Appendix C. Moments of Composite Areas

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Appendix C. Moments of Composite Areas

C.1 Centroid

The plane area characteristics have special significance in various relationships governing stress and deflection of beams, columns, and shafts. Geometric properties for most areas encountered in practice are listed in numerous reference works [Ref. C.1]. Table C.1 presents several typical cases.

A rectangle is shown with a point C (centroid) marked at its center. The breadth of the rectangle is marked "b" and the length is marked "h." An xy plane is drawn from the center point C. The length from the base to the x-axis is marked "h over 2." — Table C.1. *Properties of Some Plane Areas*

A circle shows a point C (centroid) marked at its center. An xy plane is drawn with its origin at point C of the circle. The distance from the base to the x-axis is marked as "r." — Table C.1. *Properties of Some Plane Areas*

The first step in evaluating the characteristics of a plane area is to locate the centroid of the area. The centroid is the point in the plane about which the area is equally distributed. For area A shown in Fig. C.1, the first moments about the x and y axes, respectively, are given by

\begin{matrix} Q_{x} = \int_{A} y d A, Q_{y} = \int_{A} x d A, & (C.1) \end{matrix}

$\begin{matrix} Q_{x} = \int_{A} y d A, Q_{y} = \int_{A} x d A, & (C.1) \end{matrix}$

A figure shows a plane area with centroid C in an xy plane. — Figure C.1. *Plane area A with centroid C*.

A figure shows a plane area "A" within an xy plane. The center of the plane area is marked C which is the centroid. A reference x bar y bar plane is drawn in the plane area with its origin at the centroid. The distance between the two vertical axes is marked "x bar" and the distance between two horizontal axes is marked "y bar." A point dA is marked in the plane area at a distance of "x" from the vertical axis and "y" from the horizontal axis. The point dA is at a distance "rho" from the origin of the xy plane.

These properties are expressed in cubic meters or cubic millimeters in SI units and in cubic feet or cubic inches in U.S. Customary Units. The centroid of area A is denoted by point C, whose coordinates $\bar{x}$ $\bar{x}$ and $\bar{y}$ $\bar{y}$ satisfy the following relations:

\begin{matrix} \bar{x} = \frac{Q_{y}}{A} = \frac{\int_{A} x d A}{\int_{A} d A}, \bar{y} = \frac{Q_{x}}{A} = \frac{\int_{A} y d A}{\int_{A} d A} & (C.2) \end{matrix}

$\begin{matrix} \bar{x} = \frac{Q_{y}}{A} = \frac{\int_{A} x d A}{\int_{A} d A}, \bar{y} = \frac{Q_{x}}{A} = \frac{\int_{A} y d A}{\int_{A} d A} & (C.2) \end{matrix}$

When an axis possesses an axis of symmetry, the centroid is located on that axis, as the first moment about an axis of symmetry equals zero. When there are two axes of symmetry, the centroid lies at the intersection of the two axes. If an area possesses no axes of symmetry but does have a center of symmetry, the centroid coincides with the center of symmetry.

Frequently, an area can be divided into simple geometric shapes (for example, rectangles, circles, and triangles) whose areas and centroidal coordinates are known or easily determined. When a composite area is considered as an assemblage of n elementary shapes, the resultant or total area is the algebraic sum of the separate areas, and the resultant moment about any axis is the algebraic sum of the moments of the component areas. Thus, the centroid of a composite area has the coordinates

\begin{matrix} \bar{x} = \frac{Σ A_{i} {\bar{x}}_{i}}{Σ A_{i}}, \bar{y} = \frac{Σ A_{i} {\bar{y}}_{i}}{Σ A_{i}} & (C.3) \end{matrix}

$\begin{matrix} \bar{x} = \frac{Σ A_{i} {\bar{x}}_{i}}{Σ A_{i}}, \bar{y} = \frac{Σ A_{i} {\bar{y}}_{i}}{Σ A_{i}} & (C.3) \end{matrix}$

where ${\bar{x}}_{i}$ ${\bar{x}}_{i}$ and ${\bar{y}}_{i}$ ${\bar{y}}_{i}$ represent the coordinates of the centroids of the component areas, A_i(i = 1,2,...,n).

In applying Eqs. (C.3), it is important to sketch the simple geometric forms into which the composite area is resolved, as shown in Example C.1.

Example C.1 Centroid of an Angle

Locate the centroid of the angle section depicted in Fig. C.2. The dimensions are given in millimeters.

A figure shows an area consisting of two parts which are divided into two rectangles. — Figure C.2. *Example C.1. Area consisting of two parts.*

A figure shows a cross-section of the composite area which is divided into two rectangles A subscript 1 and A subscript 2. The centroids of these two rectangles are marked C subscript 1 and C subscript 2 respectively. An xy plane is drawn for reference with its centroid marked C. The length and width of the rectangle A subscript 1 are 140 and 20 respectively. The length and width of the rectangle A subscript 2 are 120 and 20 respectively. Three more reference xy planes are drawn at the centroids C, C subscript 1 and C subscript 2. The distance between x plane and horizontal axis of centroid C subscript 1 is 70. The distance between the x plane and the vertical axis of centroid C is 130. The distance between y plane and the vertical axis of centroid C subscript 2 is 70. The distance between y plane and the vertical axis of centroid C is labeled "x bar equals 35" and the distance between x plane and horizontal axis of centroid C is labeled "y bar equals 95." The three centroids are connected together using dashed lines.

