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Solutions_2

# Hence from eq iv w whence t b from eqs 1630 v mx

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Unformatted text preview: .1Mx ) td 2 (v) Along the edge 2, x = −d /4, y = d /2. Equation (iv) then becomes σz,2 = From Eqs (i)–(iii), (v) and (vi) In DB σz,1 = − σz,2 = − In BA σz,1 = σz,2 = W td W 2 1 (−1.36My + 1.26Mx ) td 2 (vi) 0.1W 2 td 1.26W td 2 (1 − z) (1 − z) whence σz,1 (B) = − 0.05Wl td 2 0.63Wl td 2 1.85Wl td 2 0.1Wl td 2 whence σz,2 (B) = − (3.6z − 1.85l) (−1.46z + 0.1l) whence σz,1 (A) = − whence σz,2 (A) = td 2 S.16.5 By inspection the centr...
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