3/31/2014 10.6 Product of Inertia • Product of Inertia: I xy = ∫ xy dA • When the x axis, the y axis, or both are an axis of symmetry, the product of inertia is zero. • Parallel axis theorem for products of inertia: I xy = I xy + x y A 9- 1 Sample Problem 10.6 Determine the product of inertia of the right triangle (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes. SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips • Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes. 9- 2 Sample Problem 10.6 (continue) SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips x x y = h1 − dA = y dx = h1 − dx b b x xel = x yel = 12 y = 12 h1 − b Integrating dIx from x = 0 to x = b, b I xy = ∫ dI xy = ∫ xel yel dA = ∫ x 0 (12 )h2 1 − bx 2 2 dx b b x 2 x3 x 4 x x 2 x3 = h 2 ∫ − + 2 dx = h − + 2 2 b 2 b 0 4 3b 8b 0 I xy = 1 b 2h 2 24 9- 3 1 3/31/2014 Sample Problem 10.6 (continue) • Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes. x = 13 b y = 13 h With the results from part a, I xy = I x′′y ′′ + x yA I x′′y ′′ = 1 b 2h2 24 ( )(13 h)(12 bh) − 13 b 1 b 2h2 I x′′y ′′ = − 72 9- 4 10.7 Moments of Inertia about inclined axis In structural and mechanical design, it is sometimes necessary to calculate the moment of inertia with respect to a set of inclined u, v, axes when the values of θ , Ix, Iy, Ixy are known. To do this we will use transformation equations which relates the x, y, and x’, y’ coordinates. From Figure, these equations are: Note: x′ = x cosθ + y sin θ y ′ = y cosθ − x sin θ 9- 5 Moments of Inertia about inclined axis,, continue • The change of axes yields I x′ = I y′ = Ix + I y 2 Ix + I y I x′y′ = Given I x = ∫ y 2 dA I y = ∫ x 2 dA I xy = ∫ xy dA we wish to determine moments and product of inertia with respect to new axes x’ and y’. 2 Ix − I y 2 Ix − I y cos 2θ − I xy sin 2θ 2 Ix − I y − cos 2θ + I xy sin 2θ 2 + sin 2θ + I xy cos 2θ • By adding the equations for Ix’ and Ix’ we can show that the polar moment of inertia about z axis passing through point O is independent of the orientation of x’ and y; Jo = Ix’ + Iy’ = Ix + Iy • These equations show that Ix’ , Iy’ and Ix’y’ depend on the angle of the inclination, θ, of the x’, y’ axes. 9- 6 2 3/31/2014 Principal Axes and Principal Moments of Inertia We will now determine the orientation of these axes about which Ix’ , Iy’ are maximum and minimum. This particular axes are called principal axes By differentiating the first of Eqs. 10-9 with respect to θ and setting the result to zero. Thus; dI x ' dθ Therefore, at θ = θp ; By substituting for θ in Ix’ , Iy’ and Ix’y’ equations and simplifying , we obtain; 9- 7 Summary,,,,, Ix − I y …….. 10-9 (a) − cos 2θ + I xy sin 2θ 2 2 Ix − I y = sin 2θ + I xy cos 2θ 2 …….. 10-9 (a) I y′ = I x′y′ Ix + Iy cos 2θ − I xy sin 2θ I x′ = 2 Ix + Iy + 2 Ix − I y …….. 10-9 (a) ………….. 10-10 …….. 10-11 9- 8 Principal Axes and Principal Moments of Inertia Squaring Eqs. 10-9a and 10-9c and adding, it is found that; (I x ′ − I ave )2 + I x2′y ′ = R 2 where ; I ave = R = Ix + Iy 2 , Ix − I y 2 + I xy 2 • At the points A and B, Ix’y’ = 0 and Ix’ is a maximum and minimum, respectively. I max, min = I ave ± R tan 2θ m = − 2 I xy Ix − I y • The equation for θm defines two angles, 90o apart which correspond to the principal axes of the area about O. • Imax and Imin are the principal moments of inertia of the area about O. 9- 9 3 3/31/2014 10.8 Mohr’s Circle for Moments and Products of Inertia • The moments and product of inertia for an area are plotted as shown and used to construct Mohr’s circle, I ave = Ix + Iy 2 Ix − Iy 2 R = + I xy 2 • Mohr’s circle may be used to graphically or analytically determine the moments and product of inertia for any other rectangular axes including the principal axes and principal moments and products of inertia. 