Solution The composite area is divided into two rectangles, A₁ and A₂, for which the centroids are known (Fig. C.2). Taking the X,Y axes as reference, Eqs. (C.3) are applied to calculate the coordinates of the centroid. The computation is conveniently carried out in the following tabular form. Note that when an area is divided into only two parts, the centroid C of the entire area always lies on the line connecting the centroids C₁ and C₂ of the components, as indicated in Fig. C.2.

No.	Ai (mm²)	${\bar{x}}_{i}$ ${\bar{x}}_{i}$ (mm)	${\bar{y}}_{i}$ ${\bar{y}}_{i}$ (mm)	$A_{i} {\bar{x}}_{i}$ $A_{i} {\bar{x}}_{i}$ (mm³)	$A_{i} {\bar{y}}_{i}$ $A_{i} {\bar{y}}_{i}$ (mm³)
1	20(140) = 2800	10	70	28 × 10³	196 × 10³
2	20(100) = 2000	70	130	140 × 10³	260 × 10³
	ΣA_i = 4800			$Σ A_{i} {\bar{x}}_{i} = 168 \times 10^{3}$ $Σ A_{i} {\bar{x}}_{i} = 168 \times 10^{3}$	$Σ A_{i} {\bar{y}}_{i} = 456 \times 10^{3}$ $Σ A_{i} {\bar{y}}_{i} = 456 \times 10^{3}$
	$\bar{x} = \frac{Σ A_{i} {\bar{x}}_{i}}{Σ A_{i}} = \frac{168 \times 10^{3}}{4800} = 35 mm,$ $\bar{x} = \frac{Σ A_{i} {\bar{x}}_{i}}{Σ A_{i}} = \frac{168 \times 10^{3}}{4800} = 35 mm,$			$\bar{y} = \frac{Σ A_{i} {\bar{y}}_{i}}{Σ A_{i}} = \frac{456 \times 10^{3}}{4800} = 95 mm$ $\bar{y} = \frac{Σ A_{i} {\bar{y}}_{i}}{Σ A_{i}} = \frac{456 \times 10^{3}}{4800} = 95 mm$

C.2 Moments of Inertia

We now consider the second moment or moment of inertia of an area (a relative measure of the manner in which the area is distributed about any axis of interest). The moments of inertia of a plane area A with respect to the x and y axes, respectively, are defined by the following integrals:

\begin{matrix} I_{x} = \int_{A} y^{2} d A, I_{y} = \int_{A} x^{2} d A & (C.4) \end{matrix}

$\begin{matrix} I_{x} = \int_{A} y^{2} d A, I_{y} = \int_{A} x^{2} d A & (C.4) \end{matrix}$

where x and y are the coordinates of the element of area dA (Fig. C.1).

Similarly, the polar moment of inertia of a plane area A with respect to an axis through O perpendicular to the area is given by

\begin{matrix} J_{o} = \int_{A} ρ^{2} d A = I_{x} + I_{y} & (C.5) \end{matrix}

$\begin{matrix} J_{o} = \int_{A} ρ^{2} d A = I_{x} + I_{y} & (C.5) \end{matrix}$

where p is the distance from point O to the element dA, and p² = x² + y². The product of inertia of a plane area A with respect to the x and y axes is defined as

\begin{matrix} I_{x y} = \int_{A} x y d A & (C.6) \end{matrix}

$\begin{matrix} I_{x y} = \int_{A} x y d A & (C.6) \end{matrix}$

In this expression, each element of area dA is multiplied by the product of its coordinates (Fig. C.1). The product of inertia of an area about any pair of axes is zero when either of the axes is an axis of symmetry.

From Eqs. (C.4) and (C.5), it is clear that the moments of inertia are always positive quantities because coordinates x and y are squared. Their dimensions are length raised to the fourth power; typical units are meters⁴, millimeters⁴, and inches⁴. The dimensions of the product of inertia are the same as for the moments of inertia; however, the product of inertia can be positive, negative, or zero depending on the values of the product xy.

The radius of gyration is a distance from a reference axis or a point at which the entire area of a section may be considered to be concentrated but still possesses the same moment of inertia as the original area. Therefore, the radii of gyration of an area A about the x and y axes and the origin O (Fig. C.1) are defined as the quantities r_x, r_y, and ρ = r_o, respectively:

\begin{matrix} r_{x} = \sqrt{\frac{I_{x}}{A}}, r_{y} = \sqrt{\frac{I_{y}}{A}}, r_{o} = \sqrt{\frac{J_{o}}{A}} & (C.7) \end{matrix}

$\begin{matrix} r_{x} = \sqrt{\frac{I_{x}}{A}}, r_{y} = \sqrt{\frac{I_{y}}{A}}, r_{o} = \sqrt{\frac{J_{o}}{A}} & (C.7) \end{matrix}$

Substituting I_x, I_y, and J_o from Eqs. (C.7) into Eq. (C.5) results in

\begin{matrix} r_{o}^{2} = r_{x}^{2} + r_{y}^{2} & (C.8) \end{matrix}

$\begin{matrix} r_{o}^{2} = r_{x}^{2} + r_{y}^{2} & (C.8) \end{matrix}$

The radius of gyration has the dimension of length, expressed in meters.

