HW5_Soln_r1 - = 8/5 = = = 1 in = = = 1 in Modulus weighted...

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AE322 – Solution Set #5 Soln 9.1 c Determine the y* and z* for the section in figure. Since the section is homogeneous, Symmetry about z0 axis implies y*= 0
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Soln 9.2 c Idealized nonhomogeneous, straight beam cross section. Give and Compute , , . Assumption: Thickness ‘t’ is small, s that areas and distances can be approximated from the centerline. Section is nonhomogeneous, the modulus weighted centroid and area moments of inertias are to be computed. = Modulus weighted centroid calculation. Hence the m. w. centroid are given by, = = = 2.92t
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= = = 8.85t Modulus weighted area moment and product of inertia table - = 1903.533 - = 274.42 - = 6 - =
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9.4 a &d For homogeneous section show determine all the moments of inertial including product of inertia with respect to horizontal and vertical axis and determine the principle axis. Modulus weighted centroid.
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Unformatted text preview: = 8/5 = = = 1 in = = = 1 in Modulus weighted area moment and product of inertia table - = 4.2693- 1.6 = 2.67 - = 4.2693- 1.6 = 2.67 - = 0- 1.6 = -d) Principal axis The coordinate axis is to be rotated by measured in anticlockwise direction in order to make product moment of inertia zero is given by, Area moment of inertial in this rotated coordinate system(X,Y) is given by, = 4.27 = 1.07 Verification: 9.4 c &f For homogeneous section show determine all the moments of inertial including product of inertia with respect to horizontal and vertical axis and determine the principle axis Modulus weighted centroid. = 0.192 = = = = = 1.11 in Modulus weighted area moment and product of inertia table - = 0.2757- = 0.039 - = 0.0292- 0.192 = 0.0292 f) Since the cross section is symmetric about z0, the axis y0-z0 is the principal axis, hence ....
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HW5_Soln_r1 - = 8/5 = = = 1 in = = = 1 in Modulus weighted...

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