Chapter 10: Moments of Inertia

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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) 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 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.. The moment of inertia about the y-axis may be determined using the following equation. Iy = x el 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 da 10.1

2 The polar moment of inertia may be determined from the rectangular moments of inertia Ix and Iy by noting that r = x + y and that JO = r da = (x + y ) da = x da + y da JO = Iy + Ix 10. 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 = y da = (y + dy) da = (y ) da + y dy da + (dy) da The distance dy is a constant and is set at the beginning of the problem. Thus, Ix = (y ) da + dy y da + (dy) da 10.

3 (y ) da - represents the moment of inertia of area A with respect to the centroidal x -axis; thus, (y ) da = I x dy y da - represents the first moment of area A with respect to the centroidal x -axis. (dy) da = (dy) A Since the centroid is located on the axis, the value of this integral is zero: y da = y A = 0, since y = 0. Thus, Ix = I x + (dy) 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) 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) A 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 = kx A, and kx = (Ix/A) ½ where kx is known as the radius of gyration with respect to the x-axis. 10.3

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 = ky 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 A, and ko = (JO/A) ½ where ko is known as the polar radius of gyration. Also observe that kx + ky = ko. 10.4

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 el da y el = y da = b dy Solve for Ix, the moment of inertia with respect to the x-axis. Ix = y el da = h 0 y b dy = b h 0 y dy = b (y 3 /3) h 0 Ix = bh 3 /3 Thus, the moment of inertia of a rectangular area about its base is bh 3 /

6 Given: Rectangular area shown. Find: The moment of inertia of a rectangular area about an axis through the centroid. Applicable equation: Ix = y el da y el = y da = b dy Solve for Ix, the moment of inertia with respect to the x-axis. h/ Ix = y el da = h/ y h/ b dy = b h/ y dy = b (y 3 h / /3) = (b/3)[(h/) 3 (- h/) 3 ] h / = (b/3)(h 3 /8 + h 3 /8) = (b/3)(h 3 /4) Ix = bh 3 /1 Thus, the moment of inertia of a rectangular area about an axis through its centroid is bh 3 /

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) y 3 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) y 3 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 da where x el = x da = y dx Thus, Iy = x 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 ) or ( y el ) are used in the expression, and the result will always be positive. 10.7

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 el da da = x dy Iy = x el da cannot be used with the horizontal element. x el = x/ diy = (1/3) x 3 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 el da = h 0 y (b/h)(h y) dy = (b/h) h 0 y (h y) dy = (b/h) h 0 (hy y 3 ) dy = (b/h)(hy 3 /3 y 4 /4) h 0 = (b/h)(h 4 /3 h 4 /4) = (b/h)(h 4 /1) Ix = bh 3 /1 10.8

9 Now solve for Iy, the moment of inertia with respect to the y-axis. d Iy = (1/3) dy (x 3 ) Iy = (1/3) x 3 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) (h 4 /4) ] = 0 + (1/3) (b 3 /h 3 ) (h 4 /4) Iy = b 3 h/1 10.9

10 Given: The area shown. Find: Ix and Iy using a horizontal element. Applicable equations: Ix = y el da da = (a x) dy Iy = x el da cannot be used with the horizontal element. x el = ½ (x + a) diy = (1/3) x 3 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 el da = b 0 y (a - y ) dy = b 0 (a y y 5/ ) dy = a y 3 /3 (/7) y 7/ b = ab3 /3 (/7) b 7/ 0 = a b 3 /3 (/7) b 1/ b 3 Format the solution in the form f (a, b 3 ) = a b 3 /3 (/7) a b 3 b = a, a = b 1/ = 7 a b 3 /1 6 a b 3 /1 Ix = a b 3 /1 Next, solve for Iy, the moment of inertia with respect to the y-axis. Iy = x el da cannot be used with a horizontal element. d Iy = (1/3) dy x 3 must be used with a horizontal element. However, the expression d Iy = (1/3) dy x 3 is only valid when the bases of all the elemental rectangular areas touch the reference axis

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 = ba 3 /3 - The moment of inertia with respect to the y-axis for Area is Iy = (1/3) x 3 dy Since y = x, then x 3 = y 3/ = (1/3) b 0 y 3/ dy = (1/3) (/5) y 5/ ) b 0 = (1/3) (/5) b 5/ Format the solution in the form f(a 3,b) = (/15) b b 3/ b = a, b 3/ = a 3 Iy = b a 3 /15 The moment of inertia for the area is Iy = Iy1 - Iy = ba 3 /3 - ba 3 /15 Iy = ba 3 /

