangle in terms of the xy coordinate system & sum the results for the rectangles to determine Ix & Ixy for the composite area. Then use Eq. (8.2) to determine the radius of gyration kx for the composite area. (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.4 Moments of Inertia of a Composite Area Solution Choose the Parts: Determine the moments of inertia by dividing the area in 2 rectangular parts 1 & 2:

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Example 8.4 Moments of Inertia of a Composite Area Solution Determine the Moments of Inertia of the Parts: For each part, introduce a coordinate system x’y’ with its origin at the centroid of the part:

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Example 8.4 Moments of Inertia of a Composite Area Solution Use the parallel-axis theorem to determine the moment of inertia of each part about the x axis (Table 8.1): Table 8.1 Determining the moments of inertia of the parts about the x axis dy (m) A (m2) Ix’ (m4) I x = I x ′ + d 2y A (m 4 ) Part 1 2

(1)(4)

Part 2 0.5

(2)(1)

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(1)( 4) 3 21.33 1 ( 2 )( 1) 3 0.67 12 1 12

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Example 8.4 Moments of Inertia of a Composite Area Solution Sum the Results: The moment of inertia of the composite area about the x axis is: I x = ( I x )1 + ( I x ) 2 = 21.33 m 4 + 0.67 m 4 = 22.00 m 4 The sum of the areas is A = A1 + A2 = 6 m2, so the radius of gyration about the x axis is: Ix 22 m 4 k= = = 1.91 m 2 A 6m (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.4 Moments of Inertia of a Composite Area Solution Repeating this procedure, determine Ixy for each part in Table 8.2: Table 8.2 Determining the products of inertia of the parts of the xy coordinate system dx (m) dy (m) A (m2) Ix’y’ (m4) I xy = I x ′y ′ + d x d y A (m 4 ) Part 1 0.5

2

(1)(4)

0

4

Part 2 2

0.5

(2)(1)

0

2

The product of inertia of the composite area is: I xy = ( I xy )1 + ( I xy ) 2 = 4 m 4 + 2 m 4 = 6 m 4

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Example 8.4 Moments of Inertia of a Composite Area Critical Thinking 

The moments of inertia you obtain do not depend on how you divide a composite area into parts & you will often have a choice of convenient ways to divide a given area.

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Example 8.5 Moments of Inertia of a Composite Area Determine Iy & ky for the composite area in Fig. 8.5.

Fig. 8.5

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Example 8.5 Moments of Inertia of a Composite Area Strategy Divide the area into a rectangle without the semicircular cutout, a semicircle without the semicircular cutout & a circular cutout. Use a parallel-axis theorem to determine Iy for each part in terms of the xy coordinate system. Then, determine Iy for the composite area by adding the values of the rectangle & semicircle & subtracting the circular cutout. Then use Eq. (8.4) to determine the radius of gyration ky for the composite area. (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.5 Moments of Inertia of a Composite Area Solution Choose the Parts: Divide the area into a rectangle, a semicircle & the circular cutout, calling them parts 1, 2 & 3, respectively:

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Example 8.5 Moments of Inertia of a Composite Area Solution Determine the Moments of Inertia of the Parts: The moments of inertia of the parts in terms of the x’y’ coordinate systems & location of the centroid of the semicircular part are summarized in Table 8.3. Use the parallel-axis theorem to determine the moment of inertia of each part about the y axis. dx (mm) Part 1 60 Part 2

120 +

Part 3 120

4( 40) 3π

A (mm2) (120)(80) 1π 2

( 40) 2

π (20)

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2

Iy’ (mm4) 1 12

( 80)(120) 3

 π − 8 ( 40 ) 4    8 9π  1π 4

( 20)

4

I y = I y′ + d x2 A (mm 4 ) 4.608 × 107 4.744 × 107 1.822 × 107 64

Example 8.5 Moments of Inertia of a Composite Area Solution Sum the Results: The moment of inertia of the composite area about the y axis is: I y = ( I y )1 + ( I y ) 2 − ( I y ) 3 = ( 4.608 + 4.744 − 1.822 ) × 107 mm4 = 7.530 × 107 mm 4

The total area is:

A = A1 + A2 − A3 = (120 mm )( 80 mm ) + 12 π ( 40 mm ) 2 − π ( 20 mm ) 2 = 1.086 × 104 mm 2

