Calculate the moments of inertia of the shaded area about the x- and y-axes. 91 44, 44 --x 91 -6767 Dimensions in millimeters Answers: Ix= i (106) mm4 ly= (10) mm4
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- What angle (in radians) should the axes be rotated which will result to the average moment of inertia for either x or y axes? Moment of Inertial about the Y-axis = 0.667 x 106 mm4 ,and Product of Inertia, Ixy = 1.000 x 106 mm4The shaded surface shown has moment of inertia about a centroidal axis always of195.75 cm4no matter how the axis is oriented. Determine:a) dimension "a"b) Ix and Iyc) orientation of the u axis with respect to which the moment of inertia will be the maximum (angle α(alpha)) and how much willImax? (need free body diagrams)Determine the moments of inertia of the rectangular area about the x- and y-axes and find the polar moment of inertia about point O.Assume h0 = 0.37h, b0 = 0.15b.
- Determine the moment of inertia of the shaded area with respect to y-axis Where: b1=10mm; b2=42mm; h40mm and h2=25mmDetermine the moment of inertia about the U axis (Iu). (theta=35%)Determine the centroidal moment of inertia.**correction in picture L200x200x20 L 200 x 200 x 20theoretical mass: 59.7 kg/marea: 7600 mm2 I: 28.8 x 106 mm4 S (I/C) :202 x 103 mm3 r: 61.6 mmx or y: 57.4 mmAxis z r: 39.3 mmC380 x 74 theoretical mass: 74.4 kg/marea: 9480 mm2 depth: 381 mm flange width: 94 mmflange thickness: 16.5 mm web thickness: 18.2 mm axis x I: 168 x 106 mm4 axis x S (I/C) :881 x103 mm3 axis x r: 133 mm axis y I: 4.60 x 106 mm4 axis y S (I/C) :62.4 x103 mm3 axis y r: 22.0 mmx: 20.3 mm W310 x 500 theoretical mass: 500.4 kg/marea: 63700 mm2 depth: 427 mm flange width: 340 mmflange thickness: 75.1 mm web thickness: 45.1 mm axis x I: 1690 x 106 mm4 axis x S (I/C) :7910 x 103 mm3 axis x r: 163 mm axis y I: 494 x 106 mm4 axis y S (I/C) :2910 x103 mm3 axis y r: 88.1 mm
- The shaded area has the following properties:�# = 1.26 �10- ��' ; �. = 6.55 �100 ��' ; and �#. = −1.02 100 ��'Determine the moments of inertia of the area about the x’ and v’ axes if θ=30oUse Virtual Work Method to calculate the rotation at C for the frame shown in the accompanying illustration. Use EI=50.0×103kN∙m2find the moment of inertia Ixy and Iyy in regard to the given axis
- Should the moment at ED be 256.95? and i am not seeing you calculate that moment at EDetermine the moment of inertia and radius of gyration of the shaded area shown with respect to the y axis.Fig. P9.11For the angular section shown: Which of the following most nearly gives the centroidal moments of Inertia Igx and Igy. a.) None of the Above b.) Igx=3.047 x 10⁶ mm⁴; Igy=840.63 x 10³ mm⁴ c.) Igx=4.57 x 10⁶ mm⁴; Igy=868.63 x 10³ mm⁴ d.) Igx=3.47 x 10⁶ mm⁴; Igy=848.63 x 10³ mm⁴ e.) Igx=3.057 x 10⁶ mm⁴; Igy=828.63 x 10³ mm⁴