12 Given: The area shown. Find: Ix and Iy using a horizontal element. Applicable equations: Ix = y el da da = (x1 x) dy Iy = x el da cannot be used with the horizontal element. x el = ½ (x1 + x) diy = (1/3) x 3 dy will be used to determine Iy. y el = y Define da in terms of y. da = (x1 x) dy = (y y /4) dy First, solve for Ix, the moment of inertia with respect to the x-axis. Ix = y el da = 4 0 = (y 4 /4 y 5 /0) 4 0 Ix = 1.8 in 4 y (y y /4) dy = 4 0 = (y 3 y 4 /4) dy Next, solve for Iy, the moment of inertia with respect to the y-axis. Iy = x el da cannot be used with a horizontal element. d Iy = (1/3) dy x 3 must be used with a horizontal element. The area may be considered as the difference between the two areas shown below. 10.1

13 Iy = Iy1 - Iy = 4 (4 3 )/1 - (1/3) x 3 dy = 4 (4) 3 /1 (1/3) 4 0 (y /4) 3 dy = 1.33 (1/3) 4 0 (y 6 /64) dy = 1.33 (1/3) (1/64) (y 7 /7) 4 0 = Iy = 9.14 in

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 = bh 3 /1 The moment of inertia with respect to the x-axis can be determined using the Parallel Axis Theorem. Ix = I x + y A = bh 3 /1 + (h/) b h = bh 3 /1 + bh 3 /4 Ix = bh 3 /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 h 3 /1 and Iy = h b 3 /1 The centroidal moments of inertia for a rectangle with respect to the x- and y- axes are I x = Ix - y A = b h 3 /1 - (h/3) (b h/) = b h 3 /1 b h 3 /18 I x = b h 3 /36 I y = Iy - x A = h b 3 /1 - (b/3) (b h/) = h b 3 /1 h b 3 /18 I y = h b 3 /

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 A) 1 Ix = I x + ( y A) Then summing the moments of inertia for each part of the composite area Ix = Ix1 + Ix + or Ix = I x1 + ( y A) 1 + I x + ( y A) + Ix = I x i + y i Ai 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 A) 1 Iy = I y + ( x A) Then summing the moments of inertia for each part of the composite area Iy = Iy1 + Iy + or Iy = I y1 + ( x A) 1 + I y + ( x A) + Iy = I y i + x i Ai A tabular solution is illustrated for the following example

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 i Ai y i I xi I yi x 3 = 4R/3π = 4(6)/3π =.55 = y 3 y 4 = - 4R/3π = - 4()/3π = I x1 = bh 3 /1 = 3(9) 3 /1 = 18.5 I y1 = bh 3 /1 = 9(3) 3 /1 = 0.5 I x = bh 3 /36 = 6(9) 3 /36 = I y = bh 3 /36 = 9(6) 3 /36 = I x3 = R 4 = (6) 4 = I y3 = I x3 = I x4 = 0.109R 4 = 0.109() 4 = 1.74 I y4 = (π/8)r 4 = (π/8)() 4 = 6.8 Use I x4 = (negative area) Use I y4 = (negative area) The moments of inertia with respect to the x- and y-axes are determined as follow. Ix = I xi + y i Ai = = in 4 Iy = I yi + x i Ai = = in 4 The location of the centroid is determined as follows. x = x iai/ Ai = /75.99 = y = y iai/ Ai = /75.99 = 1.55 The centroidal moments of inertia for the composite area are determined as follows. I x = Ix y A = (1.55) = in 4 I y = Iy x A = (- 1.11) = in

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 (9.8/) + 1(18)( /) = y = (18) 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 = [ (9.8/ 0.48) ] + [18(1) 3 /1 + 1(18)( / 0.48) ] = ( ) + ( ) Ix = in

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 10.18

19 Example Product of Inertia by Integration Given: The area shown. Find: Ixy and I xy Applicable equations: Ixy = x el y el da Using a vertical element: da = (y1 y) dx Ixy = I xy + x y A, so I xy = Ixy - x y A x el = x Define da and y el in terms of x. y el = ½ (y1 + y) da = (y1 y) dx = (3x x 3 /3) dx y el = ½ (y1 + y) = ½ (3x + x 3 /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 0 x [½ (3x + x 3 /3)] (3x x 3 /3) dx = ½ 3 0 x (9x x 6 /9) dx = ½ 3 0 (9x 3 x 7 /9) dx = ½ (9x 4 /4 x 8 /7) 3 0 = ½ (79/4 6561/7) = ½ ( ) Ixy = in 4 Next, solve for I xy using the Parallel Axis Theorem. A = da = 3 0 A = 6.75 in (3x x 3 /3) dx = (3x / x 4 /1) 3 0 = (7/ 81/1) x A = 3 x (3x x 3 /3) dx = 0 3 (3x x 4 /3) dx = (3x 3 /3 x 5 /15) = (81/3 43/15) = x A = in