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Example 8.5 Moments of Inertia of a Composite Area Solution So the radius of gyration about the y axis is: ky =

7.530 × 107 mm 4 = = 83.3 mm 4 2 A 1.086 × 10 mm

Iy

Critical Thinking  Integration is an additive process, which is why the moments of inertia of composite areas can be determined by adding (or in the case of a cutout, subtracting) the moments of inertia of the parts (C) 2005 Pearson Education South Asia Pte Ltd

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Example 8.5 Moments of Inertia of a Composite Area Critical Thinking 

But the radii of gyration of composite areas cannot be determined by adding or subtracting the radii of gyration of the parts  This can be seen from the equations relating the moments of inertia, radii of gyration & area  For this example, we can demonstrate it numerically: the operation

( k y )1 + ( k y ) 2 − ( k y )3 =

( I y )1 A1

+

( I y )2 A2

−

( I y )3 A3

= 86.3 mm

does not yield the correct radius of gyration of the composite area. 67 (C) 2005 Pearson Education South Asia Pte Ltd

Design Example 8.6 Beam Design The equal areas in Fig. 8.6 are candidates for the cross-section of a beam. (A beam with the 2nd cross section shown is called an I-beam.) compare their moments of inertia about the x axis.

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Fig. 8.6

68

Design Example 8.6 Beam Design Strategy Obtain the moment of inertia of the square cross section from Appendix B. Divide the I-beam into 3 rectangles & use the parallel-axis theorem to determine its moment of inertia by the same procedure used in Examples 8.4 & 8.5.

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Design Example 8.6 Beam Design Solution Square Cross Section: From Appendix B, the moment of inertia of the square cross section about the x axis is: Ix =

1 (144.2 mm )(144.2 mm) 3 = 3.60 × 107 mm4 12

I-Beam Cross Section: Divide the area into the rectangular parts shown: (C) 2005 Pearson Education South Asia Pte Ltd

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Design Example 8.6 Beam Design Solution Introducing the coordinate system x’y’ with their origins at the centroids of the parts:

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Design Example 8.6 Beam Design Solution Use the parallel-axis theorem to determine the moments of inertia about the x axis (Table 8.4): dy (mm)

A (mm2)

Part 1 80

(200)(40)

Part 2 0

(40)(120)

Part 3 −80

(200)(40)

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Ix’ (mm4)

I x = I x ′ + d 2y A (mm 4 )

( 200 )( 40 ) 3 5.23 × 107 1 ( 40 )( 120 ) 3 12

1 12

1 12

7 ( 200 )( 40 ) 3 0.58 × 10

5.23 × 107

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Design Example 8.6 Beam Design Solution Their sum is: I x = ( I x )1 + ( I x ) 2 + ( I x ) 3 = ( 5.023 + 0.58 + 5.23) × 107 mm 4 = 11.03 × 107 mm4

The moment of inertia of the I-beam about the x axis is 3.06 times that of the square cross section of equal area.

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Design Example 8.6 Beam Design Design Issues 

A beam is a bar of material that supports lateral loads, meaning loads perpendicular to the axis of the bar



2 common types of beams:  Simply supported beam: a beam with pinned ends  Cantilever beam: a beam with a single, built-in support

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Design Example 8.6 Beam Design Design Issues 



The lateral loads on a beam cause it to bend & it must be stiff or resistant to bending to support them It is shown in mechanics of materials that a beam’s resistance to bending depends directly on the moment of inertia of its cross-sectional area

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Design Example 8.6 Beam Design Design Issues  The

cross sections in the figure all have the same area:

(the numbers are the ratios of the moment of inertia Ix to the value of Ix for the solid square cross section) (C) 2005 Pearson Education South Asia Pte Ltd

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Design Example 8.6 Beam Design Design Issues  However,

configuring the cross section of a beam to increase its moment of inertia can be carried too far:  The “box” beam in the figure has a value of Ix that is 4 times as large as a solid square beam of the same crosssectional area but its walls are so thin that they may “buckle”

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Design Example 8.6 Beam Design Design Issues  The

stiffness implied by the beam’s large moment of inertia is not realized because it becomes geometrically unstable  1 solution used by engineers to achieve a large moment of inertia in a relatively light beam while avoiding failure due to buckling is to stabilize its walls by filling the beam with a light material such as honeycombed metal or foamed plastic (C) 2005 Pearson Education South Asia Pte Ltd

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