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Aviation History
1941
1941 - 1483.PDF
JULY 3RD, 1941. FLIGHT I CONTINUOUS BEAMS Two Unequal Spans Let one span be less than half the overall span (L) of the beam and equal in length to KL, where K is a con- venient constant, as shown in Fig. 5. UNIFORMLY LOADED CONTINUOUS BEAM FIG. 5. ISO 10O 050 A OVER Z SPAN5. 1 L KL I'L u/ = UNIFORMLY OlSTRlBUTt C :D LOAD. Then, by the Theorem ot Three Moments, the Fixing Moment (Mt) at B is such that :— 2^ j_ = —- (K3 + (1 — k)a)4 M» = ~ (I _ 3K + 3K2) = .125 (1 - 3K + 3K2)WL We can give K various values up to .50 and so find the coefficients of M6, and this has been done in Table III and the values of M „ plotted against K in Fig. 6. Reaction at A (Ra) By moments about B, R..KL W » TOTAL UNIFORMLY DISTRIBUTED LOAO. l_ - OVERALL SPAN M, =~(KL)« w w i FIG. 6. MOMENT (Me) " COEFFICIENT »WL. MB- -O313 WU. ZO (0 1O SO -so FIG. 7. UNIFDtavlLY LOAOED CONTINUOUS BEAM OVER 2 SPANS. W - TOTAL, UNtFOteMLY DISTRIBUTED LOAD. REACTION - COEfTlCJENT x W. 62SW
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