20 y A = 3 0 [½ (3x + x 3 /3)] (3x x 3 /3) dx = ½ 3 0 (9x x 6 /9) dx = ½ (9x 3 /3 x 7 /63) 3 0 y A = 3.14 in 3 = ½ (43/3 187/63) = ½ ( ) x = x A/A = 10.80/6.75 = 1.60 y = y A/A = 3.14/6.75 = 3.43 I xy = Ixy - x y A = (1.60)(3.43) 6.75 I xy = 8.5 in 4 Alternatively, solve for Ixy, the product of inertia with respect to the x- and y- axes, using a horizontal element. da = (x x1) dy x el = ½ (x + x1) y el = y Define da and x el in terms of y. da = (x x1) dy = ( 3 x el = ½ (x + x1) = ½ ( 3 3 y - y/3) dy 3 y + y/3) Ixy = x el y el da = 9 0 ½ ( 3 3 y + y/3) y ( 3 3 y - y/3) dy = ½ 9 y [(3y) /3 y /9] dy = 0 ½ 9 0 (.08 y 5/3 y 3 /9) dy = ½ [.08 (3/8) y 8/3 y 4 /36] 9 = ½ ( ) 0 Ixy = in

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 = y da Iy = x 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 = v da = (y cos θ x sin θ) da = (y cos θ x y cos θ sin θ + x sin θ) da This equation results in the following expression. Iu = ½ (Ix + Iy) + ½ (Ix Iy) cos θ Ixy sin θ A similar expression may be developed for Iv. Iv = ½ (Ix + Iy) - ½ (Ix Iy) cos θ + Ixy sin θ Observe the following relationship. Ix + Iy = Iu + Iv 10.1

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 θ Ixy sin θ with respect to θ. The results of this operation are as follows. Imax,min = ½ (Ix + Iy) ± {[(Ix Iy )/] + Ixy } ½ tan θ = - 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. 10.

23 Examples Principal Moments of Inertia Given: Area shown. Find: Imax and Imin Part Ai x i y i Ai x i Ai y i Ai x i y i I xi I yi I xyi Ix = I xi + Ai y i = = in 4 Iy = I yi + Ai x i = = 6.96 in 4 Ixy = I xyi + Ai x i y i = 0 + ( ) = in 4 I max,min Imax,min = ½ (I x + I y ) ± {[(I x I y )/] + I xy } ½ = ½ ( ) ± {[( )/] + (- 6.56) } ½ = 8.67 ± 6.78 Imax = in 4 Ix = in 4 Imin = 1.89 in 4 Iy = 6.96 in in in 4 OK tan θ = - Ixy/(Ix Iy) tan θ = - (- 6.56)/( ) = θ = Solve for Iu using θ = 37.7 ; Iu must equal in 4 or 1.89 in 4. Iu = ½ (Ix + Iy) + ½ (Ix Iy) cos θ Ixy sin θ = ½ ( ) + ½ ( ) cos (37.7) sin (37.7) = Iu = in 4 MAX 10.3

24 Given: Area shown. (Extend previous example.) Find: a) Imax and Imin b) I max and I min From previous work: a) Ix = in 4 Iy = in 4 b) I x = in 4 I y = in 4 Part Ai x i y i I xyi I xy = b h /7 = (6) (9) /7 = I xy3 = R 4 = (6) 4 = a) Find Imax and Imin Ixy = I xyi + x i y iai = = in 4 Imax,min = ½ (Ix + Iy) ± {[(Ix Iy )/] + Ixy } ½ = ½ ( ) ± {[( )/] + (16.19) } ½ = ± Imax = in 4 Ix = in 4 Imin = in 4 Iy = in in in 4 OK tan θ = - Ixy/(Ix Iy) = - (16.19)/( ) = θ =

25 b) Find I max and I min I xy = Ixy x y A = (- 1.11) (1.55) I xy = in 4 I max,min = ½ ( I x + I y) ± {[( I x I y )/] + I xy } ½ = ½ ( ) ± {[( )/] + (56.93) } ½ = ± I max = in 4 I x = in 4 I min = in 4 I y = in in in 4 OK tan θ = - Īxy/(Īx - Īy) = - (56.93)/( ) = θ =

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