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Stiffness of Beams and Cables

Stiffness of Beams and Cables 314 AIRCRAFT ENGINEERIN G August, 1939 Transitio n of a Laterally Loaded Member with Tensile End Load from a "Beam " to a "Cable " By Huang Yushan, B.Sc. and M = ½ W (a - x) (11) Introduction 0 ONSIDE R a member carrying both lateral Similarly, M/M = ....(12) load and axial tension, which we .shall refer to as a " beam-tie." If the and at x = 0, member is very stiff, it acts as a beam ; but as A' = tanh µa/µa = tanli S/S (13) its stiffness is reduced, it tends to become a The curve K vs S is also plotted in Fig. 2. cable. The present practice is to assume that For large values of S, (13) reduces to those are beams and those are cables. No clear K = l/S (13A) demarcation between them has ye t been worked and for small values of S, out. This paper, by comparing the maximum K = 1 - S2/3 (13B) bending moment and the maximum deflection, Using the 5 per cent limit, then gives an account of the effect of variation of S = 20 for K = 0·05 stiffness on the change from beam to cable. A For practical purposes, an error of 5 per cent is and = 0·39 for K = 0·95 non-dimensional stiffness constant 5 is intro­ 2 generally allowable. So as a practical criterion, If 1 per cent accuracy is required, duced and some criteria are given for practical we may say tha t a beam-tie having " K " less application. S = 100 for K = 0·01 tha n 0·05 may be regarded as a cable, while if S = 0·17 for K = 0·99 its " K " is greater tha n 0·95 it ma y be regarded as a beam. Solving (9A) and (9B), Flat Beam-Tie Under Uniformly Distributed S = 6·2 for K = 0·05 Effect of Initial Curvature Load 1 and S = 0·35 for K = 0·95 Although the above derivations are based on For a member subjected to axial tension P If the error is limited to 1 pe r cent, the criteria a flat beam-tie, the results can be extended to and uniformly distributed lateral load w per are S = 14 for A' = 0·01 cases of beam-tie with initial curvature also. unit length, as shown in Tig. 1, the moment at and S = 0·16 for A' = 0·99 Suppose the initial curve of the uniformly an y section can be derived from the following loaded beam is equation : Flat Beam-Tie Under Concentrated Load Next consider a beam-tie under a con­ centrated load W at its middle. The bending where c is th e maximum camber at middle, then, moment becomes which yields The bending moment M is also the primary bending moment M , which is the bending moment of pure beam, plus the product of the The primary bending moment M is axial tension P and the deflection y, thus M=M , + Py (4) where M = ½ w (a2 — x2) (5) Fro m (4), th e deflection of a beam-tie is y = ( M-M ) / P (6) Therefore the ratio K is still the same as given in (7). Now, if this member approaches a pure beam, M approaches M and, if it approaches a pure Similarly if the initial curve of the con­ cable, M approaches zero. The ratio M/M , centrated loading case is assumed to be which is represented here by K, defines the y = c (x/a — 1), th e same result is found. ratio of the bending moment of the beam-tie to tha t of a pure beam. On the other hand, as Conclusion given in (6), the deflection of a pure cable is equal to — M /P. The quantity (1 — A') 0 Eased upon the above calculations, in which therefore represents the ratio of the deflection it is assumed that the deflection of the member of the beam-tie to tha t of a pure cable. is not very large, the following conclusions may Dividing (3) by (5), th e ratio K for uniformly be drawn : distributed load is given as (1) For engineering purposes, if the stiffness constant 5 of a member is higher than 10, th e member may be taken as a cable ; and on the other hand if that constant is below 0·4, it may be taken as a beam. At x = 0, where both the bending moment and (2) The maximum bending moment and the deflection are maximum : maximum deflection of a beam-tie may be estimated by means of Fig. 2, the procedure is : Let the stiffness constant S be representing µa (a) Compute the stiffness S of tha t member. (b) Find the corresponding value of K from hig. 2. The ratio K is plotted against the stiffness (c) The maximum bending moment is then constant 5 as shown in Fig. 2. equal to A' times that of a pure beam under Equation (9) can be simplified for large values th e same loading. and small values of S. It gives A", = 2/S2 for large value of S (9A) (d) The maximum deflection is equal to (1 — A') times that of a pure cable under the and = 1 — S2 for small value of S. .(9B) same loading. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Aircraft Engineering and Aerospace Technology Emerald Publishing

