APPENDIX C FLEXING OF LENGTH BARS

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1 Fexing of ength bars 83 APPENDIX C FLEXING OF LENGTH BARS C.1 FLEXING OF A LENGTH BAR DUE TO ITS OWN WEIGHT Any object ying in a horizonta pane wi sag under its own weight uness it is infinitey stiff or is supported at many points aong its ength. For ength bars this causes two probems. Firsty, if there is any sagging in the vicinity of the ends of the bar, this wi cause the two end faces to tit with respect to one another causing a bar with otherwise parae faces to appear out of parae. Secondy, since the materia of the bar no onger ies in a straight ine between the two end faces, the extra bending may cause the ength of the bar, measured as the separation between the end faces, to become shorter than in its free state. One soution is to measure the bars verticay, though this is not possibe because of three reasons. Firsty, the reevant standards state that the bars shoud be measured in a horizonta pane, supported at two points termed the Airy points (see ater), since this is how they wi be used in practice. Secondy, a bar standing verticay wi contract under its own weight, see Appendix D. Thirdy, the variation of refractive index between the top and bottom of the bar due to (i) the air pressure gradient due to the Earth s gravitationa fied and (ii) the variation in the air temperature, contributes a significant measurement uncertainty. Historica soutions such as foating the bar in mercury or supporting it on a system of 8 roers or supports [1] have been rejected as hazardous or impractica. They aso do not conform to the reevant standards. The chosen soution is to support the bar on two points whose positions are chosen to make the ends of the bar vertica and parae with each other. These are termed the Airy points of the bar and their positions are usuay engraved on the bar s surface. The position of these points wi now be derived.

2 84 Appendix C C. DERIVATION OF POSITIONS OF AIRY POINTS Consider a uniform soid bar of ength L, cross-sectiona moment of inertia I, and tota weight W. This bar is supported at points, symmetricay paced about its midde, separated by a distance S. Let the reactions at the two supports be R 1 & R as shown in figure C.1. Figure C.1 - Bar supported at two points Resoving verticay, R 1 + R = W, R 1 = R R 1 = R = W Now, spit the bar into three sections (1) to (3) as shown in figure C.1, for the foowing anaysis. In each section, bending moments = EI d y (Bernoui-Euer theory) Since the bar is uniform, EI is a constant, and as such wi be removed from the foowing equations for simpicity. In section (1) In section () In section (3) d y = Wx L d y = Wx L R (L S) 1 x d y = Wx L R (L S) 1 x R x (L + S) (C.1) (C.) (C.3) Integrating equations (C.1) (C.) and (C.3) gives, respectivey,

3 Fexing of ength bars 85 = Wx 3 6L + C 1 (C.4) = Wx 3 6L R 1 x (L S)x + C (C.5) = Wx 3 6L R 1 x (L S)x R x (L + S)x + C 3 (C.6) The sope of the bar, must be continuous at the supports therefore equating (C.4) and (C.5), and substituting x = L S gives (L S) C 1 = C R 1 8 (L S) 4 (L S) ie C 1 = C + R 1 8 One constraint is that we require vertica end faces, ie x= 0 = 0 (C.7) This impies that C 1 = 0 Substituting this resut into (C.7) and using the fact that R 1 = W gives Now, matching C = W at x = L gives C 3 = WL 6 (L S) 8 (C.8) (C.9) With C 1,C,C 3 determined, equations (C.3) (C.4) and (C.5) competey describe the bending of the bar, once S is known. To find S, is matched at the boundary between regions () and (3). In region () x= L +S = W(L + S)3 48L _ W (L + S) 8 (L S)(L + S) 4 (L S) + 8 (C.10) and in region (3)

4 86 Appendix C x= L +S = W(L + S)3 48L _ W (L + S) 8 (L S)(L + S) 4 (L + S) + 8 (L + S) 4 WL 6 (C.11) Equating (10) and (11) gives which with reduction gives W(L S) = W 16 S = L 3 + S) (L 8 WL 6 i.e. S = L 3 This is the symmetrica spacing of the Airy points, i.e. approximatey of the ength of the bar. This is ony vaid for a bar supported at two points with no additiona reference fats or other masses attached to it. Even when a bar is supported at the Airy points, its centra ength wi be different to the case where it is unsupported due to the extra curvature of the bar. Figure C. shows the difference dl in ength between a bar which is unsupported and one which rests on supports positioned a distance a away from the end faces (L - S = a). Note that supporting at the Airy positions (a = 0.11) causes a change in ength of dl = -0.4 nm, which is negigibe. The support positions corresponding to a = for which there is the minimum change in ength are termed the Besse points. Figure C. - Effect of support point position, a, on change in ength, dl, of bar from unsupported state for m bar.