C.2.1 Parallel Axis Theorem

The moment of inertia of an area with respect to any axis is related to the moment of inertia around a parallel axis through the centroid by the parallel-axis theorem, sometimes called the transfer formula. It is useful for determining the moment of inertia of an area composed of several simple shapes.

To develop the parallel-axis theorem, consider the area A depicted in Fig. C.3. Let $\bar{x}$ $\bar{x}$ and $\bar{y}$ $\bar{y}$ represent the centroidal axes of the area, parallel to the axes x and y, respectively. The distances between the two sets of axes and the origin are d_x, d_y and d_o The moment of inertia with respect to the x axis is

I_{x} = \int_{A} {(y + \bar{y})}^{2} d A = \int_{A} y^{2} d A + 2 \bar{y} \int_{A} y d A + {\bar{y}}^{2} \int_{A} d A

$I_{x} = \int_{A} {(y + \bar{y})}^{2} d A = \int_{A} y^{2} d A + 2 \bar{y} \int_{A} y d A + {\bar{y}}^{2} \int_{A} d A$

The first integral on the right side equals the moment of inertia $I_{\bar{x}}$ $I_{\bar{x}}$ about the $\bar{x}$ $\bar{x}$ axis. As y is measured from the centroid axis $\bar{x}, \int_{A} y d A$ $\bar{x}, \int_{A} y d A$ is zero. Hence,

\begin{matrix} I_{x} = I_{\bar{x}} + A {\bar{y}}^{2} = I_{\bar{x}} + A d_{y}^{2} & (C.9 a) \end{matrix}

$\begin{matrix} I_{x} = I_{\bar{x}} + A {\bar{y}}^{2} = I_{\bar{x}} + A d_{y}^{2} & (C.9 a) \end{matrix}$

Similarly,

\begin{matrix} I_{y} = I_{\bar{y}} + A {\bar{x}}^{2} = I_{\bar{y}} + A d_{x}^{2} & (C.9b) \end{matrix}

$\begin{matrix} I_{y} = I_{\bar{y}} + A {\bar{x}}^{2} = I_{\bar{y}} + A d_{x}^{2} & (C.9b) \end{matrix}$

The parallel-axis theorem is thus stated as follows: The moment of inertia of an area with respect to any axis is equal to the moment of inertia around a parallel centroidal axis, plus the product of the area and the square of the distance between the two axes.

In a like manner, a relationship may be developed connecting the polar moment of inertia J_o of an area about an arbitrary point O and the polar moment of inertia J_c about the centroid of the area (Fig. C.3):

$\begin{matrix} J_{o} = J_{c} + A d_{o}^{2} & (C.10) \end{matrix}$ $\begin{matrix} J_{o} = J_{c} + A d_{o}^{2} & (C.10) \end{matrix}$

It can be shown that the product of inertia of an area I_xy with respect to any set of axes is given by the transfer formula

\begin{matrix} I_{x y} = I_{\bar{x} \bar{y}} + A \bar{x} \bar{y} = I_{\bar{x} \bar{y}} + A d_{x} d_{y} & (C.11) \end{matrix}

$\begin{matrix} I_{x y} = I_{\bar{x} \bar{y}} + A \bar{x} \bar{y} = I_{\bar{x} \bar{y}} + A d_{x} d_{y} & (C.11) \end{matrix}$

where $I_{\bar{x} \bar{y}}$ $I_{\bar{x} \bar{y}}$ denotes the product of inertia around the centroidal axes. Note that the parallel-axis theorems, Eqs. (C.9) through (C.11), may be employed only if one of the two axes involved is a centroidal axis.

For elementary shapes, the integrals appearing in the equations of this and preceding sections can usually be evaluated easily and the geometric properties of the area obtained (Table C.1). Cross-sectional areas employed in practice can often be broken into a combination of these simple shapes.

Example C.2 Moment of Inertia for a T Section

Determine the moment of inertia of the T section shown in Fig. C.4a around the horizontal axis passing through its centroid. The dimensions are given in millimeters.

Two figures highlight the dimensions of a T section and a channel section. — Figure C.4. *Example C.2. (a) T section; (b) channel section.*

A figure shows the cross sections of a T and channel section. The T section consists of a horizontal and a vertical part. The center of the horizontal part is marked A subscript 1 and the center of the vertical section is marked A subscript 2. The distance between A subscript 1 and A subscript 2 is 40 millimeters and centroid C is marked at the center of these two points. The vertical length of the T section is 80 millimeters and the horizontal length is 60 millimeters. A horizontal axis passes through the centroid and the distance between the base (vertical part) and the horizontal axis is marked "y bar equals 50 millimeters." The distance between the base (vertical part) and point A subscript 2 is 30 millimeters. The distance between the upper surface (horizontal part) and point A subscript 1 is 10 millimeters. The second figure shows a cross section of the channel section. The centroid C is marked with a reference xy plane drawn over it. The width and height of the channel section are 60 millimeters and 80 millimeters respectively. The flange thickness is 10 millimeters and the web thickness is 20 millimeters.

