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Aviation History
1946
1946 - 0470.PDF
FLIGHT MARCH 7TH, 194(5 STRESS WITHOUT STRAI N ing Moments, or, for short, the B.M. Curve, in which, justas in the cast; of the Shear Force (S.F.) Curve, the values of the bending moments at the various sections along theblade are expressed as proportions of the Greatest Bend- ing Moment (or B.M. max.). Yes, but what is the value of the B.M. max? Lookback for one moment to Fig. 3. If the total load on the blade be imagined as concentrated into one huge lumpand balanced by some miraculous agency on the extreme tip, it is clear that then B.M. max.=W. xL, or the weighttimes the length of the arm. For this system of loading, the Shear Force Diagram would be a rectangle, and theintensity of the shear constant throughout the length. So that, if drawn to the same scale as Fig. 3, the area of theS.F. rectangle would be 8in.X5in., or 40 sq. in. But the load is not thus concentrated, and the area of the S.F.Diagram is not 40 sq. in., but only 24.72 sq. in., which is but 0.618 of the rectangular area. Hence the BiggestBending Moment is discovered to be 0.618 WL, where W is the total air load on the blade and L is 0.8 of the air-screw radius, or full blade length from the axis of rotation to the extreme tip. A few moments' agitation of the sliderule, and we can write down from Fig. 4 or 5 the value of the B.M. at all or any section along the blade. Howsimple! But even this, as will be explained presently, • will not be required of us. The Centrifugal Force Problem So we have now quite simply and easily disposed of TheGreat Bending Moment Question ; it now remauis to find an equally quick and easy way of dealing with the Centri-fugal Force Problem. Here goes, then. As a step towards the solution, a delightful but whollyimpracticable airscrew blade was imagined with a radius of iooin., a maximum chord width of ioin., and, just tomake matters as easy as possible, the r.p.m. were settled at 1,000, and the weight of the material from which theairscrew is supposed to be made at 100 lb. per cubic foot. The blade, needless to say, conforms to the "Standard"plan-form as given in Fig. x. Now nobody, especially in these days when the call is allfor economy and yet more economy, wants to waste good and expensive material. So it is all to the gain of our-selves (and also to that of the aircraft and engine!) that our airscrew scantlings should be the smallest permissible,as in this case we save both cost and weight. To achieve this laudable ambition, we can make good use of thestresses induced by Centrifugal Force, since they are what the mathematicians call '' algebraically additive '' to theBending Moment Stresses ; that is, they act as relief to the compressive stresses induced by bending whilst at the sametime, of course, adding to the tensile stresses. We must now hark back for a while to our bendingmoments. Let us suppose that the total air load 011 the blade is 1,000 lb., and that the hypothetical material ofwhich our airscrew is made has a maximum permissible working stress of 10,000 lb. per sq. in. in compression, being considerably stronger in tension. We may, if we like, think of it as being made of one of the new "improved woods," in which case our imaginary values are not wholly fantastic. Now we can go ahead. First of all, neglecting centrifugal force entirely, we can find the thicknesses at the various stations necessary to resist bending, taking the compression modulus of section to be that sanctioned by the Air Ministry in A.P.970, viz., Zc = o.o8 Ct. squared, where C is the chord of the section under consideration, and t is the thickness. This gives us results as under: Section A-A .,. B-fi .. C-C ..D-D .. E-E .. F-F .. G-G .. per cent.radius 20 30 45 60 75 90 .. . 95 C (in.) 9009-60 10 00 950 785 500 3-75 B.M. Ib./in. 49,44039.700 25,36013,300 4,760 559 99 Zcrequired 4-9443-970 25361 330 0-476 0055900099 V..08C;(in.) 2-620 2-274 1 780 1-323 08706 03738 0 1816 Now we have something to get on with, but just a reminder in passing: So far as centrifugal stress is con- cerned, it makes no difference whatsoever whether we take, in our calculations, the areas of the various sections as being simply C x t, or 0.7 C x t, or any other constant value of C x t; of course, it makes a vast difference to the total centrifugal pull exerted by the blade in its rotation, but we are not so much concerned with that here, and so, for ease of calculation, we can take the area of eaGh section as being simply C x t. I mention this point here because, although it seems rather silly to do so, I have found that occasionally there is some little difficulty in grasping it. dCF We must now find —r—or, in. other words, the rate of change of centrifugal loading in lb. per inch run of blade, plotted against radius. This, of course, will equal 4 x n2 x (1,000)" x 100 x R x A* ! The first boils down 32. part of this,4xr 32. to 1.643. Hence we get: •Section A-AB-B C-CD-D E-E F-FG-G We now Radius 20 3045 60 75 90 95 2 x 12 x 1,728. \ viz.: x (i,ooo)2 x 100 2 x 12 x 1,728 (in.) A sq. in. —Cxt 9-0 x 2-629-6 x 2 274 10 0 x 1-7809-50 x 1-323 7-85 X 0 8706 50 x 0-37383-75 x 0 1816 dCFtake these values of —;—- and set dCF " dr 1-443 x r774-5 1075 1315 1238 842 276 106 them no A t< some convenient scale as ordinates. (See Fig. 6.) In the original graph I took iin. vertical as being equivalent to 250 lb. per inch run, giving us a nice compact diagram well within the capacity of an ordinary small planimeter, but, of course, it is published here on a reduced scale. This, • Remember that we are working in inch/lb. units throughout. This accounis for the presence of 12 and 1728 in the denominator—12 to bring " g " (or 32 2) io inches per jec. per sec, and 1728 because there are 1728 cubic inches to the cubit foot, and " 100 " in the numerator is the weight of our imaginary material in Ib. per cubic foot. Fig 4. Bending moment plotted againslradius. 5 Chart of fWR vs. radius. Fig. 6. Centrifugal force and stress (first stage).
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