9 - 10 Sample Problem 10.7 For the section shown, the moments of inertia with respect to the x and y axes are Ix = 10.38 in4 and Iy = 6.97 in4. Determine (a) the orientation of the principal axes of the section about O, and (b) the values of the principal moments of inertia about O. SOLUTION: • Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles and applying the parallel axis theorem to each. • Determine the orientation of the principal axes (Eq. 9.25) and the principal moments of inertia (Eq. 9. 27). 9 - 11 Sample Problem 10.7 SOLUTION: • Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles. Apply the parallel axis theorem to each rectangle, ( I xy = ∑ I x′y′ + x yA ) Note that the product of inertia with respect to centroidal axes parallel to the xy axes is zero for each rectangle. Rectangle Area, in 2 I II III y , in. x yA, in 4 1.5 − 1.25 + 1.75 − 3.28 1 .5 x , in. 0 0 0 1.5 + 1.25 − 1.75 − 3.28 ∑ x yA = −6.56 I xy = ∑ x yA = −6.56 in 4 9 - 12 4 3/31/2014 Sample Problem 10.7 • Determine the orientation of the principal axes (Eq. 9.25) and the principal moments of inertia (Eq. 9. 27). tan 2θ m = − 2 I xy Ix − I y =− 2(− 6.56) = +3.85 10.38 − 6.97 2θ m = 75.4° and 255.4° θ m = 37.7° and θ m = 127.7° I x = 10.38 in 4 I max, min = I y = 6.97 in 4 Ix + I y 2 2 Ix − I y 2 + I xy ± 2 2 I xy = −6.56 in 4 = 10.38 + 6.97 10.38 − 6.97 2 ± + (− 6.56) 2 2 I a = I max = 15.45 in 4 I b = I min = 1.897 in 4 9 - 13 Sample Problem 10.8 SOLUTION: • Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based on the circle diameter between the points. • Based on the circle, determine the orientation of the principal axes and the principal moments of inertia. The moments and product of inertia with respect to the x and y axes are Ix = 7.24x106 mm4, Iy = 2.61x106 mm4, and Ixy = -2.54x106 mm4. • Based on the circle, evaluate the moments and product of inertia with respect to the x’y’ axes. Using Mohr’s circle, determine (a) the principal axes about O, (b) the values of the principal moments about O, and (c) the values of the moments and product of inertia about the x’ and y’ axes 9 - 14 Sample Problem 10.8 SOLUTION: • Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based on the circle diameter between the points. OC = I ave = ( 1 2 (I x + I y ) = 4.925 × 106 mm4 ) CD = 12 I x − I y = 2.315 × 106 mm4 R= 6 I x = 7.24 × 10 mm 4 I y = 2.61 × 106 mm 4 I xy = −2.54 × 106 mm 4 (CD )2 + (DX )2 = 3.437 × 106 mm4 • Based on the circle, determine the orientation of the principal axes and the principal moments of inertia. tan 2θ m = DX = 1.097 2θ m = 47.6° CD θ m = 23.8° I max = OA = I ave + R I max = 8.36 × 106 mm 4 I min = OB = I ave − R I min = 1.49 × 106 mm 4 9 - 15 5 3/31/2014 Sample Problem 10.8 • Based on the circle, evaluate the moments and product of inertia with respect to the x’y’ axes. The points X’ and Y’ corresponding to the x’ and y’ axes are obtained by rotating CX and CY counterclockwise through an angle Θ = 2(60o) = 120o. The angle that CX’ forms with the x’ axes is φ = 120o - 47.6o = 72.4o. I x ' = OF = OC + CX ′ cos ϕ = I ave + R cos 72.4o I x′ = 5.96 × 106 mm 4 I y ' = OG = OC − CY ′ cos ϕ = I ave − R cos 72.4o I y′ = 3.89 × 106 mm4 I x′y ' = FX ′ = CY ′ sin ϕ = R sin 72.4o OC = I ave = 4.925 × 106 mm4 R = 3.437 × 106 mm4 I x′y′ = 3.28 × 106 mm4 9 - 16 6
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