Solution Location of Centroid: The area is divided into component parts A₁ and A₂, for which the centroids are known (Fig. C.4a). Because the y axis is an axis of symmetry, $\bar{x} = 0$ $\bar{x} = 0$ and

\bar{y} = \frac{A_{1} {\bar{y}}_{1} + A_{2} {\bar{y}}_{2}}{A_{1} + A_{2}} = \frac{20 (60) 70 + 60 (20) 30}{20 (60) + 60 (20)} = 50 mm

$\bar{y} = \frac{A_{1} {\bar{y}}_{1} + A_{2} {\bar{y}}_{2}}{A_{1} + A_{2}} = \frac{20 (60) 70 + 60 (20) 30}{20 (60) + 60 (20)} = 50 mm$

Centroidal Moment of Inertia: Application of the transfer formula to Fig. C.4a results in

$\begin{matrix} I_{x} & = Σ (I_{\bar{x}} + A d_{y}^{2}) \\ = \frac{1}{12} (60) {(20)}^{3} + 20 (60) 20^{2} + \frac{1}{12} (20) {(60)}^{3} + 20 (60) 20^{2} \\ = 136 (10^{4}) {mm}^{4} \end{matrix}$ $\begin{matrix} I_{x} & = Σ (I_{\bar{x}} + A d_{y}^{2}) \\ = \frac{1}{12} (60) {(20)}^{3} + 20 (60) 20^{2} + \frac{1}{12} (20) {(60)}^{3} + 20 (60) 20^{2} \\ = 136 (10^{4}) {mm}^{4} \end{matrix}$

Comment Interestingly, the properties of this T section about the centroidal x axis are the same as those of the channel (Fig. C.4b). Both sections possess an axis of symmetry.

C.2.2 Principal Moments of Inertia

The moments of inertia of a plane area depend not only on the location of the reference axis, but also on the orientation of the axes about the origin. The variation of these properties with respect to axis location are governed by the parallel-axis theorem, as described in Section C.2.1. We now derive the equations for transformation of the moments and product of inertia at any point of a plane area.

The area shown in Fig C.5 has the moments and product of inertia I_x, I_y, I_xy with respect to the x and y axes defined by Eqs. (C.4) and (C.6). We need to determine the moments and product of inertia I_x′, I_y′ and I_x′y′ about axes x′, y′ making an angle θ with the original x, y axes. The new coordinates of an element dA can be expressed by projecting x and y onto the rotated axes (Fig. C.5):

\begin{matrix} x^{'} = x \cos θ + y \sin θ, y^{'} = y \cos θ - x \sin θ & (a) \end{matrix}

$\begin{matrix} x^{'} = x \cos θ + y \sin θ, y^{'} = y \cos θ - x \sin θ & (a) \end{matrix}$

A figure shows a plane area in an xy plane. — Figure C.5. *Rotation of axes*.

A figure shows a plane area "A" in an xy plane with its origin marked O. The xy plane gets slightly tilted upwards and it is represented by dashed lines. A point dA is marked on the plane area. The distance between the horizontal axis and vertical axis of the xy plane to the point dA is marked y and x respectively. The distance between the horizontal axis and vertical axis of the x bar y bar plane to the point dA is marked y bar and x bar respectively. The angle of deflection between the two planes is labeled "theta."

Then, by definition,

$\begin{matrix} I_{x^{'}} & = \int_{A} {y^{'}}^{2} d A = \int_{A} {(y \cos θ - x \sin θ)}^{2} d A \\ = \cos^{2} θ \int_{A} y^{2} d A + \sin^{2} θ \int_{A} x^{2} d A - 2 \sin θ \cos θ \int_{A} x y d A \end{matrix}$ $\begin{matrix} I_{x^{'}} & = \int_{A} {y^{'}}^{2} d A = \int_{A} {(y \cos θ - x \sin θ)}^{2} d A \\ = \cos^{2} θ \int_{A} y^{2} d A + \sin^{2} θ \int_{A} x^{2} d A - 2 \sin θ \cos θ \int_{A} x y d A \end{matrix}$

Upon substituting Eqs. (C.4) and (C.6), this expression becomes

\begin{matrix} I_{x^{'}} = I_{x} \cos^{2} θ + I_{y} \sin^{2} θ - 2 I_{x y} \sin θ \cos θ & (b) \end{matrix}

$\begin{matrix} I_{x^{'}} = I_{x} \cos^{2} θ + I_{y} \sin^{2} θ - 2 I_{x y} \sin θ \cos θ & (b) \end{matrix}$

The moment of inertia I_y′ may be readily found by substituting θ + π/2 for in the expression for I_x′. Similarly, using the definition I_x′y′ = ∫_A x′y′ dA, we obtain

$\begin{matrix} I_{x^{'} y^{'}} = (I_{x} - I_{y}) \sin θ \cos θ + I_{x y} (\cos^{2} θ - \sin^{2} θ) & (c) \end{matrix}$ $\begin{matrix} I_{x^{'} y^{'}} = (I_{x} - I_{y}) \sin θ \cos θ + I_{x y} (\cos^{2} θ - \sin^{2} θ) & (c) \end{matrix}$

