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Aviation History
1934
1934 - 1442.PDF
SUPPLEMENT TO FLIGHT 766/ 84 THE AIRCRAFT ENGINEER JULY 26, 1934 OX is greater than the angle SQX which the adjusted locus RQ makes with OX, and that AX is stiffer than XB. In other words : in passing from a part of the beam to another part less stiff the angle through which the locus is turned reduces the angle which the adjusted locus makes with the dividing line. Along OX set off in the positive direction a distance / w w \QY equal to ( — 2 5 1 and through V draw UV parallel to SQ. UV is then the final adjusted locus and cuts BjT in X2> the vertex for the length XB. Since X2 is now known, the construction can be repeated from the end B and Xs found. To do this draw X2W X2Wperpendicular to OX and divide X 2W, so that YTW = —• and join X21Y. . Along OX measure off from Y distance YZ in the positive w HIdirection equal to— 2 ——5 and through Z draw ZXXM2 Mi" parallel to X21Y to meet A^ in X,, then Xt is the vertex for the length AX, and circles on OX, and OX2 as diameters complete the diagram. In this case the number of degrees in the sector AOX corresponding to a unit of length of AX is not the same as the number of degrees corresponding to a unit of length of XB. This obtains generally : in more complicated cases a different angular scale must be used for each part of the beam into which the " jumps" of moment of inertia divide it. It is also to be borne in mind in calculating the true shear that the appropriate vertex and value of p for the segment under consideration must be used. The actual position of a vertex is no indication that it applies to a given segment. It is only by following the construction rigorously that diagrams of unusual shape can be exactly drawn. Example (4) A steel beam AB, 95 in. long, is so constructed that 45 in. of its length from A to X has a moment of inertia of .32 in.4 while the remaining length has a moment of inertia of .4 in.4 There is a uniform distributed upward loading of 7.5 lb./in. over the entire length and a compressive end load of 7,000 lb. The end moment at A is 3,500 lb./in. clockwise externally to the left, there is a moment at Bj of 3,000 lb./in. anti- clockwise externally to the right. The necessary quantities are as follows :— .V .32 x 3 x 107 = \f.OOO729 = .027 uXB = aAX = aXB = IV 7,000 ,4 X 3 X 10 .027 X 45 X 57.3 .02415 x 50 x 57-3 7-5 -7 = v 000583 = .02415 = 69.7 deg. w .000729 7-5 .000583 = 69-2 deg. = 10,270 lb./in. = 12,860 lb./in. ' The diagram is shown in Fig. 12. The lettering is the same as in Fig. 11, but in this case the moment of inertia increases as we pass through X. -."-.™; .:--•:/ v; "•/;' CASE VI '• .--._, Change of Moment in the Bay This case may occur because of a change in the cros> section of beam over a part of the spar which results in a shift of the neutral plane above or below the neutral plane of the remainder. AX //// x\ X F1G.I2 Examp/eNo.* ,— \ \ -ve \ M _ 138-3° V zrf ll \l1 V rz r 1/ /u / / -ve / 3S00 Les INS. I • Kins* "Mtttttttt i *• I.-4I11S* "ttttttJttttU-7iLBs/lBK, 7 3000 LBS INS. P«700OLBS. B It may also arise because of a load offset on its point of attachment to the beam. As this interesting case is of rather infrequent occurrence the details of the constrviction will not be given here : they have already been described in the pages of Flight: the reader is referred for these particulars to Flight Aircraft Engineering Supplement, August 26, 1932, where he will find all details. Having mastered the six cases already described he may now, in order to deal with any combination of the various types of loading, superimpose the constructions. FIG. 13 Example No. S. Example No. 5 : Beam under complex Loading Conditions. -
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