Stiffness of Beams and Cables

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Publisher
Emerald Publishing
Copyright
Copyright © Emerald Group Publishing Limited
ISSN
0002-2667
DOI
10.1108/eb030528
Publisher site
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Abstract

314 AIRCRAFT ENGINEERIN G August, 1939 Transitio n of a Laterally Loaded Member with Tensile End Load from a "Beam " to a "Cable " By Huang Yushan, B.Sc. and M = ½ W (a - x) (11) Introduction 0 ONSIDE R a member carrying both lateral Similarly, M/M = ....(12) load and axial tension, which we .shall refer to as a " beam-tie." If the and at x = 0, member is very stiff, it acts as a beam ; but as A' = tanh µa/µa = tanli S/S (13) its stiffness is reduced, it tends to become a The curve K vs S is also plotted in Fig. 2. cable. The present practice is to assume that For large values of S, (13) reduces to those are beams and those are cables. No clear K = l/S (13A) demarcation between them has ye t been worked and for small values of S, out. This paper, by comparing the maximum K = 1 - S2/3 (13B) bending moment and the maximum deflection, Using the 5 per cent limit, then gives an account of the effect of variation of S = 20 for K = 0·05 stiffness on the change from beam to cable. A For practical purposes, an error of 5 per cent is and = 0·39 for K = 0·95 non-dimensional stiffness constant 5 is intro­ 2 generally allowable. So as a practical criterion, If 1 per cent accuracy is required, duced and some criteria are given for practical we may say tha t a beam-tie having " K " less application. S = 100 for K = 0·01 tha n 0·05 may be regarded as a cable, while if S = 0·17 for K = 0·99 its " K " is greater tha n 0·95 it ma y be regarded as a beam. Solving (9A) and (9B), Flat Beam-Tie Under Uniformly Distributed S = 6·2 for K = 0·05 Effect of Initial Curvature Load 1 and S = 0·35 for K = 0·95 Although the above derivations are based on For a member subjected to axial tension P If the error is limited to 1 pe r cent, the criteria a flat beam-tie, the results can be extended to and uniformly distributed lateral load w per are S = 14 for A' = 0·01 cases of beam-tie with initial curvature also. unit length, as shown in Tig. 1, the moment at and S = 0·16 for A' = 0·99 Suppose the initial curve of the uniformly an y section can be derived from the following loaded beam is equation : Flat Beam-Tie Under Concentrated Load Next consider a beam-tie under a con­ centrated load W at its middle. The bending where c is th e maximum camber at middle, then, moment becomes which yields The bending moment M is also the primary bending moment M , which is the bending moment of pure beam, plus the product of the The primary bending moment M is axial tension P and the deflection y, thus M=M , + Py (4) where M = ½ w (a2 — x2) (5) Fro m (4), th e deflection of a beam-tie is y = ( M-M ) / P (6) Therefore the ratio K is still the same as given in (7). Now, if this member approaches a pure beam, M approaches M and, if it approaches a pure Similarly if the initial curve of the con­ cable, M approaches zero. The ratio M/M , centrated loading case is assumed to be which is represented here by K, defines the y = c (x/a — 1), th e same result is found. ratio of the bending moment of the beam-tie to tha t of a pure beam. On the other hand, as Conclusion given in (6), the deflection of a pure cable is equal to — M /P. The quantity (1 — A') 0 Eased upon the above calculations, in which therefore represents the ratio of the deflection it is assumed that the deflection of the member of the beam-tie to tha t of a pure cable. is not very large, the following conclusions may Dividing (3) by (5), th e ratio K for uniformly be drawn : distributed load is given as (1) For engineering purposes, if the stiffness constant 5 of a member is higher than 10, th e member may be taken as a cable ; and on the other hand if that constant is below 0·4, it may be taken as a beam. At x = 0, where both the bending moment and (2) The maximum bending moment and the deflection are maximum : maximum deflection of a beam-tie may be estimated by means of Fig. 2, the procedure is : Let the stiffness constant S be representing µa (a) Compute the stiffness S of tha t member. (b) Find the corresponding value of K from hig. 2. The ratio K is plotted against the stiffness (c) The maximum bending moment is then constant 5 as shown in Fig. 2. equal to A' times that of a pure beam under Equation (9) can be simplified for large values th e same loading. and small values of S. It gives A", = 2/S2 for large value of S (9A) (d) The maximum deflection is equal to (1 — A') times that of a pure cable under the and = 1 — S2 for small value of S. .(9B) same loading.

Journal

Aircraft Engineering and Aerospace TechnologyEmerald Publishing

Published: Aug 1, 1939

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