The transformation equations for the moments and product of inertia may be rewritten by introducing double-angle trigonometric relations in the following form:

\begin{matrix} I_{x^{'}} = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} - I_{y}}{2} \cos 2 θ - I_{x y} \sin 2 θ & (C .12 a) \end{matrix}

$\begin{matrix} I_{x^{'}} = \frac{I_{x} + I_{y}}{2} + \frac{I_{x} - I_{y}}{2} \cos 2 θ - I_{x y} \sin 2 θ & (C .12 a) \end{matrix}$

\begin{matrix} I_{y^{'}} = \frac{I_{x} + I_{y}}{2} - \frac{I_{x} - I_{y}}{2} \cos 2 θ + I_{x y} \sin 2 θ & (C.12 b) \end{matrix}

$\begin{matrix} I_{y^{'}} = \frac{I_{x} + I_{y}}{2} - \frac{I_{x} - I_{y}}{2} \cos 2 θ + I_{x y} \sin 2 θ & (C.12 b) \end{matrix}$

\begin{matrix} I_{x^{'} y^{'}} = \frac{I_{x} - I_{y}}{2} \sin 2 θ + I_{x y} \cos 2 θ & (C.12 c) \end{matrix}

$\begin{matrix} I_{x^{'} y^{'}} = \frac{I_{x} - I_{y}}{2} \sin 2 θ + I_{x y} \cos 2 θ & (C.12 c) \end{matrix}$

Comparing the expressions here and in Chapter 1, we observe that moments of inertia (I_x, I_y, I_x′, I_y′) correspond to the normal stresses (σ_x, σ_y, σ_x′, σ_y), the negatives of the products of inertia (–I_xy, –I_x′y′) correspond to the shear stresses (τ_xy, τ_x′y′), and the polar moment of inertia (J_o) corresponds to the sum of the normal stresses (σ_x + σ_y). Thus, Mohr’s circle analysis and the characteristics for stress apply to these properties of area.

The angle θ at which the moment of inertia I_x′ of Eq. (C.12a) assumes an extreme value may be obtained from the condition dI_x′ / dθ = 0:

\begin{matrix} (I_{x} - I_{y}) \sin 2 θ + 2 I_{x y} \cos 2 θ = 0 & (d) \end{matrix}

$\begin{matrix} (I_{x} - I_{y}) \sin 2 θ + 2 I_{x y} \cos 2 θ = 0 & (d) \end{matrix}$

Then

\begin{matrix} \tan 2 θ_{p} = - \frac{2 I_{x y}}{I_{x} - I_{y}} & (C.13) \end{matrix}

$\begin{matrix} \tan 2 θ_{p} = - \frac{2 I_{x y}}{I_{x} - I_{y}} & (C.13) \end{matrix}$

The quantity θ_p represents the two values of θ that locate the principal axes about which the principal or maximum and minimum moments of inertia occur.

When we compare Eq. (C.12c) with Eq. (d), it becomes clear that the product of inertia is zero for the principal axes. If the origin of axes is located at the centroid of the area, they are referred to as the centroidal principal axes. In Section C.2, we noted that the products of inertia relative to the axes of symmetry are zero. Thus, an axis of symmetry coincides with a centroidal principal axis.

The principal moments of inertia are determined by introducing the two values of θ_p into Eq. (C.12a). The sine and cosine of angles 2θ_p defined by Eq. (C.13) are

\begin{matrix} \cos 2 θ_{p} = \frac{I_{x} - I_{y}}{2 r}, \sin 2 θ_{p} = - \frac{I_{x y}}{r} & (e) \end{matrix}

$\begin{matrix} \cos 2 θ_{p} = \frac{I_{x} - I_{y}}{2 r}, \sin 2 θ_{p} = - \frac{I_{x y}}{r} & (e) \end{matrix}$

where $r = {{[(I_{x} - I_{y}) / 2]}^{2} + I_{x y}^{2}}^{1 / 2}$ $r = {{[(I_{x} - I_{y}) / 2]}^{2} + I_{x y}^{2}}^{1 / 2}$ . When these expressions are inserted into Eq. (C.12a), we obtain

\begin{matrix} I_{1, 2} = \frac{I_{x} + I_{y}}{2} \pm \sqrt{{(\frac{I_{x} - I_{y}}{2})}^{2} + I_{x y}^{2}} & (C.14) \end{matrix}

$\begin{matrix} I_{1, 2} = \frac{I_{x} + I_{y}}{2} \pm \sqrt{{(\frac{I_{x} - I_{y}}{2})}^{2} + I_{x y}^{2}} & (C.14) \end{matrix}$

in which I₁ and I₂ denote the maximum and minimum principal moments of inertia, respectively.

Example C.3 Principal Moments of Inertia for an Angular Section

Calculate the centroidal principal moments of inertia for the angular section shown in Fig. C.6a. The dimensions are given in millimeters.

