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Aviation History
1926
1926 - 0607.PDF
77 AUGUST 26, 1926 THE AIRCRAFT ENGINEER SUPPLEMENT TO FLIGHT 4-5 g. continued for any length of time, result in a complete loss of faculties. This loss of faculties is due to the fact that the blood is driven from the head, thus depriving the brain tissues of the necessary oxygen. To the pilot it seemed that sight was the only faculty that was lost. The flight surgeons, McCook Field are of the opinion that sight is the last faculty to be lost under these conditions, even though the pilot may be under the impression that he retains all the others. This opinion is based on the observation of men undergoing rebreather test. The acceleration which an individual can withstand for any length of time depends upon his blood pressure, the person with the higher blood pressure being able to withstand the higher acceleration. Upon the condition of the heart depends the ability of the individual to reco\er quickly from the effect of prolonged acceleration. If the heart is in good condition, there is no danger in vnder- going such a strain unless the acceleration is continued for a period in excess of 10 or 12 minutes, after which death will result. The same is true of the rebreather test; unconscious- ness will result from the deprivation of oxygen and death will result if this is continued for the same length of time." It is also interesting to note that according to this Report the physical effect of the accelerations in rough weather, not being associated with definite actions on the part of the pilot appear to be much more severe than the actual figures would indicate. (To he continv-d.) T.ABLE I.—Values of A for all values of C. Beciangv'ar Sections. SPINDLED AND HOLLOW SPARS. By Lieut.-Col. J. D. BLYTH, O.B.E., M.I.AE.E., late R.A.F. The most usual method of arriving at the maximum amount of spindling or hollowing out permissible in the case of any particular spar section is by trial and error, a process which is apt to become laborious in the absence of good guessing—or luck. The following tables have been got out with the object of making it possible to arrive with certainty at a suitable section at the first attempt. The following symbols are employed :— I = Moment of inertia. = Section modulus. = Bending moment. = Torque. = Shear force. = Ultimate compressive strength of material. = ,, tensile ,, ,, = „ shear „ „ Z M T S fc ft ft We will first consider a spar of rectangular section, laterally loaded only, which we wish to spindle to either I, Zl, or n section. B = Width of section. D = Depth of section. d = Thickness of flange. t = Thickness of web. It should be noted that in the case of the hollow Q section, t is the combined thickness of the walls forming the web. The most rapid method gives flange and web thicknesses slightly greater than are actually required, so the spar will be on the safe side. The procedure is as follows :— First find t, the thickness of the web. This is given by t = S__ 2DXext find the value of C, which is given by MC = where Z = BD In Table I, values of A are tabulated for values of C from 0 to 1-0. d = AD. c •0 •1 •3 •4 •5 •6 •7 •8 •9 1-0 0 1 2 3 4 0 -002'-004-005 -007 •017 •036 •019 -021•023 •025 5 •009 •027 6 7 •010 -012 •028 -030 8 •014 •032 •038 -040-042 -044 -046 -048 -050!-052 •056!-058 -060 •079 • 103 9 •016 •034 •054 •063 -065 -067 -069 072 -074 -076 •081 i-083 -086 •106 -109 •132;-135 -138 •111 •141 •166 169:-173 -177 •208 -213 -218 •088 •114 •145 •181 •091 -093-096 •117 •120 -123 •148 -151 -loo •185 •190 •098-101 •126 -129 •158 •194 -198 •223 -229 -235 -241 -247 -254 •2C8 -276 -285;-294 -305 •500 — — — •316-329 -345 -365 — [ — — — •162 •203 •261 •393 We get t = f X ^ = 0-42 in. The method will be made clear by taking an example and working it out. Suppose we have a spar whose section is 2 in. wide and 4 in. deep, M = 16,000 lb.-in. S = 900 lb. fc = 5,500 lb. per square inch fs = 800 lb. per square inch. The spar is to be spindled to I section. 900 800 and Z = 5-33 whence C = 0 • 545. From Table I we see that when C=-54, A=-114: and when C—-55, A=-117. Interpolating in these values we get a value in the present case, A = -116 d = "AD = -464 in. Remembering that the values found are on the high side, we will make the flanges 0-45 in. deep, and the web 0-4 in. thick. Checking the section so obtained we find that we have for bending a factor of safety of 1 • 15, and for shear a factor of safety of 1 • 14. The section found in this way will be suitable in most cases. Occasions may arise, however, when it is desired to lighten the spar as much as possible, i.e., to spindle it to the limit. In this case the procedure is a little longer. First find t as before. Next find the value of ^—, and subtract the value so found6 from the known value of M. Calling the remainder M', find the value of C, which in this case is given by where Z' = (B - Look up the value of X in Table I; and as before, d= AD. Taking as an example the spar already described. As before, t = 0 • 4 in. Then M' = M - o = 10,150 lb.-in. Z' = 4-27 C = -433 whence A = -087, and d = 0-35 in. This section has for bending a factor of safety of 1 -01. We will next consider a spar of rectangular section, later- ally loaded as before, together with an end load P. In this ease the procedure is divided into two steps, as follows. First find the section neglecting end load, in the manner already described : and let A sq. inches be the area of the section so found. This enables us to find the value of -r-. 532*
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