2 Second Moment of Area Support Tutorial PDF

Preview

Summary

Download 2 Second Moment of Area Support Tutorial PDF

Description

Q2 Calculate the second moment of area of the horizontal x-x axis of the beam cross-section below. 15 mm

70 mm 15 mm

100 mm

15 mm

Solution For a symmetrical cross section, the neutral axis or centroid is halfway from bottom to top. There is no need to use the first moment of area to locate this. Neutral axis is 50 mm from bottom. The beam can be considered as three rectangular sections with the following Ix values. Web (1), b = 0.015 mm, h = 0.07 mm. Ix = bh3/12 = 4.29x10-7 m4 Flange (2), b = 0.1 mm, h = 0.015 mm. Ix = bh3/12 = 2.81x10-8 m4 These are the I values at the neutral axis of each section. For the flanges this needs adjusted to the neutral axis of the web section as per the cross-section. Use parallel axis theorem, Ic = Ix + Ay2 to convert the flange values to the centroid of the cross section. Area of flange, A = 0.1 m x 0.015 m = 0.0015 m2. Distance from centroid of cross-section to centroid of flange, y = 35 mm + 7.5 mm = 0.0425 m. For flanges, Ic = 2.81x10-8 + 0.0015(0.0425)2 = 2.74x10-6 m4 There are two flanges and one web, therefore I = 4.29x10-7 + (2) 2.74x10-6 I = 5.90x10-6 m4...