Two figures illustrate the dimensions of an angle section and Mohr's circle for moments of inertia. — Figure C.6. *Example C.3. (a) Angle section; (b) Mohr’s circle for moments of inertia.*

A figure shows an angle section of width 120 millimeters and height 140 millimeters present along an xy plane. The thickness of the angle section is given as "t equals 20 millimeters." The vertical part of the and the horizontal part of the angle section is labeled as A subscript 1 and A subscript 2 respectively. A point C (centroid) is marked at the center of the angle section with a reference xy plane drawn over it. The xy plane with its origin at the centroid gets tilted downward at an angle "theta subscript p equals 35.54 degrees" and it is represented by dashed x bar y bar plane. The distance between the two horizontal axes is labeled "y bar equals 75 millimeters" and the distance between the two vertical axes is labeled "x bar equals 35." The second figure shows a Mohr's circle for moments of inertia drawn inside an xy plane is shown. The horizontal and the vertical axes represent the moments of inertia acting along the x and y axes. The center of the circle is marked C and the origin of the xy plane is marked O. The distance between the vertical axis and the farthest point of the circle on the horizontal axis is labeled "I subscript 1 equals 1184." The distance between the vertical axis and the shortest point of the circle on the horizontal axis is labeled "I subscript 2 equals 296." The distance between the vertical axis and center of the circle is "I subscript x plus I subscript y over 2 equals 740." A vertical dashed line xy is drawn through the origin of the circle, which makes an angle of 71.08 degrees with the horizontal axis. The point at which the dashed line cuts the circle in the 1st quadrant is marked x, which is present at a distance of "I subscript xy equals 420" from the horizontal axis and from the vertical axis it is at a distance of "I subscript x equals 884."

Solution Location of Centroid: The x, y axes are the reference axes through the centroid C (Fig. C.6a), whose location was determined in Example C.1. Again, the area is divided into rectangles A₁ and A₂.

Moments of Inertia: Applying the parallel-axis theorem, with reference to the x and y axes, we have

\begin{matrix} I_{x} & = Σ (I_{\bar{x}} + A d_{y}^{2}) \\ = \frac{1}{12} (20) {(140)}^{3} + 20 (140) 25^{2} + \frac{1}{12} (100) 20^{3} + 100 (20) {(35)}^{2} \\ = 884 (10^{4}) {mm}^{4} \\ I_{y} & = Σ (I_{\bar{y}} + A d_{x}^{2}) \\ = \frac{1}{12} (140) (20^{3}) + 140 (20) (25^{2}) + \frac{1}{12} (20) {(100)}^{3} + 20 (100) {(35)}^{2} \\ = 596 (10^{4}) {mm}^{4} \end{matrix}

$\begin{matrix} I_{x} & = Σ (I_{\bar{x}} + A d_{y}^{2}) \\ = \frac{1}{12} (20) {(140)}^{3} + 20 (140) 25^{2} + \frac{1}{12} (100) 20^{3} + 100 (20) {(35)}^{2} \\ = 884 (10^{4}) {mm}^{4} \\ I_{y} & = Σ (I_{\bar{y}} + A d_{x}^{2}) \\ = \frac{1}{12} (140) (20^{3}) + 140 (20) (25^{2}) + \frac{1}{12} (20) {(100)}^{3} + 20 (100) {(35)}^{2} \\ = 596 (10^{4}) {mm}^{4} \end{matrix}$

The product of inertia about the xy axes is obtained as described in Section C.2:

\begin{matrix} I_{x y} & = \sum (I_{\bar{x} \bar{y}} + A d_{x} d_{y}) \\ = 0 + 20 (140) (- 25) (- 25) + 0 + 100 (20) (35) (35) \\ = 420 (10^{4}) {mm}^{4} \end{matrix}

$\begin{matrix} I_{x y} & = \sum (I_{\bar{x} \bar{y}} + A d_{x} d_{y}) \\ = 0 + 20 (140) (- 25) (- 25) + 0 + 100 (20) (35) (35) \\ = 420 (10^{4}) {mm}^{4} \end{matrix}$

Principal Moments of Inertia: Equation (C.13) yields

$\begin{array}{l} \tan 2 θ_{p} = - \frac{2 (420)}{884 - 596} = - 2.917, & 2 θ_{p} = - 71.08 ° - 251.08 ° & or & - 251.08 ° \end{array}$ $\begin{array}{l} \tan 2 θ_{p} = - \frac{2 (420)}{884 - 596} = - 2.917, & 2 θ_{p} = - 71.08 ° - 251.08 ° & or & - 251.08 ° \end{array}$

Thus, the two values of θ_p are –35.54° and –125.54°. Using the first of these values, Eq. (C.12a) results in I_x′ = 1184(10⁴) mm⁴. The principal moments of inertia are, from Eq. (C.14),

$I_{1, 2} = [\frac{884 + 596}{2} \pm \sqrt{{(\frac{884 - 596}{2})}^{2} + 420^{2}}] 10^{4} = [740 \pm 444] 10^{4}$ $I_{1, 2} = [\frac{884 + 596}{2} \pm \sqrt{{(\frac{884 - 596}{2})}^{2} + 420^{2}}] 10^{4} = [740 \pm 444] 10^{4}$

or I₁ = I_x′ = 1184(10⁴) mm⁴ and I₂ = I_y′ = 296(10⁴) mm⁴. The principal axes are indicated in Fig. C.6a as the x′ and y′ axes.

