Solutions_2

Equation iv then becomes z1 1 175my 01mx td 2 v along

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Unformatted text preview: σ= + i.e. σ = 2.08x − 1.12y and at the point A where x = 66.67 mm, y = −50 mm σ (A) = 194.7 N/mm2 (tension) 400 000 × 5.67 × 105 − 800 000 × 0.8 × 105 7.81 × 1010 x −800 000 × 1.49 × 105 + 400 000 × 0.8 × 105 7.81 × 1010 y S.16.3 Initially, the section properties are determined. By inspection the centroid of area, C, is a horizontal distance 2a from the point 2. Now referring to Fig. S.16.3 and taking moments of area about the flange 23 (5a + 4a)t y = 5at (3a/2) ¯ from which y = 5a/6 ¯ 188 Solutions Manual S.16.4 Referring to Fig. S.16.4. W 2 2W 1 4 B D3 l/2 z A l/2 Fig. S.16.4 In DB Mx = −W...
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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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