Page 55 - Civil Engineering Formulas
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CASE 5. Beam Fixed Both Ends—Continuous Load, Uniformly Distributed
M (max) = WL At center: 24 WL 3 1 D (max) = EI 384 WL M 1 (max) = x: At 12 Wx 2 (L 2 – 2Lx + x 2 ) D = EIL 24 L 2 W – = Lx – x 2 + 6 2L L D W R 1 M M 1 V
At center: At supports: x: At M R x V
W 2
R = R 1 = V (max) = Wx W – V = L 2
x:
At
24
