Solutions_2

164 referring to fig s164 w 2 2w 1 4 b d3 l2 z a l2

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Unformatted text preview: = 13.33 mm ¯ The second moments of area of the cross-section are calculated using the approximations for thin-walled sections described in Section 16.4.5. Then 2 × 1003 = 5.67 × 105 mm4 Ixx = 40 × 2 × 502 + 80 × 1 × 502 + 12 2 × 403 1 × 803 Iyy = 100 × 2 × 13.332 + + 2 × 40 × 6.672 + 12 12 2 + 1 × 80 × 26.67 = 1.49 × 105 mm4 Ixy = 40 × 2(6.67)(50) + 80 × 1(26.67)(−50) = −0.8 × 105 mm4 186 Solutions Manual 40 mm y 2.0 mm 2.0 mm y C x 100 mm x 1.0 mm 80 mm Fig. S.16.2 The denominator in Eq. (16.18) is then (5.67 × 1.49 − 0.82 ) × 1010 = 7.81 × 1010 . From Eq. (16.18)...
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This note was uploaded on 03/10/2009 for the course MAE 157 taught by Professor Valdevit during the Winter '08 term at UC Irvine.

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