5 C.3 COMPENSATION FOR MASS OF WRUNG FLAT Fexing of ength bars 87 When a reference fat is wrung to one end face of a bar, this adds additiona bending and wi cause the bar supported at the Airy points to exhibit a paraeism error. Techniques for compensating for the extra mass of the fat incude suppying an additiona ifting force by means of weights or evers which effectivey cances out the weight of the fat [] or by moving the support points towards the ends of the bar [3]. The atter soution has been adopted as being easier to impement and is detaied beow. Consider the bar and reference fat (paten) shown in figure C.3. Figure C.3 - Bar supported at new support points with fat attached to one face The supports are positioned at x = and x = + a, with being the haf-ength of the bar. As before, appying Bernoui-Euer bending theory to the three regions gives three equations EI d y A = wx for 0 < x (C.13) EI d y B = wx R 1(x + ) for < x + a (C.14) EI d y = wx C R 1(x + ) R (x a ) for + a < x (C.15) Integrating (C.13) (C.14) and (C.15) and determining arbitrary constants by continuity at support points, gives EI A = wx 3 6

6 88 Appendix C EI B = wx3 6 R 1 (x + ) EI C = wx3 6 R 1 (x + ) R (x a ) This means that the ange between the end faces, α, is given simpy by Thus C, x= EIα = 4 3 w3 1 (R 1 + R )( + a ) 1 (R 1 R a )( + a ) (C.16) Now, resoving verticay, R 1 + R = W + M and taking moments about the centre of the bar gives R 1 R a = M( + p), substituting into (C.16) gives EIα = W 3 a 1a + M ( + p + )( + p a ) p { } (C.17) To check the previous derivation for the Airy points, setting M = 0, = a does indeed give the same soution for the positions of the supports. To see the effect of supporting the bar and fat at the unmodified Airy points, the excess tit of the ends of the bar can be cacuated from α = M 3EI 1+ 3p The fats are 70 mm diameter, 15 mm thick and have a density of 7800 kg m -3. This gives vaues of M = kg, p = 7.5 x 10-3 m, I = x 10-8 m 4, and for stee, Youngs moduus, E = 03 GPa. For m bar, = 1 m, this gives a vaue for α of 6.34 x 10-5 radians. Converting this to a change of ength across the face of the bar gives a vaue of 1.4 µm, or over 4 fringes. To correct this, the two supports must be moved either symmetricay, or by moving just one support. Let M W = np where n is the ratio of the cross-section of the fat to the cross-section of the bar, assuming that the bar and fat are made of the same materia, as required to minimise the phase correction.

7 Fexing of ength bars 89 From (C.17), setting α = 0, dividing by W and substituting M W = np gives 3 a 1a + np {( + p + )( + p a ) p }= 0 There are 4 soutions for the positions of the support points: the first two being nonsymmetrica and the remaining two being symmetrica and identica except for a change of sign. The non-symmetrica soutions eave one of the supports at its Airy point, and the soution of the above equation gives the position of the other support. For the symmetrica soution, both of the supports are moved outwards from their Airy points and retain their symmetrica pacing about the centre of the bar. Case (i), support is unmoved, substituting a = 3 in (C.17) 3 a np + p + ( ) + p 3 p = 0 Separating terms in np ( + p ) + p 3 p + np + p 3 = np ( + p ) + p 3 p = a 1 3 np + p 3 = 3 + np ( + p ) + p 3 p 3 np + p 3 Removing a common factor of 3 gives = np + p 3 np + p 3 ( ) + p 3 p Dividing gives

