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Aviation History
1931
1931 - 0388.PDF
SUPPLEMENT TOFLIGHT APRIL 24, 1931 THE AIRCRAFT ENGINEER measured by the length of the intercepts l1n1 and n-pi^, but 0nx 0w.2 On, On8 , _ \ B I by the ratios •- —, —-— and our first equation is sec —^ • ~* + m ) tan 6 = Sfl s« and, as usual, the second equation is p.B gjj = MB-MA + RB £AB (In this particular case, since w is zero, and there are no concentrated loads, RB = o). The value of *'B is thus found in terms of MA and MB- Bay CB would be treated similarly to get another expression for in, and the standard procedure followed. MA and MH, having been found by the above method, the actual B.M. diagram is drawn as follows :— Referring to Fig. 5, draw the sectors and boundary lines as for Fig. 4. Suppose both r»A and mB are negative. Lay off OraA and OwH in the negative directions of AA1 and BB1, and draw perpendiculars to give llx and U3(Y) respectively. IIx cuts the first boundary line/1/in fx and BJB in lx. Insert the points nx and n2 using the ratios —- and ——, as found in Fig. 4. Joint fxnx to give llit which cuts the next boundary line gxg in g,. Join g1n2 to give U3(2). The latter cuts 11,(1) in X,. Draw XaX2 perpendicular to gxg, cutting ll2 in X2. Draw X,X! perpendicular to/'/cutting llx in X,. 0X1( OX2 and OXS are the diameters for the cutting arcs to be drawn in sub-sectors 1, 2 and 3, respectively. Since w is zero, the loading arcs shrink into the point 0. The bending moments are shown shaded. In this case all the cutting arcs are below their respective loading arcs, which have shrunk into the point 0, and, therefore, the bending moment is everywhere negative. III. Combination of I and II. Suppose a change of moment of inertia, a change of distributed load w and a concentrated load all occur together at a point in the main bay. Such a point will be a boundary line between two sub-seotors in the polar diagram, and the three types of discontinuity should be dealt with in the order just given, i.e., the locus line is first rotated to account for change in I, then shifted along the boundary line, but parallel to its rotated position to account for the change of w, and finally moved along a perpendicular to the boundary line, still parallel to the rotated position to account for the concentrated load. All such movements of the locus line are recorded as points of intersection with the base line B1B, and, as before, movements of the locus line parallel to itself are measured by the lengths of the intercepts on BJB, while rotations are measured as ratios. A single example applied to a bay AB will suffice to explain the procedure. The various quantities are tabulated in Table III. Draw the sector and boundary lines and assume 0wA ositive as before (it is convenient to make this assumption Table III. always even if mA is known to be negative). Erect a per- pendicular at mA to give 11 x the first locus line which cuts BXB in Zx and Z1/ in /,. Rotate at fx to the position /^ but note that f1n1 is not the locus line 11% because the move- ments /j/a and/2/3 which account for the change of w and the concentrated load respectively, have not yet been made. A line through /3, however, parallel to f1n1 gives ZZ2, the second locus line, which cuts the next boundary line g'g in </,, and B'B in Z2. Rotate about gx to the position gxn^, measure gxg% along, and g^3 perpendicular to g'g to obtain the point ga. A line through g3 parallel to g ii^ gives the last locus line lls, cutting B'B in l3, and making the angle 0 with it. The first equation is written _ + nj,,t + mB tan 0J s H-B the intercepts nxlt, nj,a and the ratios 0n1/0l1, OM2/O?2 being measured from the figure. Note that the intercepts and ratios must be written down in the above equation in the order in which they were obtained in Fig. 6. The second equation is as usual:— P'B + SB = (MB — MA)//AB + RB and iv is obtained in terms of MA and MK. The bay CB would be treated similarly, and the values cf MA and MB would finally be obtained as described in the previous cases. The diagram is drawn as follows :— Referring to Fig. 7, draw the sectors and boundary lines as for Fig. 6, and suppose that both WJA and Wjs have been found to be positive. Lay off 0wA and OraB to represent the numerical values of mA and mv to the scale chosen, and erect perpendiculars to 0»( 1 end ()«, to give locus lines 11^ and ll3 (1) respectively. LetJ^eut B'BinZ] and/1/in/j. Insert on B^ the points ».„ Z2, «,, and I:l in such a way that the intercepts nJz, >i3ts, and the ratios Oi^/Ol^, O«2/OZ2 are equal to the values of these quantities just found in Fig. 6. Join n1f1 and through Z2 draw a parallel to give 11%, which cuts g*g in gx. Join g1n2 and through ls draw a parallel to give lls (2). This cuts ll3 (]) in X3. Through X3 draw a parallel to g3gx of Fig. 6 cutting the rotated position of ll2 (i.e. g^n^) in 22. Drop the perpendicular zg.x on gxg to cut Z/2 in X2. Through X3 draw a parallel to fsfl of Fig. 6 to cut the rotated position of 3626
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