01-02ChapGere.0030 - 5(sin u cos u(2(cos u 2 sin u 2(1 1...

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SECTION 1.8 Design for Axial Loads and Direct Shear 61 A NGLE THAT MAKES W A MINIMUM Use Eq. (2) Let S ET THE NUMERATOR 0 AND SOLVE FOR : sin 2 cos 2 sin 2 cos 2 cos 4 0 Replace sin 2 by 1 cos 2 : (1 cos 2 )(cos 2 ) 1 cos 2 cos 2 cos 4 0 Combine terms to simplify the equation: u 54.7 1 3 cos 2 u 0 cos u 1 3 sin 2 u cos 2 u sin 2 u cos 2 u cos 4 u sin 2 u cos
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Unformatted text preview: 5 (sin u cos u )(2)(cos u )( 2 sin u ) 2 (1 1 cos 2 u )( 2 sin 2 u 1 cos 2 u ) sin 2 u cos 2 u df d u 5 f 5 1 1 cos 2 u sin u cos u A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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