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Problem 9.190

1. For the 5 x 3 x -in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30 o clockwise, (b) the orientation of the principal axes

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Problem 9.190

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  1. 1 For the 5 x 3 x -in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30o clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia. 2 y 0.75 in 0.5 in 1 L5x3x 2 5 in x 1.75 in 3 in 0.5 in Problem 9.190

  2. Problem 9.190 0.75 in 0.5 in 1 For the 5 x 3 x -in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30o clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia. 2 1 L5x3x 2 5 in x 1.75 in 3 in 0.5 in y Solving Problems on Your Own 1. Draw Mohr’s circle.Mohr’s circle is completely defined by the quantities R and IAVE which represent, respectively, the radius of the circle and the distance from the origin O to the center C of the circle. These quantities can be obtained if the moments and product of inertia are known for a given orientation of the axes.

  3. Problem 9.190 0.75 in 0.5 in 1 For the 5 x 3 x -in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30o clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia. 2 1 L5x3x 2 5 in x 1.75 in 3 in 0.5 in Ixy 2q x’ x Ix , Iy y y’ Solving Problems on Your Own 2. Use the Mohr’s circle to determine moments of inertia of rotated axes.As the coordinate axes x-y are rotated through an angle q, the associated rotation of the diameter of Mohr’s circle is equal to 2q in the same sense (clockwise or counterclockwise).

  4. Problem 9.190 0.75 in 0.5 in 1 For the 5 x 3 x -in. angle cross section shown, use Mohr’s circle to determine (a) the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30o clockwise, (b) the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia. 2 1 L5x3x 2 5 in x 1.75 in 3 in 0.5 in Solving Problems on Your Own 2. Use the Mohr’s circle to determine the orientation of principal axes, and the principal moments of inertia.Points A and B where the circle intersects the horizontal axis represent the principal moments of inertia. The orientation of the principal axes is determined by the angle 2qm. Ixy 2qm x B A Ix, Iy y

  5. Problem 9.190 Solution 0.75 in 0.5 in 1 L5x3x 2 5 in x 1.75 in 3 in 0.5 in 2 1 From Fig. 9.13A: Ix = 9.45 in4, Iy = 2.58 in4 Product of inertia Ixy: Ixy = (Ixy)1 + (Ixy)2 For each rectangle Ixy = Ix’y’ + xy A andIx’y’ = 0 (symmetry) Thus: Ixy = Sxy A y 0.5 in A, in2x, inyinx y A, in4 1 1.5 0.75 -1.5 -1.6875 2 2.25 -0.5 1.0 -1.125 Ixy = Sxy A = -2.81 1 in x 1.5 in 0.75 in

  6. y Problem 9.190 Solution x 1 2 1 2 1 Ixy (in4) 2 1 2 Y (2.58, +2.81) 2.81 FX Draw Mohr’s circle. C tan 2qm = = F CF 9.45-6.02 O Ix, Iy(in4) 2qm X (9.45, -2.81) The Mohr’s circle is defined by the diameter XY where X(9.45, -2.81) and Y(2.58, 2.81). The coordinate of the center C and the radius R are calculated by: OC = Iave = (Ix + Iy) OC = (9.45 + 2.58) = 6.02 in4 R = [ (Ix - Iy)]2 + Ixy R = [ (9.45 - 2.58)]2 + (-2.82)2 R = 4.44 in4 qm = 19.6o

  7. y Problem 9.190 Solution E D C y’ Use the Mohr’s circle to determine moments of inertia of rotated axes. (a) The coordinates of X’ and Y’ give the moments and product of inertia with respect to the x’y’ axes. x 30o Ix’ = OE = OC_CE = 6.02 _ 4.44 cos 80.7o Ix’ = 5.30 in4 Iy’ = OD = OC + CD = 6.02 + 4.44 cos 80.7o Iy’ = 6.73 in4 x’ Ixy (in4) 6.02 in4 Y’ Y R = 4.44 in4 O Ix, Iy(in4) Ix’y’ = EX’ = _ 4.44 sin 80.7o 80.7o X 39.3o Ix’y’ = _ 4.38 in4 2q = 60o X’

  8. y Problem 9.190 Solution 2.81 FX tan 2qm = = CF 9.45-6.02 Ixy (in4) Y (2.58, +2.81) C F O Ix, Iy(in4) 2qm X (9.45, -2.81) b Use the Mohr’s circle to determine the orientation of principal axes, and the principal moments of inertia. a qm = 19.6o (b) The principal axes are obtained by rotating the xy axes through angle qm. x qm = 19.6o The corresponding moments of inertia are Imax and Imin : _ _ Imax, min = OC + R = 6.02 + 4.44 Imin Imax Imax = 10.46 in4 Imin = 1.574 in4 a axis corresponds to Imax b axis corresponds to Imin R = 4.44 in4

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