Comments The principal moments of inertia may also be determined readily by means of Mohr’s circle, following a procedure similar to that described in Section 1.11, as shown in Fig. C.6b. Note that the quantities indicated are expressed in cm⁴ and the results are obtained analytically from the geometry of the circle.

Example C.4 General Formulation of Moments of Inertia for an Angle Section

Derive general expressions for the centroidal moments and product of inertia for the angle section shown in Fig. C.6a. Write a computer program to obtain the centroidal principal moments of inertia.

Solution Using the X,Y axes as reference (Fig. C.6a), Eqs. (C.3) yield

\begin{matrix} \bar{x} = \frac{a (t) (t / 2) + t (b - t) (b + t) / 2}{a (t) + (b - t) t} = \frac{b^{2} + a t - t^{2}}{2 (a + b - t)} & (f) \end{matrix}

$\begin{matrix} \bar{x} = \frac{a (t) (t / 2) + t (b - t) (b + t) / 2}{a (t) + (b - t) t} = \frac{b^{2} + a t - t^{2}}{2 (a + b - t)} & (f) \end{matrix}$

\begin{matrix} \bar{y} = \frac{a (t) (a / 2) + (b - t) t (a - t / 2)}{a (t) + (b - t) t} = \frac{a^{2} + (b - t) (2 a - t)}{2 (a + b - t)} & (g) \end{matrix}

$\begin{matrix} \bar{y} = \frac{a (t) (a / 2) + (b - t) t (a - t / 2)}{a (t) + (b - t) t} = \frac{a^{2} + (b - t) (2 a - t)}{2 (a + b - t)} & (g) \end{matrix}$

The transfer formula, Eq. (C.9a), results in

\begin{matrix} \begin{matrix} I_{x} & = \sum (I_{\bar{x}} + A d_{y}^{2}) \\ = \frac{1}{12} t a^{3} + a t {(\bar{y} - \frac{a}{2})}^{2} + \frac{1}{12} (b - t) t^{3} + (b - t) t {[\bar{y} - (a - \frac{t}{2})]}^{2} \\ = \frac{t}{12} (a^{3} + b t^{2} - t^{3}) + (b t - t^{2}) {[\bar{y} - (a - \frac{t}{2})]}^{2} + a t {(\bar{y} - \frac{a}{2})}^{2} \end{matrix} & (h) \end{matrix}

$\begin{matrix} \begin{matrix} I_{x} & = \sum (I_{\bar{x}} + A d_{y}^{2}) \\ = \frac{1}{12} t a^{3} + a t {(\bar{y} - \frac{a}{2})}^{2} + \frac{1}{12} (b - t) t^{3} + (b - t) t {[\bar{y} - (a - \frac{t}{2})]}^{2} \\ = \frac{t}{12} (a^{3} + b t^{2} - t^{3}) + (b t - t^{2}) {[\bar{y} - (a - \frac{t}{2})]}^{2} + a t {(\bar{y} - \frac{a}{2})}^{2} \end{matrix} & (h) \end{matrix}$

Similarly,

$\begin{matrix} I_{y} = \frac{t}{12} [{(b - t)}^{3} + a t^{2}] + (b t - t^{2}) {[\bar{x} -^{} \frac{1}{2} (b + t)]}^{2} + a t {(\bar{x} - \frac{t}{2})}^{2} & (i) \end{matrix}$ $\begin{matrix} I_{y} = \frac{t}{12} [{(b - t)}^{3} + a t^{2}] + (b t - t^{2}) {[\bar{x} -^{} \frac{1}{2} (b + t)]}^{2} + a t {(\bar{x} - \frac{t}{2})}^{2} & (i) \end{matrix}$

\begin{matrix} I_{x y} = (b - t) t [\bar{x} - \frac{1}{2} (b + t)] [\bar{y} - (a - \frac{t}{2})] + a t (\bar{x} - \frac{t}{2}) (\bar{y} - \frac{a}{2}) & (j) \end{matrix}

$\begin{matrix} I_{x y} = (b - t) t [\bar{x} - \frac{1}{2} (b + t)] [\bar{y} - (a - \frac{t}{2})] + a t (\bar{x} - \frac{t}{2}) (\bar{y} - \frac{a}{2}) & (j) \end{matrix}$

For convenience in programming, we employ the notation

\begin{matrix} C_{1} = \frac{t}{12} (a^{3} + b t^{2} - t^{3}) & C_{2} = b t - t^{2} \\ C_{3} = \bar{y} - (a - \frac{t}{2}) & C_{4} = a t \\ C_{5} = \bar{y} - \frac{a}{2}, & C_{6} = \frac{t}{12} [{(b - t)}^{3} + a t^{2}] \\ C_{7} = \bar{x} - \frac{1}{2} (b + t) & C_{8} = \bar{x} - \frac{t}{2} \end{matrix}