8 90 Appendix C 3 = ( + p)1 np + p 3 p 3 np + p 3 3np Mutipying top and bottom by = np 3, separating factors and rearranging gives + p 3 np p 3p ( 3 + 3p ) Mutipying and coecting terms, dividing by gives np = + 6p 3 1 np 3 + 3p 1 np = 1+ 3p np 1 3 3p Thus with np / (1 + 3p / ) f ( χ) np / (1 + χ + χp / ) = 3 f ( 3) For case (ii), support 1 is unmoved, substituting = in (C.17) gives a simiar 3 soution to case (i), though because the signs of and a are reversed, the sign of the radica is aso reversed in the soution, i.e. = 3 f ( 3) For case (iii), (C.17) gives both supports are moved symmetricay, substituting = a = a in

9 Fexing of ength bars 91 3 a + np {( + p + a) ( + p a) p }= 0 Separating terms in a 3 a + np {( + p) ( + p) a }= 0 Dividing a = np 3 ( + p) 1 + np a = p 1+ np 1 np a = 1 + 3p np np a = 3 f( 0) Stricty, a = ± 3 f ( 0) though these two soutions correspond to the two choices of abeing the supports, i.e. they are the same physica soution. In summary, setting α to zero in (C.17) aows for three soutions: (i) Support remains at the Airy position, and support 1 moves to a new position a = 3, = 3 f( 3) (ii) Support 1 remains at the Airy position and support moves to a new position = 3, a = 3 f( 3) (iii) Both supports move by equa amounts to new symmetrica positions

10 9 Appendix C = a = a = 3 f( 0) where np / (1 + 3p / ) f ( χ) np / (1 + χ + χp / ) Suitabe toerances on the positioning of the supports may be cacuated by differentiating (C.17) with respect to a, this wi be performed for the symmetrica soution (case (iii)). Substituting = a = a in (C.17) gives EIα = W 3 a + M { ( + p + a) ( + p a ) p} ( ) EIα = W 3 Wa + M + p a Differentiating with respect to a gives Hence δa = EIδα = a( W + M)δa EI δα (18) a( W + M) For m bar, for a maximum vaue of δα of 1.16 x 10-6 which corresponds to the vaue of 1 µin (0.05 µm) error chosen by Wiiams, δa =.4 x 10-3, or.4 mm. This is better than the toerance for the genera case for which Wiiams cacuated a vaue of 0.7 mm. Thus the use of symmetrica support positions is preferabe, for which positioning within.4 mm is required. Thus by accurate positioning of the support positions, the additiona bending may be atered in such a way that the end faces of the bar remain vertica and parae. The effect of this additiona bending on the ength of the bar wi now be examined. C.4 EFFECT OF FLEXURE OF BARS ON THEIR LENGTH The effect on the measured ength of the bar is measured on the neutra axis of the bar which runs through the centre of the bar. For a section of the bar, ength, with

11 Fexing of ength bars 93 gradient θ the change in ength compared to the free state is given by θ, and θ =. Thus the tota change in ength aong the whoe bar is given by 0 1 It is possibe to perform this integra, substituting for from equations derived earier, but a simpe order of magnitude estimate shows that this is not required as the overa change in ength is negigibe. Since 0 1 max a maximum vaue for the change in ength due to bending may be cacuated. Figure C.4 shows the variation in the vertica position of the neutra axis of m bar with a fat wrung on, supported at the modified symmetrica Airy points and the sope of the bar. The maximum sope is seen to be 8 x 10-6 at approximatey 0.7 m from the free end of the bar. Thus the maximum change in ength of the bar is 6.4 x m (0.00 fringe), i.e. negigibe.

12 94 Appendix C Gradient, Dispacement Position Figure C.4 - Variation in vertica position and gradient (dashed ine) of the neutra pane of m bar, supported at modified Airy points REFERENCES FOR APPENDIX C [1] Rot F H Gauges and Fine Measurement (MacMian & Co.: London) (199) [] Bayer-Hems F Beigung von Endmaßen bei horizontaer Lagerung auf Scheiden PTB Mitteiungen (1967) 1/ / ], [3] Wiiams D C The paraeism of a ength bar with an end oad J. Sci. Instrum. (196)

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