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Aviation History
1919
1919 - 0679.PDF
^—~ MAY 22, 1919 Jv^^tjb THE WHIRLING OF AN AIRSCREW By J. MORRIS, B.A. 1. Introduction. IN designing an engine it is not only necessary to examine every possible source of failure and make due allowance accordingly, but it is also essential to reduce vibration to a minimum. A frequent source of danger with regard to rotating shafts is due to the fact that at certain speeds dependent on the stiffness of the shaft and its loading, the shaft tends to collapse. Such a speed is called "critical" or "whirling" speed. Further, if the shaft is run at speeds bordering on the " critical" speed dangerous vibrations will be set up. An investigation is here made into the " whirling '" speed of a shaft running in a long bearing at one end and carrying an air- regard to is claimed far at screw at the other end. It will be shown that the method adopted is applicable to the case of a shaft of any section (variable or otherwise). In particular complete solutions will be given for : (1) a uniform shaft; (2) a conically tapered shaft. « The usual assumptions with whirling will be made, but it that the method herein adopted is simpler and more general than those present in vogue. For a strict investigation the motion should be examined when the shaft is both vibrating and rotating. That speed for which vibration is about to cease is then taken as the dangerous speed. However, the same result will be arrived at if the shaft is considered as rotating with steady motion, and the speed found at which the displacements—lateral and angular— tend to become large. To simplify the treatment the latter method will be adopted in the case under consideration. 2. Notation Employed. Wa = Weight of the air-screw in pounds. I'„ •» Difference between the moments of inertia of W,, about the centre line of the shaft and an axis through its mass centre perpendicular to the former axis in lbs. (feet).2 w — Weight in pounds of that portion of the shaft from the extreme end up to the bearing. I = Length of the shaft in feet. E = Young's modulus of elasticity for the material of the shaft in pounds weight per square foot. n «= Angular velocity of the shaft (supposed constant). ft,- = Angular velocity of the shaft at which whirling takes place. N •• Whirling speed in r.p.m. Thus— 30JV 1 900 ^ " v °r ft,2 ~ ir2N2 y„ — Deflection of the shaft in feet per lb. at the point of attachment of the air-screw due to unit load (dead weight) at that point. Z„ — Slope in radians per lb. at that point due to the same load. = also the deflection in feet/(lb. foot) due to unit couple (static) at that point.* </>„ = Slope in radians/(lb. foot) at that point due to unit couple (static) at that point. ya = Deflection at the mass centre of the air-screw due to the total loading of the shaft. 8„ = Slope at that point due to the same loading. * This follows from the following reciprocal relation due to Lord Rayleigh. (See Phil. Mag., xlviii, pp. 452-436,1874, and xlix, pp. 183-185, 1875):— " For a rod (not necessarily uniform in section) if P and Q be two points on it:—A couple at P would do as much work in acting over the rotation at P due to a simple force at Q as the force at Q would do in acting over the displacement at Q due to the couple at P." W„ + w 8 —- = m„ and SHAFT = *>'.. Where g has its usual significance and is taken as 32 feet/sec2. 3. Solution of Problem. The shaft with air-screw attached is regarded as rotating with steady motion in a deflected position. There will be acting at the point of attachment of the air-screw (1) A centrifugal force— (W._+D n*ya or m an !7„ (1) where to W„ has been added - (weight of the shaft)—this is the correct proportion for a uniform shaft, and will be con sidered near enough for a non-uniform shaft. (2) A centrifugal couple— l'an2e„ - — - or -p'.,aH a Hence— y„ = maa2yaya - p'an20aza \ and e„ = W^V^ - p'a&BaQaJ or— (1 - maya&)i* + p'azan2eu = 0 \ and— - mazaa2ya + (1 + £'„4>on2)0„ = 0/ Solving equations (4) for y„ or 9„ we find— (21 (3) (4) m„p'„(y<r<t>„ - z„2)n* + (m„y„ - p'„^„)0.2 - 1 7a or = o (5) (6) Hence either ya = 8„ = o or map'„ y„<p„ - za2)n* + (may„ - p'„<pa)Ci2 -1=0 Thus the critical angle of velocity answering to " whirling " will be given by the positive root of n2 in (6). Thus— •_ 1 =»w0ya - p'q<pa + *J(maya - p'a$a)% + 4.map'a(ya<t>a - z„2) n2 _ 2 or -1 ~m"y"~ $'"*" + <J m"y«+?"$<•)* + Am"P'<>*«* £V 1 =i'[m„y« - />>« + V(«i«y. + ^>8)! + 4»»»M M N2 1800 l" This formula will give the critical speed in r.p.m. when— ya, za, <t>a are known. For a uniform shaft— Va . ^FTI, Za = T- 1 and <t>a ~ 3irEr*' irEr4 r v~Er* (8 where r is the radius of the cross section. For a conically tapered shaft— 4^_ _4l*(rl, + 2rar _^r/±yj>r"+^ ,\ y" ~ 3*E»V„*' Za ~ 6rEr„V,» and*"- 3*Er„Va» where re is the radius at the bearing and r,t the radius at the air-screw. Hence, given the requisite dimensions, we can calculate y„, za and <f>a from the expressions (8) or (9) according as the shaft is uniform in section or conically tapered. We next insert these values in the formula— 1 w\ W a + -3Jy„ -!>„ + J |(w. +^y« + r+.y +4(w* *jJF-*f i8oo£ and we have the required value of N answering to whirling. In a shaft for which ya, z„, tpa cannot be calculated simply they can be found by direct experiment. For example, if a static load of W lbs. placed at the point of attachment of the air-screw produces a deflection of yw inches and a slope B7„ degrees at that point— y- - 12Wand *• - V80W Similarly, if a static couple G lbs.-feet applied at the same point produces a deflection of 7G inches, and a slope 90 degrees at that point— 7c irflfi '• = 12G and *" " 180G 679
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