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Aviation History
1919
1919 - 0706.PDF
From the trigonometrical tables we calculate : /, $) = i.017 /« («) = 1.029 fs ») = 1.025 /, (9) = 1.25; /2 (<*>) = 1.464 /, (</>) = 1.392 /•(•) =i.<u ' U (*) = 1-275 /» W = 1-234 Then equations (13) to (16) give :— 2.58 M + 40.3 M' — 19.45 19.45 M' - 2.75 N — 39.9 1.017 M + 0.5I5M' 1.257 N + 0.732 N' And, solving these, we find M = — 9,700, M' = — 1,310, N = — 12,700, N' = — 1,350. Second Approximation.—Taking these values of the couples, we find for the corrected values olf the thrusts :— P - 291, Q = 3540, R = 357. (18) Proceeding as before we obtain :— a = 5.90 IO~8 6 = 0.266 = 15.3" io_s <p = 0.800 = 45-9° ior' + = 0.610 = 35" N' = — 51,700. N' = 63,400. = — 10,380. = — 16,900. and (3 y 17.8 19.1 /, (6) = 1.02 /s (6) = 1.034 h (e) - l-°29 /, (<p) = 1.228 f. (4>) = 1.409 f:i <p) = 1.347 /, (+) = 1.116 f, (+) = I.207 /:) (+) = 1.176 The equations for the couples are :— 2.59 M +39.2 M' — 18.4 N' = — 52,100 18.4 M' - 2.64 N -38.6N'= 61,200 1.02 M + 0.517 M' =—10,40 1.228N + 0.705 N' . = — 16,350 which give :— M = — 9540, M' = — 1,340 ; N = — 12,570, N' = — 1,360. Taking these values we find :— P = 292, Q = 3,545, R = 359, which are sufficiently near to those from which we started the second approximation (18) to show that we need not proceed any further. The bending moment diagrams for the three members are now easily plotted from (5) and (6) ; the results for the members AB and BC are shown in Fig. 3, together with the diagram for AB when the strut is pin-jointed. The maximum bending moment on the strut is 1,360 lb. ins. This gives a stress due to bending = 1,360 x 0.75 •*• 0.7 = 1,460 lbs./in.2, and.due to the thrust there is a stress = 359/5.9 —61 lbs./in.2, so that the total compression stress is 1,521 lbs./in.1 7. General Equations.—We shall now deal with a structure of any number of bays, as shown in Fig. 7, where two con secutive bays are shown. "** N ja l*X Hi* m -Nl Fl« t The notation for the bending moments is shown in the diagram ; in addition we shall use 6X, 0, ; I,, I.:,.... ; P,, P3, .. .. ; /), L for the longitudinal member of ABC ; fr, <h : l'u *J ! Qi- Q* for the A'B'C . . . . ; +1J, I]-, R]2, for the cross members such as BB'. The shearing forces at the end of each member, considered as a separate beam, are denoted in the same manner as in Fig. 6 ; X,, X'], etc., are used for the upper beam ; Yj, Yi', etc., for the lower beam ; Z15, Z'12> members. Then we have :— etc., for the cross X,= Y,= Zi2 = Wi li M,' - M, » + h w h N/ - N, 2 + /, w, /, M, - M'j X', - ' +• —--, 2 l\ , »,'/, N, - N,' Y> - 2 + h M',, - M„ Z'»- M.s - M'12 > (19) [P»l. 2 (M„_ M'„_j) + (N„ - N'„) + (M„_!, H - M'«_ .«)] (20) where [P«](l denotes the value of P„ when the structure is pin-jointed. Similarly :— Q« = [Q4, + iff (M„ - M'„) + (N„ - N'„) + (M,,.,, and R«_i - M'„_„ „)"] (21 etc., etc. If the last complete bay to the right be denoted by the suffix 1, then in the nth bay from that end, we have El 51 An Air Raid in the Caspian. DETAILS have only just been made known of a raid by British aircraft which occurred in the Caspian on May 10. From their base at Baku, the machines proceeded to Astra khan, a distance of more than 400 miles, and bombed the [R«-i,«](, + M„ - M'„ N*_, - N'„ (22) In ' h, . The equations are easily verified. Now, considering two contiguous bays, such as those shown in Fig. 7, we must equate the slopes of AB, B'B, and BC at B. Using the results (2) and (3), and paying due respect to signs, we obtain the equations ;— M, I, . ... M,' Ij w^lf 3'. M, 2 2 , . M' M , , M,,A. M',,\ >> 24I2 h (»S) 3115 A *J + 61, /• W Similarly at B we obtain :— 3i', /1 W + er, fA*l) + 24r, (23) f, (•,) N2 h = - ~KT' H vh) ~ 6IL 6l» h M M'i2A 3li2 2 2 /• /* ^ 4!,' /» <*2) h (+1! (24) (25) sufficient to And we also have the equations :— Mw- M', - M3\etc M'B- N2 - N',jetc- The systems of equations (20) to (25) obtain all the couples and end-loads. In the event of any of the members being in tension, the modifications mentioned in Section 4 must be made. If any members have no end-load, the/'s for that member all become unity. The equations can easily be extended to include cases where all the bracing members are rigidly jointed, and the upper and lower spans are unequal, as, for example, a Warren girder. If external couples are applied at the points A, B, C, etc., equations (25) must be modified accordingly. 8. The Effect of the Stretching of the Ties.—In the above work we have neglected the fact that the ties stretch, thereby allowing a vertical movement of the cross members or struts. Usually this effect is not important, but it may be allowed for, if desired, by making the following modifications to the theory. (1) In Sections 4 and 5, if 8 denote the amount by which the end of B sinks below A, neglecting the changes of length of the members AB, BC, CD, 8 is given by— Y +R 8 = EA -JP + ' (26) where A is the area of the cross-section of the tie. Then ;— xi (a) To the expression for y, (1), add -y. (b) To the expression for -£. add 5// in (2) and (3). (c) In the expressions for the reactions at the ends, to S' add PS//; from S subtract Pb/l. (d) The equations (5) to (8) for the bending moment are unaltered. (e) To the left-hand side of equations (13), (14), (15) and 8 (16) add the term j • (2) In the general equations of Section 7, the expressions for the reactions (19) must be corrected as indicated in (1) c above, and— To the first line of equations (23) and (24) add 8i2//. To the second line of equations (23) and (24) add 8,,,//, where J12 is the amount by which B sinks below C, and S0l the amount by which A sinks below B ; 812 and 8W are given by expressions similar to (26) above. m m Bolshevist fleet and arsenal there. Several hits were observed and fires were seen to break out. The Bolshevist fleet in Astrakhan consists of ten small destroyers and two submarines. The submarines have been out, but they always discreetly vanish when the British flotilla on the Caspian appears. 706
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