$\begin{matrix} C_{1} = \frac{t}{12} (a^{3} + b t^{2} - t^{3}) & C_{2} = b t - t^{2} \\ C_{3} = \bar{y} - (a - \frac{t}{2}) & C_{4} = a t \\ C_{5} = \bar{y} - \frac{a}{2}, & C_{6} = \frac{t}{12} [{(b - t)}^{3} + a t^{2}] \\ C_{7} = \bar{x} - \frac{1}{2} (b + t) & C_{8} = \bar{x} - \frac{t}{2} \end{matrix}$

and

\begin{matrix} D = \frac{1}{2} (I_{x} + I_{y}), & E = \frac{1}{2} (I_{x} - I_{y}), & F = - I_{x y} \end{matrix}

$\begin{matrix} D = \frac{1}{2} (I_{x} + I_{y}), & E = \frac{1}{2} (I_{x} - I_{y}), & F = - I_{x y} \end{matrix}$

Equations (h), (i), (j), (C.13), and (C.14) are thus

\begin{matrix} I_{x} & = C_{1} + C_{2} (C_{3}^{2}) + C_{4} (C_{5}^{2}) \\ I_{y} & = C_{6} + C_{2} (C_{7}^{2}) + C_{4} (C_{8}^{2}) \\ I_{x y} & = C_{2} + (C_{3}) C_{7} + C_{4} (C_{5}) C_{8} \\ I_{1} & = D + {(E^{2} + F^{2})}^{1 / 2} \\ I_{2} & = D - {(E^{2} + F^{2})}^{1 / 2} \\ θ_{p} & = arctan (\frac{F}{E}) \end{matrix}

$\begin{matrix} I_{x} & = C_{1} + C_{2} (C_{3}^{2}) + C_{4} (C_{5}^{2}) \\ I_{y} & = C_{6} + C_{2} (C_{7}^{2}) + C_{4} (C_{8}^{2}) \\ I_{x y} & = C_{2} + (C_{3}) C_{7} + C_{4} (C_{5}) C_{8} \\ I_{1} & = D + {(E^{2} + F^{2})}^{1 / 2} \\ I_{2} & = D - {(E^{2} + F^{2})}^{1 / 2} \\ θ_{p} & = arctan (\frac{F}{E}) \end{matrix}$

The required C program is presented in Table C.2, along with input data (from Example C.3) and output values. The program was written and tested on a digital computer.

Table C.2. C Program for Moments of Inertia

Click here to view code image

/* C Program for Moments of Inertia*/
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define pi    3.14159265

void main()
{
     FILE *fin, *fout;
     float xbar, ybar, theta_p, float a, b, t;
     float C1, C2, C3, C4, C5, C6, C7, C8, Ix, Iy, Ixy, D, E, F, I1, I2;

     fin=fopen( "m_inertia.d", "r" );
     fout=fopen( "m_inertia.o", "w" );

     fscanf(fin, "%f%f%f", &a, &b, &t);
     fprintf(fout,"input:
");
     fprintf(fout,"a=%10.5f b=%10.5f t=%10.5f
", a, b, t);

     xbar=(pow(b,2)-pow(t,2)+a*t)/(2.*(a+b-t));
     ybar=((b-t)*(2.*a-t)+pow(a,2))/(2.*(a+b-t));

     C1=t*(pow(a,3)+b*pow(t,2)-pow(t,3))/12.;
     C2=b*t-t*t;
     C3=ybar-(a-t/2.);
     C4=a*t;
     C5=ybar-a/2.;
     C6=t*(pow((b-t),3)+a*pow(t,2))/12.;
     C7=xbar-(b+t)/2.;
     C8=xbar-t/2.;
     Ix=C1+C2*pow(C3,2)+C4*pow(C5,2);
     Iy=C6+C2*pow(C7,2)+C4*pow(C8,2);
     Ixy=C2*C3*C7+C4*C5*C8;
     D=(Ix+Iy)/2.;
     E=(Ix-Iy)/2.;
     F=-Ixy;
     I1=D+sqrt(pow(E,2)+pow(F,2));
     I2=D-sqrt(pow(E,2)+pow(F,2));

     if(Iy != Ix) theta_p=(atan(F/E)/2.)*(180./pi);
     if(Iy == Ix) theta_p=45.;
     fprintf(fout,"output:
");

     fprintf(fout,"xbar=%16.5f
ybar=%16.5f
Ix=%18.5f
Iy=%18.5f
", bar, ybar, Ix, Iy);
     fprintf(fout,"Ixy=%17.5f 
I1=%18.5f 
I2=%18.5f 
theta_p=%13.5f
", Ixy, I1, I2,
     theta_p); fclose(fin);
     fclose(fout);
}
input:
a=  140.00000    b= 120.00000    t= 20.00000
output:
xbar=     35.00000
ybar=     95.00000
Ix=    8840000.00000
Iy=    5960000.00000
Ixy=    4200000.00000
I1=    11840000.00000
I2=    2960000.00000
theta_p=  –35.53768

Reference

C.1. YOUNG, W. C., BUDYNAS, R. G., and SADEGH, A. M. Roark’s Formulas for Stress and Strain, 8th ed. New York: McGraw-Hill, 2011.

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Table of Contents for Appendix C. Moments of Composite Areas

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Appendix C. Moments of Composite Areas

C.1 Centroid

C.2 Moments of Inertia

C.2.1 Parallel Axis Theorem

C.2.2 Principal Moments of Inertia

Reference

Table of Contents for
Appendix C